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by The Iceman » Sat Feb 09, 2013 10:38 pm
Difference of the total scores after taking x+1 tests and x tests must equal 90.
=> 82(x+1) - 80 x = 90 => x =4

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by sana.noor » Sat Feb 09, 2013 11:47 pm
we know the formula of average is [SUM of tests/number of test = Average --> Sum of tests = (number of tests)(average)]
origianlly 80 was the average of X tests and this average increases to 82 when an other test with score 90 is done.

X = the unknown number of test taken when the average is 80
X+1 = the unknown number of test when one more test with socre 90 is taken
Sum of the tests = 80 (X number of test) + 90
Average increases to = 82
Number of tests = X+1

Putting this in the formula

80X + 90 = (82)(x+1)
80X +90 = 82X +82
90-82 = 82X- 80X
8 = 2X
X = 4
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by Ian Stewart » Sun Feb 10, 2013 3:31 am
Or for people who use alligation for weighted average questions (I just explained that method in a different post, so I won't go over the details again) - we are combining x tests with an 80% average with 1 test with a 90% average, and getting an overall average of 82:

--80---82----------------------90--

The ratio of the blue and red distances is equal to the ratio of the number of tests of each type, so they are in a 8 to 2, or 4 to 1 ratio, so the answer is 4.
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by GMATGuruNY » Sun Feb 10, 2013 4:04 am
An alternate approach is to plug in the answers, which represent the value of x (the original number of tests).
The options for x are 2, 4, 8, 10 or 20.
When an ADDITIONAL test is included, the number of tests will be 3, 5, 9, 11, or 21.
Since the resulting average is 82, and sum = number * average, answer choice B seems the most likely: it is the only answer choice that will yield a multiple of 10 for the sum of all the scores when the additional test is included.
Sum of 5 tests with an average score of 82 = 5*82 = 410.

Answer choice B: 4
Sum of 4 tests with an average score of 80 = 4*80 = 320.
Sum of all 5 tests - sum of the first 4 tests = 410-320 = 90.
Success!
The score on the additional test is 90.

The correct answer is B.
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