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12 May 2025, 00:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
57% (02:57) correct
43%
(03:08)
wrong
based on 63
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Two internal combustion engines of the same power output were subjected to an efficiency test and it was found that one of them consumed 600 grams of petrol, while the other, which was in operation 2 hours less, consumed 384 grams. If the first engine had consumed as much petrol per hour as the second, and the second, as much as the first, then both engines would have consumed equal amounts of petrol during the same period of operation as before. How much petrol does each engine consume per hour?
(A) 36, 48 gms / hr
(B) 60, 48 gms / hr
(C) 60, 72 gms / hr
(D) 48, 72 gms / hr
(E) 30, 60 gms / hr
Gentle note to all experts and tutors: Please refrain from replying to this question until the Official Answer (OA) is revealed. Let students attempt to solve it first. You are all welcome to contribute posts after the OA is posted. Thank you all for your cooperation!
(A) 36, 48 gms / hr
(B) 60, 48 gms / hr
(C) 60, 72 gms / hr
(D) 48, 72 gms / hr
(E) 30, 60 gms / hr
Gentle note to all experts and tutors: Please refrain from replying to this question until the Official Answer (OA) is revealed. Let students attempt to solve it first. You are all welcome to contribute posts after the OA is posted. Thank you all for your cooperation!
12 May 2025, 01:59
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Let:
- Engine 1 ran for t hours and consumed 600 g of petrol.
- Engine 2 ran for t - 2 hours and consumed 384 g of petrol.
Rates of Consumption:
Had Engine 1 consumed at the rate of Engine 2:
Petrol Used = t * (384/(t-2))
Had Engine 2 consumed at the rate of Engine 1:
Petrol Used = (t-2) * (600/t)
Equating both, we have:
600/384 = (t/(t-2))^2
10/8 = t/(t-2) => t = 10
Plugging back the value of t:
- Engine 1 ran for t hours and consumed 600 g of petrol.
- Engine 2 ran for t - 2 hours and consumed 384 g of petrol.
Rates of Consumption:
| Engine | Consumption Rate (g/h) |
|---|---|
| Engine 1 | 600/t |
| Engine 2 | 384/(t-2) |
Had Engine 1 consumed at the rate of Engine 2:
Petrol Used = t * (384/(t-2))
Had Engine 2 consumed at the rate of Engine 1:
Petrol Used = (t-2) * (600/t)
Equating both, we have:
600/384 = (t/(t-2))^2
10/8 = t/(t-2) => t = 10
Plugging back the value of t:
| Engine | Consumption Rate (g/h) | Value |
|---|---|---|
| Engine 1 | 600/t | 600/10 = 60 g/h |
| Engine 2 | 384/(t-2) | 384/8 = 48 g/h |
13 May 2025, 00:01
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IC1 ran for x hrs and consumed 600 gm petrol
IC2 ran for x-2 hrs and consumed 384 gm petrol
IC1 consumption rate= 384/x-2
IC2 consumption rate= 600/x
Both have consumed same amount of petrol
384/x-2*x=600/x*(x-2)
384x^2=600x^2-2400x+2400
solving x=10,10/9
10/9 is not possible so x=10
answer is IC1= 48gm/hr, IC2=60gm/hr which is B
IC2 ran for x-2 hrs and consumed 384 gm petrol
IC1 consumption rate= 384/x-2
IC2 consumption rate= 600/x
Both have consumed same amount of petrol
384/x-2*x=600/x*(x-2)
384x^2=600x^2-2400x+2400
solving x=10,10/9
10/9 is not possible so x=10
answer is IC1= 48gm/hr, IC2=60gm/hr which is B
