ML Aggarwal Solutions for Class 7 Maths Chapter 2 Fractions and Decimals are designed by experienced teachers in a descriptive manner. These solutions help students to analyse the type of questions that would appear in the final examination. Students who aspire to score high marks are suggested to practise ML Aggarwal Solutions on a regular basis. For more conceptual knowledge, students can make use of ML Aggarwal Solutions for Class 7 Maths Chapter 2 Fractions and Decimals PDF from the link available below.
A number representing a part of a whole is known as a Fraction, and a number whose whole number and the fractional part are separated by a decimal point is known as Decimals. Chapter 2 of ML Aggarwal Solutions for Class 7 is vital, as it is continued in further classes also. The solutions help students to understand the concepts in an effective way.
ML Aggarwal Solutions for Class 7 Maths Chapter 2: Fractions and Decimals
Access ML Aggarwal Solutions for Class 7 Maths Chapter 2: Fractions and Decimals
1. What fraction of each of the following figure is shaded?
Solution:
(i) In the given figure, the shaded fraction is 2 / 8 = 1 / 4
(ii) In the given figure, the shaded fraction is 3 / 10
(iii) In the given figure, the shaded fraction is 5 / 12
(iv) In the given figure, the shaded fraction is 7 / 13
2. Evaluate the following:
(i) (4 / 3) + (7 / 8)
(ii) 
(iii) (5 / 12) + (1 / 18) – (2 / 9)
Solution:
(i) (4 / 3) + (7 / 8)
L.C.M. of 3, 8 is 24, we get,
= (32 + 21) / 24
= 53 / 24
We get,
=
(ii)
This can be written as,
= (17 / 2) – (29 / 8)
L.C.M. of 2, 8 is 8, we get,
= (68 – 29) / 8
= 39 / 8
We get,
=
(iii) (5 / 12) + (1 / 18) – (2 / 9)
L.C.M. of 12, 18, 9 is 36, we get,
= (15 + 2 – 8) / 36
= (17 – 8) / 36
= 9 / 36
Dividing both the numerator and denominator by 9, we get,
= (9 ÷ 9) / (36 ÷ 9)
We get,
= 1 / 4
3. Evaluate the following:
(i) 7 × (3 / 5)
(ii) 21 × (3 / 14)
(iii) 
× 8
(iv) 5 × 
Solution:
(i) 7 × (3 / 5)
On simplification, we get,
= 21 / 5
=
(ii) 21 × (3 / 14)
On further calculation, we get,
= 9 / 12
=
(iii)
× 8
This can be written as,
= (17 / 5) × 8
On calculating further, we get,
= 136 / 5
=
(iv) 5 ×
This can be written as,
= 5 × (27 / 4)
= 135 / 4
We get,
=
4. Find the reciprocal of each of the following:
(i) 3 / 7
(ii) 13 / 9
(iii) 8
Solution:
(i) The reciprocal of 3 / 7 is 7 / 3
(ii) The reciprocal of 13 / 9 is 9 / 13
(iii) The reciprocal of 8 is 1 / 8
5. Write the following numbers in the expanded form:
(i) 20.03
(ii) 200.03
(iii) 2.034
Solution:
(i) 20.03
The expanded form of the given decimal is shown below,
= 2 × 10 + 0 × 1 + 0 × (1 / 10) + 3 × (1 / 100)
(ii) 200.03
The expanded form of the given decimal is shown below,
= 2 × 100 + 0 × 10 + 0 × 1 + 0 × (1 / 10) + 3 × (1 / 100)
(iii) 2.034
The expanded form of the given decimal is shown below,
= 2 × 1 + 0 × (1 / 10) + 3 × (1 / 100) + 4 × (1 / 1000)
6. Find the following:
(i) 2.7 × 4
(ii) 2.71 × 5
(iii) 2.5 × 0.3
(iv) 2.3 × 4.35
(v) 238.06 × 7.5
(vi) 0.79 × 32.4
(vii) 1.07 × 0.02
(viii) 10.05 × 1.05
Solution:
(i) 2.7 × 4
= (27 / 10) × 4
= 108 / 10
= 10.8
(ii) 2.71 × 5
On calculation, we get,
= (271 / 100) × 5
= 1355 / 100
We get,
= 13.55
(iii) 2.5 × 0.3
On further calculation, we get,
= (25 / 10) × (3 / 10)
= 75 / 100
We get,
= 0.75
(iv) 2.3 × 4.35
On further calculation, we get,
= (23 / 10) × (435 / 100)
=10005 / 1000
We get,
= 10.005
(v) 238.06 × 7.5
On simplification, we get,
= (23806 / 100) × (75 / 10)
23806
× 75
_____________
119030
1666420
______________
1785450
______________
= 1785450 / 1000
We get,
= 1785.45
(vi) 0.79 × 32.4
On further calculation, we get,
= (79 / 100) × (324 / 10)
324
× 79
__________
2916
22680
________________
25596
_________________
= 25596 / 1000
We get,
= 25.596
(vii) 1.07 × 0.02
On simplification, we get,
= (107 / 100) × (2 / 100)
= 214 / 10000
We get,
= 0.0214
(viii) 10.05 × 1.05
On calculating, we get,
= (1005 / 100) × (105 / 100)
1005
×105
_________
5025
100500
______________
105525
______________
= 105525 / 10000
We get,
= 10.5525
7. Simplify the following:
(i) (3 / 5) of 
+ 
(ii) (4 / 5) × 
– 2 × (3 / 5)
(iii) {(4 / 5) + 2}{3 – (2 / 3)}
Solution:
(i) (3 / 5) of
+
This can be written as,
= (3 / 5) of (10 / 9) + (7 / 2)
= (3 / 5) × (10 / 9) + (7 / 2)
We get,
= (2 / 3) + (7 / 2)
L.C.M. of 3, 2 is 6, we get,
= (4 + 21) / 6
= 25 / 6
=
(ii) (4 / 5) ×
– 2 × (3 / 5)
This can be written as,
= (4 / 5) × (19 / 8) – 2 × (3 / 5)
We get,
= (19 / 10) – (6 / 5)
L.C.M. of 10, 5 is 10, we get,
= (19 – 12) / 10
We get,
= (7 / 10)
(iii) {(4 / 5) + 2}{3 – (2 / 3)}
On simplification, we get,
= {(4 + 10) / 5} × {(9 – 2) / 3}
= (14 / 5) × (7 / 3)
= (14 × 7) / (5 × 3)
We get,
= 98 / 15
=
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