Collectively, the five children in the Kramer family have 30 trophies.

Question

Collectively, the five children in the Kramer family have 30 trophies. If each child has at least one trophy and no two children have the same number of trophies, what is the greatest number of trophies that the child with the second-highest number of trophies could have?

A. 10
B. 11
C. 12
D. 13
E. 14

A. 10
B. 11
C. 12
D. 13
E. 14

houston1980 · Accepted Answer

Let the 5 children be named M, N, P, Q and R. assign minimum possible values starting from M up to P. Q and N will then have the highest possible values.

M = 1 trophy
N = 2 trophies
P = 3 trophies

At this point M+N+P = 1+2+3 = 6. Therefore Q +R will be 30-6 = 24. Since Q and R cannot have the same number of trophies: Q will be 11 trophies and R 13 trophies.

Q = 11 trophies (second-highest number of trophies )
R = 13 trophies

Hence   option B (11 trophies)    is the answer.

Give me +1 Kudos if you like the explanation. Thanks.

SonalSinha803 · Answer

Let , the number of trophies be in the order -1, 2, 3, x, y and we have to find x.
 1 + 2 + 3 = 6
Therefore, x + y = 24

Now, the possible values of x and y could be -

x = 5, 6, 7, 8, 9, 10, 11
Correspondingly, y = 19, 18, 17, 16, 15, 14, 13

Thus, the greatest possible value of x = 11

Hence, option B.

Please correct me if I am wrong and if I am right, please give some Kudos !!

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JeffTargetTestPrep · Answer

We can let the three children who have the least number of trophies have 1, 2, and 3 trophies, respectively, since no child can have the same number of trophies as any other sibling. Thus, these three children have a total of 1 + 2 + 3 = 6 trophies.

So we are left with a total of 30 - 6 = 24 trophies for the two children with the greatest number of trophies. Since we want the second-most number of trophies to be a large as possible, we will ensure that the greatest number of trophies is as small as possible while still retaining its role as the greatest number. We see that a 12-12 split almost works, but the two numbers are not allowed to be the same value.

Thus, we can let the child with the greatest number of trophies have 13 trophies, and hence, the child with the second most trophies has 11.

Answer: B

SHEHAB01 · Answer

5x=30
so, x=6
so at least each have 6 trophies and as no two children have the same number of trophies.So trophies may arrange as
6+7+17=30
6+8+16=30
6+9+15=30
6+10+14=30
6+11+13=30
6+12+12=30 
Therefore, the second highest number of trophies is 11
ans:B

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Sammarth · Answer

Please change the tag to Max/Min problem. This isnt a mixture problem

Bunuel · Answer

________________________
Done. Thank you!

Collectively, the five children in the Kramer family have 30 trophies.

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