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17 Jun 2020, 06:33
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A
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Difficulty:
75%
(hard)
Question Stats:
45% (01:37) correct
55%
(01:37)
wrong
based on 102
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R is the set of all integers between integers A and B, inclusive. What is the average (arithmetic mean) of R ?
(1) A + B = 20
(2) B − A = 10
Are You Up For the Challenge: 700 Level Questions
(1) A + B = 20
(2) B − A = 10
Are You Up For the Challenge: 700 Level Questions
17 Jun 2020, 06:38
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Bunuel
Set of All integers = Arithmetic Progression
CONCEPT: When terms are in Arithmetic Progression then, \(Mean = Median = \frac{First+Last}{2}\)
Question: Average = mean = ?
STatement 1: A + B = 20
If A = 1 and B = 19, i.e. R = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}
\(Mean = \frac{1+19}{2} = 10\)
This will be true for any values of A and B bnecause \(Mean = \frac{First + Last }{2}\)
SUFFICIENT
STatement 2: B − A = 10
Case 1: A = 1 and B = 11 \(Mean = \frac{1+11}{2} = 6\)
Case 1: A = 2 and B = 12 \(Mean = \frac{2+12}{2} = 7\)
Different possible values of Mean i.el
NOT SUFFICIENT
Answer: Option A
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17 Jun 2020, 07:38
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Statement 1: A+B = 20
Case 1: Let A and B = 10&10
The mean of R = 10+10/2 => 10
Case 2: let A and B = 18&2
The mean of R = (2,3,4,5........18)
=> 170/17 = 10
In any value of A and B the mean is only 10 (Sufficient)
Statement 2: B-A = 10
Case 1: if B is 10 and A is 0
Mean is 5
Case 2: B is 11 and A is 1
Mean is 6
(Insufficient)
Answer is A
Posted from my mobile device
Case 1: Let A and B = 10&10
The mean of R = 10+10/2 => 10
Case 2: let A and B = 18&2
The mean of R = (2,3,4,5........18)
=> 170/17 = 10
In any value of A and B the mean is only 10 (Sufficient)
Statement 2: B-A = 10
Case 1: if B is 10 and A is 0
Mean is 5
Case 2: B is 11 and A is 1
Mean is 6
(Insufficient)
Answer is A
Posted from my mobile device
