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

Concept: $\cos \theta \le 1$ for every angle, so a sum of three cosines can equal 3 only when each cosine equals 1.

1. $\cos A+\cos B+\cos C=3\Rightarrow \cos A=\cos B=\cos C=1$.
2. So $A=B=C=0^{\circ }$.
3. $\sin A+\sin B+\sin C=0+0+0=0$.

Trick: Sum of $n$ cosines $=n$ ⇒ every angle is $0^{\circ }$, so every sine is 0.

Answer: (C) $0$

Concept: A balance that measures 25% less gives only 750 g when it shows 1 kg, so the seller's cost falls while he charges for 1 kg.

1. Let CP of 1 kg $=₹100$; he actually gives 750 g, costing $₹75$.
2. Marked price $=100\times 1.15=115$; after 10% discount SP $=115\times 0.9=103.5$.
3. Profit $\%=\frac{103.5-75}{75}\times 100=\frac{28.5}{75}\times 100=38\%$.

Trick: Net % $=\frac{1.15\times 0.9}{0.75}-1=1.38-1=38\%$.

Answer: (D) $38\%$

Concept: Sum = average × number. The middle (8th) result = total sum − (first 7 sum + last 7 sum).

1. Sum of 15 results $=15\times 21=315$.
2. First 7: $7\times 21=147$; last 7: $7\times 20=140$.
3. 8th result $=315-147-140=28$.

Trick: First 7 equal the overall average, so the 8th $=21+7\times (21-20)=28$.

Answer: (C) $28$

Concept: In a right-angled triangle the two acute angles add up to ${90}^{\circ }$. The lower part of the figure (A, D) is drawn as the mirror image of the upper triangle in the diameter AB.

1. Upper triangle: the vertical line from C meets the horizontal line at a right angle.
2. So the angle at the left vertex $={90}^{\circ }-{50}^{\circ }={40}^{\circ }$.
3. The slanting line through A to D is the reflection of the slanting line through C in AB, so it makes the same angle with the horizontal: $x={40}^{\circ }$.

Trick: Right angle in a triangle ⇒ other angle $={90}^{\circ }-\text{given}$; mirror-image lines make equal angles.

Answer: (A) ${40}^{\circ }$

Concept: Add each year's column; the largest column total is the answer.

1. 2021: $1+1+0+0+1=3$; 2020: $0+2+0+1+0=3$.
2. 2019: $2+0+1+1+1=5$; 2018: $1+0+1+0+0=2$.
3. 2017: $1+1+0+1+1=4$. Maximum is 5 in 2019.

Trick: For a year, add the column; for a player, add the row.

Answer: (C) 2019

Concept: Marks = percentage × maximum marks. Then % more $=\frac{\text{difference}}{\text{base}}\times 100$.

1. Q: Physics $=90\%$ of $80=72$; Hindi $=54\%$ of $100=54$; total $=126$.
2. T: Geography $=88\%$ of $75=66$; History $=42\%$ of $50=21$; total $=87$.
3. % more $=\frac{126-87}{87}\times 100=\frac{39}{87}\times 100\approx 44.8\%\approx 45\%$.

Trick: Base is the 'than' quantity (T's total 87).

Answer: (A) $45$

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

1. $\frac{{\sin }^{2}A+(1+\cos A{)}^2}{\sin A(1+\cos A)}$.
2. Numerator $={\sin }^{2}A+1+2\cos A+{\cos }^{2}A=2+2\cos A=2(1+\cos A)$.
3. Value $=\frac{2(1+\cos A)}{\sin A(1+\cos A)}=\frac{2}{\sin A}=2\text{cosec}A$.

Trick: Put $A={60}^{\circ }$: $\frac{1}{\sqrt{3}}+\sqrt{3}=\frac{4}{\sqrt{3}}=2\text{cosec}{60}^{\circ }$.

Answer: (A) $2\text{cosec}A$

Concept: If HCF of 108, 72 and $5a$ is $a$, then $a$ must divide HCF(108, 72).

1. $\text{HCF}(108,72)=36$, so $a$ is a factor of 36.
2. Every factor of 36 already divides both 108 and 72.
3. So $\text{LCM}(108,72,a)=\text{LCM}(108,72)=216$.

Trick: A number that divides the others never changes their LCM.

Answer: (D) $216$

Concept: $132=4\times 3\times 11$; the number must pass the tests for 4, 3 and 11.

1. Divisible by 4: last two digits $3P$ ⇒ $P=2$ or $6$.
2. Divisible by 3: digit sum $=30+P$ ⇒ $P=0,3,6,9$. Common value: $P=6$.
3. Check 11 for 63366336: alternate sums $6+3+6+3=18$ and $3+6+3+6=18$, difference 0 ✓.

Trick: Use the test of 4 first – it leaves only two choices.

Answer: (D) $6$

Concept: Area of a circle $=\pi r^2$.

1. $\frac{22}{7}{r}^2=1386$.
2. $r^2=1386\times \frac{7}{22}=441$.
3. $r=21\text{cm}$.

Yaad rakhein: $1386=\frac{22}{7}\times {21}^2$ – a standard value worth remembering.

Answer: (D) $21\text{cm}$

Concept: Successive discounts are applied one after another on the reduced price.

1. After 10%: $700\times 0.9=630$.
2. Cash discount 20%: $630\times 0.8=504$.

Trick: Net discount $=10+20-\frac{10\times 20}{100}=28\%$; $700\times 0.72=504$.

Answer: (B) ₹504

Concept: Time $=\frac{\text{distance}}{\text{speed}}$; take speeds $2x$ and $3x$.

1. $\frac{50}{2x}-\frac{50}{3x}=10$.
2. $\frac{150-100}{6x}=10\Rightarrow \frac{50}{6x}=10\Rightarrow x=\frac{5}{6}$.
3. Difference of speeds $=3x-2x=x=\frac{5}{6}\text{m/s}$.

Trick: Speeds $2:3$ ⇒ times $3:2$; 1 part of time $=10$ s, so times are 30 s and 20 s, speeds $\frac{5}{3}$ and $\frac{5}{2}$; difference $\frac{5}{6}$.

Answer: (B) $\frac{5}{6}\text{m/s}$

Concept: X, Y, Z, W in proportion means $X:Y=Z:W$, i.e. $XW=YZ$.

1. $3\times 39=13\times Z$.
2. $Z=\frac{117}{13}=9$.

Yaad rakhein: Product of extremes = product of means.

Answer: (D) $9$

Concept: Add all males and all females, then simplify the ratio.

1. Males $=1500+1460+1105+1305=5370$.
2. Females $=1220+1320+1180+1170=4890$.
3. $5370:4890=537:489=179:163$ (divide by 10, then by 3).

Trick: Males are more, so the first term must be larger – only $179:163$ or $537:389$; check $537:389$ is not $5370:4890$.

Answer: (C) $179:163$

Concept: Net change for $+a\%$ then $-b\%$ $=a-b-\frac{ab}{100}$.

1. $25-30-\frac{25\times 30}{100}=-5-7.5=-12.5\%$.
2. Check: $100\to 125\to 125\times 0.7=87.5$.

Trick: Multiply factors: $1.25\times 0.7=0.875$ ⇒ 12.5% fall.

Answer: (D) 12.5% decreased

Concept: Successive increases multiply: final $=P(1.10)(1.22)(1.05)$.

1. $1.10\times 1.22=1.342$.
2. $1.342\times 1.05=1.4091$.
3. Overall increase $=40.91\%\approx 40.9\%$.

Trick: $10+22+\frac{10\times 22}{100}=34.2$; then $34.2+5+\frac{34.2\times 5}{100}=40.91$.

Answer: (B) $40.9\%$

Concept: Biology students = % opted × total enrolled; the male-female ratio is not needed.

1. School B: $36\%$ of $400=144$.
2. School D: $18\%$ of $800=144$.
3. Ratio $=144:144=1:1$.

Trick: $36\times 4=18\times 8$ – equal at a glance.

Answer: (C) $1:1$

Concept: Add each cricketer's row; the largest row total is the answer.

1. A: $1+0+1+0+1=3$; B: $1+0+2+1+1=5$.
2. C: $1+2+0+0+1=4$; D: $0+0+1+1+0=2$; E: $0+1+1+0+1=3$.
3. Maximum $=5$ by B.

Trick: Row sums for a person, column sums for a year.

Answer: (C) B

Concept: % change $=\frac{\text{new}-\text{old}}{\text{old}}\times 100$. Company Y: 25, 35, 35, 40, 50 (2016–2020).

1. 2017: $\frac{35-25}{25}\times 100=40\%$.
2. 2018: $0\%$; 2019: $\frac{5}{35}\times 100\approx 14.3\%$.
3. 2020: $\frac{10}{40}\times 100=25\%$. Maximum in 2017.

Trick: A small base gives a big %: the rise of 10 on 25 beats 10 on 40.

Answer: (D) 2017

Concept: ${\text{cosec}}^{2}A-{\cot }^{2}A=1$, i.e. $(\text{cosec}A+\cot A)(\text{cosec}A-\cot A)=1$.

1. $m\times (\text{cosec}A-\cot A)=1$.
2. $\text{cosec}A-\cot A=\frac{1}{m}$.

Yaad rakhein: Same pair works for $\sec A\pm \tan A$: they are reciprocals.

Answer: (A) $\frac{1}{m}$