If 5 engines consume 6 metric tonnes of coal when each is running for

Solution

Given

• 5 engines consume 6 metric tonnes of coal when each is running for 9 hours a day
• 3 engines of the former type consume as much as 4 engines of the latter type

To find

• How many metric tonnes of coal will be needed for 8 engines, each running 10 hours a day?

Approach and Working out
We have:

• Direct relation between engine and coal: more engine – more coal
• Direct relation between hours and coal: more hours per day – more coal
• Direct relation between rate of consumption and coal: more consumption – more coal

Ratio of engines = 5: 8
Ratio of hours per day = 9: 10
Ratio of consumption rate = (1/3): (1/4)
These ratios are equal to 6: n

Therefore,

• n = 8 x 10 x ¼ x 6 x 1/5 x 1/9 x 3
Or, n = 8

Thus, option A is the correct answer.

Correct Answer: Option A

Hi Payal,

Could you elaborate how you got the consumption rate?

Thanks,

Hi Bunuel,

Are these types of questions asked in actual GMAT ? If yes, I request you to please provide an easier approach solution.

Thanks,
Ankit Singh

This is a real time GMAT question of Work-Rate problem of "Joint Variation". I need to mention that I usually proceed this type of question with a very easy approach:- ( FOLLOW THE ATTACHED DIAGRAM)
Thus,

OLD Engine ( 5) --> hrs (9) ----> MT (6)
NEW Engine (8) ---> hrs (10) ---> MT (?) , let x MT

Now 3 old engines \(\equiv\ \) 4 new engines
Hence, 8 New engine \(\equiv\ \) 6 old engines
So, the format will be:
OLD Engine ( 5) --> hrs (9) ----> MT (6)
OLD Engine (6) ---> hrs (10) ---> MT (?) , let x MT
( No. of Engine inversely proportional to hrs while MT is constant and hrs is directly proportional to MT while No. of engines are constant)
So the equation will be ( see attached image)
the equation will be 5*9*x = 6*10*6 ( x= MT of coal that we need to find out as answer)
x = 8 MT
Ans A

If 5 engines consume 6 metric tonnes of coal when each is running for 9 hours a day, how many metric tonnes of coal will be needed for 8 engines, each running 10 hours a day, given that 3 engines of the former type consume as much as 4 engines of the latter type.

Assuming the rate of coal consumption of former engine kind per hour = X.
Assuming the rate of coal consumption of the latter engine kind per hour = Y.

Given \(3X\ =\ 4Y,\ X\ =\ \frac{4Y}{3}\) .
The total working hours for the former engine = 9*5 = 45 hours 45X = 6 metric tonnes.
For the latter engines the number of working hours = 80Y = 60X.

Since 45X requires 6 metric tonnes, 60X would require 8 metric tonnes.