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Source: — Data Sufficiency |
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by MartyMurray » Fri Feb 12, 2016 8:04 pm
didieravoaka wrote:In a candy dish, the ratio of red to yellow candies is 2:5, the ratio of red to green candies is 3:8, and the ratio of yellow to blue candies is 9:2. What is the maximum total number of yellow and green candies combined if the maximum combined number of red and blue candies is fewer than 85?

(A) 144

(B) 189

(C) 234

(D) 279

(E) 309
The only actual number we have is the 85 > red + blue. So to get to other actual numbers we have to calculate the ratio of red to blue, and then determine how many reds and blues there are.

R:Y = 2:5

R:G = 3:8

Y:B = 9:2

Merge the ratios by creating common numbers so that the ratios can be combined.

Merge R:Y and R:G by making R the same number. The smallest number that works for R is the least common multiple of 2 and 3, which is 6.

R:Y = 6:15

R:G = 6:16

So R:Y:G = 6:15:16

Now combine R:Y:G with Y:B by making Y the same number in each. The least common multiple of 9 and 15 is 45.

R:Y:G = 18:45:48

Y:B = 45:10

So, R:Y:G:B = 18:45:48:10.

R + B = (18 + 10)x = 28x < 85

28 * 3 = 84. So x = 3, and the max R + B = 84.

R = 3 * 10 = 30 and B = 3 * 18 = 54

So the max of Y + G = (3 * 45) + (3 * 48) = 135 + 144 = 279

The correct answer is D.
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by Matt@VeritasPrep » Sun Feb 14, 2016 10:00 pm
I think we can do it more quickly. Start with the ratios:

5r = 2y
8r = 3g
2y = 9b

Now let's max it out. Suppose r + b = 84. Since we have both r and b in terms of the same thing (2y), we can find their values in terms of each other:

5r = 9b

r = (9/5)b

and

(9/5)b + b = 84

so b = 30, r = 54.

From there, plug these into our initial ratio equations to solve:

5r = 2y
5*54 = 2y
135 = y

8r = 3g
8*54 = 3g
144 = g

so y + g = 135 + 144 = 279.

There's probably a more artful way of doing it -- this is arithmetic-heavy -- but at least it's quick.
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