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

Concept: $(a+b{)}^2+(a-b{)}^2=2(a^2+b^2)$.

1. Here $a=7x$, $b=4y$.
2. $2(49x^2+16y^2)=98x^2+32y^2$.

Trick: The $\pm 2ab$ terms cancel, so no $xy$ term can appear — reject option (B).

Answer: (A) $98x^2+32y^2$

Concept: Both items cost the same, so overall profit % = average of the two profit %.

1. $\frac{10+x}{2}=20\Rightarrow x=30\%$.
2. Check: total CP ₹$6000$, needed SP ₹$7200$; first SP ₹$3300$, second ₹$3900$ = $30\%$ profit ✓

Trick: Equal CPs → simple average of percentages.

Answer: (C) $30\%$

Concept: Convert each ratio to a decimal (or compare with $\frac{1}{2}$).

1. $\frac{41}{92}\approx 0.45$, $\frac{33}{41}\approx 0.80$, $\frac{42}{49}\approx 0.86$, $\frac{35}{47}\approx 0.74$.
2. Only $\frac{41}{92}$ is less than $\frac{1}{2}$, so it is the smallest.

Trick: Benchmark $\frac{1}{2}$: the only ratio whose antecedent is less than half the consequent wins.

Answer: (A) $41:92$

Concept: Distance is directly proportional to petrol used.

1. $28$ litres $=8\times 3.5$ litres.
2. Distance $=8\times 75.5=604$ km.

Trick: Spot the multiplier $\frac{28}{3.5}=8$ first.

Answer: (C) $604$

Concept: Effective discount $=\frac{\text{free}}{\text{bought}+\text{free}}\times 100$.

1. Customer takes $8$ items, pays for $5$.
2. Discount $=\frac{3}{8}\times 100=37.5\%$.

Trick: Buy $x$ get $y$ free → $\frac{y}{x+y}\times 100$.

Answer: (B) $37.5\%$

Concept: Here $(0.2)$ is the index (power) of $0.00243$: $(0.00243{)}^{0.2}=(0.00243{)}^{\frac{1}{5}}$, i.e. the fifth root.

1. $0.00243=\frac{243}{100000}=\frac{3^5}{{10}^5}=(0.3{)}^5$.
2. ${[(0.3{)}^5]}^{\frac{1}{5}}=0.3$.

Trick: Power $0.2=\frac{1}{5}$ → fifth root; $243=3^5$.

Answer: (A) $0.3$

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$.

1. First part: $21\text{m/s}\times 600\text{s}=12600$ m $=12.6$ km.
2. Total distance $=12.6+9+10=31.6$ km in $30$ min $=\frac{1}{2}$ h.
3. Average $=31.6\times 2=63.2\text{km/h}$.

Trick: $21\text{m/s}=21\times \frac{18}{5}=75.6\text{km/h}$; in $\frac{1}{6}$ h that is $12.6$ km.

Answer: (B) $63.2\text{km/h}$

Concept: A number ending in 0 or 5 is divisible by 5.

1. Digits of 150328 in descending order: $853310$.
2. Five times 13 $=65$; $853310-65=853245$.
3. Last digit is 5, so it is divisible by 5 (not by 2; alternating sum $=3$, so not by 11). Its digit sum $27$ also makes it divisible by 3.

Trick: Only the unit digit matters for 2 and 5: $0-5$ ends in 5.

Answer: (B) $5$

Concept: Divide by $xy$ to get linear equations in $\frac{1}{x},\frac{1}{y}$.

1. $\frac{6}{y}+\frac{7}{x}=5$ and $\frac{10}{x}-\frac{4}{y}=4$. Put $a=\frac{1}{x}$, $b=\frac{1}{y}$: $7a+6b=5$, $10a-4b=4$.
2. Multiply: $14a+12b=10$, $30a-12b=12$; add: $44a=22$, $a=\frac{1}{2}$, $b=\frac{1}{4}$.
3. So $x=2$, $y=4$.

Trick: Check options: $x=2,y=4$: $12+28=40=5\times 8$ ✓, $40-8=32=4\times 8$ ✓

Answer: (C) $2$ and $4$

Concept: $\cos$ of ${45}^{\circ },{135}^{\circ },{225}^{\circ },{315}^{\circ }$ is $\pm \frac{1}{\sqrt{2}}$; squaring removes the sign.

1. Each ${\cos }^2$ term $=\frac{1}{2}$.
2. Sum $=4\times \frac{1}{2}=2$.

Trick: All four angles are ${45}^{\circ }$ reference angles → each square is $\frac{1}{2}$.

Answer: (A) $2$

Concept: Percentage $=\frac{\text{marks}}{\text{maximum}}\times 100$ — subjects have different maximum marks.

1. 1985: History $\frac{90}{200}=45\%$, Geography $\frac{35}{100}=35\%$, Maths $\frac{88}{200}=44\%$.
2. Science $\frac{120}{200}=60\%$, English $\frac{90}{150}=60\%$.
3. Science and English are equal.

Trick: Never compare raw marks when the totals differ.

Answer: (C) Science, English

Concept: Three equal diameters fill the big diameter, so $R=3r$. Area of semicircle $=\frac{1}{2}\pi r^2$.

1. Red $=3\times \frac{1}{2}\pi r^2=\frac{3}{2}\pi r^2$.
2. Big semicircle $=\frac{1}{2}\pi (3r{)}^2=\frac{9}{2}\pi r^2$; blue $=\frac{9}{2}\pi r^2-\frac{3}{2}\pi r^2=3\pi r^2$.
3. Red : Blue $=\frac{3}{2}:3=1:2$.

Trick: Areas scale as $r^2$: big $=9$ units, red $=3$, blue $=6$ → $1:2$.

Answer: (B) $1:2$

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$ (not the average of speeds).

1. Time $=\frac{100}{50}+\frac{140}{70}=2+2=4$ h.
2. Average $=\frac{240}{4}=60\text{km/h}$.

Trick: Equal times (2 h each) → simple average $\frac{50+70}{2}=60$.

Answer: (A) $60\text{km/h}$

Concept: Longest pole in a room = body diagonal $=\sqrt{l^2+b^2+h^2}$.

1. $\sqrt{{60}^2+{30}^2+{20}^2}=\sqrt{3600+900+400}$.
2. $=\sqrt{4900}=70$ ft.

Trick: $60,30,20=10\times (6,3,2)$ and $6^2+3^2+2^2=49=7^2$.

Answer: (B) $70$ feet

Concept: In a bar graph the tallest bar shows the maximum value.

1. Bars for Jan, Mar, Apr, Jun are around or below 6000; Feb is a little higher.
2. May's bar is the tallest (about 7000+).

Yaad rakhein: Just compare the bar heights — no calculation needed.

Answer: (C) May

Concept: On melting, volume is conserved: $n\times V_{\text{sphere}}=V_{\text{cylinder}}$.

1. Sphere ($r=3$): $\frac{4}{3}\pi (3{)}^3=36\pi$.
2. Cylinder ($r=4$, $h=18$): $\pi (4{)}^2(18)=288\pi$.
3. $n=\frac{288\pi }{36\pi }=8$.

Trick: $n=\frac{3R^{2}H}{4r^3}=\frac{3\times 16\times 18}{4\times 27}=8$.

Answer: (C) $8$

Concept: Euclid's division method: HCF$(a,b)=$ HCF$(b,amodb)$.

1. $832=3\times 238+118$.
2. $238=2\times 118+2$.
3. $118=59\times 2+0$ → HCF $=2$.

Trick: $238=2\times 7\times 17$, $832=2^6\times 13$ — only 2 is common.

Answer: (C) $2$

Concept: Volume of a cuboid $=l\times b\times h$.

1. Let breadth $=b$; height $=6b$; length $=\frac{1}{2}\times 6b=3b$.
2. $3b\times b\times 6b=18b^3=23958\Rightarrow b^3=1331$.
3. $b=\sqrt[3]{1331}=11$ cm.

Trick: Check: $18\times {11}^3=18\times 1331=23958$ ✓ (with 21 cm the volume would be $166698$).

Answer: (C) $11$ cm

Concept: Fix a convenient base (Sohan $=100$) and apply the changes step by step.

1. Sohan $=100$ → Shyam $=100-30\%=70$.
2. Ram $=70+70\%$ of $70=70\times 1.7=119$.
3. Ram : Sohan $=119:100$.

Trick: $0.7\times 1.7=1.19$.

Answer: (A) $119:100$

Concept: Divisible by 16 → last four digits divisible by 16; by 11 → (sum of odd-place digits − sum of even-place digits) is $0$ or a multiple of 11.

1. Last four digits $30M6$: $3056=16\times 191$ ✓; 3016, 3036, 3076, 3096 are not multiples of 16 — so $M=5$.
2. Number $34R0503056$: odd places $3+R+5+3+5=16+R$; even places $4+0+0+0+6=10$.
3. $16+R-10=6+R$ must be $11$ → $R=5$.

Trick: Only option (C) has $M=5$, so it is chosen right after the 16-test.

Answer: (C) $5$ and $5$