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11 Aug 2021, 09:00
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C
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58% (02:18) correct
42%
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Set A consists of the four integers x, x^2, x^3 and x^4, and set B consists of the four integers x, 2x, 3x and 4x. What is the probability that a randomly selected integer from 1 to 100, inclusive, is a member of neither set A nor set B?
(1) x > 5
(2) x < 10
(1) x > 5
(2) x < 10
11 Aug 2021, 22:05
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Bunuel
Set A: {\(x, x^2,x^3,x^4\)} and set B: {\(x, 2x, 3x, 4x\)}
If x>100, then probability is 0. But x=100, will give the probability as \(\frac{1}{100}\), as only 100 will be in the sets and the remaining greater than 100.
So, we require to know the value of x.
(1) x > 5
x=-6 and x=6 will give different answers.
Or
If x>100, then probability is 0. But x=100, will give the probability as \(\frac{1}{100}\).
Insufficient
(2) x < 10
If x=1, then Set A: {\(x, x^2,x^3,x^4\)} and set B: {\(x, 2x, 3x, 4x\)}=> Set A: {\(1, 1,1,1\)} and set B: {\(1, 2, 3, 4\)}. P = \(\frac{4}{100}=\frac{1}{25}\)
If x=5, then Set A: {\(x, x^2,x^3,x^4\)} and set B: {\(x, 2x, 3x, 4x\)}=> Set A: {\(5, 25, 125, 625\)} and set B: {\(5, 10, 15, 20\)}. P = \(\frac{5}{100}=\frac{1}{20}\)
Insufficient
Combined
\(5<x<10\), so x can be any integer from 6 to 9.
SET A: The squares of all numbers from 6 to 9 are <100, but cubes are more. Thus, in Set A, each integer will give two values x and x^2 <100.
SET B: First 4 multiples of x in range 5<x<10 will all be <100. That is x, 2x, 3x, 4x will all be <100, when x lies in the range 5<x<10.
But we have to check for any repetition in any SET.
x, 2x, 3x, 4x are all different. There is one repetition of x in SET A, but x^2 will be different in case of all integers.
SO P = \(\frac{5}{100}=\frac{1}{20}\)
Sufficient
C
14 Aug 2021, 17:32
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Bunuel
We only care about the combined set which is \({x, 2x, 3x, 4x, x^2, x^3, x^4}\)
Statement 1:
If we let x = 6, the set would be {6, 12, 18, 24, 36, >100, >100}. Note we only care about how many elements are between 1 and 100, and for this set, there are 5 such elements.
If we let x = 101, none of the elements would be between 1 and 100. Insufficient.
Statement 2:
If we let x = 1, all elements are below 100. If we let x = 6, we have only 5 elements within 100. Insufficient.
Combined:
Check both extreme cases x = 6 and x = 9. Both have {<100, <100, <100, <100, <100, >100, >100}, a total of 5 elements within the range of 1 - 100. Then we know all cases follow this trend. Thus the probability of neither is always 95/100. Sufficient.
Ans: C
