GSEB Class 6 Maths Solutions Chapter 7 Fractions Exercise 7.6

 

Question 1. Solve the following fraction problems:
(a) \( \frac { 2 }{ 3 } + \frac { 1 }{ 7 } \)
(b) \( \frac { 3 }{ 10 } + \frac { 7 }{ 15 } \)
(c) \( \frac { 4 }{ 9 } + \frac { 2 }{ 7 } \)
(d) \( \frac { 5 }{ 7 } + \frac { 1 }{ 3 } \)
(e) \( \frac { 2 }{ 5 } + \frac { 1 }{ 6 } \)
(f) \( \frac { 4 }{ 5 } + \frac { 2 }{ 3 } \)
(g) \( \frac { 3 }{ 4 } – \frac { 1 }{ 3 } \)
(h) \( \frac { 5 }{ 6 } – \frac { 1 }{ 3 } \)
(i) \( \frac { 2 }{ 3 } + \frac { 3 }{ 4 } + \frac { 1 }{ 2 } \)
(j) \( \frac { 1 }{ 2 } + \frac { 1 }{ 3 } + \frac { 1 }{ 6 } \)
(k) \( 1\frac {1}{3} + 3\frac { 2 }{ 3 } \)
(l) \( 4\frac { 2 }{ 3 } + 3\frac { 1 }{ 4 } \)
(m) \( \frac { 16 }{ 5 } - \frac { 7 }{ 5 } \)
(n) \( \frac { 4 }{ 3 } – \frac { 1 }{ 2 } \)
Answer:
(a) For \( \frac { 2 }{ 3 } + \frac { 1 }{ 7 } \):
To add these, we need a common denominator. The LCM of 3 and 7 is 21.
We convert each fraction:
\( \frac { 2 }{ 3 } = \frac{2 \times 7}{3 \times 7} = \frac { 14 }{ 21 } \)
\( \frac { 1 }{ 7 } = \frac{1 \times 3}{7 \times 3} = \frac { 3 }{ 21 } \)
Now, we add them:
\( \frac { 2 }{ 3 } + \frac { 1 }{ 7 } = \frac { 14 }{ 21 } + \frac { 3 }{ 21 } = \frac { 14 + 3 }{ 21 } = \frac { 17 }{ 21 } \)
(b) For \( \frac { 3 }{ 10 } + \frac { 7 }{ 15 } \):
To add these fractions, we find a common denominator. The LCM of 10 and 15 is 30.
We convert each fraction to have this denominator:
\( \frac { 3 }{ 10 } = \frac{3 \times 3}{10 \times 3} = \frac { 9 }{ 30 } \)
\( \frac { 7 }{ 15 } = \frac{7 \times 2}{15 \times 2} = \frac { 14 }{ 30 } \)
Now, we add the converted fractions:
\( \frac { 3 }{ 10 } + \frac { 7 }{ 15 } = \frac { 9 }{ 30 } + \frac { 14 }{ 30 } = \frac { 9 + 14 }{ 30 } = \frac { 23 }{ 30 } \)
(c) For \( \frac { 4 }{ 9 } + \frac { 2 }{ 7 } \):
To add these, we need a common denominator. The LCM of 9 and 7 is 63.
We convert each fraction:
\( \frac { 4 }{ 9 } = \frac{4 \times 7}{9 \times 7} = \frac { 28 }{ 63 } \)
\( \frac { 2 }{ 7 } = \frac{2 \times 9}{7 \times 9} = \frac { 18 }{ 63 } \)
Now, we add them together:
\( \frac { 4 }{ 9 } + \frac { 2 }{ 7 } = \frac { 28 }{ 63 } + \frac { 18 }{ 63 } = \frac { 28 + 18 }{ 63 } = \frac { 46 }{ 63 } \)
(d) For \( \frac { 5 }{ 7 } + \frac { 1 }{ 3 } \):
To add these fractions, we find a common denominator. The LCM of 7 and 3 is 21.
We convert each fraction to have this denominator:
\( \frac { 5 }{ 7 } = \frac{5 \times 3}{7 \times 3} = \frac { 15 }{ 21 } \)
\( \frac { 1 }{ 3 } = \frac{1 \times 7}{3 \times 7} = \frac { 7 }{ 21 } \)
Now, we add the converted fractions:
\( \frac { 5 }{ 7 } + \frac { 1 }{ 3 } = \frac { 15 }{ 21 } + \frac { 7 }{ 21 } = \frac { 15 + 7 }{ 21 } = \frac { 22 }{ 21 } \)
This can also be written as a mixed number: \( 1\frac { 1 }{ 21 } \).
(e) For \( \frac { 2 }{ 5 } + \frac { 1 }{ 6 } \):
To add these fractions, we find a common denominator. The LCM of 5 and 6 is 30.
We convert each fraction to have this denominator:
\( \frac { 2 }{ 5 } = \frac{2 \times 6}{5 \times 6} = \frac { 12 }{ 30 } \)
\( \frac { 1 }{ 6 } = \frac{1 \times 5}{6 \times 5} = \frac { 5 }{ 30 } \)
Now, we add the converted fractions:
\( \frac { 2 }{ 5 } + \frac { 1 }{ 6 } = \frac { 12 }{ 30 } + \frac { 5 }{ 30 } = \frac { 12 + 5 }{ 30 } = \frac { 17 }{ 30 } \)
(f) For \( \frac { 4 }{ 5 } + \frac { 2 }{ 3 } \):
To add these, we need a common denominator. The LCM of 5 and 3 is 15.
We convert each fraction:
\( \frac { 4 }{ 5 } = \frac{4 \times 3}{5 \times 3} = \frac { 12 }{ 15 } \)
\( \frac { 2 }{ 3 } = \frac{2 \times 5}{3 \times 5} = \frac { 10 }{ 15 } \)
Now, we add them:
\( \frac { 4 }{ 5 } + \frac { 2 }{ 3 } = \frac { 12 }{ 15 } + \frac { 10 }{ 15 } = \frac { 22 }{ 15 } \)
This can also be written as a mixed number: \( 1\frac { 7 }{ 15 } \).
(g) For \( \frac { 3 }{ 4 } – \frac { 1 }{ 3 } \):
To subtract these, we need a common denominator. The LCM of 4 and 3 is 12.
We convert each fraction:
\( \frac { 3 }{ 4 } = \frac{3 \times 3}{4 \times 3} = \frac { 9 }{ 12 } \)
\( \frac { 1 }{ 3 } = \frac{1 \times 4}{3 \times 4} = \frac { 4 }{ 12 } \)
Now, we subtract them:
\( \frac { 3 }{ 4 } – \frac { 1 }{ 3 } = \frac { 9 }{ 12 } – \frac { 4 }{ 12 } = \frac { 9 – 4 }{ 12 } = \frac { 5 }{ 12 } \)
(h) For \( \frac { 5 }{ 6 } - \frac { 1 }{ 3 } \):
To subtract these, we need a common denominator. The LCM of 6 and 3 is 6.
We convert the second fraction:
\( \frac { 1 }{ 3 } = \frac{1 \times 2}{3 \times 2} = \frac { 2 }{ 6 } \)
Now, we subtract them:
\( \frac { 5 }{ 6 } - \frac { 1 }{ 3 } = \frac { 5 }{ 6 } – \frac { 2 }{ 6 } = \frac { 3 }{ 6 } \)
This fraction can be simplified to \( \frac { 1 }{ 2 } \).
(i) For \( \frac { 2 }{ 3 } + \frac { 3 }{ 4 } + \frac { 1 }{ 2 } \):
To add these fractions, we find a common denominator. The LCM of 2, 3, and 4 is 12.
We convert each fraction to have this denominator:
\( \frac { 2 }{ 3 } = \frac{2 \times 4}{3 \times 4} = \frac { 8 }{ 12 } \)
\( \frac { 3 }{ 4 } = \frac{3 \times 3}{4 \times 3} = \frac { 9 }{ 12 } \)
\( \frac { 1 }{ 2 } = \frac{1 \times 6}{2 \times 6} = \frac { 6 }{ 12 } \)
Now, we add the converted fractions:
\( \frac { 2 }{ 3 } + \frac { 3 }{ 4 } + \frac { 1 }{ 2 } = \frac { 8 }{ 12 } + \frac { 9 }{ 12 } + \frac { 6 }{ 12 } = \frac{8+9+6}{12} = \frac { 23 }{ 12 } \)
(j) For \( \frac { 1 }{ 2 } + \frac { 1 }{ 3 } + \frac { 1 }{ 6 } \):
To add these fractions, we find a common denominator. The LCM of 2, 3, and 6 is 6.
We convert each fraction to have this denominator:
\( \frac { 1 }{ 2 } = \frac{1 \times 3}{2 \times 3} = \frac { 3 }{ 6 } \)
\( \frac { 1 }{ 3 } = \frac{1 \times 2}{3 \times 2} = \frac { 2 }{ 6 } \)
The third fraction \( \frac { 1 }{ 6 } \) already has the common denominator.
Now, we add the converted fractions:
\( \frac { 1 }{ 2 } + \frac { 1 }{ 3 } + \frac { 1 }{ 6 } = \frac { 3 }{ 6 } + \frac { 2 }{ 6 } + \frac { 1 }{ 6 } = \frac {3+2+1}{6} = \frac { 6 }{ 6 } \)
This fraction simplifies to 1.
(k) For \( 1\frac {1}{3} + 3\frac { 2 }{ 3 } \):
Method I: Convert to improper fractions first.
\( 1\frac { 1 }{ 3 } = \frac { (1 \times 3) + 1 }{ 3 } = \frac { 4 }{ 3 } \)
\( 3\frac { 2 }{ 3 } = \frac { (3 \times 3) + 2 }{ 3 } = \frac { 11 }{ 3 } \)
Now, we add the improper fractions:
\( \frac { 4 }{ 3 } + \frac { 11 }{ 3 } = \frac { 4 + 11 }{ 3 } = \frac { 15 }{ 3 } = 5 \)
Method II: Add whole numbers and fractional parts separately.
\( 1\frac {1}{3} + 3\frac { 2 }{ 3 } = (1 + \frac{1}{3}) + (3 + \frac{2}{3}) \)
Combine the whole numbers and fractions:
\( = (1 + 3) + (\frac{1}{3} + \frac{2}{3}) \)
\( = 4 + (\frac{1+2}{3}) \)
\( = 4 + \frac{3}{3} \)
\( = 4 + 1 = 5 \)
(l) For \( 4\frac { 2 }{ 3 } + 3\frac { 1 }{ 4 } \):
First, convert the mixed numbers to improper fractions.
\( 4\frac { 2 }{ 3 } = \frac { (4 \times 3) + 2 }{ 3 } = \frac { 14 }{ 3 } \)
\( 3\frac { 1 }{ 4 } = \frac { (3 \times 4) + 1 }{ 4 } = \frac { 13 }{ 4 } \)
To add these, we find a common denominator. The LCM of 3 and 4 is 12.
Convert each improper fraction:
\( \frac { 14 }{ 3 } = \frac{14 \times 4}{3 \times 4} = \frac { 56 }{ 12 } \)
\( \frac { 13 }{ 4 } = \frac{13 \times 3}{4 \times 3} = \frac { 39 }{ 12 } \)
Now, we add them:
\( \frac { 14 }{ 3 } + \frac { 13 }{ 4 } = \frac { 56 }{ 12 } + \frac { 39 }{ 12 } = \frac{56+39}{12} = \frac { 95 }{ 12 } \)
This can also be written as a mixed number: \( 7\frac { 11 }{ 12 } \).
(m) For \( \frac { 16 }{ 5 } – \frac { 7 }{ 5 } \):
Since these fractions already have the same denominator, we can simply subtract the numerators.
\( \frac { 16 }{ 5 } – \frac { 7 }{ 5 } = \frac { 16 – 7 }{ 5 } = \frac { 9 }{ 5 } \)
This can also be written as a mixed number: \( 1\frac { 4 }{ 5 } \).
(n) For \( \frac { 4 }{ 3 } – \frac { 1 }{ 2 } \):
To subtract these, we need a common denominator. The LCM of 3 and 2 is 6.
We convert each fraction:
\( \frac { 4 }{ 3 } = \frac{4 \times 2}{3 \times 2} = \frac { 8 }{ 6 } \)
\( \frac { 1 }{ 2 } = \frac{1 \times 3}{2 \times 3} = \frac { 3 }{ 6 } \)
Now, we subtract them:
\( \frac { 4 }{ 3 } – \frac { 1 }{ 2 } = \frac { 8 }{ 6 } – \frac { 3 }{ 6 } = \frac { 8 – 3 }{ 6 } = \frac { 5 }{ 6 } \)
In simple words: To add or subtract fractions, first make sure the bottom numbers (denominators) are the same. Find the smallest common multiple (LCM) of the denominators and change both fractions to use this new common denominator. Then, you can simply add or subtract the top numbers (numerators). If the answer is an improper fraction, you can convert it to a mixed number for clarity.

Exam Tip: Always look for the Least Common Multiple (LCM) of the denominators to make calculations simpler. Remember to convert mixed numbers to improper fractions or deal with whole numbers separately before adding or subtracting.

 

Question 2. Sarita bought \( \frac {2}{5} \) metre of ribbon and Lalita \( \frac { 3 }{ 4 } \) metre of ribbon. What is the total length of the ribbon they bought?
Answer:
Length of ribbon bought by Sarita = \( \frac { 2 }{ 5 } \) metre
Length of ribbon bought by Lalita = \( \frac { 3 }{ 4 } \) metre
Total ribbon bought by them = \( \frac { 2 }{ 5 } + \frac { 3 }{ 4 } \) metre
To add these fractions, we need a common denominator. The LCM of 4 and 5 is 20.
Convert each fraction:
\( \frac { 2 }{ 5 } = \frac{2 \times 4}{5 \times 4} = \frac { 8 }{ 20 } \)
\( \frac { 3 }{ 4 } = \frac{3 \times 5}{4 \times 5} = \frac { 15 }{ 20 } \)
Now, we add the converted fractions:
Total ribbon = \( \frac { 8 }{ 20 } + \frac { 15 }{ 20 } = \frac { 8 + 15 }{ 20 } = \frac { 23 }{ 20 } \) metre
In simple words: To find the total ribbon, you add the two lengths. Make sure the fractions have the same bottom number before adding the top numbers. The result is the total length.

Exam Tip: When solving word problems involving fractions, identify the operation needed (addition, subtraction, multiplication, or division) and then apply the correct fraction rules. Always include units in your final answer.

 

Question 3. Naina was given \( 1\frac {1}{2} \) piece of cake and Najma was given \( 1\frac { 1 }{ 3 } \) piece of cake. Find the total amount of cake given to both of them.
Answer:
Amount of cake given to Naina = \( 1\frac {1}{2} \) piece
Amount of cake given to Najma = \( 1\frac { 1 }{ 3 } \) piece
Total amount of cake given to them = \( 1\frac {1}{2} + 1\frac { 1 }{ 3 } \) piece
First, convert the mixed numbers to improper fractions:
\( 1\frac { 1 }{ 2 } = \frac { (1 \times 2) + 1 }{ 2 } = \frac { 3 }{ 2 } \)
\( 1\frac { 1 }{ 3 } = \frac { (1 \times 3) + 1 }{ 3 } = \frac { 4 }{ 3 } \)
Now, find a common denominator for \( \frac { 3 }{ 2 } \) and \( \frac { 4 }{ 3 } \). The LCM of 2 and 3 is 6.
Convert each improper fraction:
\( \frac { 3 }{ 2 } = \frac{3 \times 3}{2 \times 3} = \frac { 9 }{ 6 } \)
\( \frac { 4 }{ 3 } = \frac{4 \times 2}{3 \times 2} = \frac { 8 }{ 6 } \)
Finally, add the converted fractions:
Total cake = \( \frac { 9 }{ 6 } + \frac { 8 }{ 6 } = \frac { 9 + 8 }{ 6 } = \frac { 17 }{ 6 } \) piece
In simple words: To find the total cake, change the mixed numbers into top-heavy (improper) fractions. Then, find a common bottom number for both fractions, change them, and add the top numbers.

Exam Tip: When adding mixed numbers, you can either convert them to improper fractions first or add the whole numbers and fractional parts separately. Ensure your final answer is simplified or converted back to a mixed number if appropriate.

 

Question 4. Find the missing fraction in each of the following:
(a) \( \Box - \frac { 5 }{ 8 } = \frac { 1 }{ 4 } \)
(b) \( \Box - \frac { 1 }{ 5 } = \frac { 1 }{ 2 } \)
(c) \( \frac { 1 }{ 2 } - \Box = \frac { 1 }{ 6 } \)
Answer:
(a) To find the missing fraction in \( \Box - \frac { 5 }{ 8 } = \frac { 1 }{ 4 } \), we need to add \( \frac { 5 }{ 8 } \) to \( \frac { 1 }{ 4 } \).
Missing fraction \( = \frac { 1 }{ 4 } + \frac { 5 }{ 8 } \)
To add, find the LCM of 4 and 8, which is 8.
Convert \( \frac { 1 }{ 4 } \): \( \frac { 1 }{ 4 } = \frac{1 \times 2}{4 \times 2} = \frac { 2 }{ 8 } \)
Now add:
Missing fraction \( = \frac { 2 }{ 8 } + \frac { 5 }{ 8 } = \frac { 2 + 5 }{ 8 } = \frac { 7 }{ 8 } \)
So, \( \frac { 7 }{ 8 } - \frac { 5 }{ 8 } = \frac { 1 }{ 4 } \).
(b) To find the missing fraction in \( \Box - \frac { 1 }{ 5 } = \frac { 1 }{ 2 } \), we need to add \( \frac { 1 }{ 5 } \) to \( \frac { 1 }{ 2 } \).
Missing fraction \( = \frac { 1 }{ 2 } + \frac { 1 }{ 5 } \)
To add, find the LCM of 2 and 5, which is 10.
Convert each fraction:
\( \frac { 1 }{ 2 } = \frac{1 \times 5}{2 \times 5} = \frac { 5 }{ 10 } \)
\( \frac { 1 }{ 5 } = \frac{1 \times 2}{5 \times 2} = \frac { 2 }{ 10 } \)
Now add:
Missing fraction \( = \frac { 5 }{ 10 } + \frac { 2 }{ 10 } = \frac { 5 + 2 }{ 10 } = \frac { 7 }{ 10 } \)
So, \( \frac { 7 }{ 10 } - \frac { 1 }{ 5 } = \frac { 1 }{ 2 } \).
(c) To find the missing fraction in \( \frac { 1 }{ 2 } - \Box = \frac { 1 }{ 6 } \), we need to subtract \( \frac { 1 }{ 6 } \) from \( \frac { 1 }{ 2 } \).
Missing fraction \( = \frac { 1 }{ 2 } – \frac { 1 }{ 6 } \)
To subtract, find the LCM of 2 and 6, which is 6.
Convert \( \frac { 1 }{ 2 } \): \( \frac { 1 }{ 2 } = \frac{1 \times 3}{2 \times 3} = \frac { 3 }{ 6 } \)
Now subtract:
Missing fraction \( = \frac { 3 }{ 6 } – \frac { 1 }{ 6 } = \frac { 3 – 1 }{ 6 } = \frac { 2 }{ 6 } \)
Simplify the fraction:
Missing fraction \( = \frac { 1 }{ 3 } \)
So, \( \frac { 1 }{ 2 } - \frac { 1 }{ 3 } = \frac { 1 }{ 6 } \).
In simple words: When you have a missing fraction in an equation like "something minus A equals B", you find "something" by adding A and B. If it's "A minus something equals B", you find "something" by subtracting B from A. Always find a common bottom number before adding or subtracting.

Exam Tip: To solve for a missing term in an addition or subtraction equation, remember the inverse operations: if a term is subtracted, add it to both sides; if a term is added, subtract it from both sides. Always find a common denominator before performing operations on fractions.

 

Question 5. Complete the addition-subtraction box.
Answer:
(a) Given the initial numbers in the box, we will calculate the sums and differences to fill it completely.

Calculations:
Row 1 (sum): \( \frac { 2 }{ 3 } + \frac { 4 }{ 3 } = \frac{2+4}{3} = \frac { 6 }{ 3 } = 2 \)
Row 2 (sum): \( \frac { 1 }{ 3 } + \frac { 2 }{ 3 } = \frac{1+2}{3} = \frac { 3 }{ 3 } = 1 \)
Column 1 (difference): \( \frac { 2 }{ 3 } – \frac { 1 }{ 3 } = \frac{2-1}{3} = \frac { 1 }{ 3 } \)
Column 2 (difference): \( \frac { 4 }{ 3 } – \frac { 2 }{ 3 } = \frac{4-2}{3} = \frac { 2 }{ 3 } \)
Resulting filled box:

(b) Given the initial numbers in the box, we will calculate the sums and differences to fill it completely.

Calculations:
Row 1 (sum): \( \frac { 1 }{ 2 } + \frac { 1 }{ 3 } \)
LCM of 2 and 3 is 6. \( \frac{1 \times 3}{2 \times 3} = \frac { 3 }{ 6 } \), \( \frac{1 \times 2}{3 \times 2} = \frac { 2 }{ 6 } \)
\( \frac { 3 }{ 6 } + \frac { 2 }{ 6 } = \frac { 5 }{ 6 } \)
Row 2 (sum): \( \frac { 1 }{ 3 } + \frac { 1 }{ 4 } \)
LCM of 3 and 4 is 12. \( \frac{1 \times 4}{3 \times 4} = \frac { 4 }{ 12 } \), \( \frac{1 \times 3}{4 \times 3} = \frac { 3 }{ 12 } \)
\( \frac { 4 }{ 12 } + \frac { 3 }{ 12 } = \frac { 7 }{ 12 } \)
Column 1 (difference): \( \frac { 1 }{ 2 } – \frac { 1 }{ 3 } \)
LCM of 2 and 3 is 6. \( \frac { 3 }{ 6 } - \frac { 2 }{ 6 } = \frac { 1 }{ 6 } \)
Column 2 (difference): \( \frac { 1 }{ 3 } – \frac { 1 }{ 4 } \)
LCM of 3 and 4 is 12. \( \frac { 4 }{ 12 } - \frac { 3 }{ 12 } = \frac { 1 }{ 12 } \)
Final (sum of row differences OR difference of column sums): \( \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \)
Resulting filled box:

In simple words: For these boxes, you add the numbers in each row to get the total for that row, and you subtract the bottom number from the top number in each column to get the difference for that column. Finally, you perform the operation on the sums or differences to fill the last box. Always make sure fractions have common bottom numbers first.

Exam Tip: For grid-based problems, clearly show your calculations for each cell. Double-check your LCMs and fraction conversions to avoid errors, especially when working with both addition and subtraction in the same problem.

 

Question 6. A piece of wire \( \frac {7}{8} \) metre long broke into two pieces. One-piece was \( \frac { 1 }{ 4 } \) metre long. How long is the other piece?
Answer:
Length of the original piece of wire = \( \frac {7}{8} \) m
Length of one piece = \( \frac { 1 }{ 4 } \) m
To find the length of the other piece, we subtract the length of the known piece from the total length.
Length of the other part = \( \frac {7}{8} - \frac { 1 }{ 4 } \) m
To subtract these fractions, we need a common denominator. The LCM of 8 and 4 is 8.
Convert \( \frac { 1 }{ 4 } \) to an equivalent fraction with a denominator of 8:
\( \frac { 1 }{ 4 } = \frac{1 \times 2}{4 \times 2} = \frac { 2 }{ 8 } \)
Now, subtract:
Length of the other part = \( \frac { 7 }{ 8 } – \frac { 2 }{ 8 } = \frac { 7 – 2 }{ 8 } = \frac { 5 }{ 8 } \) m
Thus, the length of the other part is \( \frac {5 }{ 8 } \) m.
In simple words: You have a total length of wire and you know one part. To find the other part, you simply subtract the known part from the total. Make sure the fractions have the same bottom number before you subtract.

Exam Tip: Word problems involving "breaking into pieces" usually indicate subtraction. Always ensure units are consistent and included in your final answer.

 

Question 7. Nandini's house is \( \frac {9}{ 10 } \) km from her school. She walked some distance and then took a bus for \( \frac { 1 }{ 2 } \) km to reach the school. How far did she walk?
Answer:
Total distance between Nandini's house and school = \( \frac { 9 }{ 10 } \) km
Distance covered by bus = \( \frac { 1 }{ 2 } \) km
To find the distance Nandini walked, we subtract the bus distance from the total distance.
Distance covered by walking = \( \frac { 9 }{ 10 } - \frac { 1 }{ 2 } \) km
To subtract these fractions, we need a common denominator. The LCM of 10 and 2 is 10.
Convert \( \frac { 1 }{ 2 } \) to an equivalent fraction with a denominator of 10:
\( \frac { 1 }{ 2 } = \frac{1 \times 5}{2 \times 5} = \frac { 5 }{ 10 } \)
Now, subtract:
Distance walked = \( \frac { 9 }{ 10 } - \frac { 5 }{ 10 } = \frac { 9 – 5 }{ 10 } = \frac { 4 }{ 10 } \) km
Simplify the fraction:
Distance walked = \( \frac { 2 }{ 5 } \) km
In simple words: To find how far Nandini walked, you take the full distance to school and subtract the part she traveled by bus. Remember to make the bottom numbers of the fractions the same before you subtract.

Exam Tip: In problems where a total distance is covered in parts, subtracting the known parts from the total gives the remaining part. Always simplify fractions to their lowest terms.

 

Question 8. Asha and Samuel have bookshelves of the same size partly filled with books. Asha's shelf is \( \frac {5}{6} \) the full and Samuel's shelf is \( \frac {2}{5} \) the full. Whose bookshelf is more full and by what fraction?
Answer:
Portion of Asha's shelf full of books = \( \frac {5}{6} \)
Portion of Samuel's shelf full of books = \( \frac { 2 }{5} \)
To compare these fractions and find the difference, we need to change them into equivalent fractions that have the same denominator.
Find the LCM of 6 and 5, which is 30.
Convert Asha's fraction:
\( \frac { 5 }{ 6 } = \frac{5 \times 5}{6 \times 5} = \frac { 25 }{ 30 } \)
Convert Samuel's fraction:
\( \frac { 2 }{ 5 } = \frac{2 \times 6}{5 \times 6} = \frac { 12 }{ 30 } \)
By comparing \( \frac { 25 }{ 30 } \) and \( \frac { 12 }{ 30 } \), it is clear that Asha's shelf ( \( \frac { 25 }{ 30 } \) ) is more full.
Now, to find the fraction by which it is more full, we subtract the smaller fraction from the larger one:
Fraction by which Asha's shelf is more full = \( \frac { 25 }{ 30 } – \frac { 12 }{ 30 } = \frac { 25 – 12 }{ 30 } = \frac { 13 }{ 30 } \)
In simple words: To see who has more books and by how much, you need to compare the fractions by giving them the same bottom number. Asha's shelf is more full, and you find the difference by subtracting the smaller fraction from the larger one.

Exam Tip: To compare fractions with different denominators, always find a common denominator (preferably the LCM) to convert them into equivalent fractions. This allows for direct comparison of the numerators.

 

Question 9. Jaidev takes \( 2\frac {1}{5} \) minutes to walk across the school ground. Rahul takes \( \frac { 7 }{ 4 } \) minutes to do the same. Who takes less time and by what fraction?
Answer:
Time taken by Jaidev = \( 2\frac {1}{5} \) minutes
Time taken by Rahul = \( \frac { 7 }{ 4 } \) minutes
To compare these times, first convert Jaidev's mixed number to an improper fraction and then both fractions to have a common denominator.
Jaidev's time in improper fraction: \( 2\frac { 1 }{ 5 } = \frac { (2 \times 5) + 1 }{ 5 } = \frac { 11 }{ 5 } \) minutes
Now we compare \( \frac { 11 }{ 5 } \) and \( \frac { 7 }{ 4 } \). The LCM of 5 and 4 is 20.
Convert both fractions:
Jaidev's time = \( \frac { 11 }{ 5 } = \frac{11 \times 4}{5 \times 4} = \frac { 44 }{ 20 } \) minutes
Rahul's time = \( \frac { 7 }{ 4 } = \frac{7 \times 5}{4 \times 5} = \frac { 35 }{ 20 } \) minutes
By comparing \( \frac { 44 }{ 20 } \) and \( \frac { 35 }{ 20 } \), we see that Rahul's time is less (35 is smaller than 44).
To find by what fraction Rahul takes less time, we subtract Rahul's time from Jaidev's time:
Difference in time = \( \frac { 44 }{ 20 } - \frac { 35 }{ 20 } = \frac { 44 - 35 }{ 20 } = \frac { 9 }{ 20 } \) minutes
Thus, Rahul takes \( \frac {9}{ 20 } \) minutes less time.
In simple words: To find out who is faster and by how much, first change all times to fractions with the same bottom number. Then, you can easily see which number is smaller (faster) and subtract to find the difference.

Exam Tip: When comparing quantities involving mixed numbers and improper fractions, always convert them to a common format (usually improper fractions) and then find a common denominator for accurate comparison and subtraction.