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Rabia must earn an average (arithmetic mean) score of S percent to [#permalink]
My Approach:

Assume:
- Total exams = 10,
- S= 50%,
- Max Marks Per exam = 100
- X = maximum percentage below S to pass.


- Now, equation is: 6 * 60 + 4 * (S-X) = 10 * 50 (as she need S% to pass the exam)
=> 360 +4(S-X) = 500 => S-X = 140/4 = 35
50 - X = 35 => X = 15.

Option B
Please correct if my approach is wrong!

Originally posted by fauji on 03 Nov 2019, 11:59.
Last edited by fauji on 03 Nov 2019, 22:29, edited 1 time in total.
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Re: Rabia must earn an average (arithmetic mean) score of S percent to [#permalink]
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anmolmakkarz17 wrote:
Rabia must earn an average (arithmetic mean) score of S percent to pass her physics course. If her average score on the first 60 percent of her assignments was (S + 10) percent, and each of her assignments is weighted equally, then what is the maximum percentage below S that she can earn on her remaining assignments and still pass the course?

A) 10%
B) 15%
C) 20%
D) 85%
E) 90%


Let the total assignments = n
—> Her total score in 60% of tests = 0.6n*(S + 10)
Let her average score in remaining 40% tests = x
—> Total score = 0.4n*x

To just pass, average score = S
—> [0.6n(S + 10) + 0.4nx]/n = S
—> 0.6S + 6 + 0.4x = S
—> 0.4x = 0.4S - 6
—> x = (0.4S - 6)/0.4 = S - 60/4
—> x = S - 15

So, she can score a 15% less than her average to still pass the test

IMO Option B

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Re: Rabia must earn an average (arithmetic mean) score of S percent to [#permalink]
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Hi All,

We're told that Rabia must earn an AVERAGE (arithmetic mean) score of S percent to pass her physics course, the average score on the first 60 percent of her assignments was (S + 10) percent, and each of her assignments is weighted equally. We're asked for the MAXIMUM percentage BELOW S that she can earn on her remaining assignments and still pass the course. Since this is a 'ratio' question, it can be approached in a number of different ways, including by TESTing VALUES.

Since 60% is 3/5, we can start by saying that there are 5 total assignments.
Let's also TEST S = 70, meaning that Rabia needs an average of 70% over the 5 assignments to pass.

On the first 3 of those assignments, she averaged (70+10) = 80 percent.... In simple terms, she needs 5(70) = 350 points and she already has 3(80) = 240 points. To put it another way, she is 3(10) = 30 points AHEAD of what she needs to pass the course. She would need another 350 - 240 = 110 points to pass.

With 2 assignments remaining, she would need to average 110/2 = 55 points to pass. That is 70 - 55 = 15 points BELOW S.

Final Answer:

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Rabia must earn an average (arithmetic mean) score of S percent to [#permalink]
Solved it logically.

She earned 10% marks each in 0.6 of her subjects and she can lose them in the remaining 0.4 of the subjects to keep the avg at S.
10% x 0.6 / 0.4 = 15%

IMHO - B
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Re: Rabia must earn an average (arithmetic mean) score of S percent to [#permalink]
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Re: Rabia must earn an average (arithmetic mean) score of S percent to [#permalink]
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