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Richard 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

Say Richard bought total of \(x\) roses, out of which \(r\) were red roses. So, the number of yellow roses were \(x-r\).

\(9r + 14(x-r)= 220\);

\(14x-5r=220\);

\(14x=5(44+r)\) --> \(x\) is a multiple of 5. Only option D satisfies that.

Answer: D.

I find this approach a bit easier. This will be better even when there are two options that are multiples of 5.

The only possible combo is 12+8 = 20

So answer D
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to start, let's imagine 10 Red roses and 10 Yellow roses. That would be 10x9 + 10x14 = 230. However, the total cost was 220. So we need to lower our calculated cost by 10, which can be done by adding 2 Red roses and reducing 2 Yellow Roses: +2*9 - 2*14 = -10.

This way, we would have 12 Red roses and 8 Yellow roses: total 20 roses.

Thanks!
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This problem can also be solved by allegation method.

Total = 220
n=? (Choose a smart number that clearly divides 220) 20 is the only such number.
Combined Average= 220/20=11

Now the allegation:

9=======14
====11====
3========2

IS THE ASSUMED NUMBER (20) is clearly divisible by (3+2) =5? YES.
That mean, this 20 can be allocated to $9 &$14 groups in 3:2 ratio (12&8)

So, total number is 20.

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Have used such logic - first buyed only expensive roses and check overbudgeting for each answer (e.g. for first answer it’s 14*16 - 220 = 4). this overbudgeting must be compensated by changing of expensive rose by cheaper (5$ for each change). So overbudgeted amount must be divisible by 5 - answer D
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Hi, What is incorrect in the below approach
Let r and y be the number of red and yellow roses respectively,
9*r + 14*y = 220
y=(220-9*r)/14
by substituting the option we can see which results in an integer value of y making the substituted value as the answer. What is wrong in this approach?
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pratiksha1998
Bunuel
Richard 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

Say Richard bought total of \(x\) roses, out of which \(r\) were red roses. So, the number of yellow roses were \(x-r\).

\(9r + 14(x-r)= 220\);

\(14x-5r=220\);

\(14x=5(44+r)\) --> \(x\) is a multiple of 5. Only option D satisfies that.

Answer: D.
Hi, What is incorrect in the below approach
Let r and y be the number of red and yellow roses respectively,
9*r + 14*y = 220
y=(220-9*r)/14
by substituting the option we can see which results in an integer value of y making the substituted value as the answer. What is wrong in this approach?

Your approach is correct in principle. Trying options by substitution is fine if done carefully. However, you must remember that the answer choices (16, 17, 19, 20, 21) represent the total number of roses (r + y), not just r. Thus, directly substituting into your rearranged formula without properly accounting for the total can easily lead to mistakes.

The solutions shown earlier avoids this confusion by working with the total number of roses directly.
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