Susan invited 13 of her friends for her birthday party and created ret

Question

Susan invited 13 of her friends for her birthday party and created return gift hampers comprising one each of $3, $4, and $5 gift certificates. One of her friends did not turn up and Susan decided to rework her gift hampers such that each of the 12 friends who turned up got $13 worth gift certificates. How many gift hampers did not contain $5 gift certificates in the new configuration?

A. 5
B. 2
C. 9
D. 3
E. 7

Show Answer

Official Answer

D

A. 5
B. 2
C. 9
D. 3
E. 7

IanStewart · Answer

With the given amounts, there are only three ways to combine the certificates to get $13:

A: $5, $5, $3
B: $5, $4, $4
C: $4, $3, $3, $3

Notice that only the collections of type C do not contain a $5 certificate, so the question is just asking how many type C collections we need to make. There are several ways to finish the question. We have 13 gift certificates of each denomination ($3, $4 and $5). So notice we can't possibly have more than 4 collections of type C, because if we did, we'd need at least 15 of the $3 certificates. So the answer is either 2 or 3, but if we use only 2 collections of type C, then we'll have used 2 of the $4 certificates. We'll have 11 of those $4 certificates left, but the remaining $4 certificates can only be used in collections of type B, and no matter how many collections of type B we make, we'll get an even number of $4 certificates, so we can't get 11 of them. So the only possible answer is '3', and we have 3 collections of type C, and thus 4 of type A to get to 13 of the $3 certificates, and lastly 5 of type B to get the right number of $4 certificates.

You could also do the problem purely algebraically. If we let A, B and C be the numbers of collections of each type above, then since we need a total of 13 of the $5 certificates, 2A + B = 13; since we need a total of 13 of the $4 certificates, 2B + C = 13, and since we need a total of 13 of the $3 certificates, A + 3C = 13. So we have these equations:

2A + B = 13
2B + C = 13
A + 3C = 13

and if we add the equations, we learn 3A + 3B + 4C = 39. But we also know there are 12 collections in total, so we know A + B + C = 12. Multiplying by 3, we know 3A + 3B + 3C = 36. Subtracting this from the equation we just got, we learn C = 3.

Regor60 · Answer

Let's call the number of $3 certificates A, 4 B, and 5, C.

To build $12 baskets, there are 3 ways, call them X, Y and Z:

X: A=0, B=2, C=1
Y: A=1, B=0, C=2
Z: A=3, B=1, C=0

So,

2X+Z<=13
3Z+Y<=13
2Y+X<=13

Subtracting the first from two times the third and multiplying by 3 yields

12Y-3Z<=39. Adding to the second equation yields

13Y<=52, or Y<=4.

Examining the second equation, the maximum Z could be is if Y=0, with Z therefore<=4.

Examining the third equation, the maximum X could be is if Y=0, with X<=6.

So, X<=6, Y<=4, Z<=4

Since X+Y+Z=12, let's try some values for Y since they are limited:

Y=0 means X+Z=12, not possible given above

Y=1 means X+Z=11, also not possible

Y=2 means X=6, Z=4 but these violate 2X+Z<=13 from before

Y=3 means X=6,Z=3 or X=5,Z=4, but either of these violate 2X+Z<=13 from before

Therefore, Y=4 and X+Z=8, which can be through 6,2;5,3 or 4,4

6,2 is excluded due to the 2X+Z limitation

Z=4 is eliminated since with Y=4, the 3Z+Y constraint is violated

So X=5,Y=4 and Z=3 is the only possibility remaining, which means 3 baskets with no $5 certificates

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dbcoupie123 · Answer

13 5s
13 4s
13 3s

two configs of $13: 5,5,3 or 5,4,4

start with combination with least number of 5s. restricted to the #of pairs of 4s.

since there are only 13 4s, then the max # of possible of the 5,4,4 combo is  six

work with the 7 remaining 5s to create max $ of 5,5,3 combos similar to above.  three

no other configs remaining that can carry the extra $5.

so max is nine hampers that include 5s. the  remaining three  cannot.

My question here now is how are the remaining 3 hampers configured such that the total is $13?

the final combo with the remaining notes is 3,3,3,4 - for which we can only create one of these.

How about the other two? we are left with one $5, seven $3, and zero $4.

Tukkebaaz · Answer

We have 3,4,5 for 13 people which is redistributed among 12 ppl (so now sum for each person is 13). We can make this by 5 5 3; 4 4 5; 3 3 3 4.

Now Qn asks to find how many sets dont have a 5 i.e. no of sets with 3334 combination. So far was the hard bit, now just solve (again dont manually solve as then the 2nd bit will be equally hard).

Since you have 3334 we know the number of sets of these will have to be less than 5 (Why? coz total is 13 pieces available).

Lets start with 4, means only 1 set of 553 will be possible wont work as we need to use up all 5’s. Try with 3, we get (5 5 3) *4 sets, (4 4 5)*5sets and (3 3 3 4)*3 sets. Total sets are 4+5+3=12 sets. Voila

Required answer is 3 (choice D)

Time taken <1:30, where I spent a little less than 1 min reading the Qn a few times and thinking of what needs to be done and how can I be smart about the approach. Not to intimidate folks but just to drive the point that thinking more can lead to solving less, more often than not on the GMAT.

Hope the suggestion and approach helps

Required answer is 3 (choice D) Time taken <1:30, where I spent a little less than 1 min reading the Qn a few times and thinking of what needs to be done and how can I be smart about the approach. Not to intimidate folks but just to drive the point that thinking more can lead to solving less, more often than not on the GMAT. Hope the suggestion and approach helps

bumpbot · Answer

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Susan invited 13 of her friends for her birthday party and created ret

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