Working With Fractions - NCERT Solutions Class 7 Maths (Ganita Prakash)

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Working With Fractions – NCERT Solutions Class 7 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 7 Maths (Ganita Prakash).

NCERT Solutions Class 7

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Working With Fractions – NCERT Solutions

Q.1: Tenzin drinks {tex}\frac{1}{2}{/tex} glass of milk every day. How many glasses of milk does he drink in a week? How many glasses of milk did he drink in the month of January?

Solution:

Glass of milk drank every day {tex}=\frac{1}{2}{/tex}
Glasses of milk drank in a week {tex}=\frac{1}{2} \times 7=\frac{7}{2}{/tex}
Days in month of January = 31
Glasses of milk drank in month of January {tex}=31 \times \frac{1}{2}=\frac{31}{2}=15 \frac{1}{2}{/tex}.

Q.2: A team of workers can make 1 km of a water canal in 8 days. So, in one day, the team can make ________ km of the water canal. If they work 5 days a week, they can make ________ km of the water canal in a week.

Solution:

The length of canal made by workers in 8 days {tex}=1 {~km}{/tex}
The length of canal made by workers in 1 day {tex}=\frac{1}{8} {~km}{/tex}
By working 5 days in a week, the length of canal made by workers {tex}=5 \times \frac{1}{8}=\frac{5}{8} {~km}{/tex}.

Q.3: Manju and two of her neighbours buy 5 litres of oil every week and share it equally among the 3 families. How much oil does each family get in a week? How much oil will one family get in 4 weeks?

Solution:

Amount of oil shared by 3 families in a week = 5 litres
Amount of oil one family got in a week {tex}=\frac{5}{3}{/tex} litres
Amount of oil one family got in 4 weeks {tex}=4 \times \frac{5}{3}=\frac{20}{3}=6 \frac{2}{3}{/tex} litres.

Q.4: Safia saw the Moon setting on Monday at 10 pm. Her mother, who is a scientist, told her that every day the Moon sets {tex}\frac{5}{6}{/tex} hour later than the previous day. How many hours after 10 pm will the moon set on Thursday?

Solution:

No. of days from Monday to Thursday = 3
Delay in moon setting time per day {tex}=\frac{5}{6}{/tex} hour
Delay in moon setting time over 3 days {tex}=3 \times \frac{5}{6}{/tex} hour {tex}=\frac{5}{2}=2.5{/tex} hours
Thus, the Moon will set 2.5 hours or 2 hours 30 minutes after 10 PM on Thursday.

Q.5: Multiply and then convert it into a mixed fraction:

{tex} 7 \times \frac{3}{5} {/tex}
{tex} 4 \times \frac{1}{3} {/tex}
{tex} \frac{9}{7} \times 6 {/tex}
{tex} \frac{13}{11} \times 6 {/tex}

Solution:

{tex}7 \times \frac{3}{5}=\frac{21}{5}=4 \frac{1}{5}{/tex}.
{tex}4 \times \frac{1}{3}=\frac{4}{3}{/tex}.
{tex}\frac{9}{7} \times 6=\frac{54}{7}{/tex}.
{tex}\frac{13}{11} \times 6=\frac{78}{11}{/tex}.

Q.6: Find the following products. Use a unit square as a whole for representing the fractions:

{tex} \frac{1}{3} \times \frac{1}{5} {/tex}
{tex} \frac{1}{4} \times \frac{1}{3} {/tex}
{tex} \frac{1}{5} \times \frac{1}{2} {/tex}
{tex} \frac{1}{6} \times \frac{1}{5} {/tex}

Solution:

{tex} \frac{1}{3} \times \frac{1}{5} {/tex}
{tex}\frac{1}{3}{/tex} (multiplier) {tex}\times \frac{1}{5}{/tex} (multiplicand)
Number of rows {tex}={/tex} Denominator of the multiplicand {tex}=5{/tex}
Number of columns {tex}={/tex} Denominator of the multiplier {tex}=3{/tex}
Thus, the whole is divided into {tex}(5 \times 3)=15{/tex} equal parts
So, {tex}\frac{1}{3} \times \frac{1}{5}=\frac{1}{3 \times 5}=\frac{1}{15}{/tex}.
{tex}\frac{1}{4} \times \frac{1}{3}{/tex}
{tex}\frac{1}{4}{/tex} (multiplier) {tex}\times \frac{1}{3}{/tex} (multiplicand)
Number of rows {tex}={/tex} Denominator of the multiplicand {tex}=3{/tex}
Number of columns {tex}={/tex} Denominator of the multiplier {tex}=4{/tex}
Thus, the whole is divided into {tex}(3 \times 4)=12{/tex} equal parts
So, {tex}\frac{1}{4} \times \frac{1}{3}=\frac{1}{4 \times 3}=\frac{1}{12}{/tex}.
{tex}\frac{1}{5} \times \frac{1}{2}{/tex}
{tex}\frac{1}{5}({/tex}multiplier{tex}) \times \frac{1}{2}({/tex}multiplicand{tex}){/tex}
Number of rows {tex}={/tex} Denominator of the multiplicand {tex}=2{/tex}
Number of columns {tex}={/tex} Denominator of the multiplier {tex}=5{/tex}
Thus, the whole is divided into {tex}(2 \times 5)=10{/tex} equal parts
So, {tex}\frac{1}{5} \times \frac{1}{2}=\frac{1}{5 \times 2}=\frac{1}{10}{/tex}.
{tex}\frac{1}{6} \times \frac{1}{5}{/tex}
{tex}\frac{1}{6}{/tex} (multiplier) {tex}\times \frac{1}{5}{/tex} (multiplicand)
Number of rows {tex}={/tex} Denominator of the multiplicand {tex}=5{/tex}
Number of columns = Denominator of the multiplier = 6
Thus, the whole is divided into {tex}(5 \times 6)=30{/tex} equal parts
So, {tex}\frac{1}{6} \times \frac{1}{5}=\frac{1}{6 \times 5}=\frac{1}{30}{/tex}.

Q.7: Find the product. Use a unit square as a whole for representing the fractions and carrying out the operations.
{tex}\frac{2}{3} \times \frac{4}{5}{/tex}

Solution:

{tex}\frac{2}{3} \times \frac{4}{5}{/tex}
First, the whole is divided into 5 rows and 3 columns creating {tex}15(5 \times 3){/tex} equal parts.
The value we get by dividing {tex}\frac{4}{5}{/tex} into 3 equal parts is {tex}\frac{4}{3 \times 5}{/tex}.
Thus, we multiply this result by 2 to get the product. This is {tex}\frac{2 \times 4}{3 \times 5}{/tex}.
So, {tex}\frac{2}{3} \times \frac{4}{5}=\frac{2 \times 4}{3 \times 5}=\frac{8}{15}{/tex}.

Q.8: Find the product. Use a unit square as a whole for representing the fractions and carrying out the operations.
{tex}\frac{1}{4} \times \frac{2}{3}{/tex}

Solution:

{tex}\frac{1}{4} \times \frac{2}{3}{/tex}
First, the whole is divided into 3 rows and 4 columns creating {tex}12(3 \times 4){/tex} equal parts.
The value we get by dividing {tex}\frac{2}{3}{/tex} into 4 equal parts is {tex}\frac{2}{4 \times 3}{/tex}.
So, {tex}\frac{1}{4} \times \frac{2}{3}=\frac{1 \times 2}{4 \times 3}=\frac{2}{12}{/tex}.

Q.9: Find the product. Use a unit square as a whole for representing the fractions and carrying out the operations.
{tex}\frac{3}{5} \times \frac{1}{2}{/tex}

Solution:

{tex}\frac{3}{5} \times \frac{1}{2}{/tex}
First, the whole is divided into 2 rows and 5 columns creating {tex}10(2 \times 5){/tex} equal parts.
The value we get by dividing {tex}\frac{1}{2}{/tex} into 5 equal parts is {tex}\frac{1}{5 \times 2}{/tex}.
Thus, we multiply this result by 3 to get the product. This is {tex}\frac{3 \times 1}{5 \times 2}{/tex}.
So, {tex}\frac{3}{5} \times \frac{1}{2}=\frac{3 \times 1}{5 \times 2}=\frac{3}{10}{/tex}.

Q.10: Find the product. Use a unit square as a whole for representing the fractions and carrying out the operations.
{tex}\frac{4}{6} \times \frac{3}{5}{/tex}

Solution:

{tex}\frac{4}{6} \times \frac{3}{5}{/tex}
First, the whole is divided into 5 rows and 6 columns creating {tex}30(5 \times 6){/tex} equal parts.
The value we get by dividing {tex}\frac{3}{5}{/tex} into 6 equal parts is {tex}\frac{3}{6 \times 5}{/tex}.
Thus, we multiply this result by 4 to get the product. This is {tex}\frac{4 \times 3}{6 \times 5}{/tex}.
So, {tex}\frac{4}{6} \times \frac{3}{5}=\frac{4 \times 3}{6 \times 5}=\frac{12}{30}{/tex}.

Q.11: A water tank is filled from a tap. If the tap is open for 1 hour, {tex}\frac{7}{10}{/tex} of the tank gets filled. How much of the tank is filled if the tap is open for

{tex}\frac{1}{3}{/tex} hour ________
{tex}\frac{2}{3}{/tex} hour ________
{tex}\frac{3}{4}{/tex} hour ________
{tex}\frac{7}{10}{/tex} hour ________
For the tank to be full, how long should the tap be running?

Solution:

Part of the tank filled in 1 hour {tex}=\frac{7}{10}{/tex}

Part of the tank filled in {tex}\frac{1}{3}{/tex} hour {tex}=\frac{1}{3} \times \frac{7}{10}=\frac{7}{30}{/tex}.
Part of the tank filled in {tex}\frac{2}{3}{/tex} hour {tex}=\frac{2}{3} \times \frac{7}{10}=\frac{14}{30}=\frac{7}{15}{/tex}.
Part of the tank filled in {tex}\frac{3}{4}{/tex} hour {tex}=\frac{3}{4} \times \frac{7}{10}=\frac{21}{40}{/tex}.
Part of the tank filled in {tex}\frac{7}{10}{/tex} hour {tex}=\frac{7}{10} \times \frac{7}{10}=\frac{49}{100}{/tex}.
Part of the tank filled in 1 hour {tex}=\frac{7}{10}{/tex}.
or Time required to fill {tex}\frac{7}{10}{/tex} of a tank {tex}=1{/tex} hour.

Time required to fill 1 tank {tex}=1 \div \frac{7}{10}=1 \times \frac{10}{7}=\frac{10}{7}{/tex} hours {tex}=1 \frac{3}{7}{/tex} hours.

Q.12: The government has taken {tex}\frac{1}{6}{/tex} of Somu’s land to build a road. What part of the land remains with Somu now? She gives half of the remaining part of the land to her daughter, Krishna, and {tex}\frac{1}{3}{/tex} it to her son Bora. After giving them their shares, she kept the remaining land for herself.

What part of the original land did Krishna get?
What part of the original land did Bora get?
What part of the original land did Somu keep for herself?

Solution:

Part of Somu’s land acquired by the government {tex}=\frac{1}{6}{/tex}
Somu’s original land {tex}=1-\frac{1}{6}=\frac{6-1}{6}=\frac{5}{6}{/tex}

Part of the land given to Krishna {tex}=\frac{1}{2}{/tex} of original land {tex}=\frac{1}{2} \times \frac{5}{6}=\frac{5}{12}{/tex}.
Part of the land given to Bora {tex}=\frac{1}{3}{/tex} of original land {tex}=\frac{1}{3} \times \frac{5}{6}=\frac{5}{18}{/tex}.
Part of the land Somu kept for herself {tex}=\frac{5}{6}-\left(\frac{5}{12}+\frac{5}{18}\right){/tex}
{tex} =\frac{5}{6}-\left(\frac{3 \times 5+5 \times 2}{36}\right) {/tex}
{tex} =\frac{5}{6}-\left(\frac{15+10}{36}\right) {/tex}
{tex} =\frac{5}{6}-\left(\frac{25}{36}\right) {/tex}
{tex} =\frac{6 \times 5-25}{36} {/tex}
{tex} =\frac{30-25}{36}=\frac{5}{36} . {/tex}

Q.13: Find the area of a rectangle of sides {tex}3 \frac{3}{4} {ft}{/tex} and {tex}9 \frac{3}{5} {ft}{/tex}.

Solution:

{tex} \text { Area of rectangle }=3 \frac{3}{4} {ft} \times 9 \frac{3}{5} {ft} =\frac{15}{4} {ft} \times \frac{48}{5} {ft} {/tex}
{tex}= \frac {15}{4} \times \frac {48}{5}{/tex} = 3 {tex}=3\times 12{/tex} = 36 sq. ft

Q.14: Tsewang plants four saplings in a row in his garden. The distance between two saplings is {tex}\frac{3}{4}{/tex}m. Find the distance between the first and last sapling.

Solution:

No. of saplings in a row in garden {tex}=4{/tex}
Distance between two saplings {tex}=\frac{3}{4}{/tex}
Distance between first and last sapling {tex}=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}=\frac{3+3+3}{4}=\frac{9}{4} {~m}=2 \frac{1}{4} {~m}{/tex}.

Q.15: Which is heavier: {tex}\frac{12}{15}{/tex} of 500 grams or {tex}\frac{3}{20}{/tex} of {tex}4 {~kg} ?{/tex}

Solution:

{tex}\frac{12}{15}{/tex} of 500 grams {tex}=\frac{12}{15} \times 500 {~g}{/tex}
{tex}=\frac {12}{15}\times 500 = 4 \times 100 = 400 g{/tex}
{tex}\frac{3}{20}{/tex} of {tex}4 {~kg}=\frac{3}{20} \times 4000 {~g}{/tex}
{tex}= \frac {3}{20} \times 4000 = 3 \times 200 = 600 g{/tex}
Hence, {tex}\frac{3}{20}{/tex} of 4 kg is heavier than {tex}\frac{12}{15}{/tex} of 500 grams.

Q.16: Is the Product Always Greater than the Numbers Multiplied?
What can you conclude about the relationship between the numbers multiplied and the product? Fill in the blanks:

When one of the numbers being multiplied is between 0 and 1, the product is
________ (greater/less) than the other number.
When one of the numbers being multiplied is greater than 1, the product is
________ (greater/less) than the other number.

Solution:

When one of the numbers being multiplied is between 0 and 1, the product is less than the other number.
Example:
Let one number be {tex}\frac{1}{4}{/tex} and other number be 100.
Product {tex}=\frac{1}{4} \times 100=25{/tex}.
Hence, the product (25) is less than the other number (100).
When one of the numbers being multiplied is greater than 1 , the product is greater than the other number.
Example:
Let one number be 10 and other number 50 .
Product = {tex}10 \times 50=500{/tex}.
Hence, the product (500) is greater than the other number (50).

Q.17: In each of the figures given below, find the fraction of the big square that the shaded region occupies.

Solution:

There are 4 triangles in half of the area of the whole square.
So, there are total 8 triangles in the whole square.
Area of each triangle {tex}=\frac{1}{8}{/tex}.
Area of the shaded region {tex}=3{/tex} triangles {tex}=3 \times \frac{1}{8}=\frac{3}{8}{/tex}.
Hence, the shaded region occupies {tex}\frac{3}{8}{/tex} area of the whole square.
There are total number of 4 red small squares in the area of the whole square.
Total number of triangles in one small square {tex}=8{/tex}
So, total number of triangles in the area of the whole square {tex}=4 \times 8=32{/tex}.
Area of each triangle {tex}=\frac{1}{32}{/tex}
Area of the shaded region {tex}=2{/tex} triangles {tex}=2 \times \frac{1}{32}=\frac{2}{32}=\frac{1}{16}{/tex}.
Hence, the shaded region occupies {tex}\frac{1}{16}{/tex} area of the whole square.

Q.18: If we assume 1 gold dinar = 12 silver drammas, 1 silver dramma = 4 copper panas, 1 copper pana {tex}=6{/tex} mashakas, and 1 pana = 30 cowrie shells,
1 copper pana {tex}=\frac{1}{48}{/tex} gold dinar {tex}\left(\frac{1}{12} \times \frac{1}{4}\right){/tex}
1 cowrie shell = ________ copper panas
1 cowrie shell = ________ gold dinar.

Solution:

1 copper pana {tex}=\frac{1}{48}{/tex} gold dinar
1 copper pana = 30 cowrie shells
or 30 cowrie shells = 1 copper pana
or 1 cowrie shell {tex}=\frac{1}{30}{/tex} copper panas.
or 1 cowrie shell {tex}=\frac{1}{30} \times \frac{1}{48}{/tex} gold dinar {tex}=\frac{1}{1440}{/tex} gold dinar.

Q.19: Evaluate: {tex}3 \div \frac{7}{9}{/tex}

Solution:

{tex}3 \div \frac{7}{9}=3 \times \frac{9}{7}=\frac{27}{7}=3 \frac{6}{7}{/tex}.

Q.20: Evaluate: {tex}\frac{4}{3} \div \frac{3}{4}{/tex}

Solution:

{tex}\frac{4}{3} \div \frac{3}{4}=\frac{4}{3} \times \frac{4}{3}=\frac{16}{9}=1 \frac{7}{9}{/tex}

Q.21: Evaluate: {tex}\frac{1}{5} \div \frac{1}{9}{/tex}

Solution:

{tex}\frac{1}{5} \div \frac{1}{9}=\frac{1}{5} \times \frac{9}{1}=\frac{9}{5}=1 \frac{4}{5}{/tex}

Q.22: Evaluate: {tex}\frac{14}{4} \div 2{/tex}

Solution:

{tex}\frac {14}{4} \div 2 = \frac {14}{4} \times \frac 12 =1 \frac 34{/tex}

Q.23: Evaluate: {tex}\frac{7}{4} \div \frac{1}{7}{/tex}

Solution:

{tex}\frac{7}{4} \div \frac{1}{7}=\frac{7}{4} \times \frac{7}{1}=\frac{49}{4}=12 \frac{1}{4}{/tex}

Q.24: Evaluate: {tex}\frac{1}{6} \div \frac{11}{12}{/tex}

Solution:

{tex}\frac 16 \div \frac {11}{12} = \frac 16 \times \frac {12}{11} = \frac {2}{11}{/tex}

Q.25: Evaluate: {tex}\frac{2}{3} \div \frac{2}{3}{/tex}

Solution:

{tex}\frac{2}{3} \div \frac{2}{3} = \frac 23 \times \frac 37=1{/tex}

Q.26: Evaluate: {tex}\frac{8}{2} \div \frac{4}{15}{/tex}

Solution:

{tex}\frac{8}{2} \div \frac{4}{15} = \frac 82 \times \frac {15}{4}=15{/tex}

Q.27: Evaluate: {tex}3 \frac{2}{3} \div 1 \frac{3}{8}{/tex}

Solution:

{tex}3 \frac{2}{3} \div 1 \frac{3}{8}=\frac{11}{3} \div \frac{11}{8}=\frac{11}{3} \times \frac{8}{11}=\frac{8}{3}=2 \frac{2}{3}{/tex}.

Q.28: Evaluate: {tex}\frac{14}{6} \div \frac{7}{3}{/tex}

Solution:

{tex}\frac{14}{6} \div \frac{7}{3} = \frac {14}{6} \times \frac 37 = 1{/tex}

Q.29: Maria bought 8 m of lace to decorate the bags she made for school. She used {tex}\frac{1}{4} {~m}{/tex} for each bag and finished the lace. How many bags did she decorate?

Options:
(1) {tex}8 \times \frac{1}{4}{/tex}
(2) {tex}\frac{1}{8} \times \frac{1}{4}{/tex}
(3) {tex}8 \div \frac{1}{4}{/tex} ✅
(4) {tex}\frac{1}{4} \div 8{/tex}

Explanation:

Lace bought {tex}=8 {~m}{/tex}
Lace used to decorate one bag {tex}=\frac{1}{4} {~m}{/tex}
No. of bags decorated {tex}=8 \div \frac{1}{4}{/tex}.

Q.30: {tex}\frac{1}{2}{/tex} metre of ribbon is used to make 8 badges. What is the length of the ribbon used for each badge?

Options:
(1) {tex}8 \times \frac{1}{2}{/tex}
(2) {tex}\frac{1}{2} \div \frac{1}{8}{/tex}
(3) {tex}8 \div \frac{1}{2}{/tex}
(4) {tex}\frac{1}{2} \div 8{/tex} ✅

Explanation:

Length of ribbon used to make 8 badges {tex}=\frac{1}{2}{/tex} metres.
Length of ribbon used to make each badge {tex}=\frac{1}{2} \div 8{/tex}.

Q.31: A baker needs {tex}\frac{1}{6} {~kg}{/tex} of flour to make one loaf of bread. He has 5 kg of flour. How many loaves of bread can he make?

Options:
(1) {tex}5 \times \frac{1}{6}{/tex}
(2) {tex}\frac{1}{6} \div 5{/tex}
(3) {tex}5 \div \frac{1}{6}{/tex} ✅
(4) {tex}5 \times 6{/tex}

Explanation:

Quantity of flour needed to make 1 loaf of bread {tex}=\frac{1}{6} {~kg}{/tex}
Total quantity of flour {tex}=5 {~kg}{/tex}
No. of loafs of bread made {tex}=5 \div \frac{1}{6}{/tex}.

Q.32: If {tex}\frac{1}{4} {~kg}{/tex} of flour is used to make 12 rotis, how much flour is used to make 6 rotis?

Solution:

If {tex}1 / 4 {~kg}{/tex} is used for 12 rotis, then amount of flour per roti {tex}=\frac{1}{4} \div 12=\frac{1}{4} \times \frac{1}{12}=\frac{1}{48} {~kg}{/tex}.
For 6 rotis: {tex}6 \times \frac{1}{48}=\frac{6}{48}=\frac{1}{8} {~kg}{/tex}.
{tex}\frac{1}{8} {~kg}{/tex} of flour is used to make 6 rotis.

Q.33: Pāțīgaṇita, a book written by Sridharacharya in the 9th century CE, mentions this problem: “Friend, after thinking, what sum will be obtained by adding together {tex}1 \div \frac{1}{6}, 1 \div \frac{1}{10}, 1 \div \frac{1}{13}, 1 \div{/tex} {tex}\frac{1}{9}{/tex}, and {tex}1 \div \frac{1}{2}{/tex}”. What should the friend say?

Solution:

{tex} \text { The sum obtained } {/tex}{tex}=\left(1 \div \frac{1}{6}\right)+\left(1 \div \frac{1}{10}\right)+{/tex}{tex}\left(1 \div \frac{1}{13}\right)+\left(1 \div \frac{1}{9}\right)+\left(1 \div \frac{1}{2}\right) {/tex}
{tex} =\left(1 \times \frac{6}{1}\right)+\left(1 \times \frac{10}{1}\right)+{/tex}{tex}\left(1 \times \frac{13}{1}\right)+\left(1 \times \frac{9}{1}\right)+\left(1 \times \frac{2}{1}\right) {/tex}
{tex} =6+10+13+9+2=40 {/tex}
Therefore, the friend should say that the sum is 40.

Q.34: Mira is reading a novel that has 400 pages. She read {tex}\frac{1}{5}{/tex} of the pages yesterday and {tex}\frac{3}{10}{/tex} of the pages today. How many more pages does she need to read to finish the novel?

Solution:

The total number of pages in the novel {tex}=400{/tex}.
The number of pages read by Mira yesterday {tex}=\frac{1}{5}{/tex} of {tex}400=\frac{1}{5} \times 400=\frac{400}{5}=80{/tex} pages.
The number of pages read by Mira today {tex}=\frac{3}{10}{/tex} of {tex}400=\frac{3}{10} \times 400=\frac{1200}{10}=120{/tex} pages.
Total pages read by Mira {tex}=80+120=200{/tex} pages.
Therefore, the number of pages she needs to finish the novel {tex}=400-200=200{/tex} pages.

Q.35: A car runs 16 km using 1 litre of petrol. How far will it go using {tex}2 \frac{3}{4}{/tex} litres of petrol?

Solution:

Distance travelled by car in 1 litre of petrol {tex}=16 {~km}{/tex}.
Distance travelled by car in {tex}2 \frac{3}{4}{/tex} litres of petrol {tex}=16 \times 2 \frac{3}{4}=16 \times \frac{11}{4}{/tex} {tex}=4 \times 11=44 {~km}{/tex}.

Q.36: Amritpal decides on a destination for his vacation. If he takes a train, it will take him {tex}5 \frac{1}{6}{/tex} hours to get there. If he takes a plane, it will take him {tex}\frac{1}{2}{/tex} hour. How many hours does the plane save?

Solution:

The time taken by the train to reach the destination {tex}=5 \frac{1}{6}{/tex} hours.
The time taken by the plane to reach the destination {tex}=\frac{1}{2}{/tex} hour.
The time saved by travelling by plane {tex}=5 \frac{1}{6}-\frac{1}{2}{/tex}
{tex} =\frac{31}{6}-\frac{1}{2}=\frac{31-3}{6}{/tex}{tex}=\frac{28}{6}=\frac{14}{3}=4 \frac{2}{3} \text { hours. } {/tex}

Q.37: Mariam’s grandmother baked a cake. Mariam and her cousins finished {tex}\frac{4}{5}{/tex} of the cake. The remaining cake was shared equally by Mariam’s three friends. How much of the cake did each friend get?

Solution:

Part of cake finished by Mariam and her cousins {tex}=\frac{4}{5}{/tex}
Part of cake left {tex}=1-\frac{4}{5}=\frac{5-4}{5}=\frac{1}{5}{/tex}
This {tex}\frac{1}{5}{/tex} part of the cake is shared equally among 3 friends.
Part each friend got {tex}=\frac{1}{5} \div 3=\frac{1}{5} \times \frac{1}{3}=\frac{1}{15}{/tex}.
Thus, each of Mariam’s three friends got {tex}\frac{1}{15}{/tex} of the cake.

Q.38: Choose the option(s) describing the product of {tex}\left(\frac{565}{465} \times \frac{707}{676}\right){/tex}:

{tex} >\frac{565}{465} {/tex}
{tex} <\frac{565}{465} {/tex}
{tex} >\frac{707}{676} {/tex}
{tex} <\frac{707}{676} {/tex}
{tex} >1 {/tex}
< 1

Solution:

{tex} \frac{565}{465}>1 \text { and } \frac{707}{676}>1 {/tex}
Since, both numbers are greater than 1 , their product is greater than both the numbers.
Therefore, (a), (c) and (e) are correct options.

Q.39: What fraction of the whole square is shaded?

Solution:

The area of the bigger square is divided into four equal smaller squares.
The number of triangles in a smaller square {tex}=4{/tex}
Therefore, the total number of triangles in the bigger square {tex}=4 \times 4=16{/tex}
Area of each triangle {tex}=\frac{1}{16}{/tex}
Area of the shaded region = area of {tex}1 \frac{1}{2}{/tex} triangle {tex}=1 \frac{1}{2} \times \frac{1}{16}=\frac{3}{2} \times \frac{1}{16}=\frac{3}{32}{/tex}.
Therefore, {tex}\frac{3}{32}{/tex} fraction of the whole square is shaded.

Q.40: A colony of ants set out in search of food. As they search, they keep splitting equally at each point and reach two food sources, one near a mango tree and another near a sugarcane field. What fraction of the original group reached each food source?

Solution:

At first point ants split into two ways. So, fraction of ants at each way {tex}=1 \div 2=\frac{1}{2}{/tex}.
At second point ants split further into two ways. So, fraction of ants at each way {tex}=\frac{1}{2} \div 2=\frac{1}{2} \times{/tex} {tex}\frac{1}{2}=\frac{1}{4}{/tex}
At third point ants split further into four ways. So, fraction of ants at each way {tex}=\frac{1}{4} \div 4=\frac{1}{4} \times \frac{1}{4}{/tex} {tex}=\frac{1}{16}{/tex}.
At fourth point ants split further into two ways. So, fraction of ants at each way {tex}=\frac{1}{16} \div 2=\frac{1}{16}{/tex} {tex}\times \frac{1}{2}=\frac{1}{32}{/tex}.
Fraction of ants reaching mango tree {tex}=\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{16}+\frac{1}{32}{/tex}{tex}=\frac{16+8+2+2+1}{32}=\frac{29}{32}{/tex}.
Fraction of ants reaching sugarcane field {tex}=\frac{1}{16}+\frac{1}{32}=\frac{2+1}{32}=\frac{3}{32}{/tex}.

Q.41: What is {tex}\left(1-\frac{1}{2}\right){/tex}?
{tex} \left(1-\frac{1}{2}\right) \times\left(1-\frac{1}{3}\right) ? {/tex}
{tex} \left(1-\frac{1}{2}\right) \times\left(1-\frac{1}{3}\right) \times\left(1-\frac{1}{4}\right) \times\left(1-\frac{1}{5}\right) ? {/tex}
{tex} \left(1-\frac{1}{2}\right) \times\left(1-\frac{1}{3}\right) \times\left(1-\frac{1}{4}\right) \times{/tex}{tex}\left(1-\frac{1}{5}\right) \times\left(1-\frac{1}{6}\right) \times\left(1-\frac{1}{7}\right) {/tex}{tex}\times\left(1-\frac{1}{8}\right) \times\left(1-\frac{1}{9}\right) \times \left(1-\frac{1}{10}\right) ? {/tex}

Solution:

{tex}\left(1-\frac{1}{2}\right)=\frac{2-1}{2}=\frac{1}{2}{/tex}.
{tex}\left(1-\frac{1}{2}\right) \times\left(1-\frac{1}{3}\right)=\left(\frac{2-1}{2}\right) \times\left(\frac{3-1}{3}\right)=\frac{1}{2} \times \frac{2}{3}=\frac{1}{3}{/tex}.
{tex} \left(1-\frac{1}{2}\right) \times\left(1-\frac{1}{3}\right) \times\left(1-\frac{1}{4}\right) \times(1 \left.-\frac{1}{5}\right){/tex}{tex}=\left(\frac{2-1}{2}\right) \times\left(\frac{3-1}{3}\right) \times\left(\frac{4-1}{4}\right) \times\left(\frac{5-1}{5}\right) {/tex}
{tex} =\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}=\frac{1}{5} {/tex}
{tex}\left(1-\frac{1}{2}\right) \times\left(1-\frac{1}{3}\right) \times\left(1-\frac{1}{4}\right){/tex}{tex} \times\left(1-\frac{1}{5}\right) \times\left(1-\frac{1}{6}\right) \times\left(1-\frac{1}{7}\right) \times\left(1-\frac{1}{8}\right) \times\left(1-\frac{1}{9}\right) \times{/tex} {tex}\left(1-\frac{1}{10}\right){/tex}
{tex} =\left(\frac{2-1}{2}\right) \times\left(\frac{3-1}{3}\right) \times\left(\frac{4-1}{4}\right) {/tex}{tex}\times\left(\frac{5-1}{5}\right) \times\left(\frac{6-1}{6}\right) \times\left(\frac{7-1}{7}\right) \times\left(\frac{8-1}{8}\right) \times\left(\frac{9-1}{9}\right) \times\left(\frac{10-1}{10}\right) {/tex}
{tex} =\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \frac{5}{6} \times \frac{6}{7} \times \frac{7}{8} \times \frac{8}{9} \times \frac{9}{10}=\frac{1}{10} {/tex}

General statement:
{tex} \left(1-\frac{1}{2}\right) \times\left(1-\frac{1}{3}\right) \times\left(1-\frac{1}{4}\right) {/tex}{tex}\times\left(1-\frac{1}{5}\right) \ldots\left(1-\frac{1}{n}\right)=\frac{1}{n}. {/tex}

Q.42: A farmer had 5 grandchildren. She distributed {tex}\frac{2}{3}{/tex} acre of land to each of her grandchildren.
How much land in all did she give to her grandchildren.

Solution:

{tex} 5 \times \frac{2}{3}=\frac{2}{3}+\frac{2}{3}+\frac{2}{3}+\frac{2}{3}+\frac{2}{3}=\frac{10}{3} . {/tex}

Q.43: 1 hour of internet time costs ₹ 8. How much will {tex}1 \frac{1}{4}{/tex} hours of internet time cost?

Solution:

{tex}1 \frac{1}{4}{/tex} hours is {tex}\frac{5}{4}{/tex} hours (converting from a mixed fraction).
Cost of {tex}\frac{5}{4}{/tex} hour of internet time {tex}=\frac{5}{4} \times 8{/tex}
{tex} =5 \times \frac{8}{4} {/tex}
{tex} =5 \times 2 {/tex}
{tex} =10 . {/tex}
It costs ₹ 10 for {tex}1 \frac{1}{4}{/tex} hours of internet time.

Q.44: Leena made 5 cups of tea. She used {tex}\frac{1}{4}{/tex} litre of milk for this. How much milk is there in each cup of tea?

Solution:

Leena used {tex}\frac{1}{4}{/tex} litres of milk in 5 cups of tea. So, in 1 cup of tea the volume of milk should be:
{tex} \frac{1}{4} \div 5 {/tex}
Writing this as multiplication, we have:
{tex} 5 \times(\text { milk per cup })=\frac{1}{4} {/tex}
We perform the division as follows as per Brahmagupta’s method:
The reciprocal of 5 (the divisor) is {tex}\frac{1}{5}{/tex}.
Multiplying this reciprocal by the dividend ({tex}\frac{1}{4}{/tex}), we get
{tex} \frac{1}{5} \times \frac{1}{4}=\frac{1}{20} {/tex}
So, each cup of tea has {tex}\frac{1}{20}{/tex} litre of milk.

Q.45: Some of the oldest examples of working with non-unit fractions occur in humanity’s oldest geometry texts, the Śhulbasūtra. Here is an example from Baudhāyana’s Śhulbasūtra (c. 800 BCE).
Cover an area of {tex}7 \frac{1}{2}{/tex} square units with square bricks each of whose sides is {tex}\frac{1}{5}{/tex} units.
How many such square bricks are needed?

Solution:

Each square brick has an area of {tex}\frac{1}{5} \times \frac{1}{5}=\frac{1}{25}{/tex} square units.
The total area to be covered is {tex}7 \frac{1}{2}{/tex} sq. units {tex}=\frac{15}{2}{/tex} sq. units.
As {tex}({/tex}Number of bricks{tex}) \times({/tex}Area of a brick{tex})={/tex} Total Area,
{tex} \text { Number of bricks }=\frac{15}{2} \div \frac{1}{25} {/tex}
The reciprocal of the divisor is 25 .
Multiplying the reciprocal by the dividend, we get
{tex} 25 \times \frac{15}{2}=\frac{25 \times 15}{2}=\frac{375}{2} . {/tex}

Q.46: This problem was posed by Chaturveda Prithūdakasvāmī (c. 860 CE) in his commentary on Brahmagupta’s book Brāhmasphuṭasiddhānta.
Four fountains fill a cistern. The first fountain can fill the cistern in a day. The second can fill it in half a day. The third can fill it in a quarter of a day. The fourth can fill the cistern in one fifth of a day. If they all flow together, in how much time will they fill the cistern?

Solution:

Let us solve this problem step by step.
In a day, the number of times –

the first fountain will fill the cistern is {tex}1 \div 1=1{/tex}
the second fountain will fill the cistern is {tex}1 \div \frac{1}{2}={/tex} ________
the third fountain will fill the cistern is {tex}1 \div \frac{1}{4}={/tex} ________
the fourth fountain will fill the cistern is {tex}1 \div \frac{1}{5}={/tex} ________

The number of times the four fountains together will fill the cistern in a day is ________ {tex}+{/tex} ________ {tex}+{/tex} ________ {tex}+{/tex} ________ {tex}=12{/tex}.
Thus, the total time needed by the four fountains to fill the cistern together is {tex}\frac{1}{12}{/tex} days.