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MCA NIMCET Number System 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 PYQMCA 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 : 2MCA NIMCET PYQ

The number (2217)₈ is equivalent to

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2021 PYQ

Solution

✅ Step 1: Convert Octal to Binary

Each octal digit is represented by 3 binary digits. Let's convert (2217)₈ to binary:

2₈ = 010, 2₈ = 010, 1₈ = 001, 7₈ = 111

Binary Representation: 010 010 001 111

✅ Step 2: Group Binary in 4-bit Sections

To convert binary to hexadecimal, group the binary digits in sets of 4, starting from the right:

010 010 001 111 becomes 0010 0100 0111

✅ Step 3: Convert Binary to Hexadecimal

Now convert each group of 4 bits into its hexadecimal equivalent:

0010 = 2
0100 = 4
0111 = 7

✅ Final Answer:

The hexadecimal equivalent of (2217)₈ is: (247)₁₆

Qus : 3MCA NIMCET PYQ

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

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MCA NIMCET Previous Year PYQMCA 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 : 4MCA 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 : 5MCA 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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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2025 PYQ

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 : 6MCA NIMCET PYQ

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

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2025 PYQ

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 : 7MCA NIMCET PYQ

What is the hexadecimal representation of the decimal number 265?

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2025 PYQ

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 : 8MCA NIMCET PYQ

The binary equivalent of (234.125)₁₀?

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 9MCA NIMCET PYQ

Determine the octal equivalent of (432267)₁₀?

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 10MCA NIMCET PYQ

One Exabyte is equal to …

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 11MCA NIMCET PYQ

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

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 12MCA NIMCET PYQ

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

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2023 PYQ

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 : 13MCA NIMCET PYQ

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

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2024 PYQ

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 : 14MCA 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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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 15MCA 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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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2024 PYQ

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 : 16MCA 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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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 17MCA 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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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2016 PYQ

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 : 18MCA 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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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 19MCA 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 PYQMCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 20MCA 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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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2016 PYQ

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 : 21MCA NIMCET PYQ

Convert the Hexadecimal number 4DF to its octal equivalent

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2013 PYQ

Solution

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

Qus : 22MCA 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 PYQMCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 23MCA NIMCET PYQ

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

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2013 PYQ

Solution

(10111)₂ = 23₁₀,

(1110)₂ = 14₁₀,

23 × 14 = 322₁₀ = (142)₁₆.

Qus : 24MCA NIMCET PYQ

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

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2022 PYQ

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 : 25MCA 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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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2022 PYQ

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 : 26MCA NIMCET PYQ

‘Floating point representation' is used to represent

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 27MCA 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 PYQMCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 28MCA 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 PYQMCA 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 : 29MCA 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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Solution

Qus : 30MCA NIMCET PYQ

Complete the next series:

2 8 12

8 40 45

40 240 246

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

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2025 PYQ

Solution

Qus : 31MCA 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 PYQMCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 32MCA 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 PYQMCA NIMCET NIMCET 2025 PYQ

Solution

Qus : 33MCA NIMCET PYQ

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

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Solution

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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Number System PYQ

Solution

Solution

✅ Step 1: Convert Octal to Binary

✅ Step 2: Group Binary in 4-bit Sections

✅ Step 3: Convert Binary to Hexadecimal

✅ Final Answer:

Solution

Solution

✅ Concept:

✅ Step 1: Understanding Binary Multiplication

✅ Step 2: Perform the Multiplication

✅ Step 3: Adding the Results

✅ Final Answer:

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Decimal to Binary Conversion

? Step-by-step:

Solution

Binary Division

Solution

Solution

Custom Number System: Position of 10A

Solution

Solution

Solution (AP method):

✅ Final Answer: 6 rows is not possible.

Solution

Solution

Solution

Let the number of children = n

✅ Final Answer: The number of children = 45

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

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.

Solution

Solution

Solution

Solution

Solution

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