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21 Jun 2021, 07:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
36% (02:58) correct
64%
(03:02)
wrong
based on 130
sessions
History
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Pipe X pours a mixture of acid and water, and pipe Y pours pure water into a bucket. After 1 hour, the bucket got filled and the concentration of acid in the bucket was noted to be 8%. If pipe Y was closed after 30 minutes and pipe X continued to pour the mixture, concentration of acid in the bucket after 1 hour would have been 10%. What is the ratio of acid to the water in the mixture coming out of pipe X?
(A) 13: 2
(B) 2 : 15
(C) 3 : 20
(D) 1 : 5
(E) 2 : 13
(A) 13: 2
(B) 2 : 15
(C) 3 : 20
(D) 1 : 5
(E) 2 : 13
DashingNightmare2
Current Student
Joined: 30 May 2021
Last visit: 19 Jun 2022
Posts: 101
Given Kudos: 77
Location: India
Concentration: Marketing, Technology
Schools: ISB '23 (A)
GMAT 1: 750 Q50 V41
GPA: 2.96
30 Jul 2021, 10:03
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Let initially there was 100 Liters of solution
=> 8 Liter of acid, w1 liter of water from solution 1 and w2 liter of water from solution 2
=>8 + w1 + w2 =100 -------- 1
Now, Pipe Y was closed after 30 mins, and acid is now 10 %
acid is still 8 liters but it now constitutes 10 % of solution => total =80 Liters solution is there
=>8+w1+ (0.5 * w2) =80 --------- 2
solving 1 and 2
0.5 *w2 =20 => w2 = 40
Putting in equation 1
8+ w1+ 40 =100 => w1=52
Hence E.
Required ratio = 8 :52 = 2:13
=> 8 Liter of acid, w1 liter of water from solution 1 and w2 liter of water from solution 2
=>8 + w1 + w2 =100 -------- 1
Now, Pipe Y was closed after 30 mins, and acid is now 10 %
acid is still 8 liters but it now constitutes 10 % of solution => total =80 Liters solution is there
=>8+w1+ (0.5 * w2) =80 --------- 2
solving 1 and 2
0.5 *w2 =20 => w2 = 40
Putting in equation 1
8+ w1+ 40 =100 => w1=52
Hence E.
Required ratio = 8 :52 = 2:13
General Discussion
30 Jul 2021, 19:28
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Bunuel
The percentage of acid will be 8% at any time as the speed of the two pipes have to be constant.
Let the quantity poured in half an hour be 100ml, and the water content be \(w_m\) in mixture and \(w_p\) in pure water.
So \(8+w_m+w_p=100\)
In next half an hour, \(8+w_m\) is poured, so we have \(100+8+w_m=108+w_m\) after one hour out of which 8+8 or 16 is acid.
% of acid = \(\frac{16}{108+w_m}=\frac{10}{100}…………160=108+w_m………w_m=52\)
Ratio of acid to \(w_m=\frac{8}{52}=\frac{2}{13}\)
E
