A shop sells two variants of chocolates - one that costs $3 and the ot

Question

A shop sells two variants of chocolates - one that costs $3 and the other that costs $5. If the shop sold $108 chocolates on a given day, how many different combinations of (number of $3 sold, number of $5 sold) exist?

A. 6
B. 7
C. 8
D. 12
E. 15

A. 6
B. 7
C. 8
D. 12
E. 15

chetan2u · Answer

Hi,
the actual method would be--
3x+5y=108
and now find values of x and y that fit in..
x=1, y=21..
x=2, y not possible ....
......

We can use number properties to home on the answer without doing all calculations.. 
3x+5y=108=3*36.
 lets concentrate on y, as all other terms are multiple of 3 
3 and 5 are COPRIME, so y should be a multiple of 3..
so y=0,3,6,..., lower int value 108/5=21.6=21..
so y can take (21-0)/3 + 1= 7+1=8
ans 8.. C

EMPOWERgmatRichC · Answer

Hi All,

The 'key' to this question is in realizing that every $15 "block" could either be "5 sets of $3" OR "3 sets of $5."

$108 COULD be made up of "ALL $3 boxes", since 108/3 = 36

By subtracting $3 from $108, we get $105, which is (7)($15).

Each of those $15 blocks is either 3s or 5s; since there are 7 blocks, there are actually eight possible outcomes (and each would include that extra $3 that we subtracted from $108)

zero 3s and seven 5s
one 3s and six 5s
two 3s and five 5s
three 3s and four 5s 
four 3s and three 5s
five 3s and two 5s
six 3s and one 5
seven 3s and zero 5s

Final Answer:  C

GMAT assassins aren't born, they're made,
Rich

Mo2men · Answer

Hi Rich,

Can you please explain more about the 'block'?

For example: two 3s and five 5s does not end up $108. It seems I miss something

Thanks

saraswatpraveen07 · Answer

Let the variant with $3 as cost be x and $5 be y. With this assumption we can make the equation 3x + 5y = 108.

The number of combinations will be different values of (x, y) which satisfy the above equation.

Now we can rewrite the equation as 5y = 108 - 3x. For y to be a integer for a value of x, 108 - 3x needs to be divisible by 5.

A number is divisible by 5 if its last digit is either 0 or 5. Therefore 108 - 3x needs to end with 0 or 5. This would be possible only when 3x ends with 3 or 8.

First value of x which satisfies this condition is x = 1 and next is x = 6. This forms a series with x= 1, 6, 11 ... till 36 .  \frac{108}{3}  =36].

8 values (Ans : C)

EMPOWERgmatRichC · Answer

Hi Mo2men,

Each "block" of 15 is either three 5s or five 3s. The list I created was a description of the types of blocks that would total 108.

Two BLOCKS of 3s and five BLOCKS of 5s = 105 and then we'd add in the extra "3" that I subtracted out at the beginning.

Here's another way to look at the overall logic:

In real basic terms, we're adding up a 'bunch of 3s' and a 'bunch of 5s' to get to 108. At first glance, you might think that there are lots of different ways to get to 108, but there really are not that many possibilities.

Here's a much simpler example to start:

What if we were trying to get to a total of $18, how could we do it?

All 3s
three 5s and one 3

There's just two options there. Since 18 = 15 + 3, you can see that the 'difference' in those two options is the "15" - it's either five 3s or three 5s.

Here's another example:

What if we were trying to get to a total of $33, how could we do it?

All 3s
three 5s and six 3s
six 5s and one 3

There's just three options there. Since 33 = 30 + 3, you can see that the 'difference' in those options is in the two "15s" - each one is either five 3s or three 5s.

This pattern continues... 48, 63, 78, 93, 108 ... each is a 'multiple of 15' + 3, so you can use the same logic with any of these totals.

GMAT assassins aren't born, they're made,
Rich

sayan640 · Answer

Bunuel  Why should we start from y= 0 ? At least 1 , 5$ chocolate needs to be sold . RIght ?

Bunuel · Answer

Why do you assume that?

pappal · Answer

3x+5y=108
for y to be an integer (108-3x) should have last digit as either 0 or 5
which is possible iff last digit of 3x is either 3 or 8
so x=1,11,21,31,6,16,26,36 giving corresponding values of y to support the equation.
so total 8 combinations.

A shop sells two variants of chocolates - one that costs $3 and the ot

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A shop sells two variants of chocolates - one that costs $3 and the ot

Prep Toolkit

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