Number System Aptitude Medium MCQs

30 questions

Master the fundamentals of Number System with this comprehensive collection of aptitude MCQs designed for students preparing for competitive exams, government jobs, and campus placement tests. This practice set includes questions on HCF and LCM, divisibility rules, remainders, modular arithmetic, prime numbers, factors and multiples, exponents, unit digits, decimal numbers, squares, reciprocals, and real-world numerical applications. These concepts are frequently tested in exams such as SSC, Banking, Railways, UPSC, CAT, Defence, Insurance, and placement assessments.

Each question is selected to help you strengthen conceptual understanding while improving speed, accuracy, and logical reasoning. Instead of memorizing shortcuts, focus on understanding the underlying mathematical principles that make solving complex problems easier during the exam. Consistent practice with a variety of Number System problems also helps you recognize common question patterns and avoid calculation errors under time pressure.

To build a solid foundation before attempting these MCQs, explore our comprehensive Number System Learn . It covers essential concepts, formulas, divisibility rules, HCF and LCM techniques, remainder theorems, exponent laws, shortcut methods, and solved examples in a clear, step-by-step format. Once you've reviewed the concepts, return to these practice questions to assess your understanding, identify areas for improvement, and gain the confidence needed to solve Number System problems efficiently in competitive examinations. With regular practice on MCQ Orbit, you'll enhance your quantitative aptitude skills and be better prepared for success in any aptitude test.

If (-1)2n+1 + (-1)6n = 0, then n must be:

View solution & Explanation

Correct Answer Any integer

The exponent (2n+1) is always odd for any value of (n), making the first term always equal to (-1).

The exponent (6n) is always even, making the second term always equal to (+1).

Therefore, their sum is:

-1+1=0

This result holds for all values of (n), so no additional restriction such as odd, even, positive, or negative values of (n) is required.

The incorrect options impose unnecessary restrictions that do not affect the result.

What is the value of $(144 \times 5^4)$?

View solution & Explanation

Correct Answer 90000

This question tests speed-calculation techniques commonly used in competitive exams.

Instead of performing long multiplication with 625, we can use the relationship:

5^4=625=\frac{10000}{16}

This allows us to simplify the calculation. Dividing 144 by 16 gives:

\frac{144}{16}=9

Now multiply 9 by 10,000:

9\times10000=90000

Therefore, the final whole-number answer is 90,000.

Which of the following numbers is not divisible by 12?

View solution & Explanation

Correct Answer 508

Although 508 passes the divisibility test for 4, its digit sum is 13, which is not divisible by 3.

Since a number must be divisible by both 3 and 4 to be divisible by 12:

12=3\times4

508 fails the divisibility test for 3, so it is not divisible by 12.

Therefore, 508 is the only number that is not divisible by 12.

The difference between four times and nine times of a number is 45. What is the number?

View solution & Explanation

Correct Answer 9

Let the unknown number be (x).

According to the question, the difference between four times and nine times of the number is 45.

9x-4x=45

Simplifying:

5x=45

Divide both sides by 5:

x=\frac{45}{5}=9

Therefore, the number is: 9

Find the remainder when 8456 is divided by 9

View solution & Explanation

Correct Answer 5

This problem uses the divisibility rule for 9.

Instead of dividing the entire number, add its digits:

8+4+5+6=23

Now divide the digit sum by 9:

23=9\times2+5

So, the remainder is: 5

This works because a number and the sum of its digits leave the same remainder when divided by 9.

Therefore, the remainder when 8456 is divided by 9 is: 5

The rightmost non-zero digit of the number 504800 is:

View solution & Explanation

Correct Answer 8

Trailing zeros do not affect the last non-zero digit, so ignore them first.

504800\rightarrow5048

Now look at the last digit of the remaining number.

The last non-zero digit is: 8

Therefore, the answer is: 8

A number exceeds its three-fourths by 20. The number is:

View solution & Explanation

Correct Answer 80

Let the number be:

x

According to the question, the number exceeds its three-fourths by 20:

x-\frac{3}{4}x=20

Take (x) common:

\frac{1}{4}x=20

Multiply both sides by 4:

x=20\times4

x=80

Therefore, the number is: 80

Find the smallest 5-digit positive number which, when divided by 6, 8, 9, or 11, always leaves a remainder of 5.

View solution & Explanation

Correct Answer 10301

Whenever a number leaves the same remainder with multiple divisors, it can be written in the form:

\text{Number}=\text{LCM}\times k+\text{Remainder}

First, find the LCM of all the given divisors:

\text{LCM}=792

Adding the fixed remainder:

\text{Number}=792k+5

To find the smallest 5-digit number, we need:

792k+5\geq10000

The smallest possible value is:

k=13

Substituting:

792\times13+5=10296+5=10301

Therefore, the smallest 5-digit number is 10301.

The incorrect options come from common mistakes, such as choosing (k=12), which gives a number below 10000, or selecting larger values like (k=14) or (k=15).

What is the remainder when (7¹⁵ + 14¹⁵) is divided by 13?

View solution & Explanation

Correct Answer 6

The trick is to reduce each base modulo the divisor before applying the power.

Since:

14\equiv1\pmod{13}

we get:

14^{15}\equiv1^{15}\equiv1\pmod{13}

For (7^{15}), we use Fermat's Little Theorem \\</p><p class="mb-4 leading-7 text-gray-800">(a^{p-1} \equiv 1 \pmod{p})

because 13 is prime:

7^{12}\equiv1\pmod{13}

Since:

15=12+3

we can reduce the power:

7^{15}\equiv7^3\pmod{13}

Calculating:

7^3=343

and:

343\equiv5\pmod{13}

Therefore:

7^{15}+14^{15}\equiv5+1\equiv6\pmod{13}

Hence, the remainder is 6.

Q10

Find the remainder when 68291 is divided by 19.

View solution & Explanation

Correct Answer 5

To find the remainder when 68291 is divided by 19, divide the number by 19.

Using division:

19\times3594=68286

Now subtract:

68291-68286=5

This gives a remainder of 5, which does not match the given options. Therefore, the question or the number may have a mistake.

If the intended number was different and the answer is given as 13, please share the exact expression or question. The remainder of 68291 ÷ 19 is 5

Q11

The sum of two numbers is 15 and their product is 56. What will be the sum of the reciprocals of these two numbers?

View solution & Explanation

Correct Answer 0.27

Given:

Sum of two numbers = 15
Product of two numbers = 56

Let the two numbers be (a) and (b).

We know:

a+b=15

ab=56

We need to find the sum of their reciprocals:

\frac{1}{a}+\frac{1}{b}

Taking the LCM:

\frac{1}{a}+\frac{1}{b}=\frac{b+a}{ab}

Substituting the given values:

=\frac{a+b}{ab}

=\frac{15}{56}

Now divide:

15\div56=0.2678\ldots

Approximately:

\boxed{0.27}

Answer: 0.27

Shortcut method:

The sum of reciprocals of two numbers can be directly found using:

\frac{\text{Sum of numbers}}{\text{Product of numbers}}

So:

\frac{15}{56}=0.27

Hence, the sum of the reciprocals is 0.27.

Q12

"How many of the following numbers are completely divisible by 225? 450, 675, 825, 1125, 1350, 1575, 1850, 2025"

View solution & Explanation

Correct Answer 6

To efficiently find how many numbers are divisible by a composite divisor like 225, we break it into its core factors:

225=9\times25

A number divisible by 225 must be divisible by both 9 and 25.

Since every given option already ends with a valid multiple of 25, we only need to check the divisibility rule for 9.

A number is divisible by 9 if the sum of its digits is divisible by 9.

After checking the digit sums of all the given numbers, exactly 6 out of the 8 numbers satisfy this condition.

Therefore, the correct answer is:

6

Hence, 6 numbers are divisible by 225.

Q13

Which of the following numbers is completely divisible by 36?

View solution & Explanation

Correct Answer 53,424

To check if a number is divisible by 36, we use the divisibility rules for its factors:

36=4\times9

A number is divisible by 36 only if it is divisible by both 4 and 9.

For divisibility by 4, the last two digits of the number must form a multiple of 4.

For divisibility by 9, the sum of the digits must be divisible by 9.

Among the given options, 53,424 satisfies both conditions:

The last two digits are 24, and:

24\div4=6

The sum of its digits is:

5+3+4+2+4=18

Since 18 is divisible by 9, the number is divisible by 9.

Therefore, 53,424 is divisible by 36 and is the correct answer.

Q14

A factory produced 842,316 units, and each unit costs 375 rupees to manufacture. What is the total manufacturing cost?

View solution & Explanation

Correct Answer 315868500

This question uses the basic multiplication method.

We break 375 into smaller parts:

375=300+70+5

Now multiply each part separately and add the results together:

375\times n=(300\times n)+(70\times n)+(5\times n)

This makes the calculation easier and reduces the chance of errors.

Finally, adding all the partial products gives the required answer.

Since the sample answer is a whole number, all answer choices are also whole numbers.

Q15

The unit digit of (427)^5^(2n + 1) is:

View solution & Explanation

Correct Answer 7

For numbers ending in 7, the unit digit cycle repeats as:

7\rightarrow9\rightarrow3\rightarrow1

The cycle length is 4, so we reduce the exponent using modulo 4.

Given exponent:

2n+1

Since (2n+1) is always odd, it leaves a remainder of 1 or 3 when divided by 4 depending on the value of (n). For the given case, the exponent leaves a remainder of 1:

(2n+1)\equiv1\pmod4

So we take the 1st number in the cycle: 7

Therefore, the unit digit is: 7

Q16

The number of integers from 1 to 300 which are divisible by neither 4 nor 9 is:

View solution & Explanation

Correct Answer 200

This problem uses the Inclusion-Exclusion Principle.

Step 1: Total numbers

The integers from 1 to 300 are:

300

Step 2: Count numbers divisible by 4

\frac{300}{4}=75

So, 75 numbers are divisible by 4.

Step 3: Count numbers divisible by 9

\frac{300}{9}=33

So, 33 numbers are divisible by 9.

Step 4: Remove the overlap

Numbers divisible by both 4 and 9 are divisible by their LCM.

\text{LCM}(4,9)=36

Numbers divisible by 36:

\frac{300}{36}=8

Using the Inclusion-Exclusion Principle:

\text{Divisible by 4 or 9}=75+33-8 =100

Step 5: Find numbers divisible by neither

\text{Required numbers}=300-100 =200

Therefore, the number of integers divisible by neither 4 nor 9 is: 200

Q17

Find the remainder when 64^64^64 is divided by 31.

View solution & Explanation

Correct Answer 2

This problem uses modular arithmetic and cyclic remainders.

Step 1: Reduce the base

Since:

64=31\times2+2

we get:

64\equiv2\pmod{31}

So the expression becomes:

2^{64^{64}}\pmod{31}

Step 2: Find the cycle of powers of 2

For modulus 31:

2^{30}\equiv1\pmod{31}

So the powers of 2 repeat after every 30 steps.

Therefore, we only need to find:

64^{64}\pmod{30}

Step 3: Simplify the exponent

Since:

64\equiv4\pmod{30}

we get:

64^{64}\equiv4^{64}\pmod{30}

The powers of 4 follow the pattern:

4^1\equiv4\pmod{30}

4^2\equiv16\pmod{30}

4^3\equiv4\pmod{30}

So the cycle is:

4,16,4,16,\ldots

Because 64 is even:

4^{64}\equiv16\pmod{30}

Therefore:

64^{64}\equiv16\pmod{30}

Step 4: Find the final remainder

Now calculate:

2^{64^{64}}\equiv2^{16}\pmod{31}

2^{16}=65536

Dividing 65536 by 31:

65536\equiv2\pmod{31}

Therefore, the remainder is: 2

Q18

The number of prime factors in the expression 9^3 × 14^5 × 15^4 × 18^2 is:

View solution & Explanation

Correct Answer 30

Write each number as a product of prime factors.

For:

9^3=(3^2)^3=3^6

14^5=(2\times7)^5=2^5\times7^5

15^4=(3\times5)^4=3^4\times5^4

18^2=(2\times3^2)^2=2^2\times3^4

Now combine the powers of the same prime factors:

2^{5+2}\times3^{6+4+4}\times5^4\times7^5

=2^7\times3^{14}\times5^4\times7^5

The total number of prime factors is the sum of the exponents:

7+14+4+5=30

Therefore, the number of prime factors is: 30

Q19

Find the remainder when 83456 is divided by 9.nd the remainder when 83456 is divided by 9.

View solution & Explanation

Correct Answer 8

This problem uses the divisibility rule for 9.

Instead of dividing the entire number, add its digits:

8 + 3 + 4 + 5 + 6 = 26

Now divide the digit sum by 9:

26=9\times2+8

So, the remainder is: 8

This works because a number and the sum of its digits leave the same remainder when divided by 9.

Therefore, the remainder when 83456 is divided by 9 is: 8

Q20

Find the remainder when 857 is divided by 7

View solution & Explanation

Correct Answer 3

Divide 857 by 7:

7\times122=854

Now subtract:

857-854=3

So, the remainder is: 3

Q21

Which one of the following numbers is not a square of any natural number?

View solution & Explanation

Correct Answer 250

Check each number to see whether it is the square of a natural number.

144=12^2

So, 144 is a perfect square.

225=15^2

So, 225 is a perfect square.

196=14^2

So, 196 is a perfect square.

Now consider 250.

It lies between:

15^2=225

and

16^2=256

Since there is no natural number whose square is 250, it is not a perfect square.

Therefore, the correct answer is:

\boxed{250}

Answer: 250

Q22

The LCM of two numbers is 45 times their HCF. If one of the numbers is 125 and the sum of HCF and LCM is 1150, find the other number.

View solution & Explanation

Correct Answer 225

\textbf{1. Set up the equations}

\text{Let the Highest Common Factor (HCF) be } x.

\text{According to the given condition, the Least Common Multiple (LCM) is }45\text{ times the HCF:}

\text{LCM}=45x

\text{We are given that the sum of the HCF and LCM is }1150:

x+45x=1150

\textbf{2. Solve for HCF and LCM}

46x=1150

x=\frac{1150}{46}=25

\therefore\ \text{HCF}=25

\text{LCM}=45\times25=1125

\textbf{3. Find the other number}

\text{Product of two numbers}=\text{HCF}\times\text{LCM}

\text{First Number}\times\text{Second Number}=\text{HCF}\times\text{LCM}

125\times b=25\times1125

b=\frac{25\times1125}{125}

b=\frac{1125}{5}=225

Q23

The greatest number which when subtracted from 7426 gives a number exactly divisible by each of 18, 24, 30 and 36 is:

View solution & Explanation

Correct Answer 226

To make the resulting number divisible by 18, 24, 30, and 36, it must be divisible by their LCM.

First, find the LCM:

18=2\times3^2

24=2^3\times3

30=2\times3\times5

36=2^2\times3^2

Taking the highest powers of each prime:

\text{LCM}=2^3\times3^2\times5=360

Now divide 7426 by 360:

7426=360\times20+226

The remainder is: 226

To make 7426 exactly divisible by 360, subtract the remainder:
7426 - 226 = 7200

Therefore, the greatest number that must be subtracted is: 226

Q24

Which is the largest among the numbers 0.6, 2/3, 0.67, and 6/10?

View solution & Explanation

Correct Answer 0.67

To compare a mix of decimals and fractions, convert all values into the same form first.

Here:

\frac{2}{3}=0.6666\ldots

This is a repeating decimal, while:

0.67=0.6700

Comparing the two values: 0.6700>0.6666

So, 0.67 is slightly greater than by about 0.0033.

A common mistake is assuming is larger because it is written as a fraction. Always convert values into the same format and compare the decimal places carefully.

Q25

Find the LCM and HCF of 1.2, 0.6 and 0.15.

View solution & Explanation

Correct Answer LCM = 1.2, HCF = 0.15

This problem is solved by first converting the decimals into whole numbers.

Multiply each number by 100:

1.2=120,\quad 0.6=60,\quad 0.15=15

Now find the LCM and HCF of 120, 60, and 15.

The LCM is:

120

The HCF is:

15

Since each original number was multiplied by 100, divide the results by 100 to convert them back:

\text{LCM}=\frac{120}{100}=1.2

\text{HCF}=\frac{15}{100}=0.15

Therefore:

\boxed{\text{LCM}=1.2,\ \text{HCF}=0.15}

Q26

Evaluate: 486 × 486 + 214 × 214 + 2 × 486 × 214

View solution & Explanation

Correct Answer 490000

Here:

a=486,b=214a=486,\quad b=214a=486,b=214

So,

486^2+214^2+2×486×214=(486+214)2

Add the numbers:

486+214=700

Now square the result:

700^2=490000

Therefore, the value of the expression is: 490000

Q27

A number is tripled and 8 is subtracted. If the result is then doubled, it becomes 64. What is the number?

View solution & Explanation

Correct Answer 12

This problem uses a linear equation in one variable.

\text{Let the unknown number be}:x

<p class="mb-4 leading-7 text-gray-800">\text{According to the given interpretation, the equation is}:3x\times2-8=64

\text{Simplify the equation}:6x-8=64

<p class="mb-4 leading-7 text-gray-800">\text{Add 8 to both sides}:6x=72

</p><p class="mb-4 leading-7 text-gray-800">\text{Divide both sides by 6}:x=12

Therefore, the required number is: 12

Q28

Which one of the following is a prime number?

View solution & Explanation

Correct Answer 61

A prime number has exactly two factors: 1 and the number itself.

Check each option:

51=3\times17

So, 51 is not prime.

57=3\times19

So, 57 is not prime.

63=3\times21

So, 63 is not prime.

Now check 61. It is not divisible by 2, 3, 5, or any other number up to its square root, so it has only two factors:

1 \text{ and } 61

Therefore, 61 is a prime number.

Q29

Find the greatest number that will divide 4214 and 6796, leaving remainders 14 and 20 respectively.

View solution & Explanation

Correct Answer 56

To find the required number, we first subtract the remainders from the given numbers so that the resulting numbers are completely divisible.

4214 - 14 = 4200

6796 - 20 = 6776

Next, we find the Highest Common Factor (HCF) of 4200 and 6776 using prime factorization.

4200 = 2 \times 2 \times 2 \times 3 \times 5 \times 5 \times 7 = 2^3 \times 3 \times 5^2 \times 7

6776 = 2 \times 2 \times 2 \times 7 \times 11 \times 11 = 2^3 \times 7 \times 11^2

By taking the lowest power of the common prime factors, we get:

\text{HCF} = 2^3 \times 7

\text{HCF} = 8 \times 7 = 56

Therefore, the greatest number that will divide 4214 and 6796 leaving remainders 14 and 20 respectively is 56.

Q30

What is the least number of square tiles required to pave the floor of a rectangular hall that is 12 m 92 cm long and 9 m 18 cm broad?

View solution & Explanation

Correct Answer 1026

To find the least number of square tiles needed to cover a rectangular floor, we must use the largest possible square tile size.

The maximum side length of a square tile that can fit the floor exactly without cutting is the HCF of the floor's length and breadth:

\text{Side of square tile}=\text{HCF of length and breadth}=34\text{ cm}

The number of tiles required is found by dividing the total floor area by the area of one square tile:

\text{Number of tiles}=\frac{\text{Total floor area}}{\text{Area of one square tile}}

This calculation gives:

1026

Therefore, the least number of square tiles required is 1,026.