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Bunuel
Both the average (arithmetic mean) and the median of a set of 7 numbers equal 20. If the smallest number in the set is 5 less than half the largest number, what is the largest possible number in the set?

(A) 40
(B) 38
(C) 33
(D) 32
(E) 30


Kudos for a correct solution.

we have 7 numbers, out of which 20 is the middle one..

_ _ _ 20 _ _ A
suppose A is the largest number.
A/2 -5 = is the smallest number
to maximize A, suppose that we have all the numbers before 20 equal to the smallest number, and everything between 20 and A is equal to 20.

we know the average, we can find the sum.
20*7 = 140.
we have 3 numbers equal to 20 -> 60total
we have 3 numbers equal to A/2 -5.

now:
3(A/2 - 5) + A = 140-60 = 80.
3A/2 - 15 + A = 80 -> multiply everything by 2, to get rid of the fractions.
3A - 30 +2A = 160
5A=190
A=190/5 = 38.

answer is B.
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Bunuel
Both the average (arithmetic mean) and the median of a set of 7 numbers equal 20. If the smallest number in the set is 5 less than half the largest number, what is the largest possible number in the set?

(A) 40
(B) 38
(C) 33
(D) 32
(E) 30


Kudos for a correct solution.

The sum of the numbers is 140.

Since the smallest number in the set is 5 less than half the largest number, we can let y = the largest number, and thus we have y/2 - 5 as the smallest number.

Since we want y as large as possible, we want to minimize the other numbers in the set while keeping a sum of 140 and a median of 20. Since we know that y/2 - 5 is the smallest number in the set, we will repeat it as many times as possible, which is 3 times. Now, the fourth number must be the median of 20. And the two numbers after the median should also be equal to 20 so that we can maximize the value of y. So now we can create the equation:

y/2 - 5 + y/2 - 5 + y/2 - 5 + 20 + 20 + 20 + y = 140

3y/2 + y + 45 = 140

3y/2 + y = 95

3y + 2y = 190

5y = 190

y = 38

Answer: B
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Bunuel
Both the average (arithmetic mean) and the median of a set of 7 numbers equal 20. If the smallest number in the set is 5 less than half the largest number, what is the largest possible number in the set?

(A) 40
(B) 38
(C) 33
(D) 32
(E) 30

I would go with a different approach seen here.

We have a distribution of: X X X 20 Y Y Z.
Y = 20, and Z = 3*(20-X)+20
In other words. We are searching for any number, that is 20 + Something divisible by 3.

40? no
38? -> Yes
18/3 = 6
Now we need to test the condition given: 38/2-5 = 14; 20-14 = 6!
We are right on the way.

Therefore we go with B)

How can we get there fast?
We just try the highest and lowest number (or close friends of them), that are easy to calculate with mind math.
So 40 -> 15 (5 Rest means -> 20+3*5 = 35 max)
->Therefore we know it must me close. More specific: We already know, that 14 rest and 3+ more with the max are the solution.

Otherwise, we need to test the lowest value, look for exponents etc. that could screw up the average-logic, and go for numbers that are tilted to the closer value we are searching for the result. (In this case -> The max, that results in 20/Average in total.
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