Test: Number System(Tier 2) - SSC CGL MCQ

Let number is N

N = 65 k + 29

Now, 

This can be written as:

∴ Remainder = 3

Given:

Numbers: 242, 634, and 358

Remainders: 2, 4, and 8 respectively

Formula Used:

The greatest number that will divide the given numbers leaving the specified remainders is the HCF (Highest Common Factor) of the differences of each number with its respective remainder.

Calculation:

First, adjust the numbers by subtracting the remainders:

242 - 2 = 240

634 - 4 = 630

358 - 8 = 350

Now, find the HCF of 240, 630, and 350.

Prime factorization:

240 = 2⁴ × 3 × 5

630 = 2 × 3² × 5 × 7

350 = 2 × 5² × 7

Common factors: 2 and 5

HCF = 2 × 5 = 10

The greatest number that will divide 242, 634, and 358 leaving remainders 2, 4, and 8 respectively is 10.

Option 3 is the correct answer.

Given:

Range of numbers: 200 to 500

Formula Used:

Number of integers divisible by a, b, and c = (Largest number in the range / LCM(a, b, c)) - (Smallest number in the range / LCM(a, b, c))

Calculation:

LCM of 3, 4, and 5 = 60

Largest number in the range = 500

Smallest number in the range = 200

Number of integers divisible by 60 up to 500:

⇒ 500 / 60 = 8 (integer part only)

Number of integers divisible by 60 up to 199:

⇒ 199 / 60 = 3 (integer part only)

Number of integers between 200 and 500 divisible by 60:

⇒ 8 - 3 = 5

There are 5 numbers between 200 and 500 that are completely divisible by 3, 4, and 5.

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:

Numbers are 625 and 250.

Formula Used:

The smallest number that will be exactly divided by both 625 and 250 is the Least Common Multiple (LCM) of the given numbers.

Calculation:

Prime factorization of 625:
625 = 5⁴

Prime factorization of 250:
250 = 2 × 5³

The LCM is found by taking the highest power of each prime that appears in the factorizations.

LCM = 2¹ × 5⁴

⇒ LCM = 2 × 625

⇒ LCM = 1250

The smallest number that will be exactly divided by 625 and 250 is 1250.

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:

Old Income Tax = 4 paise per rupee.

New Income Tax = 5 paise per rupee.

Increase in revenue = 10%.

Formula Used:

Revenue = Income Tax × Amount Taxed

Calculation:

Let the initial amount taxed be 100 units.

Initial Revenue = 4 paise × 100 = 400 paise

New Revenue = 400 paise + 10% of 400 paise

New Revenue = 400 paise + 40 paise = 440 paise

Let the new amount taxed be x units.

New Revenue = 5 paise × x

5x = 440

⇒ x = 440 / 5

⇒ x = 88

Decrease in amount taxed = 100 - 88 = 12 units

Percentage decrease in the amount taxed = (12 / 100) × 100%

Percentage decrease = 12%

The correct answer is option A.

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

Amount A = 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

Given: (49¹⁵ − 1)

Concept used:

aⁿ​​​​​​ - bⁿ is divisible by (a + b) when n is an even positive integer.

Here, a & b should be prime number.

Calculation:

(49¹⁵ − 1)

⇒ ((7²)¹⁵ − 1)

⇒ (7³⁰ − 1)

Here, 30 is a positive integer.

​According to the concept,

(7³⁰ − 1) is divisible by (7 + 1) i.e., 8.

∴ 8 is a divisor of (49¹⁵ − 1).

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

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

x² + ax + b, when divided by x - 5, leaves a remainder of 34,

⇒ 5² + 5a + b = 34

⇒ 5a + b = 9      ----(1)

Again,

x² + bx + a, when divided by x - 5, leaves a remainder of 52

⇒ 5² + 5b + a = 52

⇒ 5b + a = 27      ----(2)

From (1) + (2) we get,

⇒ 6a + 6b = 36

⇒ a + b = 6

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.

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

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