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25 Nov 2019, 01:16
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
24% (02:25) correct
76%
(02:27)
wrong
based on 209
sessions
History
Date
Time
Result
Not Attempted Yet
Between 100 and 200, how many numbers are there in which one digit is the average of the other two?
A. 11
B. 12
C. 10
D. 8
E. 9
Are You Up For the Challenge: 700 Level Questions
A. 11
B. 12
C. 10
D. 8
E. 9
Are You Up For the Challenge: 700 Level Questions
3 digits are in arithmetic progression, and Hundreds digit is always 1.
1. when common difference is 0
three digits are 1,1 and 1 {111}
2. when common difference is 1
three digits are 1,2 and 3 or 0, 1 and 2 {102, 120, 123 or 132}
3. when common difference is 2
three digits are 1,3 and 5 {153 or 135}
4. when common difference is 3
three digits are 1,4 and 7 {147 or 174}
5. when common difference is 4
three digits are 1,5 and 9 {159 or 195}
Total possible numbers= 1+2*5=11
1. when common difference is 0
three digits are 1,1 and 1 {111}
2. when common difference is 1
three digits are 1,2 and 3 or 0, 1 and 2 {102, 120, 123 or 132}
3. when common difference is 2
three digits are 1,3 and 5 {153 or 135}
4. when common difference is 3
three digits are 1,4 and 7 {147 or 174}
5. when common difference is 4
three digits are 1,5 and 9 {159 or 195}
Total possible numbers= 1+2*5=11
Bunuel
General Discussion
Updated on: 25 Nov 2019, 03:39
Originally posted by Archit3110 on 25 Nov 2019, 03:20.
Last edited by Archit3110 on 25 Nov 2019, 03:39, edited 2 times in total.
Last edited by Archit3110 on 25 Nov 2019, 03:39, edited 2 times in total.
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took >2 mins to solve this one .. did it conventional way
possible values
111, 132,153,174,195,123,135,147,159,
IMO E; 9
possible values
111, 132,153,174,195,123,135,147,159,
IMO E; 9
Bunuel
)) 
