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04 Feb 2022, 06:34
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Difficulty:
95%
(hard)
Question Stats:
38% (02:41) correct
62%
(03:03)
wrong
based on 129
sessions
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Not Attempted Yet
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
A. 5
B. 2
C. 9
D. 3
E. 7
04 Feb 2022, 08:31
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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.
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.
05 Feb 2023, 10:05
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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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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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