Get the simplified Class 6 Maths NCERT Solutions of Ganita Prakash Chapter 7 Fractions textbook exercise questions with complete explanation.
Ganita Prakash Class 6 Maths Chapter 7 Solutions Fractions
NCERT Solutions for Class 6 Maths Ganita Prakash Chapter 7 Fractions
7.1 Fractional Units and Equal Shares Figure it Out (Page No. 152 – 153)
Fill in the blanks with fractions.
Question 1.
Three guavas together weigh 1 kg. If they are roughly of the same size, each guava will roughly weigh ____kg.
Solution:
\(\frac{1}{3}\)
Question 2.
A wholesale merchant packed 1 kg of rice in four packets of equal weight. The weight of each packet is _________ kg.
Solution:
\(\frac{1}{4}\)
Question 3.
Four friends ordered 3 glasses of sugarcane juice and shared it equally. Each one drank ____________ glass of sugarcane juice.
Solution:
\(\frac{3}{4}\)
As total quantity is 3 which is to be divided into four equal parts. So, the required fraction is \(\frac{3}{4}\).
Question 4.
The bis fish weighs \(\frac{1}{2}\) kg. The small one weighs \(\frac{1}{4}\) kg. Together they weigh ____ kg.
Solution:
Given the weighs of big fish = \(\frac{1}{2}\)kg and the weighs of small fish = \(\frac{1}{4}\)kg
Total weight of both fish = \(\frac{1}{2}+\frac{1}{4}=\frac{2+1}{4}\) kg
= \(\frac{3}{4}\) kg
Question 5.
Arrange these fraction words in order of size from the smallest to the biggest in the empty box below: One and a half, three quarters, one and a quarter, half, quarter, two and a half.
Solution:
∴ The fractions from smallest to the biggest are as follows: quarter, half, three quarters one and a quarter, one and a half, two and a half.
7.2 Fractional Units as Parts of a Whole Figure it Out (Page No. 155)
Question 1.
The figure below shows different fractional units of a whole chikki. How much of a whole chikki is each piece?
Solution:
(a)
Total no. of pieces formed of given size = 12
Required fraction = \(\frac{1}{12}\)
(b)
Total no. of pieces formed of given size = 4
Required fraction = \(\frac{1}{4}\)
(c)
Total no. of pieces formed of given size = 8
Required fraction = \(\frac{1}{8}\)
(d)
Total no. of pieces formed of given size = 6
Required fraction = \(\frac{1}{6}\)
(e)
Total no. of pieces formed of given size = 8
Required fraction = \(\frac{1}{8}\)
(f)
Total no. of pieces formed of given size = 8
Required fraction = \(\frac{1}{8}\)
(g)
Total no. of pieces formed of given size = 24
Required fraction = \(\frac{1}{24}\)
(h)
Total no. of pieces formed of given size = 12
Required fraction = \(\frac{1}{12}\)
7.3 Measuring Using Fractional Units Figure it Out (Page No. 158)
Question 1.
Continue this table of \(\frac{1}{2}\) for 2 more steps.
Solution:
Question 2.
Can you create a similar table for \(\frac{1}{4}\) ?
Solution:
Yes.
Question 3.
Make \(\frac{1}{3}\) using a paper strip. Can you use this to also make \(\frac{1}{6}\)?
Solution:
Take a strip of paper.
Fold the strip into three equal parts and then open up.
Yes, we can also make \(\frac{1}{6}\) using a paper strip by folding 6 again the above strip.
Question 4.
Draw a picture and write an addition statement as above to show:
(a) 5 times \(\frac{1}{4}\) of a roti
Solution:
5 times \(\frac{1}{4}\) of a roti
= \(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\)
(b) 9 times \(\frac{1}{4}\) of a roti
Solution:
9 times \(\frac{1}{4}\) of a roti
= \(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\)
Question 5.
Match each fractional unit with the correct picture:
Solution:
7.4 Marking Fraction Lengths on the Number Line Figure it Out (Page No. 160)
Question 1.
On a number line, draw lines of length \(\frac{1}{10}\), \(\frac{3}{10}\), and \(\frac{4}{5}\).
Solution:
Divide the unit into 10 equal parts and point A represents \(\frac{1}{10}\).
Divide a unit into 10 equal parts and point B represents \(\frac{3}{10}\).
Divide a unit into 5 equal parts and point C represents \(\frac{4}{5}\).
Question 2.
Write five more fractions of your choice and mark them on the number line.
Solution:
The fractions are \(\frac{3}{5}, \frac{1}{3}, \frac{5}{7}, \frac{2}{5}\) and \(\frac{1}{8}\).
Their number line representations are:
Question 3.
How many fractions lie between 0 and 1? Think, discuss with your classmates, and write your answer.
Solution:
There are an infinite number of fractions between 0 and 1.
Example: \(\frac{3}{5}, \frac{4}{5}, \frac{7}{10} \frac{1}{2}\) etc.
Question 4.
What is the length of the pink line and black line shown below? The distance between 0 and 1 is 1 unit long, and it is divided into two equal parts. The length of each part is \(\frac{1}{2}\). So the pink line is y units long. Write the fraction that gives the length of the black line in the box.
Solution:
Length of black line is \(\frac{1}{2}\);
Length of black line is \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\)
Fraction that gives length of black line = \(\frac{3}{2}\)
Question 5.
Write the fraction that gives the lengths of the black lines in the respective boxes.
Solution:
Intext Questions
Question 1.
Here, the fractional unit is dividing a length of 1 unit into three equal parts. Write the fraction that gives the length of the pink line in the box or in your notebook. (Page 159)
Solution:
Here number line OR is divided into three equal parts OP, PQ and QR.
Hence length of pink line = OP + PQ = \(\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\)
Question 2.
Here, a unit is divided into 5 equal parts. Write the fraction that gives the length of the pink lines in the respective boxes or in your notebook.
Solution:
Here number line OT = 1 unit is divided into five equal parts OP, PQ, QR, RS and ST.
Hence length of pink line OQ = OP + PQ = \(\frac{1}{5}+\frac{1}{5}=\frac{2}{5}\)
Now, length of pink line OS = OP + PQ + QR + RS = \(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{4}{5}\)
Hence, OQ = \(\frac{2}{5}\) OS = \(\frac{4}{5}\)
Question 3.
Now, a unit is divided into 8 equal parts. Write the appropriate fractions in your notebook Solution:Here number line OH is divided into 8 equal parts OA, AB, BC, CD, DE, EF, FG and GH.
Solution:
Also, OA = \(\frac{1}{8}\), OB = \(\frac{2}{8}\), OC = \(\frac{3}{8}\), OH = \(\frac{8}{8}\) = 1
7.5 Mixed Fractions Figure it Out (Page No. 162)
Question 1.
How many whole units are there in \(\frac{7}{2}\)?
Solution:
\(\frac{7}{2}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=3+\frac{1}{2}\)
So, there are 3 whole units in \(\frac{7}{2}\).
Question 2.
How many whole units are there in \(\frac{4}{3}\) and in \(\frac{7}{3}\)?
Solution:
\(\frac{4}{3}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1+\frac{1}{3}\)
So, there are 1 whole unit in \(\frac{4}{3}\).
\(\frac{7}{3}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=2+\frac{1}{3}\)
So, there are 2 whole units in \(\frac{7}{3}\).
7.5 Mixed Fractions Figure it Out (Page No. 162)
Question 1.
Figure out the number of whole units in each of the following fractions:
(a) \(\frac{8}{3}\)
(b) \(\frac{11}{5}\)
(c) \(\frac{9}{4}\)
Solution:
(a) 2
(b) 2
(c) 2
Question 2.
Can all fractions greater than 1 be written as such mixed numbers?
Solution:
Yes.
Question 3.
Write the following fractions as mixed fractions (e.g. \(\frac{9}{2}\) = 4\(\frac{1}{2}\))
(a) \(\frac{9}{2}\)
Solution:
= 4\(\frac{1}{2}\)
(b) \(\frac{9}{5}\)
Solution:
= 1\(\frac{4}{5}\)
(c) \(\frac{21}{19}\)
Solution:
= 1\(\frac{2}{19}\)
(d) \(\frac{47}{9}\)
Solution:
= 5\(\frac{2}{9}\)
(e) \(\frac{12}{11}\)
Solution:
= 1\(\frac{1}{11}\)
(f) \(\frac{19}{6}\)
Solution:
= 3\(\frac{1}{6}\)
7.5 Mixed Fractions Figure it Out (Page No. 163)
Question 1.
Write the following mixed numbers as fractions:
(a) 3\(\frac{1}{4}\)
(b) 7\(\frac{2}{3}\)
(c) 9\(\frac{4}{9}\)
(d) 3\(\frac{1}{6}\)
(e) 2\(\frac{3}{11}\)
(f) 3\(\frac{9}{10}\)
Solution:
7.6 Equivalent Fractions Figure it Out 7.7 Simplest form of a Fraction Figure it Out (Page No. 165)
Question 1.
Are \(\frac{3}{6}, \frac{4}{8}, \frac{5}{10}\) equivalent fractions? Why?
Solution:
Here, simplest form of \(\frac{3}{6}=\frac{3 \div 3}{6 \div 3}=\frac{1}{2}\) [HCF of 3 and 6 is 3]
and simplest form of \(\frac{4}{8}\) is \(\frac{4 \div 4}{8 \div 4}=\frac{1}{2}\) [HCF of 4 and 8 is 4]
and simplest form of \(\frac{5}{10}\) is \(\frac{5 \div 5}{10 \div 5}=\frac{1}{2}\) [HCF of 5 and 10 is 5]
Hence, \(\frac{3}{6}, \frac{4}{8}, \frac{5}{10}\) are equivalent fractions.
Question 2.
Write two equivalent fractions for \(\frac{2}{6}\).
Solution:
From the fractional wall we can choose any two fractions that denote the same length as \(\frac{2}{6} \cdot \frac{2}{6}=\frac{1}{3}=\frac{3}{9}\)
Question 3.
\(\frac{4}{6}\) = ___________ = ___________ = ___________ = ___________________
(Write as many as you can)
Solution:
Here,
Intext Questions
Answer the following questions after looking at the fraction wall: [Page 164]
Question 1.
Are the lengths \(\frac{1}{2}\) and \(\frac{3}{6}\) equal?
Solution:
Yes, here lengths \(\frac{1}{2}\) and \(\frac{3}{6}\) = \(\frac{1}{2}\)
Lengths are equal.
Question 2.
Are \(\frac{2}{3}\) and \(\frac{4}{6}\) equivalent fractions? Why?
Solution:
Yes, lengths \(\frac{2}{3}\) and \(\frac{4}{6}\) = \(\frac{1}{3}\) are equivalent fraction, as they have same length.
Question 3.
How many pieces of length \(\frac{1}{6}\) will make a length of \(\frac{1}{2}\)?
Solution:
Total no.of pieces = \(\frac{\frac{1}{2}}{\frac{1}{6}}=\frac{1}{2} \times \frac{6}{1}=\frac{6}{2}\) = 3
Hence three pieces of length \(\frac{1}{6}\) will make a length of \(\frac{1}{2}\)
Question 4.
How many pieces of length \(\frac{1}{6}\) will make a length of \(\frac{1}{3}\)?
Solution:
Total no. of pieces = \(\frac{\frac{1}{3}}{\frac{1}{6}}=\frac{1}{3} \times \frac{6}{1}=\frac{6}{3}\) = 2
Hence two pieces of length \(\frac{1}{6}\) will make a length of \(\frac{1}{3}\).
7.7 Simplest form of a Fraction Figure it Out (Page No. 166)
Question 1.
Three rotis are shared equally by four children, show the division in the picture and write a fraction of how much each child gets. Also, write the corresponding division facts, addition facts, and, multiplication facts.
The fraction of roti each child gets is ___________
Division fact:
Addition fact:
Multiplication fact:
Compare your picture and answer with your classmates!
Solution:
One roti is shared as shown in the figure below:
The four shares must be equal to each other!
Similar distribution will be done for the second and third roti also.
So, each child will get \(\frac{3}{4}\) a piece of roti.
The division fact is 3 ÷ 4 = \(\frac{3}{4}\)
The addition fact is 3 = \(\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{3}{4}\)
The multiplication fact is 3 = 4 × \(\frac{3}{4}\)
Question 2.
Draw a picture to show how much each child gets when 2 rotis are shared equally by 4 children. Also, write the corresponding division facts, addition facts, and multiplication facts.
Solution:
One roti is shared as shown in the figure below:
The four shares must be equal to each other!
A similar distribution will be done for the second roti also.
So, each child will get \(\frac{1}{4}\) part from a rod.
So, the total fraction of roti received by each child from 2 rotis = \(\frac{2}{4}\) = \(\frac{1}{2}\)
The division fact is 2 ÷ 4 = \(\frac{2}{4}\)
The addition fact is = \(\frac{2}{4}+\frac{2}{4}+\frac{2}{4}+\frac{2}{4}\)
The multiplication fact is 2 = 4 × \(\frac{2}{4}\)
Question 3.
Anil was in a group where 2 cakes were divided equally among 5 children. How much cake would Anil get?
Solution:
Anil would get \(\frac{2}{5}\) part of the cake.
7.7 Simplest form of a Fraction Figure it Out (Page No. 168 – 169)
Question 1.
Find the missing numbers:
(a) 5 glasses of juice shared equally among 4 friends is the same as ____________ glasses of juice shared equally among 8 friends. So, \(\frac{5}{4}\) = \(\frac{?}{8}\).
(b) 4 kg of potatoes divided equally in 3 bags is the same as 12 kgs of potatoes divided equally in ____________ bags. So, \(\frac{4}{3}\) = \(\frac{12}{?}\).
(c) 7 rods divided among 5 children is the same as rods divided among children. So, \(\frac{7}{5}\) = ____________
Solution:
(a) Here, the amount of juice each friend gets when 5 glasses are shared among 4 friends = \(\frac{\text { number of glasses }}{\text { number of friends }}=\frac{5}{4}\)
Now to determine how many glasses of juice would be needed to give each of the 8 friends the same amount = 8 × \(\frac{5}{4}\)
= 10 glasses
So, 10 glasses of juice shared equally among 8 friends is the same as 5 glasses of juice shared equally among 4 friends.
∴ \(\frac{5}{4}=\frac{10}{8}\)
(b) Here 4 kg of potatoes divided equally in 3 bags then amount of potatoes per bag = \(\frac{4 \mathrm{~kg}}{3 \text { bags }}=\frac{4}{3}\) kg per bag
Let x is the number of bags for 12 kg of potatoes, where each bag has the same amount of potatoes then
\(\frac{12 \mathrm{~kg}}{x \text { bags }}=\frac{4}{3}\) kg per bag
⇒ 12 × 3 = 4 × x
⇒ 36 = 4x
⇒ x = \(\frac{36}{4}\)
⇒ x = 9
∴ \(\frac{4}{3}=\frac{12}{9}\)
(c) Dividing 7 rotis among 4 children gives 7 each child = \(\frac{7}{5}\) of a roti. We can find an
equivalent fraction by multiplying both the numerator and the denominator by the same number. For example, multiplying both by 2.
\(\frac{7 \times 2}{5 \times 2}=\frac{14}{10}\)
So, 7 rotis divided among 5 children is the same as 14 rotis divided among 10 children
∴ \(\frac{7}{5}=\frac{14}{10}\)
Intext Questions
Question 1.
Find equivalent fractions for the given pairs of fractions such that the fractional units are the same. (Page 172)
(a) \(\frac{7}{2}\) and \(\frac{3}{5}\)
Solution:
Given fractions are \(\frac{7}{2}\) and \(\frac{3}{5}\)
Here, the denominators are 2 and 5.
And least common multiple of 2 and 5 is 10.
Hence for both fractions let’s have same denominator of 10.
Now for \(\frac{7}{2}\) multiply both the numerator and the denominator by 5.
\(\frac{7}{2}=\frac{7 \times 5}{2 \times 5}=\frac{35}{10}\)
And for \(\frac{3}{5}\) multiply both the numerator and the denominator by 2, we get,
\(\frac{3 \times 2}{5 \times 2}=\frac{6}{10}\)
Hence, the equivalent fractions with the same denominator are:
\(\frac{35}{10}\) and \(\frac{6}{10}\)
(b) \(\frac{8}{3}\) and \(\frac{5}{6}\)
Solution:
Given fractions are \(\frac{8}{3}\) and \(\frac{5}{6}\)
Here, the denominators are 3 and 6.
And least common multiple of 3 and 6 is 6.
Now for \(\frac{8}{3}\) multiply both the numerator and the denominator by 2.
\(\frac{8}{3}=\frac{8 \times 2}{3 \times 2}=\frac{16}{6}\)
\(\frac{5}{6}\) already have a denominator 6.
Hence, the equivalent fractions with the same denominator are:
\(\frac{16}{6}\) and \(\frac{5}{6}\)
(c) \(\frac{3}{4}\) and \(\frac{3}{5}\)
Solution:
Given fractions are \(\frac{3}{4}\) and \(\frac{3}{5}\)
Here, the denominators are 4 and 5.
And least common multiple of 4 and 5 is 20.
Now for \(\frac{3}{4}\) multiply both the numerator and the denominator by 5.
\(\frac{3}{4}=\frac{3 \times 5}{4 \times 5}=\frac{15}{20}\)
And for \(\frac{3}{5}\) multiply both the numerator and the denominator by 4, we get
\(\frac{3}{5}=\frac{3 \times 4}{5 \times 4}=\frac{12}{20}\)
So, the equivalent fractions with the same denominator are:
\(\frac{15}{20}\) and \(\frac{12}{20}\)
(d) \(\frac{6}{7}\) and \(\frac{8}{5}\)
Solution:
Given fractions are \(\frac{6}{7}\) and \(\frac{8}{5}\)
Here, the denominators are 7 and 5.
And least common multiple of 7 and 5 is 35.
Now for \(\frac{6}{7}\) multiply both the numerator and the denominator by 5.
\(\frac{6}{7}=\frac{6 \times 5}{7 \times 5}=\frac{30}{35}\)
And for \(\frac{8}{5}\) multiply both the numerator and the denominator by 7, we get
\(\frac{8}{5}=\frac{8 \times 7}{5 \times 7}=\frac{56}{35}\)
So, the equivalent fractions with the same denominator are:
\(\frac{30}{35}\) and \(\frac{56}{35}\)
(e) \(\frac{9}{4}\) and \(\frac{5}{2}\)
Solution:
Given fractions are \(\frac{9}{4}\) and \(\frac{5}{2}\)
Here, the denominators are 4 and 2.
And least common multiple of 4 and 2 is 4.
Now for \(\frac{5}{2}\) multiply both the numerator and the denominator by 2.
\(\frac{5}{2}=\frac{5 \times 2}{2 \times 2}=\frac{10}{4}\)
and \(\frac{9}{4}\) already have a denominator 4
So, the equivalent fractions with the same denominator are:
\(\frac{9}{4}\) and \(\frac{10}{4}\)
(f) \(\frac{1}{10}\) and \(\frac{2}{9}\)
Solution:
Given fractions are and \(\frac{1}{10}\) and \(\frac{2}{9}\)
Here, the denominators are 10 and 9.
And least common multiple of 10 and 9 is 90.
Now for \(\frac{1}{10}\) multiply both the numerator and the denominator by 9.
\(\frac{1}{10}=\frac{1 \times 9}{10 \times 9}=\frac{9}{90}\)
And for 2 multiply both the numerator and the denominator by 10, we get
\(\frac{2}{9}=\frac{2 \times 10}{9 \times 10}=\frac{20}{90}\)
So, the equivalent fractions with the same denominator are:’
\(\frac{9}{90}\) and \(\frac{20}{90}\)
(g) \(\frac{8}{3}\) and \(\frac{11}{4}\)
Solution:
Given fractions are \(\frac{8}{3}\) and \(\frac{11}{4}\)
Here, the denominators are 3 and 4.
And least common multiple of 3 and 4 is 12.
Now for \(\frac{8}{3}\) multiply both the numerator and the denominator by 4.
\(\frac{8}{3}=\frac{8 \times 4}{3 \times 4}=\frac{32}{12}\)
And for \(\frac{11}{4}\) multiply both the numerator and the denominator by 3, we get
\(\frac{11}{4}=\frac{11 \times 3}{4 \times 3}=\frac{33}{12}\)
So, the equivalent fractions with the same denominator are:
\(\frac{32}{12}\) and \(\frac{33}{12}\)
(h) \(\frac{13}{6}\) and \(\frac{1}{9}\)
Solution:
Given fractions are \(\frac{13}{6}\) and \(\frac{1}{9}\)
Here, the denominators are 6 and 9.
And least common multiple of 6 and 9 is 18.
Now for \(\frac{13}{6}\) multiply both the numerator and the denominator by 3.
\(\frac{13}{6}=\frac{13 \times 3}{6 \times 3}=\frac{39}{18}\)
And for \(\frac{1}{9}\) multiply both the numerator and the denominator by 2, we get
\(\frac{1}{9}=\frac{1 \times 2}{9 \times 2}=\frac{2}{18}\)
So, the equivalent fractions with the same denominator are:
\(\frac{39}{18}\) and \(\frac{2}{18}\)
7.7 Simplest form of a Fraction Figure it Out (Page No. 173)
Question 1.
Express the following fractions in lowest terms:
(a) \(\frac{17}{51}\)
Solution:
\(\frac{1}{3}\)
(b) \(\frac{64}{144}\)
Solution:
\(\frac{4}{9}\)
(c) \(\frac{126}{147}\)
Solution:
\(\frac{6}{7}\)
(d) \(\frac{525}{112}\)
Solution:
\(\frac{75}{16}\)
7.8 Comparing Fractions Figure it Out (Page No. 174)
Question 1.
Compare the following fractions and justify your answers:
(a) \(\frac{8}{3}, \frac{5}{2}\)
(b) \(\frac{4}{9}, \frac{3}{7}\)
(c) \(\frac{7}{10}, \frac{9}{14}\)
(d) \(\frac{12}{5}, \frac{8}{5}\)
(e) \(\frac{9}{4}, \frac{5}{2}\)
Solution:
Question 2.
Write following fractions ascending order.
(a) \(\frac{7}{10}, \frac{11}{15}, \frac{2}{5}\)
Solution:
The given fractions are \(\frac{7}{10}, \frac{11}{15}, \frac{2}{5}\)
Let us find LCM of denominator 10, 15, 5
∴ LCM of 10, 15 and 5 = 2 × 3 × 5 = 30
Now let us make denominator of each fractions as LCM
Hence given fractions in ascending order are: \(\frac{2}{5}, \frac{7}{10} \frac{11}{5}\)
(b) \(\frac{19}{24}, \frac{5}{6}, \frac{7}{12}\)
Solution:
The given fractions are \(\frac{19}{24}, \frac{5}{6}, \frac{7}{12}\)
Here LCM of 24, 6, 12 is 24.
On arranging in ascending Order, we get
\(\frac{14}{24}, \frac{19}{24}, \frac{20}{24}\)
⇒ \(\frac{7}{12}, \frac{19}{24}, \frac{5}{6}\)
Question 3.
Write the following fractions in descending order.
(a) \(\frac{25}{16}, \frac{7}{8}, \frac{13}{4}, \frac{17}{32}\)
Solution:
(b) \(\frac{3}{4}, \frac{12}{5}, \frac{7}{12}, \frac{5}{4}\)
Solution:
7.9 Relation to Number Sequences Figure it Out (Page No. 179)
Question 1.
Add the following fractions using Brahmagupta’s method:
(a) \(\frac{2}{7}+\frac{5}{7}+\frac{6}{7}\)
Solution:
\(\frac{2}{7}+\frac{5}{7}+\frac{6}{7}\)
= \(\frac{2+5+6}{7}=\frac{13}{7}\)
=1 \(\frac{6}{7}\)
(b) \(\frac{3}{4}+\frac{1}{3}\)
Solution:
\(\frac{3}{4}+\frac{1}{3}=\frac{3}{4} \times \frac{3}{3}+\frac{1}{3} \times \frac{4}{4}\)
= \(\frac{9}{12}+\frac{4}{12}=\frac{9+4}{12}\)
= \(\frac{13}{12}\)
= 1 \(\frac{1}{12}\)
(c) \(\frac{2}{3}+\frac{5}{6}\)
Solution:
\(\frac{2}{3}+\frac{5}{6}=\frac{2}{3} \times \frac{2}{2}+\frac{5}{6}\)
\(\frac{4}{6}+\frac{5}{6}=\frac{9}{6}=\frac{3}{2}\)
= 1 \(\frac{1}{2}\)
(d) \(\frac{2}{3}+\frac{2}{7}\)
Solution:
\(\frac{2}{3}+\frac{2}{7}=\frac{2}{3} \times \frac{7}{7}+\frac{2}{7} \times \frac{3}{3}\)
= \(\frac{14}{21}+\frac{6}{21}=\frac{20}{21}\)
(e) \(\frac{3}{4}+\frac{1}{3}+\frac{1}{5}\)
Solution:
\(\frac{45}{60}+\frac{20}{60}+\frac{12}{60}\)
= \(\frac{77}{60}\)
= 1\(\frac{17}{60}\)
(f) \(\frac{2}{3}+\frac{4}{5}\)
Solution:
\(\frac{10}{15}+\frac{12}{15}=\frac{22}{15}\)
= 1\(\frac{7}{15}\)
(g) \(\frac{4}{5}+\frac{2}{3}\)
Solution:
\(\frac{12}{15}+\frac{10}{15}=\frac{22}{15}\)
= 1\(\frac{7}{15}\)
(h) \(\frac{3}{5}+\frac{5}{8}\)
Solution:
\(\frac{24}{40}+\frac{25}{40}=\frac{49}{40}\)
= \(\frac{9}{40}\)
(i) \(\frac{9}{2}+\frac{5}{4}\)
Solution:
\(\frac{18}{4}+\frac{5}{4}=\frac{23}{4}\)
= 5\(\frac{3}{4}\)
(j) \(\frac{8}{3}+\frac{2}{7}\)
Solution:
\(\frac{56}{21}+\frac{6}{21}=\frac{62}{21}\)
= 2\(\frac{20}{21}\)
(k) \(\frac{3}{4}+\frac{1}{3}+\frac{1}{5}\)
Solution:
\(\frac{45}{60}+\frac{20}{60}+\frac{12}{60}=\frac{77}{60}\)
= 1\(\frac{17}{60}\)
(l) \(\frac{2}{3}+\frac{4}{5}+\frac{3}{7}\)
Solution:
\(\frac{70}{105}+\frac{84}{105}+\frac{45}{105}=\frac{199}{105}\)
= 1\(\frac{94}{105}\)
(m) \(\frac{9}{2}+\frac{5}{4}+\frac{7}{6}\)
Solution:
\(\frac{54}{12}+\frac{15}{12}+\frac{14}{12}=\frac{83}{12}\)
= \(\frac{11}{12}\)
Question 2.
Rahim mixes \(\frac{2}{3}\) liters of yellow paint with \(\frac{3}{4}\) liters of blue paint to make green paint What is the volume of green paint he has made?
Solution:
Quantity of yellow paint added = \(\frac{2}{3}\) litres
Quantity of blue paint added = \(\frac{3}{4}\) litres
Total quantity of green paint made = \(\frac{2}{3}\) + \(\frac{3}{4}\)
LCM of 3 and 4 is 12.
\(\frac{2}{3}=\frac{2}{3} \times \frac{4}{4}=\frac{8}{12}\)
\(\frac{3}{4}=\frac{3}{4} \times \frac{3}{3}=\frac{9}{12}\)
\(\frac{8}{12}+\frac{9}{12}=\frac{8+9}{12}=\frac{17}{12}\)
So, the total quantity of paint made is \(\frac{17}{12}\) liters.
Question 3.
Geeta bought \(\frac{2}{5}\) meter of lace and Shamim bought \(\frac{3}{4}\) meter of the same lace to put a complete border on a table cloth whose perimeter is 1 meter long. Find the total length of the lace they both have bought. Will the lace be sufficient to cover the whole border?
Solution:
Length of lace bought by Geeta = \(\frac{2}{5}\) m
Length of lace bought by Shamim = \(\frac{3}{4}\) m
Total length of lace bought = \(\frac{2}{5}\) + \(\frac{3}{4}\)
LCM of 5 and 4 is 20.
\(\frac{2}{5}=\frac{2}{5} \times \frac{4}{4}=\frac{8}{20}\)
\(\frac{3}{4}=\frac{3}{4} \times \frac{5}{5}=\frac{15}{20}\)
\(\frac{8}{20}+\frac{15}{20}=\frac{23}{20}= 1 \frac{3}{20}\)
This length is more than 1 m. So, lace is more than sufficient or will be left extra after covering the border.
7.9 Relation to Number Sequences Figure it Out (Page No. 181)
Question 1.
\(\frac{5}{8}-\frac{3}{8}\)
Solution:
Given \(\frac{5}{8}-\frac{3}{8}\)
As fractional unit is same i.e., \(\frac{1}{8}\) we shall simply subtract numerators keeping fractional unit as \(\frac{1}{8}\)
Then \(\frac{5}{8}-\frac{3}{8}=\frac{5-3}{8}\)
= \(\frac{2}{8}=\frac{1}{4}\) (representing in simplest form)
Question 2.
\(\frac{7}{9}-\frac{5}{9}\)
Solution:
Given \(\frac{7}{9}-\frac{5}{9}\)
As fractional unit is same i.e., \(\frac{1}{9}\) we shall simply subtract numerators keeping fractional unit as \(\frac{1}{9}\)
\(\frac{7}{9}-\frac{5}{9}\)
= \(\frac{7-5}{9}=\frac{2}{9}\)
Question 3.
\(\frac{10}{27}-\frac{1}{27}\frac{10}{27}-\frac{1}{27}\)
Solution:
Here \(\frac{10}{27}-\frac{1}{27}\)
= \(\frac{10-1}{27}\)
= \(\frac{9}{27}=\frac{1}{3}\)
7.9 Relation to Number Sequences Figure it Out (Page No. 182)
Question 1.
Carry out the following subtractions using Brahmagupta’s method:
(a) \(\frac{8}{15}-\frac{3}{15}\)
Solution:
Given \(\frac{8}{15}-\frac{3}{15}\)
Fractional unit for both fractions is \(\frac{1}{15}\) then
\(\frac{8}{15}-\frac{3}{15}=\frac{8-3}{15}\)
= \(\frac{5}{15}=\frac{1}{3}\)
(b) \(\frac{2}{5}-\frac{4}{15}\)
Solution:
Given \(\frac{2}{5}-\frac{4}{15}\)
Here LCM of 5 and 15 is 15. Fractional unit for both fractions should be \(\frac{1}{15}\)
then \(\frac{2 \times 3}{5 \times 3}-\frac{4 \times 1}{15 \times 1}\)
= \(\frac{6}{15}-\frac{4}{15}\)
= \(\frac{6-4}{15}\)
= \(\frac{2}{15}\)
(c) \(\frac{5}{6}-\frac{4}{9}\)
Solution:
Given \(\frac{5}{6}-\frac{4}{9}\)
Hence LCM of 6 and 9 is 18. Fractional unit for both fractions should be \(\frac{1}{18}\) then
(d) \(\frac{2}{3}-\frac{1}{2}\)
Solution:
Given \(\frac{2}{3}-\frac{1}{2}\)
Here LCM of 3 and 2 is 6. Fractional unit for both fractions should be \(\frac{1}{6}\)
Question 2.
Subtract as indicated:
(a) \(\frac{13}{4}\) from \(\frac{10}{3}\)
Solution:
The denominators of the given fractions are 3 and 4. The LCM of 3 and 4 is 12.
Then \(\frac{13}{4}=\frac{13 \times 3}{4 \times 3}=\frac{39}{12}, \frac{10}{3}=\frac{10 \times 4}{3 \times 4}=\frac{40}{12}\)
Therefore, \(\frac{10}{3}-\frac{13}{4}=\frac{40}{12}-\frac{39}{12}=\frac{1}{12}\)
(b) \(\frac{18}{5}\) from \(\frac{23}{3}\)
Solution:
The denominators of the given fractions are 3 and 5.
The LCM of 3 and 5 is 15.
Then, \(\frac{23}{3}=\frac{23 \times 5}{3 \times 5}=\frac{115}{15}, \frac{18}{5}=\frac{18 \times 3}{5 \times 3}=\frac{54}{15}\)
Therefore, \(\frac{23}{3}-\frac{18}{5}=\frac{115}{15}-\frac{54}{15}=\frac{61}{15}=4 \frac{1}{15}\)
(c) \(\frac{29}{7}\) from \(\frac{45}{7}\)
Solution:
The denominators are same.
Therefore, \(\frac{45}{7}-\frac{29}{7}=\frac{16}{7}=2 \frac{2}{7}\)
Question 3.
Solve the following problems:
(a) Jaya’s school is \(\frac{7}{10}\) km from her home. She takes an auto for \(\frac{1}{2}\) km from her home daily, and then walks the remaining distance to reach her school. How much does she walk daily to reach the school?
Solution:
(a) Total distance between school and home = \(\frac{7}{10}\) km
Distance travelled in Auto = \(\frac{1}{2}\) km.
∴ Distance she walks daily to reach the school
(b) Jeevika takes \(\frac{10}{3}\) minutes to take a complete round of the park and her friend Namit takes \(\frac{13}{4}\) minutes to do the same. Who takes less time and by how much?
Solution:
Time taken by Jeevika = \(\frac{10}{3}\) minutes
and time taken by Narnit = \(\frac{13}{4}\) minutes
Now, \(\frac{10}{3} \times \frac{4}{4}=\frac{40}{12}\) and \(\frac{13}{4} \times \frac{3}{3}=\frac{39}{12}\)
Clearly, \(\frac{10}{3}\) > \(\frac{13}{4}\)
∴ Jeevika takes less ti me by \(\left(\frac{10}{3}-\frac{13}{4}\right)\) minutes
= \(\left(\frac{40}{12}-\frac{39}{12}\right)\) minutes
= \(\frac{1}{12}\) minutes.