By using Ganita Prakash Class 7 Solutions and Part 1 Chapter 8 Working with Fractions Class 7 Question Answer, students can improve their problem-solving skills.
Class 7 Maths Chapter 8 Working with Fractions Solutions
Ganita Prakash Class 7 Chapter 8 Solutions
Class 7 Maths Ganita Prakash Chapter 8 Solutions Working with Fractions
8.1. Multiplication of Fractions
Page: 176
Question 1.
Figure it Out
1. Tenzin drinks \(\frac{1}{2}\) 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:
Tenzin drinks \(\frac{1}{2}\) glass of milk each day.
Number of days in a week = 7
Number of glasses of milk Tenzin
drinks in 7 days = 7 × \(\frac{1}{2}\) = \(\frac{7}{2}\) glasses
There are 31 days in the month of January.
So, he drinks in the month of January
= 31 × \(\frac{1}{2}\) = \(\frac{31}{2}\) glasses
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:
A team of workers can make 1 km of a water canal in 8 days.
So, in one day, the team can make \(\frac{1}{8}\) km of the water canal. If they work 5 days a week, they can make \(\frac{5}{8}\) km of the water canal in a week.
3. Manju and two 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:
Here, oil purchased every week = 5 litres
Number of families who share this oil = 3
Each family gets oil in a week
= \(\frac{5}{3}\) litres
One family will get oil in 4 weeks
= \(\frac{5}{3}\) × 4 litres
= \(\frac{20}{3}\) litres
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 \(\frac{5}{6}\) hour later than the previous day. How many hours after 10 pm will the moon set on Thursday?
Solution:
Number of days after Monday to Thursday = 3
Everyday the Moon sets \(\frac{5}{6}\) hour later.
Moon sets at 10 pm. So, number of hours after 10 pm will the moon set on Thursday = \(\frac{5}{6}\) × 3 = \(\frac{5}{2}\) hours
5. Multiply and then convert it into a mixed fraction :
(a) 7 × \(\frac{3}{5}\)
(b) 4 × \(\frac{1}{3}\)
(c) \(\frac{9}{7} \times 6\)
(d) \(\frac{13}{11} \times 6\)
Solution:
(a) 7 × \(\frac{3}{5}\) = \(\frac{21}{5}\) = 4 \(\frac{1}{5}\)
(b) 4 × \(\frac{1}{3}\) = \(\frac{4}{3}\) = 1 \(\frac{1}{3}\)
(c) \(\frac{9}{7}\) × 6 = \(\frac{54}{7}\) = 7 \(\frac{5}{7}\)
(d) \(\frac{13}{11}\) × 6 = \(\frac{78}{11}\) 7 \(\frac{1}{11}\)
Page: 180-181
Question 2.
Figure it Out
1. Find the following products. Use a unit square as a whole for representing the fractions:
(a) \(\frac{1}{3}\) × \(\frac{1}{5}\)
(b) \(\frac{1}{4}\) × \(\frac{1}{3}\)
(c) \(\frac{1}{5}\) × \(\frac{1}{2}\)
(d) \(\frac{1}{6}\) × \(\frac{1}{5}\)
Now, find \(\frac{1}{12}\) × \(\frac{1}{18}\)
Solution:
(a)
(b)
(c) \(\frac{1}{5}\) × \(\frac{1}{2}\) = \(\frac{1}{10}\)
(d) \(\frac{1}{6}\) × \(\frac{1}{5}\) = \(\frac{1}{30}\)
Now, \(\frac{1}{12}\) × \(\frac{1}{18}\) = \(\frac{1}{216}\)
2. Find the following products. Use a unit square as a whole for representing the fractions and carrying out the operations.
(a) \(\frac{2}{3}\) × \(\frac{4}{5}\)
(b) \(\frac{1}{4}\) × \(\frac{2}{3}\)
(c) \(\frac{3}{5}\) × \(\frac{1}{2}\)
(d) \(\frac{4}{6}\) × \(\frac{3}{5}\)
Solution:
(a)
∴ \(\frac{2}{3}\) × \(\frac{4}{5}\) = \(\frac{8}{15}\)
(b) \(\frac{1}{4}\) × \(\frac{2}{3}\) = \(\frac{2}{12}\) = \(\frac{1}{6}\)
(c) \(\frac{3}{5}\) × \(\frac{1}{2}\) = \(\frac{3}{10}\)
(d) \(\frac{4}{6}\) × \(\frac{3}{5}\) = \(\frac{12}{30}\) = \(\frac{2}{5}\)
Page: 183-184
Question 3.
Figure it Out
1. A water tank is filled from a tap. If the tap is open for 1 hour, \(\frac{7}{10}\) of the tank gets filled. How much of tank is filled if the tap is open for
(a) \(\frac{1}{3}\) hour ______
(b) \(\frac{2}{3}\) hour ______
(c) \(\frac{3}{4}\) hour ______
(d) \(\frac{7}{10}\) hour ______
(e) For the tank to be full, how long should the tap be running?
Solution:
(a) In \(\frac{1}{3}\) hour = \(\frac{1}{3}\) × \(\frac{7}{10}\) = \(\frac{7}{30}\) of the tank.
(b) In \(\frac{2}{3}\) hour =\(\frac{2}{3}\) × \(\frac{7}{10}\)=\(\frac{14}{30}\) = \(\frac{7}{15}\) of the tank.
(c) In \(\frac{3}{4}\) hour =\(\frac{3}{4}\) × \(\frac{7}{10}\)=\(\frac{21}{40}\) of the tank.
(d) In \(\frac{7}{10}\) hour =\(\frac{7}{10}\) × \(\frac{7}{10}\)=\(\frac{49}{100}\) of the tank.
(e) For the tank to be full, how long should the tap be running = \(\frac{7}{10}\) hours
= \(\frac{10}{7}\) hours = 1 \(\frac{3}{7}\) hours.
2. For government has taken \(\frac{1}{6}\) 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 \(\frac{1}{3}\) of it to her son Bora. After giving them their shares, she keeps the remaining land for herself.
(a) What part of the of original land did Krishna get?
(b) What part of the original land did Bora get?
(c) What part of the original land did Somu keep for herself?
Solution:
Let us start with a unit square.
(a) Part of the original land Krishna got
= \(\frac{15}{36}\) = \(\frac{5}{12}\)
(b) Part of the original land Bora got = \(\frac{10}{36}\)
= \(\frac{5}{18}\)
(c) Part of the original land Somu kept for
herself = \(\frac{5}{36}\)
3. Find the area of a rectangle of sides 3 \(\frac{3}{4}\) ft and 9 \(\frac{3}{5}\)ft.
Solution:
Area of rectangle = Length × Breadth
= 3 \(\frac{3}{4}\) ft × 9 \(\frac{3}{5}\) ft
4. Tsewang plant four saplings in a row in his garden. The distance between two saplings is \(\frac{3}{4}\) m. Find the distance between the first and last saplings.
[Hint: Draw a rough diagram with four saplings with distance two saplings as \(\frac{3}{4}\) m.
Solution:
Tsewang planted 4 saplings in a row. The distance between two saplings is \(\frac{3}{4}\) m.
The distance between the first and the last sapling = 3 × \(\frac{3}{4}\) = \(\frac{9}{4}\) = 2 \(\frac{1}{4}\) m.
The distance between the first and the last sapling = 3 × \(\frac{3}{4}\) = \(\frac{9}{4}\) = 2 \(\frac{1}{4}\) m.
5. Which is heavier: \(\frac{12}{15}\) of 500 grams or \(\frac{3}{20}\) of 4 kg ?
Solution:
= 600 grams
Hence, \(\frac{3}{20}\) of 4 kg is heavier.
Page: 185
Question 4.
What can you conclude about the relationship between the numbers multiplied and the product? Fill in the blanks :
1. When one of the numbers being multiplied is between 0 and 1, the product is ______ (greater/less) than the other numbers.
Solution:
Less
2. When one of the numbers being multiplied is greater than 1, the product is ______ (greater/less) than the other number.
Solution:
Greater
8.3. Some Problems involving Fraction
Page: 193
Question 5.
In each of the figure given below, find the fraction of the big square that the shaded region occupies.
Solution:
First figure \(\frac{6}{16}\)
Second figure \(\frac{1}{16}\)
Page : (196-198)
Question 6.
Figure it Out
1. Evaluate the following:
Solution:
2. For each of the questions below, choose the expression that describes the solution. Then simplify it.
(a) Maria bought 8 m of lace to decorate the bags she made for school. She used \(\frac{1}{4}\) m for each bag and finished the lace. How many bags did she decorate?
(i) 8 × \(\frac{1}{4}\)
(ii) \(\frac{1}{8}\) × \(\frac{1}{4}\)
(iii) 8 ÷ \(\frac{1}{4}\)
(iv) \(\frac{1}{4}\) ÷ 8
(b) \(\frac{1}{2}\) meter of ribbon is used to make 8 badges. What is the length used for each badge?
(i) 8 × \(\frac{1}{2}\)
(ii) \(\frac{1}{2}\) × \(\frac{1}{8}\)
(iii) 8 ÷ \(\frac{1}{2}\)
(iv) \(\frac{1}{2}\) ÷ 8
(c) A baker needs \(\frac{1}{6}\) kg of flour to make one loaf of bread. He has 5 kg of flour. How many loaves of bread can he make?
(i) 5 × \(\frac{1}{6}\)
(ii) \(\frac{1}{6}\) × 5
(iii) 5 ÷ \(\frac{1}{6}\)
(iv) 5 × 6
Solution:
(a) (iii) 8 ÷ \(\frac{1}{4}\) = 8 × 4 = 32
(b) (iv) \(\frac{1}{2}\) ÷ 8 = \(\frac{1}{2}\) × \(\frac{1}{8}\) = \(\frac{1}{16}\)
(c) (iii) 5 ÷ \(\frac{1}{6}\) = 5 × 6 = 30
3. If \(\frac{1}{4}\) kg of flour is used to make 12 rotis, how much flour is used to make 6 rotis?
Solution:
Flour required to make 12 rotis = \(\frac{1}{4}\) kg
∴ Flour required to make 6 rotis = \(\frac{1}{2}\) of \(\frac{1}{4}\)
= \(\frac{1}{8}\) kg
4. Patiganita, a book written by Sridharacharya in the 9th century CE, mentions this problem: “Friend, after thinking, what sum will be obtained by adding together 1 ÷ \(\frac{1}{6}\), 1 ÷ \(\frac{1}{10}\), 1 ÷ \(\frac{1}{13}\), 1 ÷ \(\frac{1}{9}\) “. What should the friend say?
Solution:
1 ÷ \(\frac{1}{6}\) = 1 × \(\frac{6}{1}\) = 6
1 ÷ \(\frac{1}{10}\) = 1 × \(\frac{10}{1}\) = 10
1 ÷ \(\frac{1}{13}\) = 1 × \(\frac{13}{1}\) = 13
1 ÷ \(\frac{1}{9}\) = 1 × \(\frac{9}{1}\) = 9 and
1 ÷ \(\frac{1}{2}\) = 1 × \(\frac{2}{1}\) = 2
Now the required sum is:
6 + 10 + 13 + 9 + 2 = 40
5. Mira is reading a novel that has 400 pages. She read \(\frac{1}{5}\) of the pages yesterday and \(\frac{3}{10}\) of the pages today. How many more pages does she need to read to finish the novel?
Solution:
Total number of pages = 400
Number of pages she read yesterday = 400 × \(\frac{1}{5}\) = 80
Number of pages she read today =400 × \(\frac{3}{10}\) = 120
Total number of pages she has already read = 80 + 120 = 200
Number of more pages she need to read to finish the novel = 400-200 = 200
6. A car runs 16 km using 1 litre of petrol. How far will it go using 2 \(\frac{3}{4}\) litres of petrol?
Solution:
Distance covered using 1 litre of petrol = 16 km
Distance covered using 2 \(\frac{3}{4}\) litres of petrol
= 2 \(\frac{3}{4}\) × 16
7. Amritpal decides on a destination for his vacation. If he takes a train, it will take him 5 \(\frac{1}{6}\) hours to get there. If he takes a plane, it will take him \(\frac{1}{2}\) hour. How many hours does the plane save?
Solution:
Time taken by train = 5 \(\frac{1}{6}\) hours
Time taken by plane = \(\frac{1}{2}\) hours
∴ the plane saves = 5 \(\frac{1}{6}\)–\(\frac{1}{2}\)
= \(\frac{31}{6}\) – \(\frac{1}{2}\) = \(\frac{31}{6}\) – \(\frac{3}{6}\) = \(\frac{28}{6}\) = \(\frac{14}{3}\) hours
= 4 \(\frac{2}{3}\) hours
8. Mariam’s grandmother baked a cake. Mariam and her cousins finished \(\frac{4}{5}\) of the cake. The remaining cake was shared equally by Mariam’s three friends. How much of the cake did each friend get?
Solution:
Since, Mariam and her cousins finished \(\frac{4}{5}\) of the cake
Hence, the part of cake remaining = 1 – \(\frac{4}{5}\)
= \(\frac{1}{5}\)
Now, remaining cake is being shared by Mariam’s three friends equally
Hence, part of the cake each friend gets = \(\frac{1}{3}\) of \(\frac{1}{5}\)=\(\frac{1}{3}\) × \(\frac{1}{5}\) = \(\frac{1}{15}\)
9. Choose the option(s) describing the product of (\(\frac{565}{465}\) × \(\frac{707}{676}\)) :
(a) > \(\frac{565}{465}\)
(b) < \(\frac{565}{465}\)
(c) > \(\frac{707}{676}\)
(d) < \(\frac{707}{676}\)
(e) > 1
(f) < 1
Solution:
(a), (c) and (e)
10. What fraction of the whole square is shaded?
Solution:
\(\frac{3}{32}\)
11. A colony of ants set out in search of food. As they search, they keep splitting equally at each point (as shown in the Fig. 8.7) 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:
Fraction reached near mango tree
\(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{16}\) + \(\frac{1}{16}\) + \(\frac{1}{32}\)
= \(\frac{16+8+2+2+1}{32}\) = \(\frac{29}{32}\)
Fraction reached near sugarcane field
= \(\frac{1}{32}\)+\(\frac{1}{16}\) = \(\frac{3}{32}\)
12. What is 1 – \(\frac{1}{2}\)
Make a general statement and explain.
Solution:
The general statement is:
