Primary Mathematics Chapter 3 Fractions
Page 107 Exercise 3.10 Problem 1
We are given a figure and asked to convert it into an improper fraction.
Draw a long line between the figures and then we can see 8 triangles and in that 2 triangles are shaded region.
Therefore we can say that the improper fraction will be \(\frac{2}{8}\)
Therefore, we can say that the fraction for the given figure is \(\frac{2}{8}\)
Page 107 Exercise 3.10 Problem 2
We are asked to find the missing numerators and denominators. As we can see the given are equivalent fractions.
Given: First fraction as \(\frac{2}{3}\)
The second fraction has a numerator which is 3 times 2
So, the denominator also has to be 3 times 3
Which means \(\frac{6}{9}\)
In the third fraction, it is 4 times the first fraction i.e. \(\frac{8}{12}\)
In the fourth fraction, it is 5 times the first fraction i.e. \(\frac{10}{12}\)
In the fifth fraction, it is 7 times the first fraction i.e. \(\frac{14}{21}\)
Therefore, we can say that the completed fractions are
\(\frac{2}{3}\) = \(\frac{6}{9}\)
⇒ \(\frac{8}{12}\)
⇒ \(\frac{10}{12}\)
⇒ \(\frac{14}{21}\)
Page 107 Exercise 3.10 Problem 3
We are asked to write any two fractions that are equivalent to \(\frac{1}{4}\)
As we know equivalent fraction means multiple of the given fraction.
Therefore the equivalent fraction for the given fraction is
⇒ \(\frac{1×2}{4×2}\) = \(\frac{2}{8}\)
⇒ \(\frac{1×3}{2×3}\) = \(\frac{3}{6}\)
Therefore, equivalent fractions are \(\frac{2}{8}\), \(\frac{3}{6}\)
Page 107 Exercise 3.10 Problem 4
We are asked to find the fractions that the number line represents.
In the given number it is divided by 10 parts.
So, we can see that the P is the 2nd part of the 10 parts.
So, the fraction will be 3\(\frac{6}{10}\)
Q is in the 6th part. So, the fraction will be 3\(\frac{6}{10}\)= 3\(\frac{3}{5}\).
R is the second part between 4 and 5. So, fraction will be 4\(\frac{2}{10}\)
= 4\(\frac{1}{5}\).
Therefore, the values of P, Q,R are 3\(\frac{3}{5}\) and 4\(\frac{1}{5}\)
Page 108 Exercise 3.10 Problem 5
We are asked to find the fractions that in missing in the number line.
In the given number line it is divided by 10 parts between the numbers.
As we can see the first blank is 5 the blank of 8 blanks.
So, the fraction is 1 \(\frac{5}{8}\)
The second blank is at the 7 position in 8 parts.
So, the fraction is 1\(\frac{6}{8}\)
The last blank is the first part between 2 and 3
So, 2\(\frac{1}{8}\) is the fraction.
Therefore, the values of the blanks are 1 \(\frac{5}{8}\), 1\(\frac{6}{8}\), 2\(\frac{1}{8}\) respectively.
Page 108 Exercise 3.10 Problem 6
We are asked to find which of the given fractions is small.
Given: \(\frac{1}{2}, \frac{2}{5}, \frac{3}{4}, \frac{4}{7} .\)
Let’s find the decimal values of the given fraction.
⇒ \(\frac{1}{2}\) = 0.5
⇒ \(\frac{2}{5}\) = 0.4
⇒ \(\frac{3}{4}\) = 0.75
⇒ \(\frac{4}{7}\) = 0.5714
So,\(\frac{2}{5}\) is the smallest faction.
Therefore, we can say that a given fraction is the smallest in given fractions.
Page 108 Exercise 3.10 Problem 7
We are asked to find which of the given fractions is in its simplest form.
Given: \(\frac{3}{9}, \frac{2}{10}, \frac{5}{7}\)
In \(\frac{3}{9}\) we can see that 9 is a multiple of 3 so it is not in its simplest form.
In \(\frac{2}{10}\)we can see that 10 is a multiple of 2 so it is not in its simplest form.
In \(\frac{5}{7}\)we can see that 5 is not a multiple of 7 so it is not in its simplest form.
Therefore, we can say that \(\frac{5}{7}\) is in its simplest form.
Page 108 Exercise 3.10 Problem 8
We are asked to find which of the given fractions is smaller than\(\frac{1}{2}\)
Let us find the decimal of all the given fractions, we get
⇒ \(\frac{7}{8}\) = 0.875
⇒ \(\frac{5}{9}\) = 0.55
⇒ \(\frac{2}{5}\) = 0.4
Therefore we can say that \(\frac{2}{5}\) is smaller than \(\frac{1}{2}\) is \(\frac{1}{2}\)
Therefore, we can say that \(\frac{2}{5}\) is the fraction that is smaller than is \(\frac{1}{2}\)
Page 108 Exercise 3.10 Problem 9
We are asked to find which of the given fractions is nearest to 1
Let us find the decimal of all the given fractions, we get
⇒ \(\frac{4}{5}\) = 0.8
⇒ \(\frac{6}{7}\) = 0.85
⇒ \(\frac{11}{12}\) = 0.91
⇒ \(\frac{8}{9}\) = 0.88
Therefore we can say that \(\frac{11}{12}\) is nearest to 1
Therefore we can say that \(\frac{11}{12}\) is nearest to 1
Page 108 Exercise 3.10 Problem 10
We are asked to find which of the given fractions is greater.
Given:
⇒ \(\frac{3}{8}\) or \(\frac{7}{12}\)
Let us find the decimal of all the given fractions, we get
⇒ \(\frac{3}{8}\) = 0.37
⇒ \(\frac{7}{12}\) = 0.5
Therefore we can say that \(\frac{7}{12}\) is greater than 3\(\frac{3}{8}\)
Therefore we can say that \(\frac{7}{12}\) is greater than 3\(\frac{3}{8}\)
Page 109 Exercise 3.10 Problem 11
We are asked to arrange fractions in increasing order.
Given:
⇒ \(\frac{3}{4}, \frac{7}{6}, \frac{5}{12}\), 1
Given: \(\frac{3}{4}, \frac{7}{6}, \frac{5}{12}, 1\)
Let us find the decimal of all the given fractions, we get
⇒ \(\frac{3}{4}=0.75\)
⇒ \(\frac{7}{6}=1.16\)
⇒ \(\frac{5}{2}=0.415\)
Therefore we can say that the increasing order will be
⇒ \(\frac{5}{12}<\frac{3}{4}<1<\frac{7}{6}\)
Therefore we can say that the \(\frac{5}{12}<\frac{3}{4}<1<\frac{7}{6}\)
Page 109 Exercise 3.10 Problem 12
We are asked to arrange fractions in decreasing order.
Given: \(\frac{2}{9}, \frac{2}{7}, \frac{9}{7}, \frac{2}{3}\)
⇒ \(\frac{2}{9}\) = 0.22
⇒ \(\frac{2}{7}\) = 0.28
⇒ \(\frac{9}{7}\) = 1.2
⇒ \(\frac{2}{3}\) = 0.6
Therefore we can say that the decreasing order will be.
⇒ \(\frac{9}{7}\)>\(\frac{2}{3}\)>\(\frac{2}{7}\)>\(\frac{2}{9}\).
Therefore we can say that the decreasing order will be. \(\frac{9}{7}\)>\(\frac{2}{3}\)>\(\frac{2}{7}\)>\(\frac{2}{9}\).
Page 109 Exercise 3.10 Problem 13
We are asked to find the missing fraction that makes the equation true.
Given: ____ + \(\frac{3}{8}\)=1
Let us transfer \(\frac{3}{8}\) from LHS to RHS, we get
_______= 1− \(\frac{3}{8}\)
By solving the RHS, we get
______ = \(\frac{8-3}{8}\)
______ = \(\frac{5}{8}\)
Therefore, given blank fraction value is \(\frac{5}{8}\)
Therefore, we can say that the value of the given blank is \(\frac{5}{8}\)
Page 109 Exercise 3.10 Problem 14
We are asked to find the missing fraction that makes the equation true.
Given: 1−______ = \(\frac{5}{12}\)
Let us transfer 1 from LHS to RHS and multiply both sides with a negative sign, we get
⇒ 1−______ = \(\frac{5}{12}\)
⇒ _____ \(1-\frac{5}{12}\)
By solving the RHS, we get
______ \(\frac{12-5}{12}\)
______ \(\frac{7}{12}\)
Therefore, given a blank fraction value is \(\frac{7}{12}\)
Therefore we say that the value of a given blank is \(\frac{7}{12}\)
Page 109 Exercise 3.10 Problem 15
We are asked to write an improper fraction for each of the following.
Given: 9 – Eighths.
As we know eighth means \(\frac{1}{8}\)
So, 9 eighths is 9(\(\frac{1}{8}\)). i.e. \(\frac{9}{8}\)
Therefore, we can say that the value of nine-eighths means \(\frac{9}{8}\)
Page 109 Exercise 3.10 Problem 16
We are asked to write an improper fraction for each of the following.
Given: 11 – quarters.
As we know quarter means\(\frac{1}{4}\)
So, 11 eighths is 11(\(\frac{1}{8}\)) i.e \(\frac{11}{8}\)
Therefore, we can say that the value of nine-eighths means \(\frac{11}{8}\).
Page 109 Exercise 3.10 Problem 17
We are asked to write an improper fraction for each of the following.
Given: 7-fifths.
As we know fifths means \(\frac{1}{5}\)
So, 7 fifths is 7(\(\frac{1}{5}\)) i.e \(\frac{5}{7}\)
Therefore, we can say that the value of nine-eighths means \(\frac{5}{7}\)
Page 109 Exercise 3.10 Problem 18
We are asked to write an improper fraction for each of the following.
Given: 15 – sixths.
As we know sixths means \(\frac{1}{6}\)
So, 15 fifths is 15 (\(\frac{1}{6}\)) i.e \(\frac{15}{6}\)
Therefore, we can say that the value of 15 sixths means \(\frac{15}{6}\)
Page 110 Exercise 3.10 Problem 19
We are asked to express 2\(\frac{3}{5}\) as an improper fraction.
As we know mixed fraction contains a whole number and a fraction part we have to multiply the denominator with the whole number and we have to add this to the numerator to get the numerator of the improper fraction.
So, we get
2\(\frac{3}{5}\) = \(\frac{2×5+3}{5}\)
= \(\frac{13}{5}\)
Therefore, we can say that by converting a given mixed fraction into an improper fraction we get \(\frac{13}{5}\).
Page 110 Exercise 3.10 Problem 20
We are asked to find how many thirds are there in 3
Let us write an equivalent fraction for \(\frac{3}{1}\)
Multiply the numerator and denominator by 3, and we get
= \(\frac{9}{3}\)
So, we can say that we have 9 thirds in three.
Therefore, we can say that we have 9 thirds in three.
Page 110 Exercise 3.10 Problem 21
We are asked to find how many sixths are there in 2\(\frac{1}{6}\)
Let us write an equivalent fraction for \(\frac{2}{1}\)
Multiply numerator and denominator by 6 , we get = \(\frac{12}{6}\)
So, we can say that we have 12 sixths in three.
Therefore, we can say that we have 12 sixths in two.
Page 110 Exercise 3.10 Problem 22
We are given to convert the given improper fraction into a mixed fraction.
Given: \(\frac{27}{6}\)
To convert an improper fraction into a mixed fraction let us divide the given fraction and then we have to write the divisor as the whole number and quotient as the numerator and the remainder as the numerator.
So, for 27÷6 we get 4 as quotient and 3 as remainder.
So, Mixed fraction = 4 \(\frac{3}{6}\)
Therefore, we can say that the mixed fraction for the given improper fraction is 4 \(\frac{3}{6}\)
Page 110 Exercise 3.10 Problem 23
We are asked to find which will match \(\frac{4}{5}\).
As from \(\frac{4}{5}\)we can say that 5 is diving 4.
So, therefore we can say that the correct option is 4÷5.
Therefore, we can say that the correct option is 4÷5.
Page 110 Exercise 3.10 Problem 24
We are asked to write a fraction of 8 girls sharing 2 cakes equally.
So, we can say that 2 cakes should be shared with 8 girls equally.
It implies \(\frac{2}{8}\)
⇒ \(\frac{1}{4}\)
So, each girl gets \(\frac{1}{4}\) portion from the 2 cakes.
Therefore, we can the fraction that represents the part of each girl that the get is \(\frac{1}{4}\)
Page 110 Exercise 3.10 Problem 25
We are asked to add the given fractions.
Given:
⇒ \(\frac{2}{7}+\frac{4}{7}\)
As we can see that the denominators are the same we can add them directly, and we get
⇒ \(\frac{2}{7}+\frac{4}{7}\) = \(\frac{6}{7}\)
Therefore, by adding \(\frac{2}{7}+\frac{4}{7}\) we get\(\frac{6}{7}\).
Page 110 Exercise 3.10 Problem 26
We are asked to subtract the given fractions.
Given:
⇒ \(\frac{7}{12}-\frac{3}{12}\)
As we can see that the denominators are the same we subtract and add them directly, and we get
⇒ \(\frac{7}{12}-\frac{3}{12}\)=\(\frac{3}{12}\)
⇒ \(\frac{1}{3}\)
Therefore, by subtracting \(\frac{7}{12}-\frac{3}{12}\)=\(\frac{3}{12}\) we get\(\frac{1}{3}\)
Page 110 Exercise 3.10 Problem 27
We are asked to add the given fractions.
Given:
⇒ \(\frac{6}{9}+\frac{3}{9}\).
As we can see that the denominators are the same we add them directly, we get
⇒ \(\frac{6}{9}+\frac{3}{9}\)=\(\frac{9}{9}\)
= 1
Therefore, by adding \(\frac{6}{9}+\frac{3}{9}\) we get 1
Page 110 Exercise 3.10 Problem 28
We are asked to subtract the given fractions.
Given: 2-\(\frac{6}{11}\)
As we can see the denominators are not the same, so let us find the LCM of the two denominators, and then by solving we get
⇒ \(\frac{22−6}{11}\)= \(\frac{6}{11}\)
Therefore, by subtracting \(\frac{22−6}{11}\) we get \(\frac{6}{11}\)
Page 110 Exercise 3.10 Problem 29
We are asked to add the given fractions.
Given: \(\frac{2}{8}+\frac{3}{8}+\frac{1}{8}\)
As we can see that the denominators are the same.
So we can add them directly
⇒ \(\frac{2}{8}+\frac{3}{8}+\frac{1}{8}\)
⇒ \(\frac{2+3+1}{8}\)
⇒ \(\frac{6}{8}\)
⇒ \(\frac{3}{4}\)
There fore by adding \(\frac{2}{8}+\frac{3}{8}+\frac{1}{8} \text { we get } \frac{3}{4}\)
Page 110 Exercise 3.10 Problem 30
We are asked to subtract the given fractions.
Given: 1− \(\frac{3}{10}-\frac{5}{10}\)
As we can see that the denominators are not the same.
So we have to do LCM and then we have to solve them.
By doing LCM and solving we get
⇒ \(\frac{10-3-5}{10}\)= \(\frac{2}{10}\)
⇒ \(\frac{1}{5}\)
Therefore, by subtracting 1-\(\frac{3}{10}-\frac{5}{10}\) we get \(\frac{1}{5}\)
Page 111 Exercise 3.10 Problem 31
Given: The total weight of two bags is 2 lb.
One of them is \(\frac{1}{4}\)lb.
We are asked to find the weight of the other bag.
Let us consider the other bag weight as x.
So, from the question x+\(\frac{1}{4}\) =2
Transferring \(\frac{1}{4}\) from LHS to RHS, we get
x = 2−\(\frac{1}{4}\)
x = \(\frac{8-1}{4}\)
x = \(\frac{7}{4}\)
Therefore, we can say that the other bag weight is \(\frac{7}{4}\) lb.
Page 111 Exercise 3.10 Problem 32
Given: She used \(\frac{5}{8}\) meters rope and she has 7 meters left.
The total rope is the sum of the remaining rope and the used rope
Total rope is \(\)
⇒ \(\frac{5}{8}+7\)
⇒ \(\frac{5}{8}+\frac{7}{1}\)
⇒ \( \frac{5+7(8)}{8}\)
⇒ \(\frac{5+56}{8}\)
⇒ \(\frac{61}{8}\)
Therefore we say that the total rope is \(\frac{61}{8}\)