SSC CGL 2025 PYQs

Concept: $(\sqrt{m}+\frac{1}{\sqrt{m}}{)}^2=m+\frac{1}{m}+2$.

1. $7^2=49=m+\frac{1}{m}+2$.
2. $m+\frac{1}{m}=47$.

Trick: If $x+\frac{1}{x}=k$, then $x^2+\frac{1}{x^2}=k^2-2$; here $x=\sqrt{m}$.

Answer: (C) 47

Concept: $\sec A=2$ means $\cos A=\frac{1}{2}$, so $A={60}^{\circ }$.

1. $\tan {60}^{\circ }=\sqrt{3}$; $\csc {60}^{\circ }=\frac{2}{\sqrt{3}}$.
2. $\sqrt{3}-\frac{2}{\sqrt{3}}=\frac{3-2}{\sqrt{3}}=\frac{1}{\sqrt{3}}$.

Trick: Use a triangle: hyp = 2, base = 1 → perpendicular $=\sqrt{3}$; then $\tan =\sqrt{3}$, $\csc =\frac{2}{\sqrt{3}}$.

Answer: (B) $\frac{1}{\sqrt{3}}$

Concept: Radius $⟂$ tangent at the point of contact, so Pythagoras applies.

1. ${\text{Distance}}^2=9^2+{12}^2=81+144=225$.
2. Distance = 15 cm.

Trick: 9 : 12 : 15 is the 3-4-5 triple × 3.

Answer: (D) 15 cm

Concept: The two tangents from one external point to a circle are equal in length.

1. One tangent = 15 cm.
2. So the other is also 15 cm.

Yaad rakhein: The line joining the external point to the centre bisects the angle between the two tangents.

Answer: (A) 15 cm

Concept: Write the numerator terms as multiples of the denominator terms.

1. Numerator bases (0.09, 0.03) are 0.3 times the denominator bases (0.3, 0.1).
2. Hence numerator = $0.027\times$ (denominator).
3. Value = 0.027.

Trick: Each numerator term is 0.3 × the matching denominator term, so answer $={0.3}^3$.

Answer: (D) 0.027

Concept: $(a+b{)}^2=a^2+b^2+2ab$.

1. $(\sqrt{5}+3{)}^2=(\sqrt{5}{)}^2+3^2+2\times \sqrt{5}\times 3$
2. $=5+9+6\sqrt{5}=14+6\sqrt{5}$
3. Compare with $?+6\sqrt{5}\to ?=14$

Trick: The rational part = sum of squares only: $5+9=14$; the 2ab term is the √ part.

Answer: (D) 14

Concept: Find each square root, subtract, then multiply by $\frac{1}{2}$.

1. $\sqrt{7396}=86$ (${86}^2=7396$)
2. $\sqrt{1156}=34$ (${34}^2=1156$)
3. $(86-34)\times \frac{1}{2}=52\times \frac{1}{2}=26$

Trick: 7396 ends in 6 → root ends in 4 or 6; ${85}^2=7225$, so root is 86.

Answer: (C) 26

Concept: Convert percentages to a simple equation and solve for $\frac{A}{B}$.

1. Divide both sides by 20%: $(A+B)=3\times (A-B)$.
2. Opening the bracket: $2A=4B$.
3. So A is twice B → 2 : 1.

Trick: Use componendo-dividendo: $\frac{A+B}{A-B}=\frac{3}{1}\to A:B=(3+1):(3-1)=4:2=2:1$.

Answer: (B) 2 : 1

Concept: Value of each type = number of coins × coin value.

1. Coins: 2x, 3x, 5x
2. Value: 5(2x) + 2(3x) + 1(5x) = 21x
3. 21x = 252 → x = 12
4. Coins: 24 of ₹5, 36 of ₹2, 60 of ₹1

Trick: Option check: (C) gives 120 + 72 + 60 = ₹252 ✓.

Answer: (C) 24 coins of ₹5, 36 of ₹2, and 60 of ₹1

Concept: $\text{SP}=\text{MP}\times \frac{100-d}{100}$ and $\text{CP}=\text{SP}\times \frac{100}{100+\text{profit}\%}$.

1. $\text{SP}=800\times \frac{85}{100}=$ ₹680
2. $\text{CP}=680\times \frac{100}{120}=680\times \frac{5}{6}$
3. CP = ₹566.67

Trick: $\text{MP}:\text{CP}=(100+P\%):(100-D\%)=120:85\to \text{CP}=800\times \frac{85}{120}=$ ₹566.67.

Answer: (C) ₹566.67

Concept: Find CP from profit, then MP from the first discount, then apply the new discount.

1. 8% of CP = ₹224 → CP = ₹2800; SP = ₹3024
2. SP = 80% of MP → $\text{MP}=3024\times \frac{100}{80}=$ ₹3780
3. $33.33\%=\frac{1}{3}$ discount → new SP = $3780\times \frac{2}{3}=$ ₹2520

Trick: Directly: new SP = old $\text{SP}\times \frac{\frac{2}{3}}{\frac{4}{5}}=3024\times \frac{5}{6}=$ ₹2520.

Answer: (C) ₹2520.00

Concept: For $a{x}^3+b{x}^2+cx+d=0$, sum of roots = $-\frac{b}{a}$.

1. Here $a=1,b=-8$.
2. $x+y+z=-\frac{-8}{1}=8$

Trick: Sum of roots = negative of the $a^2$ coefficient when the leading coefficient is 1; product = $-\frac{d}{a}=4$.

Answer: (D) 8

Concept: For dated loans, count days (exclude the first day, include the last) and use $\text{time}=\frac{\text{days}}{365}$.

1. Days: Mar 16 + Apr 30 + May 31 + Jun 30 + Jul 31 + Aug 31 + Sep 15 = 184
2. $\text{SI}=\frac{6000\times 12\times 184}{100\times 365}=\frac{720\times 184}{365}\approx$ ₹362.96
3. Amount = $6000+362.96=$ ₹6362.96

Trick: 6 months would give ₹360; 184 days is a little more than half a year, so SI is just above 360 → only ₹6362.96 fits.

Answer: (B) ₹6362.96

Concept: For equal SI over the same time, $\text{principal}\propto \frac{1}{\text{rate}}$.

1. $P_1:P_2:P_3=\frac{1}{4}:\frac{1}{5}:\frac{1}{10}=5:4:2$ (LCM 20)
2. Total 11 parts = ₹4400 → 1 part = ₹400
3. 5% scheme = 4 parts = ₹1600

Trick: Check: 4% of 2000 = 5% of 1600 = 10% of 800 = ₹80 each ✓.

Answer: (B) ₹1600

Concept: Horizontal distance = height ÷ tan(angle of depression).

1. 45°: distance = $\frac{150}{\tan {45}^{\circ }}=150$ m
2. 60°: distance = $\frac{150}{\tan {60}^{\circ }}=\frac{150}{\sqrt{3}}$ m
3. Gap = $150-\frac{150}{\sqrt{3}}=150(1-\frac{1}{\sqrt{3}})$ m

Trick: Larger angle = nearer car; gap = $h(\cot {45}^{\circ }-\cot {60}^{\circ })$.

Answer: (C) 150(1 − $\frac{1}{\sqrt{3}}$) m

Concept: tan(elevation) $=\frac{\text{height}}{\text{horizontal distance}}$.

1. At A: $\tan x=\frac{20\sqrt{3}}{20}=\sqrt{3}\to x={60}^{\circ }$
2. At B: $\tan y=\frac{20\sqrt{3}}{60}=\frac{1}{\sqrt{3}}\to y={30}^{\circ }$
3. $x+y={90}^{\circ }$

Trick: If $h^2=d_1\times d_2$ ($1200=20\times 60$), the two angles are complementary → sum 90°.

Answer: (C) 90°

Concept: Volume stays the same when solids are melted and recast.

1. Two hemispheres: $2\times \frac{2}{3}\pi r^3=\frac{4}{3}\pi r^3$
2. This equals a sphere of the same radius.
3. New radius = 27 cm

Trick: Two equal hemispheres always make a sphere of the same radius — no calculation needed.

Answer: (B) 27 cm

Concept: Similar cones: $\text{volume ratio}=(\text{height ratio}{)}^3$.

1. Small top cone height : full cone height = 5 : 20 = 1 : 4
2. Volume ratio = $1^3:4^3=1:64$
3. Frustum = $64-1=63$ parts → 63 : 64

Trick: Cut at $\frac{1}{n}$ of height from top → frustum : cone = $(n^3-1):n^3$.

Answer: (A) 63 : 64

Concept: Perimeter of a regular polygon = number of sides × side.

1. Hexagon side = $\frac{24}{6}=4$ cm
2. Octagon side = $\frac{3}{4}\times 4=3$ cm
3. Octagon perimeter = $8\times 3=24$ cm

Trick: Perimeter ratio = $(8\times \frac{3}{4}):6=1:1$ → same 24 cm.

Answer: (C) 24 cm

Concept: Prism volume = base area × total height; regular hexagon area = $\frac{3\sqrt{3}}{2}{a}^2$.

1. Heights 8, 11, 14, 17 → total = $\frac{4}{2}\times (8+17)=50$ cm
2. Base area = $\frac{3\sqrt{3}}{2}\times 64=96\sqrt{3}$ cm²
3. Volume = $96\sqrt{3}\times 50=4800\sqrt{3}$ cm³

Trick: Hexagon = 6 equilateral triangles: $6\times \frac{\sqrt{3}}{4}\times 64=96\sqrt{3}$.

Answer: (B) 4800$\sqrt{3}$ cm³