Q&A: practice questions with answers

Concept: Form two equations from the two statements and solve.

1. Now: $6P-Q=11$ … (i)
2. After 10 years: $6(P+10)-3(Q+10)=27\Rightarrow 6P-3Q=-3$ … (ii)
3. (i) − (ii): $2Q=14\Rightarrow Q=7$, and $P=\frac{11+7}{6}=3$.
4. Check: $6(13)-3(17)=78-51=27$ ✓

Trick: Subtract the equations directly – the $6P$ terms cancel.

Answer: (C) $7$

Concept: Volume of a sphere $V=\frac{4}{3}\pi r^3$, so $V\propto r^3$.

1. $\frac{V_1}{V_2}={\left(\frac{r_1}{r_2}\right)}^3={\left(\frac{5}{10}\right)}^3$
2. $={\left(\frac{1}{2}\right)}^3=\frac{1}{8}$

Trick: Radius doubles → volume becomes $2^3=8$ times.

Answer: (B) $1:8$

Concept: Take total work = LCM of the days, then find each person's one-day work.

1. Total work $=\text{LCM}(18,10,4)=180$ units.
2. A $=\frac{180}{18}=10$, B $=\frac{180}{10}=18$, A+B+C $=\frac{180}{4}=45$ units/day.
3. C $=45-10-18=17$ units/day.
4. C alone: $\frac{180}{17}=10\frac{10}{17}$ days.

Trick: C's rate $=\frac{1}{4}-\frac{1}{18}-\frac{1}{10}=\frac{17}{180}$.

Answer: (C) $10\frac{10}{17}$ days

Concept: For the same distance, time is inversely proportional to speed.

1. Distance $=66\times 9=594\text{km}$.
2. Gauri's time $=\frac{594}{99}=6$ hours.

Trick: $t_2=t_1\times \frac{66}{99}=9\times \frac{2}{3}=6$ hours.

Answer: (C) 6 hours

Concept: Product of consecutive numbers $n(n+1)$ – look for $n$ near $\sqrt{182}$.

1. $\sqrt{182}\approx 13.5$, so try $13$ and $14$.
2. $13\times 14=182$ ✓
3. Greater number $=14$.

Trick: Unit digit 2 → ends like $3\times 4$ → 13 × 14.

Answer: (B) $14$

Concept: Apply the divisibility tests one by one.

1. Last digit is 5 → divisible by $5$ ✓
2. Digit sum $6+3+0+6+0+5=20$ → not divisible by 3.
3. Odd number → not divisible by 14.
4. $7\times 90086=630602$, remainder $3$ → not divisible by 7.

Trick: Last digit 0 or 5 → divisible by 5 – check this first.

Answer: (A) $5$

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

1. $r=\frac{22}{2}=11\text{cm}$.
2. $S=4\pi \times {11}^2=4\pi \times 121=484\pi {\text{cm}}^2$.

Trick: With diameter $d$: $S=\pi d^2=\pi \times {22}^2=484\pi$.

Answer: (A) $484\pi {\text{cm}}^2$

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. From the right, odd places: $7+M+4=M+11$.
2. Even places: $3+6+N=N+9$.
3. Difference $=M-N+2$; it must be $0$ (since $\pm 11$ needs $M-N=9$ or $-13$, i.e. only $M=9,N=0$, which makes N the leading zero).
4. $M-N+2=0\Rightarrow M-N=-2$.

Trick: Alternate sum from the right: $7-3+M-6+4-N=M-N+2$.

Answer: (D) $M-N=-2$

Concept: Same direction → relative speed = difference; opposite direction → sum. Distance = sum of lengths.

1. Man's speed $=9\times \frac{5}{18}=2.5\text{m/s}$; relative speed $=\frac{490}{12}\text{m/s}$.
2. Train speed $=\frac{490}{12}+2.5=\frac{520}{12}=\frac{130}{3}\text{m/s}=156\text{km/h}$.
3. Opposite train: relative speed $=156+42=198\text{km/h}=55\text{m/s}$.
4. Time $=\frac{490+500}{55}=\frac{990}{55}=18\text{s}$.

Trick: $\frac{490}{12}\text{m/s}=147\text{km/h}$; add 9 → 156 km/h directly.

Answer: (C) $18$

Concept: Apply the discounts one after another on the reduced price.

1. After 16%: ₹$70000\times 0.84=58800$.
2. Extra discount $=58800-57036=$ ₹$1764$.
3. $\text{Discount}\%=\frac{1764}{58800}\times 100=3\%$.

Trick: $\frac{57036}{58800}=0.97$ → extra discount $3\%$.

Answer: (D) $3\%$

Concept: Divide the total by the sum of ratio terms to get one part.

1. Sum of ratio terms $=13+8+9=30$.
2. One part $=\frac{402}{30}=13.4$.
3. Largest − smallest $=(13-8)$ parts $=5\times 13.4=67$.

Trick: Difference $=\frac{402\times 5}{30}=67$ directly — no need to find each share.

Answer: (C) ₹67

Concept: Actual profit compares the money received with the cost of the goods actually given.

1. Let cost of the stated 1 kg be ₹100. At 65% loss, he receives ₹35.
2. Weight 70% less means he gives only 0.3 kg, costing ₹30.
3. Profit $=35-30=5$; profit$\%=\frac{5}{30}\times 100=16.67\%$.

Trick: $\text{Profit}\%=\frac{\text{SP fraction}-\text{weight fraction}}{\text{weight fraction}}\times 100=\frac{0.35-0.30}{0.30}\times 100$.

Answer: (B) 16.67%

Concept: Find each quantity separately and subtract.

1. $65\%\text{ of }65=\frac{65\times 65}{100}=42.25$.
2. $\frac{1}{5}\text{ of }35=7$.
3. Difference $=42.25-7=35.25$.

Trick: ${65}^2=4225$, so 65% of 65 is just $42.25$.

Answer: (A) 35.25

Concept: Mode = the value that occurs most often.

1. Count repeats: 81 appears 4 times.
2. 78 appears 3 times; 65, 70, 82, 86, 90 appear twice each.
3. So the mode is 81.

Trick: Only check the option values: 81 (4), 70 (2), 65 (2), 87 (1).

Answer: (A) 81

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. N23M40 from the right: odd places 0, M, 2 → $M+2$; even places 4, 3, N → $N+7$.
2. $(M+2)-(N+7)=M-N-5$ must be 0 or $\pm 11$.
3. Taking 0: $M-N=5$ (e.g. M = 6, N = 1 gives 123640 = 11 × 11240).

Trick: $M+N=-5$ is impossible for digits; check the difference form only.

Answer: (A) M − N = 5

Concept: Equal SI for equal time ⇒ principal × rate is equal for both parts.

1. Let ₹x be at 13%; then ₹(21200 − x) at 3%.
2. $x\times 13\times 8=(21200-x)\times 3\times 8$ ⇒ $13x=63600-3x$.
3. $16x=63600$ ⇒ $x=3975$.

Trick: Principals are in inverse ratio of rates $=3:13$; part at 13% $=\frac{3}{16}\times 21200=3975$.

Answer: (A) ₹3,975

Concept: No overall profit/loss ⇒ loss on one = gain on the other.

1. Let CP of first $=x$, second $=1200-x$.
2. $10\%\text{ of }x=20\%\text{ of }(1200-x)$ ⇒ $0.1x=240-0.2x$.
3. $0.3x=240$ ⇒ $x=800$.

Trick: CPs are in ratio $20:10=2:1$, so first $=\frac{2}{3}\times 1200=800$.

Answer: (C) ₹800

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

1. $10630\times 1.25=13287.5$.
2. $13287.5\times 1.5=19931.25$.
3. $19931.25\times 0.32=6378$.

Trick: $1.25\times 1.5\times 0.32=0.6$, so answer $=10630\times 0.6=6378$.

Answer: (D) 6378

Concept: Split the fraction: $\frac{\sin A+\cos A}{\cos A}=\tan A+1$.

1. $\tan A+1=\frac{17}{12}$ ⇒ $\tan A=\frac{5}{12}$.
2. Opposite = 5, adjacent = 12 ⇒ hypotenuse $=\sqrt{25+144}=13$.
3. $\sin A=\frac{5}{13}$, $\cos A=\frac{12}{13}$.

Trick: sin and cos can never exceed 1 — options (A) and (D) are out at once; 5-12-13 is a standard triplet.

Answer: (B) $\frac{5}{13},\frac{12}{13}$

Concept: Under CI, the multiplying factor grows as a power of time: factor $=(1+r{)}^t$.

1. $(1+r{)}^8=3$.
2. $9=3^2={((1+r{)}^8)}^2=(1+r{)}^{16}$.
3. So it takes 16 years.

Trick: 3 times in 8 yrs ⇒ $3^2$ in $8\times 2$, $3^3$ in $8\times 3$ years.

Answer: (D) 16 years