Number System MCQ Quiz - Objective Question with Answer for Number System - Download Free PDF

Given:

Statements:

A) ( P ) is a perfect square,

B) ( P ) is greater than 2000.

Formula used:

If ( P ) is a perfect square, then ( P ) = ( n² ) where ( n ) is a natural number. A prime number greater than 1 can't be a perfect square.

Calculations:

Check given options with A and B:

Option 1: Both A and B together

A says ( P ) is a perfect square, hence ( P ) cannot be a prime number.

Option 2: Only B

B does not help to determine if ( P ) is a prime number or not, as it only states that ( P ) > 2000.

Option 3: Only A

A says ( P ) is a perfect square, hence ( P ) cannot be a prime number.

Option 4: Not possible to decide even with both A and B

Not true, since A alone can decide.

∴ The correct answer is option (3).

Given:

Fractions: 1/3, 7/6, 5/9, 4/27, 8/15

Formula used:

The LCM of fractions is calculated as:

LCM of fractions = LCM of numerators / HCF of denominators

Calculations:

Step 1: Identify the numerators and denominators:

Numerators: 1, 7, 5, 4, 8

Denominators: 3, 6, 9, 27, 15

Step 2: Calculate LCM of numerators:

LCM(1, 7, 5, 4, 8) = 280

Step 3: Calculate HCF of denominators:

HCF(3, 6, 9, 27, 15) = 3

Step 4: Calculate the LCM of fractions:

LCM = LCM of numerators / HCF of denominators

LCM = 280 / 3

The LCM of the given fractions is 280/3.

Given:

Intervals of tolling for the five bells: 6, 5, 7, 10, and 12 seconds.

Formula Used:

The number of times the bells toll together is determined by the least common multiple (LCM) of their intervals within the given time frame.

LCM = Least Common Multiple

Calculation:

LCM of 6, 5, 7, 10, and 12

Prime factorization:

6 = 2 × 3

5 = 5

7 = 7

10 = 2 × 5

12 = 2² × 3

LCM = 2² × 3 × 5 × 7

⇒ LCM = 4 × 3 × 5 × 7

⇒ LCM = 420 seconds

1 hour = 3600 seconds

Number of times they toll together in 1 hour:

Excluding the start:

3600 / 420 = 8.57

⇒ 8 times

The correct answer is option 2

Given:

If p and q are perfect squares, then √(p/q) is always a rational number.

Formula used:

If p = a² and q = b², where a and b are integers, then:

√(p/q) = √(a²/b²) = a/b

Calculation:

Let p = 16 (4²) and q = 9 (3²):

√(p/q) = √(16/9)

⇒ √(16/9) = √(4²/3²)

⇒ √(4²/3²) = 4/3

Since 4/3 is rational, the statement holds true.

∴ The correct answer is option 3).

Given:

GCD(a, b) = 4

LCM(a, b) = 400

Formula used:

Product of two numbers = GCD × LCM

a × b = GCD(a, b) × LCM(a, b)

Let a = 4m and b = 4n, where GCD(m, n) = 1.

Calculation:

a × b = GCD(a, b) × LCM(a, b)

⇒ 4m × 4n = 4 × 400

⇒ 16mn = 1600

⇒ mn = 100

Find pairs (m, n) such that GCD(m, n) = 1 and mn = 100:

Possible pairs are:

(m, n) = (1, 100), (100, 1), (4, 25), (25, 4).

Each pair gives distinct values of a and b:

(a, b) = (4 × 1, 4 × 100) = (4, 400)

(a, b) = (4 × 100, 4 × 1) = (400, 4)

(a, b) = (4 × 4, 4 × 25) = (16, 100)

(a, b) = (4 × 25, 4 × 4) = (100, 16)

∴ The number of pairs of (a, b) is 2.

The correct answer is option (3).

Given:

3240

Concept:

If k = ax × by, then

a, and b must be prime number 

Sum of all factors = (a0 + a1 + a2 + ….. + ax) (b0 + b1 + b2 + ….. + by)

Solution:

3240 = 2³ × 3⁴ × 5¹

Sum of factors = (20 + 21 + 22 + 23) (3⁰ + 3¹ + 3² + 3³ + 3⁴) (5⁰ + 5¹)

⇒ (1 + 2 + 4 + 8) (1 + 3 + 9 + 27 + 81) (1 + 5)

⇒ 15 × 121 × 6

⇒ 10890

∴ required sum is 10890

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

Given:

Length of timber₁ = 143 m

Length of timber2 = 78 m

Length of timber3 = 117 m

Calculation:

Greatest possible length of each plank = HCF of 143, 78 and 117

143 = 13 × 11

78 = 13 × 2 × 3

117 = 13 × 3 × 3 

HCF is 13

∴ Greatest possible length of each plank is 13 m.

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'.

GIVEN:

Four bells ring simultaneously at starting and an interval of 6 sec, 12 sec, 15 sec and 20 sec respectively.

CONCEPT:

LCM: It is a number which is a multiple of two or more numbers.

CALCULATION:

LCM of (6, 12, 15, 20) = 60

All 4 bells ring together again after every 60 seconds

Now,

In 2 Hours, they ring together = [(2 × 60 × 60)/60] times + 1 (at the starting) = 121 times

∴ In 2 hours they ring together for 121 times

Mistake Points

In these type of question we assume that we have started counting the time after first ringing. Due to this when we calculate the LCM it gives us the ringing at 2nd time not the first time. So, we needed to add 1.

Given:

Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec 

Calculation:

Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec 

Now we have to take LCM of time interval

⇒ LCM of (12, 15, 20, 30) = 60

Total seconds in 8 hours = 8 × 3600 = 28800

Number of times bell rings = 28800/60

⇒ Number of times bell rings = 480

If four bells ring together in starting

⇒ 480 + 1 

∴ The bell ringing 481 times in 8 hours.

Mistake PointsThe bells start tolling together, the first toll also needs to be counted, that is the number of times of tolling since the first time.

Given:

The number is 8¹⁰ × 9⁷ × 7⁸ 

Concept used:

If a number of the form x^a × y^b × z^c ...... and so on, then total prime factors = a + b + c ..... and so on

Where x, y, z, ... are prime numbers

Calculation:

The number 8¹⁰ × 9⁷ × 7⁸ can be written as (2³)¹⁰ × (3²)⁷ × 7⁸ 

The number can ve written as 2³⁰ × 3¹⁴ × 7⁸ 

Total number of prime factors = 30 + 14 + 8

∴ The total number of prime factors are 52

We know that,

product of two numbers = L.C.M × H.C.F of those numbers

Let the second number be x.

24 × x = 168 × 6

x = 6 × 7

x = 42

Given:

676xy is divisible by 3, 7 & 11

Concept:

When 676xy is divisible by 3, 7 &11, it will also be divisible by the LCM of 3, 7 &11. 

Dividend = Divisor × Quotient + Remainder

Calculation:

LCM (3, 7, 11) = 231

By taking the largest 5-digit number 67699 and divide it by 231.

∵ 67699 = 231 × 293 + 16

⇒ 67699 = 67683 + 16 

⇒ 67699 - 16 = 67683 (completely divisible by 231)

∴ 67683 = 676xy (where x = 8, y = 3)

(3x - 5y) = 3 × 8 - 5 × 3

⇒ 24 - 15 = 9 

∴ The required result = 9