1) If 15 men can reap the crops of a field in 28 days, in how many days will 5 men reap it?
Let 5 men can reap a field in x days
So, put the same quantities on the same side.
Now, Men and Days are inversely proportional to each other. If we increase the number of men fewer days will be required to complete the work.
Inversely proportional means
i.e., 5: 15 = 28: x
Or, x = (28*15)/ 5
Or, x = 84 days
Hence, 5 men can reap a field in 84 days.
2) If 8 men can reap 80 hectares in 24 days, how many hectares can 36 men reap in 30 days?
We can resolve the problem in 2 steps:
i. If 8 men can reap 80 hectares, then how many hectares can reap by 36 men in the same number of days
Now, the same type should be on the same side
Let the required number of hectares = x
Men and hectares are directly proportional to each other.
So, 8: 36 = 80: x
Or, x = (36*80)/8
Or, x = 360 hectares
ii. If 360 hectares in 24 days, then how many hectares can reap in 30 days?
Similarly, 24: 30 = 360: x
So, x = (360*30)/24
Or, x = 450
Hence, 36 men can reap 450 hectares in 30 days.
We know that 8 men work 24 days and reap a field of 80 hectares.
Similarly, 36 men work 30 days and reap a field of x (let) hectares.
Now, we know that men * days = total work
So, 8*24 = 192, that means total work done is equals to 192
Now, 36*30 = 1080, that means the work done by 36 men is 1080 unit.
Now, put the same unit on same side
Or, 192: 1080 = 80: x
Or, x = (1080*80)/ 192 = 450
Hence, 36 men can reap 450 hectares in 30 days.
3) A fort had arrangements for 150 boys for 45 days. After 10 days, 25 boys left the fort. Then after how much time the food will be consumed completely if the consumption of food remains the same for the remaining boys?
None of these
ATQ, 25 people left the fort after 10 days, but still, the remaining food will be consumed at the same rate.
That means if the 150 boys continue till the end, then the remaining food would last for 150 boys for (45-10) = 35 days.
Now After 10 days the remaining food will be (boys * days) 150 * 35 = 5250 unit
But the 25 boys left the for after 10 days
i.e., 125 boys will consume the 5250 unit food in x days
Now, x = 5250/125 = 42 days.
That means the remaining food will last for 42 days.
The remaining food would last for 150 boys for (45-10) = 35 days. But as 25 boys have gone out, the remaining food would last for a long period (x).
The number of boys and consumption of food are inversely proportional to each other.
That means, 125 boys: 150 boys = 35 days: x
Or, x = (150*35)/ 125 = 42 days
Hence, 125 boys require 42 days to consume the remaining food.
4) If 30 men can complete a piece of work in 27 days, in what time 18 men can do another piece of work 3 times as greater?
30 men can do a piece of work in 27 days, and we know that men* days = total work
So, we can say that the total work = 30*27 = 810 units
Now ATQ, 18 men can do work 3 times greater than 810 units.
i.e., 18 men can do a work 3*810 = 2430 units in x days
Men *days = total work (2430 units)
18 * x = 2430
x = 2430/18 = 135 days
Let the number of days required to do a work = x
Now, 30 men can do a piece of work in 27 days.
ATQ, if the work gets 3 times, then 18 men can finish the work in how many days.
Now, put the same unit on the same side.
i.e., if the work is the same as previous, then 18 men: 30 men = 27 days: x days
Or, x = (30*27)/ 18
But ATQ, the work gets 3 times greater
So, x = (30*27*3)/ 18 = 2430/18 = 135 days
Hence, 18 men can finish the other work which is 3 times greater than previous work in 135 days.
5) If 9 engines consume 24 metric tons of coal, when each is working 8 hours per day, how much coal should be available for 8 engines, each running 13 hours per day, it is given that 3 engines of the former type consume as much as 4 engines of later type.
36 metric tons
12 metric tons
52 metric tons
26 metric tons
The lesser engines, less coal consumed
More working hours, more coal consumed
Both the cases are directly proportional.
If three engines of former type consume 1 unit, 1 engine will consume 1/3 unit.
If four engines of latter type consume 1 unit, 1 engine will consume ¼ units.
And, less rate of consumption, less coal consumed.
Now, Number of engines = 9: 8
Working hours = 8:13
Therefore, rate of consumption
Let the coal consumed by 8 engines is x metric tones
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