SSC CHSL 2025 PYQs

Concept: Take LCM and use ${\sin }^{2}x+{\cos }^{2}x=1$.

1. Numerator: ${\cos }^{2}x+(1+\sin x{)}^2={\cos }^{2}x+1+2\sin x+{\sin }^{2}x=2+2\sin x$.
2. Expression = $\frac{2(1+\sin x)}{(1+\sin x)\cos x}=\frac{2}{\cos x}=2\sec x$.

Trick: Put $x=0^{\circ }$: $\frac{1}{1}+\frac{1}{1}=2$, and $2\sec 0^{\circ }=2$ – only (D) gives 2.

Answer: (D) $2\sec x$

Concept: Divide by $\sqrt{a^2+b^2}$ to get a compound-angle form; or simply test options.

1. Divide $\sqrt{3}\cos \theta +\sin \theta =1$ by 2: $\sin ({60}^{\circ }+\theta )=\frac{1}{2}$.
2. ${60}^{\circ }+\theta$ lies in ${60}^{\circ }$–${150}^{\circ }$ → ${60}^{\circ }+\theta ={150}^{\circ }$ → $\theta ={90}^{\circ }$.
3. Check: $\sqrt{3}\times 0+1=1$ ✔.

Trick: Put options: $\theta ={90}^{\circ }$ → $0+1=1$ ✔ in seconds.

Answer: (A) 90°

Concept: Wages are divided in the ratio of efficiency (work done).

1. Per day together = $800÷4=$ ₹200.
2. Woman : girl = 3 : 1 → girl gets $\frac{1}{4}$.
3. Girl's daily wage = $200\times \frac{1}{4}=$ ₹50.

Trick: Girl's share of the total = $800\times \frac{1}{4}=200$ for 4 days → ₹50/day.

Answer: (A) 50/-

Concept: Perimeter of a semicircle = $\pi r+2r$ (curved arc + diameter).

1. Rectangle perimeter = $2(16+14)=60$ cm → square side = 15 cm.
2. Diameter = 15, $r=7.5$.
3. $\pi r+2r=\frac{22}{7}\times 7.5+15=23.57+15=38.57$ cm.

Trick: Shortcut: semicircle perimeter = $\left(\frac{36}{7}\right)\times r=\frac{36}{7}\times 7.5\approx 38.57$. Option (C) 23.57 is only the arc – a trap.

Answer: (A) 38.57

Concept: $\text{TSA}=2lb+2h(l+b)$; find $(l+b)$ from $(l+b{)}^2=l^2+b^2+2lb$.

1. Base diagonal 21 → $l^2+b^2=441;lb=92$.
2. $(l+b{)}^2=441+184=625$ → $l+b=25$.
3. $\text{TSA}=2\times 92+2\times 18\times 25=184+900=1084{\text{cm}}^2$.

Trick: You never need l and b separately – only their sum and product.

Answer: (C) $1084{\mathrm{cm}}^2$

Concept: Expand $y^2$ first, then $y^3=y\times y^2$.

1. $y=2+\sqrt{2}$ → $y^2=4+4\sqrt{2}+2=6+4\sqrt{2}$.
2. $y^3=(2+\sqrt{2})(6+4\sqrt{2})=12+8\sqrt{2}+6\sqrt{2}+8=20+14\sqrt{2}$.
3. $3y=6+3\sqrt{2}$ → $y^3+3y=26+17\sqrt{2}$.

Trick: Rational part = 26 must come out positive with $+\sqrt{2}$ terms only – eliminate the options with '−' at once.

Answer: (A) $26+17\sqrt{2}$

Concept: Slant height $l=\sqrt{h^2+{\left(\frac{a}{2}\right)}^2}$; $\text{TSA}=\frac{1}{2}\times \text{base perimeter}\times l+\text{base area}$.

1. $l=\sqrt{{21}^2+{10}^2}=\sqrt{541}\approx 23.25$ cm.
2. Lateral = $\frac{1}{2}\times 80\times 23.25\approx 930{\text{cm}}^2$.
3. $\text{TSA}\approx 930+400=1330{\text{cm}}^2$.

Trick: Slant height must exceed the height 21 and be near $\sqrt{541}$ (between 23 and 24) – only (A) fits.

Answer: (A) $23.25\mathrm{cm},1330{\mathrm{cm}}^2$

Concept: Isosceles triangles from radii + exterior angle property.

1. Join OB: $OB=BC$ → $\angle BOC=\angle BCO={32}^{\circ }$.
2. Exterior $\angle OBA={32}^{\circ }+{32}^{\circ }={64}^{\circ }=\angle OAB$ ($OA=OB$).
3. In $△OAC$: $\angle AOC={180}^{\circ }-{64}^{\circ }-{32}^{\circ }={84}^{\circ }$.
4. $\angle AOD={180}^{\circ }-{84}^{\circ }={96}^{\circ }$.

Trick: In this standard figure, $\angle AOD=3\times \angle ACD=3\times {32}^{\circ }={96}^{\circ }$.

Answer: (D) 96°

Concept: Make the total parts equal (LCM) before adding.

1. Totals 6 and 11 → LCM 66.
2. Vessel 1 (5 : 1) $\times 11$ → 55 : 11; Vessel 2 (6 : 5) $\times 6$ → 36 : 30.
3. Milk : water = $(55+36):(11+30)=91:41$.

Trick: Sum of the answer's terms must be 132 ($=2\times 66$): $91+41=132$ ✔.

Answer: (C) 91 : 41

Concept: 2nd-year CI − 1st-year CI = interest on the 1st-year interest.

1. $600=R\%$ of 3000 → $R=20\%$.
2. 1st-year interest = 20% of sum = 3000 → sum = ₹15,000.
3. 40% of 15000 = ₹6,000.

Trick: $\text{Sum}=\frac{(\text{1st-year CI}{)}^2}{\text{difference}}=\frac{{3000}^2}{600}=15,000$.

Answer: (C) Rs. 6,000

Concept: CP = SP + Loss.

1. SP of 1 pen = $750÷15=$ ₹50.
2. Loss = SP of 3 pens = ₹150 → CP of 15 pens = $750+150=$ ₹900.
3. CP of 1 pen = $900÷15=$ ₹60.

Trick: Loss = SP of 3 → CP of 15 = SP of 18 pens = $18\times 50=900\to \frac{900}{15}=60$.

Answer: (D) ₹ 60

Concept: $\text{Profit share}\propto \text{capital}\times \text{months}$.

1. Period = 24 months. A: $1.5L\times 8+1.8L\times 16=12L+28.8L=40.8L$.
2. B: $1L\times 24=24L$. C: $1.2L\times 8+1L\times 16=9.6L+16L=25.6L$.
3. Ratio 408 : 240 : 256 = 51 : 30 : 32 (sum 113).
4. $A=169500\times \frac{51}{113}=1500\times 51=$ ₹76,500.

Trick: Total profit $169500=113\times 1500$ — when the total is a multiple of the ratio sum, the answer is a clean multiple of 1500.

Answer: (A) ₹76,500

Concept: Find the magnitude from the triplet, then the sign from the quadrant.

1. $\cot \theta =\frac{12}{5}$ → base 12, perpendicular 5, hypotenuse 13 (5-12-13 triplet).
2. $|\sin \theta |=\frac{5}{13}$.
3. In Quadrant III only tan and cot are positive → $\sin \theta =$ $-\frac{5}{13}$.

Trick: ASTC – 'All Students Take Coffee': Q1 all, Q2 sin, Q3 tan, Q4 cos are positive.

Answer: (B) $-\frac{5}{13}$

Concept: Make the common term (n) equal in both ratios.

1. m : n = 7 : 8, n : p = 3 : 2.
2. $m=7\times 3=21$, $n=8\times 3=24$, $p=8\times 2=16$.
3. m : n : p = 21 : 24 : 16.

Trick: Cross-multiply pattern: m : n : p = $(7\times 3):(8\times 3):(8\times 2)$.

Answer: (A) 21 : 24 : 16

Concept: For 3 years: $\text{CI}-\text{SI}=P{\left(\frac{r}{100}\right)}^2\times (3+\frac{r}{100})$.

1. $\frac{r}{100}=0.1$ → difference = $P\times 0.01\times 3.1=0.031P$.
2. $0.031P=31\Rightarrow P=$ ₹1,000.

Trick: At 10% for 3 years, $\text{CI}-\text{SI}=3.1\%$ of P — remember this ready value.

Answer: (B) ₹1,000

Concept: Ratio of consecutive yearly amounts = $(1+\frac{r}{100})$.

1. $5324÷4840=1.1\Rightarrow r=10\%$.
2. After 2 years: $P\times 1.21=4840\Rightarrow P=$ ₹4,000.

Trick: Go backwards dividing by 1.1: 4840 → 4400 → 4000.

Answer: (B) ₹4,000

Concept: The 7th number is in both groups of seven.

1. Total of 13 = $13\times 51=663$.
2. First 7 = 350, last 7 = 357; together = 707.
3. Middle number = $707-663=$ 44.

Trick: Deviation method: first group −1 each (−7), last group 0; middle $=51+(-7+0)=44$.

Answer: (A) 44

Concept: $\text{Required number}=k\times \text{LCM}+\text{remainder}$.

1. $\text{LCM}(8,14,26)=8\times 7\times 13=728$.
2. $9999÷728$ → 13 times: $728\times 13=9464$.
3. Answer = $9464+6=$ 9470 (still 4-digit).

Trick: Options ending in 4 (multiple of LCM) are traps — the remainder 6 must be added.

Answer: (D) 9470

Concept: Money saved by the price cut buys the extra items at the new price.

1. 20% of ₹300 = ₹60 is saved.
2. This ₹60 buys 4 extra notebooks.
3. New price = $60÷4=$ ₹15.

Trick: $\text{New price}=\frac{\%\text{ decrease}\times \text{total money}}{\text{extra items}}$ — one line, no equation.

Answer: (C) 15

Concept: Same distance both ways → average speed = $\frac{2ab}{a+b}$.

1. Average speed = $\frac{2\times 4\times 2}{6}=\frac{8}{3}$ km/h.
2. Total distance = $\frac{8}{3}\times 3=8$ km.
3. One way = $8÷2=$ 4 km.

Trick: Option check: 4 km → $\frac{4}{4}+\frac{4}{2}=1+2=3$ h ✓.

Answer: (A) 4 km