IBM
Company
Numerical Ability
Sequence and Series
81,162,49,98,25,50,64
Read Solution (Total 15)
-
- [81,162],[49,98],[25,50],[64,128]
81*2=162
49*2=98
25*2=50
64*2=128
....:)
- 10 years agoHelpfull: Yes(34) No(1)
- 81*2=162,49*2=98,25*2=50,64*2=128
ans- 128 - 10 years agoHelpfull: Yes(2) No(0)
- 81*2=162
49*2=98
25*2=50
64*2=128 - 10 years agoHelpfull: Yes(1) No(0)
- 128
number is odd differ by 2,
9^2,9^2*2.7^2,7^2*2,5^2,5^2*2,if number=5,then
number is even difffer by 2 but starting number first even number 8^2,8^2*2 - 10 years agoHelpfull: Yes(0) No(0)
- 81*2,49*2,25*2,64*2
- 10 years agoHelpfull: Yes(0) No(0)
- (81,162)(49,98)(25,50)(64,--)
clearly 81*2=162
49*2=98 and 25*2=50 so 64*2=128
Answer=128 - 10 years agoHelpfull: Yes(0) No(0)
- 9^2,9^2*2,7^2,7^2*2,5^2,5^2*2,8^2,8^2*2=128
ans: 128 - 10 years agoHelpfull: Yes(0) No(0)
- 81+81=162
49+49=98
Ans 64+64=128
sa - 10 years agoHelpfull: Yes(0) No(0)
- 81*2===>162
49*2===>98
25*2===>50
64*2===>128 - 10 years agoHelpfull: Yes(0) No(0)
- 128
64*2=128 - 10 years agoHelpfull: Yes(0) No(0)
- 81*2=162,
49*2=98,
25*2=50,
64*2=128.
Ans=128 - 10 years agoHelpfull: Yes(0) No(0)
- 81*2=162
49*2=98
25*2=50 then
64*2=128 the answer is 128
- 10 years agoHelpfull: Yes(0) No(0)
- 81*2=162
49*2=98
25*2=50
64*2=128 - 10 years agoHelpfull: Yes(0) No(0)
- 81*2=162
49*2=98
25*2=50
so 64*2=128 - 10 years agoHelpfull: Yes(0) No(0)
- 81*2=162,49*2=98,25*2=50,64*2=128
- 10 years agoHelpfull: Yes(0) No(0)
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