SSC CGL Number System Questions with Detailed Solutions

Given
4 × (Sum of digits) = Number formed by reversing digits
Difference between digits = 3
Formula Used
Original two-digit number = 10x + y
Reversed number = 10y + x
Solution
Let the tens digit be x and units digit be y.
Given condition: 4(x + y) = 10y + x
Expand and simplify the equation:
4x + 4y = 10y + x
3x = 6y
x = 2y
Since x = 2y, we know x is greater than y.
The second condition states the difference between digits is 3:
x - y = 3
Substitute x = 2y into the second equation:
2y - y = 3
y = 3
Now, find x:
x = 2(3) = 6
The original number is 10x + y = 10(6) + 3 = 63.
Final Answer
So the correct answer is (a)

Given
Expression: $$3^{939} ÷ 8$$​​
Formula Used
Remainder Theorem: If $$(ax + 1)^n$$​ is divided by a, the remainder is always 1.
Solution
We need to express the base 3 in terms of a power that is close to a multiple of 8. We know that $$3^2$$​= 9, and 9 divided by 8 leaves a remainder of 1.
​$$3^{939}: \\3^{939} = 3^{938 + 1} = 3^{938} × 3^1\\3^{939} = (3^2)^{469} × 3\\$$​​
Substitute $$3^2$$​ with 9:
​$$= 9^{469} × 3$$​​
When 9 is divided by 8, the remainder is 1. Therefore, for $$9^{469} ÷ 8,$$​the remainder is $$1^{469} = 1.$$​​
Now multiply this by the remaining part:
Remainder = 1 × 3 = 3
Final Answer
So the correct answer is (b)

Given:
The terms are:  $$x^{3.5} y^{1.5}, \frac{y^{2.5}}{x^{-4.5}},$$​ and $$\frac{x}{y^{-3}}$$​​
Formula Used:
​$$a^{-m} = \frac{1}{a^m}$$​​
LCM of algebraic terms is the product of all unique prime bases with their highest powers.
Solution:
First, let's simplify all the terms by removing negative exponents:
Term 1 = $$x^{3.5} y^{1.5}$$​​
Term 2 = $$y^{2.5} x^{4.5} = x^{4.5} y^{2.5}$$​​
Term 3 = $$x y^3 = x^{1} y^{3}$$​​
Now, compare the powers of x and y across all three terms.
Highest power of x = max(3.5, 4.5, 1) = 4.5
Highest power of y = max(1.5, 2.5, 3) = 3
LCM = $$x^{4.5} y^3$$​​
Rewriting this in the form of the options:
​$$x^{4.5} y^3 = x^3 × x^{1.5} × y^3 = (xy)^3 x^{1.5} = (xy)^3 \sqrt{x^3}$$​​
Final Answer
So the correct answer is (a)

Given:
A fraction multiplied by itself and divided by its reciprocal gives $$10 \frac{296}{1728}.$$​​
Formula Used:
Average = $$\frac{\text{Sum of values}}{\text{Number of values}}$$​​
Solution:
Let the original fraction be $$\frac{a}{b}.$$​​
According to the question:
​$$(\frac{a}{b} × \frac{a}{b}) ÷ \frac{b}{a} = 10 \frac{296}{1728}$$​​

​$$(\frac{a}{b})^2 × \frac{a}{b} = \frac{10 × 1728 + 296}{1728}$$​​

​$$(\frac{a}{b})^3 = \frac{17280 + 296}{1728}$$​​

​$$(\frac{a}{b})^3 = \frac{17576}{1728}$$​​

We know that 17576 is the cube of 26 and 1728 is the cube of 12.
​$$(\frac{a}{b})^3 = (\frac{26}{12})^3$$​​

​$$\frac{a}{b} = \frac{26}{12}$$​​

​$$\frac{a}{b} = \frac{13}{6}$$​​
The numerator is 13 and the denominator is 6.
We need the average of their squares:
Average = $$\frac{13^2 + 6^2}{2}$$​​

=$$\frac{169 + 36}{2}$$​​

= $$\frac{205}{2}$$​ = 102.5
Final Answer
So the correct answer is (b)

Given:
The divisors are 8, 12, and 15.
The remainders are 5, 9, and 12 respectively.
Formula Used:
Required Number = (Multiple of LCM) - Common Difference
Solution:
First, we check the difference between each divisor and its corresponding remainder.
8 - 5 = 3
12 - 9 = 3
15 - 12 = 3
The common difference is 3 in all cases.
Next, we find the least common multiple of the divisors 8, 12, and 15.
The LCM of 8, 12, and 15 is 120.
The greatest possible five-digit number is 99999.
We divide 99999 by 120 to find the greatest multiple within five digits.
99999 = 120 × 833 + 39.
The greatest five-digit multiple of 120 is found by subtracting the remainder.
Multiple is 99999 - 39 = 99960.
The required number is obtained by subtracting the common difference from this multiple.
Required Number is 99960 - 3 = 99957.
Final Answer
So the correct answer is (a)

Given:
Numbers: 4358, 4748, 5138, 5788
Remainder in each case = 3
Formula Used:
Required number = HCF of $$(N_1 - R), (N_2 - R), (N_3 - R), (N_4 - R)$$​​
Solution:
Subtract the remainder 3 from each of the given numbers:
4358 - 3 = 4355
4748 - 3 = 4745
5138 - 3 = 5135
5788 - 3 = 5785
Now, find the HCF of 4355, 4745, 5135, and 5785.
Let's find the differences between consecutive terms to simplify:
4745 - 4355 = 390
5135 - 4745 = 390
5785 - 5135 = 650
The required HCF must be a factor of these differences (390 and 650).
HCF(390, 650):
390 = 130 × 3
650 = 130 × 5
The common difference factor is 130. However, 130 is an even number and cannot divide odd numbers like 4355.
Therefore, the actual HCF is a factor of 130. Testing the options, 65 is a factor of 130 and divides all the odd numbers:
4355÷ 65 = 67
4745÷ 65 = 73
Thus, the largest number is 65.
Final Answer
So the correct answer is (c)

Given:
An 8-digit number: 130A1568.
The number is divisible by 8.
Formula Used:
A number is divisible by 8 if its last three digits form a number that is divisible by 8.
Solution:
The last three digits of the given number are 568.
Check if 568 is divisible by 8:
$$\frac{568}{8}$$​ = 71
Since 568 is already perfectly divisible by 8, the condition is satisfied regardless of the digit placed at position A.
The possible values for A are 0 to 9.
The minimum value that can be assigned to A is 0.
Final Answer
So the correct answer is (a)

Given
Number = 648XYZ
Divisible by = 4, 9, 10
Y is the smallest possible number
Solution
Since the number is divisible by 10, its last digit must be 0.
So, Z = 0.
The number becomes 648XY0.
For divisibility by 4, the last two digits (Y0) must be divisible by 4.
Possible values for Y are 0, 2, 4, 6, 8.
Since Y must be the smallest possible number, Y = 0.
The number becomes 648X00.
For divisibility by 9, the sum of digits must be a multiple of 9.
Sum of digits = 6 + 4 + 8 + X + 0 + 0 = 18 + X.
For 18 + X to be divisible by 9, the smallest single-digit value for X is 0.
So, X = 0.
Required value = X + Z - Y
= 0 + 0 - 0 = 0
Final Answer
So the correct answer is (b)

Given
Numerator increased by = 25%
Denominator decreased by = 9%
New fraction = \frac{5}{7}
Formula Used
New Fraction = $$\frac{Numerator × (100 + Increase\%)}{Denominator × (100 - Decrease\%)}$$​​
Solution
Let the original fraction be $$\frac{x}{y}$$​​

$$\frac{x × 125}{y × 91} = \frac{5}{7} \\\ \\\frac{x}{y} = \frac{5 × 91}{7 × 125} \\\ \\ \\\ \\\frac{x}{y} = \frac{1 × 13}{1 × 25} = \frac{13}{25} \\\ \\$$​​

Original fraction = $$\frac{13}{25}$$​​
Difference between denominator and numerator = 25 - 13 = 12
Final Answer
So the correct answer is (d)

Given:
Sum of the two numbers = 672
Highest Common Factor (HCF) = 28
Formula Used:
Let the numbers be 28a and 28b, where 'a' and 'b' are coprime integers.
Solution:
Since the HCF is 28, the numbers can be expressed as 28a and 28b.
According to the problem, their sum is 672:
28a + 28b = 672
Divide the entire equation by 28:
a + b = \frac{672}{28}
a + b = 24
Now, we need to find pairs of coprime numbers (a, b) that add up to 24. Coprime numbers are numbers whose only common factor is 1.
Possible pairs summing to 24:
(1, 23) -> Coprime
(2, 22) -> Not coprime (common factor 2)
(3, 21) -> Not coprime (common factor 3)
(4, 20) -> Not coprime (common factor 4)
(5, 19) -> Coprime
(6, 18) -> Not coprime (common factor 6)
(7, 17) -> Coprime
(8, 16) -> Not coprime (common factor 8)
(9, 15) -> Not coprime (common factor 3)
(10, 14) -> Not coprime (common factor 2)
(11, 13) -> Coprime
(12, 12) -> Not coprime
The coprime pairs are: (1, 23), (5, 19), (7, 17), and (11, 13).
Total number of valid pairs is 4.
Final Answer
So the correct answer is (a)

Given:
Length of courtyard = 15 m 17 cm = 1517 cm
Width of courtyard = 17 m 22 cm = 1722 cm
Formula Used:
Maximum side length of a square tile = HCF of Length and Width
Solution:
To find the largest size of the square tile, we need to calculate the Highest Common Factor (HCF) of 1517 cm and 1722 cm.
Using the division method or difference method:
Difference = 1722 - 1517 = 205
Factors of 205 are 1, 5, 41, 205.
Check divisibility by 41:
1517 = 41 × 37
1722 = 41 × 42
Since 41 divides both numbers evenly and 37 and 42 are co-prime, the HCF is 41 cm.
The side length of the largest square tile is 41 cm.
Final Answer
So the correct answer is (a)

Solution
Let the number be x.
​$$\frac{4}{15} × \frac{5}{7} × x = \frac{4}{21}x \\\ \\\\frac{4}{9} × \frac{2}{5} × x = \frac{8}{45}x \\\ \\\frac{4}{21}x - \frac{8}{45}x = 12\\\ \\$$​​
LCM of 21 and 45 is 315.
​$$\frac{60x - 56x}{315} = 12 \\\ \\\frac{4x}{315} = 12\\\ \\x = \frac{12 × 315}{4}\\\ \\x = 3 × 315 = 945\\\ \\$$​​
One third of the number = $$\frac{945}{3} = 315$$​​

Final Answer
So the correct answer is (a)

Given
Sum of the two numbers = 270
HCF of the two numbers = 18
Formula Used
Numbers can be represented as HCF × a and HCF × b, where a and b are coprime integers.
Solution
Let the two unknown numbers be 18a and 18b.
According to the given sum:
18a + 18b = 270
18(a + b) = 270
​$$a + b= \frac{270}{18} = 15$$​​
Identify pairs of positive coprime numbers (a, b) that sum to 15:
Pair 1: (1, 14)
Pair 2: (2, 13)
Pair 3: (4, 11) (Note: 3 and 12 are rejected as they share a common factor of 3)
Pair 4: (7, 8) (Note: 5 and 10 share 5; 6 and 9 share 3, so they are rejected)
Total number of valid coprime pairs formed = 4
Final Answer
So the correct answer is (b) 
Exam Hall Method: 

Given
Light 1 changes after = 32 seconds
Light 2 changes after = 48 seconds
Light 3 changes after = 76 seconds
Starting time = 7:00 AM
Formula Used
Time of simultaneous change = LCM(T1, T2, T3)
Solution
Find the LCM of 32, 48, and 76 using prime factorization:
32 = $$2^5$$​​
48 = $$2^4 × 3$$​​
76 = $$2^2 × 19$$​​
LCM = $$2^5 × 3 × 19 = 32 × 57 = 1824$$​ seconds
Convert the total seconds into minutes and seconds:
$$1824 \div 60$$​ = 30 minutes and 24 seconds
Add this duration to the given starting time:
7:00:00 AM + 30 minutes 24 seconds = 7:30:24 AM
Final Answer
So the correct answer is (b)