Mathematics Must Do Problems Of Binary Numbers For NDA

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Must Do Problems Of Binary Numbers For NDA

Q 2875745666

Binary equivalent of `182` is

(A)

`(10110110)_2`

(B)

`(111000110)_2`

(C)

`(10111001)_2`

(D)

None of these

Solution:

`therefore ( 182)_(10) = ( 10110110)_2`

Correct Answer is `=>` (A) `(10110110)_2`

Q 2337434382

Which one of the following binary numbers is the
prime number?
NDA Paper 1 2007

(A)

`111101`

(B)

`111010`

(C)

`111111`

(D)

`100011`

Solution:

By help of options (a)

`111101 = 1 xx 2^5 + 1 xx 2^4 + 1 xx 2^3 + 1 xx 2 ^2 + 0 xx 2^1 + 1 xx 2^0`

`= 32 + 16 + 8 + 4 + 0+ 1 = 61`

which is a prime number.

(b) `111010 = 1 xx 2^5 + 1 xx 2^4 + 1 xx 2^3 + 0 xx 2^2`

`1 xx 2^1 + 0 xx 2^0`

`= 32 + 16 + 8 + 0 + 2 + 0 = 58`

which is not a prime number.

(C) `111111 = 1 xx 2^5 + 1 xx 2^4 + 1 xx 2^3 + 1 xx 2^2`

`+ 1 xx 2^1 + 1 xx 2^0`

`= 32 + 16 + 8 + 4 + 2 + 1 = 63`

which is not a prime number.

(d) `100011 = 1 xx 2^5 + 0 xx 2^4 + 0 xx 2^3 + 0 xx 2^2`

`+ 1 xx 2^1 + 1 xx 2^0`

`= 32 + 0 + 0 + 0 + 2 + 1 = 35`

which is not a prime number.

Correct Answer is `=>` (A) `111101`

Q 2307134088

The number `0.0011` in binary system represents
NDA Paper 1 2009

(A)

rational number 3/8 in decimal system

(B)

rational number 1/8 in decimal system

(C)

rational number 3/16 in decimal system

(D)

rational number 5/16 in decimal system

Solution:

`(0.0011) = 0 xx 1/2 + 0 xx 1/2^2 + 1 xx 1/2^3 + 1 xx 1/2^4`

`= 0 + 0 + 1/8 + 1/(16) = 3/(16)`

Correct Answer is `=>` (C) rational number 3/16 in decimal system

Q 2875045866

The decimal number corresponding to the binary number `(111000.0101)_2` is

(A)

`(56.3275)_(10)`

(B)

`(56.3125)_(10)`

(C)

`(57.4375)_(10)`

(D)

`(57.5625)_(10)`

Solution:

`(111000.0101)_2 = 1xx2^5+1xx2^4+1xx2^3+0xx2^2+0xx2^1+0xx2^0+0xx2^(-1)+1xx2^(-2)+0xx2^(-3)+1xx2^(-4)`

` = 32+16+8+1/4+1/16`

` = 56+0.25+0.0625 = ( 56.3125)_(10)`

Correct Answer is `=>` (B) `(56.3125)_(10)`

Q 2307434388

What is the product of the binary numbers
`1001.01` and `11.1`?
NDA Paper 1 2007

(A)

`101110.011`

(B)

`100000.011`

(C)

`101110.101`

(D)

`100000.101`

Solution:

`1001.01 = 1 xx 2^3 + 0 xx 2^2 + 0 xx 2^1 + 1 xx 2^0`

`+ 0 xx 2^(-1) + 0 xx 2^(-2)`

`= 8 + 0+ 0 +1 + 0 + 1/4` .

`= (37)/4 = 9.25`

and `11.1 = 1 xx 2^1 + 1 xx 2^0 + 1 xx 26(-1)`

`= 2 + 1 + 1/2 = 7/2 = 3.5`

`:. 1001.01 xx 11.1 = 9.25 xx 3.5`

` = 32.375`

Now,`(32 )_(10) = (100000)_2`

`0.375 xx 2 = 0.75`

`0.75 xx 2 = 1.5`

`0.5 xx 2 = 1.0`

and `(0.375)_(10) = (0.011 )_2`

`:. (32.375)_(10) = (100000.011)_2`

Correct Answer is `=>` (B) `100000.011`

Q 2357134084

If `(10 x 010)_2 - (11y1)_2 = (10z11)_2` , then what are
the possible values of the binary digits `x, y` and `z`,
respectively?
NDA Paper 1 2009

(A)

`0, 0, 1`

(B)

`0, 1, 0`

(C)

`1, 1, 0`

(D)

`0, 0, 0`

Solution:

`(10x010)_2 - (11y1)_2 = (10z11)_2`

`= (2^5 xx 1 + 0 xx 2^4 + x xx 2^3 + 0 xx 2^2 + 1 xx 2^1 + 0 xx 2^0)`

`- (2^3 xx 1 + 2^2 xx 1 + y xx 2^1 + 1 xx 2^0)`

`= 2^4 xx 1 + 0 xx 2^3 + 2^2 xx z + 2^1 xx 1 + 2^0 xx 1`

`=> (34 + 8x) - (13 + 2y ) = 19 + 4z`

`=> 2 = - 8x + 2y + 4z`

`=> x = 0, y = 1 , z = 0`

Correct Answer is `=>` (B) `0, 1, 0`