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

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

1. $\text{MP}\times 0.8=9080$.
2. $\text{MP}=\frac{9080}{0.8}=11350$.

Trick: Divide by 0.8 = multiply by $\frac{5}{4}$: $9080\times \frac{5}{4}=11350$.

Answer: (B) ₹11,350

Concept: Interest is charged on the unpaid balance: $R=\frac{SI\times 100}{P\times T}$.

1. Balance (principal) $=4400-2000=2400$; paid later $2440$, so $SI=40$.
2. $T=\frac{1}{12}$ year.
3. $R=\frac{40\times 100\times 12}{2400}=20\%$ per annum.

Trick: Monthly rate $=\frac{40}{2400}=\frac{1}{60}$; yearly $=\frac{12}{60}=20\%$.

Answer: (A) 20%

Concept: The bisectors of an interior angle and of the exterior angle at another vertex meet at an angle equal to half the third angle: $\angle QSR=\frac{\angle P}{2}$.

1. Exterior $\angle R=\angle P+\angle Q$, so $\frac{\angle P+\angle Q}{2}=\frac{\angle Q}{2}+\angle QSR$.
2. Hence $\angle QSR=\frac{\angle P}{2}$.
3. $\angle P=2\times {40}^{\circ }={80}^{\circ }$.

Yaad rakhein: Internal–external bisectors: $\frac{\angle P}{2}$; both internal: ${90}^{\circ }+\frac{\angle P}{2}$; both external: ${90}^{\circ }-\frac{\angle P}{2}$.

Answer: (C) ${80}^{\circ }$

Concept: $a^3+b^3=(a+b)(a^2-ab+b^2)$ with $a=x^2$, $b=y^2$.

1. $x^4+y^4=x^2{y}^2\Rightarrow x^4-x^2{y}^2+y^4=0$.
2. $x^6+y^6=(x^2+y^2)(x^4-x^2{y}^2+y^4)$.
3. $=(x^2+y^2)\times 0=0$.

Trick: Spot the factor $a^2-ab+b^2$ of a sum of cubes.

Answer: (B) 0

Concept: Percentage $=\frac{\text{part}}{\text{total}}\times 100$.

1. School B: girls $=846$, total $=1880$.
2. $\frac{846}{1880}\times 100=45\%$.

Trick: $10\%$ of $1880=188$; $846=188\times 4.5$, so $45\%$.

Answer: (B) 45%

Concept: Opposite angles of a cyclic quadrilateral are supplementary (sum ${180}^{\circ }$).

1. In EFGH, $\angle FGH$ (at G) and $\angle HEF$ (at E) are opposite.
2. $\angle HEF={180}^{\circ }-{57}^{\circ }={123}^{\circ }$.

Trick: ${33}^{\circ }$ is the complement (${90}^{\circ }-{57}^{\circ }$) – a trap.

Answer: (B) ${123}^{\circ }$

Concept: $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$; dividing by a fraction means multiplying by its reciprocal.

1. Expression $=\frac{1}{4\times 9\times 25\times 49}\times \frac{121}{36}$.
2. Square root $=\frac{1}{2\times 3\times 5\times 7}\times \frac{11}{6}$.
3. $=\frac{1}{210}\times \frac{11}{6}=\frac{11}{1260}$.

Trick: All are perfect squares – take roots term by term: $\frac{1}{2}·\frac{1}{3}·\frac{1}{5}·\frac{1}{7}·\frac{11}{6}$.

Answer: (A) $\frac{11}{1260}$

Concept: Surface area of a hemisphere ($2\pi r^2$ or $3\pi r^2$) is proportional to $r^2$.

1. Ratio $=\frac{7^2}{4^2}$.
2. $=\frac{49}{16}$, i.e. $49:16$ (new : original).

Trick: Area ratio $=$ (radius ratio)${}^2$; volume ratio $=$ (radius ratio)${}^3$.

Answer: (B) 49 : 16

Concept: Add each column over the seven years and compare.

1. 4AH $=676$; 7AH $=694$.
2. 32AH $=114+102+75+150+135+165+160=901$.
3. 35AH $=601$; 55AH $=799$.
4. Highest $=901$ (32AH).

Trick: Eyeball first: 32AH has three entries $\ge 150$ – it clearly leads.

Answer: (A) 32 AH

Concept: Compare the column totals of the classes.

1. Totals: V $=917$, VI $=924$, VII $=904$.
2. VIII $=919$, IX $=903$, X $=910$.
3. Maximum $=924$ (class VI).

Trick: Use the given Total row – no need to re-add.

Answer: (C) VI

Concept: Apply BODMAS: 'of' first, then division/multiplication, then addition/subtraction.

1. $\frac{5}{8}$ of $184=\frac{5\times 184}{8}=115$
2. $115\times 15÷5=115\times 3=345$
3. $7\frac{5}{8}+345-(?)=0\Rightarrow (?)=352\frac{5}{8}$

Trick: 'of' is just multiplication done first; $184÷8=23$ makes it quick: $23\times 5=115$.

Answer: (D) $352\frac{5}{8}$

Concept: Read the 2015 column and pick the largest value.

1. 2015 earnings: P = 56, Q = 55, R = 68, S = 60 (₹ crore).
2. The largest of these is 68, which belongs to firm R.

Trick: Look only at the asked column; ignore the other years.

Answer: (C) R

Concept: $\text{SP}=\text{MP}\times \frac{100-d}{100}$ and $\text{CP}=\frac{\text{SP}\times 100}{100+p}$.

1. $\text{SP}=7200\times \frac{85}{100}=6120$
2. $\text{CP}=\frac{6120\times 100}{120}=5100$

Trick: Directly: $\text{CP}=7200\times \frac{85}{120}=60\times 85=5100$.

Answer: (D) ₹5,100

Concept: $\frac{{\sec }^{2}A}{{\csc }^{2}A}=\frac{1/{\cos }^{2}A}{1/{\sin }^{2}A}=\frac{{\sin }^{2}A}{{\cos }^{2}A}={\tan }^{2}A$.

1. Given $\tan A=\frac{2}{5}$.
2. $\frac{{\sec }^{2}A}{{\csc }^{2}A}={\tan }^{2}A={\left(\frac{2}{5}\right)}^2=\frac{4}{25}$

Trick: $\sec /\csc$ always simplifies to $\tan$; just square the given value.

Answer: (B) $\frac{4}{25}$

Concept: $\text{CI}=P[{(1+\frac{r}{100})}^n-1]$.

1. $(1.12{)}^3=1.404928$
2. $\text{CI}=25000\times 0.404928=10123.20$

Trick: Effective rate for 3 years at $12\%$: $12+12+12+3\times 1.44+0.1728=40.4928\%$; $40.4928\%$ of 25000 = 10123.2.

Answer: (D) ₹10,123.20

Concept: A number divisible by 2, 4, 6 and 8 must be a multiple of their LCM.

1. $\text{LCM}(2,4,6,8)=24$
2. $1351=24\times 56+7$, remainder 7.
3. Number to add $=24-7=17$ (check: $1368=24\times 57$).

Trick: Least number to ADD = LCM − remainder; least number to SUBTRACT = remainder.

Answer: (D) 17

Concept: SAS rule: two sides and the angle included between them are equal ⇒ triangles are congruent.

1. Given: AC = PR (side) and ∠A = ∠P (angle).
2. ∠A lies between sides AB and AC; ∠P lies between PQ and PR.
3. So if AB = PQ, the equal angle is the included angle ⇒ congruent by SAS.

Trick: SSS needs all 3 sides; ASS and AAA are NOT congruence rules.

Answer: (B) AB = PQ by SAS

Concept: Amount on an item = its percentage × total income.

1. Education of children $=12\%$.
2. $12\%$ of ₹1,00,000 $=\frac{12}{100}\times 100000=12000$

Trick: With total ₹1,00,000, every $1\%$ is ₹1,000 — just read the percentage.

Answer: (B) 12,000

Concept: Cost price = price after both discounts + overhead (transport).

1. Price paid $=850\times 0.8\times 0.9=612$
2. Total CP $=612+55=667$; profit $=980-667=313$
3. $\text{Profit}\%=\frac{313}{667}\times 100\approx 46.93\%$

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

Answer: (D) $46.93\%$

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

1. Let numbers be $4k$ and $3k$: $12k^2=2700\Rightarrow k^2=225\Rightarrow k=15$
2. Numbers: 60 and 45.
3. Difference $=60-45=15$ (= $k$).

Trick: For ratio $a:b$ with $a-b=1$, the difference equals $k$ itself.

Answer: (A) 15