SSC CGL 2025 PYQs

Concept: Midpoint = $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.

1. $x=\frac{1+5}{2}=3$
2. $y=\frac{2+8}{2}=5$
3. Midpoint = (3, 5)

Yaad rakhein: Midpoint is just the average of the coordinates.

Answer: (C) (3, 5)

Concept: Diagonals of an n-sided polygon = $\frac{n(n-3)}{2}$.

1. Decagon: n = 10
2. Diagonals = $\frac{10\times 7}{2}=35$

Trick: Alternatively ${}^{10}{C}_2-10=45-10=35$ (all joining lines minus sides).

Answer: (A) 35

Concept: Centroid G divides median AD as $AG:GD=2:1$.

1. $G=\frac{A+2D}{3}\to 2D=3G-A$
2. x: $2D_x=18-3=15\to D_x=7.5$
3. y: $2D_y=21-5=16\to D_y=8\to D=(7.5,8.0)$

Trick: $D=G+\frac{G-A}{2}$: $x=6+1.5$, $y=7+1$.

Answer: (C) (7.5, 8.0)

Concept: Pick x so both $(x+5)$ and $(x-4)$ become perfect squares.

1. The two squares differ by $(x+5)-(x-4)=9$.
2. Squares 25 and 16 differ by 9, and $5+4=9$ ✓.
3. $x+5=25\to x=20$

Trick: Option test: $x=20\to \sqrt{25}+\sqrt{16}=5+4=9$ ✓.

Answer: (B) 20

Concept: Transverse (internal) common tangent = $\sqrt{d^2-(r_1+r_2{)}^2}$.

1. ${12}^2={20}^2-(10+r{)}^2$
2. $(10+r{)}^2=400-144=256$
3. $10+r=16\to r=6$ cm

Trick: 12, 16, 20 is a 3-4-5 triple (×4) → sum of radii = 16 directly.

Answer: (A) 6 cm

Concept: An angle subtended by a diameter at the circumference is 90°.

1. PR is a diameter, so ∠PSR = 90°.
2. In △PRS: ∠RPS = 180° − 90° − 55° = 35°.

Trick: In a right triangle formed on a diameter, the two acute angles add to 90°: 90° − 55° = 35°.

Answer: (B) 35°

Concept: $a^2-b^2=(a-b)(a+b)$ and ${\sin }^{2}A+{\cos }^{2}A=1$.

1. ${\sin }^{4}A-{\cos }^{4}A=({\sin }^{2}A-{\cos }^{2}A)(1)=\frac{1}{2}$
2. $(1-{\cos }^{2}A)-{\cos }^{2}A=\frac{1}{2}$
3. $1-2{\cos }^{2}A=\frac{1}{2}\to {\cos }^{2}A=\frac{1}{4}$

Trick: ${\sin }^{4}A-{\cos }^{4}A=-\cos 2A=\frac{1}{2}\to \cos 2A=-\frac{1}{2}\to A={60}^{\circ },{\cos }^2{60}^{\circ }=\frac{1}{4}$.

Answer: (B) $\frac{1}{4}$

Concept: ${\sec }^{2}A=1+{\tan }^{2}A$ and $\tan A=\frac{1}{\cot A}$.

1. $\tan A=\frac{1}{x+\frac{1}{x}}$
2. ${\tan }^{2}A=(x+\frac{1}{x}{)}^{-2}$
3. ${\sec }^{2}A=1+(x+\frac{1}{x}{)}^{-2}$

Trick: ${\sec }^{2}A$ is always $>1$ and here the added part must be the reciprocal square of cot A — only (C) has the "1 +" and the power −2.

Answer: (C) 1 + (x + $\frac{1}{x}$)⁻²

Concept: If every term doubles, the sum of cubes becomes $2^3=8$ times.

1. $0.22=2\times 0.11,0.16=2\times 0.08$
2. Denominator = $8({0.11}^3+{0.08}^3)$
3. Value = $\frac{1}{8}=0.125$

Trick: Ratio of bases is 1 : 2 → answer = $(\frac{1}{2}{)}^3=\frac{1}{8}$.

Answer: (A) 0.125

Concept: If each numerator term is k times the matching denominator term, the ratio of cube sums is $k^3$.

1. $\frac{0.6}{0.2}=\frac{0.06}{0.02}=\frac{0.006}{0.002}=3$
2. Numerator = $3^3\times \text{denominator}$
3. Value = 27

Trick: Spot a constant term-by-term ratio and cube it — no cubing of decimals needed.

Answer: (C) 27

Concept: Trees in a square = $n^2$, where n = trees per row.

1. Trees used = $4112-16=4096$.
2. $n^2=4096\to n=\sqrt{4096}=64$.

Trick: $4096=2^{12}$ → √ $=2^6=64$. Or: unit digit 6 → root ends in 4 or 6; ${64}^2=4096$.

Answer: (C) 64

Concept: Convert each % into a ratio and chain them.

1. Q is 20% more than R → R : Q = 5 : 6.
2. P is 25% more than Q → Q : P = 4 : 5.
3. $\frac{P}{R}=(\frac{6}{5})\times (\frac{5}{4})=\frac{6}{4}=\frac{3}{2}\to P:R=3:2$.

Trick: Multiply the factors: $1.2\times 1.25=1.5=\frac{3}{2}$.

Answer: (A) 3 : 2

Concept: For $\sqrt{n+\sqrt{n+\sqrt{n+\dots }}}$, let x = value; then $x^2=n+x$.

1. $x^2-x-30=0$.
2. $(x-6)(x+5)=0\to x=6$ (value must be positive).

Trick: Split n as $k(k+1)$: $30=5\times 6$. With '+' signs take the larger factor (6); with '−' signs take the smaller.

Answer: (D) 6

Concept: $\text{SP}=\text{MP}\times \frac{100-d}{100}$; profit% is on CP.

1. $\text{SP}=400\times \frac{80}{100}=$ ₹320.
2. Profit = $320-200=$ ₹120.
3. Profit% = $\frac{120}{200}\times 100=60\%$.

Trick: MP is $2\times \text{CP}$; 80% of 2 = 1.6 → 60% profit straight away.

Answer: (A) 60% Profit

Concept: Use LCM as tank capacity; outlet's rate is subtracted.

1. Capacity = $\text{LCM}(15,30,60)=60$ units.
2. $A=\frac{60}{15}=4,B=\frac{60}{30}=2,C=-\frac{60}{60}=-1$ unit/min.
3. Net = 5 units/min → time = $60÷5=12$ min.

Trick: Combined time must be less than A's 15 min alone – only 12 and 10 qualify; check 12.

Answer: (A) 12 min

Concept: If $x+y+z=0$, then $x^3+y^3+z^3=3xyz$.

1. Take $x=a-b,y=b-c,z=c-a$.
2. $x+y+z=a-b+b-c+c-a=0$.
3. So the sum of cubes = $3(a-b)(b-c)(c-a)$.

Trick: Spot that the three brackets cancel out – then apply $3xyz$ directly, no expansion.

Answer: (D) 3(a − b)(b − c)(c − a)

Concept: Let the part at 6% be x; total yearly interest = ₹630.

1. $6\%$ of $x$ + $9\%$ of $(9000-x)=630$.
2. $0.06x+810-0.09x=630\to 0.03x=180$.
3. x = ₹6000.

Trick: Average rate = $\frac{630}{9000}=7\%$. Distance from 9 is 2, from 6 is 1 → shares 2 : 1 → 6% part = $\frac{2}{3}\times 9000$.

Answer: (A) ₹6000

Concept: Let Asha get x; $\text{SI}=\frac{P\times R\times T}{100}$.

1. Asha: $\frac{x\times 10\times 3}{100}=0.3x$; Meena: $\frac{(15000-x)\times 12\times 3}{100}=0.36(15000-x)$.
2. $0.3x+5400-0.36x=4950\to 0.06x=450$.
3. $x=$ ₹7,500.

Trick: Average rate = $4950÷3÷15000=11\%$, exactly midway between 10% and 12% → equal halves.

Answer: (B) ₹7,500

Concept: Ground distance is adjacent to the 60° angle; cable is the hypotenuse → use cos.

1. $\cos {60}^{\circ }=\frac{\text{base}}{\text{cable}}\to \frac{1}{2}=\frac{20}{\text{cable}}$.
2. Cable = 40 m.

Trick: In a 30-60-90 triangle the side next to 60° is half the hypotenuse.

Answer: (C) 40 m

Concept: Equate $\frac{1}{3}\pi R^{2}h$ with $\frac{4}{3}\pi r^3$.

1. $h=3R\to \text{cone volume}=\frac{1}{3}\pi R^2\times 3R=\pi R^3$.
2. $\pi R^3=\frac{4}{3}\pi r^3\to \frac{R^3}{r^3}=\frac{4}{3}$.
3. $R:r=\sqrt[3]{4}:\sqrt[3]{3}$.

Trick: Cone with $h=3R$ has volume exactly $\pi R^3$ – remember it.

Answer: (B) ∛4 : ∛3