RBSE Solutions Class 6 Maths Chapter 5 Fractions Exercise 5.5

Question 1. Solve the following.
(i) \( \frac { 6 }{ 9 } - \frac { 2 }{ 9 } \)
(ii) \( \frac { 4 }{ 5 } - \frac { 3 }{ 7 } \)
(iii) \( 5 \frac { 1 }{ 2 } - 2 \frac { 1 }{ 5 } \)
(iv) \( 8 \frac { 1 }{ 4 } - 2 \frac { 2 }{ 6 } \)
(v) \( \frac { 17 }{ 6 } - \frac { 9 }{ 4 } \)
(vi) \( \frac { 3 }{ 4 } - \left( \frac { 2 }{ 5 } + \frac { 1 }{ 4 } \right) \)
Answer:
(i) \( \frac { 6 }{ 9 } - \frac { 2 }{ 9 } = \frac { 6 - 2 }{ 9 } = \frac { 4 }{ 9 } \)
(ii) We find the Least Common Multiple (L.C.M.) of the denominators 5 and 7, which is 35.
\( \frac { 4 }{ 5 } - \frac { 3 }{ 7 } = \frac { 4 \times 7 - 3 \times 5 }{ 35 } = \frac { 28 - 15 }{ 35 } = \frac { 13 }{ 35 } \)
(iii) First, convert the mixed fractions into improper fractions.
\( 5 \frac { 1 }{ 2 } = \frac { 5 \times 2 + 1 }{ 2 } = \frac { 11 }{ 2 } \)
\( 2 \frac { 1 }{ 5 } = \frac { 2 \times 5 + 1 }{ 5 } = \frac { 11 }{ 5 } \)
The L.C.M. of 2 and 5 is 10.
\( \frac { 11 }{ 2 } - \frac { 11 }{ 5 } = \frac { 11 \times 5 - 11 \times 2 }{ 10 } = \frac { 55 - 22 }{ 10 } = \frac { 33 }{ 10 } \)
(iv) Convert mixed fractions to improper fractions.
\( 8 \frac { 1 }{ 4 } = \frac { 8 \times 4 + 1 }{ 4 } = \frac { 33 }{ 4 } \)
\( 2 \frac { 2 }{ 6 } = \frac { 2 \times 6 + 2 }{ 6 } = \frac { 14 }{ 6 } \)
The L.C.M. of 4 and 6 is 12.
\( \frac { 33 }{ 4 } - \frac { 14 }{ 6 } = \frac { 33 \times 3 - 14 \times 2 }{ 12 } = \frac { 99 - 28 }{ 12 } = \frac { 71 }{ 12 } \)
(v) The L.C.M. of 6 and 4 is 12.
\( \frac { 17 }{ 6 } - \frac { 9 }{ 4 } = \frac { 17 \times 2 - 9 \times 3 }{ 12 } = \frac { 34 - 27 }{ 12 } = \frac { 7 }{ 12 } \)
(vi) First, solve the expression inside the bracket.
\( \frac { 2 }{ 5 } + \frac { 1 }{ 4 } \)
The L.C.M. of 5 and 4 is 20.
\( \frac { 2 \times 4 + 1 \times 5 }{ 20 } = \frac { 8 + 5 }{ 20 } = \frac { 13 }{ 20 } \)
Now, subtract this from \( \frac { 3 }{ 4 } \). The L.C.M. of 4 and 20 is 20.
\( \frac { 3 }{ 4 } - \frac { 13 }{ 20 } = \frac { 3 \times 5 - 13 \times 1 }{ 20 } = \frac { 15 - 13 }{ 20 } = \frac { 2 }{ 20 } = \frac { 1 }{ 10 } \)
In simple words: To solve these, find a common bottom number (L.C.M.) for fractions, then change the top numbers accordingly before adding or subtracting. Simplify the answer if possible.

🎯 Exam Tip: Always convert mixed fractions to improper fractions before performing addition or subtraction, and remember to find the L.C.M. of denominators for unlike fractions.

Question 2. Heera gave \( \frac { 1 }{ 4 } \) litre milk to Bhavna out of her \( \frac { 3 }{ 7 } \) litre milk. How much milk is left with her?
Answer: Heera started with \( \frac { 3 }{ 7 } \) litre of milk. She gave away \( \frac { 1 }{ 4 } \) litre. To find out how much milk is left, we need to subtract the amount given from the starting amount. We find the L.C.M. of 7 and 4, which is 28.
\( \frac { 3 }{ 7 } - \frac { 1 }{ 4 } = \frac { 3 \times 4 - 1 \times 7 }{ 28 } \)
\( = \frac { 12 - 7 }{ 28 } = \frac { 5 }{ 28 } \) litre
So, \( \frac { 5 }{ 28 } \) litre of milk is left with Heera. Finding common denominators is key to solving these problems.
In simple words: Heera had some milk, and she gave a part of it away. To know what is left, subtract the given amount from the total.

🎯 Exam Tip: When subtracting fractions, always ensure you have a common denominator before subtracting the numerators. Double-check your L.C.M. calculation.

Question 3. A wooden piece is \( \frac { 9 }{ 10 } \) m long and a \( \frac { 2 }{ 5 } \) m long piece has been cut from it. What is the length of the remaining piece?
Answer: The total length of the wooden piece is \( \frac { 9 }{ 10 } \) m. A smaller piece of \( \frac { 2 }{ 5 } \) m was cut from it. To find the remaining length, we subtract the cut piece's length from the total length. The L.C.M. of 10 and 5 is 10.
Length of remaining piece \( = \frac { 9 }{ 10 } - \frac { 2 }{ 5 } \)
\( = \frac { 9 \times 1 - 2 \times 2 }{ 10 } \)
\( = \frac { 9 - 4 }{ 10 } = \frac { 5 }{ 10 } = \frac { 1 }{ 2 } \) m
So, the remaining piece of wood is \( \frac { 1 }{ 2 } \) m long. This is like cutting a part of a rope and measuring what is left.
In simple words: Subtract the length of the piece cut off from the total length of the wood. Make sure the bottom numbers of the fractions are the same before you subtract.

🎯 Exam Tip: Always make sure to simplify the final fraction to its lowest terms. Clearly state the original length, the cut length, and the calculation for the remaining length.

Question 4. Anshul drank \( \frac { 2 }{ 3 } \) of one glass of water. Find out how much water is left in the glass?
Answer: Assume the total amount of water in the glass is 1 whole glass. Anshul drank \( \frac { 2 }{ 3 } \) of the water. To find the remaining water, we subtract the amount drunk from the total amount. Thinking of '1' as '3/3' helps in subtraction.
Remaining water \( = 1 - \frac { 2 }{ 3 } \)
\( = \frac { 3 }{ 3 } - \frac { 2 }{ 3 } = \frac { 3 - 2 }{ 3 } = \frac { 1 }{ 3 } \) part
So, \( \frac { 1 }{ 3 } \) of the glass of water is left. This shows that the parts of a whole always add up to one.
In simple words: Imagine the glass is full (1 whole). If Anshul drank two-thirds, then one-third of the water is left.

🎯 Exam Tip: When dealing with "parts of a whole," remember that the whole can be represented as 1 or as a fraction where the numerator and denominator are the same (e.g., \( \frac {3}{3} \)).

Question 5. Sunil purchased \( 5 \frac { 1 }{ 2 } \) kg and Vijay purchased \( 3 \frac { 4 }{ 5 } \) kg mangoes. Find how much more mangoes did Sunil purchase.
Answer: First, convert the mixed fractions into improper fractions for easier calculation. Sunil purchased \( 5 \frac { 1 }{ 2 } \) kg of mangoes, which is \( \frac { 11 }{ 2 } \) kg. Vijay purchased \( 3 \frac { 4 }{ 5 } \) kg of mangoes, which is \( \frac { 19 }{ 5 } \) kg. To find how many more mangoes Sunil purchased, we subtract Vijay's amount from Sunil's amount. The L.C.M. of 2 and 5 is 10.
Sunil purchased more mangoes \( = \frac { 11 }{ 2 } - \frac { 19 }{ 5 } \)
\( = \frac { 11 \times 5 - 19 \times 2 }{ 10 } \)
\( = \frac { 55 - 38 }{ 10 } = \frac { 17 }{ 10 } \) kg
So, Sunil purchased \( \frac { 17 }{ 10 } \) kg more mangoes than Vijay. This helps us compare quantities easily.
In simple words: Change the mixed numbers into fractions. Then, subtract the smaller amount (Vijay's mangoes) from the larger amount (Sunil's mangoes) after finding a common bottom number.

🎯 Exam Tip: Always show the conversion of mixed fractions to improper fractions and the steps to find a common denominator clearly. This makes your work easy to follow.

Question 6. Geeta finished a race in \( \frac { 13 }{ 4 } \) minutes. Neha finished it in \( \frac { 7 }{ 2 } \) minutes. Who took less time and by how much?
Answer: Geeta finished the race in \( \frac { 13 }{ 4 } \) minutes. Neha finished it in \( \frac { 7 }{ 2 } \) minutes. To compare their times, we need to make the denominators of their fractions the same. The L.C.M. of 2 and 4 is 4. So, we convert Neha's time to have a denominator of 4.
Neha's time \( = \frac { 7 }{ 2 } = \frac { 7 \times 2 }{ 2 \times 2 } = \frac { 14 }{ 4 } \) minutes.
Now we compare: Geeta took \( \frac { 13 }{ 4 } \) minutes, and Neha took \( \frac { 14 }{ 4 } \) minutes. Since \( 13 < 14 \), Geeta took less time than Neha. To find out by how much, we subtract Geeta's time from Neha's time.
Difference in time \( = \frac { 14 }{ 4 } - \frac { 13 }{ 4 } \)
\( = \frac { 14 - 13 }{ 4 } = \frac { 1 }{ 4 } \) minute.
Geeta took \( \frac { 1 }{ 4 } \) minute less time. This demonstrates how comparing fractions helps us understand differences.
In simple words: First, make both race times have the same bottom number. Then, compare the top numbers to see who finished faster. Subtract their times to find the difference.

🎯 Exam Tip: When comparing fractions, convert them to equivalent fractions with a common denominator. The fraction with the smaller numerator will be the smaller value.

Question 7. Complete the following addition and subtraction tables.
Answer: The completed tables are as follows:
(i) Addition Table

(ii) Subtraction Table

In simple words: To fill an addition table, add the fraction from the row header to the fraction from the column header. To fill a subtraction table, subtract the column header fraction from the row header fraction. Always find the lowest common denominator first.
🎯 Exam Tip: For tables involving fractions, remember to simplify each resulting fraction to its simplest form. Organize your work clearly to avoid errors in calculation.