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28 Jan 2022, 07:45
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Difficulty:
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67% (02:25) correct
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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
Are You Up For the Challenge: 700 Level Questions
(A) 30
(B) 40
(C) 50
(D) 60
(E) 70
Are You Up For the Challenge: 700 Level Questions
28 Jan 2022, 08:12
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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
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
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
If we wanted to construct a list with this sum, it could be: 10/3, 10/3, 10/3, 10, 10, 10, 10
