Mathematics Tricks & Tips of binary number for nda

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Tricks & Tips of binary number for nda

decimal to binary conversion

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 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 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 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 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 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 2805845768

The binary number corresponding to `(13.0625)_(10)` is

(A)

`(1011.0010)_2`

(B)

`(1110.0101)_2`

(C)

`(11 01.0001 )_2`

(D)

None of these

Solution:

For integer part of `( 13.0625)_(10)`

`i . e (13)_(10)`

`therefore ( 13)_(10) = ( 1101)_2`

For fractional part of ` ( 13.0625)_(10) = ( .0625)_(10)`

`0.0625xx2 = 0.125 \ \ \ \ \ \ \ 0`

`0.125xx 2 = 0.25 \ \ \ \ \ \ \ 0`

`0.25xx2 = 0.5 \ \ \ \ \ \ \ 0`

`0.5xx2 = 1.0 \ \ \ \ \ \ \ 0`

`therefore ( 13.0625)_(10) = (1101.0001)_2`

Correct Answer is `=>` (C) `(11 01.0001 )_2`

binary to decimal conversion

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 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 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 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 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 2825845761

The number `0.0011` in binary system represents

(A)

rational number 3/8 in decirnal system

(B)

rational number 1/8 in decirnai system

(C)

rational number 3/16 in decimal system

(D)

rational number 5/16 in decimal system

Solution:

`(.0011)_2 = 0xx2^(-1) + 0xx2^(-2) +1xx2^(-3) +1xx2^(-4)`

` = 1/8+1/16 = 3/16`

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

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`

Algebra of binary numbers

Q 2845045863

What is `( 1111)_2+(1001)_2-(1010)_2` equal to ?

(A)

`(111)_2`

(B)

`(1100)_2`

(C)

`(1110)_2`

(D)

`(1010)_2`

Solution:

`( 1111)_2 = 1xx2^3+1xx2^2+1xx2^3+1xx2^0 = 15`

`(1001)_2 = 1xx2^3+0+0+1xx2^0 = 9`

`(1010)_2 = 1xx2^3+0+1xx2^1+0= 10`

`therefore 15+9-10 = 14`

`therefore (14)_(10) = (1110)_2`

Correct Answer is `=>` (C) `(1110)_2`

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 2825045861

The sum of ` ( 1011.01)_2+(1001.11)_2` is

(A)

`111011`

(B)

`10001`

(C)

`10000`

(D)

`10101`

Correct Answer is `=>` (D) `10101`