# PSEB 5th Class Maths Solutions Chapter 4 Fractions Ex 4.6

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 Fractions Ex 4.6 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 5 Maths Chapter 4 Fractions Ex 4.6

Question 1.
Find the greater fraction for each pair of the following :
(a) $$\frac{2}{5}$$, $$\frac{2}{3}$$
(b) $$\frac{7}{9}$$, $$\frac{7}{12}$$
(c) $$\frac{1}{8}$$, $$\frac{1}{4}$$
(d) $$\frac{4}{6}$$, $$\frac{4}{8}$$
(e) $$\frac{3}{7}$$, $$\frac{3}{11}$$
(f) $$\frac{7}{9}$$, $$\frac{4}{9}$$
(g) $$\frac{3}{4}$$, $$\frac{1}{4}$$
(h) $$\frac{5}{8}$$, $$\frac{7}{8}$$
Solution:
(a) $$\frac{2}{5}$$, $$\frac{2}{3}$$
The numerator of both the fractions are equal. So, the fraction whose denominator is smaller will be greater.
Hence, $$\frac{2}{3}$$ is greater than $$\frac{2}{5}$$.

(b) $$\frac{7}{9}$$, $$\frac{7}{12}$$
The numerator of both the fractions are equal. So, the fraction whose denominator is smaller will be greater.
Hence, $$\frac{7}{9}$$ is greater than $$\frac{7}{12}$$.

(c) $$\frac{1}{8}$$, $$\frac{1}{4}$$
The numerator of both the fractions are equal. So, the fraction whose denominator is smaller will be greater.
Hence, $$\frac{1}{4}$$ is greater than $$\frac{1}{8}$$

(d) $$\frac{4}{6}$$, $$\frac{4}{8}$$
The numerator of both the fractions are equal. So, the fraction whose denominator is smaller will be greater.
Hence, $$\frac{4}{6}$$ is greater than $$\frac{4}{8}$$

(e) $$\frac{3}{7}$$, $$\frac{3}{11}$$
The numerator of both the fractions are equal. So, the fraction whose denominator is smaller will be greater.
Hence, $$\frac{3}{7}$$ is greater than $$\frac{3}{11}$$.

(f) $$\frac{7}{9}$$, $$\frac{4}{9}$$
The denominator of both the fractions are equal. So, the fraction whose numerator is greater will be greater.
Hence, $$\frac{7}{9}$$ is greater than $$\frac{4}{9}$$

(g) $$\frac{3}{4}$$, $$\frac{1}{4}$$
The denominator of both the fractions are equal. So, the fraction whose numerator is greater will be greater.
Hence, $$\frac{3}{4}$$ is greater than $$\frac{1}{4}$$.

(h) $$\frac{5}{8}$$, $$\frac{7}{8}$$
The denominator of both the fractions are equal. So, the fraction whose numerator is greater will be greater.
Hence, $$\frac{7}{8}$$ is greater than $$\frac{5}{8}$$

Question 2.
Find the smaller fraction for each part of the following :
(a) $$\frac{3}{5}$$, $$\frac{3}{4}$$
(b) $$\frac{5}{8}$$, $$\frac{5}{12}$$
(c) $$\frac{7}{9}$$, $$\frac{4}{9}$$
(d) $$\frac{3}{6}$$, $$\frac{3}{8}$$
(e) $$\frac{5}{7}$$, $$\frac{5}{11}$$
(f) $$\frac{8}{12}$$, $$\frac{5}{12}$$
(g) $$\frac{9}{4}$$, $$\frac{7}{4}$$
(h) $$\frac{9}{8}$$, $$\frac{7}{8}$$
Solution:
(a) $$\frac{3}{5}$$, $$\frac{3}{4}$$
The numerator of both the fractions are equal. So, the fraction whose denominator is greater will be smaller.
Hence, $$\frac{3}{5}$$ is Smaller than $$\frac{3}{4}$$

(b) $$\frac{5}{8}$$, $$\frac{5}{12}$$
The numerator of both the fractions are equal. So, the fraction whose denominator is greater will be smaller.
Hence, $$\frac{5}{12}$$ is Smaller than $$\frac{5}{8}$$

(c) $$\frac{7}{9}$$, $$\frac{4}{9}$$
The denominator of both the fractions are equal. So, the fraction whose numerator is greater will be smaller.
Hence, $$\frac{3}{5}$$ is Smaller than $$\frac{3}{4}$$

(d) $$\frac{3}{6}$$, $$\frac{3}{8}$$
The numerator of both the fractions are equal. So, the fraction whose denominator is greater will be smaller.
Hence, $$\frac{3}{8}$$ is Smaller than $$\frac{3}{6}$$

(e) $$\frac{5}{7}$$, $$\frac{5}{11}$$
The numerator of both the fractions are equal. So, the fraction whose denominator is greater will be smaller.
Hence, $$\frac{5}{11}$$ is Smaller than $$\frac{5}{7}$$

(f) $$\frac{8}{12}$$, $$\frac{5}{12}$$
The denominator of both the fractions are equal. So, the fraction whose numerator is greater will be smaller.
Hence, $$\frac{5}{12}$$ is Smaller than $$\frac{8}{12}$$

(g) $$\frac{9}{4}$$, $$\frac{7}{4}$$
The denominator of both the fractions are equal. So, the fraction whose numerator is greater will be smaller.
Hence, $$\frac{7}{4}$$ is Smaller than $$\frac{9}{4}$$

(h) $$\frac{9}{8}$$, $$\frac{7}{8}$$
The denominator of both the fractions are equal. So, the fraction whose numerator is greater will be smaller.
Hence, $$\frac{7}{8}$$ is Smaller than $$\frac{9}{8}$$

Question 3.
Write the following in increasing (ascending) order :

(a) $$\frac{7}{12}$$, $$\frac{4}{12}$$, $$\frac{1}{12}$$, $$\frac{5}{12}$$
(b) $$\frac{5}{12}$$, $$\frac{5}{9}$$, $$\frac{5}{7}$$, $$\frac{5}{4}$$
(c) $$\frac{6}{11}$$, $$\frac{4}{11}$$, $$\frac{9}{11}$$, $$\frac{3}{11}$$
(d) $$\frac{7}{8}$$, $$\frac{7}{12}$$, $$\frac{7}{4}$$, $$\frac{7}{2}$$
(e) $$\frac{12}{15}$$, $$\frac{12}{13}$$, $$\frac{12}{17}$$, $$\frac{12}{10}$$
Solution:
(a) $$\frac{7}{12}$$, $$\frac{4}{12}$$, $$\frac{1}{12}$$, $$\frac{5}{12}$$
The denominators of all of these fractions are equal. Therefore, the fraction which has the lowest numerator, will be the smallest.
Among these fractions, $$\frac{1}{12}$$ is the smallest.
Ascending order is $$\frac{1}{12}$$, $$\frac{4}{12}$$, $$\frac{5}{12}$$ and $$\frac{7}{12}$$.

(b) $$\frac{5}{12}$$, $$\frac{5}{9}$$, $$\frac{5}{7}$$, $$\frac{5}{4}$$
The numerator of all of these fractions are equal. Therefore, the fraction whose denominator is the largest, is the smallest fraction.
Among these fractions, $$\frac{5}{12}$$ is the smallest.
Ascending order is $$\frac{5}{12}$$, $$\frac{5}{9}$$, $$\frac{5}{7}$$ and $$\frac{5}{4}$$

(c) $$\frac{6}{11}$$, $$\frac{4}{11}$$, $$\frac{9}{11}$$, $$\frac{3}{11}$$
The denominators of all of these fractions are equal. Therefore, the fraction which has the lowest numerator, will be the smallest fraction.
Among these fractions, $$\frac{3}{11}$$ is the smallest.
Ascending order is $$\frac{3}{11}$$, $$\frac{4}{11}$$, $$\frac{6}{11}$$ and $$\frac{9}{11}$$.

(d) $$\frac{7}{8}$$, $$\frac{7}{12}$$, $$\frac{7}{4}$$, $$\frac{7}{2}$$
The numerator of all of these fractions are equal. Therefore, the fraction whose denominator is the largest, is the smallest fraction.
Among these fractions, $$\frac{7}{12}$$ is the smallest.
Ascending order is $$\frac{7}{12}$$, $$\frac{7}{8}$$, $$\frac{7}{4}$$ and $$\frac{7}{2}$$

(e) $$\frac{12}{15}$$, $$\frac{12}{13}$$, $$\frac{12}{17}$$, $$\frac{12}{10}$$
The numerator of all of these fractions are equal. Therefore, the fraction whose denominator is the largest, is the smallest fraction.
Among these fractions, $$\frac{12}{17}$$ is the smallest.
Ascending order is $$\frac{12}{17}$$, $$\frac{12}{15}$$, $$\frac{12}{13}$$ and $$\frac{12}{10}$$