SSC CGL Maths Questions with Detailed Solutions

​$$\textbf{Given:} \\[4pt]\text{Perimeter of Triangle } P = 30 \text{ cm} \\[4pt]\text{Perimeter of Triangle } Q = 45 \text{ cm} \\[4pt]\text{Area of Triangle } P = 50 \text{ cm}^2 \\[8pt]\textbf{Formula Used:} \\[4pt]\text{For similar triangles,} \\[2pt]\frac{\text{Area}_1}{\text{Area}_2}=\left(\frac{\text{Side}_1}{\text{Side}_2}\right)^2 \\[8pt]\textbf{Solution:} \\[6pt]\text{Ratio of corresponding sides} \\[2pt]=\frac{\text{Perimeter of } P}{\text{Perimeter of } Q}=\frac{30}{45}=\frac{2}{3} \\[10pt]\frac{50}{\text{Area of } Q}=\left(\frac{2}{3}\right)^2=\frac{4}{9} \\[10pt]\text{Area of } Q=50 \times \frac{9}{4}=112.5 \text{ cm}^2 \\[10pt]\boxed{\text{Area of Triangle } Q = 112.5 \text{ cm}^2}$$​​

Given:
Triangle ABC with D on AB such that AD = BD = CD.
​$$\angle BAC = 25^\circ.$$​​
Concept Used:
If the median to a side is half the length of that side, the triangle is right-angled at the vertex opposite to that side. Alternatively, D is the circumcenter lying on the hypotenuse AB.
Solution:
Analyze angles.
Since AD = CD,  $$\triangle$$​ADC is isosceles.
​$$\angle ACD = \angle CAD = 25^\circ.$$​​
Since BD = CD, $$\triangle BDC$$​ is isosceles.
Let $$\angle CBD = \angle BCD = x.$$​​
Find x.
Sum of angles in \triangle ABC is $$180^\circ.$$​​
​$$\angle A + \angle B + \angle C = 180^\circ\\25^\circ + x + (25^\circ + x) = 180^\circ\\50^\circ + 2x = 180^\circ \implies 2x = 130^\circ \implies x = 65^\circ$$​​
So, $$\angle ABC = 65^\circ and \angle ACB = 25+65 = 90^\circ.$$​​
Calculate the required value.
So the correct answer is (b)

Given:
Distance = 540 km.
Time taken after meeting: $$t_A = 4 hrs, t_B = 9$$​ hrs.
Formula Used:
1. Ratio of speeds: $$\frac{S_A}{S_B} = \sqrt{\frac{t_B}{t_A}}\\$$​​
2. Total Distance  $$= (S_A + S_B) \times \text{time to meet}\\$$​​
3. Time to meet t 
Solution:
Calculate speed ratio.
​$$\frac{S_A}{S_B} = \sqrt{\frac{9}{4}} = \frac{3}{2}\\$$​​
Let $$S_A = 3x$$​ and $$S_B = 2x.$$​​
Calculate time to meet.
​$$t = \sqrt{4 \times 9} = 6 hours.$$​​
Use distance to find x.
Distance $$= (S_A + S_B) \times t\\$$​​
​$$540 = (3x + 2x) \times 6\\540 = 30x\\x = 18\\$$​​
 Find faster speed.
​$$S_A = 3(18) = 54 \text{ km/h}\\S_B = 2(18) = 36 \text{ km/h}\\$$​​
Faster is 54 km/h.
Exam-Hall Method:
Ratio 3:2. Time meet 6h. Speed sum = 540/6 = 90. 3 units + 2 units = 90 $$\implies$$​ 1 unit = 18. Faster = 3 units = 54.
So the correct answer is (b)

Concept Used:
1. Divisibility by 2: Last digit is even.
2. Divisibility by 5: Last digit is 0 or 5.
3. Divisibility by 2 and 5 (i.e., 10): Last digit must be 0.
4. Divisibility by 9: Sum of digits is divisible by 9.
Solution:
Check condition for 2 and 5 (Must end in 0).
A. 4680 (Ends in 0) - Pass
B. 7245 (Ends in 5) - Fail (Not divisible by 2)
C. 8190 (Ends in 0) - Pass
D. 3420 (Ends in 0) - Pass
Check divisibility by 9 for passing numbers.
A. 4680: 4+6+8+0 = 18 (Divisible by 9) - Yes
C. 8190: 8+1+9+0 = 18 (Divisible by 9) - Yes
D. 3420: 3+4+2+0 = 9 (Divisible by 9) - Yes
Conclusion.
Numbers A, C, and D satisfy all conditions.
So the correct answer is (b)

Given:
Solid: Pentagonal Prism.
Concept Used:
For a prism with base sides n:
Faces F = n + 2
Vertices V = 2n
Edges E = 3n
Solution:
Step 1: Determine F, V, E for Pentagonal Prism (n=5).
F = 5 + 2 = 7
V = 2(5) = 10
E = 3(5) = 15
 Check Option A.
2F + 3V - 2E = 2(7) + 3(10) - 2(15)
= 14 + 30 - 30 = 14 (True)
So the correct answer is (a)

Given:
Equation:$$\frac{(x + 2) - (3x + 5)}{2 - 3x} = \frac{1}{11}\\$$​​
Solution:
Simplify the numerator.
(x + 2) - (3x + 5) = x + 2 - 3x - 5 = -2x - 3
Solve for x.
​$$\frac{-2x - 3}{2 - 3x} = \frac{1}{11}\\$$​​
Cross-multiply:
​$$11(-2x - 3) = 1(2 - 3x)\\-22x - 33 = 2 - 3x\\-33 - 2 = -3x + 22x\\-35 = 19x \implies x = -\frac{35}{19}\\$$​​

Option C:
1 − x = 18x + 36
Solve it:
1 − 36 = 18x + x
−35 = 19x
x = −35/19
So the correct answer is (c)

Given:
Item P: SP = ₹ 510, Loss = 15%
Item Q: SP = ₹ 720, Profit = 20%
Combined SP = ₹ 1320
Formula Used:
CP  $$= \frac{SP \times 100}{100 \pm \text{Profit/Loss}\%}$$​​
Profit = SP - CP
​$$\text{Profit}\% = \frac{\text{Profit}}{CP} \times 100\\$$​​
Solution:
Calculate CP of Item P.
​$$CP_P = \frac{510 \times 100}{100 - 15} = \frac{51000}{85} = 600\\$$​​
Calculate CP of Item Q.
​$$CP_Q = \frac{720 \times 100}{100 + 20} = \frac{72000}{120} = 600\\$$​​
Calculate Total CP and Total SP.
Total CP = 600 + 600 = 1200
Total SP = 1320 (Given)
Calculate Profit Percent.
Profit = 1320 - 1200 = 120\
​$$\text{Profit}\% = \frac{120}{1200} \times 100 = 10\%\\$$​​
So the correct answer is (a)

​$$141120 = p^2 \times q^6 \times r^1 \times s^2$$​​
where p, q, r, s are prime numbers.

Find the value of (p + q + r − s).

Solution

First, find the prime factorization of 141120.

141120 = 14112 × 10
= (14112) × (2 × 5)

Now factorize 14112:

14112 = 2⁵ × 3² × 7²

So,
141120 = 2⁶ × 3² × 5¹ × 7²

Compare with given form

p² × q⁶ × r¹ × s²

Matching powers:

q⁶ = 2⁶ => q = 2
p² = 3² => p = 3
r¹ = 5¹ => r = 5
s² = 7² => s = 7

Required value

p + q + r − s
= 3 + 2 + 5 − 7
= 3
Correct option: C

Given:
Number 1:$$\left(-\frac{2}{3}\right)^{-1} = -\frac{3}{2}\\$$​​
Target Product: Reciprocal of $$\left(-\frac{5}{3}\right)^{-1}\\$$​​
Solution:
Simplify the target.
​$$\left(-\frac{5}{3}\right)^{-1} = -\frac{3}{5}\\$$​​
Reciprocal of this is $$-\frac{5}{3}.$$​​
Set up the equation.
Let the number be x.
​$$x \times \left(-\frac{3}{2}\right) = -\frac{5}{3}\\x = \left(-\frac{5}{3}\right) \div \left(-\frac{3}{2}\right)\\x = \left(-\frac{5}{3}\right) \times \left(-\frac{2}{3}\right)\\x = \frac{10}{9}\\$$​​
So the correct answer is (c)

​$$\textbf{Given:} \\ \space \, \\ \text{Initial Value} = ₹75 \\ \space \, \\ \text{First change: Increase by } 20\% \\ \space \, \\ \text{Second change: Decrease by } 20\% \\ \space \, \\ \textbf{Formula Used:} \\ \space \, \\ \text{Final Value} = \text{Initial Value} \times \left(1 + \frac{R_1}{100}\right) \times \left(1 - \frac{R_2}{100}\right) \\ \space \, \\ \textbf{Solution:} \\ \space \, \\ \text{Step 1: Increase by 20\%} \\ \space \, \\ \text{New Value} = 75 \times \left(1 + \frac{20}{100}\right) \\ \space \, \\ = 75 \times 1.2 = 90 \\ \space \, \\ \text{Step 2: Decrease by 20\%} \\ \space \, \\ \text{Final Value} = 90 \times \left(1 - \frac{20}{100}\right) \\ \space \, \\ = 90 \times 0.8 \\ \space \, \\ = 72 \\ \space \, \\ \textbf{Final Answer} \\ \space \, \\ ₹72$$​​

​$$\textbf{Given:} \\ \space \, \\ \text{Length of trains: } L_1 = 105\text{ m}, L_2 = 90\text{ m} \\ \space \, \\ \text{Speeds: } S_1 = 45\text{ km/hr}, S_2 = 72\text{ km/hr} \\ \space \, \\ \text{Direction: Opposite} \\ \space \, \\ \textbf{Concept Used:} \\ \space \, \\ \text{Relative speed in opposite direction is sum of speeds.} \\ \space \, \\ \textbf{Formula Used:} \\ \space \, \\ \text{Time} = \frac{\text{Total Distance}}{\text{Relative Speed}} \\ \space \, \\ \textbf{Solution:} \\ \space \, \\ \text{Total Distance} = L_1 + L_2 = 105 + 90 = 195\text{ m} \\ \space \, \\ \text{Relative Speed} = 45 + 72 = 117\text{ km/hr} \\ \space \, \\ \text{Convert speed to m/s:} \\ \space \, \\ 117 \times \frac{5}{18} = \frac{13 \times 5}{2} = \frac{65}{2} = 32.5\text{ m/s} \\ \space \, \\ \text{Calculate Time:} \\ \space \, \\ T = \frac{195}{32.5} \\ \space \, \\ T = 6 \text{ seconds} \\ \space \, \\ \textbf{Final Answer} \\ \space \, \\ 6 \text{ Seconds}$$​​

​$$\textbf{Given:} \\ \space \, \\ 3 \sin \theta \cos(90^\circ - \theta) = 1 \\ \space \, \\ \textbf{Formula Used:} \\ \space \, \\ \cos(90^\circ - \theta) = \sin \theta \\ \space \, \\ \sin 2\theta = 2\sin\theta\cos\theta, \quad \cos 2\theta = 1 - 2\sin^2\theta \\ \space \, \\ \textbf{Solution:} \\ \space \, \\ \text{Simplify the given equation:} \\ \space \, \\ 3 \sin \theta (\sin \theta) = 1 \\ \space \, \\ \sin^2 \theta = \frac{1}{3} \Rightarrow \sin \theta = \frac{1}{\sqrt{3}} \\ \space \, \\ \text{Calculate } \cos \theta: \\ \space \, \\ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} \\ \space \, \\ \text{Calculate } \sin 2\theta \text{ and } \cos 2\theta: \\ \space \, \\ \sin 2\theta = 2 \left(\frac{1}{\sqrt{3}}\right) \left(\frac{\sqrt{2}}{\sqrt{3}}\right) = \frac{2\sqrt{2}}{3} \\ \space \, \\ \cos 2\theta = 1 - 2\left(\frac{1}{3}\right) = \frac{1}{3} \\ \space \, \\ \text{Sum:} \\ \space \, \\ \sin 2\theta + \cos 2\theta = \frac{2\sqrt{2}}{3} + \frac{1}{3} = \frac{2\sqrt{2} + 1}{3} \\ \space \, \\ \textbf{Final Answer} \\ \space \, \\ \frac{2\sqrt{2} + 1}{3}$$​​

​$$\textbf{Given:} \\ \space \, \\ \text{Length of ladder (hypotenuse)} = 50 \text{ cm} \\ \space \, \\ \text{Angle with wall} = 45^\circ \\ \space \, \\ \textbf{Concept Used:} \\ \space \, \\ \text{In a right-angled triangle formed by the ladder, wall, and ground:} \\ \space \, \\ \cos \theta = \frac{\text{Adjacent (Height)}}{\text{Hypotenuse}} \\ \space \, \\ \textbf{Solution:}$$​​

​$$\text{Let height be } h. \\ \space \, \\ \cos 45^\circ = \frac{h}{50} \\ \space \, \\ \frac{1}{\sqrt{2}} = \frac{h}{50} \\ \space \, \\ h = \frac{50}{\sqrt{2}} = 25\sqrt{2} \text{ cm} \\ \space \, \\ \textbf{Final Answer} \\ \space \, \\ 25\sqrt{2} \text{ cm}$$​​