RD Sharma solution class 8 chapter 10 Direct and Inverse Variation Exercise 10.1

Exercise 10.1

Page-10.6

Question 1:

Explain the concept of direct variation.

Answer 1:

$$\begin{gathered} \mathrm{When} \mathrm{two} \mathrm{variables} \mathrm{are} \mathrm{connected} \mathrm{to} \mathrm{each} \mathrm{other} \mathrm{in} \mathrm{such} \mathrm{a} \mathrm{way} \mathrm{that} \mathrm{if} \mathrm{we} \mathrm{increase} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{one} \mathrm{variable}, \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{other} \mathrm{variable} \mathrm{also} \mathrm{increases} \\ \mathrm{and} \mathrm{vice}-\mathrm{versa}. \mathrm{Similarly}, \mathrm{if} \mathrm{we} \mathrm{decrease} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{one} \mathrm{variable}, \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{other} \mathrm{variable} \mathrm{also} \mathrm{decreases} \mathrm{and} \mathrm{vice}-\mathrm{versa}. \\ \mathrm{Therefore}, \mathrm{if} \mathrm{the} \mathrm{ratio} \mathrm{between} \mathrm{two} \mathrm{variables} \mathrm{remains} \mathrm{constant}, \mathrm{it} \mathrm{is} \mathrm{said} \mathrm{to} \mathrm{be} \mathrm{in} \mathrm{direct} \mathrm{variation}. \end{gathered}$$

Question 2:

Which of the following quantities vary directly with each other?
(i) Number of articles (x) and their price (y).
(ii) Weight of articles (x) and their cost (y).
(iii) Distance x and time y, speed remaining the same.
(iv) Wages (y) and number of hours (x) of work.
(v) Speed (x) and time (y) (distance covered remaining the same).
(vi) Area of a land (x) and its cost (y).

Answer 2:

(i) The number of articles is directly related to the price. Therefore, they will vary directly with each other.
(ii) The number of articles is directly related to the weight of the articles. Therefore, they will vary directly with each other.
(iii) Speed is constant. Therefore, distance and time does not vary directly.
(iv) The number of hours is directly related to the wages. Therefore, it is a direct variation.
(v) Distance is constant. Therefore, speed and time does not vary directly.
(vi) If the area of a land is large, its cost will also be high. Thus, it is a direct variation.

Thus, the respective values in (i), (ii), (iv) and (vi) vary directly with each other.

Question 3:

In which of the following tables x and y vary directly?
(i)

a	7	9	13	21	25
b	21	27	39	63	75

(ii)

a	10	20	30	40	46
b	5	10	15	20	23

(iii)

a	2	3	4	5	6
b	6	9	12	17	20

(iv)

a	12	22	32	42	52
b	13	23	33	43	53

Answer 3:

$$\begin{gathered} \mathrm{If} x \mathrm{and} y \mathrm{vary} \mathrm{directly}, \mathrm{the} \mathrm{ratio} \mathrm{of} \mathrm{the} \mathrm{corresponding} \mathrm{values} \mathrm{of} x \mathrm{and} y \mathrm{remain}s \mathrm{consta}\mathrm{nt}. \\ \left(i\right) \\ \frac{x}{y}=\frac{7}{21}=\frac{1}{3} \\ \frac{x}{y}=\frac{9}{27}=\frac{1}{3} \\ \frac{x}{y}=\frac{13}{39}=\frac{1}{3} \\ \frac{x}{y}=\frac{21}{63}=\frac{1}{3} \\ \frac{\mathrm{x}}{\mathrm{y}}=\frac{25}{75}=\frac{1}{3} \\ \mathrm{In} \mathrm{all} \mathrm{the} \mathrm{cases}, \mathrm{the} \mathrm{ratio} \mathrm{is} \mathrm{the} \mathrm{same}. \mathrm{Therefore}, x \mathrm{and} y \mathrm{vary} \mathrm{directly}. \\ \left(ii\right) \\ \frac{x}{y}=\frac{10}{5}=2 \\ \frac{x}{y}=\frac{20}{10}=2 \\ \frac{x}{y}=\frac{30}{15}=2 \\ \frac{x}{y}=\frac{40}{20}=2 \\ \frac{x}{y}=\frac{46}{23}=2 \\ \mathrm{In} \mathrm{all} \mathrm{the} \mathrm{cases}, \mathrm{the} \mathrm{ratio} \mathrm{is} \mathrm{the} \mathrm{same}. \mathrm{Therefore}, x \mathrm{and} y \mathrm{vary} \mathrm{directly}. \\ \left(iii\right) \\ \frac{x}{y}=\frac{2}{6}=\frac{1}{3} \\ \frac{x}{y}=\frac{3}{9}=\frac{1}{3} \\ \frac{x}{y}=\frac{4}{12}=\frac{1}{3} \\ \frac{x}{y}=\frac{5}{17}=\frac{5}{17} \\ \frac{x}{y}=\frac{6}{20}=\frac{3}{10} \\ \mathrm{In} \mathrm{all} \mathrm{the} \mathrm{cases}, \mathrm{the} \mathrm{ratio} \mathrm{is} \mathrm{not} \mathrm{the} \mathrm{same}. \mathrm{Therefore}, x \mathrm{and} y \mathrm{do} \mathrm{not} \mathrm{vary} \mathrm{directly}. \\ \left(iv\right) \\ \frac{x}{y}=\frac{1^2}{1^3}=1 \\ \frac{x}{y}=\frac{2^2}{2^3}=\frac{1}{2} \\ \frac{x}{y}=\frac{3^2}{3^3}=\frac{1}{3} \\ \frac{x}{y}=\frac{4^2}{4^3}=\frac{1}{4} \\ \frac{x}{y}=\frac{5^2}{5^3}=\frac{1}{5} \\ \mathrm{In} \mathrm{all} \mathrm{the} \mathrm{cases}, \mathrm{the} \mathrm{ratio} \mathrm{is} \mathrm{not} \mathrm{the} \mathrm{same}. \mathrm{Therefore}, x \mathrm{and} y \mathrm{do} \mathrm{not} \mathrm{vary} \mathrm{directly}. \\ \mathrm{Thus}, \mathrm{in} \left(i\right) and \left(ii\right), x and y \mathrm{vary} \mathrm{directly}. \end{gathered}$$

Question 4:

Fill in the blanks in each of the following so as to make the statement true:
(i) Two quantities are said to vary.... with each other if they increase (decrease) together in such a way that the ratio of the corresponding values remains same.
(ii) x and y are said to vary directly with each other if for some positive number k,...... = k.
(iii) If u = 3 v, then u and v vary .... with each other.

Answer 4:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{directly} \\ \left(\mathrm{ii}\right) x a\mathrm{nd} y a\mathrm{re} \mathrm{said} \mathrm{to} \mathrm{vary} \mathrm{directly} \mathrm{with} \mathrm{each} \mathrm{other} \mathrm{if} \frac{x}{y}=k, \mathrm{where} k \mathrm{is} \mathrm{a} \mathrm{positive} \mathrm{number} \\ \left(\mathrm{iii}\right) \mathrm{because} u = 3v, u \mathrm{and} y \mathrm{vary} \mathrm{directly} \mathrm{with} \mathrm{each} \mathrm{other} \end{gathered}$$

Question 5:

Complite the following tables given that x varies directly as y.
(i)

x	2.5	...	...	15
y	5	8	12	...

(ii)

x	5	...	10	35	25	...
y	8	12	...	...	...	32

(iii)

x	6	8	10	...	20
y	15	20	...	40	...

(iv)

x	4	9	...	...	3	...
y	16	...	48	36	...	4

(v)

x	3	5	7	9
y	...	20	28	...

Answer 5:

$$\begin{gathered} \mathrm{Here}, x \mathrm{and} y \mathrm{vary} \mathrm{directly}. \\ \therefore x = ky \\ \left(\mathrm{i}\right) x = 2.5 \mathrm{and} y = 5 \\ \mathrm{i}.\mathrm{e}., 2.5 = k \times 5 \\ \Rightarrow k = \frac{2.5}{5} = 0.5 \\ \mathrm{For} y = 8 \mathrm{and} k = 0.5, \mathrm{we} \mathrm{have}: \\ x = ky \\ \Rightarrow x = 8 \times 0.5 = 4 \\ \mathrm{For} y = 12 \mathrm{and} k = 0.5, \mathrm{we} \mathrm{have}: \\ x = ky \\ \Rightarrow x = 12 \times 0.5 = 6 \\ \mathrm{For} x = 15 \mathrm{and} k = 0.5, \mathrm{we} \mathrm{have}: \\ x = ky \\ \Rightarrow 15 = 0.5 \times y \\ \Rightarrow y = \frac{15}{0.5} = 30 \\ \left(\mathrm{ii}\right) x = 5 and y = 8 \\ \mathrm{i}.\mathrm{e}., 5 = k \times 8 \\ \Rightarrow k = \frac{5}{8} = 0.625 \\ \mathrm{For} y = 12 \mathrm{and} k = 0.625, \mathrm{we} \mathrm{have}: \\ x = ky \\ \Rightarrow x = 12 \times 0.625 = 7.5 \\ \mathrm{For} x = 10 \mathrm{and} k = 0.625, \mathrm{we} \mathrm{have}: \\ x = ky \\ \Rightarrow 10 = 0.625 \times y \\ \Rightarrow y = \frac{10}{0.625} = 16 \\ \mathrm{For} x = 35 \mathrm{and} k = 0.625, \mathrm{we} \mathrm{have}: \\ x=ky \\ \Rightarrow 35 = 0.625 \times y \\ \Rightarrow y = \frac{35}{0.625} = 56 \\ \mathrm{For} x = 25 \mathrm{and} k = 0.625, \mathrm{we} \mathrm{have}: \\ x = ky \\ \Rightarrow 25 = 0.625 \times y \\ \Rightarrow y = \frac{25}{0.625} = 40 \\ \mathrm{For} y = 32 \mathrm{and} k = 0.625, \mathrm{we} \mathrm{have}: \\ x = ky \\ \Rightarrow x = 0.625 \times 32 = 20 \\ \left(\mathrm{iii}\right) x = 6 \mathrm{and} y = 15 \\ \mathrm{i}.\mathrm{e}., 6 = k \times 15 \\ \Rightarrow k = \frac{6}{15} = 0.4 \\ \mathrm{For} x = 10 \mathrm{and} k = 0.4, \mathrm{we} \mathrm{have}: \\ y = \frac{10}{0.4} = 25 \\ \mathrm{For} y = 40 \mathrm{and} k = 0.4, \mathrm{we} \mathrm{have}: \\ x = 0.4 \times 40 = 16 \\ \mathrm{For} x = 20 \mathrm{and} k = 0.4, \mathrm{we} \mathrm{have}: \\ y = \frac{20}{0.4} = 50 \\ \left(\mathrm{iv}\right) x = 4 \mathrm{and} y =16 \\ \mathrm{i}.\mathrm{e}., 4 = k \times 16 \\ \Rightarrow k = \frac{4}{16} = \frac{1}{4} \\ \mathrm{For} x = 9 \mathrm{and} k =\frac{1}{4}, \mathrm{we} \mathrm{have}: \\ 9 = ky \\ \Rightarrow y = 4 \times 9 = 36 \\ \mathrm{For} y = 48 \mathrm{and} k = \frac{1}{4}, \mathrm{we} \mathrm{have}: \\ x = ky \\ =\frac{1}{4} \times 48 = 12 \\ \mathrm{For} y = 36 \mathrm{and} k = \frac{1}{4}, \mathrm{we} \mathrm{have}: \\ x = ky \\ = \frac{1}{4} \times 36 = 9 \\ \mathrm{For} x = 3 \mathrm{and} k = \frac{1}{4}, \mathrm{we} \mathrm{have}: \\ x = ky \\ \Rightarrow 3 = \frac{1}{4} \times y \\ \Rightarrow y = 12 \\ \mathrm{For} y = 4 \mathrm{and} k = \frac{1}{4}, \mathrm{we} \mathrm{have}: \\ x = ky \\ = \frac{1}{4} \times 4 = 1 \\ \left(\mathrm{v}\right) x = 5 \mathrm{and} y = 20 \\ \mathrm{i}.\mathrm{e}., 5 = k \times 20 \\ \Rightarrow k = \frac{5}{20} = \frac{1}{4} \\ \mathrm{For} x = 3 \mathrm{and} k = \frac{1}{4}, \mathrm{we} \mathrm{have}: \\ 3 = \frac{1}{4} \times y \\ \Rightarrow y = 4 \times 3 = 12 \\ \mathrm{For} x = 9, k = \frac{1}{4}, \mathrm{we} \mathrm{have}: \\ x = ky \\ \Rightarrow 9 = \frac{1}{4} \times y \\ \Rightarrow y = 9 \times 4 = 36 \end{gathered}$$

Page-10.7

Question 6:

Find the constant of variation from the table given below:

x	3	5	7	9
y	12	20	28	36

Set up a table and solve the following problems. Use unitary method to verify the answer.

Answer 6:

$$\begin{gathered} \mathrm{Since} \mathrm{it} \mathrm{is} \mathrm{a} \mathrm{direct} \mathrm{variation}, \frac{x}{y} = k. \\ \mathrm{For} x = 3 \mathrm{and} y = 12, \mathrm{we} \mathrm{have}: \\ k = \frac{3}{12} = \frac{1}{4} \\ \mathrm{Thus}, \mathrm{in} \mathrm{all} \mathrm{cases}, k = \frac{1}{4} \end{gathered}$$

Question 7:

Rohit bought 12 registers for Rs 156, find the cost of 7 such registers.

Answer 7:

Let the cost of 7 registers be​ Rs x.
 

Register	12	7
Cost(in Rs.)	156	x

$$\begin{gathered} \mathrm{If} \mathrm{he} \mathrm{buys} \mathrm{less} \mathrm{number} \mathrm{of} \mathrm{registers}, \mathrm{the} \mathrm{cost} \mathrm{will} \mathrm{also} \mathrm{be} \mathrm{less}. \\ \mathrm{Therefore}, \mathrm{it} \mathrm{is} \mathrm{a} \mathrm{direct} \mathrm{variation}. \\ \mathrm{We} \mathrm{get}: \\ 12:7 = 156:x \\ \Rightarrow \frac{12}{7} = \frac{156}{x} \\ \mathrm{Applying} \mathrm{cross} \mathrm{muliplication}, \mathrm{we} \mathrm{get}: \\ x = \frac{156\times 7}{12} \\ = 91 \\ \mathrm{Thus}, \mathrm{the} \mathrm{cost} \mathrm{of} 7 \mathrm{such} \mathrm{registers} \mathrm{will} \mathrm{be} \mathrm{Rs} 91. \end{gathered}$$

Question 8:

Anupama takes 125 minutes in walking a distance of 100 metre. What distance would she cover in 315 minutes?

Answer 8:

 Let the distance travelled in 315 minutes be x km.
 

Time (in minute)	125	315
Distance(in metre)	100	x

$$\begin{gathered} \mathrm{If} \mathrm{the} \mathrm{distance} \mathrm{travelled} \mathrm{is} \mathrm{more}, \mathrm{the} \mathrm{time} \mathrm{needed} \mathrm{to} \mathrm{cover} \mathrm{it} \mathrm{will} \mathrm{also} \mathrm{be} \mathrm{more}. \\ \mathrm{Therefore}, \mathrm{it} \mathrm{is} \mathrm{a} \mathrm{direct} \mathrm{variation}. \\ \mathrm{We} \mathrm{get}: \\ 125:315 = 100:x \\ \Rightarrow \frac{125}{315} = \frac{100}{x} \\ \mathrm{Applying} \mathrm{cross} \mathrm{muliplication}, \mathrm{we} \mathrm{get}: \\ x = \frac{100 \times 315}{125} \\ = 252 \\ \mathrm{Thus}, \mathrm{Anupama} \mathrm{would} \mathrm{cover} 252 \mathrm{metre} \mathrm{in} 315 \mathrm{minutes}. \end{gathered}$$

Question 9:

If the cost of 93 m of a certain kind of plastic sheet is Rs 1395, then what would it cost to buy 105 m of such plastic sheet?

Answer 9:

Length of plastic sheet (in metre)	93	105
Cost (in Rs)	1395	x

Let the cost of the plastic sheet per metre be Rs x.
$$\begin{gathered} \mathrm{If} \mathrm{more} \mathrm{sheets} \mathrm{are} \mathrm{bought}, \mathrm{the} \mathrm{cost} \mathrm{will} \mathrm{also} \mathrm{be} \mathrm{more}. \\ \mathrm{Therefore}, \mathrm{it} \mathrm{is} \mathrm{a} \mathrm{direct} \mathrm{variation}. \\ \mathrm{We} \mathrm{get}: \\ 93:105 = 1395:x \\ \Rightarrow \frac{93}{105} = \frac{1395}{x} \\ \mathrm{Applying} \mathrm{cross} \mathrm{muliplication}, \mathrm{we} \mathrm{get}: \\ x = \frac{105 \times 1395}{93} \\ = 1575 \\ \mathrm{Thus}, \mathrm{the} \mathrm{required} \mathrm{cost} \mathrm{will} \mathrm{be} \mathrm{Rs} 1,575. \end{gathered}$$

Question 10:

Suneeta types 1080 words in one hour. What is her GWAM (gross words a minute rate)?

Answer 10:

Number of words	1080	x
Time (in minute)	60	1

Let x be her GWAM.
$$\begin{gathered} \mathrm{If} \mathrm{the} \mathrm{time} \mathrm{taken} \mathrm{is} \mathrm{less}, \mathrm{GWAM} \mathrm{will} \mathrm{also} \mathrm{be} \mathrm{less}. \\ \mathrm{Therefore}, \mathrm{it} \mathrm{is} \mathrm{a} \mathrm{direct} \mathrm{variation}. \\ 1080:x = 60:1 \\ \Rightarrow \frac{1080}{x} = \frac{60}{1} \\ \mathrm{Applying} \mathrm{cross} \mathrm{muliplication}, \mathrm{we} \mathrm{get}: \\ x = \frac{1080\times 1}{60} \\ = 18 \\ \mathrm{Thus}, \mathrm{her} \mathrm{GWAM} \mathrm{will} \mathrm{be} 18. \end{gathered}$$

Question 11:

A car is travelling at the average speed of 50 km/hr. How much distance would it travel in 12 minutes?

Answer 11:

Distance (in km)	50	x
Time (in minute)	60	12

Let the distance be x km.
$$\begin{gathered} \mathrm{If} \mathrm{the} \mathrm{time} \mathrm{taken} \mathrm{is} \mathrm{less}, \mathrm{the} \mathrm{distance} \mathrm{covered} \mathrm{will} \mathrm{also} \mathrm{be} \mathrm{less}. \\ \mathrm{Therefore}, \mathrm{it} \mathrm{is} \mathrm{a} \mathrm{direct} \mathrm{variation}. \\ 50:x = 60:12 \\ \Rightarrow \frac{50}{x} = \frac{60}{12} \\ \mathrm{Applying} \mathrm{cross} \mathrm{muliplication}, \mathrm{we} \mathrm{get}: \\ x=\frac{50\times 12}{60} \\ =10 \\ \mathrm{Thus}, \mathrm{the} \mathrm{required} \mathrm{distance} \mathrm{will} \mathrm{be} 10 \mathrm{km}. \end{gathered}$$

Question 12:

68 boxes of a certain commodity require a shelf-length of 13.6 m. How many boxes of the same commodity would occupy a shelf length of 20.4 m?

Answer 12:

Number of Boxes	68	x
Shelf-length (in m)	13.6	20.4

Let x be the number of boxes that occupy a shelf-length of 20.4 m.

$$\begin{gathered} \mathrm{If} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{shelf} \mathrm{increases}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{boxes} \mathrm{will} \mathrm{also} \mathrm{increase}. \\ \mathrm{Therefore}, \mathrm{it} \mathrm{is} \mathrm{a} \mathrm{case} \mathrm{of} \mathrm{direct} \mathrm{variation}. \\ \frac{68}{x} = \frac{13.6}{20.4} \\ \Rightarrow 68 \times 20.4 = x \times 13.6 \\ \Rightarrow x = \frac{68\times 20.4}{13.6} \\ = \frac{1387.2}{13.6} \\ = 102 \\ \mathrm{Thus}, 102 \mathrm{boxes} \mathrm{will} \mathrm{occupy} \mathrm{a} \mathrm{shelf}-\mathrm{length} \mathrm{of} 20.4 \mathrm{m}. \end{gathered}$$

Question 13:

In a library 136 copies of a certain book require a shelf-length of 3.4 metre. How many copies of the same book would occupy a shelf-length of 5.1 metres?

Answer 13:

Number of copies	136	x
Length the shelf (in m)	3.4	5.1

Let x be the number of copies that would occupy a shelf-length of 5.1 m.

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{copies} \mathrm{and} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{shelf} \mathrm{are} \mathrm{in} \mathrm{direct} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ \frac{136}{x} = \frac{3.4}{5.1} \\ \Rightarrow 136 \times 5.1 = x \times 3.4 \\ \Rightarrow x = \frac{136\times 5.1}{3.4} \\ = 204 \\ \mathrm{Thus}, 204 \mathrm{copies} \mathrm{will} \mathrm{occupy} \mathrm{a} \mathrm{shelf} \mathrm{of} \mathrm{length} 5.1 \mathrm{m}. \end{gathered}$$

Question 14:

The second class railway fare for 240 km of Journey is Rs 15.00. What would be the fare for a journey of 139.2 km?

Answer 14:

Let Rs x be the fare for a journey of 139.2 km.

Distance (in km)	240	139.2
Fare (in Rs.)	15	x

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{distance} \mathrm{travelled} \mathrm{and} \mathrm{the} \mathrm{fare} \mathrm{are} \mathrm{in} \mathrm{direct} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ \frac{240}{139.2} = \frac{15}{x} \\ \Rightarrow 240 \times x = 15 \times 139.2 \\ \Rightarrow x = \frac{15 \times 139.2}{240} \\ = \frac{2088}{240} \\ = 8.7 \\ \mathrm{Thus}, \mathrm{the} \mathrm{fare} \mathrm{for} \mathrm{a} \mathrm{journey} \mathrm{of} 139.2 \mathrm{km} \mathrm{will} \mathrm{be} \mathrm{Rs} 8.70. \end{gathered}$$

Question 15:

If the thickness of a pile of 12 cardboards is 35 mm, find the thickness of a pile of 294 cardboards.

Answer 15:

Let x cm be the thickness of a pile of 294 cardboards.
 

Thickness (in cm)	3.5	x
Cardboard	12	294

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{pile} \mathrm{of} \mathrm{the} \mathrm{cardboards} \mathrm{and} \mathrm{its} \mathrm{thickness} \mathrm{are} \mathrm{in} \mathrm{direct} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ \frac{3.5}{x} = \frac{12}{294} \\ \Rightarrow 3.5 \times 294 = x \times 12 \\ \Rightarrow x = \frac{3.5 \times 294}{12} \\ = \frac{1029}{12} \\ = 85.75 \mathrm{cm} \\ \mathrm{Thus}, \mathrm{the} \mathrm{thickness} \mathrm{of} \mathrm{a} \mathrm{pile} \mathrm{of} 294 \mathrm{cardboards} \mathrm{will} \mathrm{be} 85.75 \mathrm{cm} (\mathrm{or} 857.5 \mathrm{mm}). \end{gathered}$$

Question 16:

The cost of 97 metre of cloth is Rs 242.50. What length of this can be purchased for Rs 302.50?

Answer 16:

Let x metre be the length of the cloth that can be purchased for Rs 302.50.
 

Length (in m)	97	x
Cost (in Rs)	242.50	302.50

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{cloth} \mathrm{and} \mathrm{its} \mathrm{cost} \mathrm{are} \mathrm{in} \mathrm{direct} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ \frac{97}{x} = \frac{242.50}{302.50} \\ \Rightarrow 97 \times 302.50 = x \times 242.50 \\ \Rightarrow x = \frac{97 \times 302.50}{242.50} \\ = \frac{29342.50}{242.50} \\ = 121 \\ \mathrm{Thus}, \mathrm{the} \mathrm{required} \mathrm{length} \mathrm{will} \mathrm{be} 121 \mathrm{metre}. \end{gathered}$$

Question 17:

11 men can dig $6\frac{3}{4}$ metre long trench in one day. How many men should be employed for digging 27 metre long trench of the same type in one day?

Answer 17:

Let x be the number of men required to dig a trench of 27 metre.
 

Number of men	11	x
Length (in m)	$\frac{27}{4}$	27

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{trench} \mathrm{and} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{men} \mathrm{are} \mathrm{in} \mathrm{direct} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ \frac{11}{x} = \frac{27/4}{27} \\ \Rightarrow 11 \times 27 = x \times \frac{27}{4} \\ \Rightarrow x = \frac{11 \times 27 \times 4}{27} \\ = 44 \\ \mathrm{Thus}, 44 \mathrm{men} \mathrm{will} \mathrm{be} \mathrm{required} \mathrm{to} \mathrm{dig} \mathrm{a} \mathrm{trench} \mathrm{of} 27 \mathrm{metre}. \end{gathered}$$

Question 18:

A worker is paid Rs 210 for 6 days work. If his total income of the month is Rs 875, for how many days did he work?

Answer 18:

Let x be the number of days for which the worker is paid Rs 875.
 

Income (in Rs.)	210	875
Number of days	6	x

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{income} \mathrm{of} \mathrm{the} \mathrm{worker} \mathrm{and} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{working} \mathrm{days} \mathrm{are} \mathrm{in} \mathrm{direct} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ \frac{210}{875} = \frac{6}{x} \\ \Rightarrow 210 \times x = 875 \times 6 \\ \Rightarrow x = \frac{875 \times 6}{210} \\ = \frac{5250}{210} \\ = 25 \\ \mathrm{Thus}, \mathrm{the} \mathrm{required} \mathrm{number} \mathrm{of} \mathrm{days} \mathrm{is} 25. \end{gathered}$$

Question 19:

A woker is paid Rs 200 for 8 days work. If he works for 20 days, how much will he get?

Answer 19:

 Let Rs x be the income for 20 days of work.
 

Income (in Rs)	200	x
Number of days	8	20

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{income} \mathrm{and} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{working} \mathrm{days} \mathrm{are} \mathrm{in} \mathrm{direct} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ \frac{200}{x} = \frac{8}{20} \\ \Rightarrow 200 \times 20 = 8x \\ \Rightarrow x = \frac{200 \times 20}{8} \\ = \frac{4000}{8} \\ = 500 \\ \mathrm{Thus}, \mathrm{the} \mathrm{worker} \mathrm{will} \mathrm{get} \mathrm{Rs} 500 \mathrm{for} \mathrm{working} 20 \mathrm{days}. \end{gathered}$$

Question 20:

The amount of extension in an elastic string varies directly as the weight hung on it. If a weight of 150 gm produces an extension of 2.9 cm, then what weight would produce an extension of 17.4 cm?

Answer 20:

Let x gm be the weight that would produce an extension of 17.4 cm.
 

Weight (in gm)	150	x
Length (in cm)	2.9	17.4

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{amount} \mathrm{of} \mathrm{extension} \mathrm{in} \mathrm{an} \mathrm{elastic} \mathrm{string} \mathrm{and} \mathrm{the} \mathrm{weight} \mathrm{hung} \mathrm{on} \mathrm{it} \mathrm{are} \mathrm{in} \mathrm{direct} \mathrm{variation}, we have: \\ \frac{150}{x} = \frac{2.9}{17.4} \\ \Rightarrow 17.4 \times 150 = 2.9 \times x \\ \Rightarrow x = \frac{17.4 \times 150}{2.9} \\ = \frac{2610}{2.9} \\ = 900 \\ \mathrm{Thus}, \mathrm{the} \mathrm{required} \mathrm{weight} \mathrm{will} \mathrm{be} 900 \mathrm{gm}. \end{gathered}$$

Page-10.8

Question 21:

The amount of extension in an elastic spring varies directly with the weight hung on it. If a weight of 250 gm produces an extension of 3.5 cm, find the extension produced by the weight of 700 gm.

Answer 21:

Let x cm be the extension produced by the weight of 700 gm.
 

Weight (in gm)	250	700
Length (in cm)	3.5	x

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{amount} \mathrm{of} \mathrm{extension} \mathrm{in} \mathrm{an} \mathrm{elastic} \mathrm{spring} \mathrm{varies} \mathrm{and} \mathrm{the} \mathrm{weight} \mathrm{hung} \mathrm{on} \mathrm{it} \mathrm{is} \mathrm{in} \mathrm{direct} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ \frac{250}{700} = \frac{3.5}{x} \\ \Rightarrow x \times 250 = 3.5 \times 700 \\ \Rightarrow x = \frac{3.5 \times 700}{250} \\ = \frac{2450}{250} \\ = 9.8 \\ \mathrm{Thus}, \mathrm{the} \mathrm{required} \mathrm{extension} \mathrm{will} \mathrm{be} 9.8 \mathrm{cm}. \end{gathered}$$

Question 22:

In 10 days, the earth picks up 2.6 × 10⁸ pounds of dust from the atmosphere. How much dust will it pick up in 45 days?

Answer 22:

Let the amount of dust picked up by the earth in 45 days be x pounds.

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{amount} \mathrm{of} \mathrm{dust} \mathrm{picked} \mathrm{up} \mathrm{by} \mathrm{the} \mathrm{earth} \mathrm{and} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{days} \mathrm{are} \mathrm{in} \mathrm{direct} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ \mathrm{Ratio} \mathrm{of} \mathrm{the} \mathrm{dust} \mathrm{picked} \mathrm{up} \mathrm{by} \mathrm{the} \mathrm{earth} \mathrm{in} \mathrm{pounds} = \mathrm{ratio} \mathrm{of} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{days} \mathrm{taken} \\ \Rightarrow \frac{10}{45} = \frac{2.6 \times {10}^8}{x} \\ \Rightarrow x \times 10 = 45 \times 2.6 \times {10}^8 \\ \Rightarrow x = \frac{45 \times 2.6 \times {10}^8}{10} \\ = \frac{117 \times {10}^8}{10} \\ = 11.7 \times {10}^8 \\ \mathrm{Thus}, 11.7\times {10}^8 \mathrm{pounds} \mathrm{of} \mathrm{dust} \mathrm{will} \mathrm{be} \mathrm{picked} \mathrm{up} \mathrm{by} \mathrm{the} \mathrm{earth} \mathrm{in} 45 \mathrm{days}. \end{gathered}$$

Question 23:

In 15 days, the earth picks up 1.2 × 10⁸ kg of dust from the atmosphere. In how many days it will pick up 4.8 × 10⁸ kg of dust?

Answer 23:

Let x be the number of days taken by the earth to pick up $4.8\times {10}^8$ kg of dust.
$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{amount} \mathrm{of} \mathrm{dust} \mathrm{picked} \mathrm{up} \mathrm{by} \mathrm{the} \mathrm{earth} \mathrm{and} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{days} \mathrm{are} \mathrm{in} \mathrm{direct} \mathrm{variation}, \mathrm{we} \mathrm{get}: \\ \frac{15}{x} = \frac{1.2 \times {10}^8}{4.8 \times {10}^8} \\ \Rightarrow x = 15 \times \frac{4.8}{1.2} \\ \Rightarrow x = 60 \end{gathered}$$
Thus, the required number of days will be 60.