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At the start of the day the amount of water in two identical buckets

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At the start of the day the amount of water in two identical buckets  [#permalink]

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28 Dec 2017, 20:45
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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

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Re: At the start of the day the amount of water in two identical buckets  [#permalink]

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29 Dec 2017, 05:04
Bunuel wrote:
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

A:B::3:2
A:B::3+x:2+4x = 3:10

30+10x = 6+12x
2x=24
x=12
4x was added to bucket b = 4*12 = 48
D
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Re: At the start of the day the amount of water in two identical buckets  [#permalink]

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29 Dec 2017, 09:55
(3 + x) / (2 + 4x) = 3 / 10 (because of A to B ratio)

Cross-multiplying gives 30 + 10x = 6 + 12x
Solving for x gives us, x = 12

Since 4x liters was added to bucket B, 4*12 = 48 liters (answer D)
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At the start of the day the amount of water in two identical buckets  [#permalink]

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29 Dec 2017, 14:57
Bunuel wrote:
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

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Re: At the start of the day the amount of water in two identical buckets  [#permalink]

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22 Jul 2018, 03:24
Bunuel wrote:
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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Joined: 09 Jul 2018
Posts: 5
At the start of the day the amount of water in two identical buckets  [#permalink]

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22 Jul 2018, 05:21
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
At the start of the day the amount of water in two identical buckets &nbs [#permalink] 22 Jul 2018, 05:21
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At the start of the day the amount of water in two identical buckets

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