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16 Jan 2020, 08:19
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
56% (02:27) correct
44%
(02:07)
wrong
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There are four numbers in sequence. The average of first three is 6; the average of the last three is 7 and the last number is 3 more than the first. The average of second and third number is:
A) 6
B) 6.5
C) 5.75
D) 7
E) Indeterminate
A) 6
B) 6.5
C) 5.75
D) 7
E) Indeterminate
16 Jan 2020, 20:48
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Since numbers are in a sequence, they are in some order, but repetition is allowed.
Now, the numbers are in sequence and last number is 3 more than the first means all can be consecutive integers - a, a+1, a+2, a+3
Or when repetition is there it can be a, a, a, a+3
The average of first three is 6, but the average of last three is 7.
1) Now we can easily have a sequence of consecutive numbers 5,6,7,8 satisfying all given conditions and average of 6 and 7 is 6.5
2) There could be repetitions too => 6,6,6,9. This also satisfies all conditions but the average of 6 and 6 is 6
E
16 Jan 2020, 21:41
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"The average of first three is 6" - I think to myself that perhaps all 3 numbers are 6. I am not given that they are distinct.
"the average of the last three is 7" - So I have 6, 6 and a number to give average 7. The two 6s are 1 each less than 7 so the next number will be 2 more than 7 i.e. 9.
Then the numbers become 6, 6, 6, 9. The last number is 3 more than the first so this works.
Now, indeterminate is also in the options so I will try to find if 6.5 can be the answer since it lies between 6 and 7.
5, 6.5, 6.5, 8
This is also possible.
Hence, I cannot uniquely find the average of second and third.
Answer (E)
