GSEB Class 6 Maths Solutions Chapter 7 અપૂર્ણાંક સંખ્યાઓ Exercise 7.6

 

Question 1. Solve the following:
(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 \( \frac{2}{3} + \frac{1}{7} \):
The least common multiple (LCM) of 3 and 7 is 21.

So, the LCM is \( 3 \times 7 = 21 \).
We convert the given fractions to equivalent fractions with the common denominator of 21:
\( \frac{2}{3} = \frac{2 \times 7}{3 \times 7} = \frac{14}{21} \) and \( \frac{1}{7} = \frac{1 \times 3}{7 \times 3} = \frac{3}{21} \).
Now, we add the fractions: \( \frac{2}{3} + \frac{1}{7} = \frac{14}{21} + \frac{3}{21} = \frac{14+3}{21} = \frac{17}{21} \).

(b) For \( \frac{3}{10} + \frac{7}{15} \):
The least common multiple (LCM) of 10 and 15 is 30.
So, the LCM is \( 2 \times 3 \times 5 = 30 \).
We convert the given fractions to equivalent fractions with the common denominator of 30:
\( \frac{3}{10} = \frac{3 \times 3}{10 \times 3} = \frac{9}{30} \) and \( \frac{7}{15} = \frac{7 \times 2}{15 \times 2} = \frac{14}{30} \).
Now, we add the fractions: \( \frac{3}{10} + \frac{7}{15} = \frac{9}{30} + \frac{14}{30} = \frac{9+14}{30} = \frac{23}{30} \).

(c) For \( \frac{4}{9} + \frac{2}{7} \):
The least common multiple (LCM) of 9 and 7 is 63.
So, the LCM is \( 3 \times 3 \times 7 = 63 \).
We convert the given fractions to equivalent fractions with the common denominator of 63:
\( \frac{4}{9} = \frac{4 \times 7}{9 \times 7} = \frac{28}{63} \) and \( \frac{2}{7} = \frac{2 \times 9}{7 \times 9} = \frac{18}{63} \).
Now, we add the fractions: \( \frac{4}{9} + \frac{2}{7} = \frac{28}{63} + \frac{18}{63} = \frac{28+18}{63} = \frac{46}{63} \).

(d) For \( \frac{5}{7} + \frac{1}{3} \):
The least common multiple (LCM) of 7 and 3 is 21.
We convert the given fractions to equivalent fractions with the common denominator of 21:
\( \frac{5}{7} = \frac{5 \times 3}{7 \times 3} = \frac{15}{21} \) and \( \frac{1}{3} = \frac{1 \times 7}{3 \times 7} = \frac{7}{21} \).
Now, we add the 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 fraction: \( 1\frac{1}{21} \).

(e) For \( \frac{2}{5} + \frac{1}{6} \):
The least common multiple (LCM) of 5 and 6 is 30.
We convert the given fractions to equivalent fractions with the common denominator of 30:
\( \frac{2}{5} = \frac{2 \times 6}{5 \times 6} = \frac{12}{30} \) and \( \frac{1}{6} = \frac{1 \times 5}{6 \times 5} = \frac{5}{30} \).
Now, we add the fractions: \( \frac{2}{5} + \frac{1}{6} = \frac{12}{30} + \frac{5}{30} = \frac{12+5}{30} = \frac{17}{30} \).

(f) For \( \frac{4}{5} + \frac{2}{3} \):
The least common multiple (LCM) of 5 and 3 is 15.
We convert the given fractions to equivalent fractions with the common denominator of 15:
\( \frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15} \) and \( \frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15} \).
Now, we add the fractions: \( \frac{4}{5} + \frac{2}{3} = \frac{12}{15} + \frac{10}{15} = \frac{12+10}{15} = \frac{22}{15} \).

(g) For \( \frac{3}{4} – \frac{1}{3} \):
The least common multiple (LCM) of 4 and 3 is 12.
We convert the given fractions to equivalent fractions with the common denominator of 12:
\( \frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12} \) and \( \frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12} \).
Now, we subtract the fractions: \( \frac{3}{4} - \frac{1}{3} = \frac{9}{12} – \frac{4}{12} = \frac{9-4}{12} = \frac{5}{12} \).

(h) For \( \frac{5}{6} - \frac{1}{3} \):
The least common multiple (LCM) of 6 and 3 is 6.
So, the LCM is \( 2 \times 3 = 6 \).
We convert the given fractions to equivalent fractions with the common denominator of 6:
\( \frac{5}{6} = \frac{5 \times 1}{6 \times 1} = \frac{5}{6} \) and \( \frac{1}{3} = \frac{1 \times 2}{3 \times 2} = \frac{2}{6} \).
Now, we subtract the fractions: \( \frac{5}{6} - \frac{1}{3} = \frac{5}{6} - \frac{2}{6} = \frac{5-2}{6} = \frac{3}{6} = \frac{1}{2} \).

(i) For \( \frac{2}{3} + \frac{3}{4} + \frac{1}{2} \):
The least common multiple (LCM) of 3, 4, and 2 is 12.
So, the LCM is \( 2 \times 2 \times 3 = 12 \).
We convert the given fractions to equivalent fractions with the common denominator of 12:
\( \frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12} \), \( \frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12} \), and \( \frac{1}{2} = \frac{1 \times 6}{2 \times 6} = \frac{6}{12} \).
Now, we add the fractions: \( \frac{2}{3} + \frac{3}{4} + \frac{1}{2} = \frac{8}{12} + \frac{9}{12} + \frac{6}{12} = \frac{8+9+6}{12} = \frac{23}{12} \).
This can also be written as a mixed fraction: \( 1\frac{11}{12} \).

(j) For \( \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \):
The least common multiple (LCM) of 2, 3, and 6 is 6.
So, the LCM is \( 2 \times 3 = 6 \).
We convert the given fractions to equivalent fractions with the common denominator of 6:
\( \frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6} \), \( \frac{1}{3} = \frac{1 \times 2}{3 \times 2} = \frac{2}{6} \), and \( \frac{1}{6} = \frac{1 \times 1}{6 \times 1} = \frac{1}{6} \).
Now, we add the fractions: \( \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1 \).

(k) For \( 1\frac{1}{3} + 3\frac{2}{3} \):
First, convert the mixed fractions to improper fractions:
\( 1\frac{1}{3} = \frac{1 \times 3 + 1}{3} = \frac{4}{3} \)
\( 3\frac{2}{3} = \frac{3 \times 3 + 2}{3} = \frac{11}{3} \)
Since these are like fractions (they have the same denominator), we can add them directly:
\( \frac{4}{3} + \frac{11}{3} = \frac{4+11}{3} = \frac{15}{3} = 5 \).
Alternatively, we can add the whole numbers and the fractions separately:
\( 1\frac{1}{3} + 3\frac{2}{3} = (1 + \frac{1}{3}) + (3 + \frac{2}{3}) = 1 + 3 + \frac{1}{3} + \frac{2}{3} \)
\( = 4 + \frac{1+2}{3} = 4 + \frac{3}{3} = 4 + 1 = 5 \).

(l) For \( 4\frac{2}{3} + 3\frac{1}{4} \):
First, convert the mixed fractions to improper fractions:
\( 4\frac{2}{3} = \frac{4 \times 3 + 2}{3} = \frac{14}{3} \)
\( 3\frac{1}{4} = \frac{3 \times 4 + 1}{4} = \frac{13}{4} \)
The least common multiple (LCM) of 3 and 4 is 12.
We convert the given fractions to equivalent fractions with the common denominator of 12:
\( \frac{14}{3} = \frac{14 \times 4}{3 \times 4} = \frac{56}{12} \) and \( \frac{13}{4} = \frac{13 \times 3}{4 \times 3} = \frac{39}{12} \).
Now, we add the fractions: \( \frac{14}{3} + \frac{13}{4} = \frac{56}{12} + \frac{39}{12} = \frac{56+39}{12} = \frac{95}{12} \).
This can also be written as a mixed fraction:
\( 95 \div 12 = 7 \) with a remainder of \( 11 \), so \( 7\frac{11}{12} \).
(m) For \( \frac{16}{5} - \frac{7}{5} \):
These are like fractions (they have the same denominator). We can subtract them directly:
\( \frac{16}{5} - \frac{7}{5} = \frac{16-7}{5} = \frac{9}{5} \).
This can also be written as a mixed fraction: \( 1\frac{4}{5} \).

(n) For \( \frac{4}{3} - \frac{1}{2} \):
The least common multiple (LCM) of 3 and 2 is 6.
We convert the given fractions to equivalent fractions with the common denominator of 6:
\( \frac{4}{3} = \frac{4 \times 2}{3 \times 2} = \frac{8}{6} \) and \( \frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6} \).
Now, we subtract the fractions: \( \frac{4}{3} - \frac{1}{2} = \frac{8}{6} - \frac{3}{6} = \frac{8-3}{6} = \frac{5}{6} \).
In simple words: To add or subtract fractions, first find the smallest common bottom number (LCM) for all fractions. Then, change each fraction to have that new bottom number. Once they all have the same bottom number, you can simply add or subtract the top numbers. Always simplify your answer if possible. For mixed numbers, either convert them to improper fractions first or add/subtract the whole numbers and fractions separately.

Exam Tip: Remember to always find a common denominator (LCM) before adding or subtracting fractions. Ensure to simplify your final answer to its lowest terms or convert to a mixed number if appropriate.

 

Question 2. Sarita bought \( \frac{2}{5} \) metre of ribbon and Lalita bought \( \frac{3}{4} \) metre of ribbon. What is the total length of ribbon bought by both of them?
Answer:
Length of ribbon bought by Sarita \( = \frac{2}{5} \) metre.
Length of ribbon bought by Lalita \( = \frac{3}{4} \) metre.
To find the total length of ribbon bought by both, we need to add these two lengths:
Total length \( = \frac{2}{5} + \frac{3}{4} \) metre.
The least common multiple (LCM) of the denominators 5 and 4 is 20.
Convert both fractions to have a denominator of 20:
\( \frac{2}{5} = \frac{2 \times 4}{5 \times 4} = \frac{8}{20} \)
\( \frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20} \)
Now, add the converted fractions:
\( \frac{2}{5} + \frac{3}{4} = \frac{8}{20} + \frac{15}{20} = \frac{8+15}{20} = \frac{23}{20} \) metre.
The total length of ribbon bought by both is \( \frac{23}{20} \) metres, which can also be written as \( 1\frac{3}{20} \) metres.
In simple words: To find the total ribbon length, we add the two fractions. First, we find a common bottom number for the fractions, which is 20. Then we change each fraction to use this new bottom number and add them together.

Exam Tip: For word problems involving addition of fractions, clearly state the given values, find the LCM of denominators, convert to like fractions, and then perform the addition. Express the final answer in appropriate units and simplify if possible.

 

Question 3. Nena was given \( 1\frac{1}{2} \) cake and Najma was given \( 1\frac{1}{3} \) cake. What is the total amount of cake given to both of them?
Answer:
Amount of cake given to Nena \( = 1\frac{1}{2} \).
Amount of cake given to Najma \( = 1\frac{1}{3} \).
To find the total amount of cake, we add the two mixed fractions:
Total cake \( = 1\frac{1}{2} + 1\frac{1}{3} \).
First, convert the mixed fractions to improper fractions:
\( 1\frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2} \)
\( 1\frac{1}{3} = \frac{1 \times 3 + 1}{3} = \frac{4}{3} \)
The least common multiple (LCM) of the denominators 2 and 3 is 6.
Convert both fractions to have a denominator of 6:
\( \frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6} \)
\( \frac{4}{3} = \frac{4 \times 2}{3 \times 2} = \frac{8}{6} \)
Now, add the converted fractions:
Total cake \( = \frac{3}{2} + \frac{4}{3} = \frac{9}{6} + \frac{8}{6} = \frac{9+8}{6} = \frac{17}{6} \).
The total amount of cake given to both is \( \frac{17}{6} \), which can also be written as the mixed fraction \( 2\frac{5}{6} \).
In simple words: To find the total cake, we combine the amounts Nena and Najma received. We change the mixed numbers into simple fractions, then find a common bottom number, and finally add the fractions together.

Exam Tip: When adding mixed numbers, convert them to improper fractions first to avoid errors. Always find a common denominator before adding or subtracting fractions.

 

Question 4. Fill in the blank boxes:
(a) \( \text{_} - \frac{5}{8} = \frac{1}{4} \)
(b) \( \text{_} - \frac{1}{5} = \frac{1}{2} \)
(c) \( \frac{1}{2} - \text{_} = \frac{1}{6} \)
Answer:
(a) Let the missing fraction be \(x\). So, the equation is \( x - \frac{5}{8} = \frac{1}{4} \).
To find \(x\), we move \( \frac{5}{8} \) to the other side of the equation, making it an addition:
\( x = \frac{1}{4} + \frac{5}{8} \).
The least common multiple (LCM) of the denominators 4 and 8 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, add the fractions:
\( x = \frac{2}{8} + \frac{5}{8} = \frac{2+5}{8} = \frac{7}{8} \).
So, the blank box should be filled with \( \frac{7}{8} \).

(b) Let the missing fraction be \(x\). So, the equation is \( x - \frac{1}{5} = \frac{1}{2} \).
To find \(x\), we move \( \frac{1}{5} \) to the other side of the equation, making it an addition:
\( x = \frac{1}{2} + \frac{1}{5} \).
The least common multiple (LCM) of the denominators 2 and 5 is 10.
Convert both fractions to have a denominator of 10:
\( \frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10} \)
\( \frac{1}{5} = \frac{1 \times 2}{5 \times 2} = \frac{2}{10} \)
Now, add the fractions:
\( x = \frac{5}{10} + \frac{2}{10} = \frac{5+2}{10} = \frac{7}{10} \).
So, the blank box should be filled with \( \frac{7}{10} \).

(c) Let the missing fraction be \(x\). So, the equation is \( \frac{1}{2} - x = \frac{1}{6} \).
To find \(x\), we can rearrange the equation as:
\( x = \frac{1}{2} - \frac{1}{6} \).
The least common multiple (LCM) of the denominators 2 and 6 is 6.
Convert \( \frac{1}{2} \) to an equivalent fraction with a denominator of 6:
\( \frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6} \).
Now, subtract the fractions:
\( x = \frac{3}{6} - \frac{1}{6} = \frac{3-1}{6} = \frac{2}{6} \).
Simplify the result:
\( x = \frac{1}{3} \).
So, the blank box should be filled with \( \frac{1}{3} \).
In simple words: To find the missing fraction, we use inverse operations. If a number is subtracted, we add it to the other side. If the unknown is being subtracted, we swap it with the result. Always find a common denominator before adding or subtracting fractions.

Exam Tip: When solving for a missing fraction, treat it like an algebraic variable. Rearrange the equation to isolate the unknown, and then perform the necessary addition or subtraction after finding a common denominator.

 

Question 5. Fill in the boxes for the given additions and subtractions:
(a)

(b)
Answer:
(a)
Calculations for (a):
Row 1, Column 1: \( \frac{1}{3} + \frac{2}{3} = \frac{1+2}{3} = \frac{3}{3} = 1 \)
Row 1, Column 2: \( \frac{1}{3} + \frac{4}{3} = \frac{1+4}{3} = \frac{5}{3} = 1\frac{2}{3} \)
Row 2, Column 1: \( \frac{2}{3} + \frac{2}{3} = \frac{2+2}{3} = \frac{4}{3} = 1\frac{1}{3} \)
Row 2, Column 2: \( \frac{2}{3} + \frac{4}{3} = \frac{2+4}{3} = \frac{6}{3} = 2 \)

(b)
Calculations for (b):
**Row 1: Adding \( \frac{1}{3} \) to \( \frac{1}{2} \) and \( \frac{1}{3} \)**
- For \( \frac{1}{3} + \frac{1}{2} \): The LCM of 3 and 2 is 6.
\( \frac{1}{3} = \frac{1 \times 2}{3 \times 2} = \frac{2}{6} \)
\( \frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6} \)
So, \( \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6} \).
- For \( \frac{1}{3} + \frac{1}{3} \):
So, \( \frac{1}{3} + \frac{1}{3} = \frac{1+1}{3} = \frac{2}{3} \).

**Row 2: Adding \( \frac{1}{4} \) to \( \frac{1}{2} \) and \( \frac{1}{3} \)**
- For \( \frac{1}{4} + \frac{1}{2} \): The LCM of 4 and 2 is 4.
\( \frac{1}{2} = \frac{1 \times 2}{2 \times 2} = \frac{2}{4} \)
So, \( \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4} \).
- For \( \frac{1}{4} + \frac{1}{3} \): The LCM of 4 and 3 is 12.
\( \frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12} \)
\( \frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12} \)
So, \( \frac{1}{4} + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12} \).
In simple words: To fill the table, we add the fraction from the start of each row to the fraction at the top of each column. We need to make sure the fractions have the same bottom number before we can add them together.

Exam Tip: When filling addition/subtraction tables, perform each individual operation carefully. Ensure a common denominator is found for each sum or difference before combining the numerators.

 

Question 6. A wire of length \( \frac{7}{8} \) metre is cut into two pieces. One piece is \( \frac{1}{4} \) metre long. How long is the other piece?
Answer:
Original length of the wire \( = \frac{7}{8} \) metre.
Length of one piece of wire \( = \frac{1}{4} \) metre.
To find the length of the other piece, we subtract the length of the first piece from the total length:
Length of the other piece \( = \frac{7}{8} - \frac{1}{4} \) metre.
The least common multiple (LCM) of the denominators 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 the fractions:
\( \frac{7}{8} - \frac{1}{4} = \frac{7}{8} - \frac{2}{8} = \frac{7-2}{8} = \frac{5}{8} \) metre.
Therefore, the length of the other piece of wire is \( \frac{5}{8} \) metre.
In simple words: We know the total length of a wire and how long one piece is. To find the length of the other piece, we simply take away the first piece's length from the total. We make sure both fractions have the same bottom number first.

Exam Tip: For problems involving 'remaining' or 'other piece', subtraction is usually required. Always ensure the fractions have a common denominator before performing subtraction.

 

Question 7. Nandini's house is \( \frac{9}{10} \) kilometres from her school. She walked some distance and then took a bus for \( \frac{1}{2} \) kilometre to reach the school. How much distance did she walk?
Answer:
Total distance from Nandini's house to school \( = \frac{9}{10} \) kilometres.
Distance covered by bus \( = \frac{1}{2} \) kilometre.
The remaining distance was walked by Nandini.
To find the distance she walked, we subtract the bus distance from the total distance:
Distance walked \( = \frac{9}{10} - \frac{1}{2} \) kilometres.
The least common multiple (LCM) of the denominators 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 the fractions:
\( \frac{9}{10} - \frac{1}{2} = \frac{9}{10} - \frac{5}{10} = \frac{9-5}{10} = \frac{4}{10} \).
Simplify the result:
\( \frac{4}{10} = \frac{2}{5} \) kilometres.
Nandini walked \( \frac{2}{5} \) kilometres.
In simple words: Nandini travels a total distance to school. We know how much she traveled by bus. To find how much she walked, we subtract the bus journey from the total journey. We get both fractions to have the same bottom number before subtracting.

Exam Tip: When dealing with total distances and parts covered by different modes of transport, subtraction is typically used to find the unknown part. Remember to simplify the final fractional answer.

 

Question 8. Asha and Samuel have book-shelves of the same size, full of books. Asha's book-shelf is \( \frac{5}{6} \) full of books, and Samuel's book-shelf is \( \frac{2}{5} \) full of books. Whose book-shelf is more full? By what fraction?
Answer:
Fraction of Asha's book-shelf full of books \( = \frac{5}{6} \).
Fraction of Samuel's book-shelf full of books \( = \frac{2}{5} \).
To compare these fractions, we need to find a common denominator. The least common multiple (LCM) of 6 and 5 is 30.
Convert both fractions to equivalent fractions with a denominator of 30:
For Asha: \( \frac{5}{6} = \frac{5 \times 5}{6 \times 5} = \frac{25}{30} \).
For Samuel: \( \frac{2}{5} = \frac{2 \times 6}{5 \times 6} = \frac{12}{30} \).
Now, compare the numerators: \( 25 > 12 \).
Since \( \frac{25}{30} > \frac{12}{30} \), Asha's book-shelf is more full.
To find out by what fraction it is more full, we subtract Samuel's fraction from Asha's:
Difference \( = \frac{25}{30} - \frac{12}{30} = \frac{25-12}{30} = \frac{13}{30} \).
Asha's book-shelf has \( \frac{13}{30} \) more books than Samuel's.
In simple words: We compare how full Asha's and Samuel's bookshelves are by making the bottom numbers of their fractions the same. We see that Asha's is fuller. Then, we subtract the smaller fraction from the larger one to find out by how much.

Exam Tip: When comparing fractions or finding the difference between them, always convert them to like fractions (with a common denominator) first. Clearly state both the comparison and the difference in your answer.

 

Question 9. Jaidev takes \( 2\frac{1}{5} \) minutes to walk across the school ground. Rahul takes \( \frac{7}{4} \) minutes to walk across the same ground. 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.
First, convert Jaidev's time to an improper fraction: \( 2\frac{1}{5} = \frac{2 \times 5 + 1}{5} = \frac{11}{5} \) minutes.
Now, we need to compare \( \frac{11}{5} \) and \( \frac{7}{4} \). To compare these fractions, we find a common denominator. The least common multiple (LCM) of 5 and 4 is 20.
Convert both fractions to equivalent fractions with a denominator of 20:
For Jaidev: \( \frac{11}{5} = \frac{11 \times 4}{5 \times 4} = \frac{44}{20} \).
For Rahul: \( \frac{7}{4} = \frac{7 \times 5}{4 \times 5} = \frac{35}{20} \).
Now, compare the numerators: \( 35 < 44 \).
Since \( \frac{35}{20} < \frac{44}{20} \), Rahul takes less time.
To find out by what fraction less time, we subtract Rahul's time from Jaidev's time:
Difference \( = \frac{11}{5} - \frac{7}{4} = \frac{44}{20} - \frac{35}{20} = \frac{44-35}{20} = \frac{9}{20} \) minutes.
Rahul takes \( \frac{9}{20} \) minutes less time than Jaidev.
In simple words: We compare the time taken by Jaidev and Rahul. First, we make sure both times are written as simple fractions with the same bottom number. We see who has the smaller number, which means they took less time. Then, we subtract to find the exact difference.

Exam Tip: Always convert mixed numbers to improper fractions before comparing or performing operations. When comparing quantities, ensure they are in the same units and have a common denominator for fractions. Clearly state both the comparison and the difference in your answer.