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

Exercise 12A

Page-162

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

OPEN IN YOUTUBE

Question 1:

Observe the tables given below and in each one find whether x and y are proportional:
(i)

x	3	5	8	11	26
y	9	15	24	33	78

(ii)

x	2.5	4	7.5	10	14
y	10	16	30	40	42

(iii)

x	5	7	9	15	18	25
y	15	21	27	60	72	75

Answer 1:

(i)
$Clearly, \frac{x}{y} = \frac{3}{9} = \frac{5}{15} = \frac{8}{24} = \frac{11}{33} = \frac{26}{78} = \frac{1}{3} (constant) Therefore, x and y are proportional .$

(ii)
$Clearly, \frac{x}{y} = \frac{2.5}{10} = \frac{4}{16} = \frac{7.5}{30} = \frac{10}{40} = \frac{1}{4}, while \frac{14}{42} = \frac{1}{3} i . e ., \frac{2.5}{10} = \frac{4}{16} = \frac{7.5}{30} = \frac{10}{40} is not equal to \frac{14}{42} . Therefore, x and y are not proportional .$

(iii)
$Clearly, \frac{x}{y} = \frac{5}{15} = \frac{7}{21} = \frac{9}{27} = \frac{25}{75} = \frac{1}{3}, while \frac{15}{60} = \frac{18}{72} = \frac{1}{4} i . e ., \frac{5}{15} = \frac{7}{21} = \frac{9}{27} = \frac{25}{75} is not equal to \frac{15}{60} and \frac{18}{72} . Therefore, x and y are not proportional .$

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

OPEN IN YOUTUBE

Question 2:

If x and y are directly proportional, find the values of x₁ , x₂ and y₁ in the table given below:

x	3	x₁	x₂	10
y	72	120	192	y₁

Answer 2:

$Since x and y are directly propotional, we have : \frac{3}{72} = \frac{x_{1}}{120} = \frac{x_{2}}{192} = \frac{10}{y_{1}} Now, \frac{3}{72} = \frac{x_{1}}{120} \Rightarrow x_{1} = \frac{120 \times 3}{72} = 5$

$And, \frac{3}{72} = \frac{x_{2}}{192} \Rightarrow x_{2} = \frac{3 \times 192}{72} = 8$
$And, \frac{3}{72} = \frac{10}{y_{1}} \Rightarrow y_{1} = \frac{72 \times 10}{3} = 240$
$Therefore, x_{1} = 5, x_{2} = 8 and y_{1} = 240$

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

OPEN IN YOUTUBE

Question 3:

A truck covers a distance of 510 km in 34 litres of diesel. How much distance would it cover in 20 litres of diesel?

Answer 3:

Let the required distance be x km. Then, we have:

Quantity of diesel (in litres)	34	20
Distance (in km)	510	x

Clearly, the less the quantity of diesel consumed, the less is the distance covered.
So, this is a case of direct proportion.

Now, \frac{34}{510} = \frac{20}{x} \Rightarrow \frac{1}{15} = \frac{20}{x} \Rightarrow x \times 1 = 20 \times 15 = 300

Therefore, the required distance is 300 km.

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

OPEN IN YOUTUBE

Question 4:

A taxi charges a fare of Rs 2550 for a journey of 150 km. How much would it charge for a journey of 124 km?

Answer 4:

Let the charge for a journey of 124 km be ₹x.

Price(in ₹)	2550	x
Distance(in km)	150	124

More is the distance travelled, more will be the price.
So, it is a case of direct proportion.

∴ \frac{2550}{150} = \frac{x}{124} \Rightarrow x = \frac{2550 \times 124}{150} = 2108

Thus, the taxi charges ₹2,108 for the distance of 124 km.

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

OPEN IN YOUTUBE

Question 5:

A loaded truck covers 16 km in 25 minutes. At the same speed, how far can it travel in 5 hours?

Answer 5:

Let the required distance be x km. Then, we have:
$1 h = 60 \min i . e ., 5 h = 5 \times 60 = 300 \min$ .

Distance (in km)	16	x
Time (in min)	25	300

Clearly, the more the time taken, the more will be the distance covered.

So, this is a case of direct proportion.

Now, \frac{16}{25} = \frac{x}{300} \Rightarrow x = (\frac{16 \times 300}{25}) \Rightarrow x = 192

Therefore, the required distance is 192 km.

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

OPEN IN YOUTUBE

Question 6:

If 18 dolls cost Rs 630, how many dolls can be bought for Rs 455?

Answer 6:

Let the required number of dolls be x. Then, we have:

No of dolls	18	x
Cost of dolls (in rupees)	630	455

Clearly, the less the amount of money, the less will be the number of dolls bought.
So, this is a case of direct proportion.

Now, \frac{18}{630} = \frac{x}{455} \Rightarrow \frac{1}{35} = \frac{x}{455} \Rightarrow x = \frac{455}{35} \Rightarrow x = 13

Therefore, 13 dolls can be bought for Rs 455.

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

OPEN IN YOUTUBE

Question 7:

If 9 kg of sugar costs ₹ 238.50, how much sugar can be bought for ₹ 371?

Answer 7:

Let the quantity of sugar bought for ₹371 be x kg.

Quantity(in kg)	9	x
Price(in ₹)	238.50	371

The price increases as the quantity increases. Thus, this is a case of direct proportion.

∴ \frac{9}{238.50} = \frac{x}{371} \Rightarrow x = \frac{9 \times 371}{238.50} = 14

Thus, the quantity of sugar bought for ₹371 is 14 kg.

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

OPEN IN YOUTUBE

Question 8:

The cost of 15 metres of a cloth is Rs 981. What length of this cloth can be purchased for Rs 1308?

Answer 8:

Let the length of cloth be x m. Then, we have:

Length of cloth (in metres)	15	x
Cost of cloth (in rupees)	981	1308

Clearly, more length of cloth can be bought by more amount of money.
So, this is a case of direct proportion.

Now, \frac{15}{981} = \frac{x}{1308} \Rightarrow x = \frac{15 \times 1308}{981} \Rightarrow x = 20

Therefore, 20 m of cloth can be bought for Rs 1,308.

Page-163

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

OPEN IN YOUTUBE

Question 9:

In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 15m high. If the length of the ship is 35 metres, how long is the model ship?

Answer 9:

Let x m be the length of the model of the ship. Then, we have:
$1 m = 100 cm Therefore, 15 m = 1500 cm 35 m = 3500 cm$

	Length of the mast (in cm)	Length of the ship (in cm)
Actual ship	1500	3500
Model of the ship	9	x

Clearly, if the length of the actual ship is more, then the length of the model ship will also be more.
So, this is a case of direct proportion.

Now, \frac{1500}{9} = \frac{3500}{x} \Rightarrow x = \frac{3500 \times 9}{1500} \Rightarrow x = 21 cm

Therefore, the length of the model of the ship is 21 cm.

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

OPEN IN YOUTUBE

Question 10:

In 8 days, the earth picks up (6.4 × 10⁷) kg of dust from the atmosphere. How much dust will it pick up in 15 days?

Answer 10:

Let x kg be the required amount of dust. Then, we have:

No. of days	8	15
Dust (in kg)	$6.4 \times 10^{7}$	x

Clearly, more amount of dust will be collected in more number of days.
So, this is a case of direct proportion.

Now, \frac{8}{6.4 \times 10^{7}} = \frac{15}{x} \Rightarrow x = \frac{15 \times 6.4 \times 10^{7}}{8} \Rightarrow x = 12 \times 10^{7}

Therefore, 12,00,00,000 kg of dust will be picked up in 15 days.

Q11 | Ex-12A | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

OPEN IN YOUTUBE

Question 11:

A car is travelling at the average speed of 50 km/hr. How much distance would it travel in 1 hour 12 minutes?

Answer 11:

Let x km be the required distance. Then, we have:

$1 h = 60 \min i . e ., 1 h 12 \min = (60 + 12) \min = 72 \min$

Distance covered (in km)	50	x
Time (in min)	60	72

Clearly, more distance will be covered in more time.
So, this is a case of direct proportion.

Now, \frac{50}{60} = \frac{x}{72} \Rightarrow x = \frac{50 \times 72}{60} \Rightarrow x = 60

Therefore, the distance travelled by the car in 1 h 12 min is 60 km.

Q12 | Ex-12A | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

OPEN IN YOUTUBE

Question 12:

Ravi walks at the uniform rate of 5 km/hr. What distance would he cover in 2 hours 24 minutes?

Answer 12:

Let x km be the required distance covered by Ravi in 2 h 24 min.
Then, we have:
$1 h = 60 \min i . e ., 2 h 24 \min = (120 + 24) \min = 144 \min$

Distance covered (in km)	5	x
Time (in min)	60	144

Clearly, more distance will be covered in more time.
So, this is a case of direct proportion.

Now, \frac{5}{60} = \frac{x}{144} \Rightarrow x = \frac{5 \times 144}{60} \Rightarrow x = 12

Therefore, the distance covered by Ravi in 2 h 24 min is 12 km.

Q13 | Ex-12A | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

OPEN IN YOUTUBE

Question 13:

If the thickness of a pile of 12 cardboards is 65 mm, find the thickness of a pile of 312 such cardboards.

Answer 13:

Let x mm be the required thickness. Then, we have:

Thickness of cardboard (in mm)	65	x
No. of cardboards	12	312

Clearly, when the number of cardboard is more, the thickness will also be more.
So, it is a case of direct proportion.

Now, \frac{65}{12} = \frac{x}{312} \Rightarrow x = \frac{65 \times 312}{12} \Rightarrow x = 1690

Therefore, the thickness of the pile of 312 cardboards is 1690 mm.

Q14 | Ex-12A | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

OPEN IN YOUTUBE

Question 14:

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 14:

Let x be the required number of men.

$Now, 6 \frac{3}{4} m = \frac{27}{4} m$

Then, we have:

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

Clearly, the longer the trench, the greater will be the number of men required.
So, it is a case of direct proportion.

Now, \frac{11}{\frac{27}{4}} = \frac{x}{27} \Rightarrow \frac{11 \times 4}{27} = \frac{x}{27} \Rightarrow x = 44

Therefore, 44 men should be employed to dig a trench of length 27 m.

Q15 | Ex-12A | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

OPEN IN YOUTUBE

Question 15:

Reenu types 540 words during half an hour. How many words would she type in 8 minutes?

Answer 15:

Let Reenu type x words in 8 minutes.

No. of words	540	x
Time taken (in min)	30	8

Clearly, less number of words will be typed in less time.
So, it is a case of direct proportion.

Now, \frac{540}{30} = \frac{x}{8} \Rightarrow x = \frac{540 \times 8}{30} \Rightarrow x = 144

Therefore, Reenu will type 144 words in 8 minutes.

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

Exercise 12A

Page-162

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

Question 1:

Answer 1:

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

Question 2:

Answer 2:

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

Question 3:

Answer 3:

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

Question 4:

Answer 4:

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

Question 5:

Answer 5:

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

Question 6:

Answer 6:

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

Question 7:

Answer 7:

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

Question 8:

Answer 8:

Page-163

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

Question 9:

Answer 9:

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

Question 10:

Answer 10:

Q11 | Ex-12A | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 11:

Answer 11:

Q12 | Ex-12A | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 12:

Answer 12:

Q13 | Ex-12A | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 13:

Answer 13:

Q14 | Ex-12A | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 14:

Answer 14:

Q15 | Ex-12A | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

Question 15:

Answer 15:

No comments:

Post a Comment

Contact Form