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A certain organization presents reward to some people...

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A certain organization presents reward to some people... [#permalink]

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A certain organization presents reward to some people. There are 3 kinds of reward that respectively are $125, $40, $15. If the total reward is $1,735, what is the least possible value of the number of the people who received reward?

a. 17
b. 22
c. 32
d. 47
e. 90

[Reveal] Spoiler:
A
[Reveal] Spoiler: OA

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Re: A certain organization presents reward to some people... [#permalink]

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New post 11 Dec 2012, 00:11
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To get the minimum value, one must maximize the largest price: $125
\(\frac{1735}{125}=\frac{347}{25}=13rem.110\)

Note that \(13*125=1625\) so we calculate the remaining reward \(40A + 15B = 110\)
Again to get minimum value, maximize the largest remaining reward: \($40\)

\(\frac{110}{40}=2rem.30\)
Note that \(40*2=80\) so we calculate the remaining reward \(15B = 110 - 80 =30\)


\(\frac{30}{15}=2\)

Thus, 13 + 2 + 2 = 17

Answer: A
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Re: A certain organization presents reward to some people... [#permalink]

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New post 15 Oct 2012, 20:55
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aeros232 wrote:
A certain organization presents reward to some people. There are 3 kinds of reward that respectively are $125, $40, $15. If the total reward is $1,735, what is the least possible value of the number of the people who received reward?

a. 17
b. 22
c. 32
d. 47
e. 90

[Reveal] Spoiler:
A

from question :
1735= 125A + 40B +15C
Since we have 3 variables and only other information we can infer is that each of A, B and C should be non negative integer (>=0). Hence, we need to plug in numbers to find out.
Now since the question is asking for 'least' number of rewards, best approach to start with is -finding least 'possible' number of rewards - it can happen when every award is 125. So number of reward in this case 1735/125 = 13 + some remainder. hence ans must be greater than 14.

So we can start with A=13
This gives us
1735 = 13*125 + 40B +15C
or 40B + 15C =110
or 8B + 3C =22
Do we have any combination for B and C that works for this? yes.. if B=C=2.
Hence total number of rewards = A+B+C=13+2+2 = 17

same is given in A.

At this point, some observations could also be made. There could be a doubt in mind, what if A=12 and B+C <5? but note, it is not possible because for every A reduced, difference (125) is to be filled by at least 4 of Bs and Cs (40 and 15 respectively). Second, none of the answer choices is below 17.

Hence Ans A it is.
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Re: A certain organization presents reward to some people... [#permalink]

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New post 22 Jan 2015, 10:48
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Re: A certain organization presents reward to some people... [#permalink]

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New post 04 Feb 2015, 10:08
To get the minimum number of people, we want to maximize the number of people receiving the highest award.

1735/125 = 13 remainder 110.
110/40 = 2 remainder 30.
30/15 = 2.

13 +2 +2 = 17.
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Re: A certain organization presents reward to some people... [#permalink]

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New post 25 Jul 2016, 11:58
Hello from the GMAT Club BumpBot!

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Re: A certain organization presents reward to some people...   [#permalink] 25 Jul 2016, 11:58
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