CBSE Class 7 Mathematics Fractions And Decimals Worksheet Set A

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DECIMAL (ADDITION AND SUBTRACTION)

Steps How to Add or Subtract Fractions with Different Denominators

Step 1: Given two unlike fractions where the denominators are NOT the same.

Step 2: Make the denominators the same by finding the Least Common Multiple (LCM) of their denominators. This step is exactly the same as finding the Least Common Denominator (LCD).

Step 3: Rewrite each fraction into its equivalent fraction with a denominator which is equal to the Least Common Multiple that you found in step #2. 

Step 4: Now, add or subtract the “new” fractions from step #3. Always reduce the answer to its lowest terms.

Questions based on above topics:-

Question. Compare the fraction: (i) (5/8)and (7/12)
Solution:
By cross multiplication, we have:
5 × 12 = 60 and 8 × 7 = 56
But,
60 > 56
∴ (5/8) > (7/12)

(ii) (5/9)and (11/15)
Solution:-
By cross multiplication, we have:
5 × 15 = 75 and 9 × 11 = 99
But,
75 < 99
∴ (5/9) < (11/15)

Question. Arrange the following fraction in ascending order : (3/4), (5/6), (7/9), (11/12)
Solution:
LCM of 4, 6, 9, 12 = 2× 2× 3× 3=36
Now, let us change each of the given fraction into an equivalent fraction having 36 as the denominator. [(3/4) × (9/9)] = (27/36) [(5/6) × (6/6)] = (30/36) [(7/9) × (4/4)] = (28/360 [(11/12) × (3/3)] = (33/36)
Clearly,
(27/36) < (28/36) < (30/36) < (33/36)
Hence,
(3/4) < (7/9) < (5/6) < (11/12)
Hence, the given fractions in ascending order are (3/4), (7/9), (5/6), (11/12)

Question. Arrange the following fraction in descending order : (3/4), (7/8), (7/12), (17/24)
Solution:
LCM of 4, 8, 12, 24 = 4 × 2 × 3 = 24
Now, let us change each of the given fraction into an equivalent fraction having 24 as the denominator. [(3/4) × (6/6)] = (18/24) [(7/8) × (3/3)] = (21/24) [(7/12) × (2/2)] = (14/24) [(17/24) × (1/1)] = (17/24)
Clearly,
(21/24) > (18/24) > (17/24) > (14/24)
Hence,
(7/8) > (3/4) > (17/24) > (7/12)
Hence, the given fractions in descending order are (7/8), (3/4), (17/24), (7/12)

Question. Reenu got (2/7) part of an apple while sonal got (4/5) part of it. Who got the larger part and by how much?
Solution:
From the question,
Reenu got (2/7) part of an apple
Sonal got (4/5) part of an apple
First we have to compare the given fraction (2/7) and (4/5) to know who got the larger part of the apple.
Then,
By cross multiplication, we have
2 × 5 = 10 and 4 × 7 = 28
But,
10 < 28
∴ (2/7) < (4/5)
So, Sonal got the larger part of the apple
Now,
= (4/5)-(2/7)
= [(28-10)/35]
= [18/35]
∴ Sonal got (18/35) part of the apple larger than Reenu.

Question. Find the sum: (i) (5/9)+(3/9)
Solution:
For adding two like fractions, the numerators are added and the denominator remains the same.
= (5+3)/9
= (8/9)
(ii) (8/9) + (7/12)
Solution:-
For addition of two unlike fractions, first change them to the like fractions.
LCM of 9, 12 = 36
Now, let us change each of the given fraction into an equivalent fraction having 36 as the denominator. [(8/9) × (4/4)] = (32/36) [(7/12) × (3/3)] = (21/36)
Now, add the like fractions,
= (32/36) + (21/36)
= (32+21)/36
= (53/36)
= [1(17/36)]

Question. Find the difference: (i) (5/7) – (2/7)
Solution:
The subtraction of fraction can be performed in a manner similar to that of addition.
For subtracting two like fractions, the numerators are subtract and the denominator remains the same.
= (5 – 2)/7
= (3/7)
(ii) (5/6) – (3/4)
Solution:-
For subtraction of two unlike fractions, first change them to the like fractions.
LCM of 6, 4 = 12
Now, let us change each of the given fraction into an equivalent fraction having 12 as the denominator.
= [(5/6) × (2/2)] = (10/12)
= [(3/4) × (3/3)] = (9/12)
Now,
= (10/12)-(9/12)
= [(10 – 9)/12]
= (1/12)

Question. Simplify: (2/3) + (5/6) – (1/9)
Solution:
LCM of 3, 6, 9 = 18
Now, let us change each of the given fraction into an equivalent fraction having 18 as the denominator.
= (2/3) × (6/6) = (12/18)
= (5/6) × (3/3) = (15/18)
= (1/9) × (2/2) = (2/18)
Then,
= (12/18) + (15/18) – (2/18)
= (12+ 15 – 2)/ 18
= (27-2)/ 18
= (25/18)
= [1(7/18)]

Question. Aneeta bought [3(3/4)] kg apples and [4(1/2)] kg guava. What is the total weight of fruits purchased by her?
Solution:
The total weight of fruits bought by Aneeta = [3(3/4)] + [4(1/2)]
We have,
First convert each mixed fraction into improper fraction
= [3(3/4)] = (15/4)
= [4(1/2)] = (9/2)
Then,
= (15/4) + (9/2)
LCM of 4, 2 = 4
Now, let us change each of the given fraction into an equivalent fraction having 4 as the denominator
= (15/4) × (1/1) = (15/4)
= (9/2) × (2/2) = (18/4)
= (15/4) + (18/4)
= (15 + 18)/ 4)
= (33/4)
= [8(1/4)]
The total weight of fruits purchased by Aneeta is [8(1/4)] kg

(ADDITION AND SUBTRACTION)

Question. Write each of the following as decimals:
(i) (8/100)
(ii) 20 + (9/10) + (4/100)
Solution:
(i) Given (8/100)
Mark the decimal point two places from right to left
(8/100) = 0.08

(ii) Given 20 + (9/10) + (4/100)
First convert the fractions (9/10) and (4/100) to decimals
Consider (9/10)
Mark the decimal point one place from right to left
(9/10) = 0.9
Now consider (4/100)
Mark the decimal point two places from right to left
(4/100) = 0.04
20 + (9/10) + (4/100) = 20 + 0.9 + 0.04
= 20.94

Question. Convert each of the following fractions as decimals:
(i) 0.04
(ii) 17.38
Solution:
(i) Given 0.04
Here we have to convert given decimals into fractions
0.04 can be written as (0.04/1)
Now multiply both numerator and denominator by 100 then we get
(0.04/1) = (0.04 × 100 /1 × 100)
= (4/100)
= (1/25)

(ii) Given 17.38
Here we have to convert given decimals into fractions
17.38 can be written as (17.38/1)
Now multiply both numerator and denominator by 100 then we get
(17.38/1) = (17.38 × 100 /1 × 100)
= (1738/100)
= (869/50)

Question. Express the following fractions as decimals:
(i) (23/10)
(ii) 25 (1/8)
Solution:
(i) Given (23/10)
Divide 23 by 10 we get
(23/10) = 2.3

(ii) Given 25 (1/8)
25 (1/8) can be written as
25 (1/8) = 25 + (1/8)
Consider (1/8),
Now multiply both numerator and denominator by 125 to get 1000 as denominator
25 (1/8) = 25 + (1/8) = 25 + (1 × 125 /8 × 125)
= 25 + (125/1000)
= 25 + 0.125
= 25.125

Question. Add the following:
(i) 41.8, 39.24, 5.01 and 62.6
(ii) 18.03, 146.3, .829 and 5.324
Solution:
(i) Given 41.8, 39.24, 5.01 and 62.6
41.8 + 39.2. + 5.01 + 62.6
= 148.65

(ii) Given 18.03, 146.3, 0.829 and 5.324
18.03+ 146.3+ 0.829 + 5.324
= 170.483

Question. Find the value of:
(i) 9.756 – 6.28
(ii) 48.1 – 0.37
Solution:
(i) Given 9.756 – 6.28
9.756 – 6.28
=3.476

(ii) Given 48.1 – 0.37
48.1 – 0.37
=47.73

Question. Take out of 3.547 from 7.2
Solution:
Given:- 3.547 from 7.2
7.2 – 3.547
= 3.653

Question. What is to be added to 36.85 to get 59.41?
Solution:
Given 36.85 and 59.41
Let the unknown number be x
x + 36.85 = 59.41
x = 59.41 – 36.85
x = 22.56
Hence 22.56 is to be added to 36.85 to get 59.41

Question. What is to be subtracted from 17.1 to get 2.051?
Solution:
Let the unknown number be x
Given that x is to be subtracted from 17.1 to get 2.051
17.1 – x = 2.051
17.1 – 2.051 = x
x = 17.1 – 2.051
x = 15.049

Question. By how much should 34.79 be increased to get 70.15?
Solution:
Let x be the unknown number
x + 34.79 = 70.15
x = 70.15 – 34.79
x = 35.36
35.36 should be increased to 34.79 to get 70.15

Question. By how much should 59.71 be decreased to get 34.58?
Solution:
Let x be the unknown number
59.71 – x =34.58
59.71 – 34.58 = x
x = 59.71 – 34.58
x = 25.13
25.13 should be decreased by 59.71 to get 34.58

Solve the following:

Q1. Multiply:

a) 1.25 x 20
b) 8.88 x 10000
c) 3.01 x 1100

Q 2. Evaluate:
a) 5.134 ÷ 1.7
b) 0.9257 ÷ 100
c) 2.73 ÷ 1.3

Q3. What should be added to 4.187 to get 9?

Q4. Convert 75m into cm.

Q5. What is the perimeter of a rectangle with length 0.42m & breadth 31.5cm?

Q6. Mrs. Sunita uses 3.204 litres of oil to make 9 dishes of equal proportion.

How much oil was used for each dish?

Q7. By what number 0.014 be multiplied to get 1.4?

Q8. Write in the exponential form:
a) m x m
b) 4 x 4 x y x y x y
c) 5 x 5 x 3 x 3 x 3 x 2 x 2

Please click the below link to access CBSE Class 7 Decimals Worksheet (1)