Category: ICSE Solutions

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 2 Fractions and Decimals Ex 2.5

Question 1.
Write the following numbers in the expanded form:
(i) 20.03
(ii) 200.03
(iii) 2.034
Solution:

Question 2.
Write the place value of digit 2 in the following decimal numbers:
(i) 2.56
(ii) 21.37
(iii) 10.25
(iv) 63.352
Solution:
(i) 2.56, place value of 2 = 2 × 1 = 2
(ii) 21.37, place value of 2 = 2 × 10 = 20
(iii) 10.25, place value of 2 = 2 × \(\frac { 1 }{ 10 }\) = \(\frac { 2 }{ 10 }\)
(iv) 63.352, place value of 2 = 2 × \(\frac { 1 }{ 1000 }\) = \(\frac { 2 }{ 1000 }\)

Question 3.
Convert the following decimal numbers to frac ions (in simplest form):
(i) 0.8
(ii) 0.225
(iii) 0. 0092
(iv) 3.025
Solution:

Question 4.
Convert the following decimals to mixed fractions:
(i) 5.05
(ii) 63.125
(iii) 17.075
(iv) 317.0006
Solution:

Question 5.
Convert the following fractions into decimal numbers:
(i) \(\frac { 3 }{ 5 }\)
(ii) \(\frac { 7 }{ 8 }\)
(iii) 3\(\frac { 5 }{ 16 }\)
(iv) 137\(\frac { 13 }{ 62 }\)
Solution:

Question 6.
Which is greater?
(i) 0.5 or 0.05
(ii) 7 or 0.7
(iii) 2.03 or 2.30
(iv) 0.8 or 0.88
Solution:
Which is greater
(i) 0.5 or 0.05
Converting into like decimals
0.50 and 0.05
50 > 05
0.5 > 0.05
(ii) 7 or 0.7
Comparing integral parts
7 > 0
7 > 0.7
(iii) 2.03 or 2.30
Integral numbers are same
But in decimal part
03 < 30 or 30 > 03
2.30 > 2.03
(iv) 0.8 or 0.88
Converting into like decimal
0.8 = 0.80
88 > 80
0.88 > 0.8

Question 7.
Arrange the following decimal numbers in ascending order:
(i) 38.02, 38.021, 3.802, 83.02, 38.002
(ii) 46.542, 46.452, 46.254, 46.05, 64.542, 46.0542
Solution:
(i) 38.02, 38.021, 3.802, 83.02, 38.002
Converting into like decimal (upto 3 decimals)
38.020, 38.021, 3.802, 83.020, 38.002
Now arranging in ascending order,
3.802, 38.002, 38.020, 38.021, 83.020
(ii) 46.542, 46.452, 46.254, 46.05, 64.542, 46.0542
Converting into like decimals (upto 4 decimals)
46.5420, 46.4520, 46.2540, 46.0500, 64.5420, 46.0542
Now arranging into ascending order,
46.0500, 46.0542, 46.2540, 46.4520, 46.5420, 64.5420
⇒ 46.05, 46.0542, 46.254, 46.452, 46.542, 64.542

Question 8.
Arrange the following decimal numbers in descending order:
(i) 5.6, 0.93, 1.87, 1.9, 1.78, 0.39
(ii) 71.201, 20.1, 2.01, 3.1, 2.14, 0.652
Solution:
(i) 5.6, 0.93, 1.87, 1.9, 1.78, 0.39
Converting into like decimals (up two places)
5.60, 0.93, 1.87, 1.90, 1.78, 0.39
Arranging into descending order,
5.60, 1.90, 1.87, 1.78, 0.93, 0.39
⇒ 5.6, 1.9, 1.87, 1.78, 0.93, 0.39
(ii) 71.201, 20.1, 2.01, 3.1, 2.14, 0.652
Converting into like decimals (upto 3 places)
71.201, 20.100, 2.010, 3.100, 2.140, 0.652
Arranging into descending order,
71.201, 20.010, 3.100, 2.140, 2.010, 0.652
⇒ 71.201, 20.01, 3.1, 2.14, 2.01, 0.652

Question 9.
Express as rupees using decimals:
(i) 7 paise
(ii) 77 rupees 77 paise
(iii) 235 paise
Solution:
(i) 7 paise = ₹ \(\frac { 7 }{ 100 }\) = ₹ 0.07
(ii) 77 rupees 77 paise = ₹ 77\(\frac { 77 }{ 100 }\) = ₹ 77.77
(iii) 235 paise = ₹ \(\frac { 235 }{ 100 }\) = ₹ 2.35

Question 10.
Express 5 cm in metre and kilometre.
Solution:

Question 11.
Express in kg using decimals:
(i) 200 g
(ii) 3470 g
(iii) 4 kg 8 g
Solution:
(i) 200 g = \(\frac { 200 }{ 1000 }\) = 0.2 kg
(ii) 3470 g = \(\frac { 3470 }{ 1000 }\) = 3.470 kg
(iii) 4 kg 8 g = 4\(\frac { 8 }{ 1000 }\) = 4.008 kg

Question 12.
Add:
(i) 5.765, 9.2, 3.08
(ii) 15.49, 8.3572, 0.903, 7.8
Solution:
(i) 5.765, 9.2, 3.08
Converting into like decimal (up to 3 places)
5.765, 9.200, 3.080
= 5.765 + 9.200 + 3.080
= 18.045

(ii) 15.49, 8.3572, 0.903, 7.8
Converting into like decimals (up to 4 places)
= 15.4900 + 8.3572 + 0.9030, 7.8000
= 32.5502

Question 13.
Workout the following:
(i) 72.53 – 46.782
(ii) 18.376 – 5.43 – 8.8976
(iii) 28.5 – 9.708 – 6.234
(iv) 8.2 – 4.56 – 0.7912 + 2.67
Solution:
(i) 72.53 – 46.782
= 72.530 – 46.782
= 25.748

(ii) 18.376 – 5.43 – 8.8976
= 18.3760 – 5.4300 – 8.8976 (Converting into like decimals)
= 18.3760 – 14.3276
= 4.0484

(iii) 28.5 – 9.708 – 6.234
= 28.500 – 9.708 – 6.234
= 28.500 – 15.942
= 12.558

(iv) 8.2 – 4.56 – 0.7912 + 2.67
= 8.2000 – 4.5600 – 0.7912 + 2.6700 (Converting into like decimals)
= 8.2000 + 2.6700 – 4.5600 – 0.7912
= 10.8700 – 5.3512
= 5.5188

Question 14.
(i) What number added to 3.56 gives 13.016?
(ii) What number should be subtracted from 30 to get 23.709?
(iii) What is the excess of 20.4 over 9.7403?
Solution:
(i) The required number = 13.016 – 3.56 = 13.016 – 3.560 = 9.456

(ii) The required number = 30 – 23.709 = 30.000 – 23.709 = 6.291

(iii) The required number = 20.4 – 9.7403 = 20.4000 – 9.7403 = 10.6597

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 4 Playing with Numbers Ex 4.1

Question 1.
Fill in the blanks:
(i) A number having exactly two factors is called a …..
(ii) A number having more than two factors is called a ………
(iii) 1 is neither ……… nor ………
(iv) The smallest prime number is ………
(v) The smallest odd prime number is ………
(vi) The smallest composite number is ………
(vii) The smallest odd composite number is ………
(viii)All prime numbers (except 2) are ………
Solution:
(i) A number having exactly two factors is called a prime number.
(ii) A number having more than two factors is called a composite number.
(iii) 1 is neither prime nor composite.
(iv) The smallest prime number is 2.
(v) The smallest odd prime number is 3.
(vi) The smallest composite number is 4.
(vii) The smallest odd composite number is 9.
(viii) All prime numbers (except 2) are odd numbers.

Question 2.
State whether the following statements are ture (T) or false (F):
(i) The sum of three odd numbers is an even number.
(ii) The sum of two odd numbers and one even number is an even number.
(iii) The product of two even numbers is always an even number.
(iv) The product of three odd numbers is an odd number.
(v) If an even number is divided by 2, the quotient is always an odd number.
(vi) All prime numbers are odd.
(vii) All even numbers are composite.
(viii) Prime numbers do not have any factors.
(ix) A natural number is called a composite number if it has atleast one more factor other than 1 and the number itself.
(x) Two consecutive numbers cannot be both prime.
(xi) Two prime numbers are always co-prime numbers.
Solution:
(i) False
(ii) True
(iii) True
(iv) True
(v) False
(vi) False
(vii) False
(viii) False
(ix) True
(x) False
(xi) True

Question 3.
Write all the factors of the following natural numbers:
(i) 68
(ii) 27
(iii) 210
Solution:
(i) 68
The factors of 68 are : 1, 2, 4, 17, 34, 68
(ii) 27
The factors of 27 are : 1, 3, 9, 27
(iii) 210
The factors of 210 are :
1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, 210

Question 4.
Write first six multiples of the following natural numbers:
(i) 3
(ii) 5
(iii) 12
Solution:
(i) 3
The first six multiples of 3 are
3, 6, 9, 12, 15, 18
(ii) 5
The first six multiples of 5 are
5, 10, 15, 20, 25, 30
(iii) 12
The first six multiples of 12 are
12, 24, 36, 48, 60, 72

Question 5.
Match the items in column 1 with the items in column 2:

Solution:

Question 6.
Find the common factors of :
(i) 20 and 28
(ii) 35 and 50
(iii) 56 and 120
Solution:
(i) 20 and 28
The factors of 20 are:
1, 2, 4, 5, 10, 20 The factors of 28 are:
1,2,4,7,14,28
The common factors of 20 and 28 are: 1, 2, 4

(ii) 35 and 20
The factors of 35 are:
1, 5, 7, 35
The factors of 20 are:
1, 2, 4, 5, 10, 20
The common factors of 35 and 20 are 1,5

(iii) 56 and 120
The factors of 56 are:
1, 2, 4, 7, 8, 14, 28, 56
The factors of 120 are:
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60,120
The common factors of 56 and 120 are 1, 2, 4,8

Question 7.
Find the common factors of:
(i) 4, 8, 12
(ii) 10, 30 and 45
Solution:
(i) The factors of 4 are:
1, 2, 4
The factors of 8 are:
1, 2, 4, 8
The factors of 12 are:
1,2, 3,4, 6, 12
The common factors of 4, 8, 12 are 1, 2, 4

(ii) 10, 30 and 45 .
The factor of 10 are:
1, 2, 5, 10
The factor of 30 are:
1, 2, 3, 5, 10, 15, 30
The factor of 45 are:
1, 3, 5, 9, 15, 45
The common factors of 10, 30, 45 are 1, 5

Question 8.
Write all natural numbers less than 100 which are common multiples of 3 and 4.
Solution:
Multiples of 3 are : 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87,
90, 93, 96, 99, 102, 105, 108, ……….
Multiples of 4 are : 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, ……………
∴ Common multiples of 3 and 4 are : 12, 24, 36, 48, 60, 72, 84, 96, 108, ………..
All the numbers less than 100 which are common multiples of 3 and 4 are 12, 24, 36, 48, 60, 72, 84 and 96.

Question 9.
(i) Write the odd numbers between 36 and 53.
(ii) Write the even numbers between 232 and 251.
Solution:
(i) The odd numbers between 36 and 53 are:
37, 39, 41, 43, 45, 47, 49, 51.
(ii) The even numbers between 232 and 251 are: 234, 236, 238, 240, 242, 244, 246, 248, 250.

Question 10.
(i) Write four consecutive odd numbers succeeding 79.
(ii) Write three consecutive even numbers preceding 124.
Solution:
(i) Four consecutive odd numbers succeeding 79 are : 81, 83, 85, 87.
(ii) Three consecutive even numbers preceding 124 are : 118, 120, 122.

Question 11.
What is greatest prime number between 1 and 15?
Solution:
The greatest prime number between 1 and 15 is 13.

Question 12.
Which of the following numbers are prime?
(i) 29
(ii) 57
(iii) 43
(iv) 61
Solution:
(i) 29
We have, 29 = 1 × 29
⇒ 29 has exactly two factors 1 and 29 itself.
∴ 29 is a prime number.

(ii) 57
We have, 57 = 1 × 57 = 3 × 19 = 57
∴ Factors of 57 are 1, 3, 19 and 57
⇒ 57 has more than two factors
∴ 57 is not a prime.

(iii) 43
We have, 43 = 1 × 43
⇒ 43 has exactly two factors 1 and 43 itself.
∴ 43 is a prime number.

(iv) 61
We have, 61 = 1 × 61
⇒ 61 has exactly two factors 1 and 61 itself.
∴ 61 is a prime number.

Question 13.
Which of the following pairs of numbers are co-prime?
(i) 12 and 35
(ii) 15 and 37
(iii) 27 and 32
(iv) 17 and 85
(v) 515 and 516
(vi) 215 and 415
Solution:
(i) 12 and 35
The factors of 12 are 1,2, 3,4, 6, 12
The factors of 35 are 1, 5, 7, 35
Since, the common factor of 12 and 35 is 1
∴ They are co-prime.

(ii) 15 and 37
The factors of 15 are 1, 3, 5, 15
The factors of 37 are 1, 37
The common factor of 15 and 37 is 1
∴ They are co-prime.

(iii) 27 and 32
The factors of 27 are 1, 3, 9, 27
The factors of 32 are 1, 2, 4, 8, 16, 32
Since, the common factor of 27 and 32 is 1 They are co-prime

(iv) 17 and 85
The factors of 17 are 1, 17
The factors of 85 are 1, 5, 17, 85
The common factors of 17 and 85 are 1 and 17
∴ They are not co-prime because they have more than 1 common factor.

(v) 515 and 516
The factors of 515 are 1, 5, 103, 515
The factors of 516 are 1, 2, 3, 4, 6, 12, 43, 86, 129, 172, 258, 516
Since, the common factor of 515 and 516 are 1 and 5
∴ So, they are not co-prime.

(vi) 215 and 415
The factors of 215 are 1, 5, 43, 215
The factors of 415 are : 1, 5, 83, 415
Since, the common factor of 215 and 415 are 1 and 5.
∴ So, they are not co-prime.

Question 14.
Express each of the following numbers as the sum of two odd primes:
(i) 24
(ii) 36
(iii) 84
(iv) 98
Solution:
(i) 24
⇒ 24 = 5 + 19

(ii) 36
⇒ 36 = 7 + 29

(iii) 84
⇒ 84 = 17 + 67

(iv) 98
⇒ 98 = 19 + 79

Question 15.
Express each of the following numbers as the sum of twin-primes:
(i) 24
(ii) 36
(iii) 84
(iv) 120
Solution:
(i) 24
⇒ 24 = 11 + 13

(ii) 36
⇒ 36 = 17 + 19

(iii) 84
⇒ 84 = 41 + 43

(iv) 120
⇒ 120 = = 59 + 61

Question 16.
Express each of the following numbers as the sum of three odd primes:
(i) 21
(ii) 35
(iii) 49
(iv) 63
Solution:
(i) 21
⇒ 21 = 3 + 7 + 11

(ii) 35
⇒ 35 = 5 + 11 + 19

(iii) 49
⇒ 49 = 7 + 11 + 31

(iv) 63
⇒ 63 = 7 + 13 + 43

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 2 Fractions and Decimals Ex 2.4

Question 1.
Find the reciprocal of each of the following:
(i) \(\frac { 3 }{ 7 }\)
(iii) \(\frac { 13 }{ 9 }\)
(iii) 8
Solution:
Reciprocal of:

Question 2.
Evaluate the following:


Solution:

Question 3.
How many pieces each 5\(\frac { 1 }{ 6 }\) metres long can be cut from a cloth 77\(\frac { 1 }{ 2 }\) metres long?
Solution:
Total length of cloth = 77\(\frac { 1 }{ 2 }\) m
Length of one piece = 5\(\frac { 1 }{ 6 }\) m

Question 4.
By what number should 4\(\frac { 7 }{ 8 }\) be multiplied to get 87\(\frac { 3 }{ 4 }\) ?
Solution:

Question 5.
In a hostel’s mess, each student gets \(\frac { 1 }{ 3 }\) litre of milk every day. If the total consumption of the milk is 57\(\frac { 2 }{ 3 }\) litres per day, how many students are there in the hostel?
Solution:
Every students gets \(\frac { 1 }{ 3 }\) l of milk per day
Total consumption of milk per day = 57 \(\frac { 2 }{ 3 }\) l

Question 6.
The cost of 5\(\frac { 1 }{ 4 }\) kg apples is ₹ 336. What is the rate of apples per kg?
Solution:
Cost of 5\(\frac { 1 }{ 4 }\) kg of apples = ₹ 336
Cost of one kg = 336 + 5\(\frac { 1 }{ 4 }\)
= 336 ÷ \(\frac { 21 }{ 4 }\)
= ₹ 336 × \(\frac { 4 }{ 21 }\)
= ₹ 64

Question 7.
The length of a rectangular plot of area 68\(\frac { 3 }{ 4 }\) sq. m is 12\(\frac { 1 }{ 2 }\) m, find its width.
Solution:
Area of a rectangular field = 68\(\frac { 3 }{ 4 }\) sq. m

Question 8.
If the cost of 5\(\frac { 1 }{ 2 }\) kg of sugar is ₹ 206\(\frac { 1 }{ 4 }\),then find the cost of 8\(\frac { 1 }{ 4 }\) kg of sugar.
Solution:

Question 9.
Renu completed \(\frac { 2 }{ 3 }\) part of her homework in 2 hours. How much part of her homework had she completed in 1\(\frac { 1 }{ 4 }\) hours?
Solution:
In 2 hours, homework in completed = \(\frac { 2 }{ 3 }\)
In 1 hour it will be completed

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 3 Integers Check Your Progress

Question 1.
Use the appropriate symbol < or > to fill in the following blanks:
(i) (-3 + ……… (-6) (-3) – (-6)
(ii) (-21) – (-10) ……. (-31)+ (-11)
(iii) 45 – (-11) ……….. (57) + (-4)
(iv) (-25) – (-42) …………. (-42) – (-25)
Solution:
(i) (-3 + (-6) < (-3) – (-6)
(ii) (-21) – (-10) > (-31) + (-11)
(iii) 45 – (-11) > (57) +(-4)
(iv) (-25 – (-42) > (-42) – (-25)

Question 2.
Find the value of:
(i) 12 + ( -3) + 5 – (-2)
(ii) 39 – 35 + 7-(-4) + 21
(iii) -15- (-2) – 71 – 8 + 6
Solution:
(i) 12 + (-3) + 5 – (-2)
= 12 – 3 + 5 + 2
= 9 + 7= 16

(ii) 39 – 35 + 7 – (-4) + 21
= 39 – 35 + 7 + 4 + 21
= 4 + 11 + 21
= 15 + 21 =36

(iii) -15 – (-2) – 71 – 8 + 6
= -15 + 2 – 71 – 8 + 6
= -13 – 79 + 6
= 92 + 6 = -86

Question 3.
Evaluate:
(i) |-13| – |-15|
(ii) |35 – 41| – |7-(-2)|
Solution:
(i) |-13| – |-15|
= +13 – 15 =-2

(ii) |35 – 41| – |7 – (-2)|
= 6 – 9 = -3

Question 4.
Arrange the following integers in ascending order:
-39, 35, -102, 0, -51, -5, -6, 7
Solution:
-102, -51, -39, -6, -5, 0, 7, 35

Question 5.
Find the successor and the predecessor of -199.
Solution:
Successor = -199 – 1 = -198
Predecessor = -199 – 1 = -200

Question 6.
Subtract the sum of -235 and 137 from -152.
Solution:
Sum of (-235 and 137)
= -235 + 137
= 137 – 235 = -98
Now, subtract the sum of (-235 and 137) from -152
= 152 – (-98)
= -152 + 98 = -54

Question 7.
What must be added to -176 to get -95?
Solution:
Let the number to be added = x
∴ -176 + x = -95
x = -95 + 176 = 81

Question 8.
What is the difference in height between a point 270 m above sea level and 80 m below sea level?
Solution:
Height above sea level = +270 m
Height below sea level = -80 m
Difference = +270 – (-80)
= 270 + 80 = 350 m

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 3 Integers Objective Type Questions

Mental Maths

Question 1.
Fill in the blanks:
(i) The absolute value of 0 is ………
(ii) The sum of two negative integers is always a ……….. integer.
(iii) The smallest positive integer is ……………
(iv) The largest negative integer is ………….
(v) 17 + ………….. = 0
(vi) ……………. -15 = -10
(vii) The predecessor of -99 is …………
Solution:
(i) The absolute value of 0 is 0.
(ii) The sum of two negative integers is always a negative integer.
(iii) The smallest positive integer is 1.
(iv) The largest negative integer is -1.
(v) 17 + -17 = 0
(vi) 5 – 15 = -10
(vii) The predecessor of -99 is -100.

Question 2.
State whether the following statements are true (T) or false (F):
(i) The sum of a positive integer and a negative integer is always a negative integer.
(ii) Zero is an integer.
(iii) The sum of an integer and its negative is always zero.
(iv) The sum of three integers can never be zero.
(v) |-7| < |-3|.
(vi) -20 is to the left of -21 on the number line.
(vii) The successor of -29 is -30.
(viii) 0 is greater than every negative integer.
(ix) The difference of two integers is always an integer.
(x) Additive inverse of a negative integer is always a positive integer.
Solution:
(i) The sum of a positive integer and a negative integer is always a negative integer. False
(ii) Zero is an integer. True
(iii) The sum of an integer and its negative is always zero. True
(iv) The sum of three integers can never be zero. False
(v) |-7| < |—3|. False
(vi) -20 is to the left of -21 on the number line. False
(vii) The successor of -29 is -30. False
(viii)0 is greater than every negative integer. True
(ix) The difference of two integers is always an integer. True
(x) Additive inverse of a negative integer is always a positive integer. True

Question 3.
State whether the following statements are true or false. If a statement is false, write the corresponding correct statement.
(i) -8 is to the right of-10 on the number line.
(ii) -100 is to the right of -50 on the number line.
(iii) Smallest negative integer is -1.
(iv) -26 is greater than -25.
(v) -187 is the predecessor of-188.
Solution:
(i) -8 is to the right of-10 on the number line. True
(ii) -100 is to the right of -50 on the number line. False
Correct:
-100 is to the left of-50 on the number line.
(iii) Smallest negative integer is -1. False
Correct :
Greatest negative integer is -1.
(iv) -26 is greater than -25. False
Correct:
-26 is smaller than -25.
(v) -187 is the predecessor of-188. False
Correct:
-187 is the successor of -188.

Multiple Choice Questions

Choose the correct answer from the given four options (4 to 17):
Question 4.
The integer which is 5 more than -2 is
(a) -7
(b) -3
(c) 3
(d) 7
Solution:
3 (c)

Integer 3 is 5 more than -2

Question 5.
The number of integers between -1 and 1 is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
‘0’ lies between -1 and 1
∴ -1, 0, 1 = 1 number (b)

Question 6.
The number of integers between -3 and 2 are
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
-2, -1,0, 1 lies between -3 and 2
-3, -2, -1, 0, 1,2 = 4 numbers (c)

Question 7.
The number of whole numbers between -6 and 6 is
(a) 11
(b) 10
(c) 6
(d) 5
Solution:
Number -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 lies between -6 and 6.
Among them 0, 1, 2, 3, 4, 5 are whole numbers.
∴ 6 whole numbers lies between -6 and 6 (c)

Question 8.
The greatest integer lying -10 and -15 is
(a) -10
(b) -11
(c) -14
(d) -15
Solution:
-11 (b)

Question 9.
The smallest integer lying between -10 and -15 is
(a)-10
(b) -11
(c) -14
(d) -15
Solution:
-14 (c)

Question 10.
Which of the following statement is true?
(a) |10 – 4| = |10| + |—4|
(b) Additive inverse of -5 is 5
(c) -1 lies on the right of 0 on the number line
(d) -7 is greater than -3
Solution:
Additive inverse of -5 is 5 (b)

Question 11.
Which of the following statement is false?
(a) -20 – (-5) = -15
(b) |-18| > |-13|
(c) 23 + (-31) = 8
(d) Every negative integer is less than 5
Solution:
23 + (-31) = 8
The correct answer will be
23 + (-31) = -8 (c)

Question 12.
Which of the following statements is false?
(a) (-3) + (-11) is an integer
(b) (-19)+ 13 = 13 +(-19)
(c) (-15) + 0 = -15 = 0 + (-15)
(d) Negative of-7 does not exist
Solution:
Negative of -7 does not exist, is false statement.
Negative of -7 is – (-7) = 7 (d)

Question 13.
If the sum of two integers is -17 and one of them is -9, then the other is
(a) 8
(b) -8
(c) 26
(d) -26
Solution:
-17 – (9)
= -17 + 9 = -8 (b)

Question 14.
On subtracting -7 from -4, we get
(a) 3.
(b) -3
(c) -11
(d) none of these
Solution:
-4 – (-7)
= -4 + 7 = 3 (a)

Question 15.
(-12) + 17 – (-10) is equal to
(a) -5
(b) 5
(c) 15
(d) -15
Solution:
(-12) + 17 – (-10)
= -12 + 27 = 15 (c)

Question 16.
Which of the following statements is true?
(a) -13 > – 8 – (-6)
(b) -5 – 4 > -12 + 2
(c) (-8) – 3 = (-3) – (-8)
(d) (-15) – (-22) < (-22) – (-15)
Solution:
-5 – 4 > -12 + 2
= -9 > -10
∴ -9 is always greater than -10 (b)

Question 17.
The statement “when an integer is added to itself, the sum is less than the integer” is
(a) always true
(b) never true
(c) true only when the integer is negative
(d) true when the integer is zero or positive
Solution:
true only when the integer is negative (c)

Higher Order Thinking Skills (HOTS)

Question 1.
Can the sum of successor and predecessor of an integer be an odd integer?
Solution:
No, the sum of successor and predecessor of an integer cannot be an odd integer.

Question 2.
What is the sum of all integers from -500 to 500?
Solution:
The sum of all integers from -500 to 500 = -500 + 500 = 0

Question 3.
Find two positive integers such that their product is 1,00,000 and none of them contains 0 as a digit.
Solution:
We shall find the factors of 1,00,000 to find the two positive integers such that their product is 1,00,000.

The factors of 100000 = 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5 × 5
∴ The positive integers which have the product 100000 are
(i) 2 × 2 × 2 × 2 × 2 = 32
(ii) and 5 × 5 × 5 × 5 × 5 = 3125

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 2 Fractions and Decimals Ex 2.3

Question 1.
Evaluate the following:

Solution:

Question 2.
Find:

Solution:

Question 3.
Evaluate the following:

Solution:

Question 4.
Find the value of:

Solution:

Question 5.
Which is greater:

Solution:

Question 6.
If 1 metre cloth costs ₹ 31\(\frac { 3 }{ 4 }\), find the cost of 5\(\frac { 1 }{ 2 }\) metres cloth.
Solution:

Question 7.
If the speed of a car is 105\(\frac { 1 }{ 5 }\) km/h, find the distance covered by it in 3\(\frac { 3 }{ 5 }\) hours.
Solution:
Speed of a car = 105\(\frac { 1 }{ 5 }\) km/h = \(\frac { 526 }{ 5 }\) km/h
It will covered a distance in 3\(\frac { 3 }{ 5 }\) hours

Question 8.
A car runs 16 km using 1 litre of petrol. How much distance will it cover in 2\(\frac { 3 }{ 4 }\) litres of petrol?
Solution:
A car runs in 1 litre of petrol = 16 km
It will run in 2\(\frac { 3 }{ 4 }\) litres of petrol

Question 9.
Sushant reads \(\frac { 1 }{ 3 }\) part of a book in 1 hour. How much part of the book will he read in 2\(\frac { 1 }{ 5 }\) hours?
Solution:
Sushant reads a book in 1 hour = \(\frac { 1 }{ 3 }\) part
He will read it in 2\(\frac { 1 }{ 5 }\) hours

Question 10.
An ornament is made of gold and copper and weighs 52 grams. If \(\frac { 2 }{ 13 }\) of its part is copper, find the weight of pure gold in it.
Solution:
Weight of ornament = 52 grams
In it \(\frac { 2 }{ 13 }\) part is copper
Weight of copper = 52 × \(\frac { 2 }{ 13 }\) = 8 grams
Then weight of gold = 52 grams – 8 grams = 44 grams

Question 11.
In a class of 40 students, \(\frac { 1 }{ 5 }\) of the total number of students like to study English and \(\frac { 2 }{ 5 }\) of the total number of students like to study Mathematics and the remaining like to study Science.
(i) How many students like to study English?
(ii) How many students like to study Mathematics?
(iii) What fraction of the total number of students like to study Science?
Solution:
The number of students in a class = 40
(i) Number of students who like English = \(\frac { 1 }{ 5 }\) of 40 = 8 students
(ii) Number of students who like Mathematics = \(\frac { 2 }{ 5 }\) of 40 = 16 students
(iii) Remaining students = 40 – (8 + 16) = 40 – 24 = 16 students
Number of students who like Science = 16
Fraction of students = \(\frac { 16 }{ 40 }\) = \(\frac { 2 }{ 5 }\)

Question 12.
A rectangular sheet of paper is 12\(\frac { 1 }{ 2 }\) cm long and 10\(\frac { 2 }{ 3 }\) cm wide. Find its
(i) perimeter
(ii) area
Solution:
Length of rectangular sheet of paper = 12\(\frac { 1 }{ 2 }\) cm = \(\frac { 25 }{ 2 }\) cm
and breadth = 10\(\frac { 2 }{ 3 }\) = \(\frac { 32 }{ 3 }\) cm

Question 13.
In a school, \(\frac { 25 }{ 54 }\) of the students are girls and the rest are boys. If the number of boys is 2030, find the number of girls.
Solution:
In a school,
Number of girls = \(\frac { 25 }{ 54 }\)
Then, number of boys = 1 – \(\frac { 25 }{ 54 }\) = \(\frac { 29 }{ 54 }\)
Number of boys = 2030
\(\frac { 29 }{ 54 }\) of total students = 2030
Total number of students = 2030 × \(\frac { 54 }{ 29 }\) = 3780
Number of girls = 3780 – 2030 = 1750

Question 14.
In an orchard, \(\frac { 1 }{ 5 }\) are orange grees, \(\frac { 3 }{ 13 }\) are mango trees and the rest are banana trees. If the banana trees are 148 in number, find the total number of trees in the orchard.
Solution:
In an orchard,
Orange trees = \(\frac { 1 }{ 5 }\) part
Mango trees = \(\frac { 3 }{ 13 }\) part
Banana trees = Rest

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 3 Integers Ex 3.4

Question 1.
Find the value of:
(i) 6 – 9 + 4
(ii) -5 – (-3) + 2
(iii) 7 + (-5) + (-6)
(iv) 6 – 3 – (-5)
Solution:
(i) 6 – 9 + 4
= (6 + 4) – 9 = 10 – 9= 1

(ii) -5 – (-3) + 2
= -5 + 3 + 2 = -5 + 5 = 0

(iii) 7 + (-5) + (-6)
= 7 – 5 – 6 = 2 – 6 = -4

(iv) 6 – 3 – (-5)
= 6 – 3 + 5 = 8

Question 2.
Evaluate the following:
(i) -77 + (-84) + 318
(ii) 54 + (-218) – (-76)
(iii) -121 – (-78) + (-193) + 576
(iv) -65 + (-76) – (-28) + 32
Solution:
(i) -77 + (- 84) + 318
= -77 – 84 + 318
= -(77 + 84)+ 318
= -(161) + 318
= -161 +318
= 318 – 161 = 157

(ii) 54 + (-218) – (-76)
= 54 – 218 + 76
= (54 + 76) – 218
= 130 – 218 = – 88

(iii) -121 – (-78) + (-193) + 576
= -121 + 78 – 193 + 576
= -121 – 193 + 78 + 576
= -(121 + 193) + 78 + 576
= -(314) + 654
= 654 – 314 = 340

(iv) -65 + (-76) – (-28) + 32
= -65 – 76 + 28 + 32
= -(65 + 76) + 60
= -141 + 60 = -81

Question 3.
Find the value of:
(i) 8 – 6 + (-2) – (-3) + 1
(ii) 31 + (-23) – 35 + 18 – 4 – (-3)
Solution:
(i) 8 – 6 + (-2) – (-3) + 1
= 8 – 6 – 2 + 3 + 1
=-6 – 2 + 8 + 3 + 1
= -6 – 2 + 12
=-8 + 12 = 4

(ii) 31 + (-23) – 35 + 18 – 4 – (-3)
= 31 – 23 – 35 + 18 – 4 + 3
= -23 – 35 – 4 + 31 + 18 + 3
= -23 – 35 – 4 + 52
= -62 + 52 = -10

Question 4.
Rashmi deposited ₹ 4370 in her account on Monday and then withdrew ₹ 2875 on Tuesday. Next day she deposited ₹ 1550. What was her balance on Thursday?
Solution:
Rashmi deposited in her account on Monday = ₹ 4370
Less withdrawal on Tuesday = ₹ 2875
So the Balance on Tuesday
= ₹ 4370 – ₹ 2875
= ₹ 1495
Again she deposited on Wednesday = ₹ 1550
Balance on Thursday
= ₹ 1495 + ₹ 1550 = ₹ 3045

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 3 Integers Ex 3.3

Question 1.
Evaluate the following, using the number line:
(i) 4 – (-2)
(ii) -4 – (-2)
(iii) 3 – 6
(iv) -3 – (-5)
Solution:
(i) Start from 4 on the number line.
Move 2 units to the digits we reach at 6
∴ 4 – (-2) = 4 + 2 = 6

(ii) Start from -4 on the number line.
Move 2 units to the right, we reach at -2
∴ -4 – (-2) = —4 + 2 = -2

(iii) Start from 3 on the number line.
Move 6 units to the left, we reach at -3
3 – 6 = -3

(iv) Start from -3 on the number line.
Move 5 units to the right, we reach at 2
-3 – (-5) = -3 + 5 = 2

Question 2.
Subtract :
(i) -6 from 9
(ii) 6 from -9
(iii) -6 from -9
(iv) -725 from -63
(v) -376 from 10
(vi) 92 from -620
Solution:
(i) 9 – (-6) = 9 + 6 = 15
(ii) -9 – 6 = -15
(iii) -9 – (-6) = -9 + 6 = -3
(iv) -63 – (-725) = -63 + 725 = +662
(v) 10 – (-376) = 10 + 376 = 386
(vi) -620 – 92 = -712

Question 3.
Evaluate the following:
(i) -237 – (+ 1884)
(ii) -346 – (- 1275)
(iii) -190 – (-3512)
(iv) -2718 – (+ 6827)
Solution:
(i) -237 – (+ 1884)
= -237 – 1884
= -(237 + 1884) = -2121

(ii) -346 -(- 1275)
= -346 + 1275
= 1275 – 346 = 929

(iii) -190 – (-3512)
= -190 + 3512
= 3512- 190 = 3322

(iv) 2718 – (+ 6827)
= -2718 – 6827
= -(2718 + 6827) = -9545

Question 4.
The sum of two integers is 17. If one of them is -35, find the other.
Solution:
One number = -35
Sum of two integers =17
Second number = Sum of integers – (The given number)
= 17 – (-35)
= 17 + 35 = 52

Question 5.
What must be added to -23 to get -9?
Solution:
Let the number to be added = x
∴ -23 + x = -9
∴ The required number = -9 – (-23)
= -9 + 23 = 14

Question 6.
Find the predecessor of 0.
Solution:
Predecessor of 0 = 0 – 1 = -1

Question 7.
Find the successor and the predecessor of the following integers:
(i) -31
(ii) -735
(iii) -240
Solution:
(i) Successor of -31 = -31 + 1 = -30
Predecessor of -31 = -31 – 1 = -32
(ii) Successor of -735 = -735 + 1 = -734
Predecessor of-735 = -735 – 1 = -736
(iii) Successor of -240 = -240 + 1 = -239
Predecessor of -240 = -240 – 1 = -241

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 2 Fractions and Decimals Ex 2.2

Question 1.
Evaluate the following:


Solution:

Question 2.
Simplify the following:

Solution:

Question 3.
Jaishree studies for 5\(\frac { 2 }{ 3 }\) hours daily. She devotes 2\(\frac { 4 }{ 5 }\) hours of her time for Science and Mathematics. How much time does she devote to other subjects?
Solution:
Time given to study = 5\(\frac { 2 }{ 3 }\) hours daily
Time given to Science and Maths = 2\(\frac { 4 }{ 5 }\) hours

Question 4.
Ramesh solved \(\frac { 2 }{ 7 }\) part of an exercise while Reshma solved \(\frac { 4 }{ 5 }\) of it. Who solved the lesser part? By how much?
Solution:
Ramesh solved \(\frac { 2 }{ 7 }\) part of an exercise
But Reshma solved \(\frac { 4 }{ 5 }\) of it

So, Reshma does \(\frac { 18 }{ 35 }\) part of it less then Ramesh.

Question 5.
Sonali had ₹ 35\(\frac { 3 }{ 5 }\). She got ₹ 16\(\frac { 1 }{ 15 }\) from her mother and spent ₹ 28\(\frac { 2 }{ 3 }\) on food. How much money is left with her?
Solution:
Sonali had = ₹ 35\(\frac { 3 }{ 5 }\) = ₹ \(\frac { 178 }{ 5 }\)
She gets ₹ 16\(\frac { 1 }{ 15 }\) from her mother
and spent ₹ 28\(\frac { 2 }{ 3 }\) on food
Amount left with her

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 2 Fractions and Decimals Ex 2.1

Question 1.
What fraction of each of the following figure is shaded?

Solution:
(i) Fraction is \(\frac { 2 }{ 8 }\) = \(\frac { 1 }{ 4 }\)
(ii) Fraction is \(\frac { 3 }{ 10 }\)
(iii) Fraction is \(\frac { 5 }{ 12 }\)
(iv) Fraction is \(\frac { 7 }{ 13 }\)

Question 2.
What fraction of an hour is 35 minutes?
Solution:
(i) 1 hour = 60 minutes
Fraction = \(\frac { 35 }{ 60 }\) = \(\frac { 7 }{ 12 }\)

Question 3.
Convert the following fractions into improper fractions :
(i) 2\(\frac { 7 }{ 9 }\)
(ii) 5\(\frac { 4 }{ 11 }\)
Solution:

Question 4.
Convert the following fractions into mixed fractions:
(i) \(\frac { 73 }{ 8 }\)
(ii) \(\frac { 94 }{ 13 }\)
Solution:

Question 5.
Fill in the missing numbers in the following equivalent fractions:
(i) \(\frac { 3 }{ 7 }\) = \(\frac { …. }{ 35 }\)
(ii) \(\frac { 5 }{ …. }\) = \(\frac { 30 }{ 18 }\)
(iii) \(\frac { ….. }{ 9 }\) = \(\frac { 56 }{ 72 }\)
Solution
(i) \(\frac { 3 }{ 7 }\) = \(\frac { …. }{ 35 }\)
The denominator in the second fraction is 35.
To get 35 from 7, we have to multiply 7 by 5.

(ii) \(\frac { 5 }{ …. }\) = \(\frac { 30 }{ 18 }\)
To make both fractions equal, we multiply the numerator of the first fraction by 5.
The numerator itx the first fraction is 5. To get 5 from 30, we have to divide 30 ÷ 6.
To make both fractions equal, we divide denominator of the second fraction by 6.

(iii) \(\frac { ….. }{ 9 }\) = \(\frac { 56 }{ 72 }\)
The denominator in the second fraction is 72 and the denominator in the first fraction is 9.
To get 9 from 72, we have to divide 72 ÷ 8
To make both fractions equal, we divide the numerator of the second fraction by 8.

Question 6.
Reduce the following fractions to their simplest form:
(i) \(\frac { 48 }{ 72 }\)
(ii) \(\frac { 276 }{ 115 }\)
(iii) \(\frac { 72 }{ 336 }\)
Solution:

Question 7.
Convert the following fractions into equivalent like fractions:

Solution:

Question 8.
Arrange the given fractions in descending order:

Solution:

Question 9.
Arrange the given fractions in ascending order:

Solution:

ML Aggarwal Class 7 Solutions for ICSE Maths

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 3 Integers Ex 3.2

Question 1.
Evaluate the following, using the numbers line
(i) 4 + (-5)
(ii) (-4) + 5
(iii) 7 + (-3)
(iv) -6 + (-2)
Solution:
(i) Start from 4 on the number line.
Move 5 units to the left, we reach at -1 4
∴ + (-5) = 4 – 5 = -1

(ii) Start from -4 on the number line.
Move 5 units to the right, we reach at 1
∴ (-4) + 5 = -4 + 5 = 1

(iii) Start from 7 on the number line.
Move 3 units to the left, we reach at 4
∴ 7 + (-3) = 7 – 3 = 4

(iv) Start from -6 on the number line.
Move 2 units to the left, we reach at -8
⇒ -6 + (-2) = -6 – 2 = -8

Question 2.
Evaluate the following :
(i) (-8) + (-14)
(ii) -35 + (-47)
(iii) 91 + (-48)
(iv) (-203) + 501
(v) (-36) + 29
(vi) (-131) + 97.
Solution:
(i) (-8) + (-14)
= -8 – 14 = -22

(ii) -35 + (-47)
= -35 – 47 = -82

(iii) 91 + (-48)
= 91 – 48 = 43

(iv) (-203) + 501
= -203 + 501 =298

(v) (-36) + 29
= -36 + 29 = -7

(vi) (-131) + 97
= -131 + 97 = -34

Question 3.
Evaluate the following :
(i) -1083 + (-3974)
(ii) 706 + (-394)
(iii) 1309 + (-2811)
Solution:
(i) -1083 + (-3974)
= -1083 – 3974
= -(1083 + 3974) = -5057

(ii) 706 + (-394)
= 706 – 394 = 312

(iii) 1309 + (-2811)
= -2811 + 1309 = -1502

Question 4.
Fill in the following blanks :
(i) -(-5) = ……..
(ii) -(-30) = …….
(iii) -(-539) = ………
Solution:
(i) -(-5) = 5
(ii) -(-30) = 30
(iii) -(-539) = 539

Question 5.
Write down the additive inverses of:
(i) 9
(ii) -11
(iii) -237
(iv) 567
Solution:
(i) Additive inverse of 9 = (-9) = -9
(ii) Additive inverse of -11 = -(-11) = 11
(iii) Additive inverse of -237 = -(-237)
= 237
(iv) Additive inverse of 567 = -(567)
= -567

ML Aggarwal Class 6 Solutions for ICSE Maths

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 1 Integers Check Your Progress

Question 1.
Evaluate the following:
(i) (-7) × (-9) × (-11)
(ii) (-5) × 7 × (-6) × (-8)
(iii) (-1024) ÷ 32
(iv) (-216) ÷ (-12)
Solution:
(i) (-7) × (-9) × (-11) = (63) × (-11) = -693
(ii) (-5) × 7 × (-6) × (-8) = -35 × (48) = -1680
(iii) (-1024) ÷ 32
= \(\frac { -1024 }{ 32 }\)
= -32
(iv) (-216) ÷ (-12)
= \(\frac { -216 }{ -12 }\)
= +18
= 18

Question 2.
What will be the sign of the product if we multiply 39 negative integers and 98 positive integers?
Solution:
39 is odd and 98 is even.
Product of 39 negative integers and 98 positive integers will be negative.

Question 3.
Use the sign >, < or = in the box to make the following statements true:
(i) (-15) + 38 ……… 27 + (-50)
(ii) (-13) × 0 × (-5) …….. (-7) × (-6) × 14
(iii) (-18) ÷ (-3) …….. (-10) + (-15) + 31
(iv) (-5) × (-7) × (-10) …….. (-1400) ÷ (-4)
Solution:
(i) (-15) + 38 ……. 27 + (-50)
LHS = -15 + 38 = 23
RHS = 27 – 50 = -23
(23) > (-23)

(ii) (-13) × 0 × (-5) ……. (-7) × (-6) × 14
LHS = -13 × 0 × (-5) = 0
and (-7) × (-6) × 14
RHS = 42 × 14 = 588
0 < 588

(iii) (-18) ÷ (-3) ……. (-10) + (-15) + 31
-18
LHS = -18 ÷ -3 = 6
RHS = (-10)+ (-15) + 31 = -25 + 31 = 6
6 = 6

(iv) (-5) × (-7) × (-10) ……… (-1400) ÷ (-4)
LHS = (-5) × (-7) × (-10) = 35 × (-10) = -350
RHS = (-1400) ÷ (-4) = -350
-350 < 350

Question 4.
Is {(-45) ÷ (-15)} ÷ (-3) = (-45) ÷ [(-15) ÷ (-3)]?
Solution:
(-45) ÷ (-15) ÷ (-3)
LHS = \(\frac { -45 }{ -15 }\) ÷ (-3) = 3 ÷ -3 = -1
and RHS = -45 ÷ (\(\frac { -15 }{ -3 }\))
= -45 ÷ 5
= -9
(-1) ≠ -9
No, not equal.

Question 5.
A cement company earns a profit of ₹ 8 per bag of white cement sold and a loss of ₹ 5 per bag of grey cement sold.
(i) The company sells 3000 bags of white cement and 5000 bags of grey cement in a month. What is its profit or loss?
(ii) What is the number of white cement bags it must sell to have neither profit nor loss, if the number of grey cement bags sold is 6400?
Solution:
Profit on one white cement bag = ₹ 8
and loss on one bag grey cement bag = ₹ 5
(i) On the sale of 3000 bags of white cement and 5000 bags of grey cement bags.
Sale = 3000 × 8 – 5000 × 5 = ₹ 24000 – ₹ 25000 = – ₹ 1000
There will be a loss of ₹ 1000

(ii) Loss on 6400 bags of grey cement = ₹ 6400 × 5 = 32000
In order to get a gain of ₹ 32000
The white cement bags be sold = \(\frac { 32000 }{ 8 }\) = 4000 bags

Question 6.
Simplify the following:
(i) (-7) + (-6) ÷ 2 – {(-5) × (-4) – (3 – 5)}
(ii) 11 – [7 – (5 – 3 (9 – \(\bar { 3-6 }\))}].
Solution:
(i) (-7) + (-6) ÷ 2 – {(-5) × (-4) – (3 – 5)}
= (-7) + (-6) ÷ 2 – [+20 – (-2)]
= (-7) + (-6) ÷ 2 – (20 + 2)
= (-7) + \(\frac { -6 }{ 2 }\) – 22
= (-7) + (-3) – 22
= -7 – 3 – 22
= -32

(ii) 11 – [7 – (5 – 3 (9 – \(\bar { 3-6 }\))}]
= 11 – [7 – {5 – 3 (9 – 3 + 6)}]
= 11 – [7 – {5 – 3 × 12}]
= 11 – [7 – {5 – 36}]
= 11 – [7 – {-31}]
= 11 – [7 + 31]
= 11 – 38
= -27

ML Aggarwal Class 7 Solutions for ICSE Maths