Integers MCQ Quiz - Objective Question with Answer for Integers - Download Free PDF

Given:

The first 100 natural numbers (1 to 100).

We need to count the number of times the digit '8' appears.

Calculations:

Let's list the numbers from 1 to 100 and identify where the digit '8' appears.

In the units place:

These are numbers ending in 8:

8, 18, 28, 38, 48, 58, 68, 78, 88, 98

Counting these, we find '8' appears 10 times in the units place.

In the tens place:

These are numbers where '8' is in the tens digit (from 80 to 89):

80, 81, 82, 83, 84, 85, 86, 87, 88, 89

Counting these, we find '8' appears 10 times in the tens place.

Total count:

Add the counts from the units place and the tens place:

Total = (Count in units place) + (Count in tens place)

Total = 10 + 10 = 20

∴ The digit '8' will be repeated 20 times when writing down the first 100 natural numbers.

Concept used:

A prime number is a number that is divisible only by itself and 1 (e.g. 2, 3, 5, 7, 11).

Calculation:

Prime numbers that are lesser than 20

⇒ 2, 3, 5, 7, 11, 13, 17, 19

Now, there are 8 such numbers.

∴ 8 prime numbers are there less than 20.

Given:

When the digits of a two-digit number are reversed, the value of the number is increased by 45.

The sum of the digits is 11.

Formula Used:

Let the original number be 10x + y, where x and y are the digits of the number.

Reversed number = 10y + x

Given that the reversed number is 45 more than the original number:

10y + x = 10x + y + 45

Given that the sum of the digits is 11:

x + y = 11

Calculation:

From the equation 10y + x = 10x + y + 45:

⇒ 10y + x - 10x - y = 45

⇒ 9y - 9x = 45

⇒ y - x = 5

We have two equations:

1) x + y = 11

2) y - x = 5

Adding the two equations:

⇒ (x + y) + (y - x) = 11 + 5

⇒ 2y = 16

⇒ y = 8

Substituting y = 8 in x + y = 11:

⇒ x + 8 = 11

⇒ x = 3

Original number = 10x + y

⇒ 10(3) + 8

⇒ 30 + 8

⇒ 38

The original number is 38.

A prime number is a number that has only two factors 1 and itself.

As every number is a factor of itself.

So the two-digit prime numbers which remain prime even when interchanging the digits are-

11,13,17,31,37,71,73,79,97

∴ There are "9" two-digit prime numbers are there between 10 to 100 which remain prime numbers when the order of their digits is reversed.

Given :

X and Y are natural numbers other than 1, and Y is greater than X

Calculations :

1 < X < Y 

Let X be 2 and Y be 3         (X and Y are natural numbers)

Now put value in each option to get the values 

Option(1) ⇒ XY = 2 × 3 = 6 

Option(2) ⇒ X/Y = 2/3 

Option(3) ⇒ Y/X = 3/2 

Option(4) ⇒ (X + Y)XY = 5/6 

We can see that option 1 represents the largest value among all 

∴ Option1 will be the correct choice. 

Given:-

A+B+C+D+E = Rs. 720 

E - A = 40

Concept used:-

Arithmatic progression -

a, a + d, a + 2d, a + 3d, a + 4d

n^th term(T_n) = a + (n -1)d

Calculation:- 

Let, A receive Rs. a and the difference between each consecutive person be Rs. d.

Amount_E = a + 4d

AmountA = a

According to the question,

⇒ a + 4d - a = 40

⇒ 4d = 40

⇒ d = 10

Also,

a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 720

⇒ 5a + 10d = 720

⇒ 5a + 10 × 10 = 720

⇒ 5a = 720 - 100

⇒ a = 620/5 = 124

So, Amount_B = a + d = 124 + 10 = Rs. 134

Alternate Method

Calculation:

A, B, C, D and E 

As the amount received is in AP,

Difference in an amount of two consecutive members is the same.

⇒ B – A = C – B = D – C = E – D

We have E – A = 40, 

⇒ B – A = 10, C – B = 10, D – C  = 10, E – D = 10,

Let say A received Rs. x,

Then B, C, D and E will receive,

⇒ x + 10, x + 20, x + 30, x + 40

According to the question,

⇒ x + (x + 10) + (x + 20) + (x + 30) + (x + 40) = 720

⇒ 5x + 100 = 720

⇒ 5x = 620

⇒ x = 124

B will receive = x + 10 = 124 + 10 = 134

∴ B will receive amount of Rs. 134

Given:

The sum of seven consecutive natural numbers = 1617

Calculation:

Let the numbers be n, n + 1, n + 2, n + 3, n + 4, n + 5, n + 6 respectively

⇒ 7n + 21 = 1617

⇒ 7n = 1596

⇒ n = 228

The numbers is 228, 229, 230, 231, 232, 233, 234

Out of these 229, 233 are prime numbers

∴ Required prime numbers is 2

Concept used:

Twin prime numbers are pairs of prime numbers that have a difference of exactly two.

In other words, if (p, p+2) are both prime numbers, then they are considered twin primes.

Formally, if p and p+2 are both primes, then they are known as twin primes.

For example, (3, 5), (11, 13), and (17, 19) are pairs of twin primes.

Calculation:

Twin primes are pairs of successive primes that differ by two. 

The primes from 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

Options:

(37, 41) - Difference between them is 4.

(3, 7) - The difference between them is 4.

(43, 47) - Difference between them is 4.

(71, 73) - Difference between them is 2.

Here, in the given option (71 and 73) are prime numbers and their difference is '2'.

Calculations:

\(\frac{{45}}{{53}} = \frac{1}{{a + \frac{1}{{b + \frac{1}{{c - \frac{2}{5}}}}}}}\)

\(⇒ \frac{1}{{1 \; + \;\frac{8}{{45}}}} = \frac{1}{{a\; +\; \frac{1}{{b\;+ \;\frac{1}{{c\; -\; \frac{2}{5}}}}}}}\)

\(⇒ \frac{1}{{1\; + \;\frac{1}{{5\; + \;\frac{5}{8}}}}} = \frac{1}{{a\; +\; \frac{1}{{b\;+ \;\frac{1}{{c\; -\; \frac{2}{5}}}}}}}\)

\(⇒ \frac{1}{{1\; + \;\frac{1}{{5\; + \;\frac{1}{{2 \;- \;\frac{2}{5}}}}}}} = \frac{1}{{a\; +\; \frac{1}{{b\;+ \;\frac{1}{{c\; -\; \frac{2}{5}}}}}}}\)

Compare both the sides

⇒ a = 1, b = 5 and c = 2

Value of (4a – b + 3c) = 4 × 1 – 5 + 3 × 2

⇒ 10 – 5 = 5

∴ The value of 4a – b + 3c is 5

Given:

The number of trailing zeros in 142!

Concept used:

Number of trailing zeroes in n! = Number of times n! is divisible by 10 = Highest power of 10 which divides n! = Highest power of 5 in n!

Calculation:

According to the question,

142/5 = 28

28/5 = 5

5/5 = 1

Given:

375!

Concept Used:

Number of zeros = Number of pairs of 2 × 5

Calculation:

Quotient of  375/5 = 75

Quotient of  75/5 = 15

Quotient of  15/5 = 3

Concept used:

For finding number of factors of any number we do prime factorisation of that number.

If X = p₁^a × p₂^b, then

Number of factors of X = (a + 1) × (b + 1)

where, p1, p2 → Prime factors of X

Calculation:

Now, 40 = 2³ × 5¹

⇒ Number of factors of 40 = (3 + 1) × (1 + 1) = 4 × 2 = 8

∴ The number of factors of 40 is 8.

Triangular number will satisfy following conditions,

X_n = n(n + 1) /2

⇒ X₁ = 1

⇒ X₂ = 3

⇒ X₃ = 6

⇒ X₄ = 10

⇒ X₅ = 15

⇒ A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself).

⇒ Prime Factors of 943 = 23 × 41

⇒ Prime Factors of 323 = 17 × 19

⇒ Prime factors of 713 = 23 × 31

⇒ Prime factors of 409 = 1 × 409

⇒ 60 × 18 × 15

⇒ (4 × 3 × 5) × (2 × 9) × (3 × 5)

⇒ 2³ × 3⁴ × 5²

∴ Number of factors of the expression,