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The average (arithmetic mean) of 67, 78, x, and 101
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03 Jun 2017, 02:09
03 Jun 2017, 02:09
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78% (00:52) correct
21% (00:59) wrong based on 147 sessions
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Quantity A |
Quantity B |
The average (arithmetic mean) of 67, 78, x, and 101 |
The average (arithmetic mean) of 66, 79, x, and 102 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: The average (arithmetic mean) of 67, 78, x, and 101
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27 Sep 2017, 04:41
27 Sep 2017, 04:41
The two columns can be written as \(\frac{67+78+x+101}{4}\) and \(\frac{66+79+x+102}{4}\). Using matching operations method, and after rearranging, we get column A equal to \(1+x\) and column B equal to \(2+x\), thus whatever value of X, column B is greater!
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The average (arithmetic mean) of 67, 78, x, and 101
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29 Jul 2020, 06:06
29 Jul 2020, 06:06
Carcass wrote:
Quantity A |
Quantity B |
The average (arithmetic mean) of 67, 78, x, and 101 |
The average (arithmetic mean) of 66, 79, x, and 102 |
We can solve this question using matching operations (my favorite Quantitative Comparison strategy)
Given:
QUANTITY A: The average (arithmetic mean) of 67, 78, x, and 101
QUANTITY B: The average (arithmetic mean) of 66, 79, x, and 102
Rewrite as follows:
QUANTITY A: \(\frac{67+78+x+101}{4}\)
QUANTITY B: \(\frac{66+79+x+102}{4}\)
Multiply both quantities by \(4\) to get:
QUANTITY A: \(67+78+x+101\)
QUANTITY B: \(66+79+x+102\)
Subtract \(66\) from both quantities to get:
QUANTITY A: \(1+78+x+101\)
QUANTITY B: \(79+x+102\)
Subtract \(78\) from both quantities to get:
QUANTITY A: \(1+x+101\)
QUANTITY B: \(1+x+102\)
Subtract \(101\) from both quantities to get:
QUANTITY A: \(1+x\)
QUANTITY B: \(1+x+1\)
Subtract \(x\) from both quantities to get:
QUANTITY A: \(1\)
QUANTITY B: \(1+1\)
Answer: B
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