IBM
Company
Numerical Ability
Sequence and Series
4,2,5,9,5,11,13,7,16,17,9,?
Read Solution (Total 9)
-
- 4,2,5,9,5,11,13,7,16,17,9,?
see the triplets
4,2,5 -> diff b/w 4,5 = 1
9,5,11 -> diff b/w 9,11 = 2
13,7,16 -> diff b/w 13,16 = 3
17,9,21 -> diff b/w 17,21 = 4
so the answer should be 21 - 12 years agoHelpfull: Yes(82) No(6)
- answer is 21
break the series in three series as following-
4,9,13,17
2,5,7,9
5,11,16,?
common diff is -5,-4,-4 in 1st
common diff is +3,+2,+2 in 2nd
common diff is -6,+5, x in 3rd ...so x must be +5
so, next term = 16+5 = 21 - 12 years agoHelpfull: Yes(23) No(6)
- answer is 21..
4+2=6-1=5
9+5=14-3=11
13+7=20-4=16
17+9=26-5=21..
Hence the series can be continued like this way... - 12 years agoHelpfull: Yes(10) No(16)
- Ans is 21.
logic is
(4,2,5)
4-2=2+3=5
(9,5,11)
9-4=5+6=11
(13,7,16)
13-6=7+9=16
(17,9,?)
17-8=9+12=21
- 10 years agoHelpfull: Yes(10) No(2)
- The answer is 21.
Break the series into 4 parts
4 2 5
9 5 11
13 7 16
17 9 ?
and work vertically as follows.
1st v-line 4+9=13, 13+[4]=17;
2nd v-line 2+5=7, 7+[2]=9;
3rd v-line 5+11=16, 16+[5]=21.
- 10 years agoHelpfull: Yes(7) No(0)
- 4-2=2, 9-5=4, 13-7=6, 17-9=8
5+6=11, 11+5=16, 16+4=20. ans=20 - 10 years agoHelpfull: Yes(2) No(11)
- 21,taking 3 digits(4,2,5)(9,5,11),(13,7,16),(17,9,?(21))
4+1=5,
9+2=11,
13+3=16,
17+4=21. - 8 years agoHelpfull: Yes(2) No(0)
- ANS 21
4,2,5 -> diff b/w 4,5 = 1
9,5,11 -> diff b/w 9,11 = 2
13,7,16 -> diff b/w 13,16 = 3
17,9,21 -> diff b/w 17,21 = 4
- 8 years agoHelpfull: Yes(0) No(0)
- Break the series into 4 parts
4 2 5 => (2+2,2,2+3)
9 5 11 => (4+5,5,5+6)
13 7 16 => (6+7,7,7+9)
17 9 ? => (8+9,9,9+12)
so ans is 21 - 7 years agoHelpfull: Yes(0) No(0)
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