Q&A: practice questions with answers

Concept: Evaluate percentages, roots and products first, then subtract (BODMAS).

1. $34\%$ of $50=\frac{34}{100}\times 50=17$.
2. $\sqrt[3]{512}=8$, $\sqrt{36}=6$, so $\frac{4}{6}\times 8\times 6=32$.
3. $y=17-32=-15$.

Trick: The 6 cancels with the denominator 6, leaving $4\times 8=32$ instantly.

Answer: (B) $-15$

Concept: Rates add: if P, Q, R take $p,q,r$ hours, then $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}$ when P + Q together equal R.

1. Let P take $t$ h. Then Q takes $t-15$ h and R takes $t-15-12=t-27$ h.
2. $\frac{1}{t}+\frac{1}{t-15}=\frac{1}{t-27}$.
3. Try $t=45$: $\frac{1}{45}+\frac{1}{30}=\frac{2+3}{90}=\frac{1}{18}$, and $45-27=18$ ✓.

Trick: Put the options directly: only $t=45$ makes $t-15$ and $t-27$ both divide a common LCM neatly.

Answer: (B) 45 hours

Concept: Fourth proportional of $a,b,c$ is $d=\frac{b\times c}{a}$.

1. For 10, 80, 30: $d_1=\frac{80\times 30}{10}=240$.
2. For 7, 28, 19: $d_2=\frac{28\times 19}{7}=76$.
3. Ratio $=240:76=60:19$.

Trick: Divide both by 4: $240÷4=60$, $76÷4=19$.

Answer: (B) 60:19

Concept: For $a{x}^2+bx+c$ with $a>0$, minimum value $=c-\frac{b^2}{4a}$ (at $x=-\frac{b}{2a}$).

1. $a=7,b=-7,c=22$.
2. $\frac{b^2}{4a}=\frac{49}{28}=\frac{7}{4}$.
3. Minimum $=22-\frac{7}{4}=\frac{88-7}{4}=\frac{81}{4}$.

Trick: Minimum $=\frac{4ac-b^2}{4a}=\frac{616-49}{28}=\frac{567}{28}=\frac{81}{4}$.

Answer: (B) $\frac{81}{4}$

Concept: Successive increases of $a\%$ and $b\%$ multiply: new $=P(1+\frac{a}{100})(1+\frac{b}{100})$.

1. After year 1: $290000\times \frac{115}{100}=333500$.
2. After year 2: $333500\times \frac{125}{100}=416875$.

Trick: Net change $=15+25+\frac{15\times 25}{100}=43.75\%$; $290000\times 1.4375=416875$.

Answer: (C) 4,16,875

Concept: Overall average $=\frac{\sum (\text{count}\times \text{average})}{\text{total count}}$.

1. Total spent $=75\times 493+47\times 397+15\times 465=36975+18659+6975=62609$.
2. Total members $=75+47+15=137$.
3. Average $=\frac{62609}{137}=457$.

Trick: Take 457 as base: deviations $75\times 36-47\times 60+15\times 8=2700-2820+120=0$, so the average is exactly 457.

Answer: (A) 457

Concept: First find the maximum marks from the given percentage, then compute the friend's percentage.

1. $33\%$ of maximum $=416+79=495$.
2. Maximum marks $=\frac{495\times 100}{33}=1500$.
3. Friend's percentage $=\frac{555}{1500}\times 100=37\%$.

Trick: $1\%$ of 1500 = 15; $555÷15=37$.

Answer: (B) 37%

Concept: $x^3+\frac{1}{x^3}={(x+\frac{1}{x})}^3-3(x+\frac{1}{x})$.

1. Put $x+\frac{1}{x}=-3$.
2. $(-3{)}^3-3(-3)=-27+9=-18$.

Trick: For $x+\frac{1}{x}=a$: $x^3+\frac{1}{x^3}=a^3-3a$ (and $x^2+\frac{1}{x^2}=a^2-2$).

Answer: (B) $-18$

Concept: Events repeating every $a$ and $b$ minutes coincide again after $\text{LCM}(a,b)$ minutes.

1. $12=2^2\times 3$, $18=2\times 3^2$.
2. $\text{LCM}=2^2\times 3^2=36$.
3. After 36 minutes A finishes 3 cycles and B finishes 2 cycles together.

Trick: $\text{LCM}=\frac{12\times 18}{\text{HCF}}=\frac{216}{6}=36$.

Answer: (D) 36

Concept: Area $=\pi r^2$, perimeter (circumference) $=2\pi r$.

1. Given $\pi r^2=2\times 2\pi r$.
2. Divide by $\pi r$: $r=4$ units.
3. Diameter $=2r=8$ units.

Trick: Area $=k\times$ circumference gives $r=2k$; here $k=2$, so $r=4$, $d=8$.

Answer: (B) 8 units

Concept: Equal amounts after different periods: $I(1+r{)}^{11}=K(1+r{)}^{12}$.

1. Cancel $(1.1{)}^{11}$: $I=K\times 1.1$, so $I:K=11:10$.
2. Total 21 parts $=1407$, one part $=67$.
3. Krish's share $=10\times 67=670$.

Trick: The one who invests for an extra year gets the smaller share, in ratio $100:(100+r)$ – here $10:11$.

Answer: (A) 670

Concept: Volume of the solid = volume of cube + volume of cuboid on top.

1. Cube: $x^3$. Cuboid: base $x\times x$, height $(x-4)$ → $x^2(x-4)$.
2. $x^3+x^2(x-4)=1.5x^3$.
3. $x^2(x-4)=0.5x^3\Rightarrow x-4=0.5x\Rightarrow x=8$.

Trick: The cuboid must be half the cube, so its height is half the edge: $x-4=\frac{x}{2}$.

Answer: (B) 8

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$ – never average the speeds directly.

1. Times: $\frac{810}{30}=27$ h, $\frac{492}{12}=41$ h, $\frac{988}{38}=26$ h.
2. Total time $=27+41+26=94$ h; total distance $=810+492+988=2290$ km.
3. Average speed $=\frac{2290}{94}=24\frac{34}{94}=24\frac{17}{47}$ km/h.

Trick: Divide each distance by its speed first – here all three come out as whole hours, so the only work is one final division.

Answer: (B) $24\frac{17}{47}$

Concept: Total surface area of a solid hemisphere $=2\pi r^2+\pi r^2=3\pi r^2$.

1. $3\times \frac{22}{7}\times r^2=272$
2. $r^2=\frac{272\times 7}{66}=\frac{1904}{66}=\frac{952}{33}$
3. $r=\sqrt{\frac{952}{33}}$ cm.

Trick: Solid hemisphere → $3\pi r^2$ (curved + flat base); hollow bowl → only $2\pi r^2$. Don't drop the base.

Answer: (C) $\sqrt{\frac{952}{33}}$

Concept: Two successive discounts of $a\%$ and $b\%$ leave $\frac{(100-a)(100-b)}{100}\%$ of the marked price.

1. SP of X $=500+636=1136$. After 20% and 45%: $\text{SP}=\text{MP}\times 0.8\times 0.55=0.44\text{MP}$.
2. So $0.44\text{MP}=1136$.
3. Y at 67% discount: ${\text{SP}}_Y=0.33\text{MP}=1136\times \frac{0.33}{0.44}=1136\times \frac{3}{4}=852$.
4. Profit $=852-500=352$; profit% $=\frac{352}{500}\times 100=70.4\%$.

Trick: You never need MP itself – $\frac{0.33}{0.44}=\frac{3}{4}$, so SP of Y is just $\frac{3}{4}$ of SP of X.

Answer: (D) $70.4\%$

Concept: Compare cost and selling price of one unit; $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$.

1. CP of one article $=\frac{480}{12}=$ ₹40.
2. SP of one article = ₹78, profit $=78-40=$ ₹38.
3. Profit% $=\frac{38}{40}\times 100=95\%$.

Trick: "47 dozens" is a distraction – percentage profit is the same for one article as for all of them.

Answer: (D) $95\%$

Concept: For a fixed distance, time is inversely proportional to speed.

1. Speed becomes $\frac{3}{7}$ of normal → time becomes $\frac{7}{3}$ of normal time $T$.
2. Extra time $=\frac{7}{3}T-T=\frac{4}{3}T=20$ minutes.
3. $T=20\times \frac{3}{4}=15$ minutes.

Trick: Speed ratio $3:7$ → time ratio $7:3$; the difference of 4 parts = 20 min, so 1 part = 5 min and usual time = 3 parts = 15 min.

Answer: (C) 15 minutes

Concept: $\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$ for grouped data.

1. Modal class = 30–40 (highest frequency $f_1=25$); $l=30$, $h=10$, $f_0=18$, $f_2=10$.
2. $\text{Mode}=30+\frac{25-18}{50-18-10}\times 10=30+\frac{70}{22}$
3. $=30+3.18=33.18$.

Trick: The mode always lies inside the modal class – only option (C) and (D) are between 30 and 40 and close; compute $\frac{70}{22}\approx 3.18$ to settle it.

Answer: (C) $33.18$

Concept: ${\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n$ and $a^m\times a^n=a^{m+n}$; write every term with the same base.

1. ${\left(\frac{5}{16}\right)}^{16}={\left(\frac{16}{5}\right)}^{-16}$, so LHS $={\left(\frac{16}{5}\right)}^{13-16+5}={\left(\frac{16}{5}\right)}^2$.
2. ${\left(\frac{16}{5}\right)}^2={\left(\frac{5}{16}\right)}^{-2}$.
3. Equate powers: $6m+16=-2\Rightarrow 6m=-18\Rightarrow m=-3$.

Trick: Flipping the fraction flips the sign of the exponent – convert everything to base $\frac{5}{16}$: $-13+16-5=-2$.

Answer: (B) $-3$

Concept: Express both selling prices as fractions of the cost price.

1. Let actual SP $=S$ and CP $=C$. $\frac{2}{5}S=0.76C\Rightarrow S=\frac{0.76\times 5}{2}C=1.9C$.
2. New SP $=0.84S=0.84\times 1.9C=1.596C$.
3. Profit% $=(1.596-1)\times 100=59.6\%$.

Trick: Take CP = 100: $\frac{2}{5}S=76\Rightarrow S=190$; $84\%$ of 190 = 159.6 → profit 59.6%.

Answer: (A) $59.6\%$