Practicing Ganita Prakash Class 7 Solutions and RBSE Class 7 Maths Chapter 8 Working with Fractions Solutions Question Answer helps develop logical thinking and accuracy.
RBSE Class 7 Maths Chapter 8 Working with Fractions Solutions
Ganita Prakash Class 7 Chapter 8 Solutions
Class 7 Maths Ganita Prakash Part 1 Chapter 8 Solutions
In-text Questions
Page 173
Question 1.
Aaron walks 3 kilometres in 1 hour. How far can he walk in 5 hours?
Solution:
∵ Distance covered by Aaron in 1 hour
= 3 kilometres
∴ Distance covered by Aaron in 5 hours
= 5 × 3 kilometres
= 3 + 3 + 3 + 3 + 3 kilometers
= 15 kilometres
Question 2.
Aaron’s pet tortoise walks out a much slower pace. It can walk only \(\frac {1}{4}\) kilometre in 1 hour. How far can it walk in 3 hours?
Solution:
∵ Distance covered in 1 hour = \(\frac {1}{4}\) km
∴ Distance covered in 3 hours = 3 × \(\frac {1}{4}\)km
= (\(\frac {1}{4}\) + \(\frac {1}{4}\) + \(\frac {1}{4}\))km = \(\frac {3}{4}\)km
Hence, the tortoise can walk \(\frac {3}{4}\) km in 3 hours.
Page 174
Question 1.
We saw that Aaron can walk 3 kilometres in 1 hour. How far can he walk in \(\frac {1}{5}\) hours?
Solution:
Distance covered in \(\frac {1}{5}\) hours
\(\frac {1}{5}\) × 3 km = \(\frac {3}{5}\) km.
Question 2.
How far can Aaron walk in \(\frac {2}{5}\) hours?
Solution:
Distance covered in \(\frac {2}{5}\) hours = \(\frac {2}{5}\) × 3 km.
= \(\frac {6}{5}\)km
Page 182
Question 1.
Multiply the following fractions and express the product in its lowest form :
\(\frac {12}{7}\) × \(\frac {5}{24}\).
Solution:
When multiplying fractions, we can first divide the numerator and denominator by their common factors before multiplying the numerators and denominators. This is called cancelling the common factors.
\(\frac {12}{7}\) × \(\frac {5}{24}\) = \(\frac{1 \times 5}{7 \times 2}\) = \(\frac {5}{14}\).
Page 185
Question 1.
What can you conclude about the relationship between the numbers multiplied and the product? Fill in the blanks:
(i) When one of the numbers being multiplied is between 0 and 1, the product is ________ (greater/less) than the other number.
(ii) When one of the numbers being multiplied is greater than 1, the product is ________ (greater/less) than the other number.
Solution:
(i) less
(ii) greater
Page 187
Question 1.
This is the same as \(\frac {2}{3}\) × ? = 3
Can you find the answer?
Solution:
\(\frac {2}{3}\) × ? = 3
? = 3 ÷ \(\frac {2}{3}\) = 3 × \(\frac {3}{2}\) = \(\frac {9}{2}\)
Page 188
Question 1.
What is \(\frac {2}{3}\) ÷ \(\frac {3}{5}\)
Solution:
\(\frac {2}{3}\) ÷ \(\frac {3}{5}\)
= \(\frac {2}{3}\) × \(\frac {5}{3}\)
= \(\frac {10}{9}\)
Page 191
Question 1.
Four fountains fill a cistern. The fist 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:
In a day, the number of times—
- the first fountain will fill the cistern is 1 ÷ 1 = 1 time
- the second fountain will fill the cistern is 1 ÷ \(\frac {1}{2}\) = 2 times
- the third fountain will fill the cistern is 1 ÷ \(\frac {1}{4}\) = 4 times
- the fourth fountain will fill the cistern is 1 ÷ \(\frac {1}{5}\)= 5 times
The number of times the four fountains together will fill the cistern in a day is 1 + 2 + 4 + 5 = 12. Thus, the total time needed by the four fountains to fill the cistern together is \(\frac {1}{12}\) days.
Page 193
Question 1.
In each of the figures given below, find the fraction of the big square that the shaded region occupies.
Solution:
Shaded region
= 3 × \(\frac {1}{8}\) = \(\frac {3}{8}\)
= \(\frac {1}{16}\)
Class 7 Maths Ganita Prakash Chapter 8 Solutions
Figure it Out (Pages 176-177)
Question 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:
∵ Milk drank by Tenzin every day
= \(\frac {1}{2}\) glass .
∴ Milk drank by Tenzin in a week
= \(\frac {1}{2}\) × 7 glasses
∵ There are 7 days in a week 7
= \(\frac {7}{2}\) glasses
Again, milk drank by Tenzin in the month of January = \(\frac {1}{2}\) × 31 glasses
∵ There are 31 days in the month of January 31
= \(\frac {31}{2}\) glasses
Question 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 team of workers in 8 days can make = 1 km of water canal
∴ The team of workers in 1 day can make
= \(\frac {1}{8}\) km of water canal
∴ The team of workers in 5 days of a week can make
= \(\frac {1}{8}\) × 5 km of water canal
= \(\frac {5}{8}\) km of water canal
Hence, the team of workers can make \(\frac {5}{8}\) km of water canal in a week.
Question 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:
∵ Oil shared every week equally among the 3 families = 5 litres
∴ Oil each family get in a week = \(\frac {5}{3}\) litres
∴ Oil each family gets in 4 weeks
= \(\frac {5}{3}\) × 4 litres
= 5 × \(\frac {4}{3}\) litres = \(\frac {20}{3}\) litres
Hence, one family gets \(\frac {20}{3}\) litres of oil in 4 weeks.
Question 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 from Monday to Thursday
= 3 (Monday to Tuesday, Tuesday to Wednesday, Wednesday to Thursday)
∵ Moon sets in one day = \(\frac {5}{6}\) hour later
∴ Moon sets in three days
= \(\frac {5}{6}\) × 3 hours later
= 5 × \(\frac {3}{6}\) hours later = 5 × \(\frac {1}{2}\) hours later
= \(\frac {5}{2}\) hours later
Hence, the moon will set \(\frac{5}{2}\) hours later after 10 pm on Thursday.
Question 5.
Multiply and then convert it into a mixed fraction :
(a) 7 × \(\frac {3}{5}\)
Solution:
7 × \(\frac {3}{5}\)
= \(\frac {21}{5}\) = \(\frac {20+1}{5}\)
= \(\frac {20}{5}\) + \(\frac {1}{5}\) = 4 + \(\frac {1}{5}\) = 4\(\frac {1}{5}\)
(b) 4 × \(\frac {1}{2}\)
Solution:
4 × \(\frac {1}{2}\) = \(\frac {4}{3}\) = \(\frac {3+1}{3}\)
= \(\frac {3}{3}\) + \(\frac {1}{3}\) = 1 + \(\frac {1}{3}\) = 1\(\frac {1}{3}\)
(c) \(\frac {9}{7}\) × 6
Solution:
\(\frac {9}{7}\) × 6 = \(\frac {54}{7}\)
= \(\frac {49+5}{7}\) = \(\frac {49}{7}\) + \(\frac {5}{7}\)
= 7 + \(\frac {5}{7}\) = 7\(\frac {5}{7}\)
(d) \(\frac {13}{11}\) × 6
Solution:
\(\frac {13}{11}\) × 6 = \(\frac {78}{11}\)
= \(\frac {77+1}{11}\) = \(\frac {77}{11}\) + \(\frac {1}{11}\)
= 7 + \(\frac {1}{11}\) = 7\(\frac {1}{11}\)
Figure it Out (Page 180-181)
Question 1.
Find the following products. Use a unit square as a whole for representing the fractions:
(a) \(\frac {1}{3}\) × \(\frac {1}{5}\)
Solution:
∴ \(\frac {1}{3}\) × \(\frac {1}{5}\) = \(\frac {1}{15}\)
(b) \(\frac {1}{4}\) × \(\frac {1}{3}\)
Solution:
∴ \(\frac {1}{4}\) × \(\frac {1}{3}\) = \(\frac {1}{12}\)
(c) \(\frac {1}{5}\) × \(\frac {1}{2}\)
Solution:
∴ \(\frac {1}{5}\) × \(\frac {1}{2}\) = \(\frac {1}{10}\)
(d) \(\frac {1}{6}\) × \(\frac {1}{5}\)
Solution:
∴ \(\frac {1}{6}\) × \(\frac {1}{5}\) = \(\frac {1}{30}\)
(e) \(\frac {1}{12}\) × \(\frac {1}{18}\)
Solution:
∴ \(\frac {1}{12}\) × \(\frac {1}{18}\) = \(\frac {1}{216}\)
Question 2.
Find the following products. Use a unit square as a whole for representing the fractions and carrying out the operations.
(a) \(\frac {2}{3}\) × \(\frac {4}{5}\)
Solution:
∴ \(\frac {2}{3}\) × \(\frac {4}{5}\) = \(\frac {8}{15}\)
(b) \(\frac {1}{4}\) × \(\frac {2}{3}\)
Solution:
∴ \(\frac {1}{4}\) × \(\frac {2}{3}\) = \(\frac {2}{12}\)
(c) \(\frac {3}{5}\) × \(\frac {1}{2}\)
Solution:
∴ \(\frac {3}{5}\) × \(\frac {1}{2}\) = \(\frac {3}{10}\)
(d) \(\frac {4}{6}\) × \(\frac {3}{5}\)
Solution:
∴ \(\frac {4}{6}\) × \(\frac {3}{5}\) = \(\frac {12}{30}\)
Figure it Out (Pages 183-184)
Question 1.
A water tank is filed from a tap. If the tap is open for 1 hour, \(\frac {7}{10}\) the tank gets filed. How much of the tank is filed 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) ∵ Tank filled in 1 hour = \(\frac {7}{10}\)
∴ Tank filled in \(\frac {1}{3}\) hour = \(\frac {7}{10}\) × \(\frac {1}{3}\)
= \(\frac{7 \times 1}{10 \times 3}\) = \(\frac {7}{30}\)
(b) ∵ Tank filled in 1 hour = \(\frac {7}{10}\)
∴ Tank filled in \(\frac {2}{3}\) hour = \(\frac {7}{10}\) × \(\frac {2}{3}\)
= \(\frac{7 \times 1}{5 \times 3}\) = \(\frac {7}{15}\)
(c) ∵ Tank filled in 1 hour = \(\frac {7}{10}\)
∴ Tank filled in \(\frac {3}{4}\) hour = \(\frac {7}{10}\) × \(\frac {3}{4}\)
= \(\frac{7 \times 3}{10 \times 4}\) = \(\frac {21}{40}\)
(d) ∵ Tank filled in 1 hour = \(\frac {7}{10}\)
∴ Tank filled in \(\frac {7}{10}\) hour = \(\frac {7}{10}\) × \(\frac {7}{10}\)
= \(\frac{7 \times 7}{10 \times 10}\) = \(\frac {49}{100}\)
(e) ∵ \(\frac {7}{10}\) of the tank get filled in = 1 hour
∴ 1 tank gets filled in \(\frac {10}{7}\) hours
Hence, the tap should be running for \(\frac {10}{7}\) hours.
Question 2.
The 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 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:
Part of the land may remains with Somu now
= 1 – \(\frac {1}{6}\)
= \(\frac {6}{6}\) – \(\frac {1}{6}\) = \(\frac {6-1}{6}\) = \(\frac {5}{6}\)
(a) Krishna received half of the remaining land
\(\frac {1}{2}\) × \(\frac {5}{6}\) = \(\frac {5}{12}\)
Hence, Krishna received \(\frac {5}{12}\) part.
(b) Bora received one-third of the remaining land.
\(\frac {1}{3}\) × \(\frac {5}{6}\) = \(\frac {5}{18}\)
Hence, Bora received \(\frac {5}{8}\) part.
(c) Part of the original land Somu kept for herself
\(\begin{aligned}
& =\frac{5}{6}-\left(\frac{5}{12}+\frac{5}{18}\right) \\
& =\frac{5}{6}-\frac{15+10}{36}=\frac{5}{6}-\frac{25}{36} \\
& =\frac{30-25}{36}=\frac{5}{36}
\end{aligned}\)
Question 3.
Find the area of a rectangle of sides 3\(\frac {3}{4}\)ft and 9\(\frac {3}{5}\)ft.
Solution:
Length of the rectangle = 3\(\frac {3}{4}\) ft = \(\frac {15}{4}\)ft
Breadth of the rectangle = 9\(\frac {3}{5}\)ft \(\frac {48}{5}\)ft
Area of the reactangle = Length × Breadth
= \(\frac {15}{4}\) × \(\frac {48}{5}\)
= 3 × 12 = 36 sq. ft
Question 4.
Tsewang plants 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 sapling. [Hint: Draw a rough diagram with four saplings with distance between two saplings as \(\frac {3}{4}\) m]
Solution:
Rough Diagram:
Distance between the first and last sapling
= \(\frac {3}{4}\)m + \(\frac {3}{4}\)m + \(\frac {3}{4}\)m = 3 × \(\frac {3}{4}\)m
= \(\frac {3}{1}\) × \(\frac {3}{4}\)m = \(\frac{3 \times 3}{1 \times 4}\)m = \(\frac {9}{4}\)m
Question 5.
Which is heavier: \(\frac {12}{15}\) of 500 grams or \(\frac {3}{20}\) of 4 kg?
Solution:
\(\frac {12}{15}\) of 500 grams = \(\frac {12}{15}\) × 500 grams
= \(\frac {12}{15}\) × \(\frac {500}{1}\) grams
= 4 × 100 grams = 400 grams 3 3
\(\frac {3}{20}\) of 4 kg = \(\frac {3}{20}\) of 4000 grams
∵ 1 kg= 1000 grams
= \(\frac {3}{20}\) × 4000 grams
= \(\frac {3}{20}\) × \(\frac {4000}{1}\) grams
= 3 × 200 grams = 600 grams
∵ 600 grams > 400 grams
∴ \(\frac {3}{20}\) of 4 kg is heavier than \(\frac {12}{15}\) of 500 grams
Figure it Out (Pages 196-198)
Question 1.
Evaluate the following:
(1) 3 ÷ \(\frac {7}{9}\)
(2) \(\frac {14}{4}\) ÷ 2
(3) \(\frac {2}{3}\) ÷ \(\frac {2}{3}\)
(4) \(\frac {14}{6}\) ÷ \(\frac {7}{3}\)
(5) \(\frac {4}{3}\) ÷ \(\frac {3}{4}\)
(6) \(\frac {7}{4}\) ÷ \(\frac {1}{7}\)
(7) \(\frac {8}{2}\) ÷ \(\frac {4}{15}\)
(8) \(\frac {1}{5}\) ÷ \(\frac {1}{9}\)
(9) \(\frac {1}{6}\) ÷ \(\frac {11}{12}\)
(10) 3 \(\frac {2}{3}\) ÷1 \(\frac {3}{8}\)
Solution:
\(\text { (1) } \begin{aligned}
3 \div \frac{7}{9} & =\frac{3}{1} \div \frac{7}{9}=\frac{3}{1} \times \frac{9}{7} \\
& =\frac{3 \times 9}{1 \times 7}=\frac{27}{7}
\end{aligned}\)
\((2) \frac{14}{4} \div 2=\frac{14}{4} \div \frac{2}{1}=\frac{14}{4} \times \frac{1}{2}=\frac{14 \times 1}{4 \times 2}=\frac{7}{4}\\
(3) \frac{2}{3} \div \frac{2}{3}=\frac{2}{3} \times \frac{3}{2}=\frac{2 \times 3}{3 \times 2}=1\\
(4) \frac{14}{6} \div \frac{7}{3}=\frac{14}{6} \times \frac{3}{7}=\frac{14 \times 3}{6 \times 7}=1\\
(5) \frac{4}{3} \div \frac{3}{4}=\frac{4}{3} \times \frac{4}{3}=\frac{4 \times 4}{3 \times 3}=\frac{16}{9}\\
(6) \frac{7}{4} \div \frac{1}{7}=\frac{7}{4} \times \frac{7}{1}=\frac{7 \times 7}{4 \times 1}=\frac{49}{4}\\
(7) \frac{8}{2} \div \frac{4}{15}=\frac{8}{2} \times \frac{15}{4}=\frac{8 \times 15}{2 \times 4}=15\\
(8) \frac{1}{5} \div \frac{1}{9}=\frac{1}{5} \times \frac{9}{1}=\frac{1 \times 9}{5 \times 1}=\frac{9}{5}\\
(9) \frac{1}{6} \div \frac{11}{12}=\frac{1}{6} \times \frac{12}{11}=\frac{1 \times 12}{6 \times 11}=\frac{2}{11}\)
\(\text { (10) } \begin{aligned}
3 \frac{2}{3} \div 1 \frac{3}{8}= & \frac{3 \times 3+2}{3} \div \frac{1 \times 8+3}{8} \\
& =\frac{11}{3} \div \frac{11}{8}=\frac{11}{3} \times \frac{8}{11} \\
& =\frac{11 \times 8}{3 \times 11}=\frac{8}{3} .
\end{aligned}\)
Question 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
Solution:
To find the total number of bags decorated by Maria, we need to divide the total length of the ribbon by the length of the ribbon required to decorated one bag.
i. e., 8 ÷ \(\frac {1}{4}\) = 8 × 4 = 32
Therefore, expression (iii) 8 ÷ \(\frac {1}{4}\) is correct.
(b) \(\frac {1}{2}\) meter of ribbon is used to make 8 badges. What is the length of the ribbon 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
Solution:
To find the length of ribbon required for one badge, we need to divide the total length of the ribbon by the total number of badges made.
i.e. \(\frac {1}{2}\) ÷ 8 = \(\frac {1}{2}\) × \(\frac {1}{8}\) = \(\frac {1}{16}\)
Therefore, expression (iv) \(\frac {1}{2}\) ÷ 8 is correct.
(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:
To find the total number of loaves of bread that can be made, we need to divide the total weight of flour by the weight of the flour that is required to make one loaf of bread.
i.e. 5 ÷ \(\frac {1}{6}\) = 5 × 6 = 30
Therefore, expression (iii) 5 ÷ \(\frac {1}{6}\) is correct.
Question 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 used to make 12 rotis = \(\frac {1}{4}\)kg
∴ Flour used to make 1 roti = \(\frac{1}{4 \times 12}\)kg
∴ Flour used to make 6 rotis = \(\frac{1 \times 6}{4 \times 12}\)kg
= \(\frac {1}{8}\)kg
Question 4.
Patiganita, a book written by Sridhar- acharya 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 ÷ \(\frac {1}{9}\) and 1 ÷ \(\frac {1}{2}\).
What should the friend say?
Solution:
\(1 \div \frac{1}{6}=\frac{1}{1} \times \frac{6}{1}=\frac{1 \times 6}{1 \times 1}=6\)
\(\begin{aligned}
& 1 \div \frac{1}{10}=\frac{1}{1} \times \frac{10}{1}=\frac{1 \times 10}{1 \times 1}=10 \\
& 1 \div \frac{1}{13}=\frac{1}{1} \times \frac{13}{1}=\frac{1 \times 13}{1 \times 1}=13 \\
& 1 \div \frac{1}{9}=\frac{1}{1} \times \frac{9}{1}=\frac{1 \times 9}{1 \times 1}=9 \\
& 1 \div \frac{1}{2}=\frac{1}{1} \times \frac{2}{1}=\frac{1 \times 2}{1 \times 1}=2
\end{aligned}\)
∴ Sum obtained by adding together = 6 + 10 + 13 + 9 + 2 = 40
Question 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:
400 × \(\frac {1}{2}\) = 80
400 × \(\frac {1}{2}\) = 40 × 3 = 120
∴ Total number of pages read by Mira=80 + 120 = 200
Thus, the number of pages she needs to read to finish the novel = 400 – 200 = 200
Question 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:
∵ In 1 litre of petrol, car runs= 16 km
∴ 2\(\frac {3}{4}\) litres (i.e. \(\frac {11}{4}\) litres) of petrol, car will run
= \(16 \times \frac{11}{4}=\frac{16}{1} \times \frac{11}{4}=44 \mathrm{~km}\)
Question 7.
Amritpal decides on a destination for his vacation. If he takes a train, it will take him 5\(\frac {1}{7}\) 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 the train to reach the destination = 5\(\frac {1}{6}\)hours = \(\frac {31}{6}\) hours
Time taken by the plane to reach the destination = \(\frac {1}{2}\)hour
∴ Number of hours saved by the plane
\(=\frac{31}{6}-\frac{1}{2}=\frac{31}{6}-\frac{3}{6}=\frac{31-3}{6}=\frac{28}{6}=\frac{14}{3} \text { hours }\)
Question 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:
Remaining cake = 1 – \(\frac {4}{5}\) = \(\frac {5}{5}\) – \(\frac {4}{5}\)
= \(\frac {5-4}{5}\) = \(\frac {1}{5}\)
∵ It was shared equally by three friends.
∴ Each friend got \(=\frac{1}{5} \times \frac{1}{3}=\frac{1 \times 1}{5 \times 3}=\frac{1}{15}\)
Question 9.
Choose the option(s) describing the product of \(\left(\frac{565}{465} \times \frac{707}{676}\right)\):
(a) > \(\frac {565}{465}\)
(b) < \(\frac {565}{465}\) (c) > \(\frac {707}{676}\)
(d) < \(\frac {707}{676}\) (e) > 1
(f) < 1
Solution:
Option (a), (c) and (e) are correct.
Question 10.
What fraction of the whole square is shaded?
Solution:
Fraction of the whole square shaded
\(=\frac{1}{16}+\frac{1}{32}=\frac{2}{32}+\frac{1}{32}=\frac{3}{32}\)
Question 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.) 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 = 1 ÷ 2 = \(\frac {1}{2}\)
At the second point, ants split into two ways. So, fraction of ants at each way
\(=\frac{1}{2} \div 2=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)
At the third point, ants split into four ways. So, fraction of ants at each way
\(=\frac{1}{4} \div 4=\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}\)
At the fourth point, ants split into two ways. So, fraction of ants at each way
\(=\frac{1}{16} \div 2=\frac{1}{16} \times \frac{1}{2}=\frac{1}{32}\)
Hence, a fraction of ants at the mango tree
\(\begin{aligned}
& =\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{16}+\frac{1}{32} \\
& =\frac{16+8+2+2+1}{32}=\frac{29}{32}
\end{aligned}\)
Fraction of ants near sugarcane field is
\(=\frac{1}{32}+\frac{1}{16}=\frac{1+2}{32}=\frac{3}{32}\)
Question 12.
What is 1 – \(\frac {1}{2}\)?
\(\begin{aligned}
& \left(1-\frac{1}{2}\right) \times\left(1-\frac{1}{3}\right) ? \\
& \left(1-\frac{1}{2}\right) \times\left(1-\frac{1}{3}\right) \times\left(1-\frac{1}{4}\right) \times\left(1-\frac{1}{5}\right) ? \\
& \left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{6}\right) \\
& \quad\left(1-\frac{1}{7}\right)\left(1-\frac{1}{8}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{10}\right) ?
\end{aligned}\)
Make a general statement and explain.
Solution:
\(\begin{aligned}
& 1-\frac{1}{2}=\frac{2}{2}-\frac{1}{2}=\frac{2-1}{2}=\frac{1}{2} \\
& \left(1-\frac{1}{2}\right) \times\left(1-\frac{1}{3}\right)=\frac{1}{2} \times \frac{2}{3}=\frac{1}{3} \\
& \left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right) \\
& =\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}=\frac{1}{5} \\
& \left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{6}\right) \\
& \quad\left(1-\frac{1}{7}\right)\left(1-\frac{1}{8}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{10}\right) \\
& =\frac{1}{2} \frac{2}{3} \frac{3}{4} \frac{4}{5} \frac{5}{6} \frac{6}{7} \frac{8}{8} \frac{9}{10}=\frac{1}{10}
\end{aligned}\)
General Statement:
\(\begin{array}{r}
\left(1-\frac{1}{a}\right)\left(1-\frac{1}{a+1}\right)\left(1-\frac{1}{a+2}\right)\left(1-\frac{1}{a+3}\right) \cdots \cdots \\
\left(1-\frac{1}{a+k}\right)=\frac{a-1}{a+k}
\end{array}\)
Explain: Here, we observe that in this pattern of product denominator of each term cancels the numerator or the next term and the final product is the numerator of the first term and the denominator of the last term.
It’s Puzzle Time!
Chess Puzzles — Non-attacking Queens
Chess is a popular 2-player strategy game. This game has its origins in India. It is played on an 8 × 8 chequered grid. There are 2 sets of pieces—black and white—one set for each player. Find out how each piece should move and the rules of the game.
Here is a famous chess-based puzzle. From its current position, a Queen piece can move along the horizontal, vertical or diagonal. Place 4 Queens such that no 2 queens attack each other. For example, the arrangement below is not valid as the queens are in the line of attack of each other.
Now, place 8 queens on this 8 × 8 grid so that no 2 queens attack each other!
Solution:
Four queens are placed on this 4 × 4 grid such that no 2 queens attack each other.
Now, 8 queens are placed on this 8 × 8 grid such that no 2 queens attack each other.
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