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Bill took 4 math tests, and each test received a score out

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Bill took 4 math tests, and each test received a score out

by AAPL » Fri Jul 12, 2019 3:52 am

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Magoosh

Bill took 4 math tests, and each test received a score out of 100. What was Bill's average (arithmetic mean) score for all 4 tests?

1. Bill's first 3 tests received an average score of 50.
2. Bill's last 2 tests received an average score of 50.

OA E
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Source: — Data Sufficiency |

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by Brent@GMATPrepNow » Fri Jul 12, 2019 4:59 am

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AAPL wrote:Magoosh

Bill took 4 math tests, and each test received a score out of 100. What was Bill's average (arithmetic mean) score for all 4 tests?

1. Bill's first 3 tests received an average score of 50.
2. Bill's last 2 tests received an average score of 50.

OA E
Target question: What was Bill's average (arithmetic mean) score for all 4 tests?

Statement 1: Bill's first 3 tests received an average score of 50.
Since we don't have any information about the 4th test, statement 1 is NOT SUFFICIENT

Statement 2: Bill's last 2 tests received an average score of 50.
Since we don't have any information about the first two tests, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
There are several scnenarios that satisfy BOTH statements. Here are two:
Case a: The 4 test scores, in order, are {50, 50, 50, 50}. In this case, the answer to the target question is The average score for all 4 tests is 50
Case b: The 4 test scores, in order, are {0, 50, 100, 0}. In this case, the answer to the target question is The average score for all 4 tests is 150/4
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

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by AAPL » Fri Jul 12, 2019 10:42 am

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Let \(a, b, c, d\) be the 4 scores of Bill.

We are asked \(\frac{a + b + c + d}{4} = ?\)


Statement 1: \(a + b + c = 3\cdot 50\)

Not Sufficient. \(\color{red}\chi\)

Statement 2: \(c + d = 2\cdot 50\)

Not Sufficient. \(\color{red}\chi\)

Combining \(1 + 2\)

Not Sufficient. \(\color{red}\chi\)

Hence, __E__ is the correct answer.
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