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

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Intern
Joined: 25 Oct 2016
Posts: 4
A company has three levels of bonus: 750, 1500, 7350  [#permalink]

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02 Oct 2018, 05:59
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Difficulty:

55% (hard)

Question Stats:

59% (02:27) correct 41% (02:30) wrong based on 41 sessions

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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
Math Expert
Joined: 02 Aug 2009
Posts: 7978
Re: A company has three levels of bonus: 750, 1500, 7350  [#permalink]

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02 Oct 2018, 06:11
1
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]

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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
Math Expert
Joined: 02 Aug 2009
Posts: 7978
Re: A company has three levels of bonus: 750, 1500, 7350  [#permalink]

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02 Oct 2018, 18:48
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]

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03 Oct 2018, 17:50
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.

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Re: A company has three levels of bonus: 750, 1500, 7350   [#permalink] 03 Oct 2018, 17:50
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