SSC CGL 2025 PYQs

Concept: Value of one ratio part = known amount ÷ its ratio term.

1. 9 parts = ₹27,000 → 1 part = ₹3,000.
2. $Q=7\times 3,000=$ ₹21,000.

Trick: $Q=27000\times \frac{7}{9}=3000\times 7$ — answer must be a multiple of 7 thousand.

Answer: (C) ₹21,000

Concept: Interior angle of regular n-gon = ${180}^{\circ }-\frac{{360}^{\circ }}{n}$.

1. Pentagon: $180-72={108}^{\circ }$.
2. Decagon: $180-36={144}^{\circ }$.
3. 108 : 144 = 3 : 4.

Trick: Exterior angle ($\frac{360}{n}$) is faster to find; subtract from 180.

Answer: (A) 3 : 4

Concept: Side = $\sqrt{\text{area}}$, perimeter = $4\times$ side.

1. Side = $\sqrt{196}=14$ m.
2. Perimeter = $4\times 14=56$ m.

Trick: Perimeter of a square must be a multiple of 4 — only 52 and 56 qualify; ${13}^2=169$, so 56.

Answer: (D) 56 m

Concept: Take numbers as 8x, 9x and form an equation.

1. $\frac{8x+8}{9x+8}=\frac{17}{19}$.
2. $152x+152=153x+136\to x=16$.
3. Numbers = 128 and 144.

Trick: Options must be in ratio 8 : 9 — only 128 : 144 ($=16\times 8:16\times 9$) is.

Answer: (B) 128 and 144

Concept: $\frac{P}{S}=\frac{P}{Q}\times \frac{Q}{R}\times \frac{R}{S}$.

1. $\frac{P}{S}=\frac{2}{3}\times \frac{6}{5}\times \frac{10}{7}$.
2. $=\frac{120}{105}=\frac{8}{7}$.
3. P : S = 8 : 7.

Trick: Cancel first: $\frac{2\times 6\times 10}{3\times 5\times 7}$ → $\frac{6}{3}=2$, $\frac{10}{5}=2$ → $2\times 2\times 2:7=8:7$.

Answer: (A) 8 : 7

Concept: Remaining % = 100 − given %; then scale up.

1. Green = $100-25-40=35\%$.
2. $35\%=70\to 1\%=2\to 100\%=200$.

Trick: 70 is 35% → total is just $2\times 100=200$.

Answer: (C) 200

Concept: Work stage by stage on the right base.

1. Attempted = 80% of 180 = 144; correct = 75% of 144 = 108.
2. Not attempted = 36; grace = 25% of 36 = 9.
3. Total = $108+9=117$.

Trick: Correct = $180\times (0.8\times 0.75+0.2\times 0.25)=180\times 0.65=117$.

Answer: (B) 117

Concept: $\frac{MP}{CP}=\frac{100+\text{profit}\%}{100-\text{discount}\%}$.

1. $\frac{MP}{CP}=\frac{115}{75}=\frac{23}{15}$.
2. Markup = $\frac{23-15}{15}\times 100=\frac{800}{15}=53.33\%$ (≈ 53.34%).

Trick: Let CP = 100: $SP=115=75\%$ of $MP\to MP=153.33$ → 53.33% above.

Answer: (D) 53.34%

Concept: $\text{New average}=\text{old average}+\frac{\text{new score}-\text{old average}}{\text{new count}}$.

1. Excess over old average = $110-50=60$.
2. Spread over 11 innings: $\frac{60}{11}=5.45$.
3. New average = $50+5.45=55.45$.

Trick: This deviation method avoids big totals: $50+\frac{60}{11}$.

Answer: (C) 55.45

Concept: Sum of angles of a quadrilateral $={360}^{\circ }$.

1. $P+Q+R=360-90={270}^{\circ }$.
2. Ratio parts = $2+3+4=9$ → 1 part $={30}^{\circ }$.
3. $Q=3\times 30={90}^{\circ }$.

Trick: Q is the middle ratio term → Q = average of P, Q, R = $\frac{270}{3}={90}^{\circ }$.

Answer: (A) 90°

Concept: Effective price = amount paid ÷ items received.

1. Paid = $3\times 800=$ ₹2,400.
2. Items = 4 → $2400÷4=$ ₹600.

Trick: Buy n get 1 free → price $\times \frac{n}{n+1}=800\times \frac{3}{4}=600$.

Answer: (B) ₹600

Concept: Take capacity = LCM; inlets add, outlet subtracts.

1. Capacity = LCM(10, 15, 12) = 60 units.
2. Rates: +6, +4, −5 → net 5 units/min.
3. Time = $60÷5=12$ min.

Trick: $\frac{1}{10}+\frac{1}{15}-\frac{1}{12}=\frac{6+4-5}{60}=\frac{1}{12}$ → 12 min.

Answer: (D) 12 min

Concept: Combined average = total of all values ÷ total count.

1. Juniors: $3\times 25,000=75,000$; seniors: $2\times 50,000=1,00,000$.
2. Total = 1,75,000 for 5 people → ₹35,000.

Trick: Weighted average lies nearer to the bigger group: $25,000+\left(\frac{2}{5}\right)\times 25,000=35,000$.

Answer: (C) ₹35,000

Concept: Removing a part takes milk and water in the current ratio.

1. Start: milk 60 L, water 20 L. Remove 12 L ($\frac{3}{20}$ of mix): milk −9, water −3; add 12 L water → milk 51, water 29.
2. Remove 12 L again: water $-29\times \frac{3}{20}=-4.35\to 24.65$ L; adding milk does not change water.
3. Water % = $\frac{24.65}{80}\times 100=30.81\%$.

Trick: Track only water: $20\times \frac{17}{20}+12=29$; then $29\times \frac{17}{20}=24.65$.

Answer: (B) 30.81%

Concept: $\sin =\frac{\text{perpendicular}}{\text{hypotenuse}}$, $\cot =\frac{\text{base}}{\text{perpendicular}}$.

1. Perpendicular = 35, hypotenuse = 37.
2. Base = $\sqrt{{37}^2-{35}^2}=\sqrt{72\times 2}=\sqrt{144}=12$.
3. $\cot A=\frac{12}{35}$.

Trick: Triplet 12-35-37 is standard; $a^2-b^2=(a+b)(a-b)=72\times 2$.

Answer: (B) $\frac{12}{35}$

Concept: $\pi$ radians = ${180}^{\circ }$.

1. $\frac{5\pi }{6}\times \frac{180}{\pi }=5\times 30$.
2. $={150}^{\circ }$.

Trick: Replace $\pi$ by 180 directly: $5\times \frac{180}{6}={150}^{\circ }$.

Answer: (D) 150°

Concept: If $A+B={90}^{\circ }$, then $\sin B=\cos A$.

1. $\sin A=\frac{3}{5}\to \cos A=\frac{4}{5}$ (3-4-5 triangle).
2. $\sin A\times \sin B=\sin A\times \cos A=\frac{3}{5}\times \frac{4}{5}=\frac{12}{25}$.

Trick: The product $\sin A\cos A=\frac{\sin 2A}{2}$ is always $\le \frac{1}{2}$, ruling out $\frac{25}{12}$ and $\frac{4}{5}$ at once.

Answer: (C) $\frac{12}{25}$

Concept: Orthocentre = meeting point of the three altitudes.

1. In a right triangle the two legs are perpendicular, so each leg is the altitude to the other.
2. They meet at the right-angle vertex — so the orthocentre lies there.
3. Midpoint of the hypotenuse is the circumcentre, not the orthocentre (trap A).

Yaad rakhein: Right triangle: orthocentre at 90° vertex, circumcentre at hypotenuse midpoint.

Answer: (B) At the vertex where the right angle is formed

Concept: Perpendicular from centre bisects the chord: $(\text{half-chord}{)}^2=r^2-d^2$.

1. Half-chord = $\sqrt{{29}^2-{20}^2}=\sqrt{841-400}=\sqrt{441}=21$.
2. Chord = $2\times 21=42$ cm.

Trick: 20-21-29 is a Pythagorean triplet — half-chord is 21 instantly.

Answer: (B) 42 cm

Concept: Radius $⟂$ tangent: ${\text{tangent}}^2=d^2-r^2$.

1. Tangent = $\sqrt{{20}^2-{12}^2}=\sqrt{400-144}=\sqrt{256}$.
2. $=16$ cm.

Trick: 12-16-20 = 4 × (3-4-5) triplet.

Answer: (D) 16 cm