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A company has three levels of bonus: 750, 1500, 7350

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trankimphuong
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A company has three levels of bonus: 750, 1500, 7350 [#permalink] A company has three levels of bonus: 750, 1500, 7350   02 Oct 2018, 05:59
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A company has three levels of bonus: 750, 1500, 7350. Total amount of bonus paid is 62,550. Each level of bonus has at least one reciever. At least, how many employees get bonus?

A. 9
B. 10
C. 11
D. 14
E. 15
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C
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chetan2u
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Re: A company has three levels of bonus: 750, 1500, 7350 [#permalink] A company has three levels of bonus: 750, 1500, 7350   02 Oct 2018, 06:11
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trankimphuong wrote:
A company has three levels of bonus: 750, 1500, 7350. Total amount of bonus paid is 62,550. Each level of bonus has at least one reciever. At least, how many employees get bonus?

A. 9
B. 10
C. 11
D. 14
E. 15



let the employees be x,y and z respectively..
so \(750x+1500y+7350z=62550..........75x+150y+735=6255.......5x+10y+49z=417\)
Now we are looking for the LEAST, so take minimum values of x and see if y and z come as integer
so x=1
5+10y+49z=417.........10y+49z=417-5=412...
now take z as max possible so z=8 as 49*8=392
10y+49*8=412....10y=412-392=20....y=2
here we get an integer value for y too, so x=1, y=2, and z=8 fits in and will give us LEAST number
ans = x+y+z=1+2+8=11

C
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Re: A company has three levels of bonus: 750, 1500, 7350 [#permalink] A company has three levels of bonus: 750, 1500, 7350   02 Oct 2018, 17:26
chetan2u, just so I am clear: You are saying we need to test the minimum value x so that y and z are integers (I understand the previous steps). Your explanation stops at 10y+49z=412. Could you explain the reason why you need to then test the max value z to determine y? Would love to understand the logic, and any other situations you would want to use this min/max strategy.

Thanks
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Re: A company has three levels of bonus: 750, 1500, 7350 [#permalink] A company has three levels of bonus: 750, 1500, 7350   02 Oct 2018, 18:48
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ada453 wrote:
chetan2u, just so I am clear: You are saying we need to test the minimum value x so that y and z are integers (I understand the previous steps). Your explanation stops at 10y+49z=412. Could you explain the reason why you need to then test the max value z to determine y? Would love to understand the logic, and any other situations you would want to use this min/max strategy.

Thanks



Hi

We are looking the least value of x+y+z..
The equation is 5x+10y+49z=417..
Now if you increase one z, it increases the amount by 49 which is shivaling to 10*5y or 20*5x..
So equivalent increase of 1*z is 10*y or 20*x...
Therefore we look for maximum of z
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Re: A company has three levels of bonus: 750, 1500, 7350 [#permalink] A company has three levels of bonus: 750, 1500, 7350   03 Oct 2018, 17:50
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trankimphuong wrote:
A company has three levels of bonus: 750, 1500, 7350. Total amount of bonus paid is 62,550. Each level of bonus has at least one reciever. At least, how many employees get bonus?

A. 9
B. 10
C. 11
D. 14
E. 15


Let a, b and c be the number of employees receiving bonuses of 750, 1500 and 7350, respectively. We have:

750a + 1500b + 7350c = 62,550

Since we want the least number of employees to get a bonus, we want as many employees as possible to get the largest bonus, 7350. Since 62,550/7350 ≈ 8.5, we see that c can be no more than 8.

If c = 8, then we have:

750a + 1500b + 7350 x 8 = 62,550

750a + 1500b = 3750

At this point, we can see that b can be 2 and a can be 1 so that 750a + 1500b = 3750. Therefore, the least number of employees who get a bonus is 1 + 2 + 8 = 11.

Answer: C
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Re: A company has three levels of bonus: 750, 1500, 7350 [#permalink] A company has three levels of bonus: 750, 1500, 7350   15 Mar 2023, 07:37
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To minimize the number of employees receiving a bonus, you need to maximize the number who get the highest bonus. Because there is at least one person receiving each level, start by saying just 1 person receives 750 and 1 person receives 1500. Subtract those from 62550 and you get 8 plus a remainder. Having that remainder means, for the math to work out evenly, you have to distribute one more smaller bonus to an additional person. So instead of the 10 people accounted for above (one at 750, one at 1500, and 8 at 7350), you're going to one more person to receive a smaller bonus, so 11 people. It isn't necessary to do the math out at that point.
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A company has three levels of bonus: 750, 1500, 7350 [#permalink] A company has three levels of bonus: 750, 1500, 7350   06 May 2024, 08:21
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Got this in my GMAT Focus Practice Exam 5 - maybe label accordingly.

Here is layman's way of solving it. With these types of questions you want to start from the largest.

Quick starting point for math is 7,350*10 = 73,500 > 64,800. So deduct 2*7,350 to get to 58,800 < 64,800. So 8 people at most got 7,350.

Then you need to fill 64,800 - 58,800 = 6,000 which would be simply 4*1,500 and the answer would be 12 employees but now you're not using the 750s.

Well, observe that 2*750 = 1,500 so you pretty quickly see that you have 8*7,350 + 3*1,500 + 2*750 = 64,800 or 13 employees.­
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Re: A company has three levels of bonus: 750, 1500, 7350 [#permalink] A company has three levels of bonus: 750, 1500, 7350   06 May 2024, 08:24
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jayouh wrote:
Got this in my GMAT Focus Practice Exam 5 - maybe label accordingly.

Here is layman's way of solving it. With these types of questions you want to start from the largest.

Quick starting point for math is 7,350*10 = 73,500 > 64,800. So deduct 2*7,350 to get to 58,800 < 64,800. So 8 people at most got 7,350.

Then you need to fill 64,800 - 58,800 = 6,000 which would be simply 4*1,500 and the answer would be 12 employees but now you're not using the 750s.

Well, observe that 2*750 = 1,500 so you pretty quickly see that you have 8*7,350 + 3*1,500 + 2*750 = 64,800 or 13 employees.­

Upated the tags. ­Thank you! 
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Re: A company has three levels of bonus: 750, 1500, 7350 [#permalink] A company has three levels of bonus: 750, 1500, 7350   06 May 2024, 08:31
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jayouh wrote:
Got this in my GMAT Focus Practice Exam 5 - maybe label accordingly.

Here is layman's way of solving it. With these types of questions you want to start from the largest.

Quick starting point for math is 7,350*10 = 73,500 > 64,800. So deduct 2*7,350 to get to 58,800 < 64,800. So 8 people at most got 7,350.

Then you need to fill 64,800 - 58,800 = 6,000 which would be simply 4*1,500 and the answer would be 12 employees but now you're not using the 750s.

Well, observe that 2*750 = 1,500 so you pretty quickly see that you have 8*7,350 + 3*1,500 + 2*750 = 64,800 or 13 employees.­

­Could you please post a screenshot of the question? Thanky you!
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Re: A company has three levels of bonus: 750, 1500, 7350 [#permalink] A company has three levels of bonus: 750, 1500, 7350   06 May 2024, 08:37
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Bunuel wrote:
jayouh wrote:
Got this in my GMAT Focus Practice Exam 5 - maybe label accordingly.

Here is layman's way of solving it. With these types of questions you want to start from the largest.

Quick starting point for math is 7,350*10 = 73,500 > 64,800. So deduct 2*7,350 to get to 58,800 < 64,800. So 8 people at most got 7,350.

Then you need to fill 64,800 - 58,800 = 6,000 which would be simply 4*1,500 and the answer would be 12 employees but now you're not using the 750s.

Well, observe that 2*750 = 1,500 so you pretty quickly see that you have 8*7,350 + 3*1,500 + 2*750 = 64,800 or 13 employees.­

­Could you please post a screenshot of the question? Thanky you!

­
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Re: A company has three levels of bonus: 750, 1500, 7350 [#permalink] A company has three levels of bonus: 750, 1500, 7350   06 May 2024, 09:29
if you simplify the equation, you will get.
5x + 10y +49z =417.

if z=6, 5x +10y = 123; (5x + 10y has to be a 0 or 5 at the end)
if z=7, 5x + 10y = 74;
if z=8, 5x +10y= 25; (this make sense, if x=1 and y=2 or x=3 and y=1; but least number is x=1, y=2 and z=8, total=11)

(Looking at the options, I have started with z=6)
answer check
if z=9, 5x+10y= -24 (negative, NOT POSSIBLE)
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Re: A company has three levels of bonus: 750, 1500, 7350 [#permalink] A company has three levels of bonus: 750, 1500, 7350   06 May 2024, 09:44
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jayouh wrote:
Bunuel wrote:
jayouh wrote:
Got this in my GMAT Focus Practice Exam 5 - maybe label accordingly.

Here is layman's way of solving it. With these types of questions you want to start from the largest.

Quick starting point for math is 7,350*10 = 73,500 > 64,800. So deduct 2*7,350 to get to 58,800 < 64,800. So 8 people at most got 7,350.

Then you need to fill 64,800 - 58,800 = 6,000 which would be simply 4*1,500 and the answer would be 12 employees but now you're not using the 750s.

Well, observe that 2*750 = 1,500 so you pretty quickly see that you have 8*7,350 + 3*1,500 + 2*750 = 64,800 or 13 employees.­

­Could you please post a screenshot of the question? Thanky you!

­

­Thank you very much!

That question is discussed here: 

https://gmatclub.com/forum/last-year-a- ... 29085.html

THI TOPIC IS LOCKED.
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Re: A company has three levels of bonus: 750, 1500, 7350 [#permalink]
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