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

Concept: Number of small spheres $=\frac{\text{big volume}}{\text{small volume}}={\left(\frac{D}{d}\right)}^3$.

1. $\frac{D}{d}=\frac{32}{4}=8$
2. Number $=8^3=512$

Trick: Sphere volumes scale as the cube of the diameter.

Answer: (C) $512$

Concept: Simplify the innermost brackets first, then remove the square bracket with a minus sign (all signs change).

1. $(3b+a)-(2a+b)+2a=2b+a$
2. $b+2a-(2b+a)-a=b+2a-2b-a-a$
3. $=-b$

Trick: Put $a=1,b=1$: $1+2-3-1=-1=-b$ ✓.

Answer: (C) $-b$

Concept: With SI, (balance + its interest till the end) = (sum of instalments + interest on each instalment for the months left).

1. Balance $=2500-520=1980$; 4 monthly instalments
2. Amount of 1980 for 4 months: $1980(1+\frac{25}{100}\times \frac{4}{12})=2145$
3. Instalments: $4x+x\times \frac{25}{100}\times \frac{3+2+1}{12}=4.125x$
4. $4.125x=2145\Rightarrow x=520$

Trick: Check: four instalments of ₹520 + small interest ≈ ₹2,145 – option ₹520.

Answer: (D) Rs.520

Concept: Distance by which A beats B = distance B fails to cover when A finishes.

1. Rajesh's speed $=\frac{1200}{40}=30\text{m/s}$
2. In $36$ s Rajesh covers $30\times 36=1080\text{m}$
3. Lead $=1200-1080=120\text{m}$

Trick: Beat distance = loser's speed × time gap $=30\times 4=120\text{m}$.

Answer: (B) $120\text{m}$

Concept: Total of the group = average × number; subtract the known part.

1. All 50: $50\times 30000=1500000$
2. 35 members: $35\times 37500=1312500$
3. Other 15: $\frac{187500}{15}=12500$

Trick: Deviation: $35\times 7500=262500$ extra must be balanced by 15 people: $30000-\frac{262500}{15}=12500$.

Answer: (C) 12,500

Concept: Get $\cot \theta$ from the given relation; the required expression is $1-\cot \theta$.

1. $\sin \theta =(\sqrt{2}-1)\cos \theta \Rightarrow \cot \theta =\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$
2. $\frac{\sin \theta -\cos \theta }{\sin \theta }=1-\cot \theta$
3. $=1-(\sqrt{2}+1)=-\sqrt{2}$

Trick: $\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$ (rationalise).

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

Concept: Overall profit/loss % is found on total CP: find each CP from $\text{CP}=\frac{\text{SP}\times 100}{100\pm \text{p}\%}$.

1. CP₁ $=\frac{1500\times 100}{125}=1200$; CP₂ $=\frac{2800\times 100}{50}=5600$; CP₃ $=\frac{3500\times 100}{125}=2800$
2. Total CP $=9600$, total SP $=1500+2800+3500=7800$
3. Loss $=1800$; $\frac{1800}{9600}\times 100=18.75\%$

Trick: Never average the percentages – use total CP and SP.

Answer: (D) $18.75\%$ loss

Concept: With percentages in the pie, difference $=(\text{difference of }\%)\times \text{total}$.

1. Housing $=15\%$, Transport $=5\%$
2. Difference $=10\%$
3. $10\%\text{ of }100000=10000$

Trick: Subtract percentages first, convert once.

Answer: (D) 10,000

Concept: Expand each product, add like terms, then factorise.

1. $(x-2y)(y-3x)=-3x^2+7xy-2y^2$
2. $(x+y)(x-y)=x^2-y^2$
3. $(x-3y)(2x+y)=2x^2-5xy-3y^2$
4. Sum $=0·x^2+2xy-6y^2=2y(x-3y)$

Trick: Put $x=1,y=1$: $(-1)(-2)+0+(-2)(3)=-4$, and $2y(x-3y)=2(-2)=-4$ ✓.

Answer: (A) $2y(x-3y)$

Concept: The required number is the smallest 5-digit multiple of the LCM.

1. $\text{LCM}(9,12,15,25,27)=2^2\times 3^3\times 5^2=2700$
2. $2700\times 4=10800$ is the first multiple above $10000$

Trick: Highest powers: $2^2$ (12), $3^3$ (27), $5^2$ (25).

Answer: (A) $10800$

Concept: Product of two numbers $=\text{HCF}\times \text{LCM}$.

1. Let HCF $=h$, LCM $=5h$
2. $5h^2=20480\Rightarrow h^2=4096\Rightarrow h=64$
3. LCM $=320$

Trick: Check: $64\times 320=20480$ ✓.

Answer: (C) 64 and 320

Concept: If the smallest number divides all the others, it is the HCF.

1. $250=2\times 125$, $750=6\times 125$
2. So $125$ divides all three: HCF $=125$

Trick: First test the smallest number itself.

Answer: (C) $125$

Concept: LCM = product of highest powers of all prime factors.

1. $8=2^3$, $12=2^2\times 3$, $18=2\times 3^2$
2. LCM $=2^3\times 3^2=72$

Trick: $72$ is the first multiple of $18$ that $8$ also divides.

Answer: (C) $72$

Concept: Read the right column for each organisation and add.

1. P + R in 2019 $=69+63=132$
2. Q + R in 2020 $=74+90=164$
3. $132:164=33:41$ (divide by 4)

Trick: Read the year columns carefully – mixing 2019 and 2020 is the usual slip.

Answer: (B) $33:41$

Concept: Expenditure corresponds to $3$ parts; find one part.

1. $3$ parts $=5400\Rightarrow 1$ part $=1800$
2. Income $=7\times 1800=12600$

Trick: Income $=\frac{7}{3}\times$ expenditure.

Answer: (A) ₹12,600

Concept: Turn the equality into a ratio of incomes.

1. $0.3R=0.4S\Rightarrow R=\frac{4}{3}S$
2. $0.5R=\frac{1}{2}\times \frac{4}{3}S=\frac{2}{3}S$
3. $\frac{2}{3}=66\frac{2}{3}\%$

Trick: Scale: $30\%\to 40\%$, so $50\%\to \frac{40}{30}\times 50=66\frac{2}{3}\%$.

Answer: (D) $66\frac{2}{3}\%$

Concept: Time is proportional to the work done (same person).

1. $\frac{2}{3}$ work in $12$ days $\Rightarrow$ full work in $12\times \frac{3}{2}=18$ days
2. $\frac{1}{6}$ work $=\frac{18}{6}=3$ days

Trick: Direct: $12\times \frac{1/6}{2/3}=12\times \frac{1}{4}=3$.

Answer: (A) $3$

Concept: A chord equal to the radius makes an equilateral triangle with the centre, so it subtends ${60}^{\circ }$ at the centre.

1. Angle at a point on the major arc $=\frac{{60}^{\circ }}{2}={30}^{\circ }$
2. Angle at a point on the minor arc $={180}^{\circ }-{30}^{\circ }={150}^{\circ }$ (cyclic quadrilateral)
3. Difference $={150}^{\circ }-{30}^{\circ }={120}^{\circ }$

Trick: Angle in major segment $=\frac{\theta }{2}$, minor segment $={180}^{\circ }-\frac{\theta }{2}$.

Answer: (B) ${120}^{\circ }$

Concept: The line of centres is the perpendicular bisector of the common chord, so half-chord $=12$ forms right triangles with each radius.

1. Radii: $15$ and $13$
2. $O_{1}M=\sqrt{{15}^2-{12}^2}=9$
3. $O_{2}M=\sqrt{{13}^2-{12}^2}=5$
4. $O_1{O}_2=9+5=14\text{cm}$

Trick: Triplets $(9,12,15)$ and $(5,12,13)$ appear instantly.

Answer: (B) $14\text{cm}$

Concept: Compare the 'Total' column (it equals the sum of the five courses).

1. Totals: A $10340$, B $11000$, C $10340$, D $10160$, E $8300$
2. Largest $=11000$ (city B)

Trick: Verify one row: B $=1350+2650+5100+1050+850=11000$ ✓.

Answer: (D) B