Category: OP Malhotra Solutions

Students appreciate clear and concise ICSE Class 10 Maths Solutions S Chand Chapter 18 Arithmetic Mean, Median, Mode and Quartiles Ex 18(a) that guide them through exercises.

S Chand Class 10 ICSE Maths Solutions Chapter 18 Arithmetic Mean, Median, Mode and Quartiles Ex 18(a)

Question 1.
Find the number half-way between 0.2 and 0.02.
Solution:
Half-way number (average) between 0.2 and 0.02 = \(\frac{0.2+0.02}{2}\) = \(\frac{0.22}{2}\) = 0.11

Question 2.
Find the mean of the following sets of numbers :
(a) 4, 5, 7, 8
(b) 3, 5, 0, 2, 8
(c) 2.5, 2.4, 3.5, 2.8, 2.9, 3.3, 3.6
(d) -6, -2, -1, 0, 1, 2, 5, 9
(e) First five prime numbers
(f) First eight even natural numbers
(g) All factors of 30
(h) First five multiples of 8
(i) x, x+1, x+2, x+3, x+4, x+5, x+6
Solution:
(a) 4, 5, 7, 8
Here n = 4
∴ Mean = \(\frac{\text { Sum }}{n}\) = \(\frac{4+5+7+8}{4}\) = \(\frac { 24 }{ 4 }\) = 6

(b) 3, 5, 0, 2, 8
Here n = 5
∴ Mean = \(\frac{3+5+0+2+8}{5}\) = \(\frac { 18 }{ 5 }\) = 3.6

(c) 2.5, 2.4, 3.5, 2.8, 2.9, 3.3, 3.6
Here n = 8
∴ Mean = \(\frac{2.5+2.4+3.5+2.8+2.9+3.3+3.6}{7}\) = \(\frac { 21.0 }{ 7 }\) = 3.0

(d) -6,-2,-1,0, 1,2, 5, 9
Here n = 8
∴ Mean = \(\frac{-6-2-1+0+1+2+5+9}{8}\) = \(\frac{-9+17}{8}\) = \(\frac{8}{8}\) = 1

(e) First 5 prime numbers are 2, 3, 5,7, 11
∴ Mean = \(\frac{2+3+5+7+11}{8}\) = \(\frac{28}{5}\) = 5.6

(f) First 8 even natural numbers are 2, 4, 6, 8, 10, 12, 14, 16
∴ Mean = \(\frac{2+4+6+8+10+12+14+16}{8}\) = \(\frac{72}{8}\) = 9

(g) All factors of 30 are 1, 2. 3, 5, 6, 10, 15, 30
Here n = 8
∴ Mean = \(\frac{1+2+3+5+6+10+15+30}{8}\) = \(\frac{72}{8}\) = 9

(h) First 5 multiples of 8 are 8, 16, 24, 32, 40
∴ Mean = \(\frac{8+16+24+32+40}{5}\) = \(\frac{120}{5}\) = 24

(i) x, x + 1, x + 2, x + 3, x + 4, x + 5, x + 6
Here n = 7
Mean = \(\frac{x+x+1+x+2+x+3+x+4+x+5+x+6}{7}\) = \(\frac{7 x+21}{7}\) = \(\frac{7(x+3)}{7}\) = x + 3

Question 3.
The mean of the numbers 6, y, 7, x, 14 is 8. Express y in terms of x.
Solution:
6, y, 7, x, 14
Here n = 5 and Mean = 8
Total sum = \(n \bar{x}\) = 5 × 8 = 40
∴ 6 + y + 7 + x + 14 = 40 ⇒ 7 + x + 27 = 40 ⇒ 7 = 40 – 27 – x =13 – x

Question 4.
Nisha secured 73, 86, 78 and 75 marks in four tests. What is the least number of points she should secure in her next test if she has to have an average of 80?
Solution:
Number of total tests = 5
Average of 5 tests = 80
∴ Total points, Nisha secured = 5 × 80 = 400
But her score in 4 tests = 73 + 86 + 78 + 75 = 312
∴ Score in fifth test = 400 – 312 = 88

Question 5.
A class of 10 students was given a test in Mathematics. The marks, out of 50, secured by the students were as follows :
31, 36, 27, 38, 45, 39, 32, 29, 41, 38 Find the mean score.
Solution:
Here n = 10
and scores are 31, 36, 27, 38, 45, 39, 32, 29, 41, 38
∴ Sum = 31 + 36 + 27 + 38 + 45 + 39 + 32 + 29+ 41 + 38 = 356
∴ Mean = \(\frac{356}{10}\) = 35.6

Question 6.
Find the mean of the following frequency distributions:
(a)

Weight (kg)	30	31	32	33	34
No. of students	8	10	15	8	9

(b)

Marks	20	25	30	35	40
No. of students	5	10	12	8	5

(c)

X	2	5	7	8
y	2	4	6	3

(d)

X	0.1	0.2	0.3	0.4	0.5	0.6
f	30	60	20	40	10	50

Solution:
(a)

Weight (in kg) (x)	No. of students (f)	f × x
30 31 32 33 34	8 10 15 8 9	240 310 480 264 306
Total	50	1600

Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 1600 }{ 50 }\) = 32 kg
(b)

Marks (x)	No. of students (f)	f × x
20	5	100
25	10	250
30	12	360
35	8	280
40	5	200
Total	40	1190

∴ Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 1190 }{ 40 }\) = 29.75
(c)

x	f	f × x
2	2	4
5	4	20
7	6	42
8	3	24
Total	15	90

∴ Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 90 }{ 15 }\) = 6
(d)

x	f	f × x
0.1 0.2 0.3 0.4 0.5 0.6	30 60 20 40 10 50	3 12 6 16 5 30
Total	210	72

∴ Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 72 }{ 210 }\) = \(\frac { 12 }{ 35 }\) = 0.34

Question 7.
Fill in the blanks :
While calculating the mean of the grouped data, we make the assumption that frequency in any class is centred at its
Solution:
Mean of grouped data in any class is called its class mark.

Question 8.
The frequency distribution of marks obtained by 40 students of a class is as under :
Calculate the Arithmetic mean.

Marks	0-8	8-16	16-24	24-32	32-40	40-48
Students	5	3	10	16	4	2

Solution:

Marks	Class Mark x	Frequency (f)	f × x
0-8 8-16 16-24 24-32 32-40 40-48	4 12 20 28 36 44	5 3 10 16 4 2	20 36 200 448 144 88
Total		40	936

Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 936 }{ 40 }\) = 23.4 marks

Question 9.
Find the mean of the following data:
(a)

Marks	10-14	15-19	20-24	25-29	30-34
No. of students	4	6	12	5	3

(b)

Class	0-10	11-20	21-30	31-40	41-50
Frequency	3	4	2	5	6

Solution:
(a)

Marks	Class Mark x	Frequency (f)	f × x
10-14 15-19 20-24 25-29 30-34	12 17 22 27 32	4 6 12 5 3	48 102 264 135 96
Total		30	645

Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 645 }{ 30 }\) = 21.5 marks

(b)

Class	Class Mark x	Frequency (f)	f × x
0-10 11-20 21-30 31-40 41-50	5 15.5 25.5 35.5 45.5	3 4 2 5 6	15.0 62.0 51.0 177.5 273.0
Total		20	578.5

Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 578.5 }{ 20 }\) = 28.925

Question 10.
In a class of 60 boys the marks obtained in a monthly test were as under :

Marks	10-20	20-30	30-40	40-50	50-60
Students	10	25	12	08	05

Find the mean marks of the class.
Solution:

Marks	Class Mark (x)	Frequency (f)	f × x
10-20	15	10	150
20-30	25	25	625
30-40	35	12	420
40-50	45	08	360
50-60	55	05	275
Total		60	1830

Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 1830 }{ 60 }\) = 30.5

Question 11.
Compute the mean of the following frequency table by:
(i) direct-method and
(ii) short-cut method

Class	5-10	10-15	15-20	20-25	25-30	30-35	35-40	40-45	45-50
Frequency	10	6	4	12	8	4	2	1	3

Solution:

Class	Class Marks (x)	Frequency (f)	f × x	A = 27.5 d = x – A	f × d
5-10 10-15 15-20 20-25 25-30 30-35 35-40 40-45 45-50	7.5 12.5 17.5 22.5 A=27.5 32.5 37.5 42.5 47.5	10 6 4 12 8 4 2 1 3	75.0 75.0 70.0 270.0 220.0 130.0 75.0 42.5 142.5	-20 -15 -10 -5 0 5 10 15 20	-200 -90 -40 -60 0 20 20 15 60
Total		50	1100.0		-275

(i) Direct method — Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac { 1100 }{ 50 }\) = 22

(ii) Short-cut method— Mean = A + \(\frac{\sum f d}{\sum f}\) = 27.5 + \(\frac{-275}{50}\) = 27.5 – 5.5 = 22

Question 12.
The following table gives the classification of 100 cows of a dairy farm, according to the amount of milk given by each in a dairy.

Amount of milk in	0-2	2-4	4-6	6-8	8-10	10-12	12-14	14-16	16-18
No. of cows	4	14	17	20	10	13	12	10	10

Calculate the mean correct to first place of decimal.
Solution:

Amount of milk in kg	Class Mark (x)	No. of Cows (f)	f × x
0-2	1	4	4
2-4	3	14	42
4-6	5	17	85
6-8	7	20	140
8-10	9	10	90
10-12	11	13	143
12-14	13	12	156
14-16	15	10	150
16-18	17	10	170
Total		110	980

∴ Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac{980}{110}\) = 8.9 kg

Question 13.
The weights (in grams) of 50 apples picked out at random from a consignment are given below : 82, 118, 80,110, 104, 84, 106,107, 76, 82,109,107,115, 93,187, 95,123,125, 111, 92, 86, 70,126, 78,130, 129,139,119,115,128,100,186,84, 99, 113,204, 111, 141,136,123, 90, 115, 98, 110, 78, 90, 107, 81, 131, 75.
(i) What is the range of the weights ?
(ii) Form a frequency distribution with class intervals 70-89, 90-109, and so on.
(iii) Use your frequency distribution to calculate the mean.
Solution:
(i) Maximum weight = 204 g
Minimum weight = 70 g
∴ Range = 204 – 70 = 134 g

(ii) Now forming the frequency distribution table, taking class intervals such as 70-89, 90-109, so on

Class interval	Class Marks (x)	Frequency (f)	A = 139.5 d = x – A	f × d
70-89 90-109 110-129 130-149 150-169 170-189 190-209	79.5 99.5 119.5 A = 139.5 159.5 179.5 199.5	12 14 16 5 0 2 1	-60 -40 -20 0 20 40 60	-720 -560 -320 0 0 80 60
Total	Class Marks (x)	50		-1460

Let A = 139.5

(iii) Mean = A + \(\frac{\sum f d}{\sum f}\) = 139.5 + \(\frac{-1460}{50}\) = 139.5 – 29.2 = 110.3

Question 14.
The following table gives the weekly wages of workers in a factory.

Weekly wages (in Rs.)	50-55	55-60	60-65	65-70	70-75	75-80	80-85	85-90
No. of workers	5	20	10	10	9	6	12	8

Calculate :
(i) the mean,
(ii) the number of workers getting weekly wages below Rs. 80, and
(iii) the number of workers getting Rs. 65 and more but less than Rs. 85 as weekly wages.
Solution:

Weekly wages	Class Marks (x)	Frequency (f)	A=67.5 d=x – A	f × d
50-55	52.5	5	-15	-75
55-60	57.5	20	-10	-200
60-65	62.5	10	-5	-50
65-70	A = 67.5	10	0	0
70-75	72.5	9	5	45
75-80	77.5	6	10	60
80-85	82.5	12	15	180
85-90	87.5	8	20	160
Total		80		120

(i) Mean = A + \(\frac{\sum f d}{\sum f}\) = 67.5 + \(\frac{120}{80}\) = 67.5 + 1.5 = 69
(ii) Number of workers getting wages below Rs: 80 = 60
(iii) Number of workers getting more than Rs. 65 but less them 85 = 10 + 9 + 6 + 12 = 37

Question 15.
Find the mean of the following data:

Marks obtained	Less than 10	Less than 20	Less than 30	Less than 40	Less than 50
No. of students	7	19	32	42	50

Solution:

Marks obtained	Class Mark no. (x)	No. of workers (c.f.)	(f)	fx
0-10	5	7	7	35
10-20	15	19	12	180
20-30	25	32	13	325
30-40	35	42	10	350
40-50	45	50	8	360
Total			50	1250

Mean = \(\frac{\sum f x}{\sum f^{\prime}}\) = \(\frac{1250}{50}\) = 25

Question 16.
Recast the following cumulative table into the form of an ordinary frequency distribution and determine the value of the mean :

Solution:

Marks	Class Marks	c.f	f	A=27.5 d = x – A	f × d
0-5	2.5	14	14	-25	-350
5-10	7.5	25	11	-20	-220
10-15	12.5	37	12	-15	-180
15-20	17.5	56	19	-10	-190
20-25	22.5	94	38	-5	-190
25-30	A = 27.5	112	18	0	0
30-35	32.5	130	18	5	90
35-40	37.5	144	14	10	140
40-45	42.5	148	4	15	60
45-50	47.5	150	2	20	40
Total			150		-800

Mean = A + \(\frac{\sum f d}{\sum f}\) = 27.5 + \(\frac{-800}{150}\) = 27.5 – 5.3 = 22.2

Question 17.
Find the missing frequency in the following data if arithmetic mean is 19.92.

Class	4-8	8-12	12-16	16-20	20-24	24-28	28-32	32-36	36-40
Frequency	11	13	16	14	–	9	17	6	4

Solution:
Let the missing frequency be p, then

Class	Class Marks (x)	Frequency (f)	f × x
4-8	6	11	66
8-12	10	13	130
12-16	14	16	224
16-20	18	14	252
20-24	22	p	22 p
24-28	26	9	234
28-32	30	17	510
32-36	34	6	204
36-40	38	4	152
Total		90+p	1772+22 p

Mean is given = 19.92
∴ Mean = \(\frac{\sum f x}{\sum f}\) ⇒ 19.92 = \(\frac{1772+22 p}{90+p}\) ⇒ 1772 + 22p = 1792.8 + 19.92p
22p = 19.92p = 1792.8 – 1772
2.08p = 20.8
∴ p = \(\frac{20.8}{2.08}\) = \(\frac{208 \times 100}{10 \times 208}\) = 10
∴ Missing Frequency = 10

Question 18.
Calculate the Arithmetic mean, correct to one decimal place, for the following frequency distribution of marks obtained in an Arithmetic test.

Marks	0-10	10-20	20-30	30-40	40-50
No. of students	2	5	20	8	7

Solution:

Marks	Class Marks (x)	Frequency (f)	f  × x
0-10	5	2	10
10-20	15	5	75
20-30	25	20	500
30-40	35	8	280
40-50	45	7	315
Total		42	1180

Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac{1180}{42}\) = 28.095 = 28.1 (approx)

Question 19.
The following table gives the marks scored by students in the examination:

Marks	0-5	5-10	10-15	15-20	20-25	25-30	30-35	35-40
No. of students	3	7	15	24	16	8	5	2

Calculate the mean mark, correct to two decimal places.
Solution:

Marks (Class)	Class Marks (x)	Frequency (f)	Let  A = 17.5 d = x – A	f × d
0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40	2.5 7.5 12.5 A = 17.5 22.5 27.5 32.5 37.5	3 7 15 24 16 8 5 2	-15 -10 -5 0 5 10 15 20	-45 -70 -75 0 80 80 75 40
Total		80		85

A = 17.5
Mean = A + \(\frac{\sum f d}{\sum f}\) = 17.5 + \(\frac { 85 }{ 80 }\) = 17.5 + 1.06 = 18.56

Utilizing OP Malhotra Class 10 Solutions Chapter 4 Linear Inequations in One Variable Ex 4 as a study aid can enhance exam preparation.

S Chand Class 10 ICSE Maths Solutions Chapter 4 Linear Inequations in One Variable Ex 4

Question 1.
On a bargain counter, the shopkeeper put labels on various goods showing their prices, Rs. P, where P e {real numbers}. Write a mathematical sentence for each of the following labels :
(a) more than Rs. 7.50
(b) not less than Rs. 10
(c) not more than Rs. 22
(d) less than Rs. 11
Solution:
(a) more than Rs. 7.50 : > Rs. 7.50
(b) not less than Rs. 10 : ≮ Rs. 10
(c) not more than Rs. 22 : ≯ Rs. 22
(d) less than Rs. 11 : < Rs. 11

Question 2.
You are given the following numbers :
-2.6    5.1  -3   0.4   1.2    -3.1     4.7
Fill in the blanks.
(a) A = {A : A ≤ -3} = {…..}
(b) B = {A : A ≤ 1} = {…..}
Solution:
Given numbers are ; -2.6, 5.1, -3, 0.4, 1.2, -3.1, 4.7
(a) A = {x : x ≥ -3}
= {-3, -2.6, 0.4, 1.2, 4.7, 5.1}

(b) B = {x : x ≤ 1} = {-3.1, -3, -2.6, 0.4}

Question 3.
If the replacement set is {-2, -1, +1, +2, +4, +5, +9}, what is the solution set of each of the following mathematical sentences?
(a) x + \(\frac { 3 }{ 2 }\) > \(\frac { 5 }{ 2 }\)
(b) x – 4 = – 3
(c) 2x – 5 ≥ 10
(d) 3y ÷ 2 ≤ \(\frac { 5 }{ 2 }\)
(e) 4x² = 16
Solution:
Replacement set = {-2, -1, +1, +2, +4, +5, +9}
(a) x + \(\frac { 3 }{ 2 }\) > \(\frac { 5 }{ 2 }\)
⇒ x > \(\frac { 5 }{ 2 }\) – \(\frac { 3 }{ 2 }\) ⇒ x > \(\frac { 5 – 3}{ 2 }\)
⇒ x > \(\frac { 2 }{ 2 }\) ⇒ x > 1
∴ Solution set = {+2, +4, +5, +9}

(b) x – 4 = – 3
⇒ x = – 3 + 4
⇒ x = 1
Solution set = {+1}

(c) 2x – 5 ≥ 10
⇒ 2x ≥ 10 + 5 ⇒ 2x ≥ 15
⇒ x ≥ y
⇒ A ≥ 7.5
∴ Solution set = {+9}

(d) 3y ÷ 2 ≤ \(\frac { 5 }{ 2 }\)
⇒ 3y ≤ \(\frac { 5 }{ 2 }\) x 2 ⇒ 3y ≤ 5
⇒ y ≤ \(\frac { 5 }{ 3 }\)
⇒ y ≤ 1\(\frac { 2 }{ 3 }\)
∴ Solution set = {+1, -1, -2} = {-2. -1, +1}

(e) 4x² = 16
⇒ x² = \(\frac { 16 }{ 4 }\)
⇒ x² = 4 ⇒ x² = (±2)²
∴ x = ±2,
∴ Solution set = {-2, +2}

Question 4.
List the solution set of 30 – 4 (2x – 1) < 30, given that x is a positive integer.
Solution:
30 – 4 (2x – 1) < 30
⇒ 30 – 8x + 4 < 30
⇒ 34 – 8x < 30
⇒ 34 – 30 < 8x ⇒ 4 < 8x ⇒ 8A > 4 ⇒ x > \(\frac { 4 }{ 8 }\)
⇒ x > \(\frac { 1 }{ 2 }\)
∴ A is a positive integer,
∴ Solution set = {1, 2, 3, 4, }

Question 5.
If the replacement set is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, what is the solution set of the following mathematical sentences?
(a) x + \(\frac { 4 }{ 3 }\) = \(\frac { 7 }{ 3 }\)
(b) 2x + 1 < 3 (c) x – 6 > 10 – 6
(d) x + 5 = 20
(e) 2x + 3 ≥ 17
Solution:
Replacement set = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(a) x + \(\frac { 4 }{ 3 }\) = \(\frac { 7 }{ 3 }\) ⇒ x = \(\frac { 7 }{ 3 }\) – \(\frac { 4 }{ 3 }\)
⇒ x = \(\frac { 7 – 4 }{ 3 }\) = \(\frac { 3 }{ 3 }\) ⇒ x = 1
∴ Solution set = 1

(b) 2A + 1 – 1 < 3 ⇒ 2x < 3 – 1
⇒ 2A < 2 ⇒ x < \(\frac { 2 }{ 2 }\)
⇒ x < 1
∴ Solution set = {0}

(c) x – 6 > 10 – 6
⇒ x > 10 – 6 + 6
⇒ x > 10 – 12 ⇒ x > – 2
∴ Solution set = Φ {∵ – 2 ∉ to the given set}

(d) x + 5 = 20 ⇒ x = 20 – 5
⇒ x = 15
∴ Solution set = Φ {∵ 15 ∉ to the given set}

(e) 2A + 3 ≥ 17 ⇒ 2x ≥ 17 – 3
⇒ 2A ≥ 14 ⇒ x ≥ \(\frac { 14 }{ 2 }\)
⇒ A ≥ 7
∴ Solution set = {7, 8, 9}

Question 6.
(a) x ∈ {2, 4, 6, 9} and y ∈ {4, 6, 18, 27, 54}. Form all ordered pairs (x, y) such that x is a factory and x < y. (b) Find the truth set of the inequality x > y + 2 where
(x, y) ∈ {(1, 2), (2, 3), (5, 1), (7, 3), (5, 6), (6, 5)}.
Solution:
(a) x ∈ {2, 4, 6, 9} and y ∈ {4, 6, 18, 27, 54}
∵ A is a factor of y and x < y ∴ Ordered pairs will be {(2, 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)} (b) x > y + 2 where
(A, y) ∈ {(1, 2), (2, 3), (5, 1), (7, 3), (5, 6), (6, 5)}
If (A, y) = (1, 2) then 1 > 2 + 2 ⇒ 1 > 4 which is not true
If (A, y) = (2, 3) then 2 > 3 + 2 ⇒ 2 > 5 which is not true
If (A, y) = (5, 1) then 5 > 1 + 2 ⇒ 5 > 3 which is true
∴ (5, 1) is its solution
If (x, y) = (7, 3), then 7 > 3 + 2 ⇒ 7 > 5 which is true
∴ (7, 3) is its solution
If (x, y) = (5, 6). then 5 > 6 + 2 ⇒ 5 > 8 which is not true
If (x, y) = (6, 5). then 6 > 5 + 2 ⇒ 6 > 7 which is not true
∴ Solution or true set is {(5, 1), (7, 3)}

Question 7.
Find out the truth sets of the following open sentences; replacement sets are given against them.
(i) \(\frac { 5 }{ x }\) > 7 ; {1, 2}
(ii) \(\frac { 5 }{ x }\) > 2; {1, 2, 3, 4, 5, 6}
(iii) x² = 9; {-3, -2, -1, 1, 2, 3}
(tv) x + \(\frac { 1 }{ x }\) = 2 ; {0, 1,2, 3}
(v) 3x² < 2x ; {-4, -3, -2, -1, 0, 1, 2, 3, 4}
(vi) 2 (x – 3) < 1 ; {1, 2, 3, 4, …………… 10}. Solution: (i) \(\frac { 5 }{ x }\) > 7; {1, 2}
If x = 1, then \(\frac { 5 }{ 1 }\) > 7 ⇒ 5 > 7 which is not true
If x = 2, then \(\frac { 5 }{ 2 }\) > 7 which is not true
∴ x has no solution
Solution set = Φ

(ii) \(\frac { 5 }{ x }\) > 2; {1,2, 3,4, 5,6}
If x = 1, then \(\frac { 5 }{ 1 }\) > 2 which is true 5
If x = 2, then \(\frac { 5 }{ 1 }\) > 2 which is true
If x = 3, then \(\frac { 5 }{ 2 }\) > 2 which is not true
If x = 4, then \(\frac { 5 }{ 4 }\) > 2 which is not true
If x = 5, then \(\frac { 5 }{ 5 }\) > 2 ⇒ 1 > 2 which is not true
If x = 6, then \(\frac { 5 }{ 6 }\) > 2 which is not true o
∴ Solution set = {1, 2}

(iii) x² = 9; {-3, -2, -1, 1, 2, 3}
If x = -3, then (-3)² = 9 ⇒ 9 = 9 which is true
If x = -2, then (-2)² = 9 ⇒ 4 = 9 which is not true
If x = -1, then (-1)² = 9 ⇒ 1 = 9 which is not true
If x = 1, then (1 )² – 9 ⇒ 1 =9 which is not true
If x = 2, then (2)² – 9 ⇒ 4 = 9 which is not ture
If x = 3, then (3)² = 9 ⇒ 9 which is true
∴ Solution set = {-3, 3}

(iv) x + y = 2 ; {0, 1, 2, 3}
If x = 0, then 0 + \(\frac { 1 }{ 0 }\) = 2 which is not true
If x = 1, then 1 + \(\frac { 1 }{ 1 }\) = 2 ⇒ 1 + 1 = 2 ⇒ 2 = 2 which is true
If x = 2, then 2 + \(\frac { 1 }{ 2 }\) = 2 which not true 1
If x = 3, then 3 + \(\frac { 1 }{ 3 }\) = 2 which is not true
∴ Solution set = {1}

(v) 3x² < 2x ; {- 4, – 3, – 2, – 1, 0, 1, 2, 3, 4} If x = – 4, then 3 (- 4)² < 2 x (- 4)
⇒ 3 x 16 < – 8 ⇒ 48 < – 8 which is not true If x = – 3, then 3 (- 3)² < 2 x (- 3)
⇒ 3 x 9 < – 6 ⇒ 27 < – 6 which is not true If x = – 2, then 3 (-2)² < 2 x (-2)
⇒ 3 x 4 < – 4 ⇒ 12 < – 4 which is not true If x = – 1, then 3 (-1)² < 2 x (-1)
3 x 1 < – 2 ⇒ 3 < – 2 which is not true
If x = 0. then 3 (0)² < 2 (0) ⇒ 3 x 0 < 2 x 0 ⇒ 0 < 0 which is not true
If x = 1, then 3 (1)² <2xl⇒ 3xl<2 ⇒ 3 < 2 which is not true
If x = 2, then 3 (2)² <2×2⇒ 3×4<4 ⇒ 12 < 4 which is not true
If x = 3, then 3 (3)² <2×3⇒ 3*9<6 ⇒ 27 < 6 which is not true
If x = 4, then 3 (4)² <2×4⇒ 3xl6<8 ⇒ 48 < 8 which is not true
∴ Solution set = Φ

(vi) 2 (x – 3) < 1 ; {2, 3, 4, 10}
If x = 1. then 2 (1 – 3) < 1 ⇒ 2 x (-2) < 1
⇒ – 4 < 1 which is true
If x = 2, then 2 (2 – 3) < 1 ⇒ 2 (-1) < 1
⇒ – 2 < 1 which is true
If x = 3, then 2(3 – 3) < 1
⇒ 2 x 0 < 1
If x = 4, then 2(4 – 3) < 1 ⇒ 2 x 1
⇒ 2 < 1 which is not true
If x = 5, then 2 (5 – 3) < 1 ⇒ 2 x 2 < 1
⇒ 4 < 1 which is not true
If x = 10, then 2 (10 – 3) < 1 ⇒ 2 x 7 < 1
⇒ 14 < 1 which is not true
∴ Solution set = {1, 2, 3}

Question 8.
Statement: The sum of the lengths of any two sides of a triangle is always greater than the length of its third side. Let x, x + 1, x + 2 be the lengths of the three sides of a triangle.
(i) Write down the three inequations in x, each of which represents the given statement.
(ii) List the set of possible values of x which satisfy all the three inequations obtained in your answer to part (i) above, given that x is an integer.
Solution:
It is given that in a triangle,
Sum of any two sides is greater than the third side
Let x, x + 1 and x + 2 be the lengths of the three sides of a triangle,
(i) then the three inequations can be (a) x > 1 (b) x > 2 (c) x > 3
(ii) When x > 1, then the sides will be (2, 3, 4) When x > 2, then the sides will be (3, 4, 5) and when x > 3, then the side will be (4, 5,6)
Set of solution = {2, 3, 4, }
P.Q. P is the solution set of 8x – 1 > 5x + 2 and Q is the solution set of 7x – 2 ≥ 3 (x + 6), where x ∈ N. Find the set P ∩ Q.
Solution:
P is the solution set of 8x – 1 > 5x + 2
Q is the solution set of 7x – 2 ≥ 3 (x + 6)
Where x ∈ N.
Now 8x – 1 > 5x + 2
⇒ 8x – 5x > 2 + 1 ⇒ 3x > 3
⇒ x > \(\frac { 3 }{ 3 }\) ⇒ x > 1
∴ P = {2, 3, 4, 5, 6, 7, 8, }
and 7x – 2 ≥ 3 (x + 6)
⇒ 7x – 2 ≥ 3x + 18
⇒ 7x – 3x ≥ 18 + 2 ⇒ 4x ≥ 20
⇒ x ≥ \(\frac { 20 }{ 4 }\) ⇒ x ≥ 5
∴ Q = {5, 6, 7, 8, }
∴ P ∩ Q = {5, 6, 7, 8, }

Question 9.
Answer true or false.
(a) If x + 10 = y + 14, then x > y.
(b) | – 4 | – 4 = 8.
(c) If 10 – x > 3, then x < 7. (d) If p = q + 2, then p > q.
(e) If a and b are two negative integers such that, a < b. then \(\frac { 1 }{ a }\) < \(\frac { 1 }{ a }\) (f) 3 ∈ {x : 3x – 2 > 5}.
Solution:
(a) True
∵ x + 10 = y + 14
⇒ x = y + 14 – 10
⇒ x = y + 4
∴ x > y

(b) False
∵ | – 4 | – 4 = 8
⇒ 4 – 4 = 8 and 0 = 8
Which is not possible

(c) True
∵ 10 – x > 3
∴ 10 – 3 > x ⇒ 7 > x
⇒ x < 7 (d) True ∵ p = q+ 2 ∴ p > q

(e) False
∵ a, b are two negative integers and a < b then \(\frac { 1 }{ a }\) > \(\frac { 1 }{ b }\)

(f) True
∵ 3 ∈ {x : 3x – 2 > 5}
3x – 2 ≥ 5 ⇒ 3x ≥ 5 + 2
⇒ 3x ≥ 7 ⇒ x ≥ \(\frac { 7 }{ 3 }\)
∴ Solution set = {3, 4, 5, 6, …………. } (∵ x ∈ N)

Question 10.
Find the solution of the inequation 2 ≤ 2p – 3 ≤ 5.
Hence graph the solution set on the number line given below.

Solution:
2 ≤ 2p – 3 ≤ 5
(i) 2 ≤ 2p – 3
⇒ 2 + 3 < 2p
⇒ 5 ≤ 2p
⇒ \(\frac { 5 }{ 2 }\) ≤ 2p

(ii) 2p – 3 ≤ 5
⇒ 2p ≤ 5 + 3
⇒ 2p ≤ 8
⇒ p ≤ \(\frac { 8 }{ 2 }\)
⇒ p ≤ 4
From (i) and (ii)
\(\frac { 5 }{ 2 }\) ≤ P ≤ 4 ⇒ 2\(\frac { 1 }{ 2 }\) ≤ p ≤ 4
∴ Solution set = {2\(\frac { 1 }{ 2 }\) ≤ P ≤ 4, p ∈ R}
The solution set is given below on the number line.

Question 11.
If x is a negative integer, find the solution set of \(\frac { 2 }{ 3 }\) + \(\frac { 1 }{ 3 }\) (x + 1) > 0.
Solution:
x is a negative integer
\(\frac { 2 }{ 3 }\) + \(\frac { 1 }{ 3 }\) (x + 1) > 0.
⇒ \(\frac { 2 }{ 3 }\) + \(\frac { x }{ 3 }\) + \(\frac { 1 }{ 3 }\) > 0.
⇒ \(\frac { 2 }{ 3 }\) + \(\frac { 1 }{ 3 }\) + \(\frac { x }{ 3 }\) > 0.
⇒ \(\frac { 2+1 }{ 3 }\) + \(\frac { x }{ 3 }\) > 0 ⇒ 1 + \(\frac { x }{ 3 }\) > 0
⇒ \(\frac { x }{ 3 }\) > – 1 ⇒ x > – 3
∴ Solution set = {-2, -1}

Question 12.
Write open mathematical sentences, using x for the variable whose graphs would be

Solution:
(i) Here solution set = {x : x ≤ – 2, x ∈ R}
(ii) Here solution set = {x : x ≤ 4, x ∈ R}
(iii) Here solution set = {x : 4 ≤ x ≤ 5, x ∈ N}
(iv) Here solution set = {x : 1 ≤ x ≤ 5, x ∈ N, x is odd}
(v) Here solution set = (x : x > – 2, x ∈ R}

Question 13.
Answer true or false
(i) If 2 – x < 0, then x > 2
(ii) The graph of the inequation y ≤ 2x includes the origin.
Solution:
(i) True
∵ 2 – x < 0 ⇒ 2 < x ⇒ x > 2

(ii) True
∵ y ≤ 2x and origin is (0, 0)
∴ 0 ≤ 2 x 0 ⇒ 0 ≤ 0

Question 14.
If 25 – 4x ≤ 16, find :
(i) the smallest value of x when x is a real number
(ii) the smallest value of x when x is an integer.
Solution:
25 – 4x ≤ 16
⇒ 25 – 16 ≤ 4x ⇒ 9 ≤ 4x
⇒ 4x ≥ 9 ⇒ x ≥ \(\frac { 9 }{ 4 }\)
∴ Smallest value of x when x is a real number
= \(\frac { 9 }{ 4 }\) = 2\(\frac { 1 }{ 4 }\) (∵ \(\frac { 9 }{ 4 }\) is a real number)

(ii) Smallest value of x when x is a real number = 3 (∵ 3 > \(\frac { 9 }{ 4 }\))

Question 15.
Given : x ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9}, find the values of x for which – 3 < 2x – 1 < x + 4.
Solution:
– 3 < 2x – 1 < x + 4
(i) – 3 < 2x – 1 ⇒ – 3 + 1 < 2x
⇒ – 2 < 2x ⇒ \(\frac { -2 }{ 2 }\) < x ⇒ – 1 < x

(ii) 2x – 1 < x + 4 ⇒ 2x – x < 4 + 1 ⇒ x < 5
From (i) and (ii)
– 1 < x < 5
∴ Solution set will be {1, 2, 3, 4}

Question 16.
Solve the inequality : 2x – 10 < 3x – 15.
Solution:
2x – 10 < 3x – 15
⇒ – 10 + 15 < 3x – 2x
⇒ 5 < x ⇒ x > 5
∴ Solution set = {x : x > 5}

Question 17.
Solve the inequation : 3 – 2x ≥ x – 12, given that x ∈ N.
Solution:
3 – 2x ≥ x – 12
⇒ 3 + 12 ≥ x + 2x ⇒ 15 ≥ 3x
⇒ 3x ≤ 15 ⇒ x ≤ \(\frac { 15 }{ 2 }\)
⇒ x ≤ 5
∴ Solution set = {1, 2, 3, 4, 5} (∵ x ∈ N)

Question 18.
x ∈ {real numbers} and – 1 < 3 – 2x ≤ 7, evaluate x and represent it on a number line.
Solution:
x ∈ {Real numbers}
-1 < 3 – 2x ≤ 7
(i) – 1 < 3 – 2x ⇒ – 1 – 3 < – 2x
⇒ – 4 < – 2x ⇒ 4 > 2x
⇒ \(\frac { 4 }{ 2 }\) x ⇒ 2 > x ⇒ x < 2

(ii) and 3 – 2x < 7 ⇒ – 2x ≤ 7 – 3
⇒ – 2x ≤ 4 ⇒ – x ≤ \(\frac { 4 }{ 2 }\) ⇒ – x ≤ 2
⇒ – 2 ≤ x
From (i) and (ii)
The solution is – 2 ≤ x < 2
Number line

Question 19.
Find the range of values of x which satisfies
– 2\(\frac { 2 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) < 3\(\frac { 1}{ 3 }\), x ∈ R
Graph these values of x on the number line.
Solution:
– 2\(\frac { 2 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) < 3\(\frac { 1}{ 3 }\)
⇒ \(\frac { -8 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) < \(\frac { 10}{ 3 }\)
(i) \(\frac { -8 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) + \(\frac { 1 }{ 3 }\) ⇒ \(\frac { -8 }{ 3 }\) – \(\frac { 1}{ 3 }\) ≤ x
⇒ \(\frac { -9 }{ 3 }\) ≤ x ⇒ – 3 ≤ x

(ii) x + \(\frac { 1}{ 3 }\) < \(\frac { 10 }{ 3 }\) ⇒ \(\frac { 10 }{ 3 }\) – \(\frac { 1 }{ 3 }\)
⇒ x < \(\frac { 9 }{ 3 }\) ⇒ x < 3
From (i) and (ii)
Solution is – 3 < x < 3
Number line

Question 20.
Solve and graph the solution set of :
(a) 6 ≥ 2 – x, \(\frac { x }{ 3 }\) + 2 < 3; x ∈ R
(b) \(\frac { x }{ 2 }\) < \(\frac { 6-x }{ 4 }\), \(\frac { 2-x }{ 6 }\) < \(\frac { 7-x }{ 9 }\); x ∈ R
Solution:
(a) 6 ≥ 2 – x ⇒ 6 – 2 ≥ – x
⇒ 4 ≥ – x ⇒ x ≥ – 4 ⇒ – 4 ≤ x … (i)
and \(\frac { x }{ 3 }\) + 2 < 3 ⇒ x + 6 < 9 (Multiplying by 3)
⇒ x < 9 – 6 ⇒ x < 3 … (ii),
From (i) and (ii)
∴ Solution = – 4 ≤ x < 3
Number line

(b) \(\frac { x }{ 2 }\) < \(\frac { 6-x }{ 4 }\) ⇒ 2x < 6 – x (Multiplying by 4)
⇒ 2x + x < 6 ⇒ 3x < 6
⇒ x < \(\frac { 6 }{ 3 }\) ⇒ x < 2 … (i)
and \(\frac { 2-x }{ 6 }\) < \(\frac { 7-x }{ 9 }\)
⇒ 3 (2 – x) 2 (7 – x)
(Multiplying by L.C.M. of 6, 9 = 18)
⇒ 6 – 3x < 14 – 2x
⇒ 6 – 14 < – 2x + 3x ⇒ – 8 < x … (ii)
From (i) and (ii)
– 8 < x < 2
Number line

Question 21.
Find the range of values of x which satisfy :
\(\frac { -1 }{ 3 }\) ≤ \(\frac { x }{ 2 }\) – 1\(\frac { 1 }{ 3 }\) < \(\frac { 1 }{ 6 }\); x ∈ R
Graph these values of x on the real number line.
Solution:
\(\frac { -1 }{ 3 }\) ≤ \(\frac { x }{ 2 }\) – 1\(\frac { 1 }{ 3 }\) < \(\frac { 1 }{ 6 }\)
(i) \(\frac { -1 }{ 3 }\) ≤ \(\frac { x }{ 2 }\) – 1\(\frac { 1 }{ 3 }\) ⇒ \(\frac { -1 }{ 6 }\)
Multiplying by the L.C.M. of 3, 2 = 6
– 2 < 3x – 8 ⇒ – 2 + 8 ≤ 3x
⇒ 6 ≤ 3x ⇒ \(\frac { 6 }{ 3 }\) ≤ x
⇒ 2 ≤ x …. (i)
and \(\frac { x }{ 2 }\) – 1\(\frac { 1 }{ 3 }\) < \(\frac { 1 }{ 6 }\) ⇒ \(\frac { x }{ 2 }\) – \(\frac { 4 }{ 3 }\) < \(\frac { 1 }{ 6 }\)
Multiplying by the L.C.M. of 2, 3,6 = 6
3x – 8 < 1 ⇒ 3x < 1 + 8
⇒ 3x < 9 ⇒ x < \(\frac { 9 }{ 3 }\)
⇒ x < 3 … (ii)
from (i) and (ii)
2 ≤ x ≤ 3
∴ Number line

Question 22.
Write down the range of real values of x for which the inequation x > 3 and – 2 < x < 5 are both true.
Solution:
x > 3
3 < x … (i)
and -2 < x < 5
∴ Range = {-2, -1,0, 1, 2, 3, 4} … (ii)
∴ From (i) and (ii)
Range = {3 < x < 5; x ∈ R}

Question 23.
Solve and graph the solution set of 3x – 4 > 11 or 5 – 2x < 7; x ∈ R}
Solution:
3x – 4> 11 ⇒ 3x > 11 + 4
⇒ 3x > 15 ⇒ x > \(\frac { 15 }{ 3 }\)
⇒ x > 5 … (i)
and 5 – 2x ≥ 7 ⇒ 5 – 7 > 2x
⇒ – 2 ≥ 2x ⇒ \(\frac { -2 }{ 2 }\) > x
⇒ -1 > x
⇒ x ≤ – 1, x ∈ R … (ii)
From (i) and (ii)
x > 5 or x ≤ – 1; x ∈ R
Graph of solution

Self Evaluation And Revision (LATEST ICSE QUESTIONS)

Question 1.
Solve the following inequation and graph the solution set on the number line 2x – 3 < x + 2 ≤ 3x + 5, x ∈ R.
Solution:
2x – 3 < x + 2 ≤ 3x + 5, x ∈ R 2x – 3 < x + 2
⇒ 2x – x < 2 + 3 ⇒ x < 5 … (i)
and x + 2 ≤ 3x + 5 ⇒ 2 – 5 ≤ 3x – x
⇒ – 2 ≤ 2x ⇒ – \(\frac { 3 }{ 2 }\) ≤ x … (ii)
From (i) and (ii)
Solution: \(\left\{\frac{-3}{2} \leq x<5, x \in \mathrm{R}\right\}\)
On number line

Question 2.
Solve the inequation : 12 + 1 \(\frac { 5 }{ 6 }\) x ≤ 5 + 3x, x ∈ R. Represent the solution on a number line.
Solution:

Question 3.
Solve the inequation : – 3 ≤ 3 – 2x < 9, x ∈ R. Represent your solution on a number line.
Solution:
– 3 ≤ 3 – 2x < 9, x ∈ R
– 3 ≤ 3 – 2x
– 3 – 3 ≤ – 2x
⇒ – 6 ≤ – 2x
⇒ 2x ≤ 6 ⇒ x ≤ \(\frac { 6 }{ 2 }\)
⇒ x ≤ 3 … (i)
and 3 – 2x < 9
⇒ – 2x < 9 – 3 ⇒ – 2x < 6 ⇒ x > \(\frac { 6 }{ – 2 }\) ⇒ x > – 3
⇒ – 3 < x … (ii)
From (i) and (ii)
– 3 < x ≤ 3
∴ Solution = {x : – 3 < x ≤ 3}
On the number line

Question 4.
Find the value of x, which satisfies the inequation – 2 ≤ \(\frac { 1 }{ 2 }\) – \(\frac { 2x }{ 3 }\) ≤ 1\(\frac { 5 }{ 6 }\), x ∈ N.
Graph the solution on the number line.
Solution:

Question 5.
Solve the following inequation, and graph the solution on the number line
(2x – 5 ≤ 5x + 4 < 11, x ∈ R)
Solution:
2x – 5 ≤ 5x + 4 < 11, x ∈ R
Now 2x – 5 ≤ 5x + 4
⇒ – 5 – 4 ≤ 5x – 2x ⇒ – 9 ≤ 3x
⇒ \(\frac { – 9 }{ 3 }\) ≤ x ⇒ – 3 ≤ x … (i)
and 5x + 4 < 11 ⇒ 5x < 11 – 4
⇒ 5x < 7 ⇒ x \(\frac { 7 }{ 5 }\) ⇒ x < 1.4 … (ii)
From (i) and (ii)
Solution = {- 3 < x < 1.4, x ∈ R}
On number line

Question 6.
Solve 2 ≤ 2x – 3 ≤ 5, x ∈ R and mark it on a number line.
Solution:
2 ≤ 2x – 3 ≤ 5, x ∈ R
Now 2 ≤ 2x – 3 ⇒ 2 + 3 ≤ 2x
⇒ 5 ≤ 2x ⇒ \(\frac { 5 }{ 2 }\) ≤ x ⇒ 2.5 ≤ x … (i)
and 2x – 3 ≤ 5 ⇒ 2x ≤ 5 + 3
⇒ 2x ≤ 8 ⇒ x ≤ \(\frac { 8 }{ 2 }\) ⇒ x ≤ 4 … (ii)
From (i) and (ii)
Solution = {2.5 ≤ x ≤ 4, x ∈ R}
On number line

Question 7.
Given that x ∈ I, solve the inequation and graph the solution on the number line. 3 ≥ \(\frac { x-4 }{ 2 }\) + \(\frac { x }{ 3 }\) ≥ 2.
Solution:
3 ≥ \(\frac { x-4 }{ 2 }\) + \(\frac { x }{ 3 }\) ≥ 2, x ∈ 1
Now 3 ≥ \(\frac { x-4 }{ 2 }\) + \(\frac { x }{ 3 }\) ⇒ 3 ≥ \(\frac{3 x-12+2 x}{6}\)
⇒ 18 ≥ 5x – 12 ⇒ 18 + 12 ≥ 5x
⇒ 30 ≥ 5x ⇒ \(\frac { 30 }{ 5 }\) ≥ x
6 ≥ x ⇒ x ≤ 6 … (i)
and \(\frac { x-4 }{ 2 }\) + \(\frac { x }{ 3 }\) ≥ 2
\(\frac{3 x-12+2 x}{6}\) ≥ 2 ⇒ 5x – 12 ≥ 12
⇒ 5x ≥ 12 + 12 ⇒ 5x ≥ 24
⇒ x ≥ \(\frac { 24 }{ 5 }\) ⇒ x ≥ 4.8
⇒ 4.8 ≤ x … (ii)
From (i) and (ii)
4.8 ≤ x ≤ 6
∵ x ∈ I
∴ Solution set = {5, 6}
On the number line

Question 8.
Give that x ∈ R, solve the following inequality and graph the solution on the number line inequality : – 1 ≤ 3 + 4x < 23.
Solution:
– 1 ≤ 3 + 4x < 23, x ∈ R
Now – 1 ≤ 3 + 4x
⇒ – 1 – 3 ≤ 4x
⇒ – 4 ≤ 4x
⇒ \(\frac { -4 }{ 4 }\) ≤ x ⇒ – 1 ≤ x … (i)
and 3 + 4x < 23 ⇒ 4x < 23 – 3
⇒ 4x < 20 ⇒ x < \(\frac { 20 }{ 4 }\) … (ii)
From (i) and (ii)
Solution : {-1 < x < 5, x ∈ R}
On the number line

Question 9.
Solve the following inequation and graph the solution on the number line :
-2\(\frac { 2 }{ 3 }\) ≤ x + \(\frac { 2 }{ 3 }\) < 3\(\frac { 1 }{ 3 }\); x ∈ R.
Solution:
-2\(\frac { 2 }{ 3 }\) ≤ x + \(\frac { 2 }{ 3 }\) < 3\(\frac { 1 }{ 3 }\); x ∈ R.
\(\frac { -8 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) < \(\frac { 10 }{ 3 }\)
Now \(\frac { -8 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) ⇒ \(\frac { -8 }{ 3 }\) – \(\frac { 1 }{ 3 }\) ≤ x

Question 10.
Solve the given inequation and graph the solution on the number line :
2y – 3 < y + 1 ≤ 4y + 7, y ∈ R.
Solution:
2y – 3 < y + 1 ≤ 4y + 7, y ∈ R
Now 2y – 3 < y + 1
⇒ 2y – y < 1 + 3 ⇒ y < 4 … (i)
and y + 1 ≤ 4y + 7 ⇒ 1 – 7 ≤ 4y – y
⇒ – 6 ≤ 3 y ⇒ \(\frac { -6 }{ 3 }\) ≤ y
⇒ – 2 ≤ y … (ii)
From (i) and (ii)
Solution = {- 2 ≤ y < 4, y ∈ R}
On the number line

Question 11.
Solve the inequation and represent the solution set on the number line
– 3 + x ≤ \(\frac { 8x }{ 3 }\) + 2 ≤ \(\frac { 14 }{ 3 }\) + 2x, where x ∈ I.
Solution:
Given :
– 3 + x ≤ \(\frac { 8x }{ 3 }\) + 2 ≤ \(\frac { 14 }{ 3 }\) + 2x, where x ∈ I
⇒ – 3 + x ≤ \(\frac { 8x }{ 3 }\) + 2 … (i)
and \(\frac { 8x }{ 3 }\) + 2 ≤ \(\frac { 14 }{ 3 }\) + 2x … (ii)
⇒ – 5 ≤ \(\frac { 5x }{ 3 }\) and \(\frac { 2x }{ 3 }\) ≤ \(\frac { 8x }{ 3 }\)
⇒ x ≥ – 3 and x ≤ 4
∴ – 3 ≤ x ≤ 4
Solution set = {-3, -2, -1, 0, 1,2, 3, 4}
Number line

Question 12.
Solve the following inequation and represent the solution set on the number line.
– 3 < – \(\frac { 1 }{ 2 }\) – \(\frac { 2x }{ 3 }\) ≤ \(\frac { 5 }{ 6 }\), x ∈ R.
Solution:
– 3 < – \(\frac { 1 }{ 2 }\) – \(\frac { 2x }{ 3 }\) ≤ \(\frac { 5 }{ 6 }\), x ∈ R.
⇒ – 3 < \(\frac { -3-4x }{ 6 }\) ≤ \(\frac { 5 }{ 6 }\)
⇒ – 18 < – 3 – 4x ≤ 5
⇒ – 18 + 3 < – 4x ≤ 5 + 3
⇒ – 15 < – 4x ≤ 8 ⇒ \(\frac { -15 }{ -4 }\) > x ≥ \(\frac { 8 }{ -4 }\)
⇒ \(\frac { 15 }{ 4 }\) > x ≥ – 2
⇒ 3.75 > x ≥ – 2

Question 13.
Solve the following inequation and represent the solution set on the number line
2x – 5 ≤ 5x + 4 < 11, where x ∈ 1.
Solution:

Question 14.
Solve the following inequation and represent the solution set on the number line:
4x – 19 < \(\frac { 3x }{ 5 }\) – 2 ≤ – \(\frac { 2 }{ 5 }\) + x, x ∈ R.
Solution:
we have 4x – 19 < \(\frac { 3x }{ 5 }\) – 2 ≤ – \(\frac { 2 }{ 5 }\) + x, x ∈ R.
⇒ 4x – 19 < \(\frac { 3x }{ 5 }\) – 2 and \(\frac { 3x }{ 5 }\) – 2 ≤ \(\frac { – 2 }{ 5 }\) + x, x ∈ R.
⇒ 4x – \(\frac { 3x }{ 5 }\) < 17 and – 2 + \(\frac { 2 }{ 5 }\) ≤ x – \(\frac { 3x }{ 5 }\), x ∈ R.
⇒ \(\frac { 20x – 3x }{ 5 }\) < 17 and \(\frac{-10+2}{5} \leq \frac{5 x-3 x}{5}\), x ∈ R.
⇒ \(\frac { 17x }{ 5 }\) < 17 and \(\frac { -8 }{ 5 }\) ≤ \(\frac { 2x }{ 5 }\) + x, x ∈ R.
⇒ x < 5 and – 4 ≤ x, x ∈ R.
⇒ – 4 ≤ x < 5, x ∈ R
Hence, solution set is {x: -4 ≤ x ≤ 5, x ∈ R} The solution set is represented on the number line as below:

Question 15.
Solve the following inequation, write the solution set and represent it on the number line:
\(-\frac{x}{3} \leq \frac{x}{2}-1 \frac{1}{3}<\frac{1}{6}, x \in R .\)
Solution:

Question 16.
Find the values of x, which satisfy the inequation \(-2 \frac{5}{6}<\frac{1}{2}-\frac{2 x}{3} \leq 2\), x ∈ W. Graph the solution set on the number line.
Solution:

(Solution set is shown by dots on the number line).

Question 17.
Solve the following inequation and write the solution set:
13x – 5 < 15x + 4 < 7x + 12, x ∈ R Represent the solution set on a real number line.
Solution:
13x – 5 < 15x + 4 < 7x + 12, x ∈ R
13x – 5 < 15x + 4 and 15x + 4 < 7x + 12
13x < 15x + 9 and 15x < 7x + 8
0 < 2x + 9 and 8x < 8
– 9 < 2x and x < 1
– \(\frac { 9 }{ 2 }\) < x and x < 1 9
∴ – \(\frac { 9 }{ 2 }\) < x < 1
i.e. – 4.5 < x < 1
Hence, the solution set is {x : – 4.5 < x < 1, x ∈ R}
The required line is

Question 18.
Solve the following inequation, write the solution set and represent it on the number line. – 3(x – 7) ≥ 15 – 7x > \(\frac { x+1 }{ 3 }\), x ∈ R.
Solution:
– 3(x – 7) ≥ 15 – 7x > \(\frac { x+1 }{ 3 }\)
⇒ – 3(x – 7) ≥ 15 – 7x > and 15 – 7x > \(\frac { x+1 }{ 3 }\)
⇒ – 3x + 21 ≥ 15 – 21 and 45 – 21x > x + 1
⇒ – 3x + 7x ≥ 15 – 21 and 45 – 1 > x + 2x
⇒ 4x ≥ – 6 and 44 > 22x
⇒ x ≥ \(\frac { – 3 }{ 2 }\) and 2 > x
Hence, the solution set is

Parents can use ICSE Class 10 Maths Solutions S Chand Chapter 19 Histogram and Ogive Ex 19(d) to provide additional support to their children.

S Chand Class 10 ICSE Maths Solutions Chapter 19 Histogram and Ogive Ex 19(d)

Question 1.
Find out the median by plotting the following in the form of an ogive, where ‘m’ denotes the mid-mark of a class :

m	12.5	17.5	22.5	27.5	32.5	37.5	42.5	47.5	52.5
f	2	22	19	14	3	4	6	1	1

Solution:
∵ 12.5, 17.5, etc. are class marks
∴ The corresponding classes will be 10-15, 15-20, etc.

Class	m	f	c.f
10-15	12.5	2	2
15-20	17.5	22	24
20-25	22.5	19	43
25-30	27.5	14	57
30-35	32.5	3	60
35-40	37.5	4	64
40-45	42.5	6	70
45-50	47.5	1	71
50-55	52.5	1	72

Plot the points (15, 2), (20, 24), (25, 43), (30, 57), (35, 60), (40, 64), (45, 70), (50, 71), (55, 72) on the graph and join them with free hand to get on ogive as shown.

Here n = 72 which is even
∴ Median = \(\frac { n }{ 2 }\)
= \(\frac { 72 }{ 2 }\) = 36
From 36 on 7-axis, draw a line parallel to x-axis which meets the curve at P. From P, draw a perpendicular, to x-axis which meet it at M which is median.
∴ Median = 23.5 (approx)

Question 2.
From the data given below draw an ogive and find the values of median and quartiles from the graph.

Years under	10	20	30	40	50	60
Number of persons	12	31	44	55	65	70

Solution:

Years under	10	20	30	40	50	60
Number of persons(c.f.)	12	31	44	55	65	70

Now plot the points (10, 12), (20, 31), (30, 44), (40, 55), (50, 65) and (60, 70) on the graph and join them with free hand to get an ogive as shown.

(i) Herein = 70
∴ Median = \(\frac { n }{ 2 }\) = \(\frac { 70 }{ 2 }\)th = 35th term.
From 35 (A) on y-axis, draw a line parallel to x-axis meeting the curve at P and from P, draw PL ⊥ x-axis
∴ L is the median which is = 23 years
∴Meadian = 23 years

(ii) First Quartile (Q₁) = \(\frac { n }{ 4 }\) = \(\frac { 7 0 }{ 4 }\) = 17.5
From 17.5 (B) on y-axis, draw a line parallel to x-axis meeting the curve at Q and from Q, draw QM ⊥ x-axis
∴ M is the Q₁ which is 13 years

(iii) Similarly for Q₃ = \(\frac { 3n }{ 4 }\) = \(\frac { 3×40 }{ 4 }\) = \(\frac { 210 }{ 4 }\) = 52.5
From 52.5 (C) on y-axis, draw a line parallel to x-axis meeting the curve at R and from R, RN ⊥ x-axis
Then N is the Q₃ which is = 37 years

Question 3.
From the following find out the median with the help of ogive curve :

Profit per shop less than	10	20	30	40	50	60
No. of shops	12	30	57	77	94	100

Solution:
Plot the points (10, 12), (20, 30), (30, 57), (40, 77), (50, 94) and (60, 100) on the graph and join them to get an ogive.
Here n = 100
∴ Median = \(\frac { n }{ 2 }\) =\(\frac { 100 }{ 2 }\)th = 50
From 50 (A) on y-axis, draw a line parallel to x-axis which meet the curve at P From P, draw PL ⊥ x-axis which meet it at L
L is the median which is 27.8 (approx)
∴ Median 27.8 or = 28 (approx)

Question 4.
The following are the marks obtained by 50 students in Statistics :

Marks less than	10	20	30	40	50	60
No. of students	4	10	30	40	47	50

Draw an ogive and find the median marks from it.
Solution:
Plot the points (10, 4), (20, 10), (30, 30), (40, 40), (50, 47) and (60, 50) on the graph and join them with free hand to get an ogive as shown on the graph,
Here n = 50
∴ Median = \(\frac { 50 }{ 2 }\) = 25th
From 25 on y-axis, draw a line parallel to x-axis which meets the curve at P. From P, draw a line perpendicular to x-axis which meet it at L
L is the median which is 27.8 = 28
Hence median = 28

Question 5.
100 pupils in a school have heights as tabulated below :

Height in cm	121-130	131-140	141-150	151-160	161-170	171-180
No. of pupils	12	16	30	20	14	8

Draw the ogive for the above data and from it determine the median. (Use graph paper).
Solution:
Writing the given class-interval in exclusive form,

Height (in cm) (class)	Frequency (f)	c.f.
120.5-130.5	12	12
130.5-140.5	16	28
140.5-150.5	30	58
150.5-160.5	20	78
160.5-170.5	14	92
170.5-180.5	8	100

Plot the points (130.5,12), (140.5,28), (150.5,58), (160.5,78), (170.5,92) and (180.5,100) on the graph and join them with free hand to get an ogive as shown.
Here n = 100
∴ Median =\(\frac { n }{ 2 }\) = \(\frac { 100 }{ 2 }\) =50th
∴ From 50 on y-axis, draw a line parallel to x-axis which meets the curve at P From P, draw PL ⊥x-axis
Then L is the median which is =147.8
= 148 (approx)
∴ Median =148 (approx)

Question 6.
Draw an ogive curve from the following data and find out (a) the median wage and (b) number of workers earning less than Rs. 55 per week :

Weekly wages (Rs.)	0-20	20-40	40-60	60-80	80-100
No. of workers	40	51	60	38	7

Solution:

Weekly wages	No. of workers (f)	c.f.
0-20	40	40
20-40	51	91
40-60	60	151
60-80	38	189
80-100	7	196

We plot the points (20, 40), (40, 91), (60, 151), (80, 189) and (100, 196) on the graph and join them with free hand to get an ogive as shown Here n= 196
∴ Median = \(\frac { n }{ 2 }\) = \(\frac { 196 }{ 2 }\) =98

From 98 on y-axis, draw a line parallel to x-axis is which meets the curve at P. From P, draw PL ⊥ x-axis
Then L is the median which is
∴ Median =42.3

(ii) From 55 on x-axis, draw a perpendicular on x-axis which meets the curve at Q. From Q, draw a line parallel to x-axis which meets y-axis at B
Then B =136
∴ Number of workers getting less then Rs. 55 per week =136

Question 7.
Represent the following data by a cumulative frequency curve and locate the median and quartiles graphically :

Income (Rs.)	50-60	40-50	30-40	20-30	10-20	0-10
No. of persons	100	200	200	150	100	50

Solution:
Rewriting the data in order

Income (Rs.)	0-10	10-20	20-30	30-40	40-50	50-60
No. of persons	50	100	150	200	200	100

Now writing in c.f.

Income (Rs.)	No. of persons (f)	c.f.
0-10	50	50
10-20	100	150
20-30	150	300
30-40	200	500
40-50	200	700

Plot the points (10, 50), (20, 150), (3o, 300), (40, 500), (50, 700) and (60, 800) on the graph and join them with free hand to get an ogive as shown Here n = 800
∴ Median = \(\frac { n }{ 2 }\)th = \(\frac { 800 }{ 2 }\)th = 400th term
From 400 on y-axis, draw a line parallel to x-axis which meets the curve at P. From P, draw PL ⊥ x-axis
∴ L is the median which is 35
Hence median = Rs. 35

(i) Q₁ = \(\frac { n }{ 4 }\) = \(\frac { 800 }{ 4 }\) = 200
From 200 on y-axis, draw a line parallel to x-axis which meets the curve at Q.
From Q draw QM ⊥ x-axis
∴ M is the Q₁ which is 23

(ii) Q₃ = \(\frac { 3n }{ 4 }\) = \(\frac { 3×800 }{ 4 }\) = 600
From 600 on y-axis, draw a line parallel to income in Rs. x-axis which meets the curve at R. From R, draw RN ⊥ x-axis
Then N is Q₃ which is 45
Hence Q₁ = Rs. 23 and Q₃ = Rs. 45

Question 8.
Draw the ogive for the following frequency distribution. Estimate the median from your graph :

Class	1-10	11-20	21-30	31-40	41-50	51-60	61-70
Frequency	2	5	7	12	8	7	4

Solution:
Writing the class in exclusive form, and in c.f.

Class	Frequency	c.f.
0.5-10.5	2	2
10.5-20.5	5	7
20.5-30.5	7	14
30.5-40.5	12	26
40.5-50.5	8	34
50.5-60.5	7	41
60.5-70.5	4	45

Plot the points (10.5, 2), (20.5, 7), (30.5, 14), (40.5, 26), (50.5, 34), (60.5, 41) and (70.5, 45) on the graph and join them with free hand to get an ogive as shown Here n = 45 which is odd
∴ Median = \(\frac { n + 1 }{ 2 }\)th = \(\frac { 45 + 1 }{ 2 }\) = 23th

From 23 on y-axis, draw a line parallel to x-axis which meets the curve at P.
From P, draw PL ⊥ x-axis, then L is the median which is (38.1)
Hence median = 38.1

Question 9.
Attempt this question on graph paper.

Age (yrs)	5-15	15-25	25-35	35-45	45-55	55-65	65-75
No. of casualties	6	10	15	13	24	8	7

due to accidents
(i) Construct the ‘less than’ cumulative frequency curve for the above data, using 2 cm = 10 years, on one axis and 2 cm = 10 casualties on the other.
(ii) From your graph determine.
(a) The median and
(b) the upper quartile
Solution:

Age (in years)	No. of casulaties (f)	c.f.
5-15	6	6
15-25	10	16
25-35	15	31
35-45	13	44
45-55	24	68
55-65	8	76
65-75	7	83

Plot the points (15, 6), (25, 16), (35, 31), (45, 44), (55, 68), (65, 76) and (75, 83) on the graph and join them with free hand to form an ogive (less than)

(i) Median, Here n = 83
∴ \(\frac { n }{ 2 }\) = \(\frac { 83 }{ 2 }\) = 41.5
From 41.5 on y-axis, draw a line parallel to x-axis meeting the curve at P.
From P draw a perpendicular PM on x-axis
M is the median which is 43 years.

(ii) Upper quartile (Q₃) = \(\frac { 3 }{ 4 }\) × n = \(\frac { 3 }{ 4 }\) × 83
= \(\frac { 249 }{ 4 }\) = 62.25

From 62.25 ony-axis, draw a line parallel to x-axis meeting the curve at Q. From Q, draw a perpendicular on x-axis meeting it at N.
N is the upper quardant which is 52 years
∴ Q₃ = 52 years

Question 10.
Marks scored by 400 students in an examination are as follows :

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of students	10	20	22	40	55	75	80	58	28	12

Draw an ogive and from it determine:
(i) the median mark,
(ii) pass marks if 80% of the students pass examination.
Solution:

Marks	No. of casulaties	c.f.
0-10	10	10
10-20	20	30
20-30	22	52
30-40	40	92
40-50	55	147
50-60	75	222
60-70	80	302
70-80	58	360
80-90	28	388
90-100	12	400

Plot the points (10, 10), (20, 30), (30, 52), (40, 92), (50,. 147), (60, 222), (70, 302), (80, 360), (90, 388), (100,400) on the graph and join them with free hand to form an ogive as shown below:

Here n = 400

(i) Median = \(\frac { n }{ 2 }\) = \(\frac { 400 }{ 2 }\) = 200
From 200 (A) on y-axis draw a line parallel to x-axis is meeting the curve at P. From P, draw a perpendicular on x-axis meeting it at M.
M is median which is 57.5
∴ Median = 257.5
∵ 80% students passed

(ii) No. of students passed = 80% of 400 = \(\frac { 80 }{ 100 }\) × 400 = 320
∴ No. of students who did not pass = 400 – 320 = 80
From a point 80 on y-axis, draw a line parallel to x-axis meeting the curve at Q.
From Q₁ draw a line perpendicular to x-axis meeting at N. Then N is the required pass marks.
Which is 36
∴ Pass marks = 36%

Question 11.
Draw an ogive for the following frequency distribution. Use your ogive to estimate
(i) the median
(ii) the number of students who obtained more than 75% marks (use square paper to solve this question).

Marks	0-9	10-19	20-29	30-39	40-49	50-59	60-69	70-79	80-89	90-99
No. of students	5	9	16	22	26	18	11	6	4	3

Solution:
Writing the classes in exclusive form,

Marks	No. of casulaties (f)	c.f.
0.5-9.5	5	5
9.5-19.5	9	14
19.5-29.5	16	30
29.5-39.5	22	52
39.5-49.5	26	78
49.5-59.5	18	96
59.5-69.5	11	107
69.5-79.5	6	113
79.5-89.5	4	117
89.5-99.5	3	120

Now plot the points (9.5, 5), (19.5, 14), (29.5, 30), (39.5, 52), (49.5, 78), (59.5, 96), (69.5, 107), (79.5, 113), (89.5, 117), and (99.5, 120) on the graph and join them with free hand to get an ogive as shown below :
Here n = 120

(i) Median: \(\frac { n }{ 2 }\) = \(\frac { 120 }{ 2 }\) = 60
From 60 on the y-axis draw a line parallel to x-axis meeting the curves at P. From P, draw a perpendicular on x-axis meeting it at M.
M is the median which is 43
∴ Median = 43

(ii) From 75 on the x-axis draw a line parallel toy-axis meeting the curve at Q. From Q, draw a line perpendicular to y-axis meeting at N.
Now Q = 110
Total no. of students getting more them 75% marks = 120 – 110= 10

Question 12.
In a public collection towards the erection of a memorial, 1000 people contributed sums of money varying from Rs. 1 to Rs. 100 (in units of Re. 1). The following table gives the frequency distribution of contribution :

Contribution in Rs.)	No. of people	Contribution (in Rs.)	No. of people
1-10	30	51-60	180
11-20	60	61-70	140
21-30	80	71-80	70
31-40	170	81-90	40
41-50	200	91-100	30

Using a suitable scale, draw on graph paper an ogive (cumulative frequency graph) and use it to answer the following :
(i) Estimate the median.
(ii) If it is agreed to allow only those who contributed Rs. 45 or more to attend the unveilling ceremony, what percentage would attend ?
Solution:
Writing the classes in exclusive forms

Contribution (in Rs.)	No. of people (f)	c.f.
0.5-10.5	30	30
10.5-20.5	60	90
20.5-30.5	80	170
30.5-40.5	170	340
40.5-50.5	200	540
50.5-60.5	180	720
60.5-70.5	140	860
70.5-80.5	70	930
80.5-90.5	40	970
90.5-100.5	30	1000

Plot the points (10.5, 30), (20.5, 90), (30.5, 170), (40.5, 340), (50.5, 540), (60.5, 720), (70.5, 860), (80.5, 930), (90.5, 970) and (100.5, 1000) on the graph and join them with free hand to get an ogive as shown.
Here n = 1000
∴ Median = \(\frac { n }{ 2 }\) = \(\frac { 1000 }{ 2 }\) = 500th term
From 500 on y-axis; draw a line parallel to x-axis which meets the curve at P. From P, draw PL ⊥ x-axis then L is the median which is 49
∴ Median = 49 rupees

(ii) From Rs. 45 on x-axis, draw a perpendicular which meets the curve at Q and from Q, draw a parallel line to x-axis which meets y-axis at B Then B is 440
∴ Persons who could attend the ceremony = 1000 – 440 = 560
Percentage = \(\frac { 560 × 100 }{ 1000 }\) = 56%

Self-Evaluation and Revision (LATEST ICSE QUESTIONS)

Question 1.
Draw a histogram to represent the following data :

Pocket money (in ₹)	No. of students
150-200	10
200-250	5
250-300	7
300-350	4
350-400	3

Solution:

Pocket money (in ₹)	No. of students
150-200	10
200-250	5
250-300	7
300-350	4
350-400	3

Represent the pocket money along x-axis
and number of students along y-axis and
form the histogram as shown

Question 2.
Using a graph paper, draw an Ogive for the following distribution which shows a record of the weight in kilogram of 200 students.

Weight	40-45	45-50	50-55	55-60	60-65	65-70	70-75	75-80
Frequency	5	17	22	45	51	31	20	9

Use the ogive to estimate the following :
(i) The percentage of students weighing 55 kg or more;
(ii) The weight above which the heaviest 30% of the students fall;
(iii) The number of students who are
(1) under weight
(2) over weight, if 55.70 kg is considered as the standard weight.
Solution:

Weight	Frequency	c.f.
40-45	5	5
45-50	17	22
50-55	22	44
55-60	45	89
60-65	51	140
65-70	31	171
70-75	20	191

Plot the points (45, 5), (50, 22), (55, 44), (60, 89), (65, 140), (70, 171), (75, 191) and (80, 200) on the graph and join them with free hand to get an ogive as shown
From 55 on x-axis, draw a perpendicular which meets the curve at A. From A, draw a line parallel to x-axis which meets j’-axis at D which is 44.

(i) Percentage number of students weighing 55 kg or more = 200 – 44 = 156 (From ogive)
∴ Percentage = \(\frac { 156 × 100 }{ 200 }\) = 78%

(ii) The weight above which the heaviest 30% of the students fall = \(\frac { 200 × 30 }{ 100 }\) = 60
∴ Heighest students from the graph will be 65 kg and more upto 80 kg = 31 + 20 + 9 = 60

(iii) Number of students who are under weight i.e. below 55.70 kg = 47
Over weight i.e. above 55.70 kg = 200 – 47 = 153

Question 3.
If the mean of the following distribution is 7.5 find the missing frequency f:

Variable	5	6	7	8	9	10	11	12
Frequency	20	17	f	10	8	6	7	6

Solution:

Variable (x)	Frequency (f)	f x
5 6 7 8 9 10 11 12	20 17 f 10 8 6 7 6	100 102 7 f 80 72 60 77 72
Total	74+f	563+7 f

Here mean = 7.5
But mean = \(\frac{\sum f x}{\sum f}\)
∴ 7.5 = \(\frac{563+7 f}{74+f} \)
⇒ 7.5 (74 + f) = 563 + 7f ⇒ 555 + 7.5f = 563 + 7f ⇒ 7.5f – 7f = 563 – 555
⇒ 0.5f = 8 ⇒ f = \(\frac { 8 }{ 0.5 }\) = \(\frac { 8 × 10 }{ 5 }\) = \(\frac { 8 × 2 }{ 1 }\) ⇒ f = 16

Question 4.
The median of the following observation 11, 12, 14, 18, (x + 4), 30, 32, 35, 41 arranged in ascending order is 24. Find x.
Solution:
Here n = 9 which is odd
∴ Median = \(\frac { n+ 1 }{ 2 }\)th term = \(\frac { 9 + 1 }{ 2 }\) = 5th term, which is x + 4
∴ x + 4 = 24 ⇒ x = 24 – 4 = 20
∴ x = 20

Question 5.
Find the mean of the following distribution :

Class interval	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	10	6	8	12	5	9

Solution:

Class interval	Mid value (x)	Frequency (f)	f.x.
20-30 30-40 40-50 50-60 60-70 70-80	25 35 45 55 65 75	10 6 8 12 5 9	250 210 360 660 325 675
Total		50	2480

∴ Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac{2480}{50}\) = 49.6

Question 6.
The daily wages of 160 workers in a building project are given below :

Wages in Rs.	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
No. of workers	12	20	30	38	24	16	12	8

Using a graph, draw an ogive for the above distribution. Use your ogive to estimate :
(i) the median wage of the workers;
(ii) the upper quartile wage of the workers;
(iii) the lower quartile wage of the workers;
(iv) the percentage of workers who earn more than Rs. 45 a day.
Solution:

Wages (in Rs.)	No. of workers (f)	c.f.
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80	12 20 30 38 24 16 12 8	12 32 62 100 124 140 152 160

Plot the points (10, 12), (20, 32), (30, 62), (40, 100), (50, 124), (60, 140), (70, 152) and (80, 160) on the graph and join them with free hand to get an ogive as shown (i) Here n = 160
∴ Median = \(\frac{ n }{ 2 }\) = \(\frac{ 160 }{ 2 }\) = 80th
From 80 on y-axis, draw a line parallel to x-axis meeting the curve at P. From P, draw PL perpendicular on x-axis which meets it at L.
L is median which is 35
∴ Median = Rs. 35

(ii) Upper quartile (Q₃) = \(\frac{3n}{4}\)
= \(\frac{3}{4}\) × 160 = 120
From 120 on y-axis, draw a parallel line to x-axis which meets the curve at Q. From Q, draw perpendicular on x-axis meeting it at M
M is Q₃ which is 47.50
∴ Upper quartile (Q₃) = Rs. 47.50

(iii) Lower quartile (Q₁) = \(\frac{ 1 }{ 4 }\)n = \(\frac{ 1 }{ 4 }\) × 160 = 40
From 40 on j-axis, draw a line parallel to x-axis meeting the curve at R. From R, draw a perpendicular on x-axis meeting it at N.
N is the lower quartile (Q₁) which is 22.5
∴ Lower quartile (Q₁) = Rs. 22.5

(iv) Number of workers earning more than Rs. 45 per day
From 45 on x-axis, draw a perpendicular which i.-.eets the curve at B. From B, draw a line parallel to x-axis meeting 7-axis at C which is 112
∴ Number of workers getting more than Rs. 45 = 160 – 112 = 48
Percentage = \(\frac{ 48 × 100 }{ 160 }\) = 30%

Question 7.
Find the mean of the following distribution :

Class inverval	0-10	10-20	20-30	30-40	40-50
Frequency	10	6	8	12	5

Solution:

Class interval	Mid value {x}	Frequency (f)	f.x
0-10 10-20 20-30 30-40 40-50	5 15 25 35 45	10 6 8 12 5	50 90 200 420 225
Total		41	985

∴ Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac{985}{41}\) = 24.02

Question 8.
The table below shows the distributions of the scores obtained by 120 shooters in a shooting competition. Using a graph sheet, draw in ogive for the distributions.

Scores obtained	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of shooters	5	9	16	22	26	18	11	6	4	3

Use your ogive to estimate : (i) The median (ii) The inter quartile range (iii) The number of shooters who obtained more than 75% scores.
Solution:

Score obtained (class interval)	Number of students (f)	c.F.
0-10	5	5
10-20	9	14
20-30	16	30
30-40	22	52
40-50	26	78
50-60	18	96
60-70	11	107
70-80	6	113
80-90	4	117
90-100	3	120

Plot the points (10, 5), (20, 14). (30, 30). (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117), and (100, 120) on the graph and join them with free hand to get an ogive as shown below Here n = 120

(i) Median = \(\frac{n}{2}\)th = \(\frac{120}{2}\) = 60th term
From 60 on y-axis draw a line parallel to.x-axis meeting the curve at P. From P, draw a perpendi-cular to x-axis meeting it at L. L, is median which is 43
∴ Median = 43

(ii) Inner quartile (Q₁) = \(\frac{n}{4}\)th = \(\frac{120}{4}\) = 30 th term
From 30 on y-axis draw a line parallel to x-axis meeting the curve at Q. From Q, draw perpendi-cular on x-axis meeting it at M.
M is Q₁ which is 30
Upper quartile (Q₃) = \(\frac{3n}{4}\)th = \(\frac{3×120}{4}\)th = 90th term
From 90 on y-axis, draw a line parallel to x-axis meeting the curve at R. From R draw perpendi-cular on x-axis meeting it at N. Then N is Q₃ which is 57
∴ Q₃ = 57
Now interquartile range = Q₃ – Q₁ = 57 – 30 = 27

(iii) No. of shooter getting more then 75% score
From 75 on x-axis, draw a perpendicular meeting the curve at B and from B, draw a line parallel to x-axis meeting v-axis at C.
C is 110
∴ Number of shooter getting more them 75% score = 120 – 110= 10

Question 9.
Using a graph paper, draw an Ogive for the following distribution which shows the marks obtained in the General Knowledge Paper by 100 students.

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
No. of students	5	10	20	25	15	12	4	9

(i) the median,
(ii) the number of students who score marks above 65.
Solution:

Marks	Number of students (f)	c.f.
0-10	5	5
10-20	10	15
20-30	20	35
30-40	25	60
40-50	15	75
50-60	12	87
60-70	4	91
70-80	9	100

Plot the points (10, 5), (20, 15), (30, 35), (40, 60), (50, 75), (60, 87), (70, 91), (80, 100) on the graph and join them with free hand to get an ogive as shown on the graph Here n = 100

(i) Median = \(\frac { n }{ 2 }\) = \(\frac { 100 }{ 2 }\) = 50
From 50 on y-axis, draw a line parallel to x-axis meeting the curve at P.
From P, draw a perpendi-cular on x-axis meeting it at L.
L is median which is 35
∴ Median = 35 marks

(ii) Number of students getting more than 65 marks
From 65 on x-axis, draw a perpendicular meeting the curve at Q. From Q, draw a line parallel to x-axis meeting y-axis at M. M is 90
∴ Number of students getting above 65% marks = 100 – 90 = 10

Question 10.
The weights of 50 apples were recorded as given below. Calculate the mean weight, to the nearest gram, by step deivation method.

Weight in grams	Number of apples
80-85	5
85-90	8
90-95	10
95-100	12
100-105	8
105-110	4
110-115	3

Solution:

Weight (in gm)	Mid values (x)	No. of apples (f)	A = 97.5 d = x – A	f.d.
80-85 85-90 90-95 95-100 100-105 105-110 110-115	82.5 87.5 92.5 A=97.5 102.5 107.5 112.5	5 8 10 12 8 4 3	-15 -10 -5 0 5 10 15	-75 -80 -50 0 40 40 45
Total		50		-80

Mean = A + \(\frac{\sum f d}{\sum f}\) = 97.5 + \(\frac{-80}{50}\) = 97.5 – 1.6 = 95.9

Question 11.
Find the mean, median and mode of the following distribution :
8, 10, 7, 6, 10, 11, 6, 13, 10
Solution:
Mean = \(\frac{8+10+7+6+10+11+6+13+10}{9}\) = \(\frac{81}{9}\) = 9
arranging given nos. in ascending order.
6, 6, 7, 8. 10. 10, 10, 11, 13
Median = \(\frac{n+1}{2}\) th term = \(\frac{9+1}{2}\) = 5th term which is 10
∴ Median = 10
Mode = 10 (having highest frequency 3 times)

Question 12.
The following table gives the wages of workers in a factory.

Wages in Rs.	45-50	50-55	55-60	60-65	65-70	70-75	75-80
No. of workers	5	8	30	25	14	12	6

Calculate the mean by the short-cut method.
Solution:

Wages in Rs.	No. of Workers (f)	Mid. mark x	d = (x – A)	f × d
45-50 50-55 55-60 60-65 65-70 70-75 75-80	5 8 30 25 14 12 6	47.5 52.5 57.5 62.5 = A 67.5 72.5 77.5	-15 -10 -5 0 5 10 15	-75 -80 -150 0 70 120 90
Total	Σf =100			Σ fd = -25

Mean = A + \(\frac{\sum f d}{\sum f}\) = 62.5 + \(\frac{-25}{100}\) = 62.50 – 25 = 62.25

Question 13.
Attempt this question on graph paper. Marks obtained by 200 students in an examination are given below :

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of students	5	10	14	21	25	34	36	27	16	12

Draw an Ogive for the given distribution taking 2 cm = 10 marks. On one axis and 2 cm = 20 students on the other axis. From the graph, find :
(i) The Median
(ii) The upper Quartile
(iii) Number ofstudents scoring above 65 marks;
(iv) If 10 students qualify for merit scholarship, find the minimum marks required to qualify.
Solution:

Marks	No. of Students (fi)	c.f.
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100	5 10 14 21 25 34 36 27 16 12	5 15 29 50 75 109 145 172 188 200

No. of students (n) = 200
Plots the points (10. 5), (20, 15), (30, 29), (40, 50), (50, 75), (60, 109), (70, 145), (80, 172), (90, 188), (100, 200) on the graph and join them with free hand to get an ogive.

Mark a point P on y-axis corresponding to \(\frac{N}{2}\)
i.e., = \(\frac{200}{2}\) = 100
Draw PQ || x-axis, meeting the curve at Q.
Draw QR ⊥ x-axis, then the abscissa of R is 57

(i) ∴ Median = 57 marks
(ii) Mark a point A on y-axis corresponding to \(\frac{3N}{4}\) i.e., \(\frac{600}{4}\) = 150

Draw AB || x-axis, meeting the curve at B.
Draw BC ⊥ x-axis. then the abscissa of C is 72
∴ Upper quartile = 72 marks

(iii) From the graph, we have
No. of students scoring above 65 marks are (200 – 126) = 74

(iv) Minimum marks required to qualify are 92
⇒ The students from 191 to 200 (not the 190th student) qualify for merit scholarship. From point F on the y-axis indicating 190, draw a horizontal line meeting the Ogive at G. From G, draw a vertical line meeting x-axis at FI which indicates 91 marks.
Since the 190th student does not qualify for merit scholarship, so all students next to him who score more than 91 mark qualify for the scholarship.
Hence, the minimum marks required to qulify for merit scholarship = 92.

Question 14.
In a school the weekly pocket money of 50 students is as follows.

Weekly pocket money in Rs.	40-50	50-60	60-70	70-80	80-90	90-100
No. of students	2	8	12	14	8	6

Draw an histogram and a frequency polygon on the same graph. Find the mode from this graph.
Solution:
The frequency distribution table is as under:

Weekly pocket money (in Rs.)	Class mark	No. of students
40-50 50-60 60-70 70-80 80-90 90-100	45 55 65 75 85 95	2 8 12 14 8 6

Draw the histogram by taking ‘weekly pocket money along .Y-axis and no. of students along r-axis.

The frequency polygon is obtained by joining the consecutive mid-points of the upper sides (tops) of the adjacent rectangle of the histogram by means of line segments.

Now, join the two vertices A and B of the highest rectangle, diagonally to the upper corners C and D of the adjacent rectangles on either side of the highest rectangles, by line segments AC and BD. Let AC and BD intersect at L.

From L, draw the perpendicular LM on the horizontal line-the x-axis. Then, the abscissa of the point M, i.e.; OM = 72.5 determines the mode.
Hence, the required mode is Rs. 72.50

Question 15.
The distribution given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.

Marks obtained	5	6	7	8	9	10
No. of students	3	9	6	4	2	1

Solution:

Marks obtained (xi)	No. of students (fi)	Cumulative frequency (c,f)	fi xi
5	3	3	15
6	9	12	54
7	6	18	42
8	4	22	32
9	2	24	18
10	1	25	10
Total	Σfi = 25		Σfi xi =171

Mean
\(\bar{x}\) = \(\frac{\Sigma f_i}{x_i} \Sigma f_i\) = \(\frac { 171 }{ 25 }\)
\(\bar{x}\) = 6.84
Number of terms = 25 (odd)
Mean = \(\left(\frac{25+1}{2}\right)\)th term = \(\left(\frac{26}{2}\right)\)th term = 13th term
∴ Median = 7
Mode = Marks with maximum frequency is 6
∴ Mode = 6

Question 16.
The mean of the following distribution is 52 and the frequency of class interval 30-40 is ‘f’. Find ‘f’.

Class Interval	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	5	3	f	7	2	6	13

Solution:

Class interval	Frequency (fi)	Class mark (xi)	fixi
10-20	5	15	75
20-30	3	25	75
30-40	f	35	35 f
40-50	7	45	315
50-60	2	55	110
60-70	6	65	390
70-80	13	75	975
	Σfi = 36 + f		Σfixi = 1940 + 35f

Now, \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)
⇒ 52 = \(\frac{1940+35 f}{36+f}\) ⇒ 52(36 + f) = 1940 + 35f
⇒ 1872 + 52f = 1940 + 35f ⇒ 52f – 35f = 1940 – 1872
⇒ 17f = 68
⇒ f = \(\frac{68}{17}\)
⇒ f = 4

Question 17.
The monthly income of a group of 320 employees in a company is given below :

Monthly Income in Rs.	No. of Employees
6000-7000	20
7000-8000	45
8000-9000	65
9000-10000	95
10000-11000	60
11000-12000	30
12000-13000	5

Draw an ogive of the given distribution on a graph sheet taking 2 cm = Rs. 1000 on one axis and 2 cm = 50 employees on the other axis.
From the graph to determine.
(i) The median wage
(ii) the number of employees whose income is below Rs. 8500.
(iii) If the salary of a senior employee is above Rs. 11,500, find the number of senior employees in the company.
(iv) the upper quartile.
Solution:

Monthly Income in Rs. (c)	No. of Employees (f)	c.f.
6000-7000 7000-8000 8000-9000 9000-10000 10000-11000 11000-12000 12000-13000	20 45 65 95 60 30 5	20 65 130 225 285 315 320
Total	320

Plot the point (7000, 20), (8000, 65), (9000, 130), (10,000, 225) (11000, 285) (12,000, 315), (13,000, 1320) on the graph and join them with free hand to get an ogive as shown below
Here n = 320

(i) For median wage, Take OP = \(\frac{320}{2}\) = 160
on y-axis
Draw a line PQ || x-axis and from Q,
draw QM ⊥ x-axis, abcissa of M point is 9400
⇒ Median = Rs. 9400

(ii) Take OM’ = 8500 on x-axis. Draw Q’M’ || to y-axis and P’Q’ || x-axis
Where ordinate of P’ is 90
There are approximately 90 employees whose monthly wage is below Rs. 8500

(iii) There are approximately 20 employees whose salary is above Rs. 11500.

(iv) Upper quartile

Mark a point A on y-axis on \(\frac{3N}{4}\) = \(\frac{3 × 320}{4}\) = 240 and draw a line AB || x-axis, then draw BB’ ⊥ x-axis abscissa of B’ is upper quartile i.e., Rs. 10200.

Question 18.
A Mathematics aptitude test of 50 students was recorded as follows :

Marks	50-60	60-70	70-80	80-90	90-100
No. of Students	4	8	14	19	5

Draw a histogram for the above data using a graph paper and locate the mode.
Solution:

Represent marks along x-axis and students on y-axis and draw histogram as shown in the graph. In the highest rectangle. join A to C and B to D which intersect each other at P. From P, draw PQ perpendicular on x-axis meeting it at Q
Q is the mode which is 82.5
Hence, the required mode is 82.5

Question 19.
(a) (i) Using step-deviation method, calculate the mean marks of the following distribution.
(ii) State the modal class.

Class Interval	50-55	55-60	60-65	65-70	70-75	75-80	80-85	85-90
Frequency	5	20	10	10	9	6	12	8

Solution:

C.I.	Xi	Fi	Assumed mean A=67.5 Ui = \(\frac{x_t-\mathrm{A}}{h}\), h = 5	fiui
50-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90	52.5 57.5 62.5 67.5 = A 72.5 77.5 82.5 87.5	5 20 10 10 9 6 12 8	-3 -2 -1 0 1 2 3 4	-15 -40 -10 0 9 12 36 32
Total		80		24

(i) Mean = A + \(\frac{\Sigma f_i u_i}{\Sigma f_i}\) × h
= 67.5 + \(\frac{24}{80}\) × 5
= 67.5 + 1.5 = 69
(ii) Modal class = 55 – 60 (∵ frequency of this class is maximum i.e. 20)

Question 20.
Marks obtained by 200 students in an examination are given below :

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Frequency	5	11	10	20	28	37	40	29	14	6

Draw an ogive for the given distribution taking 2 cm = 10 marks on one-axis and 2 cm = 20 students on the other axis. Using the graph, determine
(i) the median marks,
(ii) the number of students who failed if minimum marks required to pass is 40.
(iii) If scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.
Solution:

Marks	f	c.f.
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100	5 11 10 20 28 37 40 29 14 6	5 16 26 46 74 111 151 180 194 200
	N = 200

Plot the point (10, 5) (20, 16) (30, 26) (40, 46) (50, 74) (60, 111) (70, 151) (80, 180) (90, 196) (100, 200) on the graph and join them in free hand to get an ogive as shown.
From the graph it is seen that

(i) Median = \(\left(\frac{n}{2} \text { th }\right)\) term = \(\left(\frac{200}{2} \text { th }\right)\) 100 observation = 57 from graph.
(ii) No. of students who failed = 46
(iii) No. of students who secured grade one = 200 – 188 = 12

Question 21.
Marks obtained by 40 students in a short assessment is given below, where a and b are two missing data.

Marks	5	6	7	8	9
Number of students	6	a	16	13	b

If the mean of the distribution is 7.2, find a and b.
Solution:
Mean of distribution = 7.2

Marks	Number of students	fx
5	6	30
6	a	6 a
7	16	112
8	13	104
9	b	9 b
Total	36 + a + b = 40	246 + 6a + 9b

35 + a + b = 40
a + b = 40 – 35 = 5 …(i)
Mean = 7.2
∴ Total = 7.2 x 40 = 288
246 + 6a + 9b = 288
6a + 9b = 288 – 246 = 42
2a + 3b = 14 …(ii)
From (i) a + b = 5 ⇒ a = 5 – b
∴ 2(5 – b) + 3b = 14 ⇒ 10 – 2b + 3b = 14
⇒ b = 14 – 10 = 4
∴ a = 5 – 4 = 1
a = 1, b = 4

Question 22.
The following distribution represents the height of 160 students of a school.

Height (in cm)	No. of Students
140-145	12
145-150	20
150-155	30
155-160	38
160-165	24
165-170	16
170-175	12
175-180	8

Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine :
(i) The median height.
(ii) The inter quartile range.
(iii) The number of students whose height is above 172 cm.
Solution:
The cumulative frequency table may be prepared as follows :

Height (in cm)	No. of Students (f)	Cumulative frequency (c.f)
140-145	12	12
145-150	20	32
150-155	30	62
155-160	38	100
160-165	24	124
165-170	16	140
170-175	12	152
175-180	8	160

Now, we take height along x-axis and number of students along the y-axis. Now, plot the point (140, 0), (145, 12), (150, 32), (155, 62), (160, 100), (165, 124), (170, 140), (175, 152) and (180, 160). Join these points by a free hand curve to get the ogive.

(i) Here N = 160 ⇒ \(\frac { N }{ 2 }\) = 80
On the graph paper take a point A on the y-axis representing 80. Through A draw horizontal line meeting the ogive at B. From B, draw BC ⊥ x-axis, meeting the x-axis at C.
The abscissa of C is 157.5
So, median = 157.5 cm

(ii) Proceeding in the same way as we have done in above, we have, Q₁ = 152 and Q₃ = 164 – 152 = 12cm
So, inter quartile range = Q₃ – Q₁ = 164 – 152 = 12 cm

(iii) From the ogive, we see that the number of students whose height is less than 172 is 145.
No. of students whose height is above 172 cm = 160- 145 = 15

Question 23.
Find the mode and median of the following frequency distribution :

X	10	11	12	13	14	15
f	1	4	7	5	9	3

Solution:

x	f	Cumulative frequency (c.f)
10	1	1
11	4	5
12	7	12
13	5	17
14	9	26
15	3	29

Here n- 29 which is odd
∴ Median = \(\left(\frac{n+1}{2}\right)\)th term = \(\left(\frac{29+1}{2}\right)\)th term = 15th term = 13
Since, the frequency corresponding to 14 is maximum, so mode = 14.

Question 24.
The median of the following observations 11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.
Solution:
Observation are:
11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47
n = 9
∴ Median = \(\left(\frac{9+1}{2}\right)^{\text {th }}\) term
i.e. 5^th term = x + 4
∴ Median = x + 4
24 = x + 4
x = 24 – 4
x = 20
Now observation are :
11, 12, 14, (20 – 2), (20 + 4), (20 + 9), 32, 38, 47 i.e. 11, 12, 14, 18, 24,29, 32, 38, 47
∴ Mean = \(\frac{11+12+14+18+24+29+32+38+47}{9}\) = \(\frac{225}{9}\) = 25

Question 25.
Draw a histogram from the following frequency distribution and find the mode from the graph :

Class	0-5	5-10	10-15	15-20	20-25	25-30
Frequency	2	5	18	14	8	5

Solution:

Class	0-5	5-10	10-15	15-20	20-25	25-30
Frequency	2	5	18	14	8	5

∴ Mode =13.6

Steps for calculation of mode
(i) Mark the end points of the upper corner of rectangle with maximum frequency as A and B.
(ii) Mark the inner corner of adjacent rectangles as C and D.
(iii) Join AC and BD do intersect at K. From K, draw KL perpendicular to x-axis.
(iv) The value of L on x-axis represents the mode.

Question 26.
Find the mean of the following distribution by step deviation method :

Class interval	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	10	6	8	12	5	9

Solution:

C.I.	Frequency (f)	Class mark (x)	d = \(\frac{x-\mathrm{A}}{i}\) [A = 45]	f × d
20-30	10	25	-2	-20
30-40	6	35	-1	-6
40-50	8	45	0	0
50-60	12	55	1	12
60-70	5	65	2	10
70-80	9	75	3	27
	50			23

Mean = A + \(\frac{\Sigma f d}{n}\) × i = 45 + \(\frac { 23 }{ 50 }\) × 10
= 45 + 4.6 = 49.6

Question 27.
The marks obtained by 120 students in a test are given below :

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of students	5	9	16	22	26	18	11	6	4	3

Use suitable scale for ogive to estimate the following :
(i) The median.
(ii) The number of students who obtained more than 75% marks in the test.
(ii) The number of students who did not pass the test if minimum marks required to pass is 40.
Solution:

Marks	f	c.f.
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100	5 9 16 22 26 18 11 6 4 3	5 14 30 52 78 96 107 113 117 120
	120

(i) Through marks 60, draw a line segment parallel to x-axis which meets the curve at A. From A, draw a line perpendicular to x-axis meeting at B.
∴ Median = 43

(ii) Through marks 75, draw a line segment parallel to y-axis which meets the curve at D. From D, draw a line perpendicular to y-axis which meets y-axis at 110.
Number of students getting more than 75% = 120 – 110 = 10 students.

(iii) Through marks 40, draw a line segment parallel toy-axis which meets the curve at C. From C, draw a line perpendicular to y-axis which meets y-axis at 52.
∴ Number of students who did not pass = 52.

Question 28.
The numbers 6, 8,10,12, 13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x.
Solution:
Arranging the numbers in ascending order
6, 8, 10, 12, 13, x
Mean of the given numbers = \(\frac{6+8+10+12+13+x}{6}\) = \(\frac{49+x}{6}\)
Number of terms (n) = 6 which is even
∴ Median = \(\frac{\left(\frac{n}{2}\right) \text { th term }+\left(\frac{n}{2}+1\right) \text { th term }}{2}\) = \(\frac{3^{r d}+4^{t h}}{2}\) = \(\) = \(\frac{10+12}{2}\) = \(\frac{22}{2}\) = 11
According to statement,
It is given that the mean of 6, 8, 10, 12, 13, x is equal to the median of 6, 8, 10, 12, 13, x
∴ \(\frac{49+x}{6}\) = 11 ⇒ 49 + x = 66 ⇒ x = 66 – 49
x = 17

Question 29.
Calculate the mean of the distribution given below using the short cut method.

Marks	11-20	21-30	31-40	41-50	51-60	61-70	71-80
No. of students	2	6	10	12	9	7	4

Solution:

Marks	Frequency (f)	Mid Value (x)	di = xi – A A = 45.5	Ui = \(\frac{x_i-\mathrm{A}}{10}\)	F × ui
11-20	2	15.5	-30	-3	-6
21-30	6	25.5	-20	-2	-12
31-40	10	35.5	-10	-1	-10
41-50	12	45.5	0	0	0
51-60	9	55.5	10	1	9
61-70	7	65.5	20	2	14
71-80	4	75.5	30	3	12
	Σ f=50				Σfui = 7

A = 45.5
Mean = A + \(\frac{\sum f u_i}{\sum f}\) × 10 = 45.5 + \(\frac{7}{50}\) × 10 = 45.5 + 1.4 = 46.9

Question 30.
(Use a graph paper for this question.) The daily pocket expenses of 200 students in a school are given below:

Marks	Number of students (frequency)
0-5	10
5-10	14
10-15	28
15-20	42
20-25	50
25-30	30
30-35	14
35-40	12

Draw a histogram representing the above distribution and estimate the mode from the graph.
Solution:

Estimated mode = 21.5

Steps for calculation of mode
(i) Mark the end points of the upper corner of rectangle with maximum frequency as A and B.
(ii) Mark the inner corner of adjacent rectangles as C and D.
(iii) Join AC and BD do intersect at K. From K, draw KL perpendicular to x-axis.
(iv) The value of L on x-axis represents the mode.

Question 31.
The marks obtained by 100 students in a Mathematics test are given below :

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of students	3	7	12	17	23	14	9	6	5	4

Draw an ogive for the given distribution on a graph sheet.
Use a scale of 2 cm = 10 units on both axis).
Use the ogive to estimate the:
(i) median.
(ii) lower quartile.
(iii) number of students who obtained more than 85% marks in the test.
(iv) number of students who did not pass in the test if the pass percentage was 35.
Solution:

Marks	Number of students	Cumulative frequency (c.f.)
0-10	3	3
10-20	7	10
20-30	12	22
30-40	17	39
40-50	23	62
50-60	14	76
60-70	9	85
70-80	6	91
80-90	5	96
90-100	4	100

On the graph paper, we plot the following points A (10, 3), B (20, 10), C (30, 22), D, (40, 39), E (50, 62), F (60, 70), G (70, 85), H (80, 91), 1 (90, 96) and V (100, 100). Join all these points by a free hand drawing. The required ogive is shown on the graph paper given below :
Here, number of students (n) = 100, which is even.

(i) Let P be the point on Y-axis representing frequency = \(\frac { n }{ 2 }\) = \(\frac { 100 }{ 2 }\) = 50.
Through P, draw a horizontal line to meet the ogive at point Q. Through Q, draw a vertical line to meet the X-axis at T. The abscissa of the point T represents 45 marks.
Hence, median marks is 45.

(ii) Let R be the point on Y-axis representing frequency = \(\frac { n }{ 4 }\) = \(\frac { 100 }{ 4 }\) = 25.
Through R, draw a horizontal line to meet the ogive at point S. Through S, draw a vertical line to meet the X-axis at N. The abscissa of the point N represents 31 marks.
Hence, the lower quartile = 31 marks.

(iii) 85% marks = 85% of 100 = 85 marks
Let the point M on X-axis represents 85 marks.
Through M, draw a vertical line to meet the ogive at the point O. Through O draw a horizontal line to meet the Y-axis at point J. The ordinate of point J represents 95 students on Y-axis.
∴ Number of students who obtained more than 85% in the test = 100 – 95 = 5.

(iv) 35% marks = 35% of 100 = 35
Let the point K on X-axis represents 35 marks.
Through K, draw a vertical line to meet the ogive at the point L. Through L, draw a horizontal line to meet the Y-axis at point U. The ordinate of point U represents 30 students on Y-axis.
Hence, the number of students, who did not pass in the test is 30.

Question 32.
Marks obtained by 30 students in a class assessment of 5 marks is given below (next page):

Marks	0	1	2	3	4	5
No. of students	1	3	6	10	5	5

Calculate the mean, median and mode of the above distribution.
Solution:

Marks	0	1	2	3	4	5
No. of students	1	3	6	10	5	5
Cumulative Frequency	1	4	10	20	25	30

Mean = \(\frac{\sum f x}{\sum f}\)
= \(\frac{0 \times 1+1 \times 3+2 \times 6+3 \times 10+4 \times 5+5 \times 5}{1+3+6+10+5+5}\)
= \(\frac { 90 }{ 30 }\) = 3
∴ 3 is the mean
There are a total of 30 observations in the data
The median is the arithmetic mean of \(\left(\frac{n}{2}\right)^{\text {th }}\) and \(\left(\frac{n}{2}+1\right)^{\text {th }}\) observation in case of even number of observations = Arithmetic mean of \(\left(\frac{30}{2}\right)^{\mathrm{th}}\) and \(\left(\frac{30}{2}+1\right)^{\text {th }}\)
= Arithmetic mean of 15^th and 16^th observation will be the median
∴ Median = \(\frac{3+3}{2}\) = 3
Frequency is highest for the observation x_i = 3
Mode = 3

Question 33.
Calculate the mean of the following distribution:

Class Interval	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	8	5	12	35	24	16

Solution:
Consider the following distribution:

Class interval	Frequency	Class mark  xi	fi xi
0-10	Fi	5	40
10-20	8	15	75
20-30	5	25	300
30-40	12	35	1225
40-50	35	45	1080
50-60	24	55	880
Total	fi = 100		fi xi =3600

Mean = \(\frac{\sum f_i x_i}{\sum f_i}\) = \(\frac { 3600 }{ 100 }\) = 36

Question 34.
The weight of 50 workers is given below:

Weight in Kg	50-60	60-70	70-80	80-90	90-100	100-110	110-120
No. of Workers	4	7	11	14	6	5	3

Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis and 2 cm = 5 workers along the other axis. Use a graph to estimate the following:
(i) the upper and lower quartiles.
(ii) if weighing 95 kg and above is considered overweight find the number of workers who are overweight.
Solution:
The cumulative frequency table of the given distribution table is as follows:

Weith in kg	Number of workers	Cumulative frequency
50-60 60-70 70-80 80-90 90-100 100-110 110-120	4 7 11 14 6 5 3	4 11 22 36 42 47 50

Plot the points (60, 4), (70, 11), (80, 22), (90, 36), (100, 42), (110, 47) and (120, 50). Then draw a free-hand curve (ogive) as shown below:

∴ N = 50

(i) Upper quartile = \(\left(\frac{3 \times N}{4}\right)^{\text {th }}\) term
= \(\left(\frac{3 \times 50}{4}\right)^{\text {th }}\) term = 37.5^th term = 92.5 kg
Lower quartile = \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) term
= \(\left(\frac{\mathrm{50}}{4}\right)^{\text {th }}\) term = 12.5^th term = 71.5 kg

(ii) Through 95 kg mark, draw a vertical line that meets graph at point P. Through point P, draw a horizontal line which meets axis for c.f. at point A and A = 39.
⇒ Weights of 39 workers are 95 kg or below it.
∴ Number of workers who are overweight = 50 – 39 = 11

Question 35.
The mean of following number is 68. Find the value of ‘x’. 45, 52, 60, x, 69, 70, 26, 81 and 94. Hence estimate the median.
Solution:
Mean = \(\frac{\text { Sum of all observations }}{\text { Total number of observation }}\)
∴ 68 = \(\frac{45+52+60+x+69+70+26+81+94}{9}\)
∴ 68 = \(\frac{497+x}{9}\)
∴ 612 = 497 + x
∴ x = 612 – 497
∴ x = 115
Arrange the numbers in ascending order 26, 45, 52, 60, 69, 70, 81, 94, 115
Since the number of observations’ is odd, the median is \(\left(\frac{n+1}{2}\right) \text { th }\) observation
⇒ The mrdian is the \(\left(\frac{9+1}{2}\right)\) = 5th observation
Hence, the median is 69.

Question 36.
The table shows the distribution of the scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution. (Take 2 cm = 10 scores on the X-axis and 2 cm = 20 shooters on the Y-axis).

Scores	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of shooters	9	13	20	26	30	22	15	10	8	7

Use your graph to estimate the following :
(i) The median.
(ii) The interquartile range.
(iii) The number of shooters who obtained a score of more than 85%.
Solution:

Scores	f	c.f.
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100	9 13 20 26 30 22 15 10 8 7	9 22 42 68 98 120 135 145 153 160
	N = 160	c.f.

(i) Median = \(\left(\frac{n}{2}\right)^{\mathrm{th}}\) term = \(\left(\frac{160}{2}\right)^{\text {th }}\) term = 80^th term
Through mark 80 on y-axis, draw a horizontal line which meets the ogive drawn at point Q. Through Q, draw a vertical line which meets the x-axis at the mark of 43.
∴ Median = 43

(ii) Since the number of terms = 160
Lower quartile (Q₁) = \(\left(\frac{160}{4}\right)^{\text {th }}\) term = 40^th term = 28
Upper quartile (Q₃) = \(\left(\frac{3 \times 160}{4}\right)^{\text {th }}\) term = 120^th term = 60
∴ Inter-Quartile range = Q₃ – Q₁
= 60 – 28 = 32
∴ Inter-Quartile range is 32

(iii) Since 85% scores = 85% of 100 = 85
Through mark for 85 on x-axis, draw a vertical line which meets the ogive drawn at point B. Through the point B, draw a horizontal line which meets the y-axis at the mark of 150.
∴ The number of shooters who obtained more than 85% score
= 160 – 150 = 10
So, the number of shooters who obtained more than 85% score is 10.

Question 37.
The histogram below represents the scores obtained by 25 students in a Mathematics mental test. Use the data to :
(i) Frame a frequency distribution table.
(ii) To calculate mean.
(iii) To determine the Modal class.

Solution:
(i)

Class interval	Frequency
0-10	2
10-20	5
20-30	8
30-40	4
40-50	6

(ii)

Class interval	Frequency (f)	Mean value (x)	f x
0-10 10-20 20-30 30-40 40-50	2 5 8 4 6	5 15 25 35 45	10 75 200 140 270
	Σf = 25		Σfx = 695

∴ Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac{695}{25}\) = 27.8

(iii) Modal class = class with maximum frequency = 20-30

The availability of step-by-step OP Malhotra Class 10 Solutions Chapter 3 Shares and Dividends Ex 3(b) can make challenging problems more manageable.

S Chand Class 10 ICSE Maths Solutions Chapter 3 Shares and Dividends Ex 3(a)

Question 1.
Find the number of shares that can be bought and the income obtained by investing :
(a) Rs. 50 in (Re. 1) shares at Rs. 1.25 paying 8%.
(b) Rs. 240 in (Rs. 5) shares at Rs. 8, paying 9%.
Solution:
(a) Investment = Rs. 50
Face value of each share = Re. 1
Market value = Rs. 1.25
Rate of dividend = 8%
∴ Number of shares = \(\frac { 50 }{ 1.25 }\) = \(\frac{50 \times 100}{125}\)
= 40 shares
Face value of 40 shares = 40 x Re. 1 = Rs. 40
= Rs. \(\frac { 320 }{ 100 }\) = Rs. 3.20

(b) Investment = Rs. 240
Face value of each share = Rs. 5
and market value = Rs. 8
Rate of dividend = 9%
∴ Number of shares = Rs. \(\frac { 240 }{ 8 }\) = 30
Face value of 30 shares = 30 x 5 = Rs. 150
Amount of dividend = Rs. \(\frac{150 \times 9}{100}\)
= Rs. \(\frac { 1350 }{ 100 }\)
= Rs. 13.50

Question 2.
A man bought 160 (Rs. 5) shares for Rs. 360. At what price did the shares stand? At what premium or discount were they quoted?
Solution:
Number of shares = 160
Face value of each share = Rs. 5
Amount of 160 shares = Rs. 360
∴ Market value of each share = Rs.\(\frac {360}{160}\)
= Rs. \(\frac {9}{4}\) = Rs. 2.25 4
Difference of values of each share
= Rs. 5.00 – Rs. 2.25 = Rs. 2.75
∴ Market value is less than the face value
∴ The shares were quoted Rs. 2.75 at discount

Question 3.
A man sold 600 (Re. 1) shares for Rs. 750. At what price did the shares stand? At what premium or discount were they quoted?
Solution:
Number of shares = 600
Face value of each share = Re. 1
Total market value of 600 shares = Rs. 750
∴ Market value of each share = Rs. \(\frac {750}{600}\)
= Rs. \(\frac {5}{4}\) = Rs. 1.25
Difference of values of each share = Rs. 1.25. – 1.00 = 0.25
Difference of the values of each shares = Rs. 1.25 – 1.00 = 0.25
∵ Market value of each share is more than its face value
∴ The shares were quoted Rs. 0.25 at premium

Question 4.
A man buys 200 ten rupee shares at Rs. 12.50 each and receives a dividend of 8%. Find the amount invested by him and dividend received by him in cash.
Solution:
Number of shares bought = 200
Face value of each share = Rs. 10
Market value = Rs. 12.50
Rate of dividend = 8%
∴ Total amount of investment = Rs. 12.50 x 200 = Rs. 2500
and amount of face value = Rs. 10 x 200 = Rs. 2000
∴ Dividend = Rs. 2000 x \(\frac {8}{100}\) = Rs. 160

Question 5.
A man bought 500 shares, each of face value Rs. 10 of a certain business concern and during the first year after purchase received Rs. 400 as dividend on his shares. Find the rate of dividend on his shares.
Solution:
Number of shares bought = 500
Nominal value of each share = Rs. 10
Amount of dividend = Rs. 400
Total nominal value of 500 shares = Rs. 10 x 500 = Rs. 5000
∴ Dividend on Rs. 5000 = Rs. 400
and dividend on Rs. 100 = \(\frac{400 \times 100}{5000}\) = 8%

Question 6.
By purchasing ₹ 25 shares for ₹ 40 each a man gets 4 per cent profit on his investment. What rate per cent is the company paying? What is his dividend if he buys 60 shares ?
Solution:
Face value of each share = Rs. 25
Market value = Rs. 40
Profit = 4%
Profit on Rs. 25 = Rs. 4
∴ Profit on Rs. 40 = \(\frac{4 \times 40}{25}=\frac{32}{5}\) = 6.4
∴ Rate of dividend = 6.4%
Face value of 60 shares = Rs. 25 x 60 = Rs. 1500
∴ Dividend on Rs. 1500 = Rs. 1500 x \(\frac { 6.4 }{ 100 }\) = Rs. 96.0 = Rs. 96

Question 7.
Mukul invests Rs. 9000 in a company paying a dividend of 6% per annum when a share of face value 100 stands at Rs. 150. What his annual income? He sells 50% of his shares when the price rises to Rs. 200. What is his gain on this transaction?
Solution:
Investment of Mukul = Rs. 9000
Rate of dividend = 6% p.a.
Face value of each share = Rs. 100
and market value = Rs. 150
∴ Number of shares = Rs. \(\frac { 9000 }{ 150 }\) = 60
Face value of 60 shares = Rs. 100 x 60 = Rs. 6000
∴ Amount of annual dividend = Rs. \(\frac{6000 \times 6}{100}\) = Rs. 360
50% of shares which were sold = 60 x \(\frac { 50 }{ 100 }\) = 30
Market value of each share = Rs. 200
∴ Amount received = Rs. 200 x 30 = Rs. 6000
and market value of remaining 30 shares = Rs. 150 x 30 = Rs. 4500
Total amount received = Rs. 6000 + 4500 = Rs. 10500
∴ Total profit = Rs. 10500 – 9000 = Rs. 1500

Question 8.
By investing Rs. 7500 in a company paying 10% dividend, an income of Rs. 500 is received. What price is paid for each Rs. 100 share? Solution:
Investment = Rs. 7500
Rate of dividend = 10%
Total income = Rs. 500
Face value of each share = Rs. 100
∴ Total face value of shares = Rs \(\frac{500 \times 100}{10}\) = Rs. 5000
∴ Number of shares = Rs. \(\frac{5000}{100}\) = 50
Now market value of 50 share; = Rs. 7500
∴ Market value of each share
= Rs. \(\frac { 7500 }{ 500 }\) = Rs. 150

Question 9.
Arun owns 560 shares of a company. The face value of each share is Rs. 25. The company declares a dividend of 9%. Calculate:
(i) the dividend Arun would receive, and
(ii) the rate of interest, on his investment. Considering that Arun bought these shares @ ₹30 per share in the market.
Solution:
Number of shares Arun has = 560
Face value of each share = ₹ 25
Rate of dividend = 9%
Now total face value of 560 shares = ₹ 25 x ₹ 60 = ₹ 14000

(i) Total dividend received by Arun
= ₹ \(\frac{14000 \times 9}{100}\) = ₹ 1260

(ii) Market value of each share = ₹ 30
Total investment = ₹ 30 x 560 = ₹ 16800
Rate of interest on the investment
= ₹ \(\frac{1260 \times 100}{16800}\)
= 7.5% or 7\(\frac { 1 }{ 2 }\)%

Question 10.
What sum should Ashok invest in ₹ 25 shares selling at ₹36 to obtain an income of ₹720, if the dividend declared is 12%. Also find :
(i) the number of shares bought by Ashok.
(ii) the percentage return on his investment.
Solution:
Face value of each share = ₹ 25
Market value = ₹ 36
Total income = ₹ 720
(i) Rate of dividend = 12%
∴ Face value of shares = \(\frac{720 \times 100}{12}\) = ₹ 6000
and number of shares = ₹ \(\frac { 6000 }{ 25 }\) = 240
Investment by Ashok = 240 x ₹ 36 = ₹ 8640
Percentage of return = ₹ \(\frac{720 \times 100}{8640}\)
= \(\frac { 25 }{ 3 }\) = 8\(\frac { 1 }{ 2 }\)% = 8.33%

Question 11.
Mr. Sharma has 60 shares of nominal value 7100 and he decides to sell them when they are at a premium of 60%. He invests the proceeds in shares of nominal value ₹ 50, quoted at 4% discount, paying 18% dividend annually. Calculate:
(i) the sale proceeds.
(ii) the number of shares he buys.
(iii) the annual dividend from these shares.
Solution:
Number of shares = 60
Face value of each share = ₹ 100
Market value = 60% premium = ₹ 100 + 60 = ₹ 160
(i) Sale proceed = ₹ 160 x 60 = ₹ 9600

(ii) Nominal value of share purchased = ₹ 50
Market value as quoted 4% discount
= ₹ 50 – \(\frac { 50×4 }{ 100 }\)
= ₹ 50 – 2 = ₹ 48
∴ Number of shares purchased = \(\frac { 9600 }{ 48 }\) = 200

(iii) Rate of dividend = 18% p.a.
Face value of 200 shares = ₹ 50 x 200 = ₹ 10000
∴ Annual dividend = ₹ 10000 x \(\frac { 18 }{ 100 }\) = ₹ 1800

Question 12.
A man invests a sum of money in ₹ 100 shares, paying 15% dividend, quoted at 20% premium. If his annual dividend is ₹ 540, calculate:
(i) his total investment.
(ii) the rate of return on his investment.
Solution:
Rate of dividend = 15% at 20% premium
Annual dividend = ₹ 540
∴ Investment = ₹ \(\frac{540 \times(100+20)}{15}\)
= ₹ \(\frac{540 \times 120}{15}\) = ₹ 4320
and rate of returns on his investment
= \(\frac{540 \times 100}{4320}\) = \(\frac { 100 }{ 8 }\)% = 12.5 %

Question 13.
A lady holds 1800 hundred rupee shares of a company that pays 15% dividend annually. Calculate her annual dividend. If she had bought these shares at 40% premium, what percentage return would she have got on her investment? Give your answer to the nearet integer.
Solution:
Number of shares = 1800
Face value of each share = ₹ 100
Dividend rate = 15%
Market value of each share = ₹ 100 + ₹ 40 = ₹ 140
(i) ∴ Annual dividend = ₹ 1800 x 15 = ₹ 27000
(ii) Investment = ₹ 1800 x 140 = ₹ 252000
∴ Percentage return = \(\frac{27000 \times 100}{252000}\)
= \(\frac { 75 }{ 7 }\) % = 10 \(\frac { 5 }{ 7 }\) % = 11%

Question 14.
A man invests ₹ 11200 in a company paying 6% dividend when its ₹ 100 share can be bought for ₹ 140. Find:
(i) his annual income.
(ii) the percentage income on his investment.
Solution:
Investment = ₹ 11200
Rate of dividend = 6%
Face value of each share = ₹ 100
Market value of each share = ₹ 140
(i) Annual income = ₹ \(\frac{11200 \times 6}{140}\) = ₹ 480
(ii) Percentage income on his investment
= \(\frac{480 \times 100}{11200} \%=\frac{30}{7} \%=4 \frac{2}{7} \%\)

Question 15.
A company with 10,000 shares of 7100 each, declares an annual dividend of 5%.
(i) What is the total amount of dividend paid by the company?
(ii) What would be the annual income of a man, who has 72 shares in the company?
(iii) If he received only 4% of his investment, find the price he had paid for each share.
Solution:
Number of share with a company = 10000
Face value of each share = ₹ 100
Rate of annual dividend = 5%
(i) Total dividend = 10000 x 5 = ₹ 50,000
(ii) Income on 72 shares = 72 x ₹ 5 = ₹ 360
(iii) Rate of returns he receives = 4%
∴ Market value of two shares
= \(\frac{\text { Dividend } \times 100}{4}\)
= \(\frac{50,000 \times 100}{4}\)
= ₹ 1250000
Market value of each share
= ₹ \(\frac {1250000}{10000}\) = ₹ 125

Question 16.
A man invest ₹ 1680 in buying shares of nominal value ₹ 24 and selling at 12% premium. The dividend on the shares is 15% per annum.
(i) Calculate the number of shares he buys.
(ii) Calculate the dividend he receives annually. (ICSE 1999)
Solution:
Investment = ₹ 1680
Nominal value of each share = ₹ 24
and market value = ₹ 24 x \(\frac {112}{100}\) = \(\frac {2688}{100}\)
= ₹ 26.88
∴ Number of shares = \(\frac {1680}{26.88}\)
= \(\frac{1680 \times 100}{2688}\) = \(\frac {125}{2}\) shares
= 62.5 shares
Rate of dividend = 15%
Market value of each share = ₹ 100 + ₹ 12
= ₹ 112
Dividend = \(\frac{1680 \times 15}{112}\) = ₹ 225

Question 17.
A man invests ₹ 7425 on buying share of face value ₹ 90 each at a premium of 10% in a company. If he at a earns ₹ 1350 as dividend at the end of the year, find
(i) the number of shares he has in the company.
(ii) the dividend percentage per share that he received.
Solution:
Amount of dividend = ₹ 1350
Investment = ₹ 7425
Face value of each share = ₹ 90
Market value = at premium of 10%
= ₹ \(\frac{90 \times(100+10)}{100}\)
= ₹ \(\frac{90 \times 110}{100}\) = ₹ 99

(i) ∴ Number of shares = \(\frac {7425}{99}\) = 75
and face value of 75 shares = 75 x ₹ 99
= ₹ 6750

(ii) Rate of dividend = \(\frac{1350 \times 100}{6750}\) = 20%

Question 18.
Abhisheksold a certain number of shares of ₹ 20 paying 8% dividend at ₹ 18 and invested the proceeds in ₹ 10 shares, paying 12% dividend at 50% premium. If the change in his annual income is ₹ 120, find the number of shares sold by him ?
Solution:
In first case,
Face value of each share = ₹ 20
Rate of dividend = 8%
Market value of each share = ₹ 18
Let number of shares = x
∴ Sale proceed by selling x shares = ₹ 18 × x = ₹ 18x
and Face value of x shares = ₹ x × 20 = ₹ 20x
∴ Dividend = 20x × \(\frac { 8 }{ 100 }\) = ₹ \(\frac { 8 }{ 5 }\) x
In second case
Face value of each share = ₹ 10
Market value 50% at premium
= ₹ \(\frac{10 \times(100+50)}{100}=₹ \frac{10 \times 150}{100}\) = ₹ 15
Rate of dividend = 12%
Number of shares purchased = \(\frac { 18x }{ 15 }\) = \(\frac { 6 }{ 5 }\)x
Now face value of each shares = \(\frac { 6 }{ 5 }\)x × 10 = ₹ 12 x
Dividend = ₹ 12x × \(\frac { 12 }{ 100 }\) = ₹ \(\frac { 36 }{ 25 }\)x
Difference in dividend = \(\frac { 8 }{ 5 }\)x – \(\frac { 36 }{ 25 }\)x
= \(\frac{40 x-36 x}{25}\) = \(\frac { 4 }{ 25 }\)x
But actual difference of dividend = ₹ 120
∴ \(\frac { 4 }{ 25 }\)x = 120
x = \(\frac{120 \times 25}{4}\) = 750
∴ Number of shares = 750

Question 19.
A person invested ₹ 8000 and ₹ 10000 in buying shares of two companies which later on declared dividends of 12% and 8% respectively. He collects the dividends and sells out his shares at a loss of 2% and 3% respectively. Find his total earning from the above transaction.
Solution:
In first case, investment = ₹ 8000
Rate of dividend = 12%
∴ Income = ₹ \(\frac{8000 \times 12}{100}\) = ₹ 960
In second case, investment = ₹ 10000
Rate of dividend = 8%
∴ Income = ₹ \(\frac{1000 \times 8}{100}\) = ₹ 800
Total income = ₹ 960 + 800 = ₹ 1760
Total loss = ₹ \(\left(8000 \times \frac{2}{100}+10000 \times \frac{3}{100}\right)\) = ₹ 800
= ₹ 160 + ₹ 300 = ₹ 460
∴ Net gain = Total income – Loss
= ₹ 1760 – 460 = ₹ 1300

Question 20.
A person invested 20%, 30% and 25% of his savings in buying shares of three different companies A, B and C, which declared dividends of 10%, 12% and 15% respectively. If his total income on account of dividends be ₹ 2337.50, find his saving and the amount which he invested in buying shares of each company.
Solution:
Let the total savings = ₹ x
Investment in A Company = \(\frac { 20 }{ 100 }\) × x = \(\frac { x }{ 5 }\)
Investment in B Company = \(\frac { 30 }{ 100 }\) × x = \(\frac { 3 }{ 10 }\)x
and investment in C Company = \(\frac { 25 }{ 100 }\) × x = \(\frac { x }{ 4 }\)
Now income from A Company = \(\frac { x }{ 5 }\) x \(\frac { 10 }{ 100 }\)
= ₹ \(\frac { x }{ 50 }\)
Income from B Company = \(\frac { 3 }{ 10 }\)x × \(\frac { 12 }{ 1o0 }\)
= ₹ \(\frac { 9 }{ 250 }\)x
and income from C Company = \(\frac { x }{ 4 }\) x \(\frac { 15 }{ 100 }\)
= ₹ \(\frac { 3x }{ 80 }\)x
∴ Total inccme = \(\frac{x}{50}+\frac{9}{250} x+\frac{3 x}{80}\)
= \(\frac{40 x+72 x+75 x}{2000}=\frac{187}{2000}\)x
But total income = ₹ 2337.50
∴ \(\frac{187}{2000} x=2337.50 \Rightarrow x=\frac{233750 \times 2000}{100 \times 187}\)
⇒ x = 25000
∴ Total investment = ₹ 25000
Now investment in A Company
= ₹ 25000 x \(\frac { 20 }{ 100 }\) = ₹ 5000
Investment in B Company
= ₹ 25000 x \(\frac { 30 }{ 100 }\) = ₹ 7500
and investment in C Company
= ₹ 25000 x \(\frac { 25 }{ 100 }\) = ₹ 6250

Self Evaluation And Revision (Latest ICSE Questions)

Question 1.
A dividend of 9% was declared on ₹ 100 shares selling at a certain price. If the rate of return is 7\(\frac { 1 }{ 2 }\)%, calculate :
(i) the market value of the share
(ii) the amount to be invested to obtain an annual dividend of ₹ 630.
Solution:
Market price of each share = x
Face value of each share = ₹ 100
Rate of dividend = 9%
Rate of return on investment = 7\(\frac { 1 }{ 2 }\)% = \(\frac { 15 }{ 2 }\)%
(i) ∴ \(\frac{x \times 15}{100 \times 2}\) = 9 ⇒ x = \(\frac{9 \times 100 \times 2}{15}\) = ₹ 120
∴ Market value of each share = ₹ 120

(ii) Amount of dividend = ₹ 630
∴ Investment = \(\frac{630 \times 100}{\frac{15}{2}}\) = \(\frac{630 \times 100 \times 2}{15}\)
= ₹ 42 x 2 x 100 = ₹ 8400

Question 2.
A man invests ₹ 8800 in buying shares of face value of rupees hundred each at a premium of 10% in a company. If he earns ₹ 1200 at the end of the year as dividend, find
(i) the number of shares he has in the company ?
(ii) the dividend percentage per share.
Solution:
Investment = ₹ 8800
Face value of each share = ₹ 1100
Market value at a premium of 10%
= ₹ 100 + 10 = ₹ 110
Total dividend he received = ₹ 1200
(i ) Number of shares = \(\frac{\text { Investment }}{\mathrm{MV}}\)
= \(\frac { 8800 }{ 110 }\) = 80

(ii) Face value of each share = ₹ 100 x 80 = ₹ 8000
∴ Rate of dividend per share = \(\frac { 1200 }{ 8000 }\) x 100
= 15%

Question 3.
A man wants to buy 62 shares available at ₹ 132 (par value of ₹ 100).
(i) How much should he invest ₹
(ii) If the dividend is 7.5%, what will be his annual income?
(iii) If he wants to increase income by ₹ 150, how many extra shares should he buy?
Solution:
Number of shares = 62
Market value of each share = ₹ 132
Face value = ₹ 100
(i) His investment = ₹ 132 x 62 = ₹ 8184

(ii) Rate of dividend = 7.5% = \(\frac { 15 }{ 2 }\) % p.a.
Annual income = ₹ 62 x 100 x \(\frac { 15 }{ 2×100 }\)
= ₹ 465

(iii) Extra income he wants = ₹ 150
Then annual income = ₹ 465 + 150 = ₹ 615
∴ Number of shares = \(\frac{615 \times 100 \times 2}{15 \times 100}\) = 82
∴ Extra share he has to buy = 82 – 62 = 20

Question 4.
A man invests ₹ 20,020 in buying shares of nominal value ₹ 26 at 10% premium. The dividend on the shares is 15% per annum. Calculate :
(i) The number of shares he buys.
(ii) The dividend he receives annually.
(iii) The rate of interest he gets on his money.
Solution:
Investment = ₹ 20020
Nominal value of each share = ₹ 26
Market value at 10% premium
= ₹ \(\frac{26 \times(100+10)}{100}\)
= ₹ \(\frac{26 \times 110}{100}\) = ₹ \(\frac { 2860 }{ 100 }\) = ₹ 28.60
Rate of dividend = 15%

(i) Number of share he bought = \(\frac { 20020 }{ 28.60 }\)
= \(\frac{20020 \times 100}{2860}\) = 70

(ii) Total dividend per year = 700 x 26 x 15%
= \(\frac{700 \times 26 \times 15}{100}\) = ₹ 2730

(iii) Rate of interest on investment
= \(\frac{2730 \times 100}{20020}\) = 13.636 %
= 13.46%

Question 5.
A man invested ₹ 45,000 in 15% ₹ 100 shares quoted at ₹ 125. When the market value of these shares rose to ₹ 140, he sold some shares, just enough to raise ₹ 8400. Calculate :
(i) the number of shares he still holds;
(ii) the dividend due to him on these remaining shares.
Solution:
Total investment = ₹ 45000
Face value of each share = ₹ 100
Market value = ₹ 125
Rate of dividend = 15%
∴ Number of shares = \(\frac { 4500 }{ 125 }\)
He sells some shares at the rate of ₹ 140
(i) Raise his income ₹ 8400
∴ Number of shares he sells = \(\frac { 8400 }{ 140 }\) = 60
Remaining shares = 360 – 60 = 300
Dividend on remaining shares = 300 x 100 x 15%
= ₹ \(\\frac{300 \times 100 \times 15}{100}\) = ₹ 4500

Question 6.
Mr. Tewari invested ₹ 29,040 in 15%, ₹ 100 shares quoted at a premium of 20%. Calculate :
(i) The number of shares bought by Mr. Tewari.
(ii) Mr. Tewari’s income from the investment.
(iii) The percentage return on his investment.
Solution:
Investment made by Tewari = ₹ 29040
Face value of each share = ₹ 100
Market value at a premium of 20%
= ₹ 100 + 20 = ₹ 120
Rate of dividend = 15%
(i) Number of shares bought = ₹ \(\frac { 29040 }{ 120 }\) = 242
(ii) Income from investment = ₹ 242 x 100 x 15%
= ₹ 242 x 100 x \(\frac { 15 }{ 100 }\) = ₹ 3630

(iii) Percentage income on investment
= ₹ \(\frac{3630 \times 100}{29040}\) = 12.5%

Question 7.
Mr. Ram Gopal invested ₹ 8000 in 7% ₹ 100 shares at ₹ 80. After a year he sold these shares at ₹ 75 each and invested the proceeds (including his dividend) in 18%, ₹ 25 shares at ₹ 41. Find :
(i) his dividend for the first year.
(ii) his annual income in the second year.
(iii) the percentage increase in his return on his original investment.
Solution:
Investment made by Ram Gopal = ₹ 8000
Face value of each share = ₹ 100
Market value = ₹ 80
Rate of dividend = 7%
Number of shares = ₹ \(\frac { 8000 }{ 80 }\) = 100
(i) Dividend for the first year = ₹ 100 x 100 x 7%
= \(\frac{100 \times 100 \times 7}{100}\) = ₹ 700

(ii) M.V. of second year = ₹ 75
∴ Sale proceed = ₹ 100 x 75 = ₹ 7500
Total investment including dividend = ₹ 7500 + 700 = ₹ 8200
Rate of dividend in second year = 18%
M.V. = ₹ 41
Face value = ₹ 25
∴ Number of shares bought = \(\frac { 8200 }{ 41 }\) = 200
Nominal value of 200 share = ₹ 25 x 200 = ₹ 5000
∴ Dividend = ₹ 5000 x 18%
= ₹ 5000 x \(\frac { 18 }{ 100 }\) = ₹ 900

(iii) Increase in income = ₹ 900 – ₹ 700 = ₹ 200
∴ Increase percent = \(\frac{200 \times 100}{8000}\)
= \(\frac { 5 }{ 2 }\) % = 2.5%

Question 8.
Ajay owns 560 shares of a company. The face value of each share is ₹ 25. The company declares a dividend of 9%. Calculate :
(i) The dividend that Ajay will get.
(ii) The rate of interest on his investment, if Ajay had paid ₹ 30 for each share.
Solution:
Ajay has shares of a company = 560
Face value of each share = ₹ 25
Rate of dividend = 9%
(i) Face value of 560 shares = ₹ 25 x 560
= 114000
∴ Total dividend he received
= ₹ 14000 x 9%
= ₹ 14000 x \(\frac { 9 }{ 100 }\) = ₹ 1260

(ii) M.V. of each share = ₹ 30
∴ Total investment = ₹ 30 x 560 = ₹ 16800
Rate of interest on his investment = \(\frac{1260 \times 100}{16800}\) = 7.5 %

Question 9.
A company with 4000 shares of nominal value of ₹ 110 each declares an annual dividend of 15%. Calculate:
(i) The total amount of dividend paid by the company.
(ii) The annual income of Shah Rukh who holds 88 shares in the company.
(iii) If he received only 10% on his investment, find the price Shah Rukh paid for each share.
Solution:
Number of shares = 4000
Nominal value of each share = ₹ 110
Rate of dividend = 15%
(i) ∴ Total amount of dividend
= ₹ 4000 x 110 x 15%
= ₹ \(\frac{4000 \times 110 \times 15}{100}\)
= ₹ 66000

(ii) Face of 88 shares = ₹ 110 x 88 = ₹ 9680
∴ Annual income of Shah Rukh
= ₹ \(\frac { 9680×15 }{ 100 }\) = ₹ 1452

(iii) Interest on investment made by Shah Rukh = 10%
∴ Price (value) of each share paid by Shah

Question 10.
Amit Kumar invests ₹ 36,000 in buying ₹ 100 shares at ₹ 10 premium. The dividend is 15% per annum. Find
(i) the number of shares he buys.
(ii) his yearly dividend.
(iii) the percentage return on his investment.
Give your answer correct to the nearest whole number.
Solution:
Investment = ₹ 36000
Face value = ₹ 100
Premium = ₹ 20, dividend = 15%
(i) No. of shares = \(\frac { 36000 }{ 120 }\) = 300

(ii) Dividend = 15% of (100 x 300)
= \(\frac { 15 }{ 100 }\) x 30000 = ₹ 4500

(iii) Per cent of return on investment
= \(\frac { 45000 }{ 36000 }\) x 100 = \(\frac { 450 }{ 36 }\) = 12.5% = 13%

Question 11.
Vivek invests ₹ 4,500 in 8% ₹ 10 shares at ₹ 15. He sells the shares when the price rises to ₹ 30, and invests the proceeds in 12% ₹ 100 shares at ₹ 125. Calculate.
(i) the sale proceeds,
(ii) the number of ₹ 125 shares he buys,
(iii) the change in his annual income from dividend.
Solution:
(i) If price of share bought is ₹ 15, then face value of share = Rs. 10
If price of share bought is ₹ 4500, then face value of share bought
= \(\frac { 10 }{ 15 }\) x 4500 = ₹ 3000
Total face value of ₹ 10 shares = ₹ 3000 Income = 8%
= \(\frac { 8 }{ 100 }\) x 3000 = Rs. 240
By selling ₹ 10 share money received = ₹ 30
By selling Rs. 3000 shares money
= \(\frac { 30 }{ 10 }\) x 3000 = ₹ 9000

(ii) By investing ₹ 125, no. of share of ₹ 100 bought = 1
By investing ₹ 9000, no. of share of ₹ 100 bought = \(\frac { 1 }{ 125 }\) x 9000 = 72
∴ No. of ₹ 125 shares bought = 72

(iii) By investing ₹ 125 in ₹ 100 share, income = ₹ 12
By investing Rs. 9000 in ₹ 100 share, income 12
= \(\frac { 12 }{ 125 }\) x 9000 = ₹ 864
Increase in income = ₹ 864 – ₹ 240 = ₹ 624

Question 12.
Mr. Parekh invested ₹ 52,000 on ₹ 100 shares at a discount of ₹ 20 paying 8% dividend. At the end of one year he sells the shares at a premium of ₹ 20. Find :
(i) The annual dividend.
(ii) The profit earned including his dividend.
Solution:
Investment = ₹ 52000
Face value of 1 share = ₹ 100
Market value of 1 share = ₹ 100 – 20 = ₹ 80
No. of shares = \(\frac { 52000 }{ 80 }\) = 650
(i) Annual dividend = \(\frac { 8 }{ 100 }\) x 650 x 100 = ₹ 5200
(ii) S.P. of 1 share = ₹ 100 + 20 = ₹ 120
S.P. of 650 shares = ₹ 120 x 650 = ₹ 78000
C.P. of 650 shares = ₹ 100 x 650 = ₹ 65000
Profit = S.P. – C.P.
= ₹ 78000 – ₹ 52000 = ₹ 26000
Profit including dividend = ₹ 26000 + ₹ 5200 = ₹ 31200

Question 13.
A man invests ₹ 9600 on ₹ 100 shares at ₹ 80. If the company pays him 18% dividend, find :
(i) the number of shares he buys.
(ii) his total dividend.
(iii) his percentage return on the shares.
Solution:
Amount of investment = ₹ 9600
Price of one share = ₹ 80
(i) ∴ No. of shares bought = ₹ \(\frac { 9600 }{ 80 }\) = 120

(ii) Face value of 120 shares = ₹ 120 x 100 = ₹ 12000
Rate of dividend = 18%
Dividend = ₹ \(\frac { 12,000×18 }{ 100 }\) = ₹ 2160

(iii) By investing ₹ 9600, returned obtained = ₹ 2160
So, percentage return = \(\frac{2.160 \times 100}{9600}\) = 22.5%

Question 14.
Salman buys 50 shares of face value ₹ 100 available at ₹ 7132.
(i) What is his investment?
(ii) If the dividend is 7.5%, what will be his annual income?
(iii) If he wants to increase his annual income by ₹ 150, how many extra shares should he buy?
Solution:
F.V. = ₹ 100
(i) M.V. = ₹ 132, no. of shares = 50
Investment = no. of shares x MV.
= 50 x 132 = ₹ 6600

(ii) Income per share = 7.5% of F.V.
= \(\frac { 75 }{ 10×100 }\) x 100 = ₹ 7.5
∴ Annual income = 7.5 x 50 = ₹ 375

(iii) New annual income = 375 + 150 = ₹ 525
∴ No. of shares = \(\frac { 525 }{ 7.5 }\) = 70
∴ No. of extra share = 70 – 50 = 20

Question 15.
Salman invests a sum of money in ₹ 50 shares, paying 15% dividend quoted at 20% premium. If his annual dividend is ₹ 600, calculate:
(i) the number of shares he bought.
(ii) his total investment.
(iii) the rate of return on his investment.
Solution:
Nominal value = ₹ 50
Dividend on 1 share = \(\frac { 15 }{ 100 }\) x ₹ 50 = ₹ 7.50
Total Dividend to Salman = ₹ 600
(i) No. of shares Salman bought = \(\frac { 600 }{ 7.50 }\)
= \(\frac{600 \times 100}{750}\) = 80

(ii) Premium on 1 share = 50 x \(\frac { 20 }{ 100 }\) = ₹ 10
Market value of 1 share = 50 + 10 = ₹ 60
Total investment for 80 shares = 80 x 60 = ₹ 4800

(iii) Rate of return = \(\frac{\text { Total dividend }}{\text { Total investment } \times 100}\)
= \(\frac { 600 }{ 4800 }\) x 100 = 12.5%

Question 16.
Rohit invested ₹ 9,600 on ₹ 100 shares at ₹ 20 premium paying 8% dividend. Rohit sold the shares when the price rose to ₹ 160. He invested the proceeds (excluding dividend) in 10% ₹ 50 shares at ₹ 40. Find the:
(i) original number of shares.
(ii) sale proceeds.
(iii) new number of shares.
(iv) change in the two dividends.
Solution:
(i) 100 shares at ₹ 20 premium means
Nominal value of the share is ₹ 100
and its marked value = 100 + 20 = ₹ 120
Money required to buy 1 share = ₹ 120
Number of shares
= \(\frac{\text { Money invested }}{\text { Market value of } 1 \text { share }}\)
= \(\frac { 9600 }{ 120 }\) = 80 shares

(ii) Dividend on 1 share = 8% of N.V.
= 8% of 100 = 8
Total dividend on 80 shares = 80 x 8 = ₹ 640
Each share is sold at ₹ 160
∴ The sale proceeds = 80 x ₹ 160
= ₹ 12800

(iii) New investment = ₹ 12800
Divident=10%
Net value = 50
Market value = ₹ 40
∴ Number of shares = \(\frac{\text { Investment }}{\text { Market value }}\)
= \(\frac { 12800 }{ 40 }\)
= 340 shares

(iv) Now, dividend on 1 share = 10% of N.V.
= 10% of 50 = 5
∴ Dividend on 340 shares = 1600
Change in two dividends = ₹ 1600 – ₹ 640 = ₹ 960

Question 17.
Ashok invested ₹ 26,400 on 12%, ₹ 25 shares of a company. If he receives a dividend of ₹ 2,475. Find the :
(i) number of shares he bought.
(ii) market value of each share.
Solution:
Investment = ₹ 26400
Rate of divident = 12%
Divident = ₹ 2475
Face value of one share = ₹ 25
Total dividend = Number of shares x Rate of dividend x Face value of one share
2475 = Number of shares x \(\frac { 12 }{ 100 }\) x 25
Number of shares = \(\frac { 2475 }{ 3 }\) = 825
Market value of one share Investment
= \(\frac{\text { Investment }}{\text { Number of shares bought }}\)
Market value of one share = \(\frac { 26400 }{ 825 }\) = ₹ 32
Ashok bought 825 shares and market value of each share is ₹ 32.

Regular engagement with ICSE Class 10 Maths Solutions S Chand Chapter 19 Histogram and Ogive Ex 19(c) can boost students’ confidence in the subject.

S Chand Class 10 ICSE Maths Solutions Chapter 19 Histogram and Ogive Ex 19(c)

Question 1.
The daily wages of casual labour employed by a group of limited concerns are given below:

Daily wages (in rupees)	3-5	5-7	7-9	9-11	11-13	13-15
Frequency	7	10	23	51	6	3

Draw a cumulative frequency curve for the above data.
Solution:

Daily wages (in Rs.)	Frequency (f)	Cumulative frequency (c.f.)
3-5	7	7
5-7	10	17
7-9	23	40
9-11	51	91
11-13	6	97
13-15	3	100

Now we shall plot the points (5, 7), (7, 17), (9, 40), (11, 91), (13, 97), and (15, 100) on the graph and join them with free hand to get an ogive as shown.

Question 2.
Draw a cumulative frequency curve for the following data :
(a)

Marks	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	5	10	22	40	15	8

(b)

Class interval	0-5	5-10	10-15	15-20	20-25	25-30	30-35	35-40
Frequency	20	23	25	28	30	21	28	16

(c)

Class interval	10-19	20-29	30-39	40-49	50-59	60-69
Frequency	10	15	20	25	30	35

Solution:
(a)

Marks	Frequency	Cumulative frequency
0-10	5	5
10-20	10	15
20-30	22	37
30-40	40	77
40-50	15	92
50-60	8	100

Now we shall plot the points (10, 5), (20, 15), (30, 37), (40, 77), (50, 92) and (60, 100) and join them with free hand to get an ogive as shown.

(b)

Class intervals	Frequency	c.f.
0-5	20	20
5-10	23	43
10-15	25	68
15-20	28	96
20-25	30	126
25-30	21	147
30-35	28	175
35-40	16	191

Now we shall plot the points (5,20), (10,43), (15,68), (20,96), (25,126), (30,147), (35,175) and (40,191) on the graph and join them with free hand to get an ogive as shown :

(c)

Give class interval	Given actual class interval	Frequency	c.f.
10-19	9.5-19.5	10	10
20-29	19.5-29.5	15	25
30-39	29.5-39.5	20	45
40-49	39.5-49.5	25	70
50-59	49.5-59.5	30	100
60-69	595-69.5	35	135

Now we shall plot the points (19.5, 10), (29.5, 25), (39.5, 45), (49.5, 70), (59.5, 100) and (69.5, 135) on the graph and join them with free hand to get an ogive.

Question 3.
Each of the 25 students in a class was given a home assignment comprising 10 questions in mathematics. The data given below show the number of questions solved and submitted by individual students on the next day.
1, 4, 5, 6, 0, 9, 3, 2, 3, 4, 6, 4, 5, 2, 7, 5, 2, 2, 3, 5, 1, 0, 7, 6, 3
(a) Taking classes as 0-2, 2-4, 4-6, etc., make a frequency table for the above distribution.
(b) Draw an Ogive (cumulative frequency curve) to represent the given data.
Solution:

To draw its ogive, we plot the points (2, 7), (4, 12), (6, 19), (8, 24), (10, 25) on the graph and join them with free hand the ogive as shown

Question 4.
Draw an ogive from the following table :

Marks obtained	C.F.
0 and more	100
10 and more	96
20 and more	68
30 and more	26
40 and more	6

Solution:

Marks obtained	c.f.
0-10	100
10-20	96
20-30	68
30-40	26
40-50	6

Now we plot the points (10, 100), (20, 96), (30, 68), (40, 26) and (50, 6) on the graph and join them with free hand to get ogive as shown.

Question 5.
The marks secured by 50 students were as under:

14	12	13	18	11	25	32	27	28	27
3	9	37	31	5	22	13	5	22	14
10	18	40	23	28	19	18	2	13	14
22	7	46	12	36	14	17	19	35	9
12	2	49	43	7	6	10	22	3	27

Taking the size of each class-interval as 10, prepare frequency table and with its help draw a cumulative frequency curve.
Solution:
Highest marks = 49
Lowest marks = 2

Now we shall plot the points (10, 11), (20, 30), (30, 41), (40, 46) and (50, 50) on the graph and join them with free hand to get an ogive as shown.

Question 6.
What is an Ogive curve? How is it useful?
A group of 140 workers in a factory were given a work aptitude test. The distribution of their scores is given below :

Score	No. of Workers	Score	No. of Workers
10-15	2	35-40	19
15-20	7	40-45	27
20-25	9	45-50	20
25-30	15	50-55	11
30-35	26	55-60	4

Draw an Ogive curve.
Solution:
Ogive is a graph of cumulative frequency distribution which is drawn in a free hand curve.

Score	No. of Workers (f)	c.f.
10-15	2	2
15-20	7	9
20-25	9	18
25-30	15	33
30-35	26	59
35-40	19	78
40-45	27	105
45-50	20	125
50-55	11	136
55-60	4	140

Now we shall plot the points (15, 2), (20, 9), (25, 18), (30, 33), (35, 59), (40, 78), (45, 105), (50, 125), (55, 136) and (60, 140) on the graph and join them with free hand to get an ogive as shown.

Students often turn to OP Malhotra Class 10 Solutions Chapter 1 GST (Goods and Service Tax) Ex 1 to clarify doubts and improve problem-solving skills.

S Chand Class 10 ICSE Maths Solutions Chapter 1 GST (Goods and Service Tax) Ex 1

Question 1.
Mr Abdul a manufacturer, sells his product worth 2,25.000 within the state. He buys goods worth ₹ 1,20.000 within the stale. If the rate of GST is @ 12% on the raw material and @ 18% on the finished product find the amount of GST he has to pay.
Solution:
C.P. of raw material for Abdul = ₹ 120000
Rate of GST on raw material = 12%
∴ Input GST i.e. GST paid by Abdul = ₹ 120000 x \(\frac {12}{100}\) = ₹ 14,400
SP of finished goods for Abdul = ₹ 2,25,000
Rate of GST on finished goods = 18%
∴ output GST, i.e. GST received by Abdul = ₹ 225000 x \(\frac {18}{100}\) = ₹ 40500
Now, GST paid by Abdul to government = output GST – Input GST = ₹ 40500 – ₹ 14400 = ₹ 26,100

Question 2.
A shoe manufacturer purchases goods worth ₹ 90.000 from the markets within the state. He sells his product in the neighbourhood market for 78,000. If the common rate of GST @ 18%. find the GST pasable/GST’ credit for the aboie transaction.
Solution:
CP of goods purchased by manufacturer = ₹ 90000
Rate of GST = 18%
∴ Input GST i.e. GST paid by Abdul = 18% of ₹ 90000 = ₹ 90000 x \(\frac {18}{10}\) = ₹ 16,200
Now manufacturer sold the product in neighbourhood market
SP of finished goods for Abdul = ₹ 78,000
Rate GST = 18%
∴ output GST i.e. GST’ received by manufacturer = 18% of 78000 = \(\frac {18}{100}\) x 78000 = ₹ 14040.
Now GST refund claimed by manufacturer from the government = Input GST – output GST
= ₹ 6200 – ₹ 14040 = ₹ 2160

Question 3.
Mrs Lata has a leather coat manufacturer unit in state A. She buys raw materials worth ₹ 80,000 from a supplier froni state B at a discount of 10%. She sells her product worth ₹ 2,20,000 0utside the state. If the rate of CGST @ 2.5 % find the IGST payable/credit of Nirs Lata.
Solution:
Since, it is a case of interstate transactions of goods/services
∴ IGST = GST
We know, that CGST = SGST = \(\frac {1}{2}\) GST
Given, CGST = 2.5%
∴ GST = 2 x CGST = 2 x 2.5% = 5%
MP of raw material for Lata = ₹ 80,000
Discount = 10 %
CP of raw material for Lata = ₹ 80000 – 10% 0f ₹ 80000 = ₹ 80000 – 8000 = ₹ 72000
∴ Input IGST i.e. GST paid by Lata = 5% 0f ₹ 72000 = \(\frac {5}{100}\) x 72000 = ₹ 3600
Further, Lata sold the products to distributer in other state
SP of finished goods for Lata = ₹ 2,20,000
∴ output IGST i.e. GSI’ received by Lata = 5% 0f ₹ 220000 = \(\frac {5}{100}\) x 220000 = ₹ 11000
Now, IGST payable by Lata to the govt. = output GST – input GST = ₹ 11000 – ₹ 3600 = ₹ 7400

Question 4.
Mrs Salim a biscuit nianufacturer buys raw goods worth 1,40,000 froni different markets sitliin the state GST @ 5%. lic sold packet biscuits worth 2,10,500 in the markets of the neighbouring state. Rate of GST on packet biscuits is 12%. Find the amount of 1GST
payable by him.
Solution:
CP of material of biscuit for manufacturer = ₹ 1,40,000
Rate of GST 5%
∴ Input GST, i.e. GST paid by manufacturer = 5% of ₹ 1,40.000 = \(\frac {5}{100}\) x 1,40,000 = ₹ 7,000
Further, manufacturer sold the biscuits in neighbouring states.
SP of biscuits for manufacturer = ₹ 2, 10,500
Rate of GST = 12%
∴ output IGST, i.e. IGST received by manufacturer 12% of ₹ 2,10,500
= \(\frac {12}{100}\) x 2,10,500 = ₹ 25,260
IGST payable by Mr. Salirn to the govemnient = output IGST – Input GST
= ₹ 25260 – ₹ 7000 = ₹ 18260

Question 5.
A seiing machine manufacturing purchases raw materials worth ₹ 5.40.000 fur his manufacturing unit from outside the state. The rate of IGST is @ 12%. He produced two types of sewing machine, 50 type A whose base price is ₹ 5000. GST 12% and 100 type B whose base price is ₹ 1,000, GST @ 18%. He had two clients outside the state and receised orders for 10 type A and 20 type B only by client X, 10 each by client Y. Find
(i) input GST, (ii) output CST (iii) CST payable / credit
Solution:
(i) Given, CP of raw material for manufacturer = ₹ 5,40,000
Rate of GST i.e. IGST = 12%
Input IGST, i.e. IGST paid by manufacturer = 12% of ₹ 5,40.000 = \(\frac {12}{100}\) x 5,40.000 = ₹ 64.800

(ii) Manufacturer received orders of 10 Type A and 20 Type B sewing machines by client X and 10 each Type A and Type B by client Y.
Base price of Type A = 5,000 and Base price of Type B = ₹ 10000
∴ SP 0f sewing machines of Type A for manuficturer = (10 x 5000) + (10 x ₹ 5000)
= ₹ 50000 + ₹ 50000 = ₹ 100000
output IGST i.e. IGST received by manufacturer on sewing machines of Type A
= 12% of 100000 = ₹ 12000
Also SP of sewing machine of Type B for manufacturer (20 x ₹ 10000) + (10 x 10000)
= ₹ 20000 + ₹ 100000 = ₹ 300000
output IGST i.e. IGST received by manufacturer on sewing machines of Type B
= 18% of ₹ 300000 = ₹ 5400
Total output IGST = ₹ 2000 + ₹ 54000 = ₹ 66000

(iii) IGST payable by manufacturer to the government Total output IGST – Input IGST
= ₹ 66000 – ₹ 6400 = ₹ 1200

Question 6.
The sales price of a sashing machine. inclusise of GST, is ₹ 28,320. If the SCST is charged at the rate of 9% of the list price, find the list price of the washing machine.
Solution:
Let the marked price of washing machine = ₹ x
Rate of SGST = 9%
We know that, in case of intra state of transacti0n of goods and services
SGST = CGST = \(\frac {1}{2}\) x GST
∴ GST = 2 x SGST = 2 x 9% = 18%
Given, SP of washing machine inclusive 0f taxes = ₹ 28320
∴ GST Amount = ₹ \(\frac {18x}{100}\)
and final amount = ₹\(\left(x+\frac{18 x}{100}\right)=₹ \frac{118 x}{100}\)
A.T.Q
\(\frac {118x}{100}\) = ₹ 28320
⇒ x = \(\frac {28320}{118}\) x 100 = ₹ 24000
∴ The list price of washing machine = ₹ 24000

Question 7.
Radhika buys crockery having marked price ₹ 4,500. She gets a discount of 12%. If the CGST is a 6%, find the amount she is required to pay for the crockery. Also find the amount of SCST.
Solution:
MP of crockery = ₹ 4500
Discount offered = 12%
SP of crockery for Radhika = ₹ 4500 – 12% 0f ₹ 4500 = ₹ 4500 – ₹ 540 = ₹ 3960
Rate of CGST = 6 %
We know that, in case of inn state transaction of goods/services
SGST = CGST = \(\frac { 1 }{ 2 }\) x GST
∴ GST = 2 x CGST = 2 x 6% = 12%
GST paid by Radhika = 12% 0f ₹ 3960 = \(\frac { 12 }{ 100 }\) x 3960 = ₹ 475.2
∴ Final price paid by Radhika = ₹ 3960 + ₹ 4752 = ₹ 4435.20
We know that, CGST = CGST = \(\frac { 1 }{ 2 }\) x GST
∴ CGST = SGST = \(\frac { 1 }{ 2 }\) x 475.20 = ₹ 237.60

Question 8.
The list price of a bell laptop is ₹ 84.000. The dealer gives a discount of 20% on the listed price. He also offers 10% additional discount on the balance. However, GST @ 28% is charged oil Laptop. Find
(i) the taxable amount.
(ii) the total amount of GSl’ the customer has to pay.
(iii) final price has to pay to the dealer including CST.
Solution:
MP of Dell Taptop = ₹ 84.000
Discount allowed = 20%
Price after discount = ₹ 84000 – 20% 0f ₹ 84000 = ₹ 84000 – ₹ 16800 = ₹ 67200
Additional discount = 10 %
Price after addittonal discount = ₹ 67200 – 10% 0f ₹ 67200 = ₹ 67200 – ₹ 6720 = ₹ 60.480

(ii) Rate of GST = 28%
GST charged = 28% of = ₹ 60,480 = \(\frac { 28 }{ 100 }\) x ₹ 60480 = ₹ 16934.40

(iii) Final price paid including GST = ₹ 60,480 + ₹ 16934.40 = ₹ 77414.40

Question 9.
Mrs Santa buys goods worth ₹ 6.500 from Easy day store. She gets a rebate of 10% on ₹ 5000 as a member and a flat discount of Soon the remaining. If the GST is charged @ 18% find the total amount she has to pay fur the goods.
Solution:
MP of goods = ₹ 6500
Discount allowed = 10% on ₹ 5000 = ₹ 500
and flat discount allowed = ₹ 50
∴ Total discount allowed = ₹ 500 + ₹ 50 = ₹ 550
SP of goods by easyday = ₹ 6500 – ₹ 550 = ₹ 5950
Rate of GST 18%
GST charged on goods = 18% of 5950 = \(\frac { 18 }{ 100 }\) x 5950 = ₹ 1071
Total price paid by Mrs. Santa = ₹ 5950 + ₹ 1071 = ₹ 7021

Question 10.
A retailer marked up the price of his goods by 20% above the list price and offers to successive discount of 10% and % on the marked up price. GST is @ 18 % on the goods. Find the list prier of goods if the consumer pay ₹ 4617 as CGST. Also find the final price consumer has to pay for the goods.
Solution:
(i) Let list price of the article = ₹ x
According to the question, goods are marked up by 20% followed by two successive discounts of 10% and 5%
∴ Taxable amount = ₹ x \(\left(1+\frac{20}{100}\right)\left(1-\frac{10}{100}\right)\left(1-\frac{5}{100}\right)=₹ x\left(\frac{120}{100}\right)\left(\frac{90}{100}\right)\left(\frac{95}{100}\right)=₹ x \frac{513}{500} x\)
Given, GST 18%
∴ SGST = \(\frac { 1 }{ 2 }\) x GST = 9%
According to the questi0n,
\(\frac{9}{100} \times \frac{513}{500} x=4617 \Rightarrow x=\frac{4617 \times 500 \times 100}{9 \times 513}\) = ₹ 51300
∴ List price = ₹ 50000
∴ Taxable amount = ₹ \(\frac { 513 }{ 2 }\) x ₹ 50000 = ₹ 51300
Also, SGST = CGST = ₹ 4617
∴ Final price paid by the c0nsumer = ₹ 51300 + ₹ 4617 = ₹ 60534

Question 11.
Anil went to a shop to buy a bicycle costing ₹ 10,620. the rate of CGST is 9%. He asks the shopkeeper to reduce the price of the bicycle to such an extent so that he has to pay ₹ 10,620 inclusise GST. Find the reduction needed in the price of the bicycle.
Solution:
MP of the bicycle = ₹ 10620 and 1maI price paid by Anil = ₹ 10620
Let the reduced price of a bicycle aller discount be ₹ x
Given CGST = 9%, GST = 18%
∴ GST amount = ₹ \(\frac { 18x }{ 100 }\)
∴ Final price = ₹\(\left(x+\frac{18 x}{100}\right)=₹ \frac{118 x}{100}\)
A.T.Q,
\(\frac { 118x }{ 100 }\) = 10620 ⇒ x = \(\frac { 10620 }{ 118 }\) x 100 = ₹ 9000
Hence, the final price of the bicycle is ₹ 9000
∴ The reduction in the price of the bicycle = MP – Final price = ₹ 10620 – ₹ 9000 = ₹ 1620

Question 12.
The price of a mobile phone is ₹ 32,000 inclusive of 28 % GST on the list price. Vineeta asks for a discount on the list price so that after charging the GST, the final price becomes the list price. Find
(i) the amount of discount which Vineeta got from the shopkeeper,
(ii) the amount of SGST on the discounted value of the mobile.
Solution:
(i) MP of a mobile phone = ₹ 32000
Rate of GST = 28 %
Let the reduced price of a mobile phone after discount be ₹ x
∴ GST amount = ₹ \(\frac { 28x }{ 100 }\)
Final price = ₹(x + \(\frac { 28x }{ 100 }\)) = ₹ \(\frac {128x }{ 100 }\)
A.T.Q
₹ \(\frac {128x }{ 100 }\) = ₹ 3200 ⇒ x = \(\frac{32000 \times 100}{128}\) = ₹ 25000
∴ Final price charted aliei reduction = ₹ 25000
∴ Vinecta got discount from shopkeeper of = ₹ 32000 – ₹ 500 = ₹ 7000

(ii) We know that.
CGST = SGST = \(\frac { 1 }{ 2 }\) x GST = \(\frac { 1 }{ 2 }\) x 28 = 14%
SGST paid by Vineeta on mobile phone = 14% of 5004J = ₹ 25000 = ₹ 3500

Question 13.
A calculator manufacturers manufacturing cosi of a calculator is ₹ 900. GST is 18%. He manufactured 120 such calculators. He marked up each by 30% and sold to a dealer at a discount of 10%. Find the final price dealer paid for the calculators. Also find the total GST received by the State Government.
Solution:
C.P. of one calculator for manufacturer = ₹ 900
C.P. of 120 calculators 11w manufacturer = ₹ 900 x 120 = ₹ 1,08,000
∴ Input GST paid by manufacturer = 18% 0f ₹ 1,08,000 = \(\frac { 18 }{ 100 }\) x 1,08.000 = ₹ 19,440
Manufacturer marked up the price by 30%
∴ MP of 120 calculators = ₹ 1,08,000 + 30% 0f ₹ 1,08,000 = ₹ 1,08,000 + ₹ 32.400 = ₹ 1,40,400
Manufacturer offers a discount of 10 % 0n M.P.
∴ S.P. of 120 calculators = ₹ 1,40,400 – 10 % 0f ₹ 1,40,400 = ₹ 1,40,400 – ₹ 14,040 = ₹ 1,26,360
∴ output GST i.e. GST received by manufacturer = 18 % 0f ₹ 1,26,360 = \(\frac { 18 }{ 100 }\) x 1,26,360 = ₹ 22,744.80
Thus, final price paid by the dealer = ₹ 1,26,360 + ₹ 22,744.80 = ₹ 1,49,104.80
Since, it is a case of intra state transaction of goods and services
∴ CGST = SGST = \(\frac { 1 }{ 2 }\) x GST
Hence, GST received by State government i.e. SGST = \(\frac { 1 }{ 2 }\) x GST = \(\frac { 1 }{ 2 }\) x ₹ 22,744.80 = ₹ 11372.4

Question 14.
The manufacturer prooduces television sets at a CGST of ₹ 32,000. He sells it to distributor at a profit of ₹ 2,000, distributor sells it to a wholesaler at a profit of ₹ 2,500, and wholesaler sells to a retailer at a profit of ₹ 3,000. Finally retailer sells to consumer at a profit of ₹ 3,500 and rate of GST is 18 %. Find
(i) the final amount which consumer pays for the article.
(ii) total GST amount.
Solution:
Given, Cost price of the television for manufacturer = ₹ 32,000 and rate of GST =18 %
∴ GST paid by manufacturer =18 % of ₹ 32,000 = \(\frac { 18 }{ 100 }\) x 32000 = ₹ 5760
The manufacturer sells it to a distributor at a profit of ₹ 2,000
Given, Sale price of the television for manufacturer = ₹ 32,000 + ₹ 2,000 = ₹ 34,000
and rate of GST = 18 %

∴ Input GST i.e. GST paid by distributor =18 % of ₹ 34,000 = \(\frac { 18 }{ 100 }\) x 34000=₹ 6120
The distibutor sells it to a wholesaler at a profit of ₹ 2,500
Given, Sale price of the television for wholesaler = ₹ 34,000 + ₹ 2,500 = ₹ 36,500
and rate of GST = 18 %

∴ Input GST i.e. GST paid by wholesaler = 18 % of ₹ 36,500 = \(\frac { 18 }{ 100 }\) x 36500 = ₹ 6570
The wholesaler sells it to a retailer at a profit of ₹ 3,000
Given, Sale price of the television for wholesaler = ₹ 36,500 + ₹ 3,000 = ₹ 39,500
and rate of GST = 18 %

∴ Input GST i.e. GST paid by retailer = 18 % of ₹ 39,500 = \(\frac { 18 }{ 100 }\) x 39500 = ₹ 7110
The retailer sells it to a consumer at a profit of ₹ 3,500
Given, Sale price of the television for retailer = ₹ 39,500 + ₹ 3,500 = ₹ 43,000 and rate 0f GST = 18 %

∴ Input GST i.e. GST paid by consumer =18 % of ₹ 43,000 = \(\frac { 18 }{ 100 }\) x 43000= ₹ 7740
(i) Total amount paid by the consumer = S.P. of the Television + GST = ₹ 43000 + ₹ 7740 = ₹ 50740
(ii) Total GST Paid = ₹ 7740

Question 15.
A manufacturer washing machines at a cGST of ₹ 11,000. He sells to ‘ B ‘, and ‘ B ‘ sells to ‘ C ‘, ‘C’ sells it to ‘D’. The GST rate is 28 % and the profit is ₹ 1,500 at each stage of selling chain. Find
(i) the total amount of GST; and
(ii) the price including GST ‘D’ paid for the washing machine.
Solution:
(i) Given, Cost price of the washing machine for manufacturer = ₹ 11,000
and rate of GST =28 %
∴ GST paid by manufacturer = 28 %of ₹ 11,000 = \(\frac { 28 }{ 100 }\) x 11000 = ₹ 3080
The manufacturer sells it to a ‘ B ‘ at a profit of ₹ 1,500
Sale price of the washing machine for manufacturer = ₹ 11,000 + ₹ 1,500 = ₹ 12,500 and rate 0f GST = 28 %

∴ Input GST i.e. GST paid by ‘B’ = 28 % of ₹ 12,500 = \(\frac { 28 }{ 100 }\) x 12500 = ₹ 3500
The ‘ B ‘ sells it to a ‘ C ‘ at a profit of ₹ 1,500
Sale price of the washing machine for ‘C’ prime = ₹ 12,500 + ₹ 1,500 = ₹ 14,000 and rate of GST = 28 %

∴ Input GST i.e. GST paid by ‘C’ = 28 % of ₹ 14,000 = \(\frac { 28 }{ 100 }\) x 14000 = ₹ 3920
The ‘C’ sells it to a ‘D’ at a profit of ₹ 1,500
Sale price of the washing machine for ‘D’ = ₹ 14,000 + ₹ 1,500 = ₹ 15,500 and rate 0f GST = 18 %
∴ Input GST i.e. GST paid by ‘D’ =28 % of ₹ 15,500=28100 15500=₹ 4340
(i) Total amount of GST = ₹ 4340
(ii) Total amount paid by the ‘D’ including GST = S.P. 0f the washing machine + GST
= ₹ 15500 + ₹ 4340 = ₹ 19840

Question 16.
The manufacturer sold a TV to a wholesaler at a profit of ₹ 1000, whose manufacturing cost is ₹ 15.000. The holesaler sold it to a trader at a profit of ₹ 1000. If the trader sold it to the customer at profit ₹ 1500. Find
(i) Total GST collected by the State government at the rate of 28%.
(ii) The amount that the customer paid for the TV.
Solution:
(i) Given, CGST price of the TV for manufacturer = ₹ 15,000
and rate GST = 28%
∴ GST paid by manufacturer = 28% of ₹ 15,000 = \(\frac { 28 }{ 100 }\) x 15000 = ₹ 4200
The manufacturer sells it to a wholesaler at a profit of ₹ 1,000
Given, Sale price of the television for manufacturer = ₹ 15,000 + ₹ 1,000 = ₹ 16,000 and rate of GST = 28 %
∴ Input GST i.e. GST paid by wholesaler = 28 % of ₹ 16,000 = \(\frac { 28 }{ 100 }\) x 15000=₹ 4480
The wholesaler sells it to a trader at a profit of ₹ 1000
Given, Sale price of the television for wholesaler = ₹ 16,000 + ₹ 1000 = ₹ 17000
and rate of GST = 28 %

∴ Input GST i.e. GST paid by trader = 28 % of ₹ 17,000 = \(\frac { 28 }{ 100 }\) x 17000 = ₹ 4760
The trader sells it to a consumer at a profit of ₹ 1500
Given, Sale price of the television for trader = ₹ 17,000 + ₹ 1,500 = ₹ 18500 and rate 0f GST = 28 %

∴ Input GST i.e. GST paid by consumer = 28 % of ₹ 18500 = \(\frac { 28 }{ 100 }\) x 18500 = ₹ 5180
(i) Total GST collected by the State Government = GST paid by consumer = ₹ 5180
(ii) Total amount paid by the consumer = S.P. of the Television + GST = ₹ 18500 + ₹ 5180 = ₹ 23680

Question 17.
Mr T N Naim purchased sportbike for his son at a discount of 15 %, having exshowroom price ₹ 4,00,500. Insurance cover premium of the bike is 8 % of the discounted value. The CGST of accessories is ₹ 20,000. Dealer offered 5 % discount on accessories. Rate of GST is 28 %on the bike and 18 % on the insurance premium and 12 % on the accessories. Find
(i) Total amount of GST to (the nearest rupee)
(ii) Total amount (to nearest rupee) including the insurance premium paid by Mr TN Naim.
Solution:
MP of sponhike ₹ 4,00,500 Discount allowed 15%
SP of sporthike = ₹ 4,00,500 – 15% of ₹ 4,00,500 = ₹ 4,00,500 – ₹ 60.075 = ₹ 3,40,425
Insurance cover premium of the hike is 8% of the discounted value

∴ Amount of insurance cover premium 8% of ₹ 3,40.425 = \(\frac { 8 }{ 100 }\) x 3.40,425 = ₹ 27.234
CGST of accessories ₹ 20,000
Discount allowed 5%
SP of accessories = ₹ 20,000 – 5% 0f ₹ 20,000 = ₹ 20,000 – 1,000 = ₹ 19,000
Rate of GS’I’ on hike 28%

∴ Amount of GST on bike and accessories = 28% of ₹ 3.40,425 = \(\frac { 8 }{ 100 }\) x ₹ 3.40.425 = ₹ 95.3 19
Rate of GST on insurance premium 18%
∴ Amvunt of GST on insurance premium 18% of ₹ 27,234 = ₹ 4,902.12
Rate of GST on accessories = 12%

∴ Amount of GST on accessories = 12% 0f ₹ 19,000 = ₹ 2,280
(i) Total amount of GST = ₹ 95,319 + ₹ 4,902 + ₹ 2,280 = ₹ 1,02,501
(ii) Total amount paid by Mr TN Naim = ₹ 3,40,425 + ₹ 19,000 + ₹ 27,234 + ₹ 1,02,501 = ₹ 4.89,160

Question 18.
A person bought following stationery items :

Item	Quantity	Rate
Chelpark ink	10 bottles	₹ 85 each
Pens	6 dozens	₹ 200 per dozen
Erasers	8 dozens	₹ 50 per dozen
Sharpeners	10 dozens	₹ 40 per dozen
Pencils	12 dozens	₹120 per dozen

If the SGST is @ 2.5%, find the amount paid by him for his purchases. Also find the amount of GST paid by him.
Solution:
Given SGST = 2.5 %
We know that, SGST = CGST = \(\frac { 1 }{ 2 }\) GST
∴ GST = 2 x SGST = 2 x 2.5 = 5 %

Question 19.
Mrs. Gupta bought the following articles from a department store:

Item	Quantity	Rate (in ₹)	Discount	GST rate
Cosmetics	2	₹ 690	5%	18%
Tea Set	1	₹ 1500	10%	12%
Shirts	4	₹ 1500	10%	12%
Packed Dry fruits	4	₹ 800	25%	18%
Unpacked food grains	1	₹ 560	10%	NIL

Find
(i) Total amount of GST
(ii) Total bill amount
Solution:

(i) Total bill amount = ₹ 1546.98 + ₹ 1512 + ₹ 4838.40 + ₹ 2832 + ₹ 504 = ₹ 11233.38
Total Taxable amount = ₹ 1311 + ₹ 1350 + ₹ 4320 + ₹ 2400 + ₹ 504 = ₹ 9885
∴ Total amount of GST = Total bill amount – total taxable amount = ₹ 11233.38 – ₹ 9885 = ₹ 1348.38

(ii) Total bill amount = ₹ 1546.98 + ₹ 1512 + ₹ 4838.40 + ₹ 2832 + ₹ 504 = ₹ 11233.38

Question 20.
Mr. Madhukar buys following goods from Big Bazaar

Item	Cost	Rate of GST	Discount
Cosmetics	₹ 4,500	12%	10%
Tea Set	₹ 25,600	28%	25%
Shirts	₹ 3,400	5%	NIL
Packed Dry fruits	₹ 2,400	5%	NIL
Unpacked food grains	₹ 1,200	NIL	NIL

Find
(i) Total taxable amount
(ii) Total discount amount
(iii) Total GST amount
(iv) Total Bill amount
Solution:

(i) Total taxable amount = ₹ 4050 + ₹ 19200 + ₹ 3400 + ₹ 2400 = ₹ 29,050
(ii) Total Discount amount = ₹ 450 +₹ 6400 = ₹ 6850
(iii) Total bill amount of GST items = ₹ 4536 + ₹ 24576 + ₹ 3570 + ₹ 2520 = ₹ 35,202
Total amount of GST = Total bill amount – total taxable amount = ₹ 36,402 – ₹ 29,050 = ₹ 6152
(iv) Total bill amount = ₹ 4536 + ₹ 24576 + ₹ 3570 + ₹ 2520 + ₹ 1200 = ₹ 36,402

Additional Questions

Question 1.
A dealer X in Meerut (UP) sold a table for f12000 to a consumer in Agra (UP). If the GST rate is 18%, calculate
(i) the IGST
(ii) the CGST
(iii) the SGST
Solution:
Since, It is a case of Intra-slate transaction of goods and services
∴ CGST = SGST = \(\frac { 1 }{ 2 }\) x GST
Given S.P. of a table = ₹ 12,000
GST. rate = 18 %
∴ CGST = SGS.T. = \(\frac { 1 }{ 2 }\) x 18% =9%
(i) IGST = Nil
(ii) C.G.S.T. = 9% of S.I of a table = 9% of ₹ 12.000 = ₹ \(\left(\frac{9}{100} \times 12,000\right)\) = ₹ 1080
(iii) S.G.S.T = 9% of S.P. of a table = 9% of 12,000 = ₹ \(\left(\frac{9}{100} \times 12,000\right)\) = ₹ 1080

Question 2.
A wholesaler A sells a machine to a retailer B for ₹ 5000 and the retailer B sells it to a consumer at a profit of 1000. If the GST rate is 12%, calculate the tax liability of the retailer B.
Solution:
Let us assume, it is a case of intra-state transaction of goods and services
∴ CGST = SGST = \(\frac { 1 }{ 2 }\) x GST
Given, S.P. of washing machine for wholesaler = ₹ 5,000
Rate of G. S.T. = 12%
∴ C.G.S.T. = S.G.S.T. = \(\frac { 1 }{ 2 }\) x G.S.T. = \(\frac { 1 }{ 2 }\) x 12% = 6%
∴ C.G.S.T. = 6% of ₹ 5000 = \(\frac { 6 }{ 100 }\) x ₹ 5000 = ₹ 300
and S.GS.T. = 6% of ₹ 5000 = \(\frac { 6 }{ 100 }\) x ₹ 5000 = ₹ 300
Hence, input GST, for Retailer are C.GS.T. ₹ 300 and S.G.S.T = ₹ 300
The retailer B sells itto a consumer at a profit of ₹ 1000
Now, S.P. of washing machine for consumer ₹ 5000 + 1000 = ₹ 6000
∴ Output C.G.S.T. 6% of ₹ 6,000 = \(\frac { 6 }{ 100 }\) x 6,000 = ₹ 360
and Output C.GS.T. = 6% of ₹ 6,000 = \(\frac { 6 }{ 100 }\) x 6,000 = ₹ 360
Now, tax liability of retailer = (Output S.GS.T. + Output C.GS.T.) (Input S.GS.T. + Input C.G.S.T.)
= (₹ 360 + ₹ 360) – (₹ 300 + ₹ 300) = ₹ 720 – ₹ 600 = ₹ 120

Question 3.
A microwave oven having a marked price of 22000 was sold by a dealer in Patna (Bihar) toa consumer in Gaya (Bihar) at a discount of 25%. If the rate of GST is 18%, calculate the IGST, CGST and SGST charged from the consumer. Also, determine the total amount of bill.
Solution:
Since, it is a case of intra-state transaction of goods and services,
∴ CGST = SGST = \(\frac { 1 }{ 2 }\) x GST and IGST = Nil
Given, Marked price of microwave = ₹ 22,000
Discount % = 25%
Discount amount = 25% of ₹ 22000 = \(\frac { 25 }{ 100 }\) × 22000 = ₹ 5500
∴ S.P. of microwave for wholesaler = M.P. – Discount = ₹ 16500
Also, given rate of GS.T. = 18%
∴ CGST = SGST = \(\frac { 1 }{ 2 }\) x 18% = 9%
Hence, CGST = SGST 9% of ₹ 16500 = \(\frac { 9 }{ 100 }\) x 16500 = ₹ 1485
Now total bill amount S.P. of Microwave + CGST + SGST
= ₹ 16500 + ₹ 1485 + ₹ 1485 = ₹ 19470

Question 4.
A dealer in Mumbai sold a telescope to an end-user in Bangalore. The marked price of the telescope was ₹ 25000 and the dealer of tred a discount of 20%. If the rate of GST is 28%, calculate the IGST, CGST and SGST charged from the end-user. Also determine the total amount of bill,
Solution:
Since, it is a case of inter-state transaction of goods and services
∴ IGST = GST
and CGST = SGST = Nil
Given, Marked price of telescope = ₹ 25000
Discount % = 20%
Discount amount = 20% of ₹ 25000 = \(\frac { 20 }{ 100 }\) x 25000 = ₹ 5000
∴ S.P. of telescope = MP. – discount amount = ₹ (25000 – 5000) = ₹ 20000
Also, given rate of GST = 28%
∴ IGST = GST = 28%
∴ IGST = 28% of ₹ 20000 = \(\frac { 28 }{ 100 }\) x 20000 = ₹ 600
Now, total bill amount = S.P. telescope + IGST = ₹ 20000 + ₹ 5600 = ₹ 25600

Question 5.
A dealer purchased a music system from the manufacturing company for ₹ 25000 and sold it to a consumer at a profit of 20%. If the rate of GST is 18%, calculate
(i) the amounts of Input CGST and Input SGST for the dealer
(ii) the amount of GST payable by dealer to the Governmcnt
(iii) the amount that the consumer has to pay for the music system
(Assume that all transactions take place in the same state)
Solution:
Since, it is a case of intra state transaction of goods and services
∴ CGST = SGST = \(\frac { 1 }{ 2 }\) x GST
(i) Given, S.P. ola music system for manufacturer = ₹ 25,000
Rate of GST = 18%
∴ CGST = 9% of ₹ 25000 = \(\frac { 9 }{ 100 }\) x 25000 = ₹ 2250
and SGST = 9% of ₹ 25000 = \(\frac { 9 }{ 100 }\) x 25000 = ₹ 2250
Hence, Input CGST = Input SGST for dealer = ₹ 2250

(ii) Also, dealer sold music system to consumer at a profit of 20%
∴ SP of u music system t1r dealer = ₹ 25000 + 20% of ₹ 25000 = ₹ 25000 + ₹ 5000 = ₹ 30000 and Rate of GST = 18%
∴ CGST = 9% of ₹ 30000 = \(\frac { 9 }{ 100 }\) x 30000 = ₹ 2700
and SGSI = 9% of ₹ 30000 = \(\frac { 9 }{ 100 }\) x 30000 = ₹ 2700
Hence. Output CGST and Output SGST for dealer = 2700
∴ The amount of GST payable by dealer to the government
= [Output CGST + Output SGST] – [Input CGST + Input SGST]
= [ ₹ 2700 + ₹ 2700] – [₹ 2250 + ₹ 2250] = ₹ 5400 – ₹ 4500 = ₹ 900

(iii) CR of music system for consumer = S.P. of music system + output CGST + output SGST
= ₹ 30000 + ₹ 2700 + 2700 = ₹ 35400

Question 6.
A registered computer engineer provides computer maintenance services to five different companies.
He offers different discounts to different companies depending upon their payment terms.

Company	C1	C2	C3	C4	C5
Service costs (in ₹)	8200	12100	13600	8000	12500
Discount	30%	25%	20%	15%	10%

If the rate of GST is 18 %, calculate the output GST for the computer engineer.
Solution:

∴ The output GST for the computer engineer = ₹ 1033.20 + ₹ 1633.50 + ₹ 1958.40 + ₹ 1224 + ₹ 2025 = 7874.10

Question 7.
Mr. Kumar a registered dealer purchased goods worth ₹ 40000 from a dealer (within the same state). If the rate of GST is 18%.
(i) Calculate the Input COSI and the Input SUSE
(ii) If the sold these goods to Mr. Dcv (within the state) for ₹ 50000, calculate Mr Kumar’s output CGST and output SGST.
(iii) Calculate the CGS I’ and SOST payable by Mr. Kurnar.
Solution:
Since, it is a case of intra state transaction of goods/services.
∴ CGST = SGST = \(\frac { 1 }{ 2 }\) x GST
(i) Given, rate of GST 18%
Hence, CGST = SGST = 9%
SP of goods for a dealer = ₹ 40,000
∴ CGST = 9% of ₹ 40000 = \(\frac { 9 }{ 100 }\) x ₹ 40000 = ₹ 3600
and SOST = 9% of ₹ 40000 = \(\frac { 9 }{ 100 }\) x ₹ 40000 = ₹ 3600
hence. Input CGST = Input SOST for registered dealer. Mr. Kumar ₹ 3600

(ii) SP of goods for Mr. Kurnar = ₹ 50.000
∴ CGST = 9% of ₹ 5000 = \(\frac { 9 }{ 100 }\) x 5000 = ₹ 4500
and SGST = 9% of ₹ 50000 = \(\frac { 9 }{ 100 }\) x 50000 = ₹ 4500
Hence, output CGSI = Output SGST for Mr. Kumar = ₹ 4500

(iii) CGST payable by Mr. Kumar to the government = Output COST – Input COST
= ₹ 4500 – ₹ 3600 = ₹ 900
Also SGST payable by Mr. Kumar to the government = Output SGST – Input SGST
= ₹ 4500 – 3600 = ₹ 900

Question 8.
A shopkeeper buys a machine at a discount of 20% from the wholesaler. The printed price of the machine is ₹ 16000 and the raie of GST is 8%. The shopkeeper sells it to the consumer at the printed price. Calculate.
(i) the price at which the machrnc is bought by the consumer
(ii) the COST and the SGST payable by the shopkeeper to the Government assuming that all the transactions were intrastate.
Solution:
Since, it is a case of intra-state transaction of goods and services.
∴ CGST = SGST x \(\frac { 1 }{ 2 }\) x GST
Given, Printed price of the machine = ₹ 16.000
and kate of GST = 8%
and discount rate = 20%
∴ SP. of machine for wholesaler = ₹ 16000 – 20% of ₹ 16000 = ₹ 12,800
Now, CGST = 4% of ₹ 12,800 = \(\frac { 4 }{ 100 }\) x 12800 = ₹ 512
and SGST = 4 % of ₹ 12800 = \(\frac { 4 }{ 100 }\) x 12800 = ₹ 512
Hence, Input CGST = Input SGST for shopkeeper = ₹ 512

(i) SP. of the machine for shopkeeper = ₹ 16000
∴ CGST = 4% of ₹ 16000 = \(\frac { 4 }{ 100 }\) x 16000 = ₹ 640
and SGST = 4% of ₹ 16000 = \(\frac { 4 }{ 100 }\) x 16000 = ₹ 640
Hence, Output COST = Output 5GST for the shopkeeper = ₹ 640
Total amount paid by the consumer = S.P. of the machine + CGSI + SGST = ₹ 16000 + ₹ 640 + ₹ 640 = ₹ 17280

(ii) CGST payable by the shopkeeper to the government Output CGST – Input SGST
= ₹ 640 – ₹ 512 = ₹ 128
and SGST payable by the shopkeeper to the government = Output SGS – Input SGST
= ₹ 640 – ₹ 512 = ₹ 128

Question 9.
A wholesaler buys a machine from the manufacturer 11w 25000. He marks the price of the machine 20% above his cost price and sells it to a retailer at 10% discount on the marked price If the rate of GST is 18% and assuming that all transactions occur within the same state, calculate
(i) the marked price of the machine
(ii) retailers cost price inclusive of GST
(iii) the CGST and SGST payable by the wholesaler to the government
Solution:
Since, it is a case of intra-state transaction of goods and services.
∴ CGST = SGST = \(\frac { 1 }{ 2 }\) x GST
Given, Sale price of the machine for rnanutcrurcr = ₹ 25,000
and rate of GST = 18%
∴ CGST = 9% of ₹ 25,000 = \(\frac { 9 }{ 100 }\) x 25000= ₹ 2250
and SGST = 9% of ₹ 25,000 = \(\frac { 9 }{ 100 }\) x 25000 = ₹ 2250
Hence, Input CGST = Input SGST for wholesaler ₹ 2250

(i) Thc wholesaler marks the price of machine at 20 above the cost price.
∴ Marked price of the machine = ₹ 25000 + 20% of ₹ 25000
= ₹ 25000 + \(\left(\frac{20}{100} \times 25000\right)\) = ₹ 25000 + 5000 = ₹ 30000

(ii) The wholesaler further sells the machine at a discount of 10% to the retailer.
∴ SP. of the machine for wholesaler = ₹ 30000 – 10% of ₹ 3000
= ₹ 30000 – \(\left(\frac{10}{100} \times 30000\right)\) = ₹ 30000 – 3000 = ₹ 27000
∴ CGST = 9% of ₹ 2700 = \(\frac { 9 }{ 100 }\) x 27000 = ₹ 2430
and SGST = 9% of ₹ 27000 = \(\frac { 9 }{ 100 }\) x 2700 = ₹ 2430
Hence, Output CGST = Output SGST for the wholesaler = ₹ 2430
Total amount paid by the retailer = S.P. of the machine + CGST + SGST
= ₹ 27000 + ₹ 2430 + ₹ 2430 = ₹ 31860

(ii) CGST payable by the wholesaler to the government = Output CGST – Input SGST
= ₹ 2430 – ₹ 2250 = ₹ 180
and SGST payable by the shopkeeper to the government Output SGST Input SGST
= ₹ 2430 – ₹ 2250 = ₹ 180

Question 10.
A manufacturer sells a dish washer to a wholesaler for ₹ 18000. The wholesaler sells it to a dealer at a profit of ₹ 1500 and the dealer sells it to a consumer at a profit of ₹ 2500. If the rate of GST is 12% and assuming that all transactions occur within the same slate, calculate
(i) the total amount of GST received by the central and the state governments on the sale of this dish washer from the manufacturer to the consumer.
(ii) the amount paid by the consumer for the dish washer.
Solution:
(i) Since, it is a case of intra-state transaction of goods and serices
∴ CGST = SGST= \(\frac { 1 }{ 2 }\) x GST
Given, Sale prive ol’ the dish washer fur manutäcturer = ₹ 18,000
and rate of GST = 12%
∴ CGST paid by wholesaler 6% of ₹ 18,000 = \(\frac { 6 }{ 100 }\) x 18000 = ₹ 1080
and SGST paid by wholesaler = 6% of ₹ 18,000 = \(\frac { 6 }{ 100 }\) x 18000 = ₹ 1080
Hence, Input CGST = Input SGST for wholesaler = ₹ 1080
The wholesaler sells it to a dealer at a profit of ₹ 1500
∴ S.P. of the machine for wholesaler = ₹ 18000 + ₹ 1500 = ₹ 19500
∴ CGST paid by rerailer = 6% of ₹ 19500 = \(\frac { 6 }{ 100 }\) x 19500 = ₹ 1170
and SGST paid by retailer = 6% of ₹ 19500 = \(\frac { 6 }{ 100 }\) x 19500 = ₹ 1170
Hence, output CGST = output SGST for the wholesaler = ₹ 1170
and Input CGST and Input CGST for the retailer = ₹ 1170
Total amount paid by die retailer = SP. of the machine + CGST + SGST
= ₹ 19500 + ₹ 1170 + ₹ 1170 = ₹ 21840
The retailer sells it to a consumer at a profit of ₹ 2500
∴ SP of the machine for dealer = ₹ 19500 + ₹ 2500 = ₹ 22000
∴ CGST paid by consumer = 6% of ₹ 22000 = \(\frac { 6 }{ 100 }\) x 22000 = ₹ 1320
and SGST paid by consumer = 6% of ₹ 22000 = \(\frac { 6 }{ 100 }\) x 22000 = ₹ 1320
Hence. Output COST = Output SOST for the retailer ₹ 1320
The total amount received by the GST received by the Central Government on the sale of this dish
washer from the manufacturer to the consumer is ₹ 1320 and by the State Government is ₹ 1320.

(ii) Total amount paid by the consumer = S.P. of the machine – CGST + SGST
= ₹ 22000 + ₹ 1320 + ₹ 1320 = ₹ 24640

Question 11.
A shopkeeper bought an air conditioner at a discount of 20% from a wholesaler, the printed price of the air conditioner being ₹ 28000. The shopkeeper sells it to a consumer at a discount of 10% on the printed price. If the GST rate is 18%, find
(i) the CGST and SGST payable by the shopkeeper to the Government.
(ii) the total amount paid by the consumer for the air conditioner
Solution:
Since, it is a case of intra-state transaction of goods and services.
∴ CGST = SGST = \(\frac { 1 }{ 2 }\) x GST
Given, Marked price of the air conditioner for wholesaler = ₹ 28,000
Discount price = 20%
S.P. of the air conditioner for wholesaler ₹ 28,000 – 20% of ₹ 28,000 = ₹ 28,000 – ₹ 5,600 = ₹ 22,400
and Rate of GST = 18%
∴ CGST = 9% of ₹ 22.400 = \(\frac { 9 }{ 100 }\) x 22400 = ₹ 2016
and SGST = 9% of ₹ 22,400 = \(\frac { 9 }{ 100 }\) x 22400 = ₹ 2016
Hence, Input CGST = Input SGST for shopkeeper = ₹ 2016
The shopkeeper further sells the air conditioner at a discount of 10% to the consumer on the printed price.
∴ S.P. of the machine for shopkeeper = ₹ 28000 – 10 % of ₹ 28000
= ₹ 28000 – (\(\frac { 10 }{ 100 }\) x 28000) = ₹ 28ooo – 28o0 = ₹ 25200
∴ CGST = 9% of ₹ 25200 = \(\frac { 9 }{ 100 }\) x 25200 = ₹ 2268
and SGST 9% of 25200 x 25200 2268
Hence, output CGST = output SGST for the shopkeeper = ₹ 2268
(i) CGST payable by the shopkeeper to the government = Output GST – Input SGST
= ₹ 2268 – ₹ 2016 = ₹ 252
and SGST payable by the shopkeeper to the government = Output SGST – Input SGST
= ₹ 2268 – ₹ 2016 = ₹ 252

(ii) ThtaI amount paid by the retailer = S.P. of the machine + CGST + SGST
= ₹ 25200 + ₹ 2268 – ₹ 2268 = ₹ 29736

Accessing OP Malhotra Class 10 Solutions Chapter 3 Shares and Dividends Ex 3(a) can be a valuable tool for students seeking extra practice.

S Chand Class 10 ICSE Maths Solutions Chapter 3 Shares and Dividends Ex 3(a)

Question 1.
Find the cost of
(i) 200 shares of Rs. 10 each available at Rs. 2.50 premium.
(ii) 220 shares of Rs. 15 each available at Rs. 1.50 discount.
(iii) Purchasing Rs. 200 stock yielding 9% at 105.
(iv) 1000 shares of Rs. 25 each at a premium of Rs. 60.
(v) 450 shares of Rs. 100 each at the market price of Rs. 30.80.
Solution:
(i) Face value of each share = Rs. 10
∴ Market value of each share = Rs. 10 + RS. 2.50 = Rs. 12.50
∴ Market value of 200 shares = Rs. 12.50 x 200 = Rs. 2500

(ii) Face value of each share = Rs. 15
Market value = Rs. 15 – Rs. 1.50 = Rs. 13.50
Market value of 220 shares
= Rs. 13.50 x 220 = Rs. 2970

(iii) Market value of each share = Rs. 10
∴ Market value of Rs. 200 shares
= Rs. 105 x 200 = Rs. 21000

(iv) Face value of each share = Rs. 25
Market value of each share
= Rs. 60 + 25 = Rs. 85
∴ Cost of 1000 shares = Rs. 85 x 1000
= Rs. 85000

(v) Face value of each share = Rs. 100
Market value of each share = Rs. 30.80
∴ Value of 450 shares = Rs. 30.80 x 450
= Rs. 13860

Question 2.
Find the annual dividend on
(i) 450 shares with par value of Rs. 10 if the annual dividend is 7.5%.
(ii) 50 Tata Mills 4% (Rs. 6) shares at Rs. 6.25.
(iii) 180 Shri Ram Fibres 7\(\frac { 1 }{ 2 }\)% (Rs. 10) shares at 14.50.
(iv) 100 Hindustan Motors (Rs. 10) shares at Rs. 8, dividend 4% p.a.
Solution:
(i) Face value of each share = Rs. 10
∴ Value of 450 shares = Rs. 10 x 450
= Rs. 4500
Rate of dividend = 7.5%
∴ Total dividend = 7.5% of Rs. 4500
= Rs. \(\frac{4500 \times 7.5}{100}\) = Rs. 337.50

(ii) Face value of each share = Rs. 6
∴ Value of 50 shares = Rs. 6 x 50 = Rs. 300
Rate of dividend = 4%
∴ Total dividend = 4% of Rs. 300
= Rs. 300 x \(\frac { 4 }{ 100 }\) = Rs. 12

(iii) Face value of each share = Rs. 10
∴ Value of 180 shares = Rs. 10 x 180
= Rs. 1800
Rate of dividend = 7\(\frac { 1 }{ 2 }\) % = \(\frac { 15 }{ 2 }\)%
∴ Total dividend = \(\frac { 15 }{ 2 }\) % of Rs. 1800 1800×15
= Rs. \(\frac { 1800 × 15 }{ 100 × 2 }\) = Rs. 135

(iv) Face value of each share = Rs. 10
∴ Value of 100 shares = Rs. 10 x 100
= Rs. 1000
Rate of dividend = 4%
∴ Total dividend = 4% of Rs. 1000
= Rs. 1000 x \(\frac { 4 }{ 100 }\) = Rs. 40

Question 3.
Write down the sum of money obtained by selling 60 (Rs. 100) shares at 112.
Solution:
By selling one share, money is obtained = Rs. 112
∴ By selling 60 shares, money will be obtained = Rs. 112 x 60 = Rs. 6720

Question 4.
Ravi invested Rs. 6250 in shares of a company paying 6% per annum. If he bought Rs. 25 shares for Rs. 31.25 each, find his annual income from this investment.
Solution:
By investing Rs. 31.25, shares can be purchased = 1
and by investing Rs. 6250, number of shares can be purchased = \(\frac{1 \times 6250}{31.25}\) = 200
Nominal value of 200 shares at the rate of Rs. 25 per share = Rs. 25 x 200 = Rs. 5000
Rate of dividend = 6%
Total dividend = Rs. 5000 x \(\frac { 6 }{ 100 }\) = Rs. 300

Question 5.
Write down the price of 100 (Rs. 10) shares costing Rs. 2000.
Solution:
Price of 100 shares = Rs. 2000
∴ Face value of each share
= 2000 x \(\frac { 1 }{ 100 }\) = Rs. 20

Question 6.
A company declares semi-annual dividend of 5%. Krishan Lai owns 25 shares (of par value Rs. 12.50 each). How much annual dividend must he receive.
Solution:
Face value of each share = Rs. 12.50
Value of 25 shares = Rs. 12.50 x 25 = Rs. 312.50
Rate of dividend = 5% semi annually
∴ Annually dividend = 5 x 2 = 10%
∴ Total dividend = Rs. 312.50 x \(\frac { 10 }{ 100 }\)
= Rs. 31.25

Question 7.
A man bought 250 (₹ 1) shares and received from them a dividend of ₹ 20. What is the rate % of the dividend?
Solution:
Number of shares = 250
Face value of each share = Re. 1
Total dividend = Rs. 20
Face value of 250 shares = 250 x 1 = Rs. 250
∴Rate of dividend = \(\frac{\text { Total dividend } \times 100}{\text { Total face value }}\)
= \(\frac{20 \times 100}{250}\)

Question 8.
A man invests Rs. 4800 in shares of a company which was paying 8% dividend at the time when a Rs. 100 share cost Rs. 160. Find (i) his annual income from the shares, (ii) the rate of interest he gets on his investment.
Solution:
Investment = Rs. 4800
Rate of dividend = 8%
Face value of each share = Rs. 100
Market value of each share = Rs. 160
∴Number of shares = Rs. \(\frac { 4800 }{ 160 }\) = 30
and total value of 30 shares = Rs. 100 x 30 = Rs. 3000
(i) ∴ Total dividend = Rs. \(\frac { 3000×8 }{ 100 }\) = Rs. 240
(ii) Rate of interest on his investment = \(\frac{240 \times 100}{4800}\) = 5%

Question 9.
A company pays a dividend of 15% on its ten-rupee shares from which is deducted income tax at the rate of 22%. Find the annual income of a man who owns 100 shares of this company.
Solution:
Rate of dividend = Rs. 15%
Face value of each share = Rs. 10
Rate of Income Tax = 22%
Number of shares = 100
Face value of 100 shares = Rs. 10 x 100
= Rs. 1000
∴ Total dividend = Rs. 1000 x \(\frac { 15 }{ 100 }\) = Rs. 150
Rate of Income Tax = 22%
∴ Total tax deducted = Rs. 150 x \(\frac { 22 }{ 100 }\) = Rs. 33
∴ Net annual income = Rs. 150 – Rs. 33 = Rs. 117

Question 10.
A man bought 1000 shares each of face value Rs. 5 at Rs. 7 per share. At the end of the year, the company from which he
bought the shares declared a dividend of 8%. Calculate
(i) the amount of money invested by the man;
(ii) the percentage return on his outlay correct to one decimal place.
Solution:
Number of shares = 1000
Face value of each share = Rs. 5
Market value of each share = Rs. 7
Rate of dividend = 8%
(i) Total face value of 1000 shares = Rs. 5 x 1000 = Rs. 5000
∴ Amount of investment = 1000 x Rs. 7 = Rs. 7000

(ii) Total dividend = Rs. 5000 x \(\frac { 8 }{ 100 }\) = Rs. 400
∴ Rate of interest = Rs. \(\frac{400 \times 100}{7000}\)
= \(\frac { 40 }{ 7 }\) = 5.7 %
Percentage of return = 5 \(\frac { 1 }{ 3 }\) %

Question 11.
A company with 10000 shares of nominal value Rs. 100 declares an annual dividend of 8% to the share-holders.
(i) Calculate the total amount of dividend paid by the company.
(ii) Ramesh had bought 90 shares of the company at Rs. 150 per share. Calculate the dividend he receives and the percentage return on his investment.
Solution:
Number of shares = 10000
Nominal value of each share = Rs. 100
Rate of dividend = 8%
∴ Total value of 10000 shares = Rs. 100 x 10000 = Rs. 1000000
(i) Total dividend paid by the company
= Rs. 1000000 x \(\frac { 8 }{ 100 }\) = Rs. 80000

(ii) Number of shares, Ramesh has = 90
Value of 90 shares = Rs. 100 x 90 = Rs. 9000
Dividend = Rs \(\frac{9000 \times 8}{100}\) = Rs. 720 100
Market value be each share = Rs. 150
Investment on purchasing 90 shares = Rs. 150 x 90 = Rs. 13500
Percent of interest on his investment
= Rs. \(\frac{720 \times 100}{13500}\)
= \(\frac { 16 }{ 3 }\) % = 5 \(\frac { 1 }{ 3 }\) %
Percentage of return = 5 \(\frac { 1 }{ 3 }\) %

Question 12.
Aditi has 40 shares of nominal value Rs. 100 and she decides to sell them when they are at a premium of 50%. She invests the proceeds in shares of nominal value of Rs. 75 quoted at a 20% discounts paying 25% dividend annually. Calculate:
(i) the sale proceeds
(ii) the number of shares she buys;
(iii) the annual dividend from these shares.
Solution:
Number of shares Aditi has = 40
Nominal value of each share = Rs. 100
∴ Total value of 40 shares = Rs. 100 x 40 = Rs. 4000
Market price of each shares = Rs. 100 + 50 = Rs. 150
(i) Amount received on selling them = Rs. 150 x 40 = Rs. 6000
(ii) In second case, nominal value of each share = Rs. 75
Market value of each share 20% discount
= Rs. \(\frac{75 \times(100-20)}{100}\) = Rs. \(\frac{75 \times 80}{100}\)
= Rs. 60
Rate of dividend = 25% annually
(i) Number of shares be bought
= Rs. \(\frac { 6000 }{ 60 }\) = 100

(ii) Face value of 100 shares at the rate of Rs. 75 = 75 x 100

(iii) Annual dividend = Rs. \(\frac{7500 \times 25}{100}\) = Rs. 1875

Question 13.
Mr. Khanna holds 1600, Rs. 100 shares of a company that pays 12% dividend annually. Calculate his annual dividend. If he had bought these shares at 50% premium, what percentage return does he get on his investment.
Solution:
Number of shares Mr. Khanna holds = 1600
Face value of each share = Rs. 100
Annually dividend = 12%
(i) Total value of 1600 shares = Rs. 100 x 1600 = Rs. 160000
and total dividend = Rs. \(\frac{160000 \times 12}{100}\)
= Rs. 19200
Cost price of each share = Rs. 100 + 50 = Rs. 150
Total investment = 1600 x 150 = Rs. 240000
∴ Percentage of return will be
= \(\frac{19200 \times 100}{240000}\)

Question 14.
A man invests Rs. 11200 in a company paying 6% dividend per annum when its Rs. 100 shares can be bought for Rs. 140. Find (i) his annual dividend, (ii) his percentage income on this investment.
Solution:
Total investment = Rs. 11200
Rate of dividend = 6%
Face value of each share = Rs. 100
and market value = Rs. 140
∴ Number of shares purchased
= RS. \(\frac { 11200 }{ 140 }\)
Face value of 80 shares = Rs. 100 x 80 = Rs. 8000
(i) Annual dividend = Rs. 8000 x \(\frac { 6 }{ 100 }\) = Rs. 480
(ii) Percentage income on his investment
= Rs. \(\frac{480 \times 100}{11200}\) = \(\frac { 30 }{ 7 }\) = 4\(\frac { 2 }{ 7 }\) % p.a.

Question 15.
A company having a capital stock of Rs. 450000 declares a dividend of 4% semi-annually.
(a) What is the annual income of a stockholder owning 135 shares at par- value of Rs. 10?
(b) What is the total amount of dividend paid annually by the company?
Solution:
Total capital stock = Rs. 450000
Rate of dividend = 4% semi annually or 4 x 2 = 8% annually
(a) Number of shares = 135
Nominal value of each share = Rs. 10
∴ Total value of 135 shares = Rs. 10 x 135 = Rs. 1350
Dividend of Rs. 1350 = Rs. 1350 x \(\frac { 8 }{ 100 }\) = Rs. 108

(b) Total amount of dividend annually 8
= Rs. 450000 x \(\frac { 8 }{ 100 }\)
= Rs. 36000

Peer review of ICSE Class 10 Maths Solutions S Chand Chapter 19 Histogram and Ogive Ex 19(b) can encourage collaborative learning.

S Chand Class 10 ICSE Maths Solutions Chapter 19 Histogram and Ogive Ex 19(b)

Question 1.

Class-interval	0-10	10-20	20-30	30-40
Frequency	10	45	12	3

Solution:

Class-interval	Frequency
0-10	10
10-20	45
20-30	12
30-40	3

Represent class-intervals along x-axis and frequency along y-axis and draw the histogram as shown here Now inside the highest rectangle we join A to C and B to D intersecting each other at P.
From P, draw PQ perpendicular on x-axis meeting it at Q, then Q is the mode which is 15 (approx)

Question 2.

Marks	1-5	6-10	11-15	16-20	21-25
f	7	10	16	32	24

Solution:
Writing the classes in exclusive form,

Marks	f
0.5-5.5	7
5.5-10.5	10
10.5-15.5	16
15.5-20.5	32
20.5-25.5	24

Represent the marks (class interval) along x-axis and f along y-axis and draw the histogram as shown. Now inside the highest rectangle. Join A to C and B to D intersecting each other at P Through P, draw PQ perpendicular on x-axis meeting it at Q, Then Q is the mode which is 18.8 = 19 (approx)
Hence mode = 19 (approx)

Question 3.

Marks	26-30	31-35	36-40	41-45
f	18	10	5	1

Solution:
Writing the classes in exclusive form

Marks	f
25.5-30.5	18
30.5-35.5	10
35.5-40.5	5
40.5-45.5	1

Represent marks along x-axis and f along y-axis and draw the histogram as shown. Now inside the highest rectangle, join A to C and B to D which intersect each other at P. Through P, draw PQ perpendicular to x-axis meeting x-axis at Q.
∴ Q is the mode which is 29 (approx)
∴ Mode = 29 (approx)

Question 4.

Mid-value	60	64	68	72	76
Frequency	25	8	10	6	4

Solution:
∵ Mid value of each class is given
∴ Forming the classes, with these mid values, we get 58-62, 62-66, 66-70, 70-74 and 74-78

Class	Frequency
58-62	25
62-66	8
66-70	10
70-74	6
74-78	4

Represent classes along x-axis and frequency along y-axis and draw the histogram as shown. Now inside the highest rectangle, join A to C and B to D intersecting each other at P. Through P, draw PQ perpendicular on x-axis.
Q is the mode which is 60.4 = (approx)
∴ Mode = 60 (approx)

Question 5.
Draw a histogram from the following data and measure the modal value :

(a)

Size class	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	5	11	19	21	16	10

(b)

Size class	60-70	70-80	80-90	90-100
Frequency	8	6	3	1

Solution:
(a) Taking size class along x-axis and frequency along y-axis, and draw the histogram as shown
∴ Mode = 33 (approx)

(b) Taking size class along x-axis and frequency along y-axis, and draw the histogram as shown

Now inside the highest rectangle, join A to C and B to D intersecting each other at P. Through P, draw PQ perpendicular to x-axis. Q is mode which is 68 (approx)
∴ Mode = 68 (approx)

Continuous practice using ICSE Class 10 Maths Solutions S Chand Chapter 19 Histogram and Ogive Ex 19(a) can lead to a stronger grasp of mathematical concepts.

S Chand Class 10 ICSE Maths Solutions Chapter 19 Histogram and Ogive Ex 19(a)

Question 1.
Represent the following distribution of ages (in years) of 35 teachers in a school by means of a histogram.

Age (in years)	25-30	30-35	35-40	40-45	45-50
Number of Teachers	12	11	8	1	3

Solution:

Age (in years)	25-30	30-35	35-40	40-45	45-50
Number of Teachers	12	11	8	1	3

Draw the axis and represent the age (in years) along x-axis and number of teachers (frequencies) along y-axis and prepare the histogram as given under.

Question 2.
The weekly observatios on cost of living index in a certain city for a year give the following frequency table :

Cost of living index	140-150	150-160	160-170	170-180	180-190	190-200
Number of weeks	5	10	16	12	7	2

Draw histogram to represent the data.
Solution:
We represent the cost of living index along x-axis and number of weeks (frequencies) along they- axis and prepare the histogram as given below.

Question 3.
Draw a histogram for daily earning of 20 drug stores given in the following data :

Daily earnings (in Rs.)	150-200	200-250	250-300	300-350
Number of stores	14	9	3	4

Solution:
We represent daily earnings in rupees along x-axis and number of stores along y-axis and prepare the histogram as shown in the figure given here.

Draw histograms for the following distributions :

Question 4.
(a)

Marks	0-10	10-20	20-30	30-40	40-50
No. of boys	3	7	5	8	2

(b)

Money earned in Rs.	0-20	20-40	40-60	60-80	80-100	100-120	120-140
No. of students	4	18	22	14	10	8	4

(c)

Class	1-10	11-20	21-30	31-40	41-50	51-60
Frequency	7	3	5	2	6	4

(d)

Class interval	10-20	20-30	30-40	40-50	50-70	70-100
Frequency	6	10	16	10	6	3

Solution:
(a)

Marks	0-10	10-20	20-30	30-40	40-50
No. of boys	3	7	5	8	2

We represent marks on the x-axis and number of boys v-axis and prepare the histogram as given here.

(b)

Money earned (in Rs.)	0-20	20-40	40-60	60-80	80-100	100-120	120-140
No: of students.	4	18	22	14	10	8	4

We represent money earned along x-axis and no. of students on they-axis and prepare the histogram as shown here.

(c)

Class	1-10	11-20	21-30	31-40	41-50	51-60
Frequency	7	3	5	2	6	4

Arranging in exclusive form.

Class	0.5-10.5	10.5-20.5	20.5-30.5	30.5-40.5	40.5-50.5	50.5-60.5
Frequency	7	3	5	2	6	4

We represent class along x-axis and frequencies along y-axis and prepare the histogram as shown here.

(d)

Class interval	10-20	20-30	30-40	40-50	50-70	70-100
Frequency	6	10	16	10	6	3

Here we represent class intervals along x-axis and frequency alongy-axis and prepare the histogram as shown here.

Question 5.
Given below are the marks obtained by 40 students in an examination :

29	45	23	30	40	11	48	01	15	35
40	03	12	48	49	18	30	24	25	29
31	32	25	22	27	41	12	13	02	44
07	43	09	49	19	13	32	39	25	03

Taking class-intervals 1 – 10, 11 – 20, ………, 41 – 50, make a frequency table for the above distribution and draw a histogram to represent it.
Solution:
We prepare frequency distribution table as given below :

We represent class intervals (exclusive) along x-axis and frequency along y-axis and prepare the histogram to represent the above data as given below :

Question 6.
Present in the form of a frequency table the marks obtained by 50 candidates. Take the class-intervals as 11 – 20; 21 – 30… etc.

35	56	25	40	38	48	58	43	30	47
46	45	31	45	56	39	46	47	23	40
48	50	36	56	35	43	59	40	48	35
53	57	33	50	23	46	49	57	35	43
64	40	50	56	36	19	49	52	51	42

Draw a histogram for the above distribution.
Solution:
Lowest data = 19,
Highest data = 64
We prepare frequency distribution table of the given data as given below :

Now we represent exclusive class intervals along x-axis and frequency alongy-axis and prepare a histogram to represent the given data as shown below:

Question 7.
Explain the methods of draw ing histogram and frequency polygon. Following table gives the marks distribution of 160 students in a certain class.

From the above data draw a histogram and frequency polygon.
Solution:

Class intervals	No. of students c.f.	Frequency
5-15	160	8
15-25	152	12
25-35	140	15
35-45	125	20
45-55	105	24
55-65	81	32
65-75	49	26
75-85	23	18
85-95	5	5
95-105	0	0

We present class intervals along x-axis and frequency alongy-axis and prepare the histogram and frequency polygon by joining the mid-points of consecutive class intervals.

Question 8.
The number of match sticks in 40 boxes, on counting was found as given below :

44	41	42	43	47	50	51	49	43	42
40	42	44	45	49	42	46	49	45	49
45	47	48	43	43	44	48	43	46	50
43	52	46	49	52	51	47	43	43	45

Taking classes 40 – 41, 42 – 43, etc., construct the frequency distribution table for the above data. Draw a histogram to represent the above distribution.
Solution:
Highest score = 52, lowest score = 40
Now we prepare the frequency distribution table in exclusive form

Now we represent classes along x-axis and frequencies along v-axis and prepare the histogram representing the given data :