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Bunuel
At the start of the day the amount of water in two identical buckets is 3 liters in bucket A and 2 liters in bucket B. If x liters are then added to A and 4x liters are added to B so that the ratio of A to B is 3 to 10, how much water has been added to bucket B?

A. 4 liter
B. 12 liters
C 16 liters
D. 48 liters
E. 72 liters
This question changes what we normally see, but solving it is done similarly.

Usually there is a given original ratio with a multiplier and quantities to be added or subtracted that yield a new ratio.

Neither ratio here "takes" the multiplier. Instead, in the arithmetic, the multiplier accompanies the amounts added.

Original ratio: \(\frac{A}{B} =\\
\frac{3}{2}\)

Add x liters to A and 4x liters to B, to yield a new ratio
\(\frac{3 + x}{2 +\\
4x}=\frac{3}{10}\)

\(3(2 + 4x) = 10(3 + x)\)
\(6 + 12x = 30 + 10x\)
\(2x = 24\)
\(x = 12\)

How much water was added to B?
This time the multiplier corresponds with amounts added.

B got 4x liters of water
x = 12
4x = 48 liters

Answer D
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Bunuel
At the start of the day the amount of water in two identical buckets is 3 liters in bucket A and 2 liters in bucket B. If x liters are then added to A and 4x liters are added to B so that the ratio of A to B is 3 to 10, how much water has been added to bucket B?

A. 4 liter
B. 12 liters
C 16 liters
D. 48 liters
E. 72 liters

Since the resulting ratio = A:B = 3:10, and all of the values in the problem are integers, the correct answer must yield a multiple of 10 when added to the 2 original liters in B.
Only D is viable:
48+2 = 50 liters.

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A B
3 2
+x +4x
=3+x =2+4x

(3+x)/(2+4x)=3/10
Therefore 30+10x = 6+12x
24 = 2x
x=12
Therefore, 4x = 48
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