DAV Class 8 Maths Chapter 4 Worksheet 2
Direct and Inverse Variation Worksheet 2
1. In the following tables, a and b vary inversely. Fill in the missing values.
(i) \[ \begin{array}{|c|c|c|c|} \hline a & 7 & - & 28 \\ \hline b & 8 & 4 & - \\ \hline \end{array} \]
Solution
\[ \begin{array}{|c|c|c|c|} \hline a & 7 & x & 28 \\ \hline b & 8 & 4 & y \\ \hline \end{array} \] \[ \begin{align*} a \text{ and } b & \text{ vary inversely} \\ \implies a \times b &= k \text{ (constant)} \\ \\7 \times 8 &= x \times 4 \\ \\ x &= \frac{7 \times \cancel8^2}{\cancel4_1} \\ \\ \color{green}x &= \color{green} 14 \\ \\7 \times 8 &= 28 \times y \\ \\ y &= \frac{ \cancel7^1 \times \cancel8^2}{\cancel{28}_{\cancel4_1}} \\ \\ \color{green}y &= \color{green} 2 \end{align*} \]
Answer \[ \begin{array}{|c|c|c|c|} \hline a & 7 & {\color{red} 14} & 28 \\ \hline b & 8 & 4 & {\color{red} 2} \\ \hline \end{array} \]
(ii) \[ \begin{array}{|c|c|c|c|} \hline a & 2.5 & 4 & 0.5 \\ \hline b & 8 & - & - \\ \hline \end{array} \]
Solution
\[ \begin{array}{|c|c|c|c|} \hline a & 2.5 & 4 & 0.5 \\ \hline b & 8 & x & y \\ \hline \end{array} \] \[ \begin{align*} a \text{ and } b & \text{ vary inversely} \\ \implies a \times b &= k \text{ (constant)} \\ \\2.5 \times 8 &= 4 \times x \\ \\ x &= \frac{2.5 \times \cancel8^2}{\cancel{4}_1} \\ \\ x &= 2.5 \times 2 \\ \color{green}x &= \color{green} 5 \\ \\2.5 \times 8 &= 0.5 \times y \\ \\ y &= \frac{\cancel{2.5}^5 \times 8}{\cancel{0.5}_1} \\ \\ y &= 5 \times 8 \\ \color{green}y &= \color{green} 40 \end{align*} \]
Answer \[ \begin{array}{|c|c|c|c|} \hline a & 2.5 & 4 & 0.5 \\ \hline b & 8 & {\color{red} 5} & {\color{red} 40} \\ \hline \end{array} \]
(iii) \[ \begin{array}{|c|c|c|c|} \hline a & 10 & - & 12 \\ \hline b & 6 & 15 & - \\ \hline \end{array} \]
Solution
\[ \begin{array}{|c|c|c|c|} \hline a & 10 & x & 12 \\ \hline b & 6 & 15 & y \\ \hline \end{array} \] \[ \begin{align*} a \text{ and } b & \text{ vary inversely} \\ \implies a \times b &= k \text{ (constant)} \\ \\10 \times 6 &= x \times 15 \\ \\ x &= \frac{\cancel{10}^2 \times \cancel6^2}{\cancel{15}_{\cancel3_1}} \\ \\ x &= 2 \times 2 \\ \color{green}x &= \color{green} 4 \\ \\10 \times 6 &= 12 \times y \\ \\ y &= \frac{\cancel{10}^5 \times \cancel6^1}{\cancel{12}_{\cancel2_1}} \\ \\ \color{green}y &= \color{green} 5 \end{align*} \]
Answer \[ \begin{array}{|c|c|c|c|} \hline a & 10 & {\color{red} 4} & 12 \\ \hline b & 6 & 15 & {\color{red} 5} \\ \hline \end{array} \]
2. The science teacher asked the students of Class-VIII to make a project report on pollution. When 10 students work on it, the work gets finished in three days. How many students are required so that the work finishes in two days?
Solution
Let \( x \) be the number of students required to finish the work in two days. \[ \begin{array}{|c|c|} \hline \text{No. of students} & 10 & x \\ \hline \text{No. of days} & 3 & 2 \\ \hline \end{array} \]
It is a case of inverse variation. As the number of days decreases, the number of students will increase.
\[ \begin{align*} \implies a \times b &= k \text{ (constant)} \\ \\ 10 \times 3 &= x \times 2 \\ \\ x &= \frac{\cancel{10}^5 \times 3}{\cancel2_1} \\ \\ x &= 5 \times 3 \\ \color{green} x &= \color{green} 15 \end{align*} \]Answer \( \color{red} 15 \, students \) required to finish the work in 2 days.
3. Running at an average speed of 40 km/hr, a bus completes a journey in \( 4\frac{1}{2} \) hours. How much time will the return journey take if the speed is increased to 45 km/hr?
Solution
Let \( x \) be the time taken for the return journey at a speed of 45 km/hr. \[ \begin{array}{|c|c|c|} \hline \text{Speed (km/hr)} & 40 & 45 \\ \hline \text{Time (hours)} & 4\frac{1}{2} & x \\ \hline \end{array} \]
It is a case of inverse variation. As the speed increases, the time taken decreases.
\[ \begin{align*} \text{Since } a \times b &= k \text{ (constant)} \\ \\ 40 \times 4\frac{1}{2} &= 45 \times x \\ \\ \cancel{40}^{20} \times \frac{9}{\cancel2_1} &= 45 \times x \\ \\ 20 \times 9 &= 45 \times x \\ \\ x &= \frac{\cancel{20}^4 \times \cancel9^1}{\cancel{45}_{\cancel5_1}} \\ \\ \color{green} x &= \color{green} 4 \text{ hours} \end{align*} \]Answer Time taken to return from the journey at the speed of 45 km/hr \( = \color{red} 4 \, hours \)
4. Disha cycles to her school at an average speed of 12 km/hr. It takes her 20 minutes to reach the school. If she wants to reach her school in 15 minutes, what should be her average speed?
Solution
Let \( x \) be the speed required for Disha to reach the school in 15 minutes. \[ \begin{array}{|c|c|c|} \hline \text{Time (minutes)} & 20 & 15 \\ \hline \text{Speed (km/hr)} & 12 & x \\ \hline \end{array} \]
It is a case of inverse variation. As the time decreases, the speed needs to increase.
\[ \begin{align*} \text{Since } a \times b &= k \text{ (constant)} \\ \\ 12 \times 20 &= x \times 15 \\ \\ x &= \frac{\cancel{12}^4 \times \cancel{20}^4}{\cancel{15}_{\cancel3_1}} \\ \\ &= 4 \times 4 \\ \color{green} x &= \color{green} 16 \text{ km/hr} \end{align*} \]Answer Disha's required average speed to reach in 15 minutes \( = \color{red} 16 \, \text{km/hr} \)
5. If 15 men can repair a road in 24 days, then how long will it take nine men to repair the same road?
Solution
Let \( x \) be the number of days required for 9 men to repair the road. \[ \begin{array}{|c|c|c|} \hline \text{No. of men} & 15 & 9 \\ \hline \text{No. of days} & 24 & x \\ \hline \end{array} \]
It is a case of inverse variation. As the number of men decreases, the number of days required increases.
\[ \begin{align*} \text{Since } a \times b &= k \text{ (constant)} \\ \\ 15 \times 24 &= 9 \times x \\ \\ x &= \frac{\cancel{15}^5 \times \cancel{24}^8}{\cancel{9}_{\cancel3_1}} \\ \\ x &= 5 \times 8 \\ \color{green} x &= \color{green} 40 \text{ days} \end{align*} \]Answer 9 men will take \( \color{red} 40 \, \text{days} \) to repair the same road.
6. If 30 goats can graze a field in 15 days, then how many goats will graze the same field in 10 days?
Solution
Let \( x \) be the number of goats required to graze the field in 10 days. \[ \begin{array}{|c|c|c|} \hline \text{No. of goats} & 30 & x \\ \hline \text{No. of days} & 15 & 10 \\ \hline \end{array} \]
It is a case of inverse variation. As the number of days decreases, the number of goats required increases.
\[ \begin{align*} \text{Since } a \times b &= k \text{ (constant)} \\ \\ 30 \times 15 &= x \times 10 \\ \\ x &= \frac{\cancel{30}^3 \times 15}{\cancel{10}_1} \\ \\ x &= 3 \times 15 \\ \color{green} x &= \color{green} 45 \text{ goats} \end{align*} \]Answer \( \color{red} 45 \, \text{goats} \) can graze the same field in 10 days.
7. A contractor with a workforce of 420 men can complete the construction of a building in nine months. Due to a request by the owners, he was asked to complete the job in seven months. How many extra men must he employ to complete the job?
Solution
Let \( x \) be the total number of men required to complete the job in seven months. \[ \begin{array}{|c|c|c|} \hline \text{No. of men} & 420 & x \\ \hline \text{No. of months} & 9 & 7 \\ \hline \end{array} \]
It is a case of inverse variation. As the number of months decreases, the number of men required increases.
\[ \begin{align*} \text{Since } a \times b &= k \text{ (constant)} \\ \\ 420 \times 9 &= x \times 7 \\ \\ x &= \frac{\cancel{420}^{60} \times 9}{\cancel7_1} \\ \\ x &= 60 \times 9 \\ \color{green} x &= \color{green} 540 \text{ men} \\ \end{align*} \]\( \color{green} 540 \, men \) can complete the work in 7 months.
\[ \begin{align*} \text{Extra men required} &= 540 - 420 \\ & =\color{red} 120 \text{ men} \end{align*} \]Answer The contractor must employ \( = \color{red} 120 \, \text{extra men} \)
8. Uday can finish a book in 25 days if he reads 18 pages every day. How many days will he take to finish it, if he reads 15 pages every day?
Solution
Let \( x \) be the number of days Uday will take to finish the book if he reads 15 pages every day. \[ \begin{array}{|c|c|c|} \hline \text{Pages per day} & 18 & 15 \\ \hline \text{No. of days} & 25 & x \\ \hline \end{array} \]
It is a case of inverse variation. As the number of pages read per day decreases, the number of days required increases.
\[ \begin{align*} \text{Since } a \times b &= k \text{ (constant)} \\ \\ 18 \times 25 &= 15 \times x \\ \\ x &= \frac{\cancel{18}^6 \times \cancel{25}^5}{\cancel{15}_{\cancel3_1}} \\ \\ x &= 6 \times 5 \\ \color{green} x &= \color{green} 30 \text{ days} \end{align*} \]Answer If he reads 15 pages every day it will take \( \color{red} 30 \, \text{days} \) to complete the book.
9. A shopkeeper has enough money to buy 40 books, each costing ₹125. How many books can he buy if he gets a discount of ₹25 on each book?
Solution
Let \( x \) be the number of books the shopkeeper can buy if he gets a discount of ₹25 on each book. The new price per book is \( \color{green} \text{₹}125 - \text{₹}25 = \text{₹}100\). \[ \begin{array}{|c|c|c|} \hline \text{Cost per book (₹)} & 125 & 100 \\ \hline \text{No. of books} & 40 & x \\ \hline \end{array} \]
It is a case of inverse variation. As the cost per book decreases, the number of books he can buy increases.
\[ \begin{align*} \text{Since } a \times b &= k \text{ (constant)} \\ \\ 125 \times 40 &= 100 \times x \\ \\ x &= \frac{\cancel{125}^{25} \times \cancel{40}^{2}}{\cancel{100}_{\cancel{20}_1}} \\ \\ x &= 25 \times 2 \\ \color{green} x &= \color{green} 50 \text{ books} \end{align*} \]Answer Shopkeeper can buy \( \color{red} 50 \, \text{books} \) if he gets a discount of ₹25 on each book.
10. Six pumps working together empty a tank in 28 minutes. How long will it take to empty the tank if four such pumps are working together?
Solution
Let \( x \) be the time taken to empty the tank with four pumps working together. \[ \begin{array}{|c|c|c|} \hline \text{No. of pumps} & 6 & 4 \\ \hline \text{Time (minutes)} & 28 & x \\ \hline \end{array} \]
It is a case of inverse variation. As the number of pumps decreases, the time required to empty the tank increases.
\[ \begin{align*} \text{Since } a \times b &= k \text{ (constant)} \\ \\ 6 \times 28 &= 4 \times x \\ \\ x &= \frac{6 \times \cancel{28}^7}{\cancel4_1} \\ \\ x &= 6 \times 7 \\ \color{green} x &= \color{green} 42 \text{ minutes} \\ \end{align*} \]Answer It will take \( \color{red} 42 \, \text{minutes} \) to empty the tank.
11. A train moving at a speed of 75 km/hr covers a certain distance in 4.8 hours. What should be the speed of the train to cover the same distance in 3 hours?
Solution
Let \( x \) be the speed of the train required to cover the same distance in 3 hours. \[ \begin{array}{|c|c|c|} \hline \text{Time (hours)} & 4.8 & 3 \\ \hline \text{Speed (km/hr)} & 75 & x \\ \hline \end{array} \]
It is a case of inverse variation. As the time decreases, the speed must increase to cover the same distance.
\[ \begin{align*} \text{Since } a \times b &= k \text{ (constant)} \\ \\ 4.8 \times 75 &= 3 \times x \\ \\ x &= \frac{4.8 \times \cancel{75}^{25}}{\cancel3_1} \\ \\ x &= 4.8 \times 25 \\ \color{green} x &= \color{green} 120 \text{ km/hr} \end{align*} \]Answer Speed of the train must be \( \color{red} 120 \, \text{km/hr} \) to cover the same distance in 3 hours.
12. A garrison of 120 men has provision for 30 days. At the end of five days, five more men joined them. How many days can they sustain on the remaining provision?
Solution
Let \( x \) be the number of days the garrison can sustain on the remaining provision after five more men joined them. \[ \begin{array}{|c|c|c|} \hline \text{No. of men} & 120 & 125 \\ \hline \text{No. of days (remaining)} & 25 & x \\ \hline \end{array} \]
The initial provision was for 30 days, but after five days, only 25 days of provision remain for 120 men. It is a case of inverse variation. As the number of men increases, the number of days the provision lasts decreases.
\[ \begin{align*} \text{Since } a \times b &= k \text{ (constant)} \\ \\ 120 \times 25 &= 125 \times x \\ \\ x &= \frac{\cancel{120}^{24} \times \cancel{25}^1}{\cancel{125}_{\cancel5_1}} \\ \\ \color{green} x &= \color{green} 24 \text{ days} \end{align*} \]Answer Number of days they can sustain the remaining provision for 125 men \( = \color{red} 24 \, \text{days} \)