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29 Aug 2020, 05:58
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C
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:
47% (02:47) correct
53%
(02:30)
wrong
based on 349
sessions
History
Date
Time
Result
Not Attempted Yet
A certain mixture is made up of compounds A, B, and C. Compound B accounts for 70 percent of the mixture. How many more liters of compound A must be added so that the mixture is 25 percent compound A?
(1) If 3 times more compound A than is currently in the mixture were added, the mixture would be 50 percent compound A.
(2) If half of the compound B were removed from the mixture, there would be 175 liters of compound B.
(1) If 3 times more compound A than is currently in the mixture were added, the mixture would be 50 percent compound A.
(2) If half of the compound B were removed from the mixture, there would be 175 liters of compound B.
29 Aug 2020, 06:28
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Let the total volume of the mixture be x and the individual compounds volume be A,B and C
A+B+C = x
B=0.7x
Hence A+C = 0.3x
Statement 1:
Given that if 3A is added to the mixture then 50% of the mixture will be compound A, implies that
4A=\(\frac{(x+3A)}{2}\)
Solving it we get
x=5A
We still cannot deduce the additional litres required to make the mixture 25% with out knowing the value of A
Statement 2:
half of compound B is removed implying only 50% of the compound B is left in the mixture
\(\frac{0.7x}{2}\)= 175
x=500 litres
Now we know x=500, B=0.7*500=350
So A+C=150
But we still don't know that value of C to determine the additional litres of A
Hence Statement B is insufficient
Statement 1 and 2 combined
we have x=5A, x=500, B=350 and A+C=150
we can find A using x=5A and x=500
A=100
And using A+C=150, we can find that C=50
So the current composition of A is 100 litres or \(\frac{(100}{500)}\) 20% of the mixture
Now we can find the additional A required. There is no point going ahead to solve for the exact number of litres required. You can conclude that both the statement together are sufficient. Hence OA is C
For those who want to know the final answer here it is
Let y be the additional litres of compound A
25% = \(\frac{(100+y)}{(500+y)}\)
Solving it
500+y=400+4y
100=3y
y=\(\frac{100}{3}\)
Hence we need to add 33.33 litres of the compound A to attain 25% composition in the mixture
Kudos if it helps!
A+B+C = x
B=0.7x
Hence A+C = 0.3x
Statement 1:
Given that if 3A is added to the mixture then 50% of the mixture will be compound A, implies that
4A=\(\frac{(x+3A)}{2}\)
Solving it we get
x=5A
We still cannot deduce the additional litres required to make the mixture 25% with out knowing the value of A
Statement 2:
half of compound B is removed implying only 50% of the compound B is left in the mixture
\(\frac{0.7x}{2}\)= 175
x=500 litres
Now we know x=500, B=0.7*500=350
So A+C=150
But we still don't know that value of C to determine the additional litres of A
Hence Statement B is insufficient
Statement 1 and 2 combined
we have x=5A, x=500, B=350 and A+C=150
we can find A using x=5A and x=500
A=100
And using A+C=150, we can find that C=50
So the current composition of A is 100 litres or \(\frac{(100}{500)}\) 20% of the mixture
Now we can find the additional A required. There is no point going ahead to solve for the exact number of litres required. You can conclude that both the statement together are sufficient. Hence OA is C
For those who want to know the final answer here it is
Let y be the additional litres of compound A
25% = \(\frac{(100+y)}{(500+y)}\)
Solving it
500+y=400+4y
100=3y
y=\(\frac{100}{3}\)
Hence we need to add 33.33 litres of the compound A to attain 25% composition in the mixture
Kudos if it helps!
General Discussion
18 Sep 2020, 23:01
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Once again, this is NOT A 95% DIFFICULTY level question if you use the right logic.
St. 1 tells us If 3 times more compound A than is currently in the mixture were added, the mixture would be 50 percent compound A. So by simple algebra A + 3A = 50% - we don't need to solve this. We know we can find the current concentration of A but we obviously don't know the volume in Litres. INSUFFICIENT
St 2 tells us If half of the compound B were removed from the mixture, there would be 175 liters of compound B. So essentially, 35% of solution = 175 litres . So we can figure out the total volume of solution in litres. But there is no information about concentration of A, INSUFFICIENT
Combine both, from 1st we know the concentration of A and from 2nd we know the total volume of solution. I guess now we can easily find the volume of A in the solution and once we know that we can calculate how much more A would be needed to get 25% concentration. Yaaay! SUFFICIENT.
C is Correct
St. 1 tells us If 3 times more compound A than is currently in the mixture were added, the mixture would be 50 percent compound A. So by simple algebra A + 3A = 50% - we don't need to solve this. We know we can find the current concentration of A but we obviously don't know the volume in Litres. INSUFFICIENT
St 2 tells us If half of the compound B were removed from the mixture, there would be 175 liters of compound B. So essentially, 35% of solution = 175 litres . So we can figure out the total volume of solution in litres. But there is no information about concentration of A, INSUFFICIENT
Combine both, from 1st we know the concentration of A and from 2nd we know the total volume of solution. I guess now we can easily find the volume of A in the solution and once we know that we can calculate how much more A would be needed to get 25% concentration. Yaaay! SUFFICIENT.
C is Correct
