Q&A: practice questions with answers

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$.

1. Times: $\frac{370}{37}=10$ h, $\frac{390}{5}=78$ h, $\frac{720}{8}=90$ h; total $=178$ h.
2. Total distance $=370+390+720=1480\text{km}$.
3. Average speed $=\frac{1480}{178}=\frac{740}{89}=8\frac{28}{89}\text{km/h}$.

Trick: Never average the speeds; always total distance ÷ total time.

Answer: (D) $8\frac{28}{89}$

Concept: $\tan \theta =\frac{\text{Height}}{\text{Base}}$.

1. Height $=250\sqrt{3}\text{m}$, base $=250\text{m}$.
2. $\tan \theta =\frac{250\sqrt{3}}{250}=\sqrt{3}$.
3. $\tan {60}^{\circ }=\sqrt{3}\Rightarrow \theta ={60}^{\circ }$.

Trick: Height $=\sqrt{3}\times$ base → ${60}^{\circ }$; base $=\sqrt{3}\times$ height → ${30}^{\circ }$.

Answer: (B) ${60}^{\circ }$

Concept: Let cost price $=C$. Gain at ₹3,437 is 75% more than loss at ₹3,338.

1. $3437-C=\frac{175}{100}(C-3338)$.
2. $4(3437-C)=7(C-3338)\Rightarrow 13748+23366=11C\Rightarrow C=\frac{37114}{11}=3374$ (₹).
3. For 50% gain: $\text{SP}=3374\times \frac{150}{100}=5061$ (₹).

Trick: 75% more means ratio gain : loss $=7:4$, so CP divides 3338 and 3437 in $4:7$: $3338+\frac{4}{11}\times 99=3374$.

Answer: (A) 5,061

Concept: Complete the square in $x^2$ and use $a^2-b^2=(a+b)(a-b)$.

1. $x^4-10x^2+22=(x^2-5{)}^2-25+22=(x^2-5{)}^2-3$.
2. $=(x^2-5{)}^2-(\sqrt{3}{)}^2$.
3. $=(x^2-5+\sqrt{3})(x^2-5-\sqrt{3})$.

Trick: Check: the constants multiply to $25-3=22$ and add to $-10$ ✓.

Answer: (D) $(x^2-5+\sqrt{3})(x^2-5-\sqrt{3})$

Concept: Work done = rate × time; an emptying pipe has negative rate.

1. In 6 minutes A fills $\frac{6}{18}=\frac{1}{3}$ of the tank; remaining $=\frac{2}{3}$.
2. Net rate of A and B $=\frac{1}{18}-\frac{1}{27}=\frac{3-2}{54}=\frac{1}{54}$ per minute.
3. Time $=\frac{2}{3}÷\frac{1}{54}=36$ minutes.

Trick: LCM method: tank $=54$ units; A $=+3$, B $=-2$, net $+1$/min; after 6 min, $54-18=36$ units → 36 min.

Answer: (A) 36

Concept: Successive percentages multiply.

1. $2\%\text{ of }50\%\text{ of }N=\frac{2}{100}\times \frac{50}{100}\times N=\frac{N}{100}$.
2. $\frac{N}{100}$ is $1\%$ of $N$.

Trick: $2\times 50=100$, and $\frac{100}{100}=1\%$.

Answer: (C) 1%

Concept: Mode = the observation that occurs most often.

1. Frequencies: 1 → 2, 3 → 7, 4 → 1, 5 → 1, 7 → 2, 8 → 2.
2. 3 occurs most often (7 times).

Trick: Tally just the repeated values; no calculation needed.

Answer: (C) 3

Concept: Empirical formula: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $89.7=3(32)-2\text{Mean}=96-2\text{Mean}$.
2. $2\text{Mean}=96-89.7=6.3$.
3. $\text{Mean}=3.15$.

Trick: Remember “3 Median − 2 Mean” – the coefficients 3 and 2 go with median and mean.

Answer: (D) 3.15

Concept: Compare each letter of the first cluster with the corresponding letter of the second using alphabet positions.

1. NO → IJ: $N(14)-5=I(9)$, $O(15)-5=J(10)$ → $-5,-5$.
2. FG → AB and JK → EF also follow $-5,-5$.
3. PK → MJ: $P(16)-3=M(13)$, $K(11)-1=J(10)$ → $-3,-1$, a different pattern.

Trick: Consecutive letters (NO, FG, JK) going back 5 give consecutive letters again; PK is not even a consecutive pair.

Answer: (A) PK - MJ

Concept: Check each letter position of the clusters separately.

1. 1st letters: L, N, P, R → $+2$ each → T
2. 2nd letters: J, L, N, P → $+2$ each → R
3. 3rd letters: Q, S, U, W → $+2$ each → Y

Trick: All three positions move $+2$: RPW → TRY.

Answer: (B) TRY

Concept: Fix the definite clues (product and sum) first, then place the remaining people.

1. D × F = 10 → floors {2, 5}.
2. B + D = 11 with D ∈ {2, 5} → D = 5, B = 6 (D = 2 would need B = 9). So F = 2.
3. Floors left: 1, 3, 4. C immediately above E → E = 3, C = 4; A = 1.
4. Between A (1) and E (3): only F → 1 person.

Trick: Sum 11 on a 6-floor building forces the pair (5, 6) at once.

Answer: (B) 1

Concept: Look for the pattern Number–Number–Number and count the middle numbers.

1. Series: 5 $ 5 % 8 7 * 5 % & 8 @ 2 6 2 £ 3 8 1 & 5 6
2. ‘2 6 2’ → 6 qualifies.
3. ‘3 8 1’ → 8 qualifies.
4. No other number has numbers on both sides (e.g. 8 7 * fails, 5 6 is at the end).

Trick: Scan only runs of three or more consecutive digits: ‘262’ and ‘381’ – each gives one middle number.

Answer: (A) 2

Concept: A letter common to both words must take the digit common to both codes.

1. GATE and TURN share only the letter T.
2. Codes 4628 and 3567 share only the digit 6.
3. So T = 6.

Trick: Common letter ↔ common digit.

Answer: (B) 6

Concept: Find the differences between consecutive terms.

1. $139-3=136$, $136-5=131$, $131-7=124$, $124-9=115$
2. Next: $115-11=104$
3. Check: $104-13=91$ ✔

Trick: Subtract consecutive odd numbers 3, 5, 7, 9, 11, 13.

Answer: (C) $104$

Concept: Facing the centre: left = clockwise, right = anticlockwise. 7 people → 7 seats.

1. Fix I. B is third to the left of I.
2. Only I between A and G → A and G are the two neighbours of I.
3. G second to the left of H: if G were on I’s left, H would sit on I’s right, clashing with A. So A is on I’s left, G on I’s right, H two seats to G’s right.
4. Two seats remain next to B; C is not B’s neighbour, so C sits beside H and G, and D sits between A and B.
5. From D going right (anticlockwise): A, I, then G → 2 people.

Trick: Place the ‘only X between’ pair around X first; it fixes most seats.

Answer: (C) Two

Concept: Find the chain of operations in the given sets and test each option.

1. $14\to 15(+1)\to 20(+5)\to 40(\times 2)$
2. $19\to 20(+1)\to 25(+5)\to 50(\times 2)$
3. Option (D): $5\to 6(+1)\to 11(+5)\to 22(\times 2)$ ✔
4. (A) starts with $+2$, (B) ends with $+5$, (C) starts with $+2$ – all fail.

Trick: Last term = double the third term; only 11 → 22 satisfies this.

Answer: (D) 5 - 6 - 11 - 22

Concept: A letter common to both words takes the digit common to both codes.

1. BLUE and LOAD share only L.
2. 4628 and 3567 share only 6.
3. So L = 6.

Trick: Common letter ↔ common digit.

Answer: (B) 6

Concept: Look for the pattern Number–Symbol–Symbol.

1. Series: @ $ & 7 7 @ 5 7 & & 4 1 9 3 $ 7 8 & 7 € € 1
2. ‘7 & &’ → qualifies.
3. ‘7 € €’ → qualifies.
4. Others fail: ‘7 @ 5’, ‘3 $ 7’, ‘8 & 7’ are followed by numbers.

Trick: Locate pairs of adjacent symbols (& &, € €) and check the element just before the first one.

Answer: (C) 2

Concept: Track the letter cluster and the number separately.

1. Each letter moves $+2$: G→I→K→M→O, K→M→O→Q→S, O→Q→S→U→W
2. Numbers: 11, 13, 15, 17 → $+2$ → 19
3. Next term: OSW 19

Trick: Inside each cluster letters also differ by 4 (G, K, O); O, S, W fits.

Answer: (A) OSW 19

Concept: Replace letters (A = ×, B = −, C = +, D = ÷) and apply BODMAS.

1. (B): $8-3+10\times 2÷5=8-3+4=9$ ✔
2. (A): $8-3\times 10+2÷5=-21.6$
3. (C): $8\times 3+10-2÷5=33.6$; (D): $8+3-10\times 2÷5=7$

Trick: Look for the option where the ‘×2÷5’ part acts on 10 (gives 4) and the rest gives 5: $8-3=5$.

Answer: (B) 8 B 3 C 10 A 2 D 5