Q&A: practice questions with answers

Concept: If $a+b+c=0$, then $a^3+b^3+c^3=3abc$.

1. Take $a=22,b=-16,c=-6$: $a+b+c=22-16-6=0$.
2. So the value $=3abc=3\times 22\times (-16)\times (-6)$.
3. $=3\times 22\times 96=6336$.
4. Check: $10648-4096-216=6336$.

Trick: Whenever the three bases add to zero, skip cubing and just multiply: $3abc$.

Answer: (C) $6336$

Concept: Area of a parallelogram $=\text{base}\times \text{height}$.

1. Area of one tile $=8\times 6=48{\text{cm}}^2$.
2. Number of tiles $=500$.
3. Floor area $=500\times 48=24000{\text{cm}}^2$.

Trick: Multiply $48\times 5=240$ and then attach the two zeros of $500$: $24000$.

Answer: (C) $24,000$

Concept: Net filling rate = sum of filling rates − emptying rate (part per hour).

1. Rates: $\frac{1}{4}$, $\frac{1}{17}$ (fill) and $\frac{1}{17}$ (empty).
2. Net rate $=\frac{1}{4}+\frac{1}{17}-\frac{1}{17}=\frac{1}{4}$ of the cistern per hour.
3. Time for $\frac{1}{4}$ cistern $=\frac{1/4}{1/4}=1$ hour.

Trick: The 17-hour filler and 17-hour emptier cancel each other – only the 4-hour tap matters.

Answer: (D) $1$

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

1. Write ${\left(\frac{5}{22}\right)}^8={\left(\frac{22}{5}\right)}^{-8}$.
2. LHS $={\left(\frac{22}{5}\right)}^{18-8+6}={\left(\frac{22}{5}\right)}^{16}={\left(\frac{5}{22}\right)}^{-16}$.
3. Compare powers: $9m+4=-16\Rightarrow 9m=-20$.
4. $m=-\frac{20}{9}$.

Trick: Convert everything to one base first, then just add/subtract the exponents.

Answer: (A) $-\frac{20}{9}$

Concept: Profit is divided in the ratio of investments (same time period).

1. Total ratio parts $=2+3+5=10$; one part $=\frac{10000}{10}=₹1000$.
2. R gets $5$ parts, P gets $2$ parts; difference $=3$ parts.
3. Difference $=3\times 1000=₹3000$.

Trick: Find only the difference of ratio terms $(5-2=3)$ and multiply by the value of one part.

Answer: (D) ₹3,000

Concept: $\tan \theta =\frac{\text{height}}{\text{base}}$; use the two right triangles sharing the tower.

1. From Q: $\tan {60}^{\circ }=\frac{h}{x}\Rightarrow h=\sqrt{3}x$.
2. From P: $\tan {30}^{\circ }=\frac{h}{x+100}\Rightarrow x+100=\sqrt{3}h=3x$.
3. So $2x=100\Rightarrow x=50\text{m}$.
4. $h=\sqrt{3}\times 50=50\sqrt{3}\text{m}$.

Trick: For ${30}^{\circ }$ and ${60}^{\circ }$, height $=\frac{\sqrt{3}}{2}\times$ the distance moved: $\frac{\sqrt{3}}{2}\times 100=50\sqrt{3}$.

Answer: (D) $50\sqrt{3}$ metres

Concept: The change from loss % to profit % is caused by the extra selling amount.

1. Loss $10\%$ → profit $5\%$: total change $=10+5=15\%$ of CP.
2. $15\%$ of CP $=₹450$.
3. CP $=\frac{450\times 100}{15}=₹3000$.

Trick: Loss to profit → add the percentages; profit to profit → subtract them.

Answer: (A) ₹3,000

Concept: Surface area of a sphere $=4\pi r^2$.

1. Total area $=4\pi \times {10}^2=400\pi {\text{cm}}^2$.
2. Painted part $=100\%-25\%=75\%$.
3. Painted area $=\frac{3}{4}\times 400\pi =300\pi {\text{cm}}^2$.

Trick: $75\%$ of $4\pi r^2$ is simply $3\pi r^2$.

Answer: (B) $300\pi {\text{cm}}^2$

Concept: $\text{Discount}\%=\frac{\text{MP}-\text{SP}}{\text{MP}}\times 100$.

1. Let CP $=100$. MP $=100+165=265$.
2. For $112\%$ gain, SP $=212$.
3. Discount $=265-212=53$.
4. Discount $\%=\frac{53}{265}\times 100=20\%$.

Trick: Take CP $=100$; MP and SP become the percentages directly.

Answer: (A) $20\%$

Concept: Make the common term (B) equal to combine two ratios.

1. $A:B=15:17$, $B:C=14:19$. Multiply first by $14$, second by $17$.
2. $A:B:C=210:238:323$; total $=771$ parts.
3. One part $=\frac{1542}{771}=2$.
4. A $=210\times 2=420$.

Trick: Since $1542=2\times 771$, every share is just double its ratio term.

Answer: (B) $420$

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$ (never the average of the speeds).

1. Times: $\frac{120}{40}=3$ h, $\frac{120}{60}=2$ h, $\frac{100}{80}=1.25$ h.
2. Total time $=6.25$ h; total distance $=340$ km.
3. Average speed $=\frac{340}{6.25}=54.4\text{km/h}$.

Trick: Simple average of speeds $(60)$ is a trap – always divide total distance by total time.

Answer: (A) $54.4\text{km/h}$

Concept: Numbers $=8a$ and $8b$ with $a,b$ co-prime; $\text{HCF}\times \text{LCM}=$ product.

1. $8a+8b=88\Rightarrow a+b=11$.
2. LCM $=8ab=240\Rightarrow ab=30$.
3. $a+b=11,ab=30\Rightarrow a=5,b=6$.
4. Numbers $=40$ and $48$.

Trick: Check options: product must be $8\times 240=1920$; only $40\times 48=1920$ with sum $88$.

Answer: (C) 40 and 48

Concept: The centroid divides each median in the ratio $2:1$ from the vertex; median of equilateral triangle $=\frac{\sqrt{3}}{2}a$.

1. AD $=\frac{\sqrt{3}}{2}\times 24=12\sqrt{3}$ cm.
2. GD $=\frac{1}{3}\times \text{AD}$.
3. GD $=\frac{12\sqrt{3}}{3}=4\sqrt{3}$ cm.

Trick: In an equilateral triangle, GD equals the inradius $=\frac{a}{2\sqrt{3}}=\frac{24}{2\sqrt{3}}=4\sqrt{3}$.

Answer: (A) $4\sqrt{3}$

Concept: At $10\%$ for 2 years, CI $=21\%$ of the principal (${1.1}^2=1.21$).

1. $21\%$ of $y=630$.
2. $y=\frac{630\times 100}{21}=3000$.
3. Amount $=3000+630=₹3630$.

Trick: Effective rate for 2 years at $r\%$: $2r+\frac{r^2}{100}=20+1=21\%$.

Answer: (D) ₹3,630

Concept: A number is divisible by 11 if (sum of digits at odd places) − (sum at even places) is $0$ or a multiple of 11.

1. Digits from the right: $5,2,p,3,4,2,9,7$.
2. Odd places: $5+p+4+9=18+p$; even places: $2+3+2+7=14$.
3. Difference $=4+p$ must be $0$ or $11$ → $p=7$.

Trick: Pair the alternate digits quickly; since $p$ is a single digit, only $4+p=11$ is possible.

Answer: (A) $7$

Concept: Net change for successive increases $a\%$ and $b\%$ $=a+b+\frac{ab}{100}$.

1. $a=18,b=22$.
2. Net $=18+22+\frac{18\times 22}{100}=40+3.96$.
3. Total increase $=43.96\%$.

Trick: Multiplying factors works too: $1.18\times 1.22=1.4396$.

Answer: (B) $43.96\%$

Concept: Change in mean $=\frac{\text{change in total}}{\text{number of items}}$.

1. Total increases by $20-10=10$.
2. Number of values $=5$.
3. Mean increases by $\frac{10}{5}=2$ (from $20$ to $22$).

Trick: Only the change matters – you never need the other four numbers.

Answer: (D) It increases by 2

Concept: Successive discounts multiply: $\text{SP}=\text{MP}\times (1-d_1)(1-d_2)$.

1. 25 items $>20$, so both $10\%$ and extra $5\%$ apply.
2. Price per item $=3000\times 0.9\times 0.95=2565$.
3. For 25 items: $2565\times 25=₹64125$.

Trick: $10\%$ then $5\%$ = single discount $10+5-\frac{50}{100}=14.5\%$; $75000\times 0.855=64125$.

Answer: (C) ₹64,125

Concept: Profit share $\propto$ capital $\times$ time.

1. Riya $=13500\times 8=108000$; Kabir $=21000\times 6=126000$; Sana $=18000\times 12=216000$.
2. Ratio $=108:126:216=6:7:12$ (divide by $18$); total $=25$.
3. Kabir $=\frac{7}{25}\times 12600=₹3528$.

Trick: Reduce the products to the smallest ratio first – here $6:7:12$ – before dividing the profit.

Answer: (A) ₹3,528

Concept: $\text{Weighted mean}=\frac{\sum w_i{x}_i}{\sum w_i}$.

1. $\frac{3\times 60+2\times 70+5M}{10}=68$.
2. $180+140+5M=680\Rightarrow 5M=360$.
3. $M=72$.

Trick: Deviation method: Physics $-8\times 3=-24$, Chemistry $+2\times 2=+4$; Maths must give $+20$, i.e. $+4$ each → $72$.

Answer: (C) $72$