ML Aggarwal Class 7 ICSE Maths Model Question Paper 3

ML Aggarwal Class 7 ICSE Maths Model Question Paper 3

(Based on Chapters 1 to 9)
Time allowed: 2\(\frac { 1 }{ 2 }\) Hours
Maximum Marks: 90

General Instructions

• All questions are compulsory.
• The question paper consists of 29 questions divided into four sections A, B, C and D.
• Section A comprises of 8 questions of 1 mark each.
• Section B comprises of 6 questions of 2 marks each.
• Section C comprises of 10 questions of 4 marks each and
• Section D comprises of 5 questions of 6 marks each.
• Question numbers 1 to 8 in Section A is multiple choice questions where you are to select one correct option out of the given four.

Section – A

Question numbers 1 to 8 are of 1 mark each.
Question 1.
The number of integers between -16 and 5 is
(a) 19
(b) 20
(c) 21
(d) 22
Solution:
Number of integer between -16 and 5 is -15 to 4 = 20 (b)

Question 2.
50 m 5 cm is the same as
(a) 50.5 m
(b) 50.05 m
(c) 50.005 m
(d) 5.05 m
Solution:
50 m 5 cm = 50.05 m (b)

Question 3.

Solution:

Question 4.
The number 5,540,000,000,000 in the scientific notation can be written as:
(a) 554 × 10¹⁰
(b) 55.4 × 10¹¹
(c) 5.54 × 10¹²
(d) 5.54 × 10¹¹
Solution:
5,540,000,000,000 = 5.54 × 10¹² (c)

Question 5.
The number of unlike terms in the expression 5x²y – 2xy² – 2yx² + 3y(xy + y²) + 7 is
(a) 3
(b) 4
(c) 5
(d) 6
Solution:
Number of unlike terms
5x²y – 2xy² – 2yx² + 3y(xy + y²) + 7
= 5x²y – 2xy² – 2yx² + 3xy² + 3y³ + 7
= 3x²y + xy² + 3y³ + 7 = 4 (b)

Question 6.
x = -2 is a solution of the equation
(a) 2x + 5 = 9
(b) 3x – 1 = 5
(c) 4x + 3 = 1
(d) 5x + 12 = 2
Solution:
5x + 12 = 5 × (-2) + 12 = -10 + 12 = 2 (d)

Question 7.
The ratio of the number of girls to the number of boys in a class is 5 : 4. If there are 16 boys in the class, then the number of students in the class is
(a) 20
(b) 32
(c) 36
(d) 45
Solution:
Ratio of girls to boys = 5 : 4
Number of boys = 16
Then number of students = \(\frac { (5+4) }{ 4 }\) × 16
= \(\frac { 9 }{ 4 }\) × 16 = 36 (c)

Question 8.
If 12% of a number is 9, then the number is
(a) 36
(b) 48
(c) 60
(d) 75
Solution:
12% of a number = 9
Number = \(\frac { 9\times 100 }{ 12 }\) = 75 (d)

Section – B

Question numbers 9 to 14 are of 2 marks each.
Question 9.
Using suitable properties, evaluate:
238 × (-44) + (-238) × 56.
Solution:
238 × (-44) + (-238) × 56
= 238 × (-44) – (238) × 56
= 238 (-44 – 56)
= 238 × (-100)
= -23800

Question 10.
State whether each of the following statement is true or false for the sets P and Q where P = {letters of TITLE} and Q = {letters of LITTLE}
(i) P ↔ Q
(ii) P = Q
Solution:
P = {letters of TITLE} = {T, I, L, E}
Q = {letters of LITTLE} = {L, I, T, E}
(i) P ↔ Q – True
(ii) P = Q – True

Question 11.
Evaluate: -3\(\frac { 3 }{ 8 }\) – (-2\(\frac { 1 }{ 6 }\))
Solution:

Question 12.
Simplify and express in the exponential form: (4³ × 3⁶) ÷ (16 × 9²).
Solution:

Question 13.
If I earn ₹ 75000 per month and spend ₹ 40000 per year for helping poor students then find the ratio of the money spent on helping poor students and the annual income.
Solution:
Earnings per month = ₹ 75000
Expenditure = ₹ 40000
Annual income = ₹ 75000 × 12 = ₹ 900000
Ratio in the money spent to the annual income = 40000 : 900000 = 2 : 45

Question 14.
If ₹ 4000 amounts to ₹ 5000 in 2 years, find the rate of simple interest per annum.
Solution:
Principal (P) = ₹ 4000
Amount (A) = ₹ 5000
S.I. = A – P = ₹ 5000 – ₹ 4000 = ₹ 1000
Time (T) = 2 years

Section – C

Question numbers 15 to 24 are of 4 marks each.
Question 15.
Simplify:

Solution:

Question 16.
Vikram’s monthly salary is ₹ 12750. He spends \(\frac { 1 }{ 5 }\) of his salary on food and out of the remaining, he spends \(\frac { 1 }{ 4 }\) on rent and \(\frac { 1 }{ 6 }\) on the education of children. Find
(i) how much he spends on each item?
(ii) how much money is still left with him?
Solution:
Monthly salary of Vikram = ₹ 12750
Spent on food = \(\frac { 1 }{ 5 }\) of 12750 = ₹ 2550
Remaining salary = ₹ 12750 – ₹ 2550 = ₹ 10200
Spent on rent = \(\frac { 1 }{ 4 }\) of ₹ 10200 = ₹ 2550
Spent on Education = \(\frac { 1 }{ 6 }\) × ₹ 10200 = ₹ 1700
Amount still left with him = ₹ 10200 – ₹ (2550 + 1700)
= ₹ 10200 – ₹ 4250
= ₹ 5950

Question 17.
Insert five rational numbers between \(\frac { -2 }{ 5 }\) and \(\frac { -1 }{ 3 }\)
Solution:

Question 18.
Afzal can walks 5\(\frac { 3 }{ 4 }\) km in one hour. How much distance will he cover in 2 hours 40 minutes? What are the health advantages of having a brisk walk?
Solution:
In one hour, Afzal can walk = 5\(\frac { 3 }{ 4 }\) = \(\frac { 23 }{ 4 }\)

Question 19.
If a vehicle covers a distance of 57.72 km in 3.7 litres of petrol. How much distance will it cover in one litre of petrol?
Solution:
In 3.7 litres of petrol a vehicle can travel = 57.72 km
It will cover in 1 litre = \(\frac { 57.72 }{ 3.7 }\)
= 15.6 km

Question 20.
The perimeter of a triangle is 5 – 3x + 7x² and two of its sides are 2x² + 3x – 2 and 3x² – x + 3. Find the third side of the triangle.
Solution:
Perimeter of a triangle = 5 – 3x + 7x²
Two sides are 2x² + 3x – 2 and 3x² – x + 3
Third side = (5 – 3x + 7x²) – (2x² + 3x – 2 + 3x² – x + 3)
= 7x² – 3x + 5 – (5x² + 2x + 1)
= 7x² – 3x + 5 – 5x² – 2x – 1
= 2x² – 5x + 4

Question 21.

Solution:

Question 22.
Solve the equation: 3(2x – 1) – 2(2 – 5x) = 1
Solution:
3(2x – 1) – 2(2 – 5x) = 1
⇒ 6x – 3 – 4 + 10x = 1
⇒ 16x – 7 = 1
⇒ 16x = 1 + 7 = 8
⇒ x = \(\frac { 8 }{ 16 }\) = \(\frac { 1 }{ 2 }\)

Question 23.
If 74% of the population of a village is illiterate and the number of literate people is 2158, then find the population of the village.
Solution:
In a village, population of illiterate person = 74%
Literate person = 100 – 74 = 26%
26% of population = 2158
Total population = \(\frac { 2158\times 100 }{ 26 }\)
= 8300 people

Question 24.
Simplify:

Solution:

Section – D

Question numbers 25 to 29 are of 6 marks each.
Question 25.
If we represent the distance above the ground by a positive rational number and that below the ground by a negative rational number, then answer the following question:
An elevator descends into a mine shaft at the rate of 4\(\frac { 3 }{ 4 }\) metre per minute. If it begins to descend from 7\(\frac { 1 }{ 2 }\) metre above the ground, what will be its position after 18 minutes from the ground?
Solution:
Speed of elevator into the mine = 4\(\frac { 3 }{ 4 }\) = \(\frac { 19 }{ 4 }\) m/min
Time taken = 18 m
Total distance covered = \(\frac { 19 }{ 4 }\) × 18 = \(\frac { 171 }{ 2 }\) m = 85.5 m
Distance above the ground = 7.5 m
Distance below the ground = 85.7 – 7.5 = 78 m
It is below the ground.
It will be = -78 m

Question 26.
In a competition, the question paper consists of 25 questions. 4 marks are awarded for every correct answer, 2 marks are deducted for every incorrect answer and no marks for not attempting a question. If Vaishali scored 58 marks and got 17 correct answers, how many questions she attempted incorrectly? How many questions she did not attempt?
Solution:
Total number of questioins = 25
For correct answer = 4 marks per question
Deduction of 2 marks for incorrect answer
No marks for unattemtped question
Vaishali got 58 marks and got 17 correct questions
Let x be the questions of wrong answer
17 × 4 – 2x + (25 – 17 – x) × 0 = 58
68 – 58 = 2x + 0
2x = 10
x = 5
Number of wrong answer questions = 5
and number of questions not attempted = 25 – 17 – 5 = 3

Question 27.
Divide ₹ 216000 into two parts such that one-fourth of one part is equal to one-fifth of the other part. Find the two parts.
Solution:
Amount = ₹ 216000
Let \(\frac { 1 }{ 4 }\) of one part = \(\frac { 1 }{ 5 }\) of second part = x
First part = 4x
Second part = 5x
4x + 5x = 216000
⇒ 9x = 216000
⇒ x = \(\frac { 216000 }{ 9 }\) = 24000
First part = ₹ 24000 × 4 = ₹ 96000
Second part = ₹ 24000 × 5 = ₹ 120000

Question 28.
If a table is sold for ₹ 437 at a loss of 8%, find its cost price. At what price must it be sold to gain 10%?
Solution:
S.P. of table = ₹ 437
Loss = 8%

Question 29.
Solve the inequality:
3 – 2x ≥ x – 10, x ∈ N.
Also, represent its solution set on the number line.
Solution:
3 – 2x ≥ x – 10, x e N
⇒ -2x – x ≥ -10 – 3
⇒ -3x ≥ -13
⇒ 3x ≤ 13
⇒ x ≤ \(\frac { 13 }{ 3 }\)
⇒ x ≤ 4\(\frac { 1 }{ 3 }\)
x = {4, 3, 2, 1}

ML Aggarwal Class 7 Solutions for ICSE Maths