Q&A: practice questions with answers

Concept: $81=3^4$; divide directly (digit-sum test of 9 is necessary but not sufficient).

1. Digit sums: 5265 → 18, 5854 → 22, 4678 → 25, 5232 → 12; only 5265 is divisible by 9.
2. $5265÷81=65$ exactly ($81\times 65=5265$).

Trick: First drop options failing the 9-test, then divide.

Answer: (A) $5265$

Concept: Compare SP and CP of the same number of articles.

1. ${\text{SP}}_{20}={\text{CP}}_6\Rightarrow \frac{\text{SP}}{\text{CP}}=\frac{6}{20}$.
2. SP $<$ CP, so it is a loss.
3. $\text{Loss}\%=\frac{20-6}{20}\times 100=70\%$.

Trick: $\text{Loss}\%=\frac{\text{difference}}{\text{goods sold}}\times 100=\frac{14}{20}\times 100$.

Answer: (D) $70\%$ loss

Concept: Joining 5 cubes of edge $a$ gives a cuboid $5a\times a\times a$.

1. $a^3=10941048\Rightarrow a=222\text{cm}$.
2. Lateral surface area $=2(l+b)h=2(5a+a)a=12a^2$.
3. $12\times {222}^2=12\times 49284=591408{\text{cm}}^2$.

Trick: Cube root: ends in 8 → unit digit 2; $10941\approx {22}^3$ → $222$.

Answer: (A) $5,91,408$

Concept: Substitute $u=x^{-2}$ to get a quadratic.

1. $u^2-10u+9=0\Rightarrow (u-1)(u-9)=0$.
2. $u=1\Rightarrow x^2=1\Rightarrow x=\pm 1$.
3. $u=9\Rightarrow x^2=\frac{1}{9}\Rightarrow x=\pm \frac{1}{3}$.

Trick: Put $x=\frac{1}{3}$: $81-90+9=0$ ✔.

Answer: (D) $\pm 1,\pm \frac{1}{3}$

Concept: $\tan A=\frac{\sin A}{\cos A}$ and $\text{cosec}A=\frac{1}{\sin A}$.

1. $\sqrt{3}\frac{\sin A}{\cos A}=3\sin A\Rightarrow \cos A=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$.
2. $\text{cosec}A\times \tan A=\frac{1}{\sin A}\times \frac{\sin A}{\cos A}=\sec A=\sqrt{3}$.
3. $2\sqrt{3}\times \sqrt{3}=6$.

Trick: $\text{cosec}A\times \tan A=\sec A$ always.

Answer: (A) $6$

Concept: A change in gain% equals the change in SP as a percent of CP.

1. Extra gain $=82\%-57\%=25\%$ of CP.
2. $25\%$ of CP $=$ ₹$238$.
3. CP $=\frac{238\times 100}{25}=$ ₹$952$.

Trick: Difference of percents ↔ difference of amounts.

Answer: (B) $952$

Concept: Split 112 as $100+12$.

1. $81\times 100=8100$.
2. $81\times 12=972$.
3. $8100+972=9072$.

Trick: Unit digit $1\times 2=2$ and $81\times 112\approx 9000$ → only 9072 fits.

Answer: (A) $9072$

Concept: Form two equations: number of notes and total value.

1. $x+y=34$ ($x$ = ₹50 notes, $y$ = ₹100 notes).
2. $50x+100y=3100\Rightarrow x+2y=62$.
3. Subtract: $y=28$, so $x=6$.

Trick: If all 34 were ₹100 notes: ₹3400; shortfall ₹300 ÷ ₹50 $=6$ notes of ₹50.

Answer: (A) $6$

Concept: Empirical relation: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. Mode $=3\times 53-2\times 42$.
2. $=159-84=75$.

Trick: $\text{Mo}=3\text{Md}-2\text{M}$.

Answer: (B) $75$

Concept: Arrange in order; for odd $n$ the median is the ${\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Ascending: 12, 19, 24, 56, 70, 71, 77, 80, 88, 98, 99.
2. $n=11$, so median $=$ 6th value.
3. 6th value $=71$.

Trick: Odd count → no averaging needed; options 70.5 and 71.5 are traps.

Answer: (A) $71$

Concept: $x^3+y^3=(x+y{)}^3-3xy(x+y)$.

1. $(x+y{)}^3={14}^3=2744$.
2. $3xy(x+y)=3\times 45\times 14=1890$.
3. $x^3+y^3=2744-1890=854$.

Trick: Here $x=9,y=5$: $729+125=854$.

Answer: (C) $854$

Concept: Upstream $=B-S$, Downstream $=B+S$.

1. $B=9k,S=4k$ → upstream $5k$, downstream $13k$.
2. $5k=\frac{27.9}{39}\times 60=\frac{1674}{39}\text{km/h}$, so $13k=\frac{1674}{15}=111.6\text{km/h}$.
3. Upstream time $=\frac{18.9\times 39}{1674}\approx 0.44$ h; downstream $=\frac{51.3}{111.6}\approx 0.46$ h.
4. Total $=0.9$ hours.

Trick: In minutes: upstream $\frac{18.9}{27.9}\times 39=26.4$ min; downstream $=26.4\times \frac{51.3}{18.9}\times \frac{5}{13}=27.6$ min; total 54 min $=0.9$ h.

Answer: (B) $0.9$

Concept: Consecutive odd numbers differ by 2; for $n$ such numbers, largest − smallest $=2(n-1)$.

1. Average $=30$, so the six numbers are $25,27,29,31,33,35$ (centred on 30).
2. Largest − smallest $=35-25=10$.
3. Check with formula: $2(6-1)=10$.

Trick: The average is not even needed: difference $=2(n-1)$ for consecutive odd/even numbers.

Answer: (C) $10$

Concept: Convert the ratio of number of coins into a ratio of values, then find one unit.

1. Let coins be $16k,8k,20k$.
2. Value in ₹: $16k\times 0.5+8k\times 0.25+20k\times 0.1=8k+2k+2k=12k$.
3. $12k=384\Rightarrow k=32$.
4. Number of 10 paisa coins $=20\times 32=640$.

Trick: Value ratio $=8:2:2$ (12 parts $=$ ₹384, 1 part $=$ ₹32); 10 paisa share $=$ ₹64, so coins $=\frac{64}{0.1}=640$.

Answer: (C) 640

Concept: If $DE\parallel AC$, then $△BDE~△BAC$ and the ratio of areas $=$ (ratio of sides)${}^2$.

1. $BA=BD+DA=12+11=23\text{cm}$.
2. $\frac{\text{ar}(BDE)}{\text{ar}(BAC)}={\left(\frac{12}{23}\right)}^2=\frac{144}{529}$.
3. Trapezium $ADEC=529-144=385$ parts.
4. Required ratio $=144:385$.

Trick: Small triangle : trapezium $=a^2:(b^2-a^2)$ where $a=BD$, $b=BA$.

Answer: (D) $144:385$

Concept: For half-yearly compounding, rate is halved and time (in periods) is doubled: $CI=P[{(1+\frac{r}{100})}^n-1]$.

1. Rate per half-year $=\frac{2}{2}=1\%$, periods $=2$.
2. ${\left(1.01\right)}^2-1=0.0201$.
3. $CI=670000\times 0.0201=13467$.

Trick: For 2 periods at $r\%$, effective rate $=2r+\frac{r^2}{100}=2.01\%$.

Answer: (B) ₹13,467

Concept: Correct sum $=$ wrong sum $-$ wrong value $+$ correct value.

1. Wrong sum $=10\times 40=400$.
2. Correct sum $=400-32+43=411$.
3. Correct average $=\frac{411}{10}=41.1$.

Trick: Change in average $=\frac{43-32}{10}=1.1$, so $40+1.1=41.1$.

Answer: (B) $41.1$

Concept: $(a+b{)}^2-(a-b{)}^2=4ab$.

1. $(a+b{)}^2={24}^2=576$.
2. $4ab=576-268=308$.
3. $ab=\frac{308}{4}=77$.

Trick: Product $=\frac{(\text{sum}{)}^2-(\text{difference}{)}^2}{4}$.

Answer: (A) $77$

Concept: If $a:b::b:c$, then $b^2=ac$.

1. $b^2=a\times 4a=4a^2\Rightarrow b=2a$.
2. $18=2a\Rightarrow a=9$.
3. $c=4a=36$.

Trick: $c=4a$ means $b=2a$ and $c=2b$ – directly $c=2\times 18=36$.

Answer: (A) $36$

Concept: Count days (exclude the first day, include the last) and use $SI=\frac{P\times R\times T}{100}$ with $T$ in years.

1. 2024 is a leap year: February days after 9th $=29-9=20$, March $=31$, April $=10$; total $=61$ days.
2. $T=\frac{61}{366}$ year.
3. $SI=\frac{3000\times 6\times 61}{100\times 366}=\frac{10980}{366}=30$.

Trick: Note $366=6\times 61$, so $\frac{61}{366}=\frac{1}{6}$ year → $SI=\frac{3000\times 6}{100\times 6}=30$.

Answer: (C) ₹30