Q&A: practice questions with answers

Concept: Take total work = LCM of days; add work in 2-day cycles, then finish the remainder.

1. Work = $\text{LCM}(6,20)=60$ units; Sanjay = 10/day, Tannu = 3/day.
2. One cycle (Tannu + Sanjay) = 13 units; 4 cycles = 8 days → 52 units.
3. Left 8 units: day 9 Tannu does 3 → 5 left; day 10 Sanjay needs $\frac{5}{10}=0.5$ day.
4. Total = $8+1+0.5=9.5$ days.

Trick: Always check who starts — the starter works on the odd day after the full cycles.

Answer: (A) 9.5

Concept: If gain = SP of x articles out of n sold, then CP of n = SP of (n − x).

1. SP of 154 − CP of 154 = SP of 66 ⇒ CP of 154 = SP of 88.
2. $\frac{\text{SP}}{\text{CP}}=\frac{154}{88}=\frac{7}{4}$.
3. Profit % = $\frac{7-4}{4}\times 100=75\%$.

Trick: Profit % = $\frac{x}{n-x}\times 100=\frac{66}{88}\times 100=75\%$.

Answer: (B) 75%

Concept: $P=\text{SI}\times 100÷(R\times T)$.

1. $R\times T=6\times 2=12$, i.e. SI is 12% of P.
2. $P=387\times 100÷12=38700÷12=3225$.

Trick: 12% of P = 387 ⇒ 1% = 32.25 ⇒ P = 3225; or check options: $3225\times 0.12=387$.

Answer: (B) 3225

Concept: Solve brackets from the innermost outwards.

1. $\frac{13-11}{2}=\frac{2}{2}=1$
2. $5÷5\times 1=1$
3. $(16\times 12)\times 1=192$

Trick: If the curly bracket simplifies to 1, the answer is just $16\times 12$.

Answer: (A) 192

Concept: Weighted average = $\Sigma (\text{students}\times \text{average})÷\text{total students}$.

1. $20\times 40=800$, $25\times 34=850$, $15\times 26=390$ → total 2040.
2. Students = 60.
3. Average = $2040÷60=34$.

Trick: Take 34 as base: deviations $20\times (+6)+25\times 0+15\times (-8)=120-120=0$ ⇒ average = 34.

Answer: (C) 34

Concept: For equal distances, average speed = $\frac{2xy}{x+y}$.

1. Onward speed = $60÷(\frac{45}{60})=60\times \frac{4}{3}=80$ km/h.
2. Return speed = 40 km/h.
3. Average = $2\times 80\times 40÷120=6400÷120=53.33$ km/h.

Trick: Average speed for equal distances is always below the simple mean (60) — eliminate 56.67 and 56.33 quickly.

Answer: (D) 53.33

Concept: For 3 years: $\text{CI}-\text{SI}=P\times {\left(\frac{R}{100}\right)}^2\times (3+\frac{R}{100})$.

1. ${\left(\frac{10}{100}\right)}^2\times (3+0.1)=0.01\times 3.1=0.031$
2. $0.031\times P=124$
3. $P=124÷0.031=4000$

Trick: At 10%, 3-year difference = 3.1% of P; at 2 years it is 1% of P.

Answer: (A) 4000

Concept: $\text{Discount }\%=\text{free items}÷\text{total items taken}\times 100$.

1. Customer gets 6 items, pays for 5.
2. Discount = 1 out of 6 = $\frac{1}{6}\times 100=16.67\%$.

Trick: Divide by total items received (5 + 1), never by items paid for.

Answer: (D) 16.67%

Concept: $\text{Average speed}=\text{total distance}÷\text{total time}$.

1. Let distance = 5 units: 3 units at 84, 2 units at 56.
2. Time = $\frac{3}{84}+\frac{2}{56}=\frac{1}{28}+\frac{1}{28}=\frac{1}{14}$.
3. Average = $5÷(\frac{1}{14})=70$ km/h.

Trick: Choose distance = LCM-friendly value (e.g. 840 km: 504 km in 6 h, 336 km in 6 h → $\frac{840}{12}=70$).

Answer: (A) 70

Concept: $V=\frac{1}{3}\pi r^{2}h$; convert height to cm first.

1. $h=0.12\text{ m}=12\text{ cm}$.
2. $168\pi =\frac{1}{3}\times \pi \times r^2\times 12=4\pi r^2\Rightarrow r^2=42$.
3. So $r=\sqrt{42}$ cm (≈ 6.48 cm).

Trick: Keep units same before using the formula; $\frac{1}{3}\times 12=4$ makes it $r^2=168÷4$.

Answer: (C) $\sqrt{42}\mathrm{cm}$

Concept: $A_1(1+r{)}^n=A_2(1+r{)}^{n+1}\Rightarrow A_1:A_2=(1+r):1$.

1. $\text{Ajay}\times {1.1}^{14}=\text{Prakash}\times {1.1}^{15}\Rightarrow \text{Ajay}=1.1\times \text{Prakash}$.
2. Ajay : Prakash = 11 : 10, total 21 parts = ₹1218.
3. 1 part = 58 ⇒ Prakash = $10\times 58=$ ₹580.

Trick: One extra year at 10% → ratio 11 : 10 directly; Prakash = $\frac{1218\times 10}{21}$.

Answer: (B) 580

Concept: If SP of a items = CP of b items, profit % = $\frac{b-a}{a}\times 100$.

1. 30 SP = 50 CP ⇒ SP : CP = 5 : 3.
2. Profit = 2 on CP 3 → $\frac{2}{3}\times 100=66.67\%$.

Trick: $\frac{50-30}{30}\times 100=66.67\%$ — divide by the 'SP' side count.

Answer: (C) 66.67%

Concept: Make the common term equal, then share the total.

1. M : S = 2 : 1 = 4 : 2 and M : C = 4 : 3.
2. M : S : C = 4 : 2 : 3 (9 parts) = 234 ⇒ 1 part = 26.
3. Maths = $4\times 26=104$.

Trick: Maths must be a multiple of $4\times 26$; also answer must be divisible by 4 → only 104 and 68 fit, and 68 is too small.

Answer: (B) 104

Concept: Do ÷ and × from left to right before −.

1. $32÷8=4$, then $4\times 5=20$
2. $2\times 3=6$
3. $20-6=14$

Trick: Never subtract $5-2$ first; that common slip gives wrong option values.

Answer: (C) 14

Concept: Use ratio parts: difference of parts = difference of ages.

1. 9 − 4 = 5 parts = 10 years ⇒ 1 part = 2.
2. Now: A = 8, B = 18.
3. After 4 years: 12 : 22 = 6 : 11.

Trick: Age difference never changes — verify: 22 − 12 = 10 years.

Answer: (C) 6 : 11

Concept: For fixed distance, $\text{time}\propto \frac{1}{\text{speed}}$.

1. Speed becomes $\frac{1}{4}$ ⇒ time becomes 4 times.
2. Extra time = $4T-T=3T=45$ min.
3. T = 15 min.

Trick: At $\frac{1}{n}$ of usual speed, usual time = delay ÷ $(n-1)$ = $45÷3=15$.

Answer: (C) 15

Concept: If a% of x = y, then $x=y\times \frac{100}{a}$.

1. $x=336\times 100÷2$
2. $x=336\times 50=16800$

Trick: 2% is $\frac{1}{50}$, so just multiply by 50.

Answer: (C) 16800

Concept: Multiply the successive factors on a base of 100.

1. $100\times 0.75=75$
2. $75\times 0.90=67.5$; $67.5\times 0.70=47.25$
3. $47.25\times 1.60=75.6$
4. Change = $100-75.6=24.4\%$ decrease.

Trick: Overall factor $0.75\times 0.9\times 0.7\times 1.6=0.756$ → 24.4% less.

Answer: (C) Decreases by 24.4%

Concept: For two numbers, $\text{product}=\text{HCF}\times \text{LCM}$.

1. $\text{LCM}=5184÷24$
2. LCM = 216

Trick: LCM must be a multiple of the HCF (24): $216=24\times 9$ ✓; 324, 125, 343 are not.

Answer: (D) 216

Concept: Clear the small bracket, then the curly bracket, then multiply.

1. $65÷5=13$
2. $\frac{16}{8}=2$; $\frac{14}{2}\times (6-4)=7\times 2=14$
3. $\{2+14\}=16\Rightarrow 13\times 16=208$

Trick: Answer = 13 × something → must be a multiple of 13: only 208 ($13\times 16$) fits.

Answer: (B) 208