SSC CGL 2025 PYQs

Concept: $\frac{1}{a-b}+\frac{1}{a+b}=\frac{2a}{a^2-b^2}$.

1. Here $a=\sqrt{5},b=\sqrt{2}$.
2. Numerator = $2\sqrt{5}$; denominator = $5-2=3$.
3. Sum = $\frac{2\sqrt{5}}{3}$.

Trick: The $\sqrt{2}$ parts cancel, so only $\sqrt{5}$ can appear in the answer.

Answer: (D) $\frac{2\sqrt{5}}{3}$

Concept: $1+{\cot }^{2}A={cosec}^{2}A$.

1. ${\cot }^{2}A={cosec}^{2}A-1$.
2. $=y^2-1$.

Trick: Check with $A={45}^{\circ }$: $cosec=\sqrt{2}$, $y^2-1=1={\cot }^2{45}^{\circ }$ ✓.

Answer: (C) y² − 1

Concept: The angle between two tangents and the angle at the centre are supplementary.

1. Each radius–tangent angle = 90°.
2. Quadrilateral sum: 90 + 90 + 70 + θ = 360.
3. θ = 110°.

Trick: Directly: angle at centre = 180° − 70° = 110°.

Answer: (B) 110°

Concept: (Tangent)² = external part × whole secant.

1. Let $\text{LM}=x\Rightarrow \text{LN}=x+15$.
2. $x(x+15)=100\Rightarrow x^2+15x-100=0$.
3. $(x+20)(x-5)=0\Rightarrow x=5$ cm.

Trick: Try options: $5\times 20=100$ ✓ instantly.

Answer: (C) 5 cm

Concept: Take the common factor out of numerator and denominator.

1. Numerator = ${0.3}^3(1+{0.1}^3)$; denominator = ${0.6}^3(1+{0.1}^3)$.
2. Ratio = $(\frac{0.3}{0.6}{)}^3=(\frac{1}{2}{)}^3$.
3. $=\frac{1}{8}=0.125$.

Trick: Each denominator term is exactly double the numerator term ⇒ answer is $(\frac{1}{2}{)}^3$.

Answer: (D) 0.125

Concept: Treat $\sqrt{a}$ and $\sqrt{b}$ as unknowns, solve by adding and subtracting, then square.

1. $\sqrt{a}+\sqrt{b}=7$ and $\sqrt{a}-\sqrt{b}=1$
2. Add: $2\sqrt{a}=8\to \sqrt{a}=4\to a=16$
3. Subtract: $2\sqrt{b}=6\to \sqrt{b}=3\to b=9$

Trick: Options (A) and (D) give $a+b=7$ – a trap for those who forget the roots; the answer must be perfect squares, 16 and 9.

Answer: (C) a = 16, b = 9

Concept: Start from the smallest 6-digit number and test whether it is a perfect square.

1. Smallest 6-digit number = 100000.
2. ${316}^2=99856$ and ${317}^2=100489$, so 100000 lies between two squares → not a perfect square.
3. Since it is already the smallest 6-digit number, it is the answer.

Trick: ${10}^5$ has an odd power of 10 (odd number of zeros) – it can never be a perfect square.

Answer: (B) 100000

Concept: If x% of p = y% of q, then p : q = y : x.

1. $37.5\%=\frac{3}{8}$, $25\%=\frac{1}{4}$
2. $\frac{3}{8}p=\frac{1}{4}q\to \frac{p}{q}=\frac{1}{4}\times \frac{8}{3}=\frac{2}{3}$
3. p : q = 2 : 3

Trick: Cross-swap the percentages: p : q = 25 : 37.5 = 2 : 3.

Answer: (D) 2 : 3

Concept: Find the cost price per kg of the mixture (weighted average), then compare with SP.

1. CP of 8 kg = $3\times 40+5\times 50=120+250=$ ₹370 → ₹46.25 per kg
2. SP per kg = ₹55.50 → profit = $55.50-46.25=$ ₹9.25
3. Gain% = $\frac{9.25}{46.25}\times 100=20\%$

Trick: SP is clearly above both ₹40 and ₹50, so it must be a gain – remove both loss options; then $46.25\times 1.2=55.50$ confirms 20%.

Answer: (B) 20% Gain

Concept: Each discount is applied on the price left after the previous one.

1. After 30% off: $800\times \frac{70}{100}=$ ₹560
2. After 20% more: $560\times \frac{80}{100}=$ ₹448
3. Final SP = ₹448

Trick: Net discount of 30% and 20% = $30+20-\frac{30\times 20}{100}=44\%$ → $800\times 0.56=$ ₹448.

Answer: (C) ₹448

Concept: Take CP = 100; discount% is always on the marked price.

1. CP = 100 → MP = 125, SP = 120
2. Discount = $125-120=5$
3. Discount% = $\frac{5}{125}\times 100=4\%$

Trick: Formula: $(100+\text{markup})(100-d)=100(100+\text{profit})$ → $125(100-d)=12000\to 100-d=96\to d=4\%$.

Answer: (B) 4%

Concept: Use $a^3+b^3=(a+b{)}^3-3ab(a+b)$.

1. Let $a+b=s$: $s^3-3\times 15\times s=152\to s^3-45s=152$
2. Try $s=8$: $512-360=152$ ✓
3. So $a+b=8$ (e.g. $a=5$, $b=3$)

Trick: With $ab=15$ test the factor pair 5 and 3: $5^3+3^3=125+27=152$ → $a+b=8$ in seconds.

Answer: (D) 8

Concept: $R=\frac{SI\times 100}{P\times T}$.

1. SI = $4020-3000=$ ₹1020
2. $R=\frac{1020\times 100}{3000\times 5}=\frac{102000}{15000}$
3. $R=6.8\%$

Trick: Total growth = $\frac{1020}{3000}=34\%$ in 5 years → $34÷5=6.8\%$ per year.

Answer: (D) 6.8%

Concept: Same time for both, so add the yearly interests and divide the total interest by it.

1. Yearly SI on ₹8000 at 6% = ₹480
2. Yearly SI on ₹5000 at 8% = ₹400 → total ₹880 per year
3. T = $2640÷880=3$ years

Trick: 2640 is exactly $3\times 880$ – check divisibility of the total by the yearly interest straight away.

Answer: (D) 3 years

Concept: The ramp is the hypotenuse; height is opposite the 30° angle, so use sin.

1. $\sin {30}^{\circ }$ = height/ramp $=\frac{1}{2}$
2. $\frac{2.5}{\text{ramp}}=\frac{1}{2}$
3. Ramp = $2.5\times 2=5$ m

Trick: In a 30° right triangle, the hypotenuse is always twice the side opposite 30°.

Answer: (C) 5 m

Concept: With the same height h, distance = $\frac{h}{\tan \theta }$.

1. From A: $\tan {30}^{\circ }=\frac{h}{DA}\to DA=h\sqrt{3}$
2. From B: $\tan {45}^{\circ }=\frac{h}{DB}\to DB=h$
3. $DA:DB=h\sqrt{3}:h=\sqrt{3}:1$

Trick: Distance ratio = $\cot {30}^{\circ }:\cot {45}^{\circ }=\sqrt{3}:1$ – no need to assume h.

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

Concept: Surface area of sphere = $4\pi r^2$.

1. $4\times \frac{22}{7}\times r^2=154$
2. $r^2=154\times \frac{7}{88}=12.25$
3. $r=3.5$ cm → diameter = 7 cm

Trick: $4\pi r^2=\pi d^2$, so $\frac{22}{7}{d}^2=154\to d^2=49\to d=7$ directly.

Answer: (C) 7 cm

Concept: Cone volume ∝ $r^{2}h$.

1. New cone: radius $\frac{r}{2}$, height $\frac{h}{4}$
2. New volume = ${\left(\frac{1}{2}\right)}^2\times \frac{1}{4}=\frac{1}{16}$ of original
3. Unused = $1-\frac{1}{16}=\frac{15}{16}$

Trick: Scale factors multiply: ${\left(\frac{1}{2}\right)}^2\times \frac{1}{4}=\frac{1}{16}$ used → $\frac{15}{16}$ left.

Answer: (B) $\frac{15}{16}$

Concept: A regular hexagon = 6 equilateral triangles: area = $6\times \frac{\sqrt{3}}{4}{a}^2=\frac{3\sqrt{3}}{2}{a}^2$.

1. Hexagon area = $\frac{3\sqrt{3}}{2}{a}^2$
2. Square area = $a^2$
3. Ratio = $\frac{3\sqrt{3}}{2}:1=3\sqrt{3}:2$

Trick: Hexagon area $\approx 2.6a^2$, more than the square's $a^2$ – so the ratio must exceed 1; only $3\sqrt{3}:2$ (≈ 2.6) and 3 : 1 qualify, and $2.6\ne 3$.

Answer: (A) 3$\sqrt{3}$ : 2

Concept: Volume of prism = base area × height; layers with the same base simply add their heights.

1. Base area = $\frac{\sqrt{3}}{4}\times 6^2=9\sqrt{3}$ cm²
2. Total height = $1+2+3+4+5=15$ cm
3. Volume = $9\sqrt{3}\times 15=135\times 1.732=233.82$ cm³

Trick: $135\times 1.732$: $135\times 1.7=229.5$, plus $135\times 0.032=4.32$ → 233.82.

Answer: (B) 233.82 cm³