4 Mark Question

Question 14 Marks

Raghu has enough money to buy 75 machines worth Rs 200 each. How many machines can he buy if he gets a discount of Rs 50 on each machine?

Answer

Let x be the number of machines he can buy if a discount of Rs. 50 is offered on each machine.

Cost of a cycle (in Rs.)	500	625
Number of cycles	25	x

Since Raghu is getting a discount of Rs 50 on each machine, the cost of each machine will get decreased by Rs. 50
If the price of a machine is less, he can buy more number of machines.
It is a case of inverse variation. Therefore, we have:
$75\times200 =\text{x}\times150$
$\Rightarrow\text{x}=\frac{75\times200}{150}$
$=\frac{15000}{150}$
$=100$
$\therefore$ The number of machines he can buy is 100.

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Question 24 Marks

It x and y vary inversely, fill in the following blanks:

x	9	...	81	243
y	27	9	...	1

Answer

Since x and y vary inversely, we have:
xy = k
For x = 9 and y = 27
9 × 27 = k
⇒ k = 243
For y = 9 and k = 243, we have:
xy = k
$\Rightarrow9\text{x}=243$
$\Rightarrow\text{y}=\frac{243}{9}$
$=27$
For x = 81 and k = 243, we have:
xy = k
$\Rightarrow81\text{y}=243$
$\Rightarrow\text{y}=\frac{243}{81}$
$=3$

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Question 34 Marks

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

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

Answer

Here, x and y vary directly.$\therefore\text{x}=\text{ky}$
x = 5 and y = 8
i.e., 5 = k × 8
$\Rightarrow\text{k} =\frac{5}{8}=0.625$
For y = 12 and k = 0.625, we have:
x = ky
⇒ x = 12 × 0.625 = 7.5
For x = 10 and k = 0.625, we have:
x = ky
⇒ 10 = 0.625 × y
$\Rightarrow\text{y}=\frac{10}{0.625}=16$
For x = 35 and k = 0.625, we have:
x = ky
⇒ 35 = 0.625 × y
$\Rightarrow\text{y}=\frac{35}{0.625}=56$
For x = 25 and k = 0.625, we have:
x = ky
⇒ 25 = 0.625 × y
$\Rightarrow\text{y}=\frac{25}{0.625}=40$
For y = 32 and k = 0.625, we have:
x = ky
⇒ x = 0.625 × 32 = 20

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Question 44 Marks

A car can finish a certain journey in 10 hours at the speed of 48km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance?

Answer

Let the increased speed be x km/h.

Time (in h)	10	8
Speed (km/h)	48	x + 48

Since speed and time taken are in inverse variation, we get:
$10\times48 = 8(\text{x+48})$
$\Rightarrow480=8\text{x}+384$
$\Rightarrow8\text{x}=480-384$
$\Rightarrow8\text{x}=96$
$=12$
Thus, the speed should be increased by 12km/h.

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Question 54 Marks

1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldiers were transferred to another fort and thus the food lasted now for 32 more days. How many soldiers left the fort?

Answer

It is given that after 4 days, out of 28 days, the fort had enough food for 1200 soldiers for (28 - 4 = 24) days.
Let x be the number of soldiers who left the fort .

Number of soldiers	1200	1200-x
Number of days for which food lasts	24	32

Since the number of soldiers and the number of days for which the food lasts are in inverse variation, we have:
$1200\times24 =(1200-\text{x})\times32$
$\Rightarrow\frac{1200\times24}{32}=1200-\text{x}$
$\Rightarrow900=1200-\text{x}$
$\Rightarrow\text{x}=1200-900$
$=300$
Thus, 300 soldiers left the fort.

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Question 64 Marks

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

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

Answer

Here, x and y vary directly.
$\therefore\text{x}=\text{ky}$
x = 4 and y = 16
i.e., 4 = k × 16
$\Rightarrow\text{k} =\frac{4}{16}=\frac{1}{4}$
For x = 9 and $\text{k}=\frac{1}{4},$ we have:
9 = ky
⇒ y = 4 × 9 = 36
For y = 48 and $\text{k}= \frac{1}{4},$ we have:
x = ky
$=\frac{1}{4}\times48=12$
For y = 36 and $\text{k}=\frac{1}{4},$ we have:
x = ky
$=\frac{1}{4}\times36 =9$
For x = 3 and $\text{k}=\frac{1}{4},$ we have:
x = ky
$\Rightarrow3 =\frac{1}{4}\times\text{y}$
⇒ y = 12
For x = 4 and $\text{k}=\frac{1}{4},$ we have:
x = ky
$=\frac{1}{4}\times4=1$

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