myhelper: RS Aggarwal solution class 8 chapter 12 Direct and Inverse Proportions Exercise 12C

Exercise 12C

Q1 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 1:

If 14 kg of pulses cost ₹ 882, what is the cost of 22 kg of pulses?
(a) ₹ 1254
(b) ₹ 1298
(c) ₹ 1342
(d) ₹ 1386

Answer 1:

Let 22 kg of pulses cost ₹x.

Quantity(in kg)	14	22
Price(in ₹)	882	x

As the quantity increases, the price also increases. So, it is a case of direct proportion.

∴ \frac{14}{882} = \frac{22}{x} \Rightarrow x = \frac{22 \times 882}{14} = 1386

Thus, the cost of 22 kg of pulses is ₹1,386.

Hence, the correct answer is option (d).

Q2 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 2:

If 8 oranges cost ₹ 52, how many oranges can be bought for ₹ 169?
(a) 13
(b) 18
(c) 26
(d) 24

Answer 2:

Let the number of oranges that can be bought for ₹169 be x.

Quantity	8	x
Price(in ₹)	52	169

As the quantity increases the price also increases. So, this is a case of direct proportion.

∴ \frac{8}{52} = \frac{x}{169} \Rightarrow x = \frac{8 \times 169}{52} = 26

Thus, 26 oranges can be bought for ₹169.

Hence, the correct answer is option (c).

Q3 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 3:

Tick (✓) the correct answer:
A machine fills 420 bottles in 3 hours. How many bottles will it fill in 5 hours?
(a) 252
(b) 700
(c) 504
(d) 300

Answer 3:

(b) 700

Let x be the number of bottles filled in 5 hours.

No. of bottles	420	$x$
Time (h)	3	5

More number of bottles will be filled in more time.

Now, \frac{420}{3} = \frac{x}{5} \Rightarrow x = \frac{420 \times 5}{3} \Rightarrow x = 700

Therefore, 700 bottles would be filled in 5 h.

Q4 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 4:

Tick (✓) the correct answer:
A car is travelling at a uniform speed of 75 km/hr. How much distance will it cover in 20 minutes?
(a) 25 km
(b) 15 km
(c) 30 km
(d) 20 km

Answer 4:

(a) 25 km

Let x km be the required distance.
Now, 1 h = 60 min

Distance (in km)	75	$x$
Time (in min)	60	20

Less distance will be covered in less time.

Now, \frac{75}{60} = \frac{x}{20} \Rightarrow x = \frac{75 \times 20}{60} \Rightarrow x = 25 km

Q5 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 5:

Tick (✓) the correct answer:
The weight of 12 sheets of a thick paper is 40 grams. How many sheets would weight 1 kg?
(a) 480
(b) 360
(c) 300
(d) none of these

Answer 5:

No. of sheets	12	$x$
Weight (in g)	40	1000

Now, \frac{12}{40} = \frac{x}{1000} \Rightarrow x = \frac{12 \times 1000}{40} \Rightarrow x = 300

Page-167

Q6 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 6:

Tick (✓) the correct answer:
A pole 14 m high casts a shadow of 10 m. At the same time, what will be the height of a tree, the length of whose shadow is 7 metres?
(a) 20 m
(b) 9.8 m
(c) 5 m
(d) none of these

Answer 6:

(b) 9.8 m
Let x m be the height of the tree.

Height of object	14	$x$
Length of shadow	10	7

The more the length of the shadow, the more will be the height of the tree.

Now, \frac{14}{10} = \frac{x}{7} \Rightarrow x = \frac{14 \times 7}{10} \Rightarrow x = 9.8

Therefore, a 9.8 m tall tree will cast a shadow of length 7 m.

Q7 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

OPEN IN YOUTUBE

Question 7:

Tick (✓) the correct answer:
A photograph of a bacteria enlarged 50000 times attains a length of 5 cm. The actual length of bacteria is
(a) 1000 cm
(b) 10⁻³ cm
(c) 10⁻⁴ cm
(d) 10⁻² cm

Answer 7:

(c) $10^{- 4} cm$
Let x cm be the actual length of the bacteria.
The larger the object, the larger its image will be.
$Now, \frac{x}{1} = \frac{5}{50000} = 10^{- 4} cm$

Hence, the actual length of the bacteria is $10^{- 4} cm$ .

Q8 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

OPEN IN YOUTUBE

Question 8:

Tick (✓) the correct answer:
6 pipes fill a tank in 120 minutes, then 5 pipes will fill it in
(a) 100 min
(b) 144 min
(c) 140 min
(d) 108 min

Answer 8:

(b) 144 min
Let x min be the time taken by 5 pipes to fill the tank.

No. of pipes	6	5
Time (in min)	120	$x$

Now, 6 \times 120 = 5 \times x \Rightarrow x = 144

Therefore, 5 pipes will take 144 min to fill the tank.

Q9 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

OPEN IN YOUTUBE

Question 9:

Tick (✓) the correct answer:
3 persons can build a wall in 4 days, then 4 persons can build it in
(a) $5 \frac{1}{3}$ days
(b) 3 days
(c) $4 \frac{1}{3}$ days
(d) none of these

Answer 9:

(b) 3 days
Let x be number of days taken by 4 persons to build the wall.

No. of persons	3	4
No. of days	4	$x$

More number of persons will take less time to build the wall.
So, it is a case of inverse proportion.

Now, 3 \times 4 = 4 \times x \Rightarrow x = 3

Therefore, 4 persons can build the wall in 3 days.

Q10 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

OPEN IN YOUTUBE

Question 10:

Tick (✓) the correct answer:
A car takes 2 hours to reach a destination by travelling at 60 km/hr. How long will it take while travelling at 80 km/hr?
(a) 1 hr 30 min
(b) 1 hr 40 min
(c) 2 hrs 40 min
(d) none of these

Answer 10:

(a) 1 h 30 min
Let x h be the time taken by the car travelling at 80 km/hr.

Speed (km/h)	60	80
Time (in h)	2	$x$

The greater the speed, the lesser will be the time taken . So, it is a case of inverse proportion . Now, 60 \times 2 = 80 \times x \Rightarrow x = \frac{120}{80} \Rightarrow x = 1.5 Therefore, the car will take 1 h 30 \min to reach its destination if it travels at a speed of 80 km / h .

RS Aggarwal solution class 8 chapter 12 Direct and Inverse Proportions Exercise 12C

Exercise 12C

Q1 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 1:

Answer 1:

Q2 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 2:

Answer 2:

Q3 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 3:

Answer 3:

Q4 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 4:

Answer 4:

Q5 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 5:

Answer 5:

Page-167

Q6 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 6:

Answer 6:

Q7 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 7:

Answer 7:

Q8 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 8:

Answer 8:

Q9 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 9:

Answer 9:

Q10 | Ex-12C | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 10:

Answer 10:

No comments:

Post a Comment

Contact Form