Q&A: practice questions with answers

Concept: After d% discount, SP = $(100-d)\%$ of MP.

1. SP = 74% of MP → MP = $2886\times \frac{100}{74}$ = ₹3,900.
2. Discount = $3900-2886$ = ₹1,014.
3. Check: $26\%\text{ of }3900=1014$ ✓

Trick: Discount : SP = $26:74=13:37$ → $2886÷37=78$; $78\times 13$ = ₹1,014.

Answer: (C) ₹1,014

Concept: Total = average × number; the common person is counted twice in the two pair-totals.

1. $A+K+M=46\times 3=138$.
2. $A+K=80$, $K+M=86$ → sum = 166.
3. $K=166-138=28$ kg.

Trick: Common person = (sum of pair totals) − (total of all three).

Answer: (B) 28

Concept: Mixing two groups: the ratio of their numbers comes from alligation.

1. Men 41, Women 25, overall 32.
2. Men : Women = $(32-25):(41-32)=7:9$.
3. Women = $96\times \frac{9}{16}=54$.

Trick: Alligation avoids equations: differences from the mean, crossed, give the ratio directly.

Answer: (A) 54

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

1. $z=\sqrt{xy}\to xy=z^2$.
2. Required = $\sqrt{x^{2}z\times y^{2}z}=\sqrt{x^2{y}^2{z}^2}=xyz$.
3. $xyz=z^2\times z=z^3$.

Trick: Put numbers: $x=1$, $y=4\to z=2$; $\sqrt{1\times 2\times 16\times 2}=\sqrt{64}=8=2^3=z^3$.

Answer: (C) $z^3$

Concept: Solve innermost brackets first, then ÷/×, then +/−.

1. $(4-2)=2$; $21÷7=3$.
2. $\frac{63}{9}=7$; $\frac{14}{2}=7$ → $7+7\times 2=7+14=21$.
3. $3\times 21=63$.

Trick: Common slip: doing $(7+7)\times 2=28$ gives 84 – multiply before adding.

Answer: (B) 63

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

1. $9\times 49=441$.
2. $\sqrt{441}=21$.

Trick: Both are perfect squares: $\sqrt{9}\times \sqrt{49}=3\times 7=21$ – no multiplication needed.

Answer: (D) 21

Concept: Expenditure = Price × Consumption.

1. New price factor = 1.60; new expenditure factor = 1.44.
2. Consumption factor = $1.44÷1.60=0.90$.
3. Reduction = 10%.

Trick: Use multipliers: $\frac{144}{160}=\frac{9}{10}$ → consumption falls by $\frac{1}{10}=10\%$.

Answer: (C) 10%

Concept: Opposite directions: relative speed = sum of speeds; distance = sum of lengths.

1. Distance = $250+250=500$ m; time 25 s.
2. Relative speed = $\frac{500}{25}=20$ m/s = $20\times \frac{18}{5}=72$ km/h.
3. Other train = $72-45=27$ km/h.

Trick: Convert m/s to km/h by $\times \frac{18}{5}$; km/h to m/s by $\times \frac{5}{18}$.

Answer: (C) 27 km/hr

Concept: Total SP = Total CP + Profit; then find the rate per unit.

1. CP of 400 pears = $4\times 1200$ = ₹4,800.
2. SP = $4800+1000$ = ₹5,800.
3. Per pear = $\frac{5800}{400}$ = ₹14.50; per dozen = $14.5\times 12$ = ₹174.

Trick: 400 pears = $\frac{100}{3}$ dozen → $5800÷(\frac{100}{3})=58\times 3$ = ₹174.

Answer: (A) 174

Concept: For 2 years, $\text{CI}-\text{SI}$ = interest on one year's SI.

1. One-year SI = $\frac{40}{2}$ = ₹20.
2. $\text{CI}-\text{SI}=4$ → rate = $\frac{4}{20}\times 100=20\%$.
3. $P=20\times \frac{100}{20}$ = ₹100.

Trick: Formula: $P=(\text{one-year SI}{)}^2÷(\text{CI}-\text{SI})=\frac{{20}^2}{4}$ = ₹100.

Answer: (A) 100

Concept: The largest measure that divides both quantities exactly is their HCF.

1. $765=3^2\times 5\times 17$.
2. $833=7^2\times 17$.
3. HCF = 17 litres.

Trick: Check options directly: $765÷17=45$ and $833÷17=49$ – both exact.

Answer: (D) 17 litres

Concept: To cross a tunnel, distance = train length + tunnel length.

1. Length to cover: $496+664=1160$ m.
2. 18 km/h → $18\times \frac{5}{18}$ → 5 m/s.
3. 1160 m at 5 m/s → 232 s.

Trick: 18 km/h = 5 m/s exactly – a handy pair to remember.

Answer: (A) 232

Concept: Take total work = LCM of days; add work cycle by cycle.

1. A = 12 days, B = 6, C = 3 → total = 12 units; A = 1, B = 2, C = 4 units/day.
2. 3-day cycle (B, A, C) = 7 units.
3. Day 4 (B): 9; day 5 (A): 10; remaining 2 units by C at 4/day = 0.5 day.
4. Total = 5.5 days.

Trick: After one cycle, keep a running total day by day – never multiply cycles blindly when the cycle doesn't divide the work.

Answer: (A) 5.5

Concept: If amounts are equal, the shares are in the inverse ratio of their growth factors.

1. Mahesh's money grows 7 years, David's 6 years.
2. $M\times (1.1{)}^7=D\times (1.1{)}^6\to D=1.1M\to M:D=10:11$.
3. $D=25200\times \frac{11}{21}$ = ₹13,200.

Trick: Only the extra year matters: one year's growth = $\frac{11}{10}$, so the younger-wait share is 11 parts out of 21.

Answer: (D) 13200

Concept: LCM = product of highest powers of all primes present.

1. $36=2^2\times 3^2$; $63=3^2\times 7$; $372=2^2\times 3\times 31$; $126=2\times 3^2\times 7$.
2. LCM = $2^2\times 3^2\times 7\times 31=252\times 31=7812$.

Trick: LCM must be divisible by 36 → must be even and digit-sum a multiple of 9: only 7812 ($7+8+1+2=18$) works.

Answer: (D) 7812

Concept: Do ÷ and × first, carefully applying sign rules.

1. (−60) ÷ (−15) = +4.
2. (−2) × 6 = −12.
3. −9 − 4 − 12 = −25.

Trick: Same signs in ÷/× give +, different signs give −; then '− (+4)' becomes −4.

Answer: (A) -25

Concept: SP at loss = $(100-L)\%$ of CP; SP at profit = $(100+P)\%$ of CP.

1. CP = $20240\times \frac{100}{92}$ = ₹22,000.
2. New SP = $22000\times \frac{112}{100}$ = ₹24,640.

Trick: Skip CP: New SP = $20240\times \frac{112}{92}=220\times 112$ = ₹24,640.

Answer: (C) 24,640

Concept: CSA of cylinder = $2\pi rh$.

1. $2\times \frac{22}{7}\times 7\times 9.5$.
2. = $44\times 9.5=418$ cm².

Trick: When $r=7$ and $\pi =\frac{22}{7}$, $2\pi r=44$ – just multiply by h.

Answer: (D) $418{\mathrm{cm}}^2$

Concept: New fraction = $\text{original}\times \frac{110}{90}=\text{original}\times \frac{11}{9}$.

1. $\text{Original}\times \frac{11}{9}=\frac{11}{45}$.
2. $\text{Original}=\frac{11}{45}\times \frac{9}{11}=\frac{9}{45}=\frac{1}{5}$.

Trick: Divide by the factor $\frac{11}{9}$: the 11s cancel at sight, leaving $\frac{9}{45}=\frac{1}{5}$.

Answer: (B) $\frac{1}{5}$

Concept: $\text{SI}=\frac{P\times R\times T}{100}\to P=\frac{\text{SI}\times 100}{R\times T}$.

1. $R\times T=2\times 4=8$.
2. $P=\frac{360\times 100}{8}$ = ₹4,500.

Trick: Total rate over the period = 8%; $8\%=360$ → $1\%=45$ → $100\%=4500$.

Answer: (D) 4500