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21 Jul 2003, 18:32
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A cask is filled with a mixture of 2 liquids A and B in the proportion A:B :: 5:3. When 16 gallons of the mixture are drawn off and the acsk filled with liquid B the proportion become A:B:: 3:5. How many gallons does the cask hold?
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21 Jul 2003, 23:12
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initially, X gallons is the capacity
liquid A) 5X/8 - five parts out of 8 given
liquid B) 3X/8
-16 gallons of mixture (according to the initial proportion of 5/3)
liquid A) 5X/8 - 10
liquid B) 3X/8 - 6
+16 gallons of liquid B
liquid A) 5X/8 - 10
liquid B) 3X/8 - 6 +16 = 3X/8 +10
A new proportion
5X/8 - 10
------------ = 3/5
3X/8 +10
solve for X, X=40
liquid A) 5X/8 - five parts out of 8 given
liquid B) 3X/8
-16 gallons of mixture (according to the initial proportion of 5/3)
liquid A) 5X/8 - 10
liquid B) 3X/8 - 6
+16 gallons of liquid B
liquid A) 5X/8 - 10
liquid B) 3X/8 - 6 +16 = 3X/8 +10
A new proportion
5X/8 - 10
------------ = 3/5
3X/8 +10
solve for X, X=40
22 Jul 2003, 09:01
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Sure, 40 is the answer. Just to share how i did this, made a table with two columns, one for substance A and and for B. I started with the known (16 gal of B) then added the mixture until I saw the correct ratio:
A B
0 16
5 19
10 22
15 25
We see that 15 and 25 are in 3:5 ratio and add up to 40. Perhaps this is not the cleanest way, and definitely would not work for "messy" numbers, but as the GMAT is generally clean, it is OK here.
A B
0 16
5 19
10 22
15 25
We see that 15 and 25 are in 3:5 ratio and add up to 40. Perhaps this is not the cleanest way, and definitely would not work for "messy" numbers, but as the GMAT is generally clean, it is OK here.
