Elitmus
Exam
Numerical Ability
Arithmetic
1024*128+256*2+10 when base is 4 how many 2 will occur
option a)2 b)4 c)6 d)8
Read Solution (Total 9)
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- 1024=1*(4^3)+0*(4^2)+2*(4)+4*(4^0)=76;
similarly,128=1*16+2*4+8*1=32;
256=2*16+5*4+6*1=58;
2=2*1=2;
10= 1*4+0*1=4;
result= (76*32+58*2+4)=(2432+116+4)=2552
number of 2 in 2552 is 2 then
a) 2 is answer - 10 years agoHelpfull: Yes(13) No(10)
- first multiply all the no. the use the basic principle of base conversion means divde the whole no. to the 4 and taking remainder..................
- 10 years agoHelpfull: Yes(4) No(3)
- b option.. 4 ans
- 10 years agoHelpfull: Yes(3) No(2)
- ans is b beause
- 10 years agoHelpfull: Yes(0) No(3)
- ans. is (d)8
- 10 years agoHelpfull: Yes(0) No(1)
- Answer is 4 and is correct .. 100% sure
- 9 years agoHelpfull: Yes(0) No(0)
- those who supporting option d..please clarify with the valid statement..cos am also intution with option (d) and (b) as well
- 9 years agoHelpfull: Yes(0) No(0)
- answer is two just by basic conversion law - solve it to decimal and than convert it to base 4
- 9 years agoHelpfull: Yes(0) No(0)
- 1028=4^5
128=2x4^3
1028*128=2*4^8
256*2=2*4^4
10=2*4^1+2*4^0
now combine all this
2*4^8+2*4^4+2*4^1+2*4^0
so ans is a) 2 - 9 years agoHelpfull: Yes(0) No(0)
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