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04 May 2022, 09:00
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
64% (01:28) correct
36%
(01:19)
wrong
based on 66
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Two groups in the class have passed the test. If the average grade for the first group was 68.2, and the average grade for the second group was 73.5, what was the average grade in two groups?
(1) The first group had 20 more students than the second one
(2) The first group has 3 times more students than the second one
(1) The first group had 20 more students than the second one
(2) The first group has 3 times more students than the second one
16 May 2022, 08:53
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Bunuel
Let the number of students in group 1 be x and in group 2 be y
Let n be the average grade in 2 group
n = \(\frac{68.2x + 73.5y}{x+y}\)
= \(\frac{68.2x + 68.2y + 5.3y}{x+y}\)
= 68.2 + \(\frac{5.3y}{x+y}\)
We have to find \(\frac{y}{x+y}\)
Statement 1 - x = y+20, \(\frac{y}{x+y}\) = \(\frac{y}{2y+20}\)
Insufficient
Statement 2 - x = 3y, \(\frac{y}{x+y}\) = \(\frac{y}{4y}\) = \(\frac{1}{4}\)
Sufficient
Option B
04 Jun 2022, 12:54
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This problem took me quite while to solve (2 mins and 58 seconds), so if anyone could tell me where I could've cut a corner (or where I wasn't correct), then that would be appreciated.
First I noted that the Average of Group 1 (A1) = 68.2, which is equal to the Sum of Group 1 Scores (S1) divided by the Number of scores in Group 1 (n1) - the standard Average = Sum / number of terms formula. I did the same with group 2, noting that A2 = 73.5 = S2 / n2
Then I rearranged the subject of the equations: S1 = 68.2*n1 and S2 = 73.5*n2
I assumed that the average that the question is asking for is the average calculated by: (S1+S2)/(n1+n2) to get the total average across groups.
(1) tells us that n1 = n2+20 so let's try and use substitution into the equation above:
(68.2(n2+20)+73.5*n2) / 2*n2+20 is what we end up with, as you can see, we cannot solve this for n2 because we end up with some coefficient number 'x' and some constant 'c' such that: (x*n2+c) / 2*n2 + 20, we cannot simplify this such that the "n2" terms cancel out, so this is NO GOOD.
(2) tells us that n1=3*n2
so let's sub this in
So we end up with an average = (68.2*3n2 + 73.5*n2) / 4*n2
Here, we can already see that the 'n2' terms will cancel out with one another, so we can indeed determine an average that is a number (doesn't involve a variable, just a pure number).
If interested the calculation yields an answer of 69.525 (but we don't need to know this for a DS question)
So answer is B
Any thoughts on how I could've done this faster or whether this method is even half-decent?
First I noted that the Average of Group 1 (A1) = 68.2, which is equal to the Sum of Group 1 Scores (S1) divided by the Number of scores in Group 1 (n1) - the standard Average = Sum / number of terms formula. I did the same with group 2, noting that A2 = 73.5 = S2 / n2
Then I rearranged the subject of the equations: S1 = 68.2*n1 and S2 = 73.5*n2
I assumed that the average that the question is asking for is the average calculated by: (S1+S2)/(n1+n2) to get the total average across groups.
(1) tells us that n1 = n2+20 so let's try and use substitution into the equation above:
(68.2(n2+20)+73.5*n2) / 2*n2+20 is what we end up with, as you can see, we cannot solve this for n2 because we end up with some coefficient number 'x' and some constant 'c' such that: (x*n2+c) / 2*n2 + 20, we cannot simplify this such that the "n2" terms cancel out, so this is NO GOOD.
(2) tells us that n1=3*n2
so let's sub this in
So we end up with an average = (68.2*3n2 + 73.5*n2) / 4*n2
Here, we can already see that the 'n2' terms will cancel out with one another, so we can indeed determine an average that is a number (doesn't involve a variable, just a pure number).
If interested the calculation yields an answer of 69.525 (but we don't need to know this for a DS question)
So answer is B
Any thoughts on how I could've done this faster or whether this method is even half-decent?
