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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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New post 15 Oct 2012, 18:37
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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


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

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New post 10 Dec 2012, 23:11
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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, 19:55
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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


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 04 Feb 2015, 09:08
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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 02 Jul 2017, 11:16
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




Is there a way to solve this question in which i do not have to divide 1735 by 125 and then move forward towards the solution?
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Re: A certain organization presents reward to some people...  [#permalink]

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New post 19 Feb 2018, 01:31
nishantt7 wrote:
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




Is there a way to solve this question in which i do not have to divide 1735 by 125 and then move forward towards the solution?



The way I did it was I saw that to get the least possible value of number of people, I had to get the most number of $125 people and minimise the rest of the two cash values, knowing what's left after taking the $40 and $15 into consideration must be a multiple of 125.

40 + 15 = 55.
1735 - 55 = 1680 clearly not a multiple of 125

2*(40 + 15) = 110
1735 - 110 = 1625 A-ha!

1625 / 125 = 13

$125 | $40 | $15
So 13 + 2 + 2 --> 17
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Re: A certain organization presents reward to some people...  [#permalink]

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New post 21 Feb 2018, 13:20
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


We can let a, b, and c = the number of $125, $40, and $15 reward recipients, respectively, and create the equation:

125a + 40b + 15c = 1,735

25a + 8b + 3c = 347

To minimize the sum a + b + c, we want first to make a as large as possible, then b as large as possible and then c as large as possible.

Since 347/25 = 13 R 22, we can let a = 13. (Notice that 25 x 13 = 325 and 347 - 325 = 22.)

Since 22/8 = 2 R 6, we can let b = 2. (Notice that 8 x 2 = 16 and 22 - 16 = 6.)

Finally we can see that c = 2 since 6/3 = 2.

Thus, the minimum number of people is 13 + 2 + 2 = 17.

Answer: A
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Re: A certain organization presents reward to some people... &nbs [#permalink] 21 Feb 2018, 13:20
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