Mathematics
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Previous Year questions Of Binary Number For NDA

Previous Year questions
Q 2771267126

If the number 235 in decimal system is converted into binary system , then what is the resulting number ?
NDA Paper 1 2016
(A)

`(11110011)_2`

(B)

`(11101011)_2`

(C)

`(11110101)_2`

(D)

`(11011011)_2`

Solution:

See image.
Correct Answer is `=>` (B) `(11101011)_2`
Q 2210367210

What is `(1000000001) _(2) - (0.0101)_( 2)` equal to?
NDA Paper 1 2015
(A)

`(512.6775)_(10)`

(B)

`(512.6875)_(10)`

(C)

`(512.6975)_(10)`

(D)

`(512.0909)_(10)`

Solution:

`(1000000001)_(2) - (0.0101)_(2)`

`= (2^(9) + 2^(0)) - (1 xx 2^(- 2) + 1 xx 2^(- 4) )`

`= (512 + 1)- (1/4 + 1/(16))`

` = 513- 5/(16) = 513 - 0.3125 = (512.6875)_(10)`
Correct Answer is `=>` (B) `(512.6875)_(10)`
Q 2303523448

The number 251 in decimal system is expressed in binary system by
NDA Paper 1 2014
(A)

`11110111`

(B)

`11111011`

(C)

`11111101`

(D)

`11111110`

Solution:

`:. (251)_2 = (text(1 1 1 1 1 0 1 1))_10`

which is the required decimal system.
Correct Answer is `=>` (B) `11111011`
Q 2377823786

The decimal representation of the number `(1011)_2`
in binary system is
NDA Paper 1 2012
(A)

`5`

(B)

`7`

(C)

`9`

(D)

`11`

Solution:

`(1011)_2 = 1 xx 2^3 + 0 xx 2^2 + 1 xx 2^1 + 1 xx 2°`

`= 8 + 0 + 2 + 1 = 11`
Correct Answer is `=>` (D) `11`
Q 2307823788

The decimal number `(57.375)_(10)` when converted
to binary number takes the form
NDA Paper 1 2012
(A)

`(111001.011)_2`

(B)

`(100111.110)_2`

(C)

`(110011.101)_2`

(D)

`(111011.011)_2`

Solution:

`:. (57)_(10) = (111001)_2`

Now,

`0.375 xx 2 = 0.75`

` 0.75 xx 2 = 1.5`

`0.5 xx 2 = 1.0`

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

`:. (57.375)_(10) = (111001.011)_2`
Correct Answer is `=>` (A) `(111001.011)_2`
Q 2317823789

The number `292` in decimal system is expressed in
binary system by
NDA Paper 1 2012
(A)

`100001010`

(B)

`100010001`

(C)

`100100100`

(D)

`101010000`

Solution:

`:. (292)_(10) = (100100100)_2`
Correct Answer is `=>` (C) `100100100`
Q 2337023882

What is the decimal number representation of the
binary number `(11101.001)_2` ?
NDA Paper 1 2012
(A)

`30.125`

(B)

`29.025`

(C)

`29.125`

(D)

`28.025`

Solution:

`(11101.001)_2 = 1 xx 2^4 + 1 xx 2^3 + 1 xx 2^2 + 0 xx 2^1`

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

`= 16 + 8+ 4+ 0+ 1 + 0 + 0 + 1/8`

`= 29 + 1/8 = (233)/8 = (29.125)_(10)`
Correct Answer is `=>` (C) `29.125`
Q 2367123985

What is the equivalent binary number of the
decimal number `13.625`?
NDA Paper 1 2010
(A)

`1101.111`

(B)

`1111.101`

(C)

`1101.101`

(D)

1111.111

Solution:

`13.625`

`0.625 xx 2 = 1.250`

`0.250 xx 2 = 0.5`

`0.5 xx 2 = 1`

`:. 13.625 = 1101.101`
Correct Answer is `=>` (C) `1101.101`
Q 2307123988

What is the value of

` ( (0.101)_2 ^((11 )_2) + (0.011)_2 ^((11 ) 2) )/( (0.101)_2^((10)_2) - (0.101)_2^((01 )_2) (0.011)_2 ^((01)_2) + (0.011)_2^((10)_2))` ?

NDA Paper 1 2010
(A)

`(0.001 )_2`

(B)

`(0.01 )_2`

(C)

`(0.1 )_2`

(D)

`(1 )_2`

Solution:

`(0.101)_2 = 2^(-1) xx 1 + 2^(-2) xx 0 + 2^(-3) xx 1`

`= 1/2 + 0 + 1/8 = 5/8`

and `(0.011)_2 = 0 xx 2^(-1) + 1 xx 2^(-2) + 1 xx 2^(-3)`

`= 0 + 1/4 + 1/8 = 3/8`

Also, `(11)_2 = 1 xx 2^1 + 1 xx 2^0 = 3`

`(10)_2 = 1 xx 2^1 + 0 xx 2^0 = 2`

and `(01)_2 = 0 xx 2^1 + 2^0 xx 2^0 = 1`

`:. ( (0.101)_2 ^((11 )_2) + (0.011)_2 ^((11 ) 2)) /( (0.101)_2^((10)_2) - (0.101)_2^((01 )_2) (0.011)_2 ^((01)_2) + (0.011)_2^((10)_2))`

` = ((5/8)^3 + (3/8)^3)/((5/8)^2 - (5/8) (3/8) + (3/8)^2)`

` = 5/8 + 3/8 = 8/8 = (1)_(10) = (1)_2`

` [ ∵ a^3 + b^3 = ( a + b) (a^2 - ab + b^2)]`
Correct Answer is `=>` (D) `(1 )_2`
Q 2327134081

If `x = (1101)_2` and `y = (110)_2` , then what is the
value of `x^2 - y^2`?
NDA Paper 1 2009
(A)

`(1000101)_2`

(B)

`(10000101)_2`

(C)

`(10001101)_2`

(D)

`(10010101)_2`

Solution:

Given,

` x = (1101)_2 = 1 xx 2^3 + 1 xx 2^2 + 0 xx 2^1 + 1 xx 2^0`

`= 8 + 4 + 0 + 1 = 13`

and `y = (110)_2 = 1 xx 2^2 + 1 xx 2^1 + 0 xx 2^0`

`= 4 + 2+ 0 = 6`

` :. x^2 - y^2 = (13)^2 - (6)^2 = 169 - 36 = 133`

`:. 133 = (10000101)_2`
Correct Answer is `=>` (B) `(10000101)_2`
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`
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 2327234181

What is the binary equivalent of decimal number
`(0.8125)_(10)`?
NDA Paper 1 2009
(A)

`(0.1101)_2`

(B)

`(0 1001)_2`

(C)

`(0.1111)_2`

(D)

`(0.1011)_2`

Solution:

` ∵ (0.1101)_2 = 1 xx 2^(-1) + 1 xx 2^(-2) + 0 xx 2^(-3) + 1 xx 2^(-4)`

` = 1/2 + 1/4 - 1/(16) = (13)/(16)`

`= (0.8125)_(10)`

Hence, `(0.8125)_(10) = (0.1101 )_2`

Alternate Method

`0.8125 xx 2 = 1.625`

`0.6125 xx 2 = 1.25`

`0.25 xx 2 = 0.50`

`0.5 xx 2 = 1.0`

`=> (0.8125)_(10) = (0.1101)_2`
Correct Answer is `=>` (A) `(0.1101)_2`
Q 2387234187

What is the decimal equivalent of `(101.101)_2` ?
NDA Paper 1 2009
(A)

`(5.225)_(10)`

(B)

`(5.525)_(10)`

(C)

`(5.625)_(10)`

(D)

`(5.65)_(10)`

Solution:

`(101.101)_2 = 1 xx 2^2 + 0 xx 2^1 + 1 xx 2^0 + 1 xx 2^(-1)`

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

`= 4 + 0 + 1 + 1/2 + 0 + 1/8`

`= (40 + 4 + 1)/8 = (45)/8 = (5.625)_(10)`
Correct Answer is `=>` (C) `(5.625)_(10)`
Q 2327334281

The binary number `0.111111 ...` (where, the digit
`1` is recurring) is equivalent in decimal system to
which one of the following?
NDA Paper 1 2008
(A)

`1/(10)`

(B)

`(11)/(10)`

(C)

`1`

(D)

`(10)/(11)`

Solution:

`0.111111...`

`= 2^(-1) + 2^(-2) + 2^(-3) + 2^(-4) + 2^(-5) + 2^(-6) + ...`

` = 1/2 + 1/4 + 1/8 + 1/(16) + 1/(32) + 1/(64) + ....`

`= (32 + 16 + 8 + 4 + 2 + 1)/(64) + ...`

`= (63)/(64) + ... = 0.9843 + ... = 1`
Correct Answer is `=>` (C) `1`
Q 2367334285

The difference of two numbers `10001100` and
`1101101 ` in binary system is expressed in decimal
system by which one of the following?
NDA Paper 1 2008
(A)

`27`

(B)

`29`

(C)

`31`

(D)

`33`

Solution:

`10001100 = 1 xx 2^7 + 0 xx 2^6 + 0 xx 2^5 + 0 xx 2^4`

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

`= 128 + 0 + 0 + 0 + 8 + 4 + 0 + 0 = 140`

and `1101101 = 1 xx 2^6 + 1 xx 2^5 + 0 xx 2^4 + 1 xx 2^3`

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

`= 64 + 32 + 0+ 8 + 4 + 0+ 1`

`= 109`

`:.` Required difference `= 140 - 109 = 31`
Correct Answer is `=>` (C) `31`
Q 2387334287

The multiplication of the number `(10101 )_2` by
`(1101 )_2` yields which one of the following?
NDA Paper 1 2007
(A)

`(100011001)_2`

(B)

`(100010001)_2`

(C)

`(110010011)_2`

(D)

`(100111001)_2`

Solution:

`∵ (10101)_2 = 2^4 xx 1 + 0 xx 2^3 + 1 xx 2^2 + 0 xx 2^1`

`+ 1 xx 2^0`

`= 16 + 0 + 4 + 0 + 1 = 21`

and `(1101)^2 = 1 xx 2^3 + 1 xx 2^2 + 0 xx 2^1 + 1 xx 2^0`

`= 8 + 4 + 0 + 1 = 13`

`:. (10101)_2 xx (1101)_2 = 21 xx 13 = 273`

`= (100010001)_2`
Correct Answer is `=>` (B) `(100010001)_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 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`
 
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