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Bob can spend a maximum of $56 on hamburgers and shakes. Hamburgers co

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Bob can spend a maximum of $56 on hamburgers and shakes. Hamburgers co  [#permalink]

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New post 21 Jan 2019, 00:18
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A
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60% (00:38) correct 40% (02:36) wrong based on 13 sessions

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Bob can spend a maximum of $56 on hamburgers and shakes. Hamburgers cost $5 apiece, and shakes cost $4 apiece. If Bob purchases both hamburgers and shakes, what is the maximum number of hamburgers he can purchase?

A 7
B 8
C 9
D 10
E 11

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Re: Bob can spend a maximum of $56 on hamburgers and shakes. Hamburgers co  [#permalink]

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New post 21 Jan 2019, 04:01
Bunuel wrote:
Bob can spend a maximum of $56 on hamburgers and shakes. Hamburgers cost $5 apiece, and shakes cost $4 apiece. If Bob purchases both hamburgers and shakes, what is the maximum number of hamburgers he can purchase?

A 7
B 8
C 9
D 10
E 11


Trail and Error method is perfect one for this question.

we need to maximize the number of Hamburgers.

Total = 56..

price / hamburgers = 5 and price / shakes = 4.

5x + 4y = 56.

For value 11, 10, 9 of x , y won't be integers.

5*11 = 55

5*10 = 50

5* 9 = 45

5*8 = 40.

we start with 11 as we want to maximize the number of hamburgers. So , we need to spend max balance on hamburgers.

5*8 + 4y = 56
4y = 16

y = 4. integer value.

Thus , max hamburgers will be 8.

B is the correct answer.
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Re: Bob can spend a maximum of $56 on hamburgers and shakes. Hamburgers co  [#permalink]

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New post 21 Jan 2019, 06:34
Bunuel wrote:
Bob can spend a maximum of $56 on hamburgers and shakes. Hamburgers cost $5 apiece, and shakes cost $4 apiece. If Bob purchases both hamburgers and shakes, what is the maximum number of hamburgers he can purchase?

A 7
B 8
C 9
D 10
E 11


\(5h + 4s ≤ 56\)

Now try to substitute the value of h from the given options such that Bob can purchase both hamburgers and shakes...

Test option (B) 5*8 + 4*4 = 56

Thus, maximum number of hamburgers that Bob can purchase is (B) 8
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Bob can spend a maximum of $56 on hamburgers and shakes. Hamburgers co  [#permalink]

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New post 21 Jan 2019, 07:43
Bunuel wrote:
Bob can spend a maximum of $56 on hamburgers and shakes. Hamburgers cost $5 apiece, and shakes cost $4 apiece. If Bob purchases both hamburgers and shakes, what is the maximum number of hamburgers he can purchase?

A 7
B 8
C 9
D 10
E 11



case 1 :
considering that Bob would be spending all 56$
so
5*8+4*4 = 56
IMO B

Case 2
spend< 56$
then 5*10 +4*1 = 54
IMO D

max burgers D 10
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Bob can spend a maximum of $56 on hamburgers and shakes. Hamburgers co   [#permalink] 21 Jan 2019, 07:43
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