MCA NIMCET Number System Previous Year Question Paper and Solution with PDF

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MCA NIMCET Number System PYQ

Qus : 1 MCA NIMCET PYQ

A group of 630 children is arranged in rows for a group photograph. Each row contains three fewer children than the row in front of it. What number of rows is not possible?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Let the first row have $n$ children and total rows be $r$. Then the numbers per row form an AP with difference $-3$ and sum $630$:

$\displaystyle \frac{r}{2}\big(2n-3(r-1)\big)=630$ $ \;\Rightarrow\; 1260=r\big(2n-3(r-1)\big)$.

Hence $r\mid1260$. Also the last row must be positive: $n-3(r-1)>0$.

Testing the options (and ensuring $n$ is integer and last term positive):

$r=4$: $1260/4=315$, $2n=315+3\cdot3=324 \Rightarrow n=162$, last term $=162-9=153>0$ ✓
$r=5$: $1260/5=252$, $2n=252+3\cdot4=264 \Rightarrow n=132$, last term $=132-12=120>0$ ✓
$r=6$: $1260/6=210$, $2n=210+3\cdot5=225 \Rightarrow n=112.5$ (not integer) ✗

Therefore, the impossible number of rows is $\boxed{6}$.

Qus : 2 MCA NIMCET PYQ

What are the values of R₁ and R₂ respectively in the expression (235)_R1 = (565)₁₀ = (1065)_R2

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Solution

Qus : 3 MCA NIMCET PYQ

The number (2217)₈ is equivalent to

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Solution

$(2217)_8 \rightarrow (?)_{16}$

Octal to Binary (har digit = 3 bits)

$2 \rightarrow 010$
$2 \rightarrow 010$
$1 \rightarrow 001$
$7 \rightarrow 111$

$(2217)_8 = (010010001111)_2$

Binary to Hex (right se 4-4 group karo)

$\underbrace{0100}_{4}\ \underbrace{1000}_{8}\ \underbrace{1111}_{F}$

$= (48F)_{16} = (048F)_{16}$

$\boxed{\text{Answer : } (048F)_{16}}$ ✓

Qus : 4 MCA NIMCET PYQ

What is the equivalent decimal value of the unsigned binary number 10101.10101?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

We separate the integer and fractional parts.

Integer part: $10101_2 = 1\times2^4 + 0\times2^3 + 1\times2^2 + 0\times2^1 + 1\times2^0 $$= 16 + 0 + 4 + 0 + 1 = 21$

Fractional part: $.10101_2 = 1\times2^{-1} + 0\times2^{-2} + 1\times2^{-3} + 0\times2^{-4} + 1\times2^{-5}$

= 0.5 + 0 + 0.125 + 0 + 0.03125

= 0.65625

Therefore, $10101.10101_2 = 21 + 0.65625 = {21.65625}$

Qus : 5 MCA NIMCET PYQ

The binary multiplication 00*11 will give

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Solution

✅ Concept:

Binary multiplication is done similarly to decimal multiplication, where each bit is multiplied individually. Let's break down $ 00_2 \times 11_2 $ step by step.

✅ Step 1: Understanding Binary Multiplication

We multiply each digit in the first binary number by each digit in the second binary number. The multiplication follows the same rules as decimal multiplication but with only 0's and 1's.

✅ Step 2: Perform the Multiplication

Let's multiply the two binary numbers:

First step: $ 0 \times 1 = 0 $
Second step: $ 0 \times 1 = 0 $
Third step: $ 0 \times 1 = 0 $

✅ Step 3: Adding the Results

Since all the results are 0, the final multiplication result is also 0.

✅ Final Answer:

The result of $ 00_2 \times 11_2 $ is: 0

Qus : 6 MCA NIMCET PYQ

The multiplication of the following two binary numbers $1100$ and $1011$ in binary notation is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2026 PYQ

Solution

First convert both binary numbers into decimal.

$1100_2=12$

$1011_2=11$

Now multiply:

$12\times 11=132$

Convert $132$ into binary:

$132=128+4$

$128=2^7$ and $4=2^2$

So,

$132=10000100_2$

Therefore,

$1100_2\times 1011_2=10000100_2$

Qus : 7 MCA NIMCET PYQ

The arithmetic mean of 2¹⁰ and 2²⁰ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 8 MCA NIMCET PYQ

In 8-bit two's complement arithmetic, compute the result of the following addition: A = 10011001, B = 11010111. What is the resulting 8-bit binary value?

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Solution

Let’s first convert both binary numbers into decimal (two’s complement form):

$A = 10011001_2$ → negative because MSB = 1 → invert bits: $01100110_2 = 102_{10}$ → add 1 → $103_{10}$ So, $A = -103$.
$B = 11010111_2$ → MSB = 1 (negative) → invert bits: $00101000_2 = 40_{10}$ → add 1 → $41_{10}$ So, $B = -41$.

Add them:

$A + B = -103 + (-41) = -144$

Now, in 8-bit two’s complement, the range is $-128$ to $+127$. Since $-144$ is out of range, overflow occurs.

But let’s compute the 8-bit result (ignoring overflow):

$10011001 + 11010111 =$ 10011001
+ 11010111
= 101110000 (9 bits)

Drop the carry beyond 8 bits → 01110000.

Therefore, the resulting 8-bit binary value is:

✅ Result = 01110000₂

(Overflow occurred, actual signed result would have been −144, but the 8-bit stored value is +112.)

Qus : 9 MCA NIMCET PYQ

Which of the following statements about ASCII and Unicode is correct?

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Solution

Solution:

Option (1) – True: Unicode was designed to be backward compatible with ASCII. The first 128 Unicode characters (U+0000 to U+007F) are identical to ASCII, ensuring compatibility.
Option (2) – False: ASCII uses 7 bits per character, while Unicode uses up to 16 bits or more (e.g., UTF-16, UTF-32 encodings).
Option (3) – False: Unicode and ASCII share the same first 128 characters; they are not completely different.
Option (4) – False: Unicode can represent far more characters than ASCII, not fewer.

✅ Correct Answer: (1) Unicode is backward compatible with ASCII and includes all ASCII characters in its encoding.

Qus : 10 MCA NIMCET PYQ

What is the hexadecimal representation of the decimal number 265?

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Solution

Convert decimal 265 to hexadecimal:

Divide by 16 repeatedly:

$265 ÷ 16 = 16$ remainder $9$
$16 ÷ 16 = 1$ remainder $0$
$1 ÷ 16 = 0$ remainder $1$

Now write remainders from last to first → $1\,0\,9$

✅ Hexadecimal representation = 0x109

Correct Option: 0x109

Qus : 11 MCA NIMCET PYQ

How many numbers from 1 to 100 are such each of which is divisible by 8 and whose at least one digit is 8?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Multiples of 8 up to 100:

8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96

Now check which contain digit 8:

Count = 4

Qus : 12 MCA NIMCET PYQ

The decimal equivalent of the hexadecimal operation A10 + B21 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 13 MCA NIMCET PYQ

Multiplication of 111₂ by 101₂ is

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Solution

Qus : 14 MCA NIMCET PYQ

The result of multiplication of the numbers (10101)₂ and (11101)₂ in hexadecimal form is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 15 MCA NIMCET PYQ

The binary equivalent of (531.53125)₁₀ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 16 MCA NIMCET PYQ

The binary equivalent of (234.125)₁₀?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 17 MCA NIMCET PYQ

Determine the octal equivalent of (432267)₁₀?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 18 MCA NIMCET PYQ

One Exabyte is equal to …

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Solution

Qus : 19 MCA NIMCET PYQ

The logic XOR operation of (4AC0)₁₆ and (B53F)₁₆ results

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 20 MCA NIMCET PYQ

If all digits 6 in the numbers from 1 to 100 are replaced by 9, by how much does the algebraic sum change?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Each digit 6 → 9, increase = 3. Count how many times digit 6 appears from 1 to 100. Two-digit numbers (10–99): Tens place = 6 → numbers 60–69 → 10 times Units place = 6 → appears once in each decade (16, 26, ..., 96) → 9 times Total digit 6 occurrences = $10 + 9 = 19$ Additionally, number 6 itself contributes 1 more occurrence. So total 6s = $19 + 1 = 20$ Increase in sum = $20 \times 3 = 60$ But this does not match options → Check again carefully. Correct counting: Tens place occurrences: 60–69 → 10 Units place occurrences: 6, 16, 26, 36, 46, 56, 66, 76, 86, 96 → 10 Total = 20 occurrences. Increase = $20 \times 3 = 60$ But total sum change involves positional value: A “6” in tens place increases number by 30 A “6” in ones place increases by 3 Count tens-place 6s = 10 → total = $10 \times 30 = 300$ Count ones-place 6s = 10 → total = $10 \times 3 = 30$ Total change = $300 + 30 = 330$

Qus : 21 MCA NIMCET PYQ

Find the unit digit of $(13647)^{3265}$.

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Solution

Unit digit depends on unit digit of base: 7.

Powers of 7 cycle every 4:

$7^1 = 7$

$7^2 = 49$ → 9

$7^3 = 343$ → 3

$7^4 = 2401$ → 1

Cycle: 7, 9, 3, 1

Find exponent mod 4: $3265 \bmod 4 = 1$

So unit digit = 7.

Qus : 22 MCA NIMCET PYQ

Equivalent of the decimal number (25.375)10 in binary form

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Solution

Decimal to Binary Conversion

? Given: Convert (25.375)₁₀ to binary.

? Step-by-step:

Integer Part (25):
Divide by 2 repeatedly:

25 ÷ 2 = 12 remainder 1  
12 ÷ 2 = 6  remainder 0  
6  ÷ 2 = 3  remainder 0  
3  ÷ 2 = 1  remainder 1  
1  ÷ 2 = 0  remainder 1

→ Binary: 11001

Fractional Part (0.375):
Multiply by 2 repeatedly:

0.375 × 2 = 0.75   → 0  
0.75 × 2  = 1.5    → 1  
0.5 × 2   = 1.0    → 1

→ Binary: .011

✅ Final Binary Answer: (25.375)₁₀ = (11001.011)₂

Qus : 23 MCA NIMCET PYQ

Find the value of $x$, if: $\left( 2^{\frac{1}{\log_x 4}} \right) \left( 2^{\frac{1}{\log_x 16}} \right) \left( 2^{\frac{1}{\log_x 256}} \right) \cdots = 2$

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Solution

We are given: $\left( 2^{\frac{1}{\log_x 4}} \right) \left( 2^{\frac{1}{\log_x 16}} \right) \left( 2^{\frac{1}{\log_x 256}} \right) \cdots = 2$ Using the identity $a^{\frac{1}{\log_b a}} = b$ : $4 = x^2,\; 16 = x^4,\; 256 = x^8$ So, $2^{\frac{1}{\log_x 4}} = 2^{\frac{1}{\log_x x^2}} = 2^{1/2} = 2^{\frac12}$ $2^{\frac{1}{\log_x 16}} = 2^{\frac{1}{\log_x x^4}} = 2^{1/4} = 2^{\frac14}$ $2^{\frac{1}{\log_x 256}} = 2^{\frac{1}{\log_x x^8}} = 2^{1/8} = 2^{\frac18}$ Thus the infinite product is: $2^{\frac12} \cdot 2^{\frac14} \cdot 2^{\frac18} \cdot 2^{\frac1{16}} \cdots$ Combine the exponents: $2^{\left( \frac12 + \frac14 + \frac18 + \frac1{16} + \cdots \right)}$ The geometric series: $\frac12 + \frac14 + \frac18 + \frac1{16} + \cdots = 1$ Therefore, the product equals: $2^1 = 2$ Hence the value of $x$ is: $\boxed{\frac12}$

Qus : 24 MCA NIMCET PYQ

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Solution

Qus : 25 MCA NIMCET PYQ

A teacher gave a student the task of adding ‘N’ natural numbers starting from 1. After a while, the student reported his result as 700. The teacher replied that the result was wrong. The student realized he had added one number twice. Find the sum of digits of the number which the student had added twice.

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Solution

Sum 1 to N = N(N+1)/2.

A nearby perfect sum around 700 is 702 because 37×38/2 = 703.

If actual should be 703 and student got 700 → he added a number 3 twice less? No.

Student's mistake makes his sum 700 = correct sum + repeated number.

Try 704 (for N = 37): 703 + x = 700 (impossible)

Try N = 36: sum = 666 → 666 + x = 700 → x = 34

Sum of digits of 34 = 7

Qus : 26 MCA NIMCET PYQ

Sum of all 3-digit numbers (digits non-zero) where all digits are perfect squares.

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Solution

Digits that are perfect squares (non-zero): 1, 4, 9 Three-digit numbers using {1,4,9}: Total = 3 × 3 × 3 = 27 3×3×3=27 numbers Sum of digits = 1 + 4 + 9 = 14 1+4+9=14 Each digit appears 9 times in each position (hundreds, tens, units). Total sum = 100 × 9 × 14 + 10 × 9 × 14 + 1 × 9 × 14 100×9×14+10×9×14+1×9×14 = 111 × 9 × 14 111×9×14 = 999 × 14 999×14 = 13986

Qus : 27 MCA NIMCET PYQ

When you reverse the digits of the number 13, the number increases by 18. How many other two digit numbers increase by 18 when their digits are reversed?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Let the two-digit number be $10a + b$. Reversed number $= 10b + a$. Given increase is $18$: $(10b + a) - (10a + b) = 18$ $9(b - a) = 18$ $b - a = 2$ Possible digit pairs $(a,b)$ are: $(1,3), (2,4), (3,5), (4,6), (5,7), (6,8), (7,9)$ Total such numbers $= 7$.

Qus : 28 MCA NIMCET PYQ

Assuming all numbers are in 2’s complement representation, which of the following numbers is divisible by $11111011$?

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Solution

Qus : 29 MCA NIMCET PYQ

The sum of 11010 + 01111 equals

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

11010

+01111

101001

Qus : 30 MCA NIMCET PYQ

The quotient, if the binary number 11010111 is divided by 101, is___

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Solution

Binary Division

Question: What is the quotient when 11010111 is divided by 101 in binary?

Step 1: Convert to decimal:

11010111 = 215
101 = 5

Step 2: Divide: 215 ÷ 5 = 43

Step 3: Convert 43 to binary = 101011

✅ Final Answer: 101011

Qus : 31 MCA NIMCET PYQ

An eight bit byte is capable of representing how many different characters?

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Solution

Number of combinations in 8 bits = 2⁸ = 256

Qus : 32 MCA NIMCET PYQ

A treasure chest has less than 100 gold coins. The number of coins is

i) One more than a multiple of 3

ii) Two more than a multiple of 4

iii) Three more than a multiple of 5 and

iv) Four more than a multiple of 6

How many coins are there in the chest?

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Solution

Qus : 33 MCA NIMCET PYQ

Consider an arbitrary number system with independent digits as 0,1 and A. If we generate first few numbers in sequence as 00, 01, 0A, 10, 11, 1A and if this process is continued to generate the numbers, then the position of 10A is ________

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Solution

Custom Number System: Position of 10A

Given digits: 0, 1, A (base-3)

Convert 10A to base-10:

1 → 1
0 → 0
A → 2
10A = 1×9 + 0×3 + 2 = 11

List of numbers in sequence:

00
01
0A
10
11
1A
A0
A1
AA
100
101
10A

✅ Final Answer: 12

Qus : 34 MCA NIMCET PYQ

A family has several children. Each boy in this family has as many sisters as brothers, but each girl has twice as many brothers as sisters. How many brothers and sisters are there?

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Solution

Qus : 35 MCA NIMCET PYQ

A Group of 630 children are seated in n rows for a group photo session. Each row contains three less children than the row in front of it. Which one of the following number of rows is not possible?

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Solution

Solution (AP method):

Total children = 630. Let rows = n, first row = a, common difference d = -3.

Sum of n terms: $S_n=\frac{n}{2}[2a+(n-1)d]=630$

$\Rightarrow 1260 = n\,[2a-3(n-1)] \;\Rightarrow\; 2a=\frac{1260}{n}+3(n-1)$

$\Rightarrow a=\frac{1}{2}\left(\frac{1260}{n}+3n-3\right)$. For a valid arrangement, a must be a positive integer and last row $a-3(n-1)>0$.

n = 3: $a=\tfrac{1}{2}(420+9-3)=\tfrac{1}{2}\cdot 426=213$ ✔️ integer (valid)
n = 4: $a=\tfrac{1}{2}(315+12-3)=\tfrac{1}{2}\cdot 324=162$ ✔️ integer (valid)
n = 5: $a=\tfrac{1}{2}(252+15-3)=\tfrac{1}{2}\cdot 264=132$ ✔️ integer (valid)
n = 6: $a=\tfrac{1}{2}(210+18-3)=\tfrac{1}{2}\cdot 225=112.5$ ❌ not integer (invalid)

✅ Final Answer: 6 rows is not possible.

Qus : 36 MCA NIMCET PYQ

405 sweets were distributed equally among a group of children such that the number of sweets received by each child is one-fifth of the number of children. The number of children in the group is

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Solution

Qus : 37 MCA NIMCET PYQ

What is the missing number in the series 4, 7, 11, 18, 29, 47, __________, 123, 199?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 38 MCA NIMCET PYQ

A bus starts from its depot filled to seating capacity. It stops at a point A where 1/6th of the passengers alight and 10 board the bus. At point B, 1/5th of the passengers alight and 3 board the bus. At point C which is the last stop, all the 55 passengers alight. The capacity of the bus is

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Solution

Let the number of children = n

Total sweets = 405

Each child gets = $ \frac{405}{n} $ sweets

Given: Each child’s sweets = $ \tfrac{1}{5} $ of the number of children

⇒ $ \frac{405}{n} = \frac{n}{5} $

Cross multiply:

$ 405 \times 5 = n^2 $
⇒ $ n^2 = 2025 $
⇒ $ n = \sqrt{2025} = 45 $

✅ Final Answer: The number of children = 45

Qus : 39 MCA NIMCET PYQ

Convert the Hexadecimal number 4DF to its octal equivalent

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Solution

(4DF)₁₆ = 4×16² + 13×16 + 15 = 1247₁₀.
Converting 1247₁₀ to octal gives (2337)₈.

Qus : 40 MCA NIMCET PYQ

If all the 6's are replaced by 9's, then the algebraic sum of all the numbers from 1 to 100 (both inclusive)

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 41 MCA NIMCET PYQ

The answer of the operation $(10111)₂ \times (1110)₂ in hex equivalent is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

(10111)₂ = 23₁₀,

(1110)₂ = 14₁₀,

23 × 14 = 322₁₀ = (142)₁₆.

Qus : 42 MCA NIMCET PYQ

The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively is

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Solution

The first number = (1657 - 6) = 1651

Second number = (2037 - 5) = 2032

Taking the HCF of two numbers 1651, 2032 we get 127.

So, if we divide 1657 and 2037 by 127 we will get remainders 6 and 5 respectively.

Qus : 43 MCA NIMCET PYQ

Suppose the largest n bit number requires ‘d’ digits in decimal representation. Which of the following relations between ‘n’ and ‘d’ is approximately correct

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Solution

n bits binary number required d -decimal digits.

So, ${10}^d{\gt}{2}^n$

Take on both side

$\log _{10}({10}^d)\gt{\log _{10}({2}^n)}^{}$

$d{\gt}n\log _{10}(2)$

Qus : 44 MCA NIMCET PYQ

‘Floating point representation' is used to represent

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 45 MCA NIMCET PYQ

Three players – Aalu, Kachaalu and Bhalu were playing poker. At least one cheated. Exactly one always spoke the truth, one always lied, and one alternated between truth and lie. Which of the following can never be true?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Checking all consistent combinations of cheaters and truth-tellers shows that Bhalu cannot be the person who always speaks the truth.

Qus : 46 MCA NIMCET PYQ

Let $x,y,z$ be distinct integers. $x$ and $y$ are odd positive integers and $z$ is an even positive integer. Which of the following cannot be true?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Odd − even = odd, so $(x-z)$ is odd. Square of an odd number is odd, not even. Hence option (A) cannot be true.

Qus : 47 MCA NIMCET PYQ

If $(12x)_3 = (123)_x$ then the value of $x$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Convert $(12x)_3$ to decimal: $1\cdot 3^2 + 2\cdot 3 + x = 9 + 6 + x = 15 + x$ Convert $(123)_x$ to decimal: $1\cdot x^2 + 2\cdot x + 3$ Equate: $15 + x = x^2 + 2x + 3$ Bring all to one side: $x^2 + x - 12 = 0$ Factor: $(x + 4)(x - 3) = 0$ So $x = 3$ or $x = -4$ But base $x$ must be > 3 (because digits 1,2,3 appear). Allowed value = 3 (but digit '3' cannot appear in base 3). So no valid base exists.

Qus : 48 MCA NIMCET PYQ

The last digit of the number obtained by multiplying the numbers 81 x 82 x 83 x 84 x 85 x 86 x 87 x 88 x 89 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Last digit of 82 x 85 = 0

Qus : 49 MCA NIMCET PYQ

If $\log_3 2,\ \log_3(2^x - 5),\ \log_3\!\left(2^x - \dfrac{7}{2}\right)$ are in arithmetic progression, then the value of $x$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 50 MCA NIMCET PYQ

Each of A, B and C is a different digit among 1 to 9. How many different values of the sum of A, B and C are possible, if ABA X AA=ACCA ?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 51 MCA NIMCET PYQ

Let S be the set of integers x such that

(i) 100 ≤ x ≤ 200

(ii) x is odd and

(iii) x is divisible by 3 but not 7

How many elements does S contain?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 52 MCA NIMCET PYQ

A certain street has 1000 buildings. A sign-maker is contracted to number the houses from 1 to 1000.

How many zeroes will be needed?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Zeros in numbers:

From 1–999 → $189$ zeros

Number 1000 adds $3$ more zeros

Total $=189+3=192$, which is not listed.

Qus : 53 MCA NIMCET PYQ

A two digit number is such that the product of its digits is 12. If 36 is added to the number, the digits interchange their places. What is the number?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Let the two-digit number be 10x + y, where x = tens digit, y = units digit.

Step 1: Product condition

x × y = 12 … (1)

Step 2: Interchange condition

If digits interchange, new number = 10y + x

Given: (10x + y) + 36 = (10y + x)

⇒ 9x – 9y = –36

⇒ x – y = –4 … (2)

Step 3: Solve equations

From (2): x = y – 4

Put in (1): (y – 4) × y = 12

⇒ y² – 4y – 12 = 0

⇒ (y – 6)(y + 2) = 0

⇒ y = 6 (since digit can’t be –2)

Then x = y – 4 = 2

Number = 10x + y = 10×2 + 6 = 26

✅ Final Answer: The number is 26.

Qus : 54 MCA NIMCET PYQ

Consider the equation (43)_x= (y3)₈ where x and y are unknown. The number of possible solutions is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 55 MCA NIMCET PYQ

Complete the next series:

2 8 12

8 40 45

40 240 246

.... .... ....

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Each row uses a multiplier that increases by 1: 4, 5, 6, 7.
Middle = Left × multiplier;

Right = Middle + multiplier.
So for row 4:

Left = previous row’s

Middle = 240 → Middle = 240×7 = 1680

Right = 1680+7 = 1687.

Qus : 56 MCA NIMCET PYQ

How many numbers between 1 and 1000 are divisible by all numbers 2, 3, 4, 5 and 6?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 57 MCA NIMCET PYQ

Choose the correct option for the remainder when X = 1! + 2! + 3! + ...........+ 100! is divided by 24

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 58 MCA NIMCET PYQ

Consider a sequence formed by concatenating the positive integers in increasing order: 12345678910111213 ... Which among the following is the closest to the 2028th digit in this infinite sequence?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

1-digit numbers: 1 to 9 → 9 digits
2-digit numbers: 10 to 99 → 90 × 2 = 180 digits
Total till 99 → 9 + 180 = 189 digits
Remaining → 2028 − 189 = 1839 digits (in 3-digit range)
Each 3-digit number has 3 digits → 1839 ÷ 3 = 613 numbers (approx.)
Last full number before 2028th digit → 100 + 612 = 712
712 → digits are 7, 1, 2
Hence, the 2028th digit = 2

Qus : 59 MCA NIMCET PYQ

Given $\sqrt{(224)_r} = (13)_r$, then the value of the radix is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 60 MCA NIMCET PYQ

What type of errors are not detected by assembler?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 61 MCA NIMCET PYQ

Insert the missing number: 16, 33, 65, 131, 261, ?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Number System PYQ

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✅ Concept:

✅ Step 1: Understanding Binary Multiplication

✅ Step 2: Perform the Multiplication

✅ Step 3: Adding the Results

✅ Final Answer:

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Decimal to Binary Conversion

? Step-by-step:

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11010

+01111

101001

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Binary Division

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Custom Number System: Position of 10A

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Solution (AP method):

✅ Final Answer: 6 rows is not possible.

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Let the number of children = n

✅ Final Answer: The number of children = 45

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Let the two-digit number be 10x + y, where x = tens digit, y = units digit.

Step 1: Product condition

Step 2: Interchange condition

Step 3: Solve equations

✅ Final Answer: The number is 26.