Aptitude Time and Work Test Paper 8

Answer: A

Explanation:

Apply formula: [Man1 * day1* hour1*]/ work1 = [Man2 * day2* hour2*]/ work2

ATQ, target = 120 days

M1 = 100, M2 =?
D1 = 45, D2 = 120-45 = 75
Let total work = 1
W1 = ¼, W2= 1-1/4 = ¾
Assume H1 = H2 = 1

Now, apply the formula

[M1 * D1 * H1]/ W1 = [M2 * D2 * H2]/ W2
Now, [H1 = H2 = 1]
So, [M1 * D1 * W2] = [M2 * D2 * W1]

Or, (100 * 45 * 3/4) = (M2 * 75 * ¼)
Or, M2 = 180

So, M2 - M1 = 180 -100 = 80 extra workers will be required.

Answer: B

Explanation:

Note: Assume the total work = LCM of the given days

Take the LCM of days = LCM of (20 and 30) = 60
Let the total work = 60

A's one day work = 60/20 = 3 unit
B's one day work = 60/30= 2 unit

If (A+B) work, then they can finish the work in 60/5 = 12 days.

But ATQ, B left the job 5 days before
Assume that B has not left the work than B's 5-day work also included in the total work.
i.e., B's 5 day work = 5 * 2 = 10 units

Now, the total work = 60+10 = 70units
If (A+B)'s one day efficiency = 5

So, to complete the work it requires = 70/5 = 14 days.

Answer: A

Explanation:

Note: Assume the total work = LCM of the given days

Take the LCM of days = LCM of (18, 27, and 36) = 108
Let the total work = 108

A's one day work = 108/18 = 6 units
B's one day work = 108/27= 4 units
C's one day work = 108/36= 3 units

ATQ, A works only for 8 days.
So, A's total work done = 8 * 6 = 48
So, remaining work = 108-48 = 60 work done by (B and C).

If (B+C) work till last, they can finish the work in 60/7= 8.57 days.

But ATQ, B left the job 6 days before it is completed.
Assume that B is not left the work than B's 6-day work is also included in the total work.
So, B's 6 day work = 6 * 4 = 24 units

Now, the total remaining work = 60+24 = 84units.

(B+C)'s one day efficiency = 7

So, to complete the work it requires = 84/7 = 12 days.

Answer: A

Explanation:

ATQ, (A+B) can complete a work in 4 days......................... (i)
Now, the efficiency of A decreases by 30% and B's efficiency is increased by 10%.
i.e., suppose A's and B's efficiency = 1 per day
Then, A = 0.7 and B = 1.1
Or, (0.7 A + 1.1 B) can finish a work in 5 days...................... (ii)

The work will be the same.

Total work = man* days

So, equation 1 = equation 2

(A+B)*4 = (0.7 A + 1.1 B)* 5
Or, 4A + 4B = 3.5A + 5.5B
Or, 0.5A = 1.5B
Or, A: B = 3: 1

That means the efficiency ratio of A and B = 3: 1
Or, total work = Efficiency* days
Or, Total work = (3+1) * 4 = 16

ATQ, the efficiency of A remains same and B's efficiency gets 1[1/2] of the original efficiency.
i.e., A: B = 3: [1 * 1[1/2]]
Or, A: B = 3: 3/2

Now, (A+B)'s one day work = 3+ 3/2 = 9/2

So, days requires = Total work/ efficiency of (A+B)
Or, days = 16/ (9/2), or, 32/9 = 3[5/9]
Hence, 3[5/9] days are required to complete the work.

Answer: B

Explanation:

ATQ, M = 200, D = 50
M1 = 200, D1 = (50-10) = 40, M2 = 200+50=250, D2=?

Apply the formula:

[M1 * D1 * H1]/ W1 = [M2 *D2 * H2]/ W2

But here W1, W2, H1, and H2 are not present so assume these to 1.

So, M1 * D1 = M2 * D2

After 10 days, 200 students can eat for 40 days, but on the other hand, 50 more students join the hostel. So, 250 men can eat it for D2 days
So, 200 * 40 = 250 * D2
D2 = 8000/250

So, the food will continue for 32 days.