Xavier Aptitude Test 2016 Paper

Question : 63 of 103

Marks: +1, -0

Two numbers in the base system B are 206

1_B

and 60

1_B

. The sum of these two numbers in decimal system is 432. Find the value of 101

0_B

in decimal system.

Solution:

(c)

\ (2061)_{\ {B}}=\ (2 × \ {B}^{3}+0 \×{B}^{2}+6 ×\ {B}+1\ )_{10}

= 2B

^{3}

+ 6B + 1 ....(1)

(601)_B

= 6 × B

^{2}

+ 1
= 6B

^{2}

+ 1 .......(2)

\∴ (2061)_{\ {B}}+(601)_{\ {B}}=2 \ {B}^{3}+6 \ {B}^{2}+6 \ {B}+2

432=2\ (\ {B}^{3}+3 \ {B}^{2}+3 \ {B}+1\ )

(B + 1)

^{3 }

= 216
(B + 1)

^{3}

= 63
B = 5
∴ 5 (1010) 5 = 1 × 5

^{3}

+ 1 × 5= 125 + 5 = (130)

^{10}

Go to Question:

Prev Question

Next Question