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Intern  Joined: 15 Oct 2013
Posts: 13
Concentration: Marketing, Finance
Schools: McCombs '16
In a candy dish the ratio of red to yellow candies is 2:5, the ratio  [#permalink]

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11 00:00

Difficulty:   55% (hard)

Question Stats: 68% (03:04) correct 32% (03:38) wrong based on 218 sessions

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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 ot 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

Originally posted by mightymk on 03 Apr 2015, 06:43.
Last edited by Bunuel on 03 Apr 2015, 07:19, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 58427
Re: In a candy dish the ratio of red to yellow candies is 2:5, the ratio  [#permalink]

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mightymk 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 ot 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 ratio of red to yellow candies is 2:5 --> R:Y = 2:5 = 18:45;
The ratio of red to green candies is 3:8 --> R:G = 3:8 = 18:48;
The ratio of yellow to blue candies is 9:2 --> Y:B = 9:2 = 45:10;

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

The maximum combined number of red and blue candies is fewer than 85: R + B < 85 --> R + B = 18x + 10x < 85 --> 28x < 85 --> the maximum value of x is 3.

Max Y + G = 45x + 48x = 93x = 93*3 = 279.

Hope it's clear.
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Intern  Joined: 15 Oct 2013
Posts: 13
Concentration: Marketing, Finance
Schools: McCombs '16
Re: In a candy dish the ratio of red to yellow candies is 2:5, the ratio  [#permalink]

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Bunuel wrote:
mightymk 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 ot 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 ratio of red to yellow candies is 2:5 --> R:Y = 2:5 = 18:45;
The ratio of red to green candies is 3:8 --> R:G = 3:8 = 18:48;
The ratio of yellow to blue candies is 9:2 --> Y:B = 9:2 = 45:10;

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

The maximum combined number of red and blue candies is fewer than 85: R + B < 85 --> R + B = 18x + 10x < 85 --> 28x < 85 --> the maximum value of x is 3.

Max Y + G = 45x + 48x = 93x = 93*3 = 279.

Hope it's clear.

yup got it . thanks bunnel
Intern  Joined: 08 Jan 2015
Posts: 3
Re: In a candy dish,the ratio of red to yellow candies is 2:5  [#permalink]

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Given that Red is to yellow is 2:5, Red is to green is 3:8 and Yellow is to blue is 9:2.

Therefore, the total number of red, yellow, blue and green balls will be 18x, 45x, 10x and 48x respectively, where x is a constant.

If the combined number of red and blue balls is fewer than 85, i.e. max 84 balls, then the maximum number of yellow and green balls will be 279.

(10x+18x) < 85.

28x < 85, i.e. 28x <= 84 (Since number of balls cannot be in fraction). Thus, x<=3.

(45x+48x) = 93x.

Max (93x) = 279.

Veritas Prep GMAT Instructor Joined: 23 Oct 2013
Posts: 144
In a candy dish,the ratio of red to yellow candies is 2:5  [#permalink]

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Fun question. This problem is all about least common multiples. It helps to initially lay out the ratios individual, r:y = 2:5, r:g = 3:8, and y:b = 9:2. Then combine them. First combine the r:y:g by finding the least common multiple between the red in the r:y and r:g ratios. In the first ratio, the red is 2, and in the second, the red is 3, so the LCM is 6. Therefore multiply the first ratio by 3 and the second by 2 to get - r:y:g = 6:15:16. You do this in order to make the red a common number so that you can combine the ratios.

Then we have to use the yellow to incorporate the last ratio. The yellow in the ratio that we just did is 15, and in the y:b ratio is 9, therefore the LCM is 45. Multiply the ratio that we currently have by 3, and the y:b ratio by 5, to get r:y:g:b = 18:45:48:10. The r+b candies are thus a multiple of 28 (18+10). The highest multiple of 28 that is still below 85 is 84 (28*3). Therefore we must multiply the y+g multiplier (45+48 = 93) times 3 in order to get to 279 yellow and green candies, or answer choice D.

I hope this helps!
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Intern  Joined: 18 Jan 2015
Posts: 13
Re: In a candy dish,the ratio of red to yellow candies is 2:5  [#permalink]

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Many thanks Brandon!Didn't think it could be solved using LCMs!

How would you rate this question?Is this a 700+ question?
Director  G
Joined: 23 Jan 2013
Posts: 525
Schools: Cambridge'16
Re: In a candy dish,the ratio of red to yellow candies is 2:5  [#permalink]

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1
what is max of y+g if r+b<85?

need to find r+b, to do so multiply r:y and y:b in numbers which are in 9/5 ratio
r:y - 2:5 | *27
r:g - 3:8
y:b - 9:2 |*15

r+b=54+30=84<85, done

y+g=135+8*(54/3)=279

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Re: In a candy dish the ratio of red to yellow candies is 2:5, the ratio  [#permalink]

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mightymk 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 ot 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

we can rearrange to below:
R 18
Y 45
G 48
B 10

so 18x+10x<85. x must be an integer. 28x<85. x can be 3, and 28x=74, which is true. if x is 4, then the condition is not

then: 45*3 + 48*3 = 135+144 = 279
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Re: In a candy dish the ratio of red to yellow candies is 2:5, the ratio  [#permalink]

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mightymk 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 ot 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

Since the ratio of red to yellow candies is 2:5 and the ratio of yellow to blue candies is 9:2, we must multiply each ratio by a different constant such that the numbers for yellow candy are the same in each ratio:

Multiplying the first ratio by 9, we see that the ratio of red to yellow candies is 18:45. If we multiply the second ratio by 5, we obtain a ratio of yellow to blue candies of 45:10. Now that the ratio numbers for the yellow candies is the same (45), we can directly relate red to blue candies as a ratio of 18:10, or 9:5.

Thus, we can create the equation:

9x + 5x < 85

13x < 85

x < 6.54

Since x is an integer, the maximum integer value of x is 6. Therefore, we have a maximum of 9(6) = 54 red candies. Since the ratio of red to yellow candies is 2:5 and 2 x 27 = 54, we have a maximum of 5 x 27 = 135 yellow candies. Likewise, since the ratio of red to green candies is 3:8 and 3 x 18 = 54, we have a maximum of 8 x 18 = 144 green candies. Therefore, we have a maximum of 135 + 144 = 279 yellow and green candies.

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