SSC CGL 2025 PYQs

Concept: Average of an AP = $\frac{\text{first term}+\text{last term}}{2}$.

1. Smallest 3-digit multiple of 19: $19\times 6=114$.
2. Largest: $19\times 52=988$ ($19\times 53=1007$).
3. Average = $\frac{114+988}{2}=\frac{1102}{2}=551$.

Trick: Average must itself be a multiple of 19 ($19\times 29=551$) — 552, 752, 352 fail.

Answer: (B) 551

Concept: Take each container = 100 units.

1. Milk = $30+50=80$ units out of 200.
2. Water = $200-80=120$ units.
3. Water% = $\frac{120}{200}\times 100=60\%$.

Trick: Equal volumes → average the water %: $\frac{70\%+50\%}{2}=60\%$.

Answer: (D) 60%

Concept: Form the bill before and after the swap.

1. Blue = ₹p/dozen, red = ₹2p; blue ordered = x dozen.
2. Original bill = $16p+xp$; delivered bill = $2xp+8p$.
3. $2x+8=1.4(16+x)\to 0.6x=14.4\to x=24$.
4. Red : Blue = 8 : 24 = 1 : 3.

Trick: Test option 1 : 3 ($x=24$): $16+24=40\to 48+8=56=1.4\times 40$ ✓.

Answer: (A) 1 : 3

Concept: Total cost must be recovered from the sellable items only.

1. Required SP = $900\times \frac{140}{100}=$ ₹1,260.
2. Only $300-90=210$ flowers can be sold.
3. So each must fetch $1260÷210=$ ₹6.

Trick: $\frac{1260}{210}=\frac{126}{21}=6$ — cancel the zero first.

Answer: (B) ₹6

Concept: Find CP from the loss, then the new SP after discount.

1. Z = 80% of CP → $\text{CP}=\frac{Z}{0.8}=1.25Z$.
2. New SP = $2.6Z\times 60\%=1.56Z$.
3. Profit = $0.31Z\to \frac{0.31}{1.25}\times 100=24.8\%$.

Trick: Take Z = 100: CP = 125, SP = 156 → profit $\frac{31}{125}=24.8\%$.

Answer: (B) 24.8%

Concept: Milk stays constant; only water changes.

1. Milk = 3 parts in both ratios, so 1 part means the same amount before and after.
2. Water: 5 → 7 parts, increase 2 parts = 8 L → 1 part = 4 L.
3. Milk = $3\times 4=12$ L.

Trick: Same milk term (3) in both ratios → compare water parts directly.

Answer: (A) 12 litres

Concept: Wages are shared in proportion to work done.

1. Total work = $\text{LCM}(12,18)=36$ units; Ram = 3/day, Shyam = 2/day.
2. In 5 days: Ram 15, Shyam 10 → 25 units.
3. Ghanshyam did $36-25=11$ units → $\frac{11}{36}\times 3600=$ ₹1,100.

Trick: ₹3600 ÷ 36 units = ₹100 per unit → 11 units = ₹1,100.

Answer: (D) ₹1,100

Concept: Liquid drawn out has the same ratio as the mixture.

1. Let total = 10x: A = 5x, B = 3x, C = 2x.
2. 10 L removed = 5 L A, 3 L B, 2 L C.
3. Then +5 L A → A = 5x; B = 3x − 3.
4. 5x − (3x − 3) = 25 → 2x = 22 → x = 11 → total = 110 L.

Trick: A loses 5 and gets 5 back — so A is unchanged; only B drops by 3.

Answer: (D) 110

Concept: Find A's time, then use $T=\frac{ab}{a+b}$.

1. A : B = 5 : 4 and B = 18 h → A = $18\times \frac{5}{4}=22.5$ h.
2. Together = $\frac{22.5\times 18}{22.5+18}=\frac{405}{40.5}=10$ h.

Trick: Efficiency ratio = 4 : 5 (inverse); B's 5 units/h × 18 h = 90 units; together 9 units/h → 10 h.

Answer: (B) 10 hours

Concept: All meet at the start after LCM of their lap times.

1. Lap time = $\frac{3}{\text{speed}}$: X = 12 min, Y = 18 min, Z = 15 min.
2. $\text{LCM}(12,18,15)=180$ min = 3 h.
3. In 6 hours (9 a.m.–3 p.m.) the 3-hour cycle gives the start moment plus two more (noon, 3 p.m.) → 3 times, counting the start.

Trick: Lap times in minutes = $\frac{180}{\text{speed}}$ (since $3km\times 60$).

Answer: (C) 3 times

Concept: Area of ring = $\pi (R^2-r^2)$.

1. r = 7 m, $R=7+3=10$ m.
2. Area = $\frac{22}{7}\times (100-49)=\frac{22}{7}\times 51=\frac{1122}{7}\approx 160.28$ m².

Trick: $R^2-r^2=(R+r)(R-r)=17\times 3=51$.

Answer: (A) 160.28 m²

Concept: For perpendicular lines, $m_1\times m_2=-1$.

1. $y=(\frac{1}{3})x+7$ → $m_1=\frac{1}{3}$.
2. $m_2=-1÷(\frac{1}{3})=-3$.

Trick: Perpendicular slope = flip the fraction and change the sign: $\frac{1}{3}\to -3$.

Answer: (A) −3

Concept: Degrees = radians $\times \frac{180}{\pi }$.

1. $3.5\times \frac{180}{\pi }=\frac{630}{\pi }$.
2. $630÷3.1416\approx {200.5}^{\circ }$ → nearest option 200.56°.

Trick: 1 radian ≈ 57.3°, so 3.5 rad ≈ 200° — only option (D) is near.

Answer: (D) 200.56°

Concept: Sector area = $\frac{1}{2}{r}^2\theta$ (θ in radians).

1. $\frac{1}{2}\times 25\times \frac{\pi }{3}=\frac{25\pi }{6}$.
2. $25\times 3.1416÷6\approx 13.09$ cm².

Trick: $\frac{\pi }{3}={60}^{\circ }$ = $\frac{1}{6}$ of the circle → $\frac{\pi r^2}{6}=\frac{78.54}{6}\approx 13.09$.

Answer: (A) 13.09 cm²

Concept: If sin x = cos y (acute angles), then x + y = 90°.

1. A + (2A − 30°) = 90°.
2. 3A = 120° → A = 40°.

Trick: Check: sin 40° = cos(80° − 30°) = cos 50° ✓.

Answer: (A) 40°

Concept: Area ratio = $(\text{ratio of any corresponding lengths}{)}^2$.

1. Altitudes 3 : 4.
2. Areas = $3^2:4^2=9:16$.

Trick: Side, altitude, median, perimeter – all in the same ratio; area is its square.

Answer: (B) 9 : 16

Concept: $(b+1{)}^2+(b-1{)}^2=2(b^2+1)$.

1. $b=\sqrt{7}\to b^2=7$.
2. Value = $2(7+1)=16$.

Trick: The cross terms +2b and −2b cancel, so $\sqrt{7}$ never needs expanding.

Answer: (D) 16

Concept: Write each surd as (perfect square) × 2.

1. $\sqrt{98}=7\sqrt{2},\sqrt{32}=4\sqrt{2},\sqrt{50}=5\sqrt{2}$.
2. $7\sqrt{2}+4\sqrt{2}-5\sqrt{2}=6\sqrt{2}$.

Trick: Just add the outside numbers: $7+4-5=6$.

Answer: (B) 6$\sqrt{2}$

Concept: A chord through the centre is the longest one — the diameter.

1. Any chord not passing through the centre is shorter than the diameter.
2. Diameter = 15 cm → maximum chord = 15 cm.

Yaad rakhein: No chord can exceed the diameter, so options above 15 cm are impossible.

Answer: (A) 15 cm

Concept: Bring sin and cos to one side, form $\tan A$.

1. Move $\cos A$ to the right: $\sin A=\cos A(\sqrt{2}-1)$.
2. Divide by $\cos A$: $\tan A$ equals $\sqrt{2}-1$, hence $\cot A=\frac{1}{\sqrt{2}-1}$.
3. Rationalise: $\frac{\sqrt{2}+1}{2-1}=\sqrt{2}+1$.

Trick: $(\sqrt{2}-1)(\sqrt{2}+1)=1$, so they are reciprocals of each other.

Answer: (C) $\sqrt{2}$ + 1