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Richard bought a number of red roses and yellow roses on...

Expert replies
by AAPL » Tue Mar 13, 2018 5:25 am
Richards bought a number of red roses and yellow roses on February 14th. Each red rose costs $9, and each yellow rose costs $14. If Richard spent a total of exactly $220, how many roses did Richard buy?

A. 16
B. 17
C. 19
D. 20
E. 21

The OA is D.

I solved this PS question writing an equation as follow,

9r + 14*(x - r) = 220, where r is the number of red roses and x is the total number of roses that Richard bought.

Then,
$$14x-5r=220\ \Rightarrow 14x=5\left(44+r\right)$$
That's mean x its a multiple of 5. Only option D satisfies that.

Is there a strategic approach to this PS question? Can any experts help, please? Thanks!
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Source: — Problem Solving |

by GMATinsight » Tue Mar 13, 2018 7:01 am
AAPL wrote:Richards bought a number of red roses and yellow roses on February 14th. Each red rose costs $9, and each yellow rose costs $14. If Richard spent a total of exactly $220, how many roses did Richard buy?

A. 16
B. 17
C. 19
D. 20
E. 21

The OA is D.
9R +14Y = 220
@Y = 8, R = 12

i.e. R+Y = 20

Answer: Option D
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by Scott@TargetTestPrep » Wed Mar 14, 2018 3:28 pm
AAPL wrote:Richards bought a number of red roses and yellow roses on February 14th. Each red rose costs $9, and each yellow rose costs $14. If Richard spent a total of exactly $220, how many roses did Richard buy?

A. 16
B. 17
C. 19
D. 20
E. 21
We can let r = the number of red roses and y = the number of yellow roses and create the equation:

9r + 14y = 220

9r = 220 - 14y

9r = 2(110 - 7y)

r = 2(110 - 7y)/9

Because r must be an integer, we see that (110 - 7y) must be a multiple of 9. When y = 8, 110 - 7(8) = 54 is a multiple of 9. Thus y = 8 and r = 2(54)/9 = 12 and r + y = 12 + 8 = 20 is the number of roses Richard bought.

Answer: D

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