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

Question

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

A 7
B 8
C 9
D 10
E 11

KSBGC · Answer

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.

Abhishek009 · Answer

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

Archit3110 · Answer

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

nitin_garg · Answer

We want to find the maximum number of hamburgers ($h). To do this, we need to spend as little as possible on shakes ($S) while still satisfying the requirement that he buys at least one.
 Step-by-Step Calculation 
1. Minimize the Shakes
Since we want the most hamburgers, we will assume Bob buys the minimum number of shakes required, which is 1.

{Cost of Shakes} = 1 x $4 = $4
2. Calculate Remaining Budget
Subtract the cost of the single shake from the total budget.

$56 - $4 = $52
3. Calculate Maximum Hamburgers
Divide the remaining budget by the cost of one hamburger $5) to see how many he can afford.

$52 / 5 = 10.4
Since Bob cannot buy a fraction of a hamburger, we round down to the nearest whole number.
Maximum Hamburgers = 10

i.e. Answer is D

shivani1351 · Answer

While, I had initially marked my answer as 8, I realised my mistake and then ended up getting the correct answer as 10.

Here's what I got incorrect in my first approach and how I rectified it:

First Approach:  5H + 4S = 56 (Notice I have equated the amount to 56 as that's the max. amount that he can spend)
H = (56-4S)/5
S min = 1 (as he purchases both, hamburgers and shakes)

When S = 1, H = (56-4)/5 which is not a positive integral value, hence we drop this as the number of hamburgers has to be a positive interger.

Similarly, for S = 2 and 3, we obtain fractional values of H, which are invalid.

When S = 4, H = (56-16)/5 = 8

Hence, I marked the answer as 8 in my first attempt. HOWEVER, NOTICE: WE HAD EQUATED THE AMOUNT TO 56, WHICH IS WRONG! 
It is wrong because we did not account for the cases when amount is less than 56, for example - what about the number of hamburgers when the amount = 55 or 54 or 53 or so on?

To solve this in the best possible way, use the Trial and Error Method using S min:
5H + 4S = 56 (S min = 1)

Let's try for H = 11 and S = 1 => 5*11 + 4*1 => 55+ 4 = 59 (invalid as it is greater than 56)
Now, let's try for H = 10 => 5*10 + 4*1 = 54 (valid as it is lesser than 56)

Here, if you observe, we get H = 10 (as opposed to 8 in the first approach), hence the correct answer is 10. (Option D)

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

Prep Toolkit

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