IBM
Company
Logical Reasoning
Number Series
1/3,1/5,3/25,9/125,_
a)125/27,b)81/125,c)12/625,d)27/625
Read Solution (Total 10)
-
- every term is multiplication of 3/5.
ans = 27/625
- 10 years agoHelpfull: Yes(37) No(2)
- d)27/625
solutn..
1/3
1/5=1/3* 3/5
3/25=1/5*3/5
9/125=3/25*3/5
so
27/625=9/125*3/5 - 10 years agoHelpfull: Yes(4) No(1)
- 1/5, 3/5^2,3^2/5^3,3^3/5^4 so the answer is 27/625
- 10 years agoHelpfull: Yes(2) No(4)
- each term is multiplication of 3/5.
ans = 27/625
- 10 years agoHelpfull: Yes(1) No(0)
- 1/3*3/5=1/5
1/5*3/5=3/25
3/25*3/5=27/625
each num is multiplied by 3/5 - 10 years agoHelpfull: Yes(1) No(2)
- d)3^-1/5^0,3^0/5^1,3^1/5^2....3^3/5^4
- 10 years agoHelpfull: Yes(1) No(1)
- 27/625
- 10 years agoHelpfull: Yes(1) No(1)
- 1/5=sqreroot of(1/3*3/25)
3/25=squre root of(1/5*9/125)
9/125=squre root of(3/25*27/625)
So ans is 27/625 - 10 years agoHelpfull: Yes(0) No(1)
- 1/3*3/5=1/5,1/5*3/5=3/25.........9/125*3/5=27/625
- 10 years agoHelpfull: Yes(0) No(0)
- ans is d)27/625
each term is mul by 3/5 - 9 years agoHelpfull: Yes(0) No(0)
IBM Other Question
1,1,4,8,14,42,_
a)84,b)57,c)51,d)48
21,22,21,22,23,22,23,24,23,24,_