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12 May 2020, 02:49
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A
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Difficulty:
85%
(hard)
Question Stats:
44% (02:30) correct
56%
(03:17)
wrong
based on 84
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Not Attempted Yet
If x is an integer between 0 and 10 is y less than the average of x and 10 ?
(1) \(10-y < y - \frac{x +10}{2}\)
(2) y is 4 times as large as x
Are You Up For the Challenge: 700 Level Questions
(1) \(10-y < y - \frac{x +10}{2}\)
(2) y is 4 times as large as x
Are You Up For the Challenge: 700 Level Questions
12 May 2020, 03:05
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Bunuel
Question: Is y < (x+10)/2?
Statement 1: \(10-y < y - \frac{x +10}{2}\)
i.e. 2y > x+30/2
i.e. \(y > \frac{x+30}{4}\)
at x = 1, y > 7.75 i.e. y > (x+10)/2
at x = 9, y > 9.75 i.e. y > (x+10)/2
i.e. Always NO to question
SUFFICIENT
Statement 2: y is 4 times as large as x
i.e. y = 4x
i.e. We need to compare 4x and (x+10)/2
at x = 1, y = 4 and (x+10)/2= 5.5 i.e .y < (x+10)/2
at x = 2, y = 8 and (x+10)/2= 6 i.e .y > (x+10)/2
NOT SUFFICIENT
Answer: Option A
12 May 2020, 05:38
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Statement 1-
\(10-y < y - \frac{x +10}{2}\)
\(2y > 10+ \frac{x +10}{2}\)
\(y> \frac{10}{2} + \frac{x +10}{4}\)
\(y> \frac{10}{2} + \frac{x}{4} + \frac{10}{4}\)
\(y>\frac{10}{2} +\frac{ x}{2} - \frac{x}{4} +\frac{10}{4}\)
\(y>\frac{x +10}{2} + \frac{10-x}{4}\)
Since \(0≤x≤10\), \(\frac{10-x}{4}\) is always greater than or equal to 0
y>[average of x and 10]+non-negative number
Sufficient
Statement 2- y=4x
1. x=1; y=4; \( \frac{x+10}{2} = 5\) (Yes)
2. x=2; y=8; \(\frac{x+10}{2} = 6\) (No)
Insufficient
\(10-y < y - \frac{x +10}{2}\)
\(2y > 10+ \frac{x +10}{2}\)
\(y> \frac{10}{2} + \frac{x +10}{4}\)
\(y> \frac{10}{2} + \frac{x}{4} + \frac{10}{4}\)
\(y>\frac{10}{2} +\frac{ x}{2} - \frac{x}{4} +\frac{10}{4}\)
\(y>\frac{x +10}{2} + \frac{10-x}{4}\)
Since \(0≤x≤10\), \(\frac{10-x}{4}\) is always greater than or equal to 0
y>[average of x and 10]+non-negative number
Sufficient
Statement 2- y=4x
1. x=1; y=4; \( \frac{x+10}{2} = 5\) (Yes)
2. x=2; y=8; \(\frac{x+10}{2} = 6\) (No)
Insufficient
Bunuel
