A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. What part of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?A. 1/5
B. 2/7
B. 7/11
C. 6/7
E. 7/8
Let the total mixture be 1.
Initial water = 3/8
Initial syrup = 5/8
Suppose fraction x of the mixture is drawn off and replaced with water.
After drawing off x, the syrup left is:
5/8 * (1 - x)
For the final mixture to be half syrup:
5/8 * (1 - x) = 1/2
5(1 - x) = 4
1 - x = 4/5
x = 1/5
Answer: A.
SwethaReddyL
Bunuel /
CrackverbalGMAT - just a quick question on the formal used here
a = Final Volume in the container
after the replacement of quantity b.
FQ/IQ = [1-x/c], isn't the c here not the total capacity?
this is how i approached but didn't get the answer,
since half and half the final quantity should be 4 and 4
acc to formula,
4/5 = [1-x/8]
4/5 = 8-x/8
32 = 40-5x
5x = 8
x = 8/5 is what i got
a little help please - the mistake is taking c as 8 i think but i wanna know the gap in my logic
thanks in advance,
swetha
In the formula,
\(\frac{FQ}{IQ}=1-\frac{b}{a}\)
\(b\) is the amount of mixture drawn off and replaced.
\(a\) is the total quantity of the mixture in the vessel after replacement. Since the vessel remains full, this is also the total capacity of the vessel.
So using 8 as the total quantity is fine.
Your equation gives:
\(\frac{4}{5}=1-\frac{x}{8}\)
so \(x=\frac{8}{5}\). But this \(x\) is the
amount drawn off, not the fraction of the whole mixture drawn off.
Since the whole mixture is 8 parts, the fraction drawn off is:
\(\frac{8/5}{8}=\frac{1}{5}\)