Q&A: practice questions with answers

Concept: Use $(a-b)(a+b)=a^2-b^2$ to pair the brackets in the denominator.

1. $(x-\frac{1}{3})(x+\frac{1}{3})=x^2-\frac{1}{9}=\frac{9x^2-1}{9}$.
2. $(1-\frac{x}{3})(1+\frac{x}{3})=1-\frac{x^2}{9}=\frac{9-x^2}{9}$.
3. Expression $=\frac{(x^2-9)(9x^2-1)\times 9\times 81}{(9x^2-1)(9-x^2)}$.
4. Since $\frac{x^2-9}{9-x^2}=-1$, value $=-729$.

Trick: Put $x=0$: $\frac{(-9)(-1)}{(-\frac{1}{3})(1)}\times \frac{9}{\frac{1}{3}\times 1}=(-27)\times 27=-729$.

Answer: (A) $-729$

Concept: For two numbers: Product of numbers $=\text{HCF}\times \text{LCM}$.

1. $168\times x=8\times 4200=33600$.
2. $x=\frac{33600}{168}=200$.

Trick: Check: HCF$(168,200)=8$ ✓.

Answer: (B) $200$

Concept: The car is treated as a point, so length of bridge = distance covered = speed × time (same units).

1. Time $=4.6\text{min}=\frac{4.6}{60}\text{h}$.
2. Distance $=48\times \frac{4.6}{60}=0.8\times 4.6=3.68\text{km}$.

Trick: $48\text{km/h}=0.8\text{km}$ per minute; $0.8\times 4.6=3.68$.

Answer: (D) $3.68\text{km}$

Concept: Let the second number be $y$; then the first is $5y$.

1. $5y\times y=2000\Rightarrow y^2=400\Rightarrow y=20$.
2. First number $=100$; sum $=100+20=120$.

Trick: Sum $=6y$, so it must be a multiple of $6$ – only $120$ fits.

Answer: (A) $120$

Concept: Find each pipe's full-tank time, then add filling rates and subtract the emptying rate.

1. P: $\frac{2}{7}$ in $18\text{h}$ → full in $63\text{h}$. Q: $\frac{5}{9}$ in $25\text{h}$ → full in $45\text{h}$.
2. In $3\text{h}$: $\frac{3}{63}+\frac{3}{45}=\frac{1}{21}+\frac{1}{15}=\frac{4}{35}$ of tank filled.
3. R empties $\frac{4}{35}$ in $8\text{h}$ → full tank in $70\text{h}$.
4. Together: $\frac{1}{63}+\frac{1}{45}-\frac{1}{70}=\frac{10+14-9}{630}=\frac{15}{630}=\frac{1}{42}$ → $42$ hours.

Trick: Use LCM $630$ as total work: P $=10$, Q $=14$, R $=-9$ units/h → net $15$; $\frac{630}{15}=42$.

Answer: (B) $42$ hours

Concept: $\text{SI}=\frac{P\times R\times T}{100}$; first find the principal from the given interest, then use it with the new rate and time.

1. $4\frac{2}{5}=\frac{22}{5}$ years: $8778=\frac{P\times 7.6\times 22}{100\times 5}=P\times 0.3344$
2. $P=\frac{8778}{0.3344}=26250$, i.e. ₹26,250
3. New SI $=\frac{26250\times 4.8}{100}\times \frac{58}{7}=1260\times \frac{58}{7}=10440$
4. Amount $=26250+10440=36690$

Trick: Amount always ends in the same digit as $P+\text{SI}$; here $250+440=690$ points straight to (C).

Answer: (C) ₹36,690

Concept: If $8A=6B=3C=k$, then $A:B:C=\frac{1}{8}:\frac{1}{6}:\frac{1}{3}$.

1. Multiply by LCM 24: $A:B:C=3:4:8$
2. Total parts $=3+4+8=15$
3. Share of A $=\frac{3}{15}\times 35200=7040$

Trick: For $aA=bB=cC$, the ratio is $\frac{1}{a}:\frac{1}{b}:\frac{1}{c}$ — reciprocals of the multipliers.

Answer: (A) ₹7,040

Concept: Profit = total selling price of the unbroken bulbs − total cost of all bulbs.

1. Cost price of 86 bulbs = ₹86
2. Bulbs left $=86-14=72$
3. SP $=72\times 3.20=230.40$
4. Profit $=230.40-86=144.40$

Trick: Broken items add nothing to SP but their cost stays in CP.

Answer: (A) ₹144.40

Concept: Using a weight 4% less means he gives 96 g for every 100 g charged. Overall factor $=\frac{\text{SP factor}}{\text{goods given}}$.

1. Price factor $=1.22$ (22% profit claimed)
2. Goods given for 1 unit charged $=0.96$
3. Overall factor $=\frac{1.22}{0.96}=1.2708$
4. Profit $\approx 27.08\%\approx 27\%$

Trick: $\text{Profit}\%=\frac{(100+p)\times 100}{100-e}-100=\frac{12200}{96}-100\approx 27\%$

Answer: (C) $27\%$

Concept: $\text{Average}=\frac{\text{Sum of observations}}{\text{Number of observations}}$

1. Sum $=40+49+56+75+38=258$ kg
2. Average $=\frac{258}{5}=51.6$ kg

Trick: Take 50 as base: deviations $-10,-1,+6,+25,-12$ sum to $+8$; $50+\frac{8}{5}=51.6$.

Answer: (B) 51.6 kg

Concept: A cube has 6 equal square faces, so total surface area $=6a^2$.

1. $a=9$ cm, one face $=9^2=81{\text{cm}}^2$
2. TSA $=6\times 81=486{\text{cm}}^2$

Trick: 729 is the volume $9^3$ and 324 is the lateral area $4a^2$ — do not confuse them.

Answer: (C) $486{\text{cm}}^2$

Concept: Margin = (winner% − loser%) of valid votes; valid votes are a percentage of polled votes, which are a percentage of registered voters.

1. Let registered voters $=N$; polled $=0.82N$
2. Valid $=0.96\times 0.82N=0.7872N$
3. Margin $=(60\%-40\%)$ of valid $=0.2\times 0.7872N=0.15744N$
4. $0.15744N=1968\Rightarrow N=12500$

Trick: Check options: only $12500\times 0.82\times 0.96\times 0.2$ gives a whole number, 1968.

Answer: (C) 12500

Concept: $\text{SI}=\frac{P\times R\times T}{100}$, where rate and time must be in the same unit (here, months).

1. $P=1200$, $R=4\%$ per month, $T=5$ months
2. SI $=\frac{1200\times 4\times 5}{100}=240$

Trick: 4% per month for 5 months = 20% in total; $20\%$ of 1200 = 240.

Answer: (D) ₹240

Concept: Successive discounts multiply: $\text{SP}=\text{MP}\times (1-\frac{d_1}{100})(1-\frac{d_2}{100})$.

1. After 5%: $6500\times 0.95=6175$
2. After 4%: $6175\times 0.96=5928$

Trick: Net discount $=5+4-\frac{5\times 4}{100}=8.8\%$; $6500\times 0.912=5928$.

Answer: (C) ₹5,928

Concept: Price paid = printed price − discount = printed price $\times \frac{100-d}{100}$.

1. Discount $=15\%$ of 2400 $=360$
2. Price paid $=2400-360=2040$

Trick: $2400\times 0.85=2040$; note the trap 2,004 (digits swapped).

Answer: (D) ₹2,040

Concept: Test each odd number for divisibility by primes up to $\sqrt{343}\approx 18.5$ (2, 3, 5, 7, 11, 13, 17).

1. 333 $=9\times 37$, 335 ends in 5, 339 $=3\times 113$, 341 $=11\times 31$, 343 $=7^3$
2. 337 is not divisible by 3, 7, 11, 13 or 17 → prime
3. Even numbers are not prime; so only one prime: 337

Trick: 341 looks prime but $11\times 31=341$ — a classic trap.

Answer: (B) 1

Concept: Evaluate powers and roots first (BODMAS), then add/subtract left to right.

1. $2^2=4$, $\sqrt{{10}^2}=10$, $9\sqrt{16}=9\times 4=36$
2. $4+10-36-17=-39$

Trick: $\sqrt{a^2}=a$ for positive $a$ — no need to square 10.

Answer: (A) $-39$

Concept: If a number increased by $r\%$ gives $y$, then the number $=\frac{y\times 100}{100+r}$.

1. $1.6x=3560$
2. $x=\frac{3560}{1.6}=2225$

Trick: Check: $2225+60\%\text{ of }2225=2225+1335=3560$.

Answer: (A) 2225

Concept: Take total work = LCM of the days; then work per day (efficiency) of C = combined − A − B.

1. Total work $=\text{LCM}(17,12,3)=204$ units
2. A $=\frac{204}{17}=12$, B $=\frac{204}{12}=17$, A+B+C $=\frac{204}{3}=68$ units/day
3. C $=68-12-17=39$ units/day
4. C alone $=\frac{204}{39}=\frac{68}{13}=5\frac{3}{13}$ days

Trick: $\frac{1}{C}=\frac{1}{3}-\frac{1}{17}-\frac{1}{12}=\frac{13}{68}$, answer is just the reciprocal.

Answer: (C) $5\frac{3}{13}$ days

Concept: Mean proportional of $a$ and $b$ is $\sqrt{ab}$.

1. $\sqrt{225\times 4}=\sqrt{900}=30$

Trick: $\sqrt{225}\times \sqrt{4}=15\times 2=30$.

Answer: (C) 30