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20 votes
20 votes
how many ounces of pure chocolate must be added to 12 oz of chocolate syrup that is 50% chocolate to make a syrup that is 75% chocolate ​

User Spencer Cole
by
2.8k points

2 Answers

14 votes
14 votes

Answer:

12

Step-by-step explanation:

.5 * 12 = 6

(6 + x) /( 12 + x)


(x+6)/(x+12) = .75

x + 6 = .75(x+12)

x = 12

User Mrabat
by
2.6k points
10 votes
10 votes

Answer: 12

============================================

Step-by-step explanation:

We have 12 oz of 50% chocolate syrup, which means we have 0.50*12 = 6 oz of pure chocolate from this bottle. We add on x more ounces of pure chocolate from the second batch. So we have 6+x ounces of pure chocolate.

This is out of 12+x ounces of mixed syrup (chocolate + other stuff) overall.

The fraction (6+x)/(12+x) represents the proportion of pure chocolate to the mixed syrup batch. Set this equal to 0.75 and solve for x.

(6+x)/(12+x) = 0.75

6+x = 0.75(12+x)

6+x = 0.75(12)+0.75(x)

6+x = 9+0.75x

x-0.75x = 9-6

0.25x = 3

x = 3/0.25

x = 12

So we must add 12 ounces of pure chocolate to the mix

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We can think of this problem in terms of colored blocks.

Let's say a brown block represents a unit of pure chocolate, while a blue block is something else (say sugar or whatever other ingredients).

Currently we have 6 brown blocks (the 6 oz of pure chocolate) out of 12 blocks overall (6 brown + 6 blue).

If we add on 12 more brown blocks, then we get 6+12 = 18 brown blocks overall out of 12+12 = 24 blocks (brown+blue).

Then notice how 18/24 = 0.75 = 75% to represent the fact that 75% of the blocks are brown. So 75% of the syrup is pure chocolate.

This thought experiment is one way to help verify the answer.

User Thomas Junk
by
3.0k points