GSEB Class 6 Maths Solutions Chapter 7 Fractions Exercise 7.5

Question 1. Write these fractions appropriately as additions or subtractions:
Answer:
(a) The figures show an addition of fractions. The first figure represents \( \frac { 1 }{ 5 } \), the second represents \( \frac { 2 }{ 5 } \), and their sum is \( \frac { 3 }{ 5 } \). So, the equation is \( \frac { 1 }{ 5 } + \frac { 2 }{ 5 } = \frac { 3 }{ 5 } \).
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(b) These figures represent a subtraction. A full circle (1) with five segments has three of its five segments removed, leaving two segments. So, the equation is \( 1 - \frac { 3 }{ 5 } = \frac { 2 }{ 5 } \).
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(c) The figures show an addition of fractions. The first figure represents \( \frac { 2 }{ 6 } \), the second represents \( \frac { 3 }{ 6 } \), and their sum is \( \frac { 5 }{ 6 } \). So, the equation is \( \frac { 2 }{ 6 } + \frac { 3 }{ 6 } = \frac { 5 }{ 6 } \).
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In simple words: Look at the pictures to see how many parts are shaded in each. Then, write the fraction that shows those shaded parts. Add them together or take them away as the picture tells you, and find the total.

Exam Tip: When dealing with pictorial representations of fractions, always count the total number of equal parts and the number of shaded parts to correctly identify the fraction.

Question 2. Solve:
(a) \( \frac { 1 }{ 18 } + \frac { 1 }{ 18 } \)
(b) \( \frac { 8 }{ 15 } + \frac { 3 }{ 15 } \)
(c) \( \frac { 7 }{ 7 } - \frac { 5 }{ 7 } \)
(d) \( \frac { 1 }{ 22 } + \frac { 21 }{ 22 } \)
(e) \( \frac { 12 }{ 15 } - \frac { 7 }{ 15 } \)
(f) \( \frac { 5 }{ 8 } + \frac { 3 }{ 8 } \)
(g) \( 1 - \frac { 2 }{ 3 } \)
(h) \( \frac { 1 }{ 4 } + \frac { 0 }{ 4 } \)
(i) \( 3 - \frac { 12 }{ 5 } \)
Answer:
(a) \( \frac { 1 }{ 18 } + \frac { 1 }{ 18 } = \frac { 1+1 }{ 18 } = \frac { 2 }{ 18 } = \frac { 1 }{ 9 } \)
(b) \( \frac { 8 }{ 15 } + \frac { 3 }{ 15 } = \frac { 8+3 }{ 15 } = \frac { 11 }{ 15 } \)
(c) \( \frac { 7 }{ 7 } - \frac { 5 }{ 7 } = \frac { 7-5 }{ 7 } = \frac { 2 }{ 7 } \)
(d) \( \frac { 1 }{ 22 } + \frac { 21 }{ 22 } = \frac { 1+21 }{ 22 } = \frac { 22 }{ 22 } = 1 \)
(e) \( \frac { 12 }{ 15 } - \frac { 7 }{ 15 } = \frac { 12-7 }{ 15 } = \frac { 5 }{ 15 } = \frac { 1 }{ 3 } \)
(f) \( \frac { 5 }{ 8 } + \frac { 3 }{ 8 } = \frac { 5+3 }{ 8 } = \frac { 8 }{ 8 } = 1 \)
(g) Since 1 can be written as \( \frac { 3 }{ 3 } \), we have \( 1 - \frac { 2 }{ 3 } = \frac { 3 }{ 3 } - \frac { 2 }{ 3 } = \frac { 3-2 }{ 3 } = \frac { 1 }{ 3 } \)
(h) \( \frac { 1 }{ 4 } + \frac { 0 }{ 4 } = \frac { 1+0 }{ 4 } = \frac { 1 }{ 4 } \)
(i) Since 3 can be written as \( \frac { 3 }{ 1 } \), to get a denominator of 5, we multiply the numerator and denominator by 5: \( \frac { 3 }{ 1 } \times \frac { 5 }{ 5 } = \frac { 15 }{ 5 } \). Therefore, \( 3 - \frac { 12 }{ 5 } = \frac { 15 }{ 5 } - \frac { 12 }{ 5 } = \frac { 15-12 }{ 5 } = \frac { 3 }{ 5 } \).
In simple words: When you add or subtract fractions that have the same bottom number (denominator), you just add or subtract the top numbers (numerators) and keep the bottom number the same. If the bottom numbers are different, you first make them the same.

Exam Tip: Remember to always simplify your final fraction to its lowest terms by dividing both the numerator and denominator by their greatest common divisor.

Question 3. Shuhham painted \( \frac { 2 }{ 3 } \) of the wall space in his room. His sister Madhavi helped and painted \( \frac { 1 }{ 3 } \) of the wall space. How much did they paint together?
Answer: The portion of the wall Shubham painted was \( \frac { 2 }{ 3 } \). The portion of the wall Madhavi painted was \( \frac { 1 }{ 3 } \). To find out how much they painted jointly, we add their portions: \( \frac { 2 }{ 3 } + \frac { 1 }{ 3 } = \frac { 2+1 }{ 3 } = \frac { 3 }{ 3 } \) or 1. Thus, Shubham and Madhavi together painted the complete wall.
In simple words: Shubham painted two-thirds of the wall, and Madhavi painted one-third. If you add these parts together, they painted the whole wall.

Exam Tip: When adding fractions with the same denominator, only add the numerators. If the sum is \( \frac{3}{3} \), \( \frac{4}{4} \), etc., always simplify it to 1, meaning the whole task is complete.

Question 4. Fill in the missing fractions.
(a) \( \frac { 7 }{ 10 } - \boxed{ \phantom{X} } = \frac { 3 }{ 10 } \)
(b) \( \boxed{ \phantom{X} } - \frac { 3 }{ 21 } = \frac { 5 }{ 21 } \)
(c) \( \boxed{ \phantom{X} } - \frac { 3 }{ 6 } = \frac { 3 }{ 6 } \)
(d) \( \boxed{ \phantom{X} } + \frac { 5 }{ 27 } = \frac { 12 }{ 27 } \)
Answer:
(a) The missing fraction is less than \( \frac { 7 }{ 10 } \) by \( \frac { 3 }{ 10 } \). So, the missing fraction is \( \frac { 7 }{ 10 } - \frac { 3 }{ 10 } = \frac { 7-3 }{ 10 } = \frac { 4 }{ 10 } \) or \( \frac { 2 }{ 5 } \).
So, \( \frac { 7 }{ 10 } - \frac { 4 }{ 10 } = \frac { 3 }{ 10 } \).
(b) The missing fraction is more than \( \frac { 3 }{ 21 } \) by \( \frac { 5 }{ 21 } \). The sum of \( \frac { 3 }{ 21 } \) and \( \frac { 5 }{ 21 } \) must equal the missing fraction. So, the missing fraction is \( \frac { 3 }{ 21 } + \frac { 5 }{ 21 } = \frac { 3+5 }{ 21 } = \frac { 8 }{ 21 } \).
So, \( \frac { 8 }{ 21 } - \frac { 3 }{ 21 } = \frac { 5 }{ 21 } \).
(c) The missing fraction is more than \( \frac { 3 }{ 6 } \) by \( \frac { 3 }{ 6 } \). The sum of \( \frac { 3 }{ 6 } \) and \( \frac { 3 }{ 6 } \) equals the missing fraction. So, the missing fraction is \( \frac { 3 }{ 6 } + \frac { 3 }{ 6 } = \frac { 3+3 }{ 6 } = \frac { 6 }{ 6 } = 1 \).
So, \( \frac { 6 }{ 6 } - \frac { 3 }{ 6 } = \frac { 3 }{ 6 } \).
(d) The missing fraction is found by subtracting \( \frac { 5 }{ 27 } \) from \( \frac { 12 }{ 27 } \). So, the missing fraction is \( \frac { 12 }{ 27 } - \frac { 5 }{ 27 } = \frac { 12-5 }{ 27 } = \frac { 7 }{ 27 } \).
So, \( \frac { 7 }{ 27 } + \frac { 5 }{ 27 } = \frac { 12 }{ 27 } \).
In simple words: To find the missing piece in a fraction problem, think about whether you need to add or subtract the known fractions. If you have a total and one part, subtract to find the other part. If you have two parts that add up to a missing total, add them.

Exam Tip: Treat missing fraction problems like simple algebra. For example, if \( A - X = B \), then \( X = A - B \). If \( X - A = B \), then \( X = A + B \).

Question 5. Javed was given \( \frac { 5 }{ 7 } \) of a basket of oranges. What fraction of oranges was left in the basket?
Answer: Let the entire basket of oranges be represented by 1. The portion of oranges given to Javed was \( \frac { 5 }{ 7 } \). To find the portion of oranges remaining in the basket, we subtract the given portion from the total: \( 1 - \frac { 5 }{ 7 } \). Since 1 can be expressed as \( \frac { 7 }{ 7 } \) (as they are equivalent fractions), we have \( \frac { 7 }{ 7 } - \frac { 5 }{ 7 } = \frac { 7-5 }{ 7 } = \frac { 2 }{ 7 } \). Therefore, \( \frac { 2 }{ 7 } \) of the oranges were left in the basket.
In simple words: If Javed took 5 out of 7 parts of the oranges, then 2 parts out of 7 were still left.

Exam Tip: When a problem refers to a "whole" or "entire" quantity, represent it as the fraction \( \frac{n}{n} \) where n is the denominator of the other fraction in the problem, to make subtraction or addition easier.