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Fish181
Joined: 13 Dec 2023
Last visit: 10 Aug 2026
Posts: 127
Status:Applying in R1 of 2024 to t15
Affiliations: University of Tennessee
Location: United States (CO)
Concentration: Strategy, Finance
GMAT Focus 1: 605 Q76 V84 DI80 
GMAT Focus 2: 615 Q78 V86 DI78
GPA: 3.62
WE:Analyst (Consumer Packaged Goods)
GMAT Focus 2: 615 Q78 V86 DI78
Posts: 127
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Difficulty:
75%
(hard)
Question Stats:
67% (02:39) correct
33%
(02:54)
wrong
based on 1098
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Last Year a company gave bonuses to a number of employees, but only in the three amounts of $750, $1500, and $7,350. If the total amount of the bonuses was $64800 and each of the three amounts was given to at least one employee, what is the fewest number of bonuses that the company could have given to employees last year?
A 10
B 11
C 12
D 13
E 14
GMAT-Club-Forum-u0hw72po.png [ 36.51 KiB | Viewed 10425 times ]
A 10
B 11
C 12
D 13
E 14
Attachment:
GMAT-Club-Forum-u0hw72po.png [ 36.51 KiB | Viewed 10425 times ]
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Fish181
Assume x $750 bonuses, y $1500 bonuses, and z $7,350 bonuses were given. Then, we'd have:
750x + 1,500y + 7,350z = 64,800
Reduce by 150:
5x + 10y + 49z = 432
Our goal is to minimize x + y + z. To achieve this, we need to maximize the number of the largest bonus, z. The value of z cannot be 9 or higher, so we try z = 8. In this case, we have:
5x + 10y + 392 = 432
5x + 10y = 40
5x + 10y = 40
Now, let's maximize the number of the next largest bonus, y. Since x > 0, the maximum value of y is 3, making x = 2.
Therefore, x + y + z = 2 + 3 + 8 = 13.
Answer: D.
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General Discussion
Fish181
Three amounts of Bonus are $750, $1500, and $7,350
Total Bonus = $64800
i.e $750a + $1500b + $7,350c = $64800
Now we need to find \((a+b+c)_{Min} = ?\)
i.e $750a + $1500b + $7,350c = $64800
i.e. $75a + $150b + $7,35c = $6480
i.e. 5a + 10b + 49c = 432
IMPORTANT: Now to minimize a+b+c, the value of c MUST be Maximized so that values of a and b are the least
Also, 5a + 10b = 432 - 49c
IMPORTANT: also, 5a+10b must be multiple of 5 so we need to fund a multiple of 49 (i.e. 49c) which when subtracted from 432 leaves a result that is divisible by 5
IMPORTANT: so the unit digit of 49c must be either 2 or 7
for max value of x=8, 49x = 49*8 = 392
i.e. 5a + 10b = 432 - 49c = 432 - 392 = 40
i.e. a+2b = 8
now, for Max value of b, b = 3 and then a = 2
so \((a+b+c)_{Min} = (2+3+8) = 13\)
Answer: Option D
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sayan640
Because 49 * 9 = 441 > 432.
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Bunuel
Hi Bunuel
Can you explain this line "Our goal is to minimize x + y + z. To achieve this, we need to maximize the number of the largest bonus, z"?
I did not understand this fact that in order to minimise x+y+z, we need to maximise z!
Freddy12
We have 750x + 1,500y + 7,350z = 64,800
The strategy to minimize the total number of bonuses (x + y + z) by maximizing z is based on efficiency. Since z represents the highest bonus amount, giving more of these reduces the total number of bonuses needed to reach the sum of $64,800. Each higher bonus contributes significantly more to the total, meaning fewer total bonuses are needed.
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Fish181
We need the minimum number of bonuses to let's try to maximize the number of the big bonus of $7350.
We know that total amount was 64,800.
7350 * 10 will be too much. Even 7350 * 9 will be much greater than 63000.
Hence maximum we can have 8 bonuses of the amount 7350.
7350*8 = 58,800
Leftover amount = 64,800 - 58,800 = 6000
Now we go on to the next big amount of 1500. We can give 4 bonuses of 1500 but we must give at least 1 bonus of each amount. Hence we give 3 bonuses of $1500 each and 2 bonuses of $750 each.
Total number of bonuses = 8 + 3 + 2 = 13
Answer (D)
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VivekSri
Computer Science students will know that this is the greedy method of solving the coin changing problem! (Ofc you don't need to know any of that to solve this)
But note that the greedy soln need not always be optimal XD
But note that the greedy soln need not always be optimal XD
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