Syntel
Company
Numerical Ability
Number System
When numbers are written in base b, we have 12 x 25 = 333, the value of b is?
a) 8
b) 6
c) None
d) 7
Read Solution (Total 7)
-
- ans d) 7
12 ,25 and 333 in decimal form can be written as (1*b^1 + 2*b^0) , (2*b^1 + 5*b^0), (3*b^2+3*b^1 + 3*b^0)
12 *25 = 333
(1*b^1 + 2) * (2*b^1 + 5*) = (3*b^2+3*b^1 + 3)
b^2+ 9b + 10 = 3b^2 + 3b + 3
b^2 -6b -7 = 0
b^2 - 7b + b - 7 =0
(b-7)(b+1)=0
b=7 or b=-1
base cannot be a negative value so b=7 - 9 years agoHelpfull: Yes(20) No(3)
- option d 7.
- 9 years agoHelpfull: Yes(3) No(1)
- (1*b^1+2)*(2*b^1+5)=(3*b^2+3*b+3)
(b+2)(2b+5)=3b^2+3b+3
b^2 -6b -7=0
b=7 or b=-1 - 9 years agoHelpfull: Yes(3) No(0)
- ans d) 7
12 ,25 and 333 in decimal form can be written as (1*b^1 + 2*b^0) , (2*b^1 + 5*b^0), (3*b^2+3*b^1 + 3*b^0)
12 *25 = 333
(1*b^1 + 2) * (2*b^1 + 5*) = (3*b^2+3*b^1 + 3)
b^2+ 9b + 10 = 3b^2 + 3b + 3
b^2 -6b -7 = 0
b^2 - 7b + b - 7 =0
(b-7)(b+1)=0
b=7&-1
- 9 years agoHelpfull: Yes(1) No(2)
- answer is 7
- 8 years agoHelpfull: Yes(0) No(0)
- Ans is 7 [(1*b^1+2)(2*b^1+5)=(3*b^2+3b^1+3)]
- 8 years agoHelpfull: Yes(0) No(0)
- 7
(b+2)(2b+5)=3b^2+3b+3 - 8 years agoHelpfull: Yes(0) No(0)
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