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
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