# Aptitude Chain Rule Test Paper 3

11) A tower 17.5 m high casts a shadow of 40.25 m. What is the height of the building which casts a shadow 28.75 m long under similar conditions?

1. 10 m
2. 17.2 m
3. 12.5 m
4. 21.25 m

Explanation:

Let the height of the building is x
Now, the shadow ratio = building ratio
Height is directly proportional to shadow, so:
40.25: 28.75 = 17.5: x
Now, x = (28.75* 17.5)/ 40.25 = 12.5 m

12) A contractor engaged 100 labors for completing a certain project within a fixed time. Whenof the time limit had expired, he found that of the originally assigned work had been already completed. The number of laborers who could be removed by the contractor for this project so that the project could be completed by the remaining laborers within the given time limit would be

1. 50
2. 20
3. 30
4. 40

Explanation:

Let the work be 4 units and the time be 5 days, this assumption is taken on the basis of the denominator part of the given fraction.
That means, 3 units of work are completed in 3 days and the 100 men working per day.
Therefore 1 unit of the work is done in 1 day by 100 men......... (i)
Now the remaining one unit of the work should be done in 2 days (5 days ? 3 days)
But by equation one, to complete 1 unit work, one day required by 100 men.
If we reduce the men working per day by 50%, then double time is required to finish the remaining work.
So, 50% reduction of men = 100/2 = 50 men
Therefore (100- 50) = 50 men must be removed.

13) One army camp had ration for 560 soldiers for 20 days, 560 soldiers reported for the camp, and after 12 days, 112 soldiers were sent to another camp. For how many days, the remaining soldiers can stay in the camp without getting any new ration?

1. 10 days
2. 12 days
3. 16 days
4. 14 days

Explanation:

After 12 days, there was ration for 560 soldiers for 8 days.
Remaining persons = (560-112) = 448
Less soldiers, more days (inverse proportion)
Let the x is the required number of days
Then, 448: 560 = 8: x
Or, x = (560*8)/ 448 = 10
Hence, the required number of days is 10.

14) If I would have been twice as efficient as today, I would have finished work in 12 days. If my efficiency is reduced to one-third of what it is at present, in how many days, I would be able to finish the work?

1. 18
2. 8
3. 72
4. 52

Explanation:

Let I finish work in x days.
With double efficiency, the time taken = x/2 days
That means when the efficiency was double (2x), then the time taken to finish the work is 12 days
Now, with the present efficiency time taken to complete the work = 12*2 =24 days
With one-third efficiency, the days required to finish the work = 3 x 24 = 72 days, as efficiency is inversely proportional to days.
Hence, when the efficiency gets one-third, then the work will be finished in 72 days.

15) Pervez and Sunny can complete a piece of work in 20 and 15 days respectively. They worked together for 6 days, after which Sunny was replaced by Ashu. If the work would be finished in next 4 days, find the number of days in which Ashu alone could complete the work.

1. 45
2. 40
3. 56
4. 36

Explanation:

Pervez can complete the work in 20 days
Sunny can complete the same job in 15 days
Let the total work = LCM of both
Therefore, the LCM of 20 and 15 is 60, that means total work = 60 units
Now, Pervez's one-day work efficiency = 60/20 = 3 units
Sunny's one-day work efficiency = 60/15 = 4 units
ATQ, (Pervez+Sunny) together works for 6 days
That means (Pervez+Sunny) have done 7*6 = 42 units work in 6 days
Now, the remaining work will be 60-42 = 18 units, and it will finish in 4 days.
Now, Sunny is replaced with Ashu:
So, per days work required to be finished = 18 units / 4 days = 4.5 units
But, we know that Pervez's one-day work = 3 units, so Ashu's one-day work = 4.5-3 = 1.5 units
Therefore, Ashu alone can finish the total work in 60 units /1.5 days = 40 days

Aptitude Chain Rule Test Paper 1
Aptitude Chain Rule Test Paper 2
Aptitude Chain Rule Test Paper 3