Q&A: practice questions with answers

Concept: Time $=\frac{\text{Distance}}{\text{Speed}}$. Let Abhay's speed be $a$ and Sameer's time be $t$.

1. $\frac{156}{a}=t+5$ … (1)
2. With double speed: $\frac{156}{2a}=\frac{78}{a}=t-8$ … (2)
3. (1) − (2): $\frac{78}{a}=13\Rightarrow a=6\text{km/h}$.
4. Check: Abhay 26 h, Sameer 21 h, doubled speed 13 h $=21-8$ ✓.

Trick: Doubling speed halves the time: half of Abhay's time is saved, and that saving $=5+8=13$ h, so full time $=26$ h → $156÷26=6$.

Answer: (C) 6 km/hr

Concept: BODMAS: do division and multiplication before addition and subtraction; watch the signs.

1. $(-60)÷(-10)=6$.
2. $(-2)\times 6=-12$.
3. $(-9)-6+(-12)=-9-6-12=-27$.

Trick: Negative ÷ negative = positive; then simply add all negatives: $9+6+12=27$.

Answer: (A) −27

Concept: Write both numbers in terms of one variable and use the product.

1. Second $=x$, first $=5x$.
2. $5x^2=10125\Rightarrow x^2=2025\Rightarrow x=45$.
3. Sum $=x+5x=6x=270$.

Trick: $2025={45}^2$ is worth remembering (also a year!).

Answer: (D) 270

Concept: CI of the 2nd year = interest on (P + 1st year interest) $=P\times \frac{R}{100}\times (1+\frac{R}{100})$.

1. 2nd-year interest $=P\times 0.1\times 1.1=0.11P$.
2. $0.11P=5214\Rightarrow P=\frac{5214}{0.11}=47400$.

Trick: At 10%, 1st year interest $=10\%$ of P, 2nd year $=11\%$ of P. So $P=5214\times \frac{100}{11}$.

Answer: (D) ₹47,400

Concept: With a false weight the trader gives only 85% of the goods but charges the price of 100%, plus 3% profit on that price.

1. Cost of goods actually given $=0.85$ (per unit of true price).
2. Money received $=1.03$.
3. Profit $\%=\frac{1.03-0.85}{0.85}\times 100=\frac{18}{85}\times 100\approx 21.18\%\approx 21\%$.

Trick: Profit $\%=\frac{(100+p)-(100-w)}{100-w}\times 100=\frac{103-85}{85}\times 100$.

Answer: (C) $21\%$

Concept: For the same period, profit is shared in the ratio of investments.

1. Brijesh's share $=\frac{2200}{17600}=\frac{1}{8}$ of profit, so Kushal's $=\frac{7}{8}$.
2. Investment ratio K : B $=7:1$; total $=63300$, so 1 part $=\frac{63300}{8}=7912.5$.
3. Difference $=7-1=6$ parts $=6\times 7912.5=47475$.

Trick: Difference $=\frac{7-1}{8}\times 63300=\frac{3}{4}\times 63300=47475$.

Answer: (C) ₹47,475

Concept: Pipes working together: add their one-hour work.

1. $\frac{1}{13}+\frac{1}{29}+\frac{1}{10}=\frac{290+130+377}{3770}=\frac{797}{3770}$.
2. Time $=\frac{3770}{797}$ hours.
3. $797\times 4=3188$, remainder $3770-3188=582$, so time $=4\frac{582}{797}$ hours.

Trick: Estimate first: C alone takes 10 h, so together it must be well below 10 h; and $\frac{3770}{797}\approx 4.7$, so only $4\frac{582}{797}$ fits.

Answer: (C) $4\frac{582}{797}$ hours

Concept: Overall average $=\frac{\text{total runs}}{\text{total matches}}$.

1. Runs in 23 matches $=23\times 42=966$.
2. Runs in next 10 $=10\times 14=140$.
3. Average $=\frac{966+140}{33}=\frac{1106}{33}\approx 33.52$.

Trick: Deviation method: $42-\frac{10\times (42-14)}{33}=42-\frac{280}{33}=42-8.48=33.52$.

Answer: (A) 33.52

Concept: Savings = Income − Expenditure; apply each percentage change to its own base.

1. Savings $=26\%$ of 60500 $=15730$; expenditure $=60500-15730=44770$.
2. New income $=1.17\times 60500=70785$; new expenditure $=1.5\times 44770=67155$.
3. New savings $=70785-67155=3630$.
4. Change $=3630-15730=-12100$ → decrease of ₹12,100.

Trick: Change in savings = change in income − change in expenditure $=10285-22385=-12100$.

Answer: (C) decrease by ₹12,100

Concept: Profit share ∝ investment × time.

1. Sharma: $4000\times 6=24000$; Gupta: $8000\times 8=64000$; Sinha: $6000\times 8=48000$.
2. Ratio $=24:64:48=3:8:6$ (total 17).
3. Gupta's share $=\frac{8}{17}\times 34000=16000$.

Trick: Cancel the 1000s and common factors early: $(4\times 6):(8\times 8):(6\times 8)=3:8:6$.

Answer: (C) ₹16,000

Concept: Check the letter gaps inside each pair: 1st→2nd, 2nd→3rd, 3rd→4th letter.

1. GI-LE: G +2 I, I +3 L, L −7 E.
2. JL-OH: J +2 L, L +3 O, O −7 H; MO-RK: M +2 O, O +3 R, R −7 K.
3. DF-IC: D +2 F, F +3 I, but I −6 C → breaks the pattern.

Trick: Write positions (G7, I9, L12, E5) and compare differences; the last gap is the giveaway.

Answer: (D) DF-IC

Concept: Swap the signs first, then solve with BODMAS.

1. After swapping: $117-16+10÷2\times 2+68-10÷5+9$.
2. $10÷2\times 2=10$ and $10÷5=2$.
3. $117-16+10+68-2+9=186$.

Trick: Rewrite the whole expression once with new signs before calculating — avoids sign slips.

Answer: (D) 186

Concept: Find the gap between consecutive letters (positions, counting cyclically after Z).

1. INS: I(9) +5 N(14) +5 S(19).
2. TYD: T(20) +5 Y(25) +5 D(4, after Z); UZE: U(21) +5 Z(26) +5 E(5).
3. QBA: Q(17) → B(2) → A(1) — not +5, +5.

Trick: Use the EJOTY landmarks (5, 10, 15, 20, 25) to count +5 jumps fast.

Answer: (B) QBA

Concept: Number positions from the top (1) to the bottom (7) and place the fixed clues first.

1. Only F above A → F at 1, A at 2.
2. Two boxes between A and D → D at 5. Only B below C → C at 6, B at 7.
3. E and G fill 3, 4; G is not immediately above D (not 4) → G 3, E 4.
4. Between E (4) and B (7): D and C → 2 boxes.

Trick: 'Only X above Y' puts X at the top and Y second — always start from such clues.

Answer: (D) 2

Concept: Track each letter position separately.

1. 1st letters: R, O, L, I → −3 each → F.
2. 2nd letters: U, R, O, L → −3 → I.
3. 3rd letters: M, J, G, D → −3 → A.

Trick: All three positions move −3 together, so just shift ILD back by 3: FIA.

Answer: (B) FIA

Concept: 'Some A are B' can be reversed to 'Some B are A'. A link through a 'No' statement gives no definite 'Some' conclusion.

1. II: 'Some kites are hats' → 'Some hats are kites' (conversion) → follows.
2. I: Hats and ants never meet, but kites outside hats may or may not be ants → not definite → does not follow.

Yaad rakhein: Some + No ⇒ only 'Some A are not C'; never 'Some A are C'.

Answer: (B) Only conclusion (II) follows.

Concept: Rearrange the digits as asked, then pick positions.

1. Digits 7, 2, 3, 1, 4, 5, 6 in ascending order: 1234567.
2. 2nd from left = 2; 2nd from right = 6.
3. Sum $=2+6=8$.

Trick: Ascending order of 1–7 is just 1234567 — no writing needed.

Answer: (C) 8

Concept: Number positions from the bottom (1) to the top (7).

1. Only three below H → H at 4. Two boxes between H and O → O at 1 or 7.
2. Only N above G → N at 7, G at 6; so O at 1.
3. M above F above P in 2, 3, 5 → P 2, F 3, M 5.
4. Second from bottom = P.

Trick: When a clue fixes the top two (N, G), the 'either-end' box (O) is forced to the other end.

Answer: (B) P

Concept: Check ratios of consecutive terms.

1. $1200\times 1=1200$, $\times 1.5=1800$, $\times 2=3600$, $\times 2.5=9000$.
2. Multipliers rise by 0.5, so next $=9000\times 3=27000$.

Trick: When terms grow fast, try multiplication factors: 1, 1.5, 2, 2.5, 3.

Answer: (D) 27000

Concept: Drop the symbols, keep only the digits in order.

1. Digits from left: 1, 6, 3, 7, 2, 6, 2, 9.
2. Third from left = 3.

Trick: No need to rewrite all — just count digits from the left until the third one.

Answer: (B) 3