Q&A: practice questions with answers

Concept: Distance to cross a tunnel = train length + tunnel length.

1. $\text{Distance}=702+238=940$ m
2. $\text{Speed}=36\times \frac{5}{18}=10\text{m/s}$
3. $\text{Time}=940÷10=94$ s

Trick: km/h → m/s: multiply by $\frac{5}{18}$; $36\text{km/h}=10\text{m/s}\text{ and }72\text{km/h}=20\text{m/s}$ are worth memorising.

Answer: (A) 94

Concept: Alligation works on cost prices, so first find the cost price of the mixture.

1. $\text{Mixture CP}=5\times \frac{100}{125}=$₹4 per kg
2. Cheap (3.80) : Dear (4.60) = $(4.60-4):(4-3.80)=0.60:0.20=3:1$
3. 1 part = 60 kg → cheap sugar = $3\times 60=180$ kg

Trick: 25% gain → $\text{CP}=\text{SP}\times \frac{4}{5}$; ₹5 → ₹4 in one glance.

Answer: (B) 180 kg

Concept: Take total work as Q's days (56 units, Q = 1 unit/day) and count each person's working days.

1. 37.5 days: Q works on odd days 1–37 = 19 days → 19 units.
2. R works on 18 full days + half of day 38 = 18.5 days → $56-19=37$ units.
3. R's rate = $37÷18.5=2$ units/day.
4. Half work = 28 units → $28÷2=14$ days.

Trick: In alternate-day questions, a '.5' day always belongs to the person whose turn it is next.

Answer: (C) 14 days

Concept: Third proportional to a and b is c where a : b = b : c, so $c=\frac{b^2}{a}$.

1. $c=(0.52{)}^2÷0.13$
2. Note $0.52=4\times 0.13$, so $c=0.52\times 4$
3. $c=2.08$

Trick: $b÷a=4$, so $c=b\times 4$ – no squaring of decimals needed.

Answer: (D) 2.08

Concept: Do ÷ and × first, keeping sign rules: (−)÷(−) = +, (−)×(+) = −.

1. (−60) ÷ (−12) = +5
2. (−3) × 8 = −24
3. −9 − 5 − 24 = −38

Trick: Collect all negatives at the end: 9 + 5 + 24 = 38, then put the minus sign.

Answer: (A) -38

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

1. $R\times T=8\times 3=24\%$
2. $P=486\times 100÷24$
3. $P=48600÷24=$₹2025

Trick: 24% of P = 486 → P must end so that ×24 gives …86: check options – $2025\times 24=48600✓$.

Answer: (D) 2025

Concept: LCM = product of the highest power of every prime present.

1. $18=2\times 3^2;72=2^3\times 3^2;144=2^4\times 3^2;184=2^3\times 23$
2. $\text{LCM}=2^4\times 3^2\times 23$
3. $=144\times 23=3312$

Trick: 144 already covers 18 and 72; just find $\text{LCM}(144,184)=144\times 23$.

Answer: (C) 3312

Concept: Literate females = total literates − literate males.

1. $\text{Females}=340000-217000=123000$
2. Total literates = 11% of 340000 = 37400; males = 11% of 217000 = 23870
3. Literate females = 13530 → $\frac{13530}{123000}\times 100=11\%$

Trick: If the whole group and one part both have 11%, the other part must also have 11% – no calculation.

Answer: (D) 11%

Concept: Division and multiplication come before addition and subtraction.

1. $10÷5=2$
2. $3\times 3=9$
3. $16+2-9=9$

Trick: Never go left-to-right blindly: $16+10=26$ first would give a wrong answer.

Answer: (B) 9

Concept: Convert averages into totals; the common person is counted twice in pair sums.

1. $C+V+M=3\times 46=138$
2. $C+V=80,V+M=88$ → sum 168
3. $V=168-138=30$ kg

Trick: Common member = (sum of pair totals) − (group total).

Answer: (B) 30

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

1. $(\frac{5}{100}{)}^2=\frac{1}{400};3+\frac{1}{20}=\frac{61}{20}$
2. $122=P\times \frac{61}{8000}$
3. $P=122\times 8000÷61=$₹16000

Trick: Difference per ₹8000 is ₹61 at 5%; $122=2\times 61$ → $P=2\times 8000$.

Answer: (A) 16000

Concept: Count how many cubes fit along each edge and multiply.

1. $36÷4=9$
2. $16÷4=4;12÷4=3$
3. $9\times 4\times 3=108$

Trick: Volume check: $(36\times 16\times 12)÷4^3=6912÷64=108$ – works because each edge is divisible by 4.

Answer: (C) 108

Concept: Solve from the innermost bracket outwards.

1. Outer part: $60÷5=12$.
2. Inside: $\frac{56}{4}$ gives 14, and $\frac{14}{3}\times (9-6)=\frac{14}{3}\times 3$ gives 14.
3. Bracket total 28; so $12\times 28=336$.

Trick: Answer = 12 × something → must be a multiple of 12; only 336 is.

Answer: (B) 336

Concept: They meet at the start point after LCM of their one-round times.

1. Round times: $\frac{150}{20}=7.5$ s, $\frac{150}{15}=10$ s, $\frac{150}{12}=12.5$ s
2. $\text{LCM}(\frac{15}{2},\frac{10}{1},\frac{25}{2})=\frac{\text{LCM}(15,10,25)}{\text{HCF}(2,1,2)}=\frac{150}{1}=150$ s
3. Check: 20, 15 and 12 rounds – all whole numbers.

Trick: LCM of fractions = LCM of numerators ÷ HCF of denominators.

Answer: (C) 150 sec

Concept: Gain = SP − CP of the same quantity; equate and compare.

1. Gain = SP of 68 chairs → CP of 153 = SP of ($153-68$) = SP of 85
2. So $85\text{SP}=153\text{CP}\to \frac{\text{SP}}{\text{CP}}=\frac{9}{5}$
3. $\text{Profit}\%=\frac{9-5}{5}\times 100=80\%$

Trick: $\text{Profit}\%=\frac{\text{gain items}}{\text{total}-\text{gain items}}\times 100$ = $\frac{68}{85}\times 100=80\%$.

Answer: (A) 80%

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

1. $(\frac{10}{100}{)}^2=\frac{1}{100};3+\frac{1}{10}=\frac{31}{10}$
2. $186=P\times \frac{31}{1000}$
3. $P=186\times 1000÷31=$₹6000

Trick: At 10%, 3-year difference = 3.1% of P → $186÷0.031=6000$.

Answer: (D) 6000

Concept: $\text{SP}=\text{CP}\times \frac{100+\text{profit}\%}{100}$.

1. $\text{SP}=540\times \frac{120}{100}$
2. $=540\times \frac{6}{5}$
3. = ₹648

Trick: 10% of 540 = 54, so 20% = 108 → $540+108=648$.

Answer: (A) 648

Concept: Find cost before and after for a fixed quantity, then % change.

1. Old cost per unit = $0.6\times 30+0.4\times 30=$₹30
2. New = $0.6\times 40+0.4\times 20=24+8=$₹32
3. Change = +2 on 30 → $\frac{2}{30}\times 100=6.67\%$ increase

Trick: Weighted change = $0.6\times (+10)+0.4\times (-10)=+2$ on ₹30 base.

Answer: (D) increase; 6.67

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

1. 4% of x = 60
2. $x=60\times 100÷4$
3. $x=1500$

Trick: 4% = 60 → 1% = 15 → 100% = 1500.

Answer: (C) 1500

Concept: Equivalent discount = $a+b-\frac{ab}{100}$ (never just a + b).

1. $12+18=30$
2. $12\times \frac{18}{100}=2.16$
3. $30-2.16=27.84\%$

Trick: Answer must be a little less than the sum (30%) – so 30% is the trap, 27.84% fits.

Answer: (B) 27.84%