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04 Feb 2021, 04:15
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
58% (03:21) correct
42%
(03:23)
wrong
based on 150
sessions
History
Date
Time
Result
Not Attempted Yet
Taps A and C can fill a tank in 5 and 20 hrs respectively. Taps B and D can drain a full tank in 10 and 40 hrs. respectively. If A, B, C and D are opened in a gap of 2 hrs. in that order the tank gets full in :
A. 28/5 hrs
B. 34/5 hrs
C. 36/5 hrs
D. 38/5 hrs
E. 39/5 hrs
A. 28/5 hrs
B. 34/5 hrs
C. 36/5 hrs
D. 38/5 hrs
E. 39/5 hrs
04 Feb 2021, 04:26
Kudos
Bookmarks
Bunuel
Let, tank capacity = LCM (5, 20, 10 and 40) units = 40 units
(Inlet does +ve work while outlet does -ve work)
Work done by Tap A in 1 hour \(= \frac{40}{5} = +8\) Units
Work done by Tap C in 1 hour \(= \frac{40}{20} = +2\) Units
Work done by Tap B in 1 hour \(= \frac{40}{10} = -4\) Units
Work done by Tap D in 1 hour \(= \frac{40}{40} = -1\) Units
Let, A works for T hours
then B works for (T-2) hours
then C works for (T-4) hours
then D works for (T-6) hours
Total work of A, B, C and D = 8T - 4(T-2) + 2*(T-4) - 1*(T-6) = 40
i.e. 8T - 4(T-2) + 2*(T-4) - 1*(T-6) = 40
i.e. 5T +6 = 40
i.e. T = 34/5 Hours = 6.8 hours
Answer: Option B
General Discussion
18 Feb 2021, 14:33
Kudos
Bookmarks
Bunuel
Solution:
The rates for taps A, B, C, and D are 1/5, 1/10, 1/20, and 1/40, respectively.
We can let x be the number of hours tap A is opened (notice that this is also the number of hours it takes to fill the tank completely since tap A is opened for the entire duration of the tank’s filling). Then tap B is opened for (x - 2) hours, tap C for (x - 4) hours, and tap D for (x - 6) hours. We can create the equation:
(1/5)(x) - (1/10)(x - 2) + (1/20)(x - 4) - (1/40)(x - 6) = 1
Multiplying the equation by 40, we have:
8x - 4x + 8 + 2x - 8 - x + 6 = 40
5x = 34
x = 34/5
Answer: B
