CBSE Class 7 Mathematics Fractions And Decimals Worksheet Set 01

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

Question 1. Multiply:
a) 1.25 x 20
b) 8.88 x 10000
c) 3.01 x 1100
Answer:
a) Let us find the product of \( 1.25 \) and \( 20 \):
\( 1.25 \times 20 = 1.25 \times 2 \times 10 = 2.5 \times 10 = 25 \)

b) To multiply \( 8.88 \) by \( 10000 \), shift the decimal point four places to the right:
\( 8.88 \times 10000 = 88800 \)

c) Let us find the product of \( 3.01 \) and \( 1100 \):
\( 3.01 \times 1100 = 3.01 \times 100 \times 11 = 301 \times 11 = 3311 \)
In simple words: Multiplying by numbers with zeros like 20, 10000, and 1100 is easier when we shift the decimal point to the right.

Exam Tip: When you multiply a decimal by a multiple of 10, first move the decimal point based on the number of zeros to simplify your work.

Question 2. Evaluate:
a) 5.134 ÷ 1.7
b) 0.9257 ÷ 100
c) 2.73 ÷ 1.3
Answer:
a) To solve \( 5.134 \div 1.7 \), multiply both numbers by 10 to make the divisor a whole number:
\( 51.34 \div 17 = 3.02 \)

b) To divide \( 0.9257 \) by \( 100 \), move the decimal point two places to the left:
\( 0.9257 \div 100 = 0.009257 \)

c) To solve \( 2.73 \div 1.3 \), multiply both numbers by 10:
\( 27.3 \div 13 = 2.1 \)
In simple words: To divide decimals, move the point to make the bottom number a whole number first. If you divide by 100, just move the point two places left.

Exam Tip: Always make the divisor a whole number before starting long division to keep your decimal placement correct.

Question 3. What should be added to 4.187 to get 9?
Answer: Let the required number to be added be \( x \).
We can write the equation as:
\( 4.187 + x = 9 \)
To find \( x \), subtract \( 4.187 \) from \( 9 \):
\( x = 9.000 - 4.187 = 4.813 \)
So, the number is \( 4.813 \).
In simple words: Subtract the smaller decimal from the whole number 9 to find the missing value.

Exam Tip: Put extra zeros after the decimal point in the whole number (like writing 9 as 9.000) to keep your columns aligned for subtraction.

Question 4. Convert 75m into cm.
Answer: We know that one meter is equal to 100 centimeters:
\( 1\text{ m} = 100\text{ cm} \)
To convert meters into centimeters, multiply the value by 100:
\( 75\text{ m} = 75 \times 100\text{ cm} = 7500\text{ cm} \)
So, \( 75\text{ m} \) is equal to \( 7500\text{ cm} \).
In simple words: Multiply the number of meters by 100 to get the answer in centimeters.

Exam Tip: Remember the basic rule: multiply by 100 to go from meters to centimeters, and divide by 100 to go from centimeters to meters.

Question 5. What is the perimeter of a rectangle with length 0.42m & breadth 31.5cm?
Answer: First, convert the length and breadth into the same units.
Let us convert the length into centimeters:
\( \text{Length} = 0.42\text{ m} = 0.42 \times 100\text{ cm} = 42\text{ cm} \)
Now, the breadth is \( 31.5\text{ cm} \).

We can find the perimeter of the rectangle using the formula:
\( \text{Perimeter} = 2 \times (\text{Length} + \text{Breadth}) \)
\( \text{Perimeter} = 2 \times (42\text{ cm} + 31.5\text{ cm}) = 2 \times 73.5\text{ cm} = 147\text{ cm} \)

In meters, this is:
\( 147\text{ cm} = \frac{147}{100}\text{ m} = 1.47\text{ m} \)
In simple words: Make sure the units are the same before calculating. Convert meters to centimeters, add the sides, and multiply by 2.

Exam Tip: Never add measurements with different units. Always convert them to match first, or you will get the wrong answer.

Question 6. Mrs. Sunita uses 3.204 litres of oil to make 9 dishes of equal proportion. How much oil was used for each dish?
Answer: We can find the quantity of oil used for each dish by dividing the total oil by the total number of dishes:
\( \text{Oil used for each dish} = \frac{3.204\text{ litres}}{9} \)
Let us perform the division:
\( 3.204 \div 9 = 0.356\text{ litres} \)
So, \( 0.356\text{ litres} \) of oil was used for each dish.
In simple words: Divide the total amount of oil by 9 to find the share for one dish.

Exam Tip: Write down your division steps carefully and align the decimal point in the quotient directly above the dividend.

Question 7. By what number 0.014 be multiplied to get 1.4?
Answer: Let the unknown multiplier be \( y \).
We can write this as:
\( 0.014 \times y = 1.4 \)
To find \( y \), divide the product by \( 0.014 \):
\( y = \frac{1.4}{0.014} \)
Multiply both the top and bottom by 1000 to clear the decimals:
\( y = \frac{1400}{14} = 100 \)
So, the required multiplier is 100.
In simple words: To change \( 0.014 \) to \( 1.4 \), we need to move the decimal point two places to the right, which means multiplying by 100.

Exam Tip: Count the number of places the decimal point shifts. A shift of two places to the right always means multiplying by 100.

Question 8. 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
Answer:
To write these in exponential form, count the number of times each base is multiplied:
a) \( m \) is multiplied 2 times:
\( m \times m = m^2 \)

b) 4 is multiplied 2 times, and \( y \) is multiplied 3 times:
\( 4 \times 4 \times y \times y \times y = 4^2 \times y^3 = 4^2y^3 \)

c) Group the identical numbers together and write them in ascending order of their bases:
2 is multiplied 2 times, 3 is multiplied 3 times, and 5 is multiplied 2 times:
\( 2^2 \times 3^3 \times 5^2 \)
In simple words: Count how many times each letter or number appears, then write that count as a power at the top.

Exam Tip: When writing exponential forms with multiple bases, arrange the terms in ascending order of their base numbers (like 2, then 3, then 5).

Question 9. Find the value of:
a) 2² x 3²
b) 6² x 10³
c) (-3) x (-2)⁴
Answer:
Let us work out the value of each term first:
a) \( 2^2 = 4 \) and \( 3^2 = 9 \)
\( 4 \times 9 = 36 \)

b) \( 6^2 = 36 \) and \( 10^3 = 1000 \)
\( 36 \times 1000 = 36000 \)

c) Here, \( (-2)^4 = (-2) \times (-2) \times (-2) \times (-2) = 16 \) (a negative base with an even exponent becomes positive).
Now multiply:
\( -3 \times 16 = -48 \)
In simple words: Calculate the power values first. Then, multiply the resulting numbers to get the final answer.

Exam Tip: Remember that any negative base raised to an even power becomes positive, whereas with an odd power it remains negative.

Question 10. Find the value of:
a) (5x)³ , where x = -2/5
b) (-ab)² , where a=2 and b=-1
Answer:
a) Substitute the value of \( x \) into the expression:
\( 5x = 5 \times \left(-\frac{2}{5}\right) = -2 \)
Now find the cube of \( -2 \):
\( (5x)^3 = (-2)^3 = -2 \times -2 \times -2 = -8 \)

b) First, substitute the values of \( a \) and \( b \) to find \( ab \):
\( ab = 2 \times (-1) = -2 \)
Now find \( -ab \):
\( -ab = -(-2) = 2 \)
Now find the square of this value:
\( (-ab)^2 = (2)^2 = 4 \)
In simple words: Replace the letters with their given numbers, simplify what is inside the bracket, and then calculate the power.

Exam Tip: Be extra careful with signs when substituting negative values. Two negative signs next to each other always become positive.