Q&A: practice questions with answers

Concept: Take tank capacity $=$ LCM of the times and add the hourly units.

1. Capacity $=\text{LCM}(5,6,12)=60$ units
2. m $=12$, n $=10$, p $=5$ units/hour
3. Together $=27$ units/hour
4. Time $=\frac{60}{27}=\frac{20}{9}=2\frac{2}{9}$ hours

Trick: $\frac{1}{5}+\frac{1}{6}+\frac{1}{12}=\frac{12+10+5}{60}=\frac{27}{60}$, invert it.

Answer: (D) $2\frac{2}{9}$ hours

Concept: Value of each coin group $=$ number of coins $\times$ coin value.

1. Let coins be $18k,12k,10k$
2. Value: $0.50\times 18k+0.25\times 12k+0.10\times 10k=9k+3k+k=13k$
3. $13k=130\Rightarrow k=10$
4. 10 paise coins $=10\times 10=100$

Trick: Value ratio $9:3:1$ → ₹130 splits as ₹90 : ₹30 : ₹10; ₹10 in 10-paise coins $=100$ coins.

Answer: (C) $100$

Concept: Sum $=$ average $\times$ count; remove the discarded numbers and divide by the new count.

1. Sum of 50 numbers $=50\times 34=1700$
2. After removing 34 and 22: $1700-56=1644$
3. New average $=\frac{1644}{48}=34.25$

Trick: Removing 34 (= average) changes nothing; removing 22 (12 below) raises average by $\frac{12}{48}=0.25$.

Answer: (C) $34.25$

Concept: Use $1+{\tan }^{2}A={\sec }^{2}A$ and $1+{\cot }^{2}A={\csc }^{2}A$.

1. Numerator $=3(1+{\tan }^{2}A)\cot A=3{\sec }^{2}A\cot A$
2. Denominator $={\csc }^{2}A\sec A$
3. Ratio $=\frac{3\sec A\cot A}{{\csc }^{2}A}=3·\frac{1}{\cos A}·\frac{\cos A}{\sin A}·{\sin }^{2}A$
4. $=3\sin A$

Trick: Check with $A={45}^{\circ }$: $\frac{6·1}{2\sqrt{2}}=\frac{3}{\sqrt{2}}=3\sin {45}^{\circ }$ ✓.

Answer: (C) $3\sin A$

Concept: Correct sum $=$ wrong sum $-$ wrong value $+$ correct value.

1. Wrong sum $=10\times 40=400$
2. Correct sum $=400-31+50=419$
3. Correct average $=\frac{419}{10}=41.9$

Trick: Average changes by $\frac{50-31}{10}=1.9$ → $40+1.9=41.9$.

Answer: (A) $41.9$

Concept: If $\frac{x}{11}=\frac{275}{x}$, cross-multiply to get $x^2$.

1. $x^2=11\times 275=3025$
2. $x=\sqrt{3025}=55$ (positive)

Trick: $3025$ ends in 25 → square of a number ending in 5: ${55}^2=3025$.

Answer: (A) $55$

Concept: Compare each letter of the first cluster with the matching letter of the second.

1. M→Q, J→N, G→K, T→X: every letter moves $+4$.
2. Check: C→G, Z→D, W→A, J→N, again $+4$ each.
3. OLIV: O→S, L→P, I→M, V→Z gives SPMZ.

Trick: Check just the first two letters (O+4 = S, L+4 = P) to reject options A and D quickly.

Answer: (B) SPMZ

Concept: Fix the definite clues first (top and bottom), then place the rest.

1. Only D above F → D at 1 (top), F at 2.
2. Only B below E → B at 7 (bottom), E at 6.
3. Three boxes between D and A → A at 5.
4. Y not immediately above A → Y not at 4, so Y at 3 and C at 4.
5. Between C (4) and B (7): A and E, i.e. two boxes.

Trick: "Only X above Y" means X is at the very top and Y is just below it.

Answer: (C) Two

Concept: Change every digit by the given rule, then compare the largest and smallest digits.

1. Digits 4 1 2 3 9 5 7.
2. Even digits +2: 4→6, 2→4. Odd digits −1: 1→0, 3→2, 9→8, 5→4, 7→6.
3. New number 6042846; greatest digit 8, smallest 0.
4. Difference $=8-0=8$.

Trick: Only the extremes matter: largest odd 9 becomes 8, smallest odd 1 becomes 0.

Answer: (D) 8

Concept: Look at the differences between consecutive terms.

1. $356-333=23,333-300=33$: differences rise by 10.
2. Next difference 43: $300-43=257$.
3. Check: $257-53=204$ and $204-63=141$ ✓.

Trick: Differences 23, 33, 43, 53, 63 form an AP with common difference 10.

Answer: (C) 257

Concept: Find the shift between the two clusters of a pair and apply the same on the right side.

1. Try # = SAG: S+5 = X, A+5 = F, G+5 = L gives XFL, so the rule is +5.
2. Right side: K+5 = P, S+5 = X, Y+5 = D (after Z we restart at A), so % = PXD.

Trick: Y+5: Z, A, B, C, D. Option D (PXO) fails on the last letter, option B (TAG) on the first.

Answer: (C) # = SAG, % = PXD

Concept: In such codes the digits may be jumbled; the letter common to both words gets the digit common to both codes.

1. CARD and DOGS share only the letter D.
2. 4681 and 3219 share only the digit 1.
3. So D = 1.

Trick: Common letter ↔ common digit.

Answer: (A) 1

Concept: Code each element from the table, then check which condition applies to the first and last elements.

1. Elements: 5, 6, #, 4, +. Direct codes: 5 = G, 6 = J, # = W, 4 = B, + = E.
2. Condition (i): first element 5 is a number and last element + is a symbol → interchange the codes of first and last.
3. Condition (ii) fails (5 is not divisible by 2); (iii) fails (first is not a symbol).
4. Code: E J W B G.

Trick: Always test the conditions only on the first and last elements; the middle ones are coded directly.

Answer: (A) EJWBG

Concept: Remove all symbols and write only the digits in their order.

1. Digits left to right: 1 5 3 4 2 7 9 6 8.
2. From the right: 8 (1st), 6 (2nd), 9 (3rd), 7 (4th).

Trick: After dropping symbols, count from the right end of the digit-only list.

Answer: (D) 7

Concept: Check the gap inside each cluster and between the first letters of the two clusters.

1. RV – TB: R+4 = V; R+2 = T; T+8 = B (V+6 = B).
2. MQ – OW: M+4 = Q; M+2 = O; Q+6 = W. EI – GO: E+4 = I; E+2 = G; I+6 = O.
3. GK – JQ: G+4 = K but G+3 = J (not +2), so it breaks the pattern.

Trick: Pattern: (+4) – (+2, +6). Only GK – JQ has +3 between first letters.

Answer: (B) GK – JQ

Concept: Find the highest and lowest numbers, then work with their first digits.

1. Numbers: 411, 559, 916, 247, 769.
2. Highest = 916 (first digit 9); lowest = 247 (first digit 2).
3. $9-2=7$.

Trick: For highest/lowest of three-digit numbers, compare the first digits first.

Answer: (A) 7

Concept: All face north, so left and right are the viewer's left and right. Place F first.

1. Only two people left of F → F at position 3.
2. Three people between F and A → A at 7.
3. One person between A and O → O at 5.
4. D immediately left of C → D at 1, C at 2. E not a neighbour of A → E at 4, B at 6.
5. Row: D C F E O B A; only A is to the right of B.

Trick: "Only two to the left" fixes an exact position (3rd from left); start with such clues.

Answer: (B) One

Concept: Track the first, second and third letters of each cluster separately (positions A = 1 … Z = 26).

1. First letters: G, Q, A, K → each +10 (cyclic) → next K+10 = U.
2. Second letters: U, E, O, Y → +10 → Y+10 = I.
3. Third letters: R, B, L, V → +10 → V+10 = F.
4. Next cluster: UIF.

Trick: +10 cyclic is the same as −16; e.g. Y(25)+10 = 35 − 26 = 9 = I.

Answer: (D) UIF

Concept: Compare letter by letter.

1. S→Z, U→B, Z→G, D→K: each letter +7.
2. Check: U→B, W→D, B→I, F→M ✓ (+7).
3. PRWA: P+7 = W, R+7 = Y, W+7 = D, A+7 = H → WYDH.

Trick: Options A and C fail at the second letter (R+7 = Y); B fails at the third.

Answer: (D) WYDH

Concept: Total = rank from top + rank from bottom − 1 (the person is counted twice).

1. Rank from top = 28, rank from bottom = 64.
2. Total $=28+64-1=91$.

Trick: Add both ranks and subtract 1.

Answer: (D) 91