Hello,
Can you please assist with this:
In a certain test, each of the students in a college class was given one of three
scores: 50, 60, or 70. In order to calculate the weighted average of the
scores in the class each score was multiplied by the percentage of students
who got each score and the products were then added. If the percentage of
students who got 50, 60, or 70, were, respectively, W1, W2, and W3, and the
weighted average was less than 60, then which of the following must be true?
I) W1 < W3
II) W1 > W3
III) W2 = 0
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III
OA: B
I tried to solve as follows:
Weighted Average = [ (50)(Percentage of students who scored 50) + (60) (Percentage of students who scored 60) + (70) (Percentage of students who scored 70) ]/ [ (Percentage of students who scored 50) + (Percentage of students who scored 60) + (Percentage of students who scored 70) ]
Since W1 = (Percentage of students who scored 50) , W2 = (Percentage of students who scored 60) and
W3 = (Percentage of students who scored 70), I was wondering if it is correct to say:
W1 + W2 + W3 = 100
i.e. Is the following correct:
Weighted Average = [ 50 (W1) + 60 (W2) + 70 (W3) ]/ 100
or should it be:
Weighted Average = [ 50 (W1) + 60 (W2) + 70 (W3) ]/ (W1 + W2 + W3)
Thanks for your help,
Sri
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Test scores - Weighted averages
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gmattesttaker2
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Hi Sri,
It sounds like you're making this question more complex than it needs to be. Here's an easy way to think of it.
You're going to average a group of numbers; each of the numbers will be 50 or 60 or 70. It's possible that a number will appear more than once. We're told at the end of the prompt that the weighted average is less than 60...
If we had one of each type of number (one 50, one 60 and one 70, then the average would be EXACTLY 60). Since we're told that the average is less than 60, that means there MUST BE more 50s than 70s. There could be any number of 60s.
The numbers W1, W2 and W3 refer to the percentage of students who got 50, 60 or 70 respective. Combining this terminology with the prior information, we know that...
W1 MUST BE greater than W3; W2 could be any option (biggest, smallest or in the middle). With those deductions, the correct answer must be B
---------------
Using your approach....
Weighted Average = [50(percentage who scored 50) + 60(percentage who scored 60) + 70(percentage who scored 70)] / 1
We divide by 1 because that accounts for 100% of the students.
Since we're dealing with percentages, W1 + W2 + W3 = 1.00
GMAT assassins aren't born, they're made,
Rich
It sounds like you're making this question more complex than it needs to be. Here's an easy way to think of it.
You're going to average a group of numbers; each of the numbers will be 50 or 60 or 70. It's possible that a number will appear more than once. We're told at the end of the prompt that the weighted average is less than 60...
If we had one of each type of number (one 50, one 60 and one 70, then the average would be EXACTLY 60). Since we're told that the average is less than 60, that means there MUST BE more 50s than 70s. There could be any number of 60s.
The numbers W1, W2 and W3 refer to the percentage of students who got 50, 60 or 70 respective. Combining this terminology with the prior information, we know that...
W1 MUST BE greater than W3; W2 could be any option (biggest, smallest or in the middle). With those deductions, the correct answer must be B
---------------
Using your approach....
Weighted Average = [50(percentage who scored 50) + 60(percentage who scored 60) + 70(percentage who scored 70)] / 1
We divide by 1 because that accounts for 100% of the students.
Since we're dealing with percentages, W1 + W2 + W3 = 1.00
GMAT assassins aren't born, they're made,
Rich
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Brent@GMATPrepNow
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We can use the answer choices to our advantage by trying to come up with a mixture of students that yields a total weighted average that's less than 60%.gmattesttaker2 wrote: In a certain test, each of the students in a college class was given one of three
scores: 50, 60, or 70. In order to calculate the weighted average of the
scores in the class each score was multiplied by the percentage of students
who got each score and the products were then added. If the percentage of
students who got 50, 60, or 70, were, respectively, W1, W2, and W3, and the
weighted average was less than 60, then which of the following must be true?
I) W1 < W3
II) W1 > W3
III) W2 = 0
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III
Consider this case:
- 98 percent of the students got a 50 score. So, W1 = 98
- 1 percent of the students got a 60 score. So, W2 = 1
- 1 percent of the students got a 70 score. So, W3 = 1
Since almost all of the students got 50 on the test, we can see that the weighted average for the entire group will be close to 50, which means the total weighted average is definitely less than 60.
Now that we have a mixture that meets the given conditions, let's start checking the statements:
I) W1 < W3
In our example, W1 > W3, so statement I need not be true.
This allows us to ELIMINATE A, C, D and E
So, the answer must be B
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Let us plot the three averages with respective numbers as below
50----------60---------70
W1 W2 W3
The weighted average is less than 60. Let us mark a point that denotes this average.
50--------X-60---------70
W1 W2 W3
As the weighted average is below 60, the scale tilts towards the left side i.e there more W1s to bring the average down.
This disproves Statement 1 thereby removing choices A,C,D and E leaving B as the only choice.
Answer is B
50----------60---------70
W1 W2 W3
The weighted average is less than 60. Let us mark a point that denotes this average.
50--------X-60---------70
W1 W2 W3
As the weighted average is below 60, the scale tilts towards the left side i.e there more W1s to bring the average down.
This disproves Statement 1 thereby removing choices A,C,D and E leaving B as the only choice.
Answer is B
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Just in case anyone didn't catch this part with all the math going on, a key trick used in all the solutions above is noticing that the answer choices almost all contain (i). Trying (i) is thus a great place to start: if it works, we can eliminate B right away, and if it doesn't, the answer MUST be B. (What a great deal!)
