RS Aggarwal Class 8 Math Twelfth Chapter Direct Inverse Proportions Exercise 12B Solution

RS Aggarwal Class 8 Math Twelfth Chapter Direct Inverse Proportions Exercise 12B Solution

EXERCISE 12B

(1) Observe the tables given below and in each case find whether x and y are inversely proportional:

(i)

x	6	10	14	16
y	9	15	21	24

Solution: Clearly, we have:

xy = 6 × 9 ≠ 10 × 15 ≠ 14 × 21 ≠ 16 × 24.

∴ x and y are not inversely proportional.

(ii)

x	5	9	15	3	45
y	18	10	6	30	2

Solution: Clearly, we have:

xy = 5 × 18 = 9 × 10 = 15 × 6 = 3 × 30 = 45 × 2 = 90 = constant.

∴ x and y are inversely proportional.

(iii)

x	9	3	6	36
y	4	12	9	1

Solution: xy = 9 × 4 = 3 × 12 = 36 × 1 ≠ 6 × 9.

∴ x and y are not inversely proportional.

(2) If x and y are inversely proportional, find the values of x₁ , x₂, y₁ and y₂ in the table given below:

x	8	x1	16	x2	80
y	y1	4	5	2	y2

 

Solution: Since x and y are inversely proportional, we must have xy = constant.

∴ 8 × y₁ = x₁ × 4 = 16 × 5 = x₂ × 2 = 80 × y₂.

Now, 16 × 5 = 8 × y₁ ⇒ 8y₁ = 80 ⇒y₁ = 10.

16 × 5 = x₁ × 4 ⇒ 4x₁ = 80 ⇒ x₁ = 20.

16 × 5 = x₂ × 2 ⇒ 2x₂ = 80 ⇒ x₂ = 40.

16 × 5 = 80 × y₂ ⇒ 80y₂ = 80 ⇒ y₂ = 1.

∴ x₁ = 20, x₂ = 40, y₁ = 10 and y₂ = 1.

(3) If 35 men can reap a field in 8 days, in how many days can 20 men reap the same field?

Solution: Let the required time be x days.

Number of men	35	20
Time (In day)	8	x

Less men will be finish reap the field more days. So, it is case of inverse proportion.

∴ 35 × 8 = 20 × x

⇒ 20x = 280

⇒ x = 14

∴ Required number of days =14.

(4) 12 men can dig a pond in 8 days. How many men can dig it in 6 days.

Solution: Let the required number of men be x.

Number of men	12	x
Time (in days)	8	6

For finishing the work in less time, more workers will be needed. So it is case of inverse proportion.

∴ 12 × 8 = x × 6

⇒ 6x = 96

⇒ x = 16

∴ Required number of workers = 16.

(5) 6 cows can graze a field in 28 days. How long would 14 cows take to graze the same field?

Solution: Let the required number of time be x days.

Number of cows	6	14
Time (In days)	28	x

For finish grazing with more cows take less time. So, it is case of inverse proportion.

∴ 6 × 28 = 14 × x

⇒ 14x = 168

⇒ x = 12

∴ Required number of days = 12 .

(6) A car takes 5 hours to reach a destination by travelling at the speed of 60 km/hr. How long will it take when the car travels at the speed of 75 km/hr?

Solution: The car travel 5 hours at the speed 60 km/hr = (5 × 60) = 300 km.

Now, For 75 km traveled the car take 1 hour

 (7) A factory requires 42 machines to produce a given number of articles in 56 days. How many machines would be required to produce the same number of articles in 48 days?

Solution: Let the required number of machine be x.

Number of machine	42	x
Time (in days)	56	48

∴ 42 ×56 = x × 48

⇒ 48x = 2352

⇒ x = 49

∴ Required number of machine is 49.

(8) 7 taps of the same size fill a tank in 1 hour 36 minutes. How long will 8 taps of the same size take to fill the tank?

Solution: Let the required time be x minutes.

Here, 1 hour 36 minutes = 96 minutes

Number of taps	7	8
Time (in minutes)	96	x

For fill the tank more taps take less time. So, it is case of inverse proportion.

∴ 7 × 96 = 8 × x

⇒ 8x = 672

⇒ x = 84

∴ Required time to fill the tank be 8 taps = 84 minutes = 1 hour 24 minutes.

(9) 8 taps of the same size fill a tank in 27 minutes. If two taps go out order, how long would the remaining taps take to fill the tank?

Solution: Let the required time be x minutes.

Number of taps	8	8 – 2 = 6
Time (in minutes)	27	x

Less tap take more time to fill the tank. So, it is case of inverse proportion.

∴ 8 × 27 = 6 × x

⇒ 6x = 216

⇒ x = 36

∴ Required time to fill the tank = 36 minutes.

(10) A farmer has enough food to feed 28 animals in his cattle for 9 days. How long would food last, if there were 8 more animals in his cattle?

Solution: Let the required number of days be x.

Time (in days)	9	x
Number of animals	28	(28+8) = 36

More animals will have food for less time. So, it is case of inverse proportion.

∴ 9 × 28 = 36 × x

⇒ 36x = 252

⇒ x = 7

∴ Required number of days = 7.

(11) A garrison of 900 men had food provisions for 42 days. However, a reinforcement of 500 men arrived. For how many days will the food last now?

Solution: Let the required time be x days.

Number of men	900	(900+500) = 1400
Time (in days)	42	x

More men will have food for less time. So, it is case of inverse proportion.

∴ 900 × 42 = 1400 × x

⇒ 1400x = 37800

⇒ 14x = 378

⇒ x = 27

∴ Required number of days = 27.

(12) In a hostel, 75 students had food provision for 24 days. If 15 students leave the hostel, for how many days would the food provision last?

Solution: Let the required number of days be x.

Number of students	75	(75 – 15) = 60
Time (in days)	24	x

Less number of students will have food for more days. So, it is case of inverse proportion.

∴ 75 × 24 = 60 × x

⇒ 60x = 1800

⇒ 6x = 180

⇒ x = 30

∴ The required number of days = 30.

(13) A school has 9 periods a day each of 40 minutes duration. How long would each period be, if the school has 8 periods a day, assuming the number of the school hours to be the same?

Solution: If duration of each period 40 minutes. Therefore, durations of 9 periods = (9 × 40) = 360 minutes

∴ Therefore required time of each periods if there are 8 periods in the school= (360 ÷ 8) = 45 minutes.

 (14) If x and y vary inversely and x = 15 when y = 6, find y when x = 9.

Solution: We can write,

xy = 15 × 6

⇒ 9y = 90

⇒ y = 10

(15) If x and y vary inversely and x = 18 when y = 8, find x when y = 16.

Solution: We can write,

xy = 18 × 8

⇒ 16x = 144

⇒ x = 9