Q&A: practice questions with answers

Concept: $a^m\times a^n=a^{m+n}$ and ${\left(\frac{b}{a}\right)}^n={\left(\frac{a}{b}\right)}^{-n}$.

1. Write everything with base $\frac{8}{21}$: ${\left(\frac{21}{8}\right)}^{12}={\left(\frac{8}{21}\right)}^{-12}$, ${\left(\frac{21}{8}\right)}^{14}={\left(\frac{8}{21}\right)}^{-14}$.
2. LHS $={\left(\frac{8}{21}\right)}^{-12+14-14}={\left(\frac{8}{21}\right)}^{-12}$.
3. Equate powers: $9m+20=-12$.
4. $9m=-32\Rightarrow m=-\frac{32}{9}$.

Trick: ${\left(\frac{8}{21}\right)}^{14}$ and ${\left(\frac{21}{8}\right)}^{14}$ cancel, leaving only ${\left(\frac{21}{8}\right)}^{12}$.

Answer: (B) $-\frac{32}{9}$

Concept: If $40\%$ is saved, expenditure is $100\%-40\%=60\%$ of income.

1. $60\%$ of income $=480$ (₹).
2. Income $=\frac{480\times 100}{60}$.
3. Income $=800$ (₹).

Trick: $60\%\to 480$, so $10\%\to 80$ and $100\%\to 800$.

Answer: (A) 800

Concept: $38=2\times 19$, so the number must be even and divisible by 19.

1. 1495 and 2423 are odd – rejected.
2. $2620÷19=137.89$… not exact.
3. $1938=38\times 51$ ✓ (since $38\times 50=1900$, $+38=1938$).

Trick: Check even first, then divide by 19.

Answer: (A) 1938

Concept: Work rate $=\frac{1}{\text{time}}$; rates of people working together add up.

1. C's rate $=\frac{1}{19}-\frac{1}{57}=\frac{3-1}{57}=\frac{2}{57}$.
2. B's rate $=\frac{1}{28}-\frac{2}{57}=\frac{57-56}{1596}=\frac{1}{1596}$.
3. So B alone takes $1596$ hours.

Trick: $28\times 57=1596$ – the LCM is directly the answer since the numerator becomes 1.

Answer: (C) 1596

Concept: For fixed distance, time $\propto \frac{1}{\text{speed}}$.

1. At $\frac{2}{5}$ of the speed, time becomes $\frac{5}{2}T$.
2. Delay $=\frac{5}{2}T-T=\frac{3}{2}T=48$ min.
3. $T=48\times \frac{2}{3}=32$ min.

Trick: Speed $\frac{a}{b}$ → usual time $=\frac{\text{delay}\times a}{b-a}=\frac{48\times 2}{3}=32$.

Answer: (B) 32 minutes

Concept: Median of $n$ (odd) observations $=$ the ${\left(\frac{n+1}{2}\right)}^{\text{th}}$ value after arranging in order.

1. Ascending order: 11, 32, 45, 53, 56, 79, 81, 87, 94, 98, 99.
2. $n=11$, so median $=$ the $\frac{11+1}{2}=6^{\text{th}}$ value.
3. $6^{\text{th}}$ value $=79$.

Trick: Count 5 values smaller and 5 larger – the one in the middle is the median.

Answer: (D) 79

Concept: Swap the signs first, then apply BODMAS.

1. After swapping: $51\times 12-162÷3+104$.
2. $=612-54+104$.
3. $=662$.

Trick: Rewrite the full expression with new signs before calculating.

Answer: (A) 662

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

1. Let the numbers be $3x$ and $16x$: $3x\times 16x=48x^2=432\Rightarrow x^2=9$, $x=3$.
2. Numbers are 9 and 48; HCF $=3$, LCM $=144$.
3. $\frac{1}{144}+\frac{1}{3}=\frac{1+48}{144}=\frac{49}{144}$.

Trick: With co-prime ratio $a:b$, HCF $=x$ and LCM $=abx$.

Answer: (C) $\frac{49}{144}$

Concept: A rhombus has four equal sides; its diagonals are generally of different lengths.

1. The diagonals bisect each other (true for every parallelogram).
2. They cut each other at right angles (${90}^{\circ }$).
3. They are equal only in the special case of a square, so in general they are not equal.

Trick: Equal diagonals → rectangle/square; perpendicular diagonals → rhombus.

Answer: (B) The diagonals of a rhombus are not equal, intersect at right angles and bisect each other.

Concept: $1{\text{m}}^3=1000$ litres and $V=l\times b\times h$.

1. Volume $=\frac{70000}{1000}=70{\text{m}}^3$.
2. $4\times b\times 10=70$.
3. $b=\frac{70}{40}=1.75\text{m}$.

Trick: Divide litres by 1000 to get cubic metres first.

Answer: (A) $1.75\text{m}$

Concept: $y\propto (x+3)\Rightarrow y=k(x+3)$.

1. $8=k(1+3)\Rightarrow k=2$.
2. At $x=2$: $y=2(2+3)$.
3. $y=10$.

Trick: Ratio method: $\frac{y_2}{8}=\frac{5}{4}\Rightarrow y_2=10$.

Answer: (A) 10

Concept: Selling price after successive gains $=\text{CP}\times (1+r_1)(1+r_2)(1+r_3)$.

1. $\text{CP}\times 1.05\times 1.10\times 1.15=11300$.
2. $1.05\times 1.10\times 1.15=1.32825$.
3. $\text{CP}=\frac{11300}{1.32825}\approx 8507.43$ (₹).

Trick: Rough check: total gain ≈ 33%, and $11300÷1.33\approx 8500$ – only option B is close.

Answer: (B) ₹8,507.43

Concept: For $a{x}^2+bx+c=0$, $D=b^2-4ac$: $D>0$ real distinct, $D=0$ real equal, $D<0$ complex.

1. $a=3$, $b=2$, $c=5$.
2. $D=2^2-4\times 3\times 5=4-60=-56$.
3. $D<0$ → roots are complex (imaginary).

Trick: If $a$ and $c$ have the same sign and $4ac>b^2$, the roots are imaginary. A quadratic never has more than 2 roots.

Answer: (A) Complex Roots (Imaginary Roots)

Concept: Successive change: multiply by $(1-\frac{12}{100})$ and then $(1+\frac{20}{100})$.

1. After $12\%$ decrease: $8000\times 0.88=7040$.
2. After $20\%$ increase: $7040\times 1.2=8448$.

Trick: Net change $=-12+20-\frac{12\times 20}{100}=+5.6\%$; $8000\times 1.056=8448$.

Answer: (D) 8,448

Concept: Apply BODMAS: bracket first, then division/multiplication, then addition.

1. Bracket: $74+12=86$.
2. $\frac{36}{9}=4$, so $4\times 86=344$.
3. $344+99=443$.

Trick: Last digit check: $4\times 6=24\to 4$, $4+9=13\to 3$ – only 443 ends in 3.

Answer: (D) 443

Concept: Combined average $=\frac{n_1{a}_1+n_2{a}_2}{n_1+n_2}$.

1. Total weight of A $=22\times 40=880\text{kg}$.
2. Total weight of B $=44\times 34=1496\text{kg}$.
3. Average $=\frac{880+1496}{66}=\frac{2376}{66}=36\text{kg}$.

Trick: Students are in ratio $1:2$, so the average divides the gap 6 in ratio $2:1$ from 34: $34+2=36$.

Answer: (B) 36

Concept: Combine ratios using the common term $b$.

1. $a:b=3:4$ and $b:c=4:7$; $b$ is already 4 in both, so $a:b:c=3:4:7$.
2. Let $a=3k$, $b=4k$, $c=7k$.
3. $\frac{a+b+c}{c}=\frac{14k}{7k}=2$.

Trick: Sum of ratio terms $÷$ c's term $=\frac{14}{7}=2$.

Answer: (D) 2

Concept: Convert men into women using $2M=9W$, then count work in woman-days.

1. Total work $=9\times 14=126$ woman-days.
2. 2 men for 9 days $=9\text{ women}\times 9=81$ woman-days.
3. Remaining $=126-81=45$ woman-days.
4. Women needed in 9 days $=\frac{45}{9}=5$.

Trick: Replace '2 men' by '9 women' everywhere – the answer is just $\frac{9\times (14-9)}{9}=5$.

Answer: (C) 5

Concept: $\text{SP}=\text{MP}\times (1-\text{discount})=\text{CP}\times (1+\text{profit})$.

1. Let CP $=100$. SP with $68\%$ profit $=168$.
2. $\text{MP}\times \frac{84}{100}=168\Rightarrow \text{MP}=200$.
3. Mark-up $X=\frac{200-100}{100}\times 100=100$.

Trick: $\frac{\text{MP}}{\text{CP}}=\frac{100+\text{P}\%}{100-\text{D}\%}=\frac{168}{84}=2$ → $100\%$ above.

Answer: (B) 100

Concept: For fixed distance, time ∝ $\frac{1}{\text{speed}}$.

1. Speed becomes $\frac{2}{5}$ ⇒ time becomes $\frac{5}{2}$ of usual time T.
2. Extra time $=\frac{5}{2}T-T=\frac{3}{2}T=18$ min.
3. $T=\frac{18\times 2}{3}=12$ min.

Trick: Usual time $=\frac{\text{late}}{\frac{5}{2}-1}=\frac{18}{1.5}=12$.

Answer: (A) 12 minutes