DAV Class 8 Maths Chapter 4 Worksheet 1
Direct and Inverse Variation Worksheet 1
1. Fill in the missing terms in the following tables, if x and y vary directly:
(i) \[ \begin{array}{|c|c|c|c|c|c|} \hline x & 6 & 8 & 12 & - \\ \hline y & 18 & - & - & 63 \\ \hline \end{array} \]
Solution
\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 6 & 8 & 12 & c \\ \hline y & 18 & a & b & 63 \\ \hline \end{array} \] \[ \begin{align*} \text{Given } x \text{ and } y & \text{ vary directly} \\ \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{6}{18} &= \frac{8}{a} \\ \\ 6 \times a &= 8 \times 18 \\ \\ a &= \frac{8 \times \cancel{18}^3}{\cancel{6}_1} \\ \\ \color{green}a &= \color{green} 24 \\ \\\frac{6}{18} &= \frac{12}{b} \\ \\ 6 \times b &= 12 \times 18 \\ \\ b &= \frac{12 \times \cancel{18}^3}{\cancel{6}_1} \\ \\ \color{green} b &= \color{green} 36 \\ \\\frac{6}{18} &= \frac{c}{63} \\ \\ 6 \times 63 &= c \times 18 \\ \\ c &= \frac{\cancel{3}^1 \times 63}{\cancel{18}_3} \\ \\ c &= \frac{63}{3} \\ \\ \color{green} c &= \color{green} 21 \end{align*} \]
Answer \[ \begin{array}{|c|c|c|c|c|c|} \hline x & 6 & 8 & 12 & {\color{red} 21} \\ \hline y & 18 & {\color{red} 24} & {\color{red} 36} & 63 \\ \hline \end{array} \]
(ii) \[ \begin{array}{|c|c|c|c|c|c|} \hline x & 10 & - & 30 & 46 \\ \hline y & 5 & 10 & - & - \\ \hline \end{array} \]
Solution
\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 10 & p & 30 & 46 \\ \hline y & 5 & 10 & q & r \\ \hline \end{array} \] \[ \begin{align*} \text{Given } x \text{ and } y & \text{ vary directly} \\ \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{10}{5} &= \frac{p}{10} \\ \\ 10 \times 10 &= 5 \times p \\ \\ p &= \frac{10 \times \cancel{10}^2}{\cancel{5}_1} \\ \\ \color{green}p &= \color{green} 20 \\ \\\frac{10}{5} &= \frac{30}{q} \\ \\ 10 \times q &= 5 \times 30 \\ \\ q &= \frac{5 \times \cancel{30}^3}{\cancel{10}_1} \\ \\ \color{green}q &= \color{green} 15 \\ \\\frac{10}{5} &= \frac{46}{r} \\ \\ 10 \times r &= 5 \times 46 \\ \\ r &= \frac{\cancel{5}^1 \times 46}{\cancel{10}_2} \\ \\ r &= \frac{46}{2} \\ \\ \color{green}r &= \color{green} 23 \end{align*} \]
Answer \[ \begin{array}{|c|c|c|c|c|c|} \hline x & 10 & {\color{red} 20} & 30 & 46 \\ \hline y & 5 & 10 & {\color{red} 15} & {\color{red} 23} \\ \hline \end{array} \]
(iii) \[ \begin{array}{|c|c|c|c|c|c|} \hline x & 6 & 8 & 10 & - \\ \hline y & 15 & 20 & - & 40 \\ \hline \end{array} \]
Solution
\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 6 & 8 & 10 & s \\ \hline y & 15 & 20 & t & 40 \\ \hline \end{array} \] \[ \begin{align*} \text{Given } x \text{ and } y & \text{ vary directly} \\ \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{6}{15} &= \frac{10}{t} \\ \\ 6 \times t &= 15 \times 10 \\ \\ t &= \frac{\cancel{15}^5 \times \cancel{10}^5}{\cancel{6}_{\cancel2_1}} \\ \\ \color{green}t &= \color{green} 25 \\ \\\frac{6}{15} &= \frac{s}{40} \\ \\ 6 \times 40 &= s \times 15 \\ \\ s &= \frac{\cancel6^2 \times \cancel{40}^8}{\cancel{15}_{\cancel3_1}} \\ \\ \color{green}s &= \color{green} 16 \end{align*} \]
Answer \[ \begin{array}{|c|c|c|c|c|c|} \hline x & 6 & 8 & 10 & {\color{red} 16} \\ \hline y & 15 & 20 & {\color{red} 25} & 40 \\ \hline \end{array} \]
2. Five bags of rice weigh 150 kg. How many such bags of rice will weigh 900 kg?
Solution
Let \( x \) be number of bags that weigh 900 kg. \[ \begin{array}{|c|c|c|} \hline \text{No. of bags } & 5 & x \\ \hline Weight \,\, (kg) & 150 & 900 \\ \hline \end{array} \]
It is a case of direct variation. As the weight of bags increases, the number of bags will also increase.
\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{5}{150} &= \frac{x}{900} \\ \\ 150 \times x &= 5 \times 900 \\ \\ x &= \frac{5 \times \cancel{900}^6}{\cancel{150}_1} \\ \\ \color{green}x &= \color{green} 30 \end{align*} \]Answer Number of bags that will weigh 900 kg \( = \color{red} 30 \ bags \)
3. The cost of 3 kg of sugar is ₹ 105. What will be the cost of 15 kg of sugar?
Solution
Let \( x \) be the cost of 15 kg of sugar. \[ \begin{array}{|c|c|c|} \hline \text{Weight of sugar (kg)} & 3 & 15 \\ \hline \text{Cost (₹)} & 105 & x \\ \hline \end{array} \]
It is a case of direct variation. As the weight of sugar increases, the cost will also increase.
\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{3}{105} &= \frac{15}{x} \\ \\ 3 \times x &= 105 \times 15 \\ \\ x &= \frac{105 \times \cancel{15}^5}{\cancel3_1} \\ \\ x &= 105 \times 5 \\ \color{green} x &= \color{green} 525 \end{align*} \]Answer Cost of 15 kg of sugar \( = \color{red} \text{₹} 525 \)
4. A motor boat covers 20 km in 4 hours. What distance will it cover in 7 hours (speed remaining the same)?
Solution
Let \( x \) be the distance covered in 7 hours. \[ \begin{array}{|c|c|c|} \hline \text{Time (hours)} & 4 & 7 \\ \hline \text{Distance (km)} & 20 & x \\ \hline \end{array} \]
It is a case of direct variation. As the time increases, the distance covered will also increase.
\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{4}{20} &= \frac{7}{x} \\ \\ 4 \times x &= 20 \times 7 \\ \\ x &= \frac{\cancel{20}^5 \times 7}{\cancel{4}_1} \\ \\ \color{green} x &= \color{green} 35 \end{align*} \]Answer Distance covered in 7 hours \( = \color{red} 35 \, \text{km} \)
5. A motor bike travels 210 km on 30 litres of petrol. How far would it travel on 7 litres of petrol?
Solution
Let \( x \) be the distance traveled on 7 litres of petrol. \[ \begin{array}{|c|c|c|} \hline \text{Petrol (litres)} & 30 & 7 \\ \hline \text{Distance (km)} & 210 & x \\ \hline \end{array} \]
It is a case of direct variation. As the amount of petrol decreases, the distance covered will also decrease.
\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{30}{210} &= \frac{7}{x} \\ \\ 30 \times x &= 210 \times 7 \\ \\ x &= \frac{\cancel{210}^7 \times 7}{\cancel{30}_1} \\ \\ \color{green} x &= \color{green} 49 \end{align*} \]Answer Distance traveled on 7 litres of petrol \( = \color{red} 49 \, \text{km} \)
6. If 12 women can weave 15 metres of cloth in a day, how many metres of cloth can be woven by 20 women in a day?
Solution
Let \( x \) be the amount of cloth woven by 20 women in a day. \[ \begin{array}{|c|c|c|} \hline \text{No. of women} & 12 & 20 \\ \hline \text{Cloth (metres)} & 15 & x \\ \hline \end{array} \]
It is a case of direct variation. As the number of women increases, the amount of cloth woven will also increase.
\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{12}{15} &= \frac{20}{x} \\ \\ 12 \times x &= 15 \times 20 \\ \\ x &= \frac{\cancel{15}^5 \times \cancel{20}^5}{\cancel{12}_{\cancel3_1}} \\ \\ \color{green} x &= \color{green} 25 \end{align*} \]Answer Metres of cloth woven by 20 women in a day \( = \color{red} 25 \, metres \)
7. If the weight of five sheets of paper is 20g, how many sheets of the same paper would weigh 2.5 kg?
Solution
Let \( x \) be the number of sheets that weigh 2.5 kg (which is 2500 grams). \[ \begin{array}{|c|c|c|} \hline \text{Number of sheets} & 5 & x \\ \hline \text{Weight (g)} & 20 & 2500 \\ \hline \end{array} \]
It is a case of direct variation. As the weight of the sheets increases, the number of sheets will also increase.
\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{5}{20} &= \frac{x}{2500} \\ \\ 5 \times 2500 &= 20 \times x \\ \\ x &= \frac{5 \times 2500}{20} \\ \\ x &= \frac{5 \times \cancel{2500}^{125}}{\cancel{20}_1} \\ \\ & = 125 \times 5 \\ \color{green} x &= \color{green} 625 \end{align*} \]Answer Number of sheets that would weigh 2.5 kg \( = \color{red} 625 \, sheets \)
8. Reena types 600 words in four minutes. How much time will she take to type 3,150 words?
Solution
Let \( x \) be the time taken to type 3150 words. \[ \begin{array}{|c|c|c|} \hline \text{Number of words} & 600 & 3150 \\ \hline \text{Time (minutes)} & 4 & x \\ \hline \end{array} \]
It is a case of direct variation. As the number of words increases, the time taken will also increase.
\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{600}{4} &= \frac{3150}{x} \\ \\ 600 \times x &= 4 \times 3150 \\ \\ x &= \frac{\cancel{4}^1 \times \cancel{3150}^{21}}{\cancel{600}_{\cancel{150}_1}} \\ \\ \color{green} x &= \color{green} 21 \end{align*} \]Answer Time taken to type 3,150 words \( = \color{red} 21 \, \text{minutes} \)
9. Rohan takes 14 steps in covering a distance of 2.8 m. What distance would he cover in 150 steps?
Solution
Let \( x \) be the distance covered in 150 steps. \[ \begin{array}{|c|c|c|} \hline \text{Number of steps} & 14 & 150 \\ \hline \text{Distance (m)} & 2.8 & x \\ \hline \end{array} \]
It is a case of direct variation. As the number of steps increases, the distance covered will also increase.
\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{14}{2.8} &= \frac{150}{x} \\ \\ 14 \times x &= 2.8 \times 150 \\ \\ x &= \frac{\cancel{2.8}^{0.2} \times 150}{\cancel{14}_1} \\ \\ x &= 0.2 \times 150 \\ \color{green} x &= \color{green} 30 \end{align*} \]Answer Distance covered in 150 steps \( = \color{red} 30 \, metres \)
10. A dealer finds that 48 refined oil cans (of 5 litres each) can be packed in eight cartons of the same size. How many such cartons will he require to pack 216 cans?
Solution
Let \( x \) be the number of cartons required to pack 216 cans. \[ \begin{array}{|c|c|c|} \hline \text{No. of oil cans} & 48 & 216 \\ \hline \text{No. of cartons} & 8 & x \\ \hline \end{array} \]
It is a case of direct variation. As the number of oil cans increases, the number of cartons required will also increase.
\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{48}{8} &= \frac{216}{x} \\ \\ 48 \times x &= 8 \times 216 \\ \\ x &= \frac{\cancel{8}^1 \times \cancel{216}^{36}}{\cancel{48}_{\cancel6_1}} \\ \\ \color{green} x &= \color{green} 36 \end{align*} \]Answer Number of cartons required to pack 216 cans \( = \color{red} 36 \ cartons \)
11. The total cost of 15 newspapers is ₹37.50. Find the cost of 25 newspapers.
Solution
Let \( x \) be the cost of 25 newspapers. \[ \begin{array}{|c|c|c|} \hline \text{No. of newspapers} & 15 & 25 \\ \hline \text{Cost (₹)} & 37.50 & x \\ \hline \end{array} \]
It is a case of direct variation. As the number of newspapers increases, the total cost will also increase.
\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{15}{37.50} &= \frac{25}{x} \\ \\ 15 \times x &= 37.50 \times 25 \\ \\ x &= \frac{\cancel{37.50}^{12.50} \times \cancel{25}^5}{\cancel{15}_{\cancel3_1}} \\ \\ x &= 12.50 \times 5 \\ \color{green} x &= \color{green} 62.50 \end{align*} \]Answer Cost of 25 newspapers \( = \color{red} \text{₹}62.50 \)
12. A labourer gets ₹ 675 for nine days of work. How many days should he work to get ₹ 900?
Solution
Let \( x \) be the number of days the labourer should work to get ₹900. \[ \begin{array}{|c|c|c|} \hline \text{No. of days} & 9 & x \\ \hline \text{Earnings (₹)} & 675 & 900 \\ \hline \end{array} \]
It is a case of direct variation. As the earnings increase, the number of days worked will also increase.
\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{9}{675} &= \frac{x}{900} \\ \\ 9 \times 900 &= 675 \times x \\ \\ x &= \frac{\cancel{9}^1 \times \cancel{900}^{12}}{\cancel{675}_{\cancel{75}_1}} \\ \\ \color{green} x &= \color{green} 12 \end{align*} \]Answer Number of days the labourer works to get ₹ 900 \( = \color{red} 12 \ days \)