List R contains 5 numbers...CAT 6

[AbigailS228]

List R contains five numbers that have an average value of 55. If the median of the numbers in the list is equal to the mean and the largest number is equal to 20 more than two times the smallest number, what is the smallest possible value in the list?

a)35
b)30
c)25
d)20
e)15

The way I solved this is with two equations. The avg formula is (L+F)/2 (correct?) so I did (L+F)/2=55 AND L=2S+20. Then I solved for S and got 30.

Why does this way not achieve the correct answer (which is c)?

[tim]

Unless the numbers are equally spaced, you can't just average the first and last terms to get the overall average. You would have to add all five numbers and divide by 5.

[RonPurewal]

If you're using a formula for "the smallest possible value", then that's ALWAYS trouble.

Formulas are for situations with definite outcomes. If a problem asks for "the value" of something, then you might be able to use formulas.
For "the largest value" or "the smallest value", on the other hand, formulas generally won't work, since formulas don't spit out entire ranges of values.
You'll have to manipulate the actual numerical values, in some way or another.

Tim (above) has pointed out the specific reason why your formulas don't work. (I don't understand what "f", "s", and "l" stand for, but apparently Tim does.)

[RonPurewal]

But, quite apart from that, you should've realized that something was fishy with using formulas here.

Thought experiment:
Say the list is a ≤ b ≤ c ≤ d ≤ e.
Right now, the problem is asking for the smallest possible value of "a".
Say the problem were to ask for the largest possible value of "a".
That should definitely have a different answer!
But, if you were to use this approach, it would come out the same. Which means it doesn't work.
(:

[AbigailS228]

Yes I totally understand that now. I was thinking evenly spaced sets, and I didn't realize it was a minimize/maximize problem. And also, L = last and F/S=first/small (dunno why I changed the variable name halfway through that was weird..).

Definitely understand now. Thanks to both of you!

[tim]

You're welcome!

[RonPurewal]

Yes I totally understand that now. I was thinking evenly spaced sets, and I didn't realize it was a minimize/maximize problem. And also, L = last and F/S=first/small (dunno why I changed the variable name halfway through that was weird..).

Definitely understand now. Thanks to both of you!

From the highlighted part, I'd suggest that you are not paying enough attention to the goal of the problem.
This question asks for "the smallest possible value"--a clear, explicit indicator that it's a max/min problem. If you're aware of this goal, it's essentially impossible to think this is an "evenly spaced sets" problem... so, I conclude, you must have been unaware of the goal.

Try to figure out WHY you were unaware of the goal, and take steps to fix the problem. E.g., perhaps you could try reading ahead to the goal of the problem before you read all of the other given information.

[AdhiK295]

lets assume the number is arrange as follow:

a, b, 55, c, 2a + 20
a <=b <= 55
55 <= c <= 2a+20

we are being told that average is 55, hence sum of all 5 numbers is 275. The sum of the smallest and the largest numbers is 3a+20.

since we are asked about the minimum a, then we have to find the maximum sum b+c, which is satisfied when b=55 and c=2a+20

we now then have to try 1 by 1 of the solution, in which B+C must suit description above

if a=15, 2a+20 = 50 (impossible since 2a+20 < 55)

if a=20, 2a+20 = 60, b+c = 275-55-80 = 140 (impossible since average of b & c is 70 , greater than 2a+20)

if a=25, 2a+20 = 70, b+c = 275-55-95 = 125 (satisfied when b=55 and c=70)

Therefore the answer is C (a=25)

[tim]

This is a good plan. With a fixed sum, if you want to minimize one thing, you should maximize the others (and vice versa).

[RonPurewal]

lets assume the number is arrange as follow:

a, b, 55, c, 2a + 20
a <=b <= 55
55 <= c <= 2a+20

we are being told that average is 55, hence sum of all 5 numbers is 275. The sum of the smallest and the largest numbers is 3a+20.

since we are asked about the minimum a, then we have to find the maximum sum b+c, which is satisfied when b=55 and c=2a+20

we now then have to try 1 by 1 of the solution, in which B+C must suit description above

if a=15, 2a+20 = 50 (impossible since 2a+20 < 55)

if a=20, 2a+20 = 60, b+c = 275-55-80 = 140 (impossible since average of b & c is 70 , greater than 2a+20)

if a=25, 2a+20 = 70, b+c = 275-55-95 = 125 (satisfied when b=55 and c=70)

Therefore the answer is C (a=25)

do you have a question? if so, please clarify. thanks.

[drtfyghujd403]

Could someone explain how to solve this problem with algebraic approach? Thank you

[RonPurewal]

there isn't a purely algebraic approach, because the concepts of median, order, and maximiziation/minimization are outside the scope of algebra.

[RonPurewal]

... and that ^^ is exactly the reason why these concepts feature so prominently on this exam.

this exam is formulated specifically to punish people who seek "an algebraic approach" to everything.

it's designed to reward flexible thinking instead.
sometimes algebra will work; sometimes it won't. nothing will work every time.

[RahulG728]

Consider the following set: (30, 55, 55, 55, 80). It follows all the conditions given in the question and makes the answer choice 30.

Experts, please comment.

Thanks!

[tim]

This does not make the answer 30. It makes 30 one possible value for the lowest number on the list. It does not make 30 the smallest POSSIBLE value, as described in the problem. This is why it is so important to pay careful attention to the wording of the problem, but more than that, it is important to understand the underlying math. The only way you could get 30 as an answer here would be if you didn't pay attention to the wording AND you fundamentally misunderstood the underlying mathematical concept that allows for multiple sets of integers with this same property. You'll definitely want to be careful about both of these in the future.