Consider these two possible scenarios.

CASE #1: The 10,000 numbers consist of 6,000 20's, one 62, one 74 and 3998 80's
Since 6,000 of the 10,000 numbers are less than 62, we can see that 62 is in the 60th percentile (since 6,000/10,000 = 60%)
Since 6,001 of the 10,000 numbers are less than 74, we can see that 74 is in the 60.01th percentile (since 6,001/10,000 = 60.01%)
So, in this case, n = 60.01
We get:
QUANTITY A: 60.01
QUANTITY B: 70
In this case QUANTITY B IS GREATER


CASE #2: The 10,000 numbers consist of 6,000 20's, one 62, 2999 63's, one 74 and 999 80's
Since 6,000 of the 10,000 numbers are less than 62, we can see that 62 is in the 60th percentile (since 6,000/10,000 = 60%)
Since 9,000 of the 10,000 numbers are less than 74, we can see that 74 is in the 90th percentile (since 9,000/10,000 = 90%)
So, in this case, n = 90
We get:
QUANTITY A: 90
QUANTITY B: 70
In this case QUANTITY A IS GREATER

Answer: D

Cheers,
Brent
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we are not told that 10 000 numbers evenly distributed, so I believe we cannot draw conclusion about n, therefore answer should be D

Ye. It is D. Kudo.

Regards

Can anyone elaborate please.

Amazing explanation. Thanks.

Regards

Thanks Domenico!

Cheers,
Brent
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Answer is D
What we have:
We have 10,000 numbers, from 20 to 80: 20,…………., 80
*The sequence is ordered, if not this question was meaningless.
Number 62 is the 60th percentile
(for non-native speakers like me, the percentile is like percentage, thus 60th percentile means if we have 100 number then 60th, now we have 10,000, so the 60th percentile is (10,000/100)*60= 60 * 100 = 6,000
-62 is 6,000th number in the sequence, so there are 6,000 numbers less than 62 in the sequence and there are 4,000 numbers more than 62 in the sequence.
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Number 74 in the nth percentile
Just like the above explanations, 74 is the n*100th number in the sequence, and as it is more than 62, it should be among those 4,000 numbers.
But we can’t perceive which one of them it can be.

Contravention of A:
62 is 6,000th number in the sequence, consider 74 is one before the last number, so it is the 9999th number, the numbers before it should be between 60 and 74 and the number between it. Here n is 99.99% and more than 70.

Contravention of B:
62 is 6,000th number in the sequence, consider 74 is right after it, so it is the 6,001st number, the numbers after it should be between 74 and 80. Here n is 6,001* 0.01 = 60.01% which is less than 70

Contravention of C:
This situation is also possible, but it can be wrong by A and B.
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If this question said "Normal Distribution", the answer would still be D, isn't that correct?

Yes, I believe the answer would still be D, since we don't know the mean or the standard deviation.
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Ok, thank you