SSC CGL Number System Questions with Detailed Solutions

Given:

Find the units digit of $$23^{65} \times 36^{94} \times 88^{77}.$$​​

Solution:

​$$23^{65}$$​ has the same units digit as $$3^{65}$$​. The 3-cycle is [3,9,7,1] repeating every 4 powers.

After every block of 4 powers, the pattern restarts. 64 powers make 16 full blocks, so the next the 6$$5^\text{th}$$​ is the 1st item of the cycle

units digit = 3.

​$$36^{94}$$​ has the same units digit as 6$$^{94}$$​. Any power of 6 ends in 6.

​$$88^{77}$$​ has the same units digit as 8$$^{77}$$​. The 8-cycle is [8,4,2,6] repeating every 4 powers.

76 powers make 19 full blocks, so the next the 77$$^\text{th}$$​ is the 1st item of the cycle

units digit = 8.

Now multiply the units digits:
$$3 \times 6 \times 8 → 3 \times 6 = 18 (\text{units} 8); (8 \times 8 = 64$$​) (units 4).

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:
The product of two relatively prime (coprime) integers equals 180.
Formula Used:
Number of distinct coprime pairs for a number with n unique prime factors is $$2^{n-1}.$$​​
Solution:
Performing the prime factorization of 180:
180 = $$2^2 × 3^2 × 5$$​​
The distinct prime factors are 2, 3, and 5.
To form valid coprime pairings, the individual prime powers must act as indivisible blocks.
The total count of unique prime factors, n, is 3.
Applying the formula to find the number of pairs:
$$2^{3-1} = 2^2$$​ = 4
Final Answer
So the correct answer is (d)

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
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
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:
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:
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)