Q&A: practice questions with answers

Concept: SI = $\frac{P\times R\times T}{100}$, so $P=\frac{SI\times 100}{R\times T}$.

1. SI = ₹700, R = 8%, T = 2 years
2. $P=\frac{700\times 100}{8\times 2}=\frac{70000}{16}$
3. $P=$ ₹4375

Trick: In 2 years at 8%, total interest = 16% of P. So $16\%=700\to 1\%=43.75\to 100\%=$ ₹4375.

Answer: (A) 4375

Concept: % increase = (increase/original value) × 100.

1. Girls = 24, boys = 36
2. Increase needed = $36-24=12$
3. % increase = $\frac{12}{24}\times 100=50\%$

Trick: 12 is exactly half of 24, so the increase is 50%.

Answer: (D) 50%

Concept: Form an equation from the statement and solve it.

1. Let the number = n. Then $\frac{n}{3}+10=65$
2. $\frac{n}{3}=55\to n=165$
3. Sum of digits = $1+6+5=12$

Trick: Work backwards: $65-10=55$, then $55\times 3=165$.

Answer: (D) 12

Concept: The difference between two people's ages never changes with time.

1. Difference after 10 years = 20 → present difference $F-S=20$
2. $F+S=50$
3. Adding: $2F=70\to F=35$, $S=15$
4. S : F = 15 : 35 = 3 : 7

Trick: Sum 50 and difference 20 → bigger = $\frac{50+20}{2}=35$, smaller = $\frac{50-20}{2}=15$.

Answer: (A) 3:7

Concept: A number divisible by both 13 and 17 (co-primes) is divisible by $13\times 17=221$.

1. Test each option with 221
2. $21658÷221=98$ exactly ($221\times 98=21658$)
3. Others leave remainders, e.g. $22372=221\times 101+51$

Trick: $221\times 100=22100$. Subtract 221 twice: $22100-442=21658$ — matches option C.

Answer: (C) 21658

Concept: Curved surface area of a cylinder = $2\pi rh$.

1. $2\times \left(\frac{22}{7}\right)\times r\times 3=7900$
2. $\frac{132}{7}r=7900$
3. $r=7900\times \frac{7}{132}=\frac{55300}{132}\approx 418.94$ cm

Trick: $132\times 419=55308$, slightly more than 55300, so r is just below 419 → 418.94.

Answer: (C) 418.94

Concept: Work per hour = $\frac{1}{\text{time}}$; subtract combined rates to get a single person's rate.

1. $A=\frac{1}{63}$, $A+C=\frac{1}{36}$ → $C=\frac{1}{36}-\frac{1}{63}=\frac{7-4}{252}=\frac{1}{84}$
2. $B+C=\frac{1}{48}$ → $B=\frac{1}{48}-\frac{1}{84}=\frac{7-4}{336}=\frac{1}{112}$
3. B alone takes 112 hours

Trick: Use LCM total work = 1008 units: $A=16$, $A+C=28$ → $C=12$, $B+C=21$ → $B=9$; $\frac{1008}{9}=112$.

Answer: (A) 112

Concept: Find the rate per minute, then multiply by total minutes.

1. Rate = $\frac{40}{5}=8$ pages per minute
2. 3 hours = 180 minutes
3. Pages = $8\times 180=1440$

Trick: In 1 hour = 12 lots of 5 min → $40\times 12=480$ pages; $\times 3=1440$.

Answer: (B) 1440

Concept: Use $(x+y)(x-y)=x^2-y^2$ with ${\sec }^{2}A-{\tan }^{2}A=1$.

1. Expression = $[\sec A+(\tan A-1)][\sec A-(\tan A-1)]$
2. $={\sec }^{2}A-(\tan A-1{)}^2={\sec }^{2}A-{\tan }^{2}A+2\tan A-1$
3. $=1+2\tan A-1=2\tan A$
4. $2\tan A=3$ (given)

Trick: This expression always simplifies to $2\tan A$ — so just read the given value: 3.

Answer: (A) $3$

Concept: $\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}$ and $a^2+b^2=(a+b{)}^2-2ab$.

1. $a^2+b^2={12}^2-2\times 8=144-16=128$
2. $\frac{a}{b}+\frac{b}{a}=\frac{128}{8}=16$

Trick: Direct formula: $\frac{a}{b}+\frac{b}{a}=\frac{(a+b{)}^2}{ab}-2=\frac{144}{8}-2=18-2=16$.

Answer: (C) 16

Concept: Centroid divides each median in 2 : 1; median of equilateral triangle = $\frac{\sqrt{3}}{2}\times$ side.

1. Median = $\frac{\sqrt{3}}{2}\times 36=18\sqrt{3}$ cm
2. $AG=\frac{2}{3}\times 18\sqrt{3}=12\sqrt{3}$ cm

Trick: In an equilateral triangle, AG = circumradius = $\frac{\text{side}}{\sqrt{3}}$ = $\frac{36}{\sqrt{3}}=12\sqrt{3}$.

Answer: (D) $12\sqrt{3}$

Concept: Solve innermost bracket first; inside a bracket, ÷ before −.

1. $18÷6=3$ → $12-3=9$ → $9÷3=3$
2. $90÷5=18$ → $18-3=15$
3. $27-15=12$
4. $38-12=26$

Trick: Remember the order: innermost ( ) → { } → [ ], and division before subtraction in each.

Answer: (A) 26

Concept: $\text{SP}=\text{MP}\times (1-d_1)(1-d_2)$; profit % = $\frac{\text{profit}}{\text{CP}}\times 100$.

1. SP of X = $3400+323=$ ₹$3723=\text{MP}\times 0.9\times 0.5=0.45\text{MP}$
2. SP of Y = $\text{MP}\times 0.72=3723\times \frac{0.72}{0.45}=3723\times 1.6=$ ₹5956.80
3. Profit on Y = $5956.80-3400=2556.80$
4. Profit % = $\frac{2556.80}{3400}\times 100=75.2\%$

Trick: No need to find MP: $\frac{\text{SP}(Y)}{\text{SP}(X)}=\frac{0.72}{0.45}=1.6$, so $\text{SP}(Y)=1.6\times 3723$.

Answer: (B) 75.2%

Concept: When B catches A, both have covered the same distance from the start.

1. B's distance = $8\times 2=16$ km
2. A walked (t + 2) hours: $5(t+2)=16$
3. $t+2=3.2$ → $t=1.2$ hours

Trick: Relative speed 3 km/h for 2 h closes 6 km = A's head start $5t$ → $t=\frac{6}{5}=1.2$.

Answer: (B) 1.2

Concept: If $a^m=a^n$ (same base), then $m=n$. (Question reads: ${74}^{0.45}=x$, ${74}^{0.12}=y$ and $x^z=y^2$.)

1. $x^z=({74}^{0.45}{)}^z={74}^{0.45z}$
2. $y^2=({74}^{0.12}{)}^2={74}^{0.24}$
3. $0.45z=0.24$ → $z=\frac{24}{45}=\frac{8}{15}\approx 0.53$

Trick: Equate powers directly: $0.45z=2\times 0.12$ → $z=\frac{0.24}{0.45}\approx 0.53$.

Answer: (B) 0.53

Concept: Work done = rate × time; remaining work ÷ rate of the person left.

1. A + B per day = $\frac{1}{10}+\frac{1}{15}=\frac{1}{6}$
2. In 3 days = $\frac{3}{6}=\frac{1}{2}$ work done
3. Remaining $\frac{1}{2}$ by A: $\frac{1}{2}÷\frac{1}{10}=5$ days

Trick: Total work = 30 units: A = 3, B = 2 per day. 3 days → 15 units; left $15÷3=5$ days.

Answer: (B) 5 days

Concept: Profit is shared in the ratio of (capital × time).

1. A : B : C = $2\times 6:3\times 4:4\times 3=12:12:12=1:1:1$
2. C's share = $\frac{1,32,000}{3}=$ ₹44,000

Trick: All products are 12 — equal shares, so just divide profit by 3.

Answer: (D) ₹44,000

Concept: New member = old member + (number × change in average).

1. Increase in total = $6\times 4$ months = 24 months = 2 years
2. New girl = $12+2=14$ years

Trick: Replacement shortcut: new = old + n × increase → $12+6\times \frac{1}{3}=14$.

Answer: (C) 14 years

Concept: Overall profit % = sum of (share of goods × profit/loss %).

1. 44% at +47%: $0.44\times 47=+20.68$
2. Remaining 56%: 25% of it = 14% at +22%: $0.14\times 22=+3.08$
3. Rest 42% at −38%: $0.42\times 38=-15.96$
4. $x=20.68+3.08-15.96=7.8$

Trick: Take total goods worth ₹100: profits 20.68 + 3.08, loss 15.96 → net ₹7.8 on ₹100.

Answer: (C) 7.8

Concept: Area of trapezium = $\frac{1}{2}\times$ (sum of parallel sides) × height.

1. Sum of parallel sides = $17+29=46$ cm
2. Area = $\frac{1}{2}\times 46\times 15=23\times 15=345{\text{cm}}^2$

Trick: Average of parallel sides (23) × height (15) = 345.

Answer: (D) 345 cm2