Both the average (arithmetic mean) and the median of a set of 7 number

Question

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.

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

Bunuel · Accepted Answer

MANHATTAN GMAT OFFICIAL SOLUTION: Out of the letters A through E, which one is your favorite? You may be thinking, “Huh? What a weird question. I don’t have a favorite.” I don’t have one in the real world either, but I do for the GMAT, and you should, too. When you get stuck, you’re going to need to be able to let go, guess, and move on. If you haven’t been able to narrow down the answers at all, then you’ll have to make a random guess—in which case, you want to have your favorite letter ready to go. If you have to think about what your favorite letter is, then you don’t have one yet. Pick it right now. I’m serious. I’m not going to continue until you pick your favorite letter. Got it? From now on, when you realize that you’re lost and you need to let go, pick your favorite letter immediately and move on. Don’t even think about it. (This assumes, of course, that your favorite letter is still in the mix. If you were able to narrow down the answers, and you crossed off your favorite letter, then obviously don’t pick that one!) Okay, let’s solve this thing. What did you do first? gmat1.png I glanced at the beginning of the text and saw that it said average (arithmetic mean). I then glanced down to the answers to see how precise I was going to have to be with calculations. The answers are very close together, so estimation isn’t going to work. I’m going to have to do real math. Okay, time to read the problem and jot down the given info. avg.png Because the problem gives the median, arrange the numbers in order from smallest to largest (this is a general requirement whenever calculating or displaying a median). Hmm. That second sentence of the problem is going to take some work. It’s a relationship between the first and last number…but I don’t know the actual value of either one. Time to set a variable: let’s call something x. My natural inclination would be to call the smallest number x, but I’ve learned to ask myself one important question before arbitrarily assigning that variable: what am I trying to find? In this case, they asked for the largest possible number, so it’s better to call the last number x. That way, I’ll be solving for the thing that they want; I’m less likely to make a mistake and accidentally solve for x = the smallest number. If the largest number is x, then the smallest is (1/2)x – 5: first.png What next? If the average is 20 and there are 7 numbers, then the sum must be (Average)(number of terms) = (20)(7) = 140. Pause for a moment and look at everything you’ve got. What’s the best plan from here? gmat2.png They want to maximize the largest number, that x. Only one number is set, the middle one (20). All 7 numbers have to add up to 140. So what needs to happen in order to make x as large as possible? Since the sum is fixed, the first 6 numbers need to be minimized in order to get the last number to be as large as possible. This step is the key to all max/min problems: figure out what you can influence and either maximize or minimize (as needed) to get to the desired outcome. Look back at your notes to remind yourself of the restrictions. The fifth and sixth numbers cannot be less than the median of 20. They could equal the median, though, so set these values to 20: last.png That leaves a sum of 140 – 20(3) = 80 for the remaining 4 numbers. Now, what about the second and third values? Again, they can’t be smaller than the first one, but they could equal it. Set all three equal to (1/2)x – 5. Check it out: you now have a way to express each of the remaining four numbers in terms of that single variable x. Time to set up an equation and solve! gmat.png The largest possible number in the set is 38. The correct answer is (B).

RAHULSAINI · Answer

Assume the last number be X

So first number would be X/2-5

We know that Sum = Average * No of Terms

So Sum = 20* 7 = 140

To maximise the Last Digit we have to Minimise all six digits

Arrange the Digits in acceding order

X/2-5 ____ ____ 20 ____ ____ X

To minimise we have to choose 2nd and 3rd digit as x/2-5 and 5th and 6th as 20.

Sum it up and equate to 140

we get X=38

So B is the Answer

mvictor · Answer

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.

ScottTargetTestPrep · Answer

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

Shin0099 · Answer

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.

Both the average (arithmetic mean) and the median of a set of 7 number

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Both the average (arithmetic mean) and the median of a set of 7 number

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