Q&A: practice questions with answers

Concept: Write both present ages with one variable and use the given condition.

1. Let son's present age $=S$. Eight years ago: $F-8=2(S-8)+11\Rightarrow F=2S+3$.
2. After $x$ years: $F+x=2(S+x)\Rightarrow 2S+3+x=2S+2x$.
3. $\Rightarrow x=3$ years.

Trick: If $F=2S+k$ now, the father becomes exactly twice the son after $k$ years. Here $k=3$.

Answer: (A) $3$

Concept: $\text{SI}=\frac{P\times R\times T}{100}$

1. $P=400,R=4\%,T=6.5$ years.
2. $\text{SI}=\frac{400\times 4\times 6.5}{100}=4\times 26=104$.

Trick: $4\%$ of ₹400 = ₹16 per year; $16\times 6.5=104$.

Answer: (D) ₹$104$

Concept: Convert everything into one day's work of one person.

1. Total work $=9\times 16=144$ man-days $=5\times 18=90$ woman-days.
2. 1 man's 1-day work $=\frac{1}{144}$, 1 woman's $=\frac{1}{90}$.
3. $2$ women $+4$ men per day $=\frac{2}{90}+\frac{4}{144}=\frac{1}{45}+\frac{1}{36}=\frac{4+5}{180}=\frac{1}{20}$.
4. Time $=20$ days.

Trick: Take total work $=\text{LCM}(144,90)=720$: man $=5$, woman $=8$ units/day; $4\times 5+2\times 8=36$; $720÷36=20$.

Answer: (D) $20$ days

Concept: Successive percentage changes multiply: $P_3=P\times (1+\frac{r_1}{100})(1+\frac{r_2}{100})(1-\frac{r_3}{100})$.

1. After year 1: $15250\times 1.25=19062.5$.
2. After year 2: $19062.5\times 1.6=30500$.
3. After year 3: $30500\times 0.43=13115$.

Trick: $1.25\times 1.6=2$, so just do $15250\times 2\times 0.43=13115$.

Answer: (A) $13,115$

Concept: Simplify the complex fraction (inner bracket) first, then add/subtract.

1. $1+\frac{4}{9}=\frac{13}{9}$, so $\frac{1}{\frac{13}{9}}=\frac{9}{13}$.
2. $\frac{2}{5}-\frac{4}{5}=-\frac{2}{5}$.
3. $\frac{9}{13}-\frac{2}{5}=\frac{45-26}{65}=\frac{19}{65}$.

Trick: Denominators 13 and 5 → answer must have 65 below; only option (A) has it.

Answer: (A) $\frac{19}{65}$

Concept: In compound interest, the 2^nd year's interest is earned on (principal + 1^st year's interest).

1. 1^st year interest $=20\%$ of $P=0.2P$; amount $=1.2P$.
2. 2^nd year interest $=20\%$ of $1.2P=0.24P$.
3. $0.24P=5916\Rightarrow P=\frac{5916}{0.24}=24650$.

Trick: 2^nd-year CI $=P\times \frac{R}{100}\times (1+\frac{R}{100})$; check options: $24650\times 0.24=5916$.

Answer: (A) ₹$24,650$

Concept: Angle sum of a triangle $={180}^{\circ }$; an exterior angle equals the sum of the two opposite interior angles.

1. In right $△BDC$: $\angle C={180}^{\circ }-{90}^{\circ }-{55}^{\circ }={35}^{\circ }$.
2. In $△AEC$: $\angle AEC={180}^{\circ }-{56}^{\circ }-{35}^{\circ }={89}^{\circ }$.
3. $\angle AEB={180}^{\circ }-{89}^{\circ }={91}^{\circ }$ (linear pair on BC).

Trick: Exterior angle: $\angle AEB=\angle EAC+\angle C={56}^{\circ }+{35}^{\circ }={91}^{\circ }$.

Answer: (D) ${91}^{\circ }$

Concept: $V=\frac{4}{3}\pi r^3$, so volume is proportional to $r^3$.

1. New radius $=\frac{r}{2}$.
2. $\frac{V_1}{V_2}=\frac{r^3}{(\frac{r}{2}{)}^3}=\frac{r^3}{\frac{r^3}{8}}=8$.
3. Ratio $=8:1$.

Trick: Radius ratio $a:b$ → volume ratio $a^3:b^3$; $2:1\Rightarrow 8:1$.

Answer: (B) $8:1$

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

1. Times: $\frac{364}{91}=4$ h, $\frac{441}{63}=7$ h, $\frac{648}{81}=8$ h; total $=19$ h.
2. Total distance $=364+441+648=1453\text{km}$.
3. Average speed $=\frac{1453}{19}\approx 76.47\text{km/h}$.

Trick: Every distance divides exactly by its speed – find the hours first, never average the speeds.

Answer: (A) $76.47$

Concept: $(a^m{)}^n=a^{mn}$; equal bases ⇒ equal exponents.

1. $x^z=({73}^{0.51}{)}^z={73}^{0.51z}$.
2. $y^{10}=({73}^{0.73}{)}^{10}={73}^{7.3}$.
3. $0.51z=7.3\Rightarrow z=\frac{7.3}{0.51}\approx 14.31$.

Trick: $z=\frac{730}{51}$; $51\times 14=714$, remainder $16$ → $14.31$.

Answer: (C) $14.31$

Concept: Make the decimal places equal, find the LCM of the whole numbers, then put the decimal back.

1. Write both with 3 decimal places: $3.900$ and $0.169$ → $3900$ and $169$ (thousandths).
2. $3900=2^2\times 3\times 5^2\times 13$, $169={13}^2$.
3. $\text{LCM}=2^2\times 3\times 5^2\times {13}^2=50700$.
4. Put back 3 decimal places: $50700÷1000=50.7$.

Trick: LCM of decimals is a multiple of both numbers – check: $50.7÷3.9=13$ and $50.7÷0.169=300$ ✓.

Answer: (D) $50.7$

Concept: Use ${\sec }^{2}A=\frac{1}{{\cos }^{2}A}$ and $1+{\cot }^{2}A={\csc }^{2}A$.

1. $\frac{1}{{\sec }^{2}A}={\cos }^{2}A$.
2. $\frac{1}{1+{\cot }^{2}A}=\frac{1}{{\csc }^{2}A}={\sin }^{2}A$.
3. Sum $={\cos }^{2}A+{\sin }^{2}A=1$.

Trick: Put $A={45}^{\circ }$: $\frac{1}{2}+\frac{1}{2}=1$ instantly.

Answer: (B) $1$

Concept: One-day work of C = combined one-day work − (A's + B's one-day work).

1. One-day work: A $=\frac{1}{17}$, B $=\frac{1}{12}$, A+B+C $=\frac{1}{2}$.
2. C $=\frac{1}{2}-\frac{1}{17}-\frac{1}{12}=\frac{102-12-17}{204}=\frac{73}{204}$.
3. Days for C $=\frac{204}{73}=2\frac{58}{73}$ days.

Trick: All three together take 2 days, so C alone must take a little more than 2 days – only option (D) fits.

Answer: (D) $2\frac{58}{73}$ days

Concept: Successive discounts $a\%$ and $b\%$ = single discount $a+b-\frac{ab}{100}$ %.

1. Single 32% discount on ₹1000 = ₹320.
2. Successive: $20+12-\frac{20\times 12}{100}=29.6\%$ → ₹296.
3. Difference $=320-296=$ ₹24.

Trick: Difference = $\frac{ab}{100}\%$ of MP $=2.4\%$ of ₹1000 = ₹24.

Answer: (B) ₹24

Concept: Overall average = total runs ÷ total matches.

1. First 23 matches: $23\times 42=966$.
2. Next 10 matches: $10\times 16=160$.
3. Average $=\frac{966+160}{33}=\frac{1126}{33}=34.12$.

Trick: Deviation method: $42-\frac{10\times (42-16)}{33}=42-7.88=34.12$.

Answer: (B) $34.12$

Concept: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$ per pen.

1. CP of one pen $=\frac{180}{12}=$ ₹15.
2. SP of one pen = ₹19, profit = ₹4.
3. $\text{Profit}\%=\frac{4}{15}\times 100=26.67\%$.

Trick: $\frac{1}{15}=6.67\%$, so $\frac{4}{15}=26.67\%$.

Answer: (B) 26.67%

Concept: For equal distances at speeds $x$ and $y$: average speed $=\frac{2xy}{x+y}$.

1. $x=20$, $y=30$ (same route both ways).
2. Average $=\frac{2\times 20\times 30}{20+30}=\frac{1200}{50}=24\text{km/h}$.

Trick: Average speed for equal distances is always a little less than the simple mean ($25$).

Answer: (A) $24$

Concept: Increasing by 100% means the number becomes double.

1. $x+100\%\text{ of }x=2x=2770$.
2. $x=\frac{2770}{2}=1385$.

Trick: +100% = ×2, so just halve the result.

Answer: (C) $1385$

Concept: Product of two numbers = HCF × LCM.

1. Other number $=\frac{13\times 1170}{130}$.
2. $=\frac{15210}{130}=117$.
3. Check: $\text{HCF}(130,117)=13$ ✓

Trick: Other number must be a multiple of HCF 13: $117=13\times 9$.

Answer: (C) $117$

Concept: Simplify the bracket, expand, then combine like terms.

1. $\frac{6}{3}=2$, so first part $=3(2x^2-23x+18)=6x^2-69x+54$.
2. Second part $=6x^2+24x-90$.
3. Subtract: $6x^2-69x+54-6x^2-24x+90=-93x+144$.

Trick: The $x^2$ terms cancel; watch the sign: $-6\times (-15)=+90$.

Answer: (D) $-93x+144$