GSEB Class 7 Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.3

Question 1. Find
(i) \( \frac { 1 }{ 4 } \) of (a) \( \frac {1}{4} \) (b) \( \frac { 3 }{ 4 } \) (c) \( \frac { 4 }{ 3 } \)
(ii) \( \frac { 1 }{ 7 } \) of (a) \( \frac { 2 }{9} \) (b) \( \frac { 6 }{ 5 } \) (c) \( \frac { 3 }{ 10 } \)
Answer:
(i) (a) \( \frac { 1 }{ 4 } \) of \( \frac { 1 }{ 4 } \)
\( = \frac { 1 }{ 4 } \times \frac { 1 }{ 4 } \)
\( = \frac { 1 \times 1 }{ 4 \times 4 } = \frac { 1 }{ 16 } \)

(b) \( \frac {1}{4} \) of \( \frac { 3 }{ 5 } \)
\( = \frac { 1 }{ 4 } \times \frac { 3 }{ 5 } \)
\( = \frac { 1 \times 3 }{ 4 \times 5 } = \frac { 3 }{ 20 } \)

(c) \( \frac { 1 }{ 4 } \) of \( \frac { 4 }{ 3 } \)
\( = \frac { 1 }{ 4 } \times \frac { 4 }{ 3 } \)
\( = \frac { 1 \times 4 }{ 4 \times 3 } = \frac { 1 }{ 3 } \)

(ii) (a) \( \frac {1}{7} \) of \( \frac { 2 }{ 9 } \)
\( = \frac { 1 }{7} \times \frac { 2 }{ 9 } \)
\( = \frac { 1 \times 2 }{ 7 \times 9 } = \frac { 2 }{ 63 } \)

(b) \( \frac {1}{7} \) of \( \frac {6}{5} \)
\( = \frac { 1 }{7} \times \frac { 6 }{ 5 } \)
\( = \frac { 1 \times 6 }{ 7 \times 5 } = \frac { 6 }{ 35 } \)

(c) \( \frac {1}{7} \) of \( \frac { 3 }{ 10 } \)
\( = \frac { 1 }{ 7 } \times \frac { 3 }{ 10 } \)
\( = \frac { 1 \times 3 }{ 7 \times 10 } = \frac { 3 }{ 70 } \)
In simple words: To find "of" between fractions, you just multiply the fractions together. Multiply the top numbers (numerators) and the bottom numbers (denominators) separately to get your answer.

Exam Tip: Remember that "of" in mathematics often implies multiplication. Always simplify your fractions to their lowest form after multiplying.

Question 2. Multiply and reduce to lowest form (if possible):
(i) \( \frac { 2 }{ 3 } \times 2\frac { 2 }{ 3 } \)
(ii) \( \frac { 2 }{7} \times \frac {7}{9} \)
(iii) \( \frac { 3 }{ 8 } \times \frac { 6 }{ 4 } \)
(iv) \( \frac {9}{ 5 } \times \frac { 3 }{ 5 } \)
(v) \( \frac {1}{ 3 } \times \frac { 15 }{ 8 } \)
(vi) \( \frac {11}{ 2 } \times \frac { 3 }{ 10 } \)
(vii) \( \frac {4}{5} \times \frac{12}{7} \)
Answer:
(i) \( \frac{ 2 }{ 3 } \times 2\frac{ 2 }{ 3 } = \frac{ 2 }{ 3 } \times \frac{ 8 }{ 3 } = \frac{ 2 \times 8 }{ 3 \times 3 } = \frac{ 16 }{ 9 } = 1\frac{ 7 }{ 9 } \)
(ii) \( \frac{ 2 }{ 7 } \times \frac{ 7 }{ 9 } = \frac{ 2 \times 7 }{ 7 \times 9 } = \frac{ 2 \times 1 }{ 1 \times 9 } = \frac{ 2 }{ 9 } \)
(iii) \( \frac{ 3 }{ 8 } \times \frac{ 6 }{ 4 } = \frac{ 3 \times 6 }{ 8 \times 4 } = \frac{ 3 \times 3 }{ 4 \times 4 } = \frac{ 9 }{ 16 } \)
(iv) \( \frac{ 9 }{ 5 } \times \frac{ 3 }{ 5 } = \frac{ 9 \times 3 }{ 5 \times 5 } = \frac{ 27 }{ 25 } = 1\frac{ 2 }{ 25 } \)
(v) \( \frac{ 1 }{ 3 } \times \frac{ 15 }{ 8 } = \frac{ 1 \times 15 }{ 3 \times 8 } = \frac{ 1 \times 5 }{ 1 \times 8 } = \frac{ 5 }{ 8 } \)
(vi) \( \frac{ 11 }{ 2 } \times \frac{ 3 }{ 10 } = \frac{ 11 \times 3 }{ 2 \times 10 } = \frac{ 33 }{ 20 } = 1\frac{ 13 }{ 20 } \)
(vii) \( \frac{ 4 }{ 5 } \times \frac{ 12 }{ 7 } = \frac{ 4 \times 12 }{ 5 \times 7 } = \frac{ 48 }{ 35 } = 1\frac{ 13 }{ 35 } \)
In simple words: To multiply fractions, you just multiply the numbers on top and the numbers on the bottom. If you have a mixed number, first change it into an improper fraction. After you multiply, check if you can make the fraction simpler by dividing both top and bottom numbers by a common factor.

Exam Tip: Always convert mixed fractions to improper fractions before multiplying. Look for opportunities to cancel common factors diagonally or vertically before performing the final multiplication to simplify calculations.

Question 3. Multiply and reduce to lowest form and convert into a mixed fraction:
(i) \( \frac { 2 }{ 5 } \times 5\frac { 1 }{ 4 } \)
(ii) \( 6\frac { 2 }{ 5 } \times \frac {7}{9} \)
(iii) \( \frac { 3 }{ 2 } \times 5\frac {1}{3} \)
(iv) \( \frac {5}{6} \times 2\frac { 3 }{7} \)
(v) \( 3\frac { 2 }{ 5 } \times 4\frac { 4 }{ 7 } \)
(vi) \( 2\frac { 3 }{ 5 } \times 3 \)
(vii) \( 3\frac { 4 }{ 7 } \times \frac { 3 }{ 5 } \)
Answer:
(i) \( \frac{ 2 }{ 5 } \times 5\frac{ 1 }{ 4 } = \frac{ 2 }{ 5 } \times \frac{ 21 }{ 4 } = \frac{ 2 \times 21 }{ 5 \times 4 } = \frac{ 1 \times 21 }{ 5 \times 2 } = \frac{ 21 }{ 10 } = 2\frac{ 1 }{ 10 } \)
(ii) \( 6\frac{ 2 }{ 5 } \times \frac{ 7 }{ 9 } = \frac{ 32 }{ 5 } \times \frac{ 7 }{ 9 } = \frac{ 32 \times 7 }{ 5 \times 9 } = \frac{ 224 }{ 45 } = 4\frac{ 44 }{ 45 } \)
(iii) \( \frac{ 3 }{ 2 } \times 5\frac{ 1 }{ 3 } = \frac{ 3 }{ 2 } \times \frac{ 16 }{ 3 } = \frac{ 3 \times 16 }{ 2 \times 3 } = \frac{ 1 \times 8 }{ 1 \times 1 } = 8 \)
(iv) \( \frac{ 5 }{ 6 } \times 2\frac{ 3 }{ 7 } = \frac{ 5 }{ 6 } \times \frac{ 17 }{ 7 } = \frac{ 5 \times 17 }{ 6 \times 7 } = \frac{ 85 }{ 42 } = 2\frac{ 1 }{ 42 } \)
(v) \( 3\frac{ 2 }{ 5 } \times 4\frac{ 4 }{ 7 } = \frac{ 17 }{ 5 } \times \frac{ 32 }{ 7 } = \frac{ 17 \times 32 }{ 5 \times 7 } = \frac{ 544 }{ 35 } = 15\frac{ 19 }{ 35 } \)
(vi) \( 2\frac{ 3 }{ 5 } \times 3 = \frac{ 13 }{ 5 } \times \frac{ 3 }{ 1 } = \frac{ 13 \times 3 }{ 5 \times 1 } = \frac{ 39 }{ 5 } = 7\frac{ 4 }{ 5 } \)
(vii) \( 3\frac{ 4 }{ 7 } \times \frac{ 3 }{ 5 } = \frac{ 25 }{ 7 } \times \frac{ 3 }{ 5 } = \frac{ 25 \times 3 }{ 7 \times 5 } = \frac{ 5 \times 3 }{ 7 \times 1 } = \frac{ 15 }{ 7 } = 2\frac{ 1 }{ 7 } \)
In simple words: When you need to multiply fractions that include mixed numbers, always start by changing those mixed numbers into improper fractions. Then, multiply the top numbers together and the bottom numbers together. Finally, if your answer is an improper fraction, change it back into a mixed number.

Exam Tip: Converting improper fractions to mixed fractions is crucial for these types of questions. Remember, a mixed fraction represents a whole number part and a fractional part.

Question 4. Which is greater:
(i) \( \frac { 2 }{7} \) of \( \frac { 3 }{ 4 } \) or \( \frac { 3 }{ 5 } \) of \( \frac { 5 }{ 8 } \)?
(ii) \( \frac { 1 }{ 2 } \) of \( \frac {6}{7} \) or \( \frac { 2 }{ 3 } \) of \( \frac { 3 }{ 7 } \)?
Answer:
(i) Comparing \( \frac {2}{7} \) of \( \frac { 3 }{ 4 } \) and \( \frac {3}{5} \) of \( \frac { 5 }{ 8 } \)
First, calculate each part:
\( \frac { 2 }{7} \) of \( \frac { 3 }{ 4 } \)
\( = \frac { 2 }{7} \times \frac { 3 }{ 4 } \)
\( = \frac { 1 \times 3 }{ 7 \times 2 } \)
\( = \frac { 3 }{ 14 } \)

And \( \frac { 3 }{ 5 } \) of \( \frac { 5 }{ 8 } \)
\( = \frac { 3 }{5} \times \frac {5}{8} \)
\( = \frac {3 \times 1}{1 \times 8} \)
\( = \frac { 3 }{ 8 } \)
To compare \( \frac { 3 }{ 14 } \) and \( \frac { 3 }{ 8 } \), we find a common denominator. The LCM of 14 and 8 is 56.
So, \( \frac { 3 }{ 14 } = \frac { 3 \times 4 }{ 14 \times 4 } = \frac { 12 }{ 56 } \)
And \( \frac { 3 }{ 8 } = \frac { 3 \times 7 }{ 8 \times 7 } = \frac { 21 }{ 56 } \)
Now, comparing the fractions with common denominators:
\( \frac {21}{ 56 } > \frac { 12 }{ 56 } \)
Therefore, \( \frac { 3 }{ 5 } \) of \( \frac {5}{8} \) is greater than \( \frac {2}{7} \) of \( \frac { 3 }{ 4 } \).

(ii) Comparing \( \frac { 1 }{ 2 } \) of \( \frac { 6 }{ 7 } \) or \( \frac { 2 }{ 3 } \) of \( \frac { 3 }{ 7 } \):
First, calculate each part:
\( \frac { 1 }{ 2 } \) of \( \frac {6}{ 7 } \)
\( = \frac { 1 }{ 2 } \times \frac { 6 }{ 7 } \)
\( = \frac { 1 \times 3 }{ 1 \times 7 } \)
\( = \frac { 3 }{ 7 } \)

And, \( \frac {2}{3} \) of \( \frac { 3 }{ 7 } \)
\( = \frac { 2 }{ 3 } \times \frac { 3 }{7} \)
\( = \frac {2 \times 1 }{ 1 \times 7 } \)
\( = \frac { 2 }{ 7 } \)
Now, compare the results: \( \frac { 3 }{ 7 } \) and \( \frac { 2 }{ 7 } \). Since both fractions have the same denominator, we just compare their numerators.
\( 3 > 2 \)
\( \implies \frac { 3 }{ 7 } > \frac { 2 }{ 7 } \)
Therefore, \( \frac {1}{2 } \) of \( \frac {6}{7} \) is greater than \( \frac { 2 }{ 3 } \) of \( \frac { 3 }{ 7 } \).
In simple words: To find which fraction is bigger, first work out the value of each "of" phrase by multiplying the fractions. Then, to compare them, make sure they both have the same bottom number (denominator). Once the bottom numbers are the same, the fraction with the bigger top number (numerator) is the larger one.

Exam Tip: When comparing fractions, finding a common denominator is essential. The least common multiple (LCM) helps simplify this process and avoid larger numbers.

Question 5. Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is \( \frac { 3 }{ 4 } \) m. Find the distance between the first and the last sapling.
Answer:
Number of saplings = 4
When 4 saplings are planted in a row, there are 3 spaces between them.
(First sapling) ---- \( \frac{3}{4} \) m ---- (Second sapling) ---- \( \frac{3}{4} \) m ---- (Third sapling) ---- \( \frac{3}{4} \) m ---- (Fourth sapling)
Distance between two adjacent saplings = \( \frac { 3 }{ 4 } \) m
Therefore, the distance between the 1st and the last (4th) sapling is the sum of these 3 distances.
\( = 3 \times \frac { 3 }{ 4 } \) m
\( = \frac { 9 }{ 4 } \) m
\( = 2\frac { 1 }{ 4 } \) m
The total distance between the first and the last sapling is \( 2\frac { 1 }{ 4 } \) m.
In simple words: If you plant four little trees in a line, there will be three spaces between them. Since each space is \( \frac{3}{4} \) meter long, you multiply \( \frac{3}{4} \) by 3 to find the total length from the first tree to the last.

Exam Tip: For problems involving objects in a row, remember that the number of gaps is always one less than the number of objects. Visualizing the setup can prevent common errors.

Question 6. Lipika reads a book for \( 1\frac { 3 }{ 4 } \) hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
Answer:
Number of days she reads = 6
Reading hours per day (for one day) = \( 1\frac { 3 }{ 4 } \) hours
Therefore, the total number of reading hours needed to complete the book is calculated by multiplying the daily reading time by the number of days.
\( = 6 \times 1\frac { 3 }{ 4 } \)
\( = 6 \times \frac {7}{ 4 } \) hours (converting the mixed number to an improper fraction)
\( = \frac {3 \times 7 }{ 2 } \) hours (simplifying by dividing 6 and 4 by 2)
\( = \frac {21}{ 2 } \) hours
\( = 10\frac {1}{ 2 } \) hours
In total, Lipika needed \( 10\frac{1}{2} \) hours to read the entire book.
In simple words: Lipika reads for \( 1\frac{3}{4} \) hours each day and finishes her book in 6 days. To find out her total reading time, multiply the hours she reads daily by the number of days she spends reading.

Exam Tip: When calculating total quantities over multiple units (like hours over days), always ensure the units are consistent and perform multiplication. Convert mixed numbers to improper fractions first for easier calculations.

Question 7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using \( 2\frac { 3 }{ 4 } \) litres of petrol.
Answer:
Distance covered with 1 litre of petrol = 16 km
Therefore, the distance covered with \( 2\frac { 3 }{ 4 } \) litres of petrol is determined by multiplying the distance per litre by the total litres.
\( = 16 \times 2\frac { 3 }{ 4 } \) km
\( = 16 \times \frac {11}{ 4 } \) km (changing the mixed number to an improper fraction)
\( = \frac { 16 \times 11}{4} \) km
\( = \frac {4 \times 11 }{ 1 } \) km (simplifying by dividing 16 and 4 by 4)
\( = 44 \) km
The car will travel 44 km using \( 2\frac{3}{4} \) litres of petrol.
In simple words: A car can go 16 kilometers on one liter of gas. To find out how far it can go on \( 2\frac{3}{4} \) liters, you simply multiply 16 by \( 2\frac{3}{4} \).

Exam Tip: This is a direct proportionality problem. Always set up the calculation as (rate) x (quantity) to find the total outcome. Unit conversion and simplification are key steps.

Question 8.
(a) (i) Provide the number in the box \( \boxed{\phantom{\frac{5}{10}}} \), such that \( \frac { 2 }{ 3 } \times \boxed{\phantom{\frac{5}{10}}} = \frac { 10 }{ 30 } \).
(ii) The simplest form of the number obtained in \( \boxed{\phantom{\frac{5}{10}}} \) is \( \boxed{\phantom{\frac{1}{2}}} \).
(b) (i) Provide the number in the box \( \boxed{\phantom{\frac{8}{15}}} \), such that \( \frac { 3 }{ 5 } \times \boxed{\phantom{\frac{8}{15}}} = \frac { 24 }{ 75 } \).
(ii) The simplest form of the number obtained in \( \boxed{\phantom{\frac{8}{15}}} \) is \( \boxed{\phantom{\frac{8}{15}}} \).
Answer:
(a)
(i) We have: \( \frac { 2 }{ 3 } \times \frac { 5 }{ 10 } = \frac { 10 }{ 30 } \), so the required number in the box is \( \frac { 5 }{ 10 } \).
The numerator \( 2 \times 5 = 10 \). The denominator \( 3 \times 10 = 30 \).
(ii) The simplest form of \( \frac{ 5 }{ 10 } \) is \( \frac{ 5 \div 5 }{ 10 \div 5 } = \frac{ 1 }{ 2 } \).

(b)
(i) We have: \( \frac { 3 }{ 5 } \times \frac { 8 }{ 15 } = \frac { 24 }{ 75 } \), so the required number in the box is \( \frac {8}{15} \).
The numerator \( 3 \times 8 = 24 \). The denominator \( 5 \times 15 = 75 \).
(ii) The simplest form of \( \frac { 8 }{ 15 } \) is \( \frac { 8 }{ 15 } \) itself, as 8 and 15 have no common factors other than 1.
In simple words: To find the missing fraction, look at how the top numbers multiply to give the new top number, and how the bottom numbers multiply to give the new bottom number. Then, simplify the fraction you found by dividing both the top and bottom by their biggest common factor.

Exam Tip: When finding a missing fraction in a multiplication, treat the numerators and denominators separately. Always reduce the final fraction to its simplest form by dividing by the greatest common divisor.