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The four largest numbers in a set of seven numbers have a mean of 10.

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Bunuel
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The four largest numbers in a set of seven numbers have a mean of 10. [#permalink] New post   28 Jan 2022, 07:45
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The four largest numbers in a set of seven numbers have a mean of 10. The four smallest numbers in the same set have a mean of 5. What is the least possible sum of the seven numbers?

(A) 30
(B) 40
(C) 50
(D) 60
(E) 70


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C
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The four largest numbers in a set of seven numbers have a mean of 10. [#permalink] New post   28 Jan 2022, 08:12
Lets assume the numbers arranged in the ascending order are as follows

a .. b .. c .. d .. e .. f .. g

We know that

a + b + c + d = 20

d + e + f + g = 40

Adding both

a + b + c + d + d + e + f + g = 60

Taking one of the 'd' to the other side

a + b + c + d + e + f + g = 60 - d

Thus to minimize the sum, we have to maximize the value of d

We know that the sum of last four digits is 40, so the value of d will be maximum when d = 10

Therefore,

a + b + c + d + e + f + g = 60 - 10

= 50

IMO C
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Re: The four largest numbers in a set of seven numbers have a mean of 10. [#permalink] New post   28 Jan 2022, 08:39
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The four largest numbers sum to 40, and the four smallest sum to 20. So if we add those eight numbers, the sum is 60, but we've now added the median twice. So the sum of the seven numbers is 60 - m, where m is the median. Now the median can't be larger than 10, because then the four largest numbers would all be greater than 10 and could never have an average of 10. So m is at most 10, and the sum 60-m is at least 50.

If we wanted to construct a list with this sum, it could be: 10/3, 10/3, 10/3, 10, 10, 10, 10
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The four largest numbers in a set of seven numbers have a mean of 10. [#permalink] New post   28 Jan 2022, 11:11
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Bunuel wrote:
The four largest numbers in a set of seven numbers have a mean of 10. The four smallest numbers in the same set have a mean of 5. What is the least possible sum of the seven numbers?

(A) 30
(B) 40
(C) 50
(D) 60
(E) 70


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Let the numbers (smallest to largest) are: a, b, c, d, e, f, g

d + e + f + g = 40
a + b + c + d = 20

Add: a + b + c + d + d + e + f + g = 60

=> a + b + c + d + e + f + g = 60 - d

We need to figure out the possible values of d

To minimize the sum of the 7 numbers, we need to maximize d:
Maximum value of d = 10 (since the largest 4 numbers have a mean 10, and d is the least of those 4 numbers)

Thus, the minimum sum of the 7 numbers = 60 - 10 = 50
Answer C


Note:
To maximize the sum of the 7 numbers, we need to minimize d:
Minimum value of d = 5 (since the smallest 4 numbers have a mean 5, and d is the largest of those 4 numbers)
Thus, the maximum sum of the 7 numbers = 60 - 5 = 55
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Re: The four largest numbers in a set of seven numbers have a mean of 10. [#permalink] New post   07 Feb 2022, 15:42
IanStewart wrote:
The four largest numbers sum to 40, and the four smallest sum to 20. So if we add those eight numbers, the sum is 60, but we've now added the median twice. So the sum of the seven numbers is 60 - m, where m is the median. Now the median can't be larger than 10, because then the four largest numbers would all be greater than 10 and could never have an average of 10. So m is at most 10, and the sum 60-m is at least 50.

If we wanted to construct a list with this sum, it could be: 10/3, 10/3, 10/3, 10, 10, 10, 10


I've never seen a more logical solution in my life.

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Re: The four largest numbers in a set of seven numbers have a mean of 10. [#permalink]
07 Feb 2022, 15:42
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