Exercise 13.2 Direct and Inverse Proportions- NCERT Solutions Class 8
Direct and Inverse Proportions
Exercise 13.2
Question 1
Which of the following are in inverse proportion?
(i) The number of workers on a job and the time to complete the job.
(ii) The time taken for a journey and the distance travelled in a uniform speed.
(iii) Area of cultivated land and the crop harvested.
(iv) The time taken for a afixed journey and the speed of the vehicle.
(v) The population of a country and the area of land per person.
Solution
Reasoning:
If two quantities are related in such a way that an increase in one quantity leads to a corresponding proportional decrease in the other, then such a variation is called inverse proportion.
Steps:
(i) The number of workers on a job and the time to complete the job.
Ans: As the number of workers increase; the job will take less time to complete. Hence, they are inversely proportional.
(ii) The time taken to a journey and the distance travelled in a uniform speed.
Ans: For more time, more distance to travel. Hence, they are not inversely proportional.
(iii) Area of cultivated land and the crop harvested.
Ans: More area of land cultivated, more crop to harvest .Hence, they are not inversely proportional.
(iv) The time taken for a afixed journey and the speed of the vehicle.
Ans: If speed is increased, it will take less time to complete the fixed journey. Hence, they are inversely proportional.
(v) The population of a country and the area of land per person.
Ans: If the population of a country increases, then the area of land per person will be decreased. Hence, they are inversely proportional.
Question 2
In a television game show, the prize money \(₹ 1,00,000\) is to be divided equally amongst the winners. Complete the table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners?
Number of winners |
\(1\) |
\(2\) |
\(4\) |
\(5\) |
\(8\) |
\(10\) |
\(20\) |
Prize for each winner (in ₹) |
\(1,00,000\) |
\(50,000\) |
\( \ldots \) |
\( \ldots \) |
\( \ldots \) |
\( \ldots \) |
\( \ldots \) |
Solution
What is known?
Amount for \(1\) winner and \(2\) winners.
What is unknown?
Prize amount for \(4, 5, 8, 10,\) and \(20\) winners.
Reasoning:
Two numbers \(x\) and \(y\) are said to vary in inverse proportion if
\[\begin{align}xy = {\rm{ }}k,{\rm{ }}\,\, x{\rm{ }} = {\rm{ }}\frac{1}{y}k\end{align}\]
Where, \(k\) is a constant.
\[\begin{align}{x_1}\;{y_1} = {x_2}\;{y_2}\end{align}\]
Steps:
If winners increase the prize amount will decrease.
\[\begin{align}{x_1}\;{y_1} &= {x_3}\;{y_3}\\1 \times 100000 &= \,4 \times {y_3}\\{y_3} &= \frac{{1 \times 100000}}{4} \\ \therefore {y_3} &= 25000\\ \\ 1 \times 100000 &= 5 \times {y_4}\\{y_4} &= \frac{{1 \times 100000}}{5} \\ \therefore {y_4} & = 20000\\ \\1 \times 100000 &= 8 \times {y_5}\\{y_5} &= \frac{{1 \times 100000}}{8} \\ \therefore {y_5} & = 12,500\\ \\1 \times 100000 &= 10 \times {y_6}\\{y_6} &= \frac{{1 \times 100000}}{{10}} \\ \therefore {y_6} & = 10000\\ \\1 \times 100000 &= 20 \times {y_7}\\{y_7} &= \frac{{1 \times 100000}}{{20}} \\ \therefore {y_7} & = 5000\end{align}\]
No. of winners |
\(4\) |
\(5\) |
\(8\) |
\(10\) |
\(20\) |
Prize for winners |
\(25,000\) |
\(20,000\) |
\(12,500 \) |
\(10,000\) |
\(5,000\) |
Question 3
Rehman is making a wheel using spokes. He wants to fix equal spokes n such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.
(i) Are the number of spokes and the angle formed between the pairs of consecutive spokes in inverse proportion?
(ii) Calculate the angle between a pair of consecutive spokes on a wheel with \(15\) spokes.
(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is \(40^\circ?\)
Number of spokes | \(4 \) | \(6\) | \(8\) | \(10\) | \(12\) |
Angle between a pair of consecutive spokes |
\(90^\circ\) | \(60^\circ\) | \(\dots\) | \(\dots\) | \(\dots\) |
Solution
What is known?
(i) Number of spokes \(4\) and \(6.\)
(ii) Angle between the spokes is \(90⁰\) and \(60⁰.\)
What is unknown?
Angle between the spokes for spoke \(8, 10\) and \(12.\)
Reasoning:
Two numbers \(x\) and \(y\) are said to vary in inverse proportion if,
\[\begin{align}x y = {\rm{ }}k,{\rm{ }}\,\,x{\rm{ }} = {\rm{ }}\frac{1}{y}k\end{align}\]
Where \(k\) is a constant.
\[{x_1}{y_1} = {x_2}{y_2}\]
Steps:
(i) If the number of spoke increases, then the angle between the spoke decreases. Hence it is in inverse variation.
(ii)
\[\begin{align}{x_1}{y_1} &= {x_3}{y_3}\\4 \times {90^{\rm{o}}} &= 8 \times {y_3}\\{y_3} &= \frac{{4 \times {{90}^{\rm{o}}}}}{8} = {45^{\rm{o}}}\\{x_1}{y_1} &= {x_4}{y_4}\\4 \times {90^{\rm{o}}} &= 10 \times {y_4}\\{y_4} &= \frac{{4 \times 90^\circ }}{{10}} = {36^{\rm{o}}}\\{x_1}{y_1} &= {x_5}{y_5}\\4 \times {90^{\rm{o}}} &= 12 \times {y_5}\\{y_5} &= \frac{{4 \times {{90}^{\rm{o}}}}}{{12}} = {30^{\rm{o}}}\end{align}\]
Number of spokes |
\(8\) |
\(10\) |
\(12\) |
Angle between spokes |
\(30^\circ\) |
\(36^\circ\) |
\(45^\circ\) |
(i) Are the number of spokes and the angle formed between the pairs of consecutive spokes in inverse proportion?
Ans: Yes, number of spokes and the angle formed between the pairs of consecutive spokes in inverse proportion, because for \(4\) spokes the angle is \(40\) but for \(8\) spokes it is \(45.\)
(ii) Calculate the angle between a pair of consecutive spokes on a wheel with \(15\) spokes. More the number of spokes,less the angle between them.
\[\begin{align}{x_1}{y_1} &= {x_2}{y_2}\\4 \times {90^{\rm{o}}} &= 15 \times {y_2}\\{y_2}& = \frac{{4 \times {{90}^{\rm{o}}}}}{{15}}\\ &= {24^{\rm{o}}}\end{align}\]
The angle between the pair of consecutive spokes on a wheel with \(15\) spokes is \({24^{\rm{o}}}.\)
(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is \({40^{\rm{o}}}.\)
\[\begin{align}{x_1}{y_1} &= {x_2}{y_2}\\4{\rm{ }} \times {\rm{ }}{90^o} &= {\rm{ }}{x_2} \times 40\\{x_2} &= \frac{{4 \times 90^\circ }}{{40}}\\ &= 9\end{align}\]
If the angle between a pair of consecutive spokes is \({40^{\rm{o}}},\) then spokes on the wheel is \(9.\)
Question 4
If a box of sweets is divided into \(24\) children, they will get \(5\) sweets each. How many would each get, if the number of children is reduced to \(4\)?
Solution
What is Known?
\(24\) children get \(5\) sweets each.
What is Unknown?
No. of sweets getting by \(20\) children. \([24 – 4 = 20]\)
Reasoning:
Two numbers \(x\) and \(y\) are said to vary in inverse proportion if
\[\begin{align}xy = {\rm{ }}k,{\rm{ }}\,\,x{\rm{ }} = {\rm{ }}\frac{1}{y}k\end{align}\]
Where \(k\) is a constant.
\[\begin{align}{x_1}{y_1} = {x_2}{y_2}\end{align}\]
Steps:
If the number of children will decrease, the number of sweets they get will increase. Hence it is the inverse proportion.
\[\begin{align}{x_1}{y_1} &= {x_2}{y_2}\\24 \times 5 &= 20 \times {y_2}\\{y_2} &= \frac{{24 \times 5}}{{20}}\\{y_2}& = 6\end{align}\]
Hence \(20\) children will get \(6\) sweets each.
Question 5
A farmer has enough food to feed \(20\) animals in his cattle for \(6\) days. How long would the food last if there were \(10\) more animals in his cattle?
Solution
Reasoning:
Two numbers \(x\) and \(y\) are said to vary in inverse proportion if
\[\begin{align}xy = {\rm{ }}k,{\rm{ }}x{\rm{ }} = {\rm{ }}\frac{1}{y}k\end{align}\]
Where \(k\) is a constant.
\[\begin{align}{x_1}{y_1} = {x_2}{y_2}\end{align}\]
What is Known:
In a cattle, food is for \(20\) animals for \(6\) days.
What is Unknown:
The number of days for \(30\) animals \((20 + 10 = 30)\)
Steps:
If the number of animals increases the number of days they can be fed will decrease, so it is in the inverse proportion.
\[\begin{align}{x_1}{y_1}& = {x_2}{y_2}\\20 \times 6 &= 30 \times {y_2}\\{y_2} &= \frac{{20 \times 6}}{{30}}\\{y_2} &= 4\end{align}\]
The food would last for \(4\) days for \(30\) animals.
Question 6
A contractor estimates that \(3\) persons could rewire Jaswinder’s house in \(4\) days. If he uses \(4\) persons instead of \(3\), how long should they take to complete the job?
Solution
What is Known?
\(3 \) persons can rewire the house in \(4\) days.
What is Unknown?
Number of days when \(4\) persons do the job.
Reasoning:
Two numbers \(x\) and \(y\) are said to vary in inverse proportion if
\[\begin{align}xy = {\rm{ }}k,{\rm{ }}x{\rm{ }} = {\rm{ }}\frac{1}{y}k\end{align}\]
Where \(k\) is a constant.
\[\begin{align}{x_1}{y_1} = {x_2}{y_2}\end{align}\]
Steps:
If the number of persons doing the job increases the day will increase so it is inverse proportion.
\[\begin{align}{x_1}{y_1} &= {x_2}{y_2}\\3 \times 4 &= 4{\rm{ }} \times {\rm{ }}{y_2}\\{y_2} &= \frac{{3\; \times \;4}}{4}\\{y_2} &= 3\end{align}\]
\(4\) persons will take \(3\) days to complete the job.
Question 7
A batch of bottles was packed in \(25\) boxes with \(12\) bottles in each box. If the same batch is packed using \(20\) bottles in each box, how many boxes would be filled?
Solution
What is Known?
\(25\) boxes were packed with \(12\) bottles in each box.
What is Unknown?
No. of boxes when \(20\) bottles are packed with each box.
Reasoning:
Two numbers \(x\) and \(y\) are said to vary in inverse proportion if
\[\begin{align}xy = {\rm{ }}k,{\rm{ }}x{\rm{ }} = {\rm{ }}\frac{1}{y}k\end{align}\]
Where \(k\) is a constant.
\[{x_1}{y_1} = {x_2}{y_2}\]
Steps:
If the no. of bottles in each box increases, no. of boxes will decrease. Hence, it is Inverse proportion.
\[\begin{align}{x_1}{y_1} &= {x_2}{y_2}\\12 \times 25 &= 20 \times {y_2}\\{y_2} &= \frac{{12 \times 25}}{{20}}\\{y_2} &= 15\end{align}\]
Hence \(15\) boxes will be filled with \(20\) bottles in each box.
Question 8
A factory requires \(42\) machines to produce a given number of articles in \(63\) days. How many machines would be required to produce the same number of articles in \(54\) days?
Solution
Reasoning:
Two numbers \(x\) and \(y\) are said to vary in inverse proportion if
\[\begin{align}xy = {\rm{ }}k,{\rm{ }}x{\rm{ }} = {\rm{ }}\frac{1}{y}k\end{align}\]
Where \(k\) is a constant.
\[{x_1}{y_1} = {x_2}{y_2}\]
What is Known:
\(42\) machines to produce a given number of articles in \(63\) days.
What is Unknown:
Machines required for producing same no. of articles in \(54\) days.
Steps:
If the number of days decreases the machine required will increase. So, it is an inverse proportion.
\[\begin{align} x_1 y_1 &= x_2 y_2 \\42 \times 63 &= 54 \times y_2 \\{y_2} &= \frac{{42 \times 63}}{{54}}\\y_2 &= 49\end{align}\]
\(49\) machines will be required to produce the same number of articles in \(54\) days.
Question 9
A car takes \(2\) hours to reach a destination by travelling at a speed of \(60\,\rm{ km/hour.}\) How long will it take when the car travels at a speed of \(80\,\rm{km/hour}?\)
Solution
What is Known?
\(2\) hours to reach at a speed of \(60\,\rm{km/hour.}\)
What is Unknown?
If the speed is \(80\,\rm{km/hour},\) the number of hours to reach.
Reasoning:
Two numbers \(x\) and \(y\) are said to vary in inverse proportion if
\[\begin{align}xy = {\rm{ }}k,{\rm{ }}x{\rm{ }} = {\rm{ }}\frac{1}{y}k\end{align}\]
Where \(k\) is a constant.
\[\begin{align}{x_1}{y_1} = {x_2}{y_2}\end{align}\]
Steps:
If the speed increases, the time required to reach will decrease. Hence, it is inverse proportion.
\[\begin{align}{x_1}{y_1} &= {x_2}{y_2}\\60 \times 2 &= 80 \times {y_2}\\{y_2} &= \frac{{60 \times 2}}{{80}}\\{y_2} &= \frac{3}{2}\\{y_2} &= 1\frac{1}{2}\;{\rm{hours}}\end{align}\]
Hence \(\begin{align}1\frac{1}{2}\end{align}\) hours are required to reach the destination if the speed is increased to \(80\,\rm{km/hour.}\)
Question 10
Two persons could fit new windows in a house in \(3\) days.
(i) One of the persons fell ill before the work started. How long would the job take now?
(ii) How many persons would be needed to fit the windows in one day?
Solution
(i) One of the persons fell ill before the work started. How long would the job take now?
What is Known?
\(2\) Persons can fit the windows in \(3\) days.
What is Unknown?
One man \(\left( {2 - 1} \right)\) can fit the window in how many days?
Reasoning:
Two numbers \(x\) and \(y\) are said to vary in direct proportion if
\[\begin{align}xy = {\rm{ }}k,{\rm{ }}x{\rm{ }} = {\rm{ }}\frac{1}{y}k\end{align}\]
Where \(k\) is a constant.
\[\begin{align}{x_1}{y_1} = {x_2}{y_2}\end{align}\]
Steps:
If the no. of men working decreases, the number of days will increase.
\[\begin{align}{x_1}{y_1}& = {x_2}{y_2}\\2 \times 3& = 1 \times {y_2}\\{y_2} &= \frac{{2 \times 3}}{1}\\{y_2} &= 6\end{align}\]
Hence the job will be completed in \(6\) days.
(ii) How many persons would be needed to fit the windows in one day?
What is Known?
\(2\) Persons can fit the windows in \(3\) days.
What is Unknown?
The persons need to fix the window in \(1\) day.
Steps:
If the number of days decreases, then the persons needed will be increase. Hence it is inverse proportion.
\[\begin{align}{x_1}{y_1} &= {x_2}{y_2}\\3 \times 2 &= 1 \times {y_2}\\{y_2} &= \frac{{3 \times 2}}{1}\\{y_2} &= 6\end{align}\]
\(6\) persons will need to fix the window in one day.
Question 11
A school has \(8\) periods a day each of \(45\) minutes of duration. How long would each period be, if the school has \(9\) periods a day assuming the number of school hours to be the same?
Solution
Reasoning:
Two numbers \(x\) and \(y\) are said to vary in inverse proportion if
\[\begin{align}xy = {\rm{ }}k,{\rm{ }}x{\rm{ }} = {\rm{ }}\frac{1}{y}k\end{align}\]
Where \(k\) is a constant.
\[\begin{align}{x_1}{y_1} = {x_2}{y_2}\end{align}\]
What is Known?
\(8\) periods of day each of \(45\) min.
What is Unknown?
\(9\) periods in a day, how many minutes each period will be?
Steps:
If the number of period increase, then the duration of the period will decrease. So, it is inverse variation.
\[\begin{align}{x_1}{y_1} &= {x_2}{y_2}\\8 \times 45 &= 9 \times {y_2}\\{y_2} &= \frac{{8 \times 45}}{9}\\{y_2} &= 40\end{align}\]
Hence, each period would be \(40\) minutes long.
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