Difficult Average Problem

Let total sum of numbers for List L be L, total sum of numbers for List M be M.

Qn states :-

(X + Y)/48 = 1.5*(X/43)

solving for Y we get Y = 29x/43

Qn is asking for Ave of (List M/Ave of List L) * 100%

Ave of List M = (29X/43)/5

Ave of List L = X/43

Thus X and 43 are cancelled out for what qn is asking, we are left with:-

(29/5) * 100% = 580%

Ans is E

so the avaage of numbers in list M is (58-10) or 48 greater then avarage of numbers in list L.
converting this in percentage we get the ans as
[spoiler]D 480%[/spoiler]

Whats OA

Sorry, I actually don't have the official answer. Can any of the experts please verify?

Thanks.

if you say that avg of L is x, you can view the list as 43 X's.

x,x,x,x,x........ 43 times.


Now it is given, the 5 elements of M increase the avg by 50%, i.e each X becomes 1.5X. So, the combined list is

1.5X,1.5X,1.5X,1.5X,1.5X,.......48 times.

That means, the 5 elements in M increase the total by 43* .5 X + 5 * 1.5 X = 29 X . Their avg = (29/5)X.


The original avg is X. Hence, the required % is 29/5 * 100 = 580.

(edited after Stuart's post. Thanks Stuart!)

the increase is : 580 - 100 = 480 %

pkw209 wrote:Hi guys,

Could someone please provide an explanation to this problem? Thanks!




a) 85%

b) 240%

c) 350%

d) 480%

e) 580%

A percent question with unspecified values - let's pick numbers.

To make life simple, let's let list L be {100, 100, 100, ...., 100}, i.e. just 43 100s.

So, the average of L is 100 and the sum of L is 4300.

The average of combined lists L and M is 50% greater, so the combined average is 150.

There are 48 numbers total, so the combined sum is 150*48 = 7200.

48 numbers total means that List M has 5 numbers.

7200 - 4300 = 2900, so those 5 numbers must sum to 2900.

Average = sum/# of terms:

2900/5 = 580

So, M has an average of 580.

Now let's be super careful: we're temped to pick (E) 580%, but the question asks:

"What percent greater than the average..."

To find percent change:

(new amount - old amount)/(old amount) = (580 - 100)/100 = 480

So, the average of M is 480% greater than L: choose (D).

Stuart Kovinsky wrote:

pkw209 wrote:Hi guys,

Could someone please provide an explanation to this problem? Thanks!




a) 85%

b) 240%

c) 350%

d) 480%

e) 580%

A percent question with unspecified values - let's pick numbers.

To make life simple, let's let list L be {100, 100, 100, ...., 100}, i.e. just 43 100s.

So, the average of L is 100 and the sum of L is 4300.

The average of combined lists L and M is 50% greater, so the combined average is 150.

There are 48 numbers total, so the combined sum is 150*48 = 7200.

48 numbers total means that List M has 5 numbers.

7200 - 4300 = 2900, so those 5 numbers must sum to 2900.

Average = sum/# of terms:

2900/5 = 580

So, M has an average of 580.

Now let's be super careful: we're temped to pick (E) 580%, but the question asks:

"What percent greater than the average..."

To find percent change:

(new amount - old amount)/(old amount) = (580 - 100)/100 = 480

So, the average of M is 480% greater than L: choose (D).

I spent so much time on this question. Thats a brilliant explanation Stuart, you made it seem so easy. But do tell me, when do you use the picking numbers strategy?

eaakbari wrote:
I spent so much time on this question. Thats a brilliant explanation Stuart, you made it seem so easy. But do tell me, when do you use the picking numbers strategy?

There are a few common situations on the GMAT in which picking numbers is very effective:

1) percent questions with unspecified values.

When you see a % question with unspecified values, you almost always want to pick 100.

2) other questions with unspecified values.

If the choices are fractions, pick the common denominator of all the fractions in the question stem. If not, pick a number that's both permissible (follows the constraints you're given) and manageable (relatively easy to work with).

3) variables in the answer choices and multiple variables in the question.

Pick numbers for the variables that appear in the choices, then work your way through the word problem (or solve the equation) with those numbers.

Then you plug the numbers back into the choices to get a match.

4) odd/even questions.

Since odds and evens follow strict rules, once you pick one set of numbers, you never have to try another.