SSC CHSL 2025 PYQs

Concept: Take total work = LCM of days; find what is left after P works alone.

1. Work = 60 units; P = 3/day, Q = 2/day.
2. P in 8 days = 24 → left 36.
3. Together 5/day → $36÷5=7.2$ days.

Trick: Fraction way: P does $\frac{8}{20}=\frac{2}{5}$; remaining $\frac{3}{5}÷(\frac{1}{20}+\frac{1}{30}=\frac{1}{12})=\frac{36}{5}=7.2$.

Answer: (A) 7.2 days

Concept: $\text{SP}=\text{MP}\times (1-\text{discount})$; $\text{CP}=\text{SP}÷(1+\text{profit})$.

1. $\text{SP}=8000\times \frac{75}{100}=$ ₹6000.
2. $\text{CP}=6000\times \frac{100}{120}=$ ₹5000.

Trick: One line: $\text{CP}=8000\times \frac{3}{4}\times \frac{5}{6}=$ ₹5000. (Options A and D both read ₹5,000; the key marks D.)

Answer: (D) ₹5,000

Concept: In SI, the difference of amounts = interest for the extra years.

1. 2 years' interest = $9880-8840=1040$ → 1 year = 520.
2. Principal = $8840-7\times 520=8840-3640=5200$.
3. Rate = $\frac{520}{5200}\times 100=10\%$.

Trick: After getting yearly SI, rate = yearly SI ÷ P × 100 – no formula juggling needed.

Answer: (C) 10%

Concept: If $x=\sqrt{20+\sqrt{20+\dots }}$, then $x=\sqrt{20+x}$.

1. Square: $x^2=20+x$ ⇒ $x^2-x-20=0$.
2. $(x-5)(x+4)=0$ ⇒ $x=5$ (x must be positive).

Trick: For $\sqrt{N+\sqrt{N+\dots }}$: write $N=n(n-1)$; the value is n. Here $20=5\times 4\Rightarrow 5$.

Answer: (A) 5

Concept: Removing part of a mixture removes acid in the same ratio; added water adds no acid.

1. Acid = 80% of 120 = 96 L.
2. Removed acid = 80% of 16 = 12.8 L → left 83.2 L.
3. New % = $\frac{83.2}{120}\times 100=69.33\%$.

Trick: Concentration × (1 − removed/total): $80\times \frac{104}{120}=80\times \frac{13}{15}=69.33\%$.

Answer: (C) 69.33%

Concept: The perpendicular from the centre bisects the chord; use Pythagoras.

1. Half chord = 12 cm, distance = 9 cm.
2. $r^2={12}^2+9^2=144+81=225\Rightarrow r=15$ cm.

Trick: 9-12-15 is the 3-4-5 triplet × 3 – answer without squaring.

Answer: (A) 15 cm

Concept: Volume stays the same; number of small spheres = ${\left(\frac{R}{r}\right)}^3$.

1. $\frac{R}{r}=\frac{8}{2}=4$.
2. Number = $4^3=64$.

Trick: For similar solids, volume scales as the cube of the length ratio – no π calculation needed.

Answer: (C) 64

Concept: Volume of sphere = volume of cylinder of the same radius.

1. $\frac{4}{3}\pi r^3=\pi r^{2}h$.
2. $h=\frac{4r}{3}\Rightarrow \frac{h}{r}=\frac{4}{3}$.

Trick: Remember: a cylinder of the same radius holding a sphere's volume is always $\frac{4}{3}$ of the radius tall ($\frac{2}{3}$ of the diameter).

Answer: (A) $\frac{4}{3}$

Concept: Read the expression as $(4.5{)}^2+\sqrt{2.25}-(6.5÷1.3)$ and simplify each part.

1. $(4.5{)}^2=20.25$.
2. $\sqrt{2.25}=1.5$.
3. $6.5÷1.3=5$.
4. $20.25+1.5-5=16.75$.

Trick: Last digits check: $\mathrm{.25}+\mathrm{.5}=\mathrm{.75}$ – only option D ends in .75.

Answer: (D) 16.75

Concept: Expression: $\frac{1}{\sqrt{7}-\sqrt{6}}+\frac{1}{\sqrt{7}+\sqrt{6}}$. Multiply by conjugates; $(\sqrt{7}{)}^2-(\sqrt{6}{)}^2=1$.

1. $\frac{1}{\sqrt{7}-\sqrt{6}}=\sqrt{7}+\sqrt{6}$.
2. $\frac{1}{\sqrt{7}+\sqrt{6}}=\sqrt{7}-\sqrt{6}$.
3. Sum = $2\sqrt{7}$.

Trick: $\frac{1}{a-b}+\frac{1}{a+b}=\frac{2a}{a^2-b^2}=\frac{2\sqrt{7}}{1}=2\sqrt{7}$.

Answer: (A) $2\sqrt{7}$

Concept: Line through (x₁, y₁) with slope m: y − y₁ = m(x − x₁).

1. y − 8 = −3(x − 6).
2. y = −3x + 18 + 8 = −3x + 26.

Trick: Plug (6, 8) into the options: −18 + 26 = 8 ✓ – only A passes with slope −3.

Answer: (A) $y=-3x+26$

Concept: Real and distinct roots ⇔ discriminant $b^2-4ac>0$.

1. $k^2-4\times 1\times 36>0\Rightarrow k^2>144$.
2. So $k>12$ or $k<-12$.
3. Among the options only $k>12$ guarantees this ($k<-11$ includes −11.5, which fails).

Trick: For $x^2+kx+c$, distinct real roots need $|k|>2\sqrt{c}=2\times 6=12$.

Answer: (A) $k>12$

Concept: Expression: $\left[\frac{1+{\cot }^{2}x}{1+{\tan }^{2}x}\right]\times \left[\frac{\csc x-\cot x}{\csc x+\cot x}\right]$.

1. $\frac{1+{\cot }^{2}x}{1+{\tan }^{2}x}=\frac{{\csc }^{2}x}{{\sec }^{2}x}={\cot }^{2}x$.
2. Write $\csc x\pm \cot x=\frac{1\pm \cos x}{\sin x}$ → second factor = $\frac{1-\cos x}{1+\cos x}$.
3. Product = ${\cot }^{2}x\frac{1-\cos x}{1+\cos x}$.

Trick: Put $x={60}^{\circ }$: first factor $\frac{1}{3}$, second $\frac{1/2}{3/2}=\frac{1}{3}$ → product $\frac{1}{9}$; only option D gives $\frac{1}{9}$.

Answer: (D) $\frac{{\cot }^2\left(x\right)\left[1-\cos \right(x\left)\right]}{1+\cos \left(x\right)}$

Concept: ${\sin }^2\theta =1-{\cos }^2\theta$ (given $\cos \theta =\frac{8}{27}$).

1. ${\cos }^2\theta =\frac{64}{729}$.
2. ${\sin }^2\theta =\frac{729-64}{729}=\frac{665}{729}$.

Trick: Only the numerator matters: $729-64=665$. Any option above 1 ($\frac{793}{729}$) is impossible.

Answer: (A) $\frac{665}{729}$

Concept: $\cot B=1$ means $B={45}^{\circ }$; and $2\sin B\cos B=\sin 2B$.

1. The right angle is at C, so B is acute; $\cot B=1\Rightarrow B={45}^{\circ }$.
2. $\sin {45}^{\circ }=\cos {45}^{\circ }=\frac{1}{\sqrt{2}}$.
3. $2\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=1$.

Trick: $2\sin B\cos B=\sin 2B=\sin {90}^{\circ }=1$.

Answer: (C) 1

Concept: Identify the athlete who made India's Olympic debut in fencing.

1. C.A. Bhavani Devi (Tamil Nadu) was the first Indian ever to qualify for Olympic fencing – women's sabre, Tokyo 2020 (held 2021).
2. Sindhu = badminton, Manu Bhaker = shooting, Dipa Karmakar = gymnastics.

Trick: Match sport to name: only one option is not a badminton/shooting/gymnastics star.

Answer: (A) Bhavani Devi

Concept: The Bar Council of India's head in 2025 was Manan Kumar Mishra.

1. Manan Kumar Mishra, senior advocate, has been BCI Chairman for several consecutive terms and held the post in May 2025.
2. Kapil Sibal, Dushyant Dave and Harish Salve are well-known senior lawyers, not the BCI chairman then.

Yaad rakhein: BCI is a statutory body created under the Advocates Act, 1961; it regulates advocates and legal education.

Answer: (B) Manan Mishra

Concept: NITI Aayog's Women Entrepreneurship Platform (WEP) runs partner-led programmes for women entrepreneurs.

1. 'EmpowHER Biz – Sapno Ki Udaan' was started by WEP together with the retail chain New Shop.
2. It helps women set up and run their own retail outlets through training and support.

Yaad rakhein: WEP was launched by NITI Aayog in 2018 on three pillars: Iccha Shakti, Gyaan Shakti and Karma Shakti.

Answer: (A) New Shop

Concept: The mission is about building common map/land data infrastructure, not satellites or free GPS.

1. Budget 2025-26 proposed a National Geospatial Mission to develop foundational geospatial data and infrastructure.
2. It aids urban planning, land-record modernisation and infrastructure projects (e.g. PM Gati Shakti).

Trick: Options with 'only' (D) or narrow freebies (C) are rarely the aim of a national mission – pick the broad framework.

Answer: (A) Establishing a unified geospatial framework to support governance, planning, and development

Concept: Global Findex is the World Bank's survey on financial inclusion (bank/mobile accounts).

1. Global Findex 2025 put India's adult account ownership at about 89%, far above the 2011 level.
2. Jan Dhan accounts, Aadhaar e-KYC and digital payments drove the rise.

Trick: 'Account ownership' = financial inclusion = Findex ('Fin' + 'dex' from index).

Answer: (C) Global Findex 2025