A certain organization presents reward to some people...

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.

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

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

I see a lot of calculations by others, well ideally we do not even need to solve this question

1735 = 125x + 40y + 15z, divide both sides by HCF= 5
327 = 28x + 8y + 3z
28x = even term
8y = even term
3z = odd term (since, even + even + odd can only result in an odd no. 327)
that means, z has to be odd no. Now since the questions asks the minimum therefore it has to be 17. In order to be double sure we can use various methods.

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