Please help me out

[tass@]

There was a leakage in the container of the refined oil. If 11 kg oil is leaked out per day then it would have lasted for 50 days, If the leakage was 15 kg per day, then it would have lasted for only 45 days. For how many days would the oil have lasted if there was no leakage and it was completely used for eating purpose

1) 72 2) 84 3)105 4)64

[Mathsbuddy]

This is a tough one to explain without drawing graphs, but I'll give it a go.
What you need to spot in the question is that the oil diminishes due to 2 factors:

Reduction Rate (R) = Leak rate (L) + Eating Rate (E)

Another equation is:
Starting Amount (A) = R x time (T)

So (L1 + E)*T1 = (L2 + E)*T2
Substituting L1 =11, T1 = 50, L2 = 15, T2 = 45
gives E = 25 (litres per day)

As A = RT
Then A = (L1 + E)*T1 = 1800 (litres)

With no leaks, then using an eating rate of E = 25 litres a day only:

T = 1800/25 = 72 (days)

Answer 1.

[GMATGuruNY]

There was a leakage in the container of the refined oil. If 11 kg oil is leaked out per day then it would have lasted for 50 days, If the leakage was 15 kg per day, then it would have lasted for only 45 days. For how many days would the oil have lasted if there was no leakage and it was completely used for eating purpose

1) 72 2) 84 3)105 4)64

The problem should indicate that the same amount of oil is required each day.

Approach 1:
Let x = the total amount of oil without leakage.

If 11kg leaks out every day for 50 days, the total amount of leakage = 11*50 = 550.
Thus, over the course of 50 days, the total amount of oil used for eating purposes = x-550.
Implication:
(x - 550) is sufficient oil to last for 50 days.

If 15kg leaks out every day for 45 days, the total amount of leakage = 15*45 = 675.
Thus, over the course of 45 days, the total amount of oil used for eating purposes = x-675.
Implication:
(x - 675) is sufficient oil to last for 45 days.

Since (x - 550) is sufficient oil to last for 50 days, while (x - 675) is sufficient oil to last for 45 days, we can set up the following proportion:
(x-550)/50 = (x-675)/45.

Solving for x, we get:
(x-550)/10 = (x-675)/9
9x - 4950 = 10x - 6750
1800 = x.

Thus, the amount of oil sufficient for 50 days = x-550 = 1800-550 = 1250.
Since 1250kg is sufficient for 50 days, the number of kg required per day = 1250/50 = 25.
Since 25kg is required per day, the total amount of oil without leakage -- 1800 kg -- is sufficient to last for the following number of days:
1800/25 = 72 days.

The correct answer is A.

Approach 2:
Let x = the amount of oil required each day.

Case 1: Oil lasts for 50 days
Since x kg are required each day, the total amount of oil used over the course of 50 days = 50x.
Since 11kg is leaked each day, the total amount of leakage over the course of 50 days = 11*50 = 550.
Thus, the total amount of oil in the container = (amount used) + (amount leaked) = 50x + 550.

Case 2: Oil lasts for 45 days
Since x kg are required each day, the total amount of oil used over the course of 45 days = 45x.
Since 15kg is leaked each day, the total amount of leakage over the course of 45 days = 15*45 = 675.
Thus, the total amount of oil in the container = (amount used) + (amount leaked) = 45x + 675.

Since the total amount of oil must be the same in each case, we get:
50x + 550 = 45x + 675
5x = 125
x = 25.

Thus, the total amount of oil in the container = 50x + 550 = 50(25) + 550 = 1800kg.
Since 25kg is required per day, the total amount of oil in the container is sufficient to last for the following number of days:
1800/25 = 72 days.