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Bunuel
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If the average (arithmetic mean) of six different numbers is 25, how m Updated on: 17 Jun 2026, 20:31
Originally posted by Bunuel on 12 Nov 2014, 09:48.
Last edited by Bunuel on 17 Jun 2026, 20:31, edited 1 time in total.
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C
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If the average (arithmetic mean) of six different numbers is 25, how many of the numbers are greater than 25?

(1) None of the six numbers is greater than 50.

(2) Three of the six numbers are 7, 8, and 9, respectively.
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C
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If the average (arithmetic mean) of six different numbers is 25, how m 12 Nov 2014, 16:56
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Bunuel

Tough and Tricky questions: Statistics.



If the average (arithmetic mean) of six different numbers is 25, how many of the numbers are greater than 25?


(1) None of the six numbers is greater than 50.

(2) Three of the six numbers are 7, 8, and 9, respectively.

Kudos for a correct solution.
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The sum of 6 numbers = 25*6 = 150

(1) None of the 6 numbers is greater than 50.

Let's assume the numbers are

1, 2, 3, 47,48,49. This satisfies the condition and we have 3 numbers greater than 25.
However, we can also have something like, 19, 20, 26, 27, 28, 30; in which case, we have 4 numbers greater than 25. Hence, this is INSUFFICIENT.


(2) Since, three of the numbers are 7, 8, 9. The sum of other three numbers has to be 150-24 = 126.
So, the other three could be any combination of numbers adding up to 126.
For e.g. 1, 41, 84; with 2 nos. greater than 25, or 41, 42, 43; with 3 numbers greater than 25. Thus, this is INSUFFICIENT.

Considering, (1) & (2) together, we need 3 numbers less than 50 that add up to 126. 126/50 leaves a remainder of 26.

Since, all the numbers are different the least of the three remaining numbers has to be greater than 26. Thus, we will have three numbers greater than 25. Hence, SUFFICIENT.

Answer (C)
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If the average (arithmetic mean) of six different numbers is 25, how m 12 Nov 2014, 19:56
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Sum of the numbers = 6*25 =150

Statement 1: None of the six numbers is greater than 50
Clearly insufficient, since while all the numbers can be 25, the numbers could also be more than 25 in other cases.

Statement 2: Three of the six numbers are 7, 8, and 9, respectively
the sum of other three number = 150 -24 = 126. This is also insufficient because all of three numbers could be greater than 25 or one or two could be greater than 25.

Combining 1 and 2
We know that all the numbers has to be greater than 25. The least value of these numbers could be 26, in which case the 3 numbers would be 26,50,50.

Therefore the answer should be C
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If the average (arithmetic mean) of six different numbers is 25, how m 13 Nov 2014, 00:03
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Bunuel

Tough and Tricky questions: Statistics.



If the average (arithmetic mean) of six different numbers is 25, how many of the numbers are greater than 25?


(1) None of the six numbers is greater than 50.

(2) Three of the six numbers are 7, 8, and 9, respectively.

Kudos for a correct solution.
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Given that sum of 6 different numbers =150 so how many numbers greater than 25

St 1 says no greater than 50...

Case 1 : Six nos can be 1,2,3,48,47, and 49 (so 3 nos greater than 25)
Case 2: 71,73,1,2,3 ( 2nos greater not sufficient)

St 2 3 nos sum is 24 so...sum of other 3 nos is 126..

Again we can have 42,41 and 43 or we can have 63,62 and 1

Combining we get sum of three nos 126 and no number greater than 50...

We will have 48,49 and 29 ( 3nos)
Also sum of 2 nos cannot be greater than 100 so the third nos will have to be greater than 26.1 in any case

Ans C
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If the average (arithmetic mean) of six different numbers is 25, how m 13 Nov 2014, 09:04
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Bunuel

Tough and Tricky questions: Statistics.



If the average (arithmetic mean) of six different numbers is 25, how many of the numbers are greater than 25?


(1) None of the six numbers is greater than 50.

(2) Three of the six numbers are 7, 8, and 9, respectively.

Kudos for a correct solution.
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Official Solution:


We must determine how many numbers in a set of six distinct numbers are greater than 25. Since the average of the six numbers is 25, these six numbers must sum to \(6 \times 25 = 150\).

Statement 1 tells us that none of the six numbers is greater than 50. Many combinations of numbers satisfy this statement. For example, the six numbers could be 19, 20, 21, 22, 23, and 45. In this case, only 1 number is greater than 25. On the other hand, if the six numbers are 1, 2, 3, 47, 48, and 49, then 3 of the numbers are greater than 25. Since we cannot find a unique value, statement 1 alone is NOT sufficient to answer the question. Eliminate answer choices A and D. The correct answer choice is B, C, or E.

Statement 2 tells us that three of the six numbers are 7, 8, and 9. This means that the other three numbers must sum to \(150 - (7 + 8 + 9) = 150 - 24 = 126\). However, since there is no limit to what any of the numbers can be, we can have different combinations. For example, the three numbers could be 1, 2, and 123. In this case, only 1 number is greater than 25. But if the three numbers are 41, 42, and 43, then all 3 are greater than 25. Statement 2 alone is also NOT sufficient. Eliminate answer choice B. The correct answer choice is either C or E.

Both statements together tell us that three of the numbers are 7, 8, and 9, that the other three numbers must sum to 126, and that none of the numbers is greater than 50. If all three remaining numbers are 25 or less (remember, the numbers must be different), then these remaining numbers can sum to, at maximum, \(25 + 24 + 23 = 72\). If two numbers are 25 or less, and one is 50 (the upper bound for numbers), the sum is at most \(25 + 24 + 50 = 99\). If one number is 25 or less, and the other two numbers are 50 and 49, then the maximum sum is \(25 + 50 + 49 = 124\). In order to reach 126, all three numbers MUST be greater than 25. Both statements together are sufficient.


Answer: C.
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If the average (arithmetic mean) of six different numbers is 25, how m 17 Jun 2026, 19:47
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Bunuel


Official Solution:


We must determine how many numbers in a set of six distinct numbers are greater than 25. Since the average of the six numbers is 25, these six numbers must sum to \(6 \times 25 = 150\).

Statement 1 tells us that none of the six numbers is greater than 50. Many combinations of numbers satisfy this statement. For example, the six numbers could be 19, 20, 21, 22, 23, and 45. In this case, only 1 number is greater than 25. On the other hand, if the six numbers are 1, 2, 3, 47, 48, and 49, then 3 of the numbers are greater than 25. Since we cannot find a unique value, statement 1 alone is NOT sufficient to answer the question. Eliminate answer choices A and D. The correct answer choice is B, C, or E.

Statement 2 tells us that three of the six numbers are 7, 8, and 9. This means that the other three numbers must sum to \(150 - (7 + 8 + 9) = 150 - 24 = 126\). However, since there is no limit to what any of the numbers can be, we can have different combinations. For example, the three numbers could be 1, 2, and 123. In this case, only 1 number is greater than 25. But if the three numbers are 41, 42, and 43, then all 3 are greater than 25. Statement 2 alone is also NOT sufficient. Eliminate answer choice B. The correct answer choice is either C or E.

Both statements together tell us that three of the numbers are 7, 8, and 9, that the other three numbers must sum to 126, and that none of the numbers is greater than 50. If all three remaining numbers are 25 or less (remember, the numbers must be different), then these remaining numbers can sum to, at maximum, \(25 + 24 + 23 = 72\). If two numbers are 25 or less, and one is 50 (the upper bound for numbers), the sum is at most \(25 + 24 + 50 = 99\). If one number is 25 or less, and the other two numbers are 50 and 49, then the maximum sum is \(25 + 50 + 49 = 124\). In order to reach 126, all three numbers MUST be greater than 25. Both statements together are sufficient.


Answer: C.
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Am I wrong at noticing that since the question does not mention that all the numbers have to be positive integers, the numbers can be negative or fractions too?
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If the average (arithmetic mean) of six different numbers is 25, how m 17 Jun 2026, 20:53
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Albatross26

Am I wrong at noticing that since the question does not mention that all the numbers have to be positive integers, the numbers can be negative or fractions too?
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You are right: the question does not say the numbers are positive integers, so the numbers can be negative, fractions, decimals, etc.

But that does not affect the conclusion when the statements are taken together.

With statement (2), the remaining three numbers must sum to 126.

With statement (1), none of them can be greater than 50.

So if even one of the remaining three numbers were 25 or less, the largest possible sum of those three would be:

25 + 50 + 50 = 125

But the required sum is 126. Therefore, all three remaining numbers must be greater than 25.

This reasoning does not require the numbers to be positive or integers.
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If the average (arithmetic mean) of six different numbers is 25, how m 22 Jun 2026, 09:50
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a1,a2,a3,a4,a5,a6
avg is 25.
so sum is =150

find - any num greater than 25?

S-1
each of the six num are less than or equal to 50.
that gives us so many different combination. so we cant reach definite ans here.
for example, 40+41+42+ 1+2+0 = 126. so 3 num.
59+60+0+1+2+3. so 2 num.

not sufficient.

S-2
a1=7,a2=8,a3=9
7+8+9+a4+a5+a6= 150
a4+a5+a6= 126

here there can be different combnination.
not sufficient.

S-1&2

we know value of 3 num. also we know all num are less than or equal to 50.
so the remaining three num need to add up to 126.
here all three num must be in 40's. if there are in 20's or 30's, its not possible to get sum 126.
so it gives definite ans that 3 num are greater than 25.

choice C

Bunuel
If the average (arithmetic mean) of six different numbers is 25, how many of the numbers are greater than 25?

(1) None of the six numbers is greater than 50.

(2) Three of the six numbers are 7, 8, and 9, respectively.
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