SSC CPO Paper I 2020, 2022, 2023, 2024 PYQs

Concept: Do ÷ and × from left to right before + and −, separately in numerator and denominator.

1. Numerator: $3÷3\times 2=2$, $6÷3=2$ → $2-2+7=7$
2. Denominator: $5÷5\times 6÷6=1$ → $1+3+3=7$
3. Value $=\frac{7}{7}=1$

Trick: Left to right: $5÷5\times 6÷6=1\times 6÷6=1$, not $5÷30÷6$.

Answer: (A) 1

Concept: Use $a^3+b^3=(a+b)(a^2-ab+b^2)$ and ${\sin }^{2}A+{\cos }^{2}A=1$.

1. ${\sin }^{3}A+{\cos }^{3}A=(\sin A+\cos A)({\sin }^{2}A+{\cos }^{2}A-\sin A\cos A)$
2. $=(\sin A+\cos A)(1-\sin A\cos A)$
3. Divide by $(\sin A+\cos A)$: $1-\sin A\cos A$

Trick: Check $A=0^{\circ }$: expression $=\frac{1}{1}=1$, and $1-\sin 0\cos 0=1$ ✓.

Answer: (C) $1-\sin A\cos A$

Concept: Selling price $=\text{MP}\times \frac{100-\text{discount}\%}{100}$.

1. SP $=5980\times \frac{82}{100}$
2. $=4903.6$
3. To the nearest rupee: ₹4,904

Trick: $18\%$ of $5980=1076.4$; $5980-1076.4=4903.6$.

Answer: (D) 4,904

Concept: Split the mixture into milk and water, add the new quantities, and match the target ratio.

1. Milk $=\frac{5}{6}\times 18=15$ L, water $=3$ L
2. After adding 3 L milk: milk $=18$ L
3. For $9:2$: water $=\frac{18\times 2}{9}=4$ L
4. Water to add $=4-3=1$ L

Trick: Milk 18 = 9 parts → 1 part = 2 L → water 2 parts = 4 L.

Answer: (C) 1

Concept: Crossing a man (a point) in a train moving the same way: length = relative speed × time, relative speed = difference.

1. Relative speed $=60-42=18\text{km/h}$
2. $18\times \frac{5}{18}=5\text{m/s}$
3. Length $=5\times 36=180$ m

Yaad rakhein: Same direction → subtract speeds; opposite → add.

Answer: (C) 180 m

Concept: Find the grand total over five years and compare $20\%$ of it with each yearly total.

1. Grand total $=115+108+149+102+101=575$ thousand
2. $20\%$ of $575=115$ thousand
3. Year with total 115 thousand = 1985

Trick: $20\%=\frac{1}{5}$, so look for the year equal to the average of the five totals ($575÷5=115$).

Answer: (A) 1985

Concept: Effective CP includes cartage. Target SP = CP × 1.42; the last lot must cover what is still missing.

1. Total CP $=100\times 19+100=₹2000$, i.e. ₹20 per kg; target SP $=2000\times 1.42=₹2840$
2. 60 kg at $50\%$: $60\times 30=₹1800$; 20 kg at $40\%$: $20\times 28=₹560$
3. 10 kg at $25\%$: $10\times 25=₹250$; total so far ₹2610
4. Last 10 kg must fetch $2840-2610=₹230$ → ₹23 per kg
5. Profit $=\frac{23-20}{20}=0.15$, i.e. $15\%$

Trick: Weighted-profit check: $\frac{60(50)+20(40)+10(25)+10x}{100}=42\Rightarrow x=15$.

Answer: (B) 0.15 (i.e. 15%)

Concept: Apply BODMAS from the innermost bracket outwards; simplify the denominator separately.

1. $\frac{1}{4}÷\frac{15}{20}=\frac{1}{4}\times \frac{4}{3}=\frac{1}{3}$; $\frac{1}{3}+2=\frac{7}{3}$; $3\times \frac{7}{3}=7$
2. $\{3+7-18\times \frac{1}{3}\}=10-6=4$; numerator $=2\times 4=8$
3. Denominator $=2(4+5)=18$
4. Value $=\frac{8}{18}=\frac{4}{9}$

Trick: $\frac{15}{20}=\frac{3}{4}$ first — makes the division instant.

Answer: (C) $\frac{4}{9}$

Concept: With a false weight, the trader gives less goods for the same price: SP of 1 unit of goods $=\frac{1+\text{profit}}{1-\text{shortfall}}$ × CP.

1. Take CP of 1 kg = ₹100; he sells at $32\%$ profit, i.e. charges ₹132
2. But he gives only $0.8$ kg (CP ₹80)
3. Gain $=\frac{132-80}{80}\times 100=65\%$

Trick: $\frac{132}{80}=1.65$ → gain $65\%$.

Answer: (D) 65%

Concept: Actual activity = percentage × total activities of that city.

1. R in B $=30\%\times 500=150$
2. R in C $=40\%\times 400=160$
3. Difference $=160-150=10$

Yaad rakhein: Percentages of different totals can't be compared directly — convert to actual numbers first.

Answer: (B) 10

Concept: Successive discounts multiply; then CP $=\frac{\text{SP}\times 100}{100+\text{profit}\%}$.

1. SP $=500\times 0.9\times 0.8=₹360$
2. CP $=\frac{360\times 100}{125}$
3. CP $=₹288$

Trick: $10\%$ and $20\%$ successive = $28\%$ single discount ($10+20-\frac{10\times 20}{100}$).

Answer: (C) ₹288

Concept: BODMAS: divisions and multiplications first, then additions and subtractions left to right.

1. $72÷8=9$, $210÷15=14$, $2\times 5=10$
2. $9-14-5+10$
3. $=0$

Trick: Group positives and negatives: $(9+10)-(14+5)=19-19=0$.

Answer: (B) 0

Concept: In a triangle $A+B+C={180}^{\circ }$, so $\frac{B+C}{2}={90}^{\circ }-\frac{A}{2}$ and $\text{cosec}\frac{B+C}{2}=\sec \frac{A}{2}$.

1. Expression $=(\tan \frac{A}{2}+\sec \frac{A}{2})(\tan \frac{A}{2}-\sec \frac{A}{2})$
2. $={\tan }^2\frac{A}{2}-{\sec }^2\frac{A}{2}$
3. $=-1$ (since ${\sec }^{2}x-{\tan }^{2}x=1$)

Yaad rakhein: $(a+b)(a-b)=a^2-b^2$ with ${\sec }^2-{\tan }^2=1$.

Answer: (A) −1

Concept: 'Added' means the 'New' column; sum it for 2019–2022 for both categories.

1. Professors added $=100+150+180+200=630$
2. Associate professors added $=110+140+190+220=660$
3. Difference $=660-630=30$

Trick: Subtract year-wise: $10-10+10+20=30$ — no need for full totals.

Answer: (A) 30

Concept: Write the percentage as a fraction: $S=\frac{75}{100}T$.

1. $S=\frac{3}{4}T$
2. $\frac{S}{T}=\frac{3}{4}$
3. $S:T=3:4$

Yaad rakhein: $75\%=\frac{3}{4}$ — memorise common percentage-fraction pairs.

Answer: (D) 3 : 4

Concept: Solve the innermost bracket first (multiplication before subtraction).

1. $1.5\times 3.02=4.53$; $4.78-4.53=0.25$
2. $0.25\times 0.25=0.0625$; $15+0.0625=15.0625$
3. $22.0625-15.0625=7.0$

Trick: Notice the matching decimal tails $\mathrm{.0625}$ — they cancel to give a whole number.

Answer: (D) 7.0

Concept: Meshing gears move the same arc length (teeth) at the contact point, so diameter × revolutions is equal.

1. $15\times 105=21\times n$
2. $n=\frac{15\times 105}{21}$
3. $n=75$

Trick: Revolutions are inversely proportional to diameter: $105\times \frac{15}{21}=105\times \frac{5}{7}=75$.

Answer: (B) 75

Concept: Innermost bracket first, then multiplication, then the outer subtraction.

1. $9-3\times 2=3$
2. $0.5\times 3=1.5$; $5+1.5=6.5$
3. $13.5-6.5=7$

Trick: Inside the bracket, multiply before subtracting: $9-6$, not $6\times 2$.

Answer: (B) 7

Concept: Canvas for a tent = curved surface area $=\pi rl$, with $l=\sqrt{r^2+h^2}$.

1. $2\pi r=30\pi \Rightarrow r=15$ ft
2. $l=\sqrt{{15}^2+{20}^2}=\sqrt{625}=25$ ft
3. CSA $=\pi \times 15\times 25=375\pi$ sq. ft.

Trick: 15–20–25 is the 3–4–5 triplet scaled by 5.

Answer: (A) $375\pi$ sq. ft.

Concept: 'Twice as good' means double efficiency. Total work = combined efficiency × days.

1. Tarun $=2$ units/day, Tripti $=1$ unit/day
2. Work $=(2+1)\times 16=48$ units
3. Tripti alone $=\frac{48}{1}=48$ days

Trick: Tripti's share is $\frac{1}{3}$ of combined rate → she takes $3\times 16=48$ days.

Answer: (A) 48