Q&A: practice questions with answers

Concept: SP $=\text{CP}\times (1+\text{markup})\times (1-\text{discount})$.

1. CP $=100$, MP $=130$.
2. SP $=130\times 0.9=117$.
3. Profit $=17\%$.

Trick: Net $=30-10-\frac{30\times 10}{100}=17\%$.

Answer: (A) 17%

Concept: Form two equations and solve.

1. Let smaller $=s$, larger $=54-s$.
2. $3s-2(54-s)=6\Rightarrow 5s=114\Rightarrow s=22.8$.
3. Larger $=31.2$; difference $=31.2-22.8=8.4$.

Trick: $3s-2l=6$ and $2s+2l=108$ → add: $5s=114$ directly.

Answer: (A) 8.4

Concept: Net change for $+a\%$ then $-b\%$ is $a-b-\frac{ab}{100}$.

1. Net $=30-20-6=+4\%$.
2. $4\%$ of original $=40\Rightarrow$ original $=1000$.
3. Check: $1000\to 1300\to 1040$, which is ₹40 more.

Trick: $1.3\times 0.8=1.04$ → a 4% rise; $40÷0.04=1000$.

Answer: (A) ₹1,000

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

1. $c=\frac{{120}^2}{240}=\frac{14400}{240}=60$.
2. Check: $240:120=120:60=2:1$.

Trick: 120 is half of 240, so the next term is half of 120 = 60.

Answer: (C) 60

Concept: Write every base as a power of 3 and use $(a^m{)}^n=a^{mn}$.

1. ${27}^{\frac{2}{3}}=(3^3{)}^{\frac{2}{3}}=3^2=9$; $9^{-\frac{1}{2}}=\frac{1}{3}$. Numerator $=9\times \frac{1}{3}=3$.
2. ${81}^{\frac{1}{4}}=3$; $3^{-1}\times 3=1$. Denominator $=1$.
3. Value $=\frac{3}{1}=3$.

Trick: In powers of 3: numerator $3^{2-1}=3^1$, denominator $3^{-1+1}=3^0$ → $3^{1-0}=3$.

Answer: (B) 3

Concept: Fix the definite positions first (counting from the top), then place the rest.

1. J is 4th from top. Only one box above L → L = 2nd, I (immediately below L) = 3rd.
2. K is above J → the only free place above J is 1st, so K = 1st.
3. Positions 5, 6, 7 remain for F, G, H. Only one box between G and F → G, F at 5 and 7; H = 6th.
4. K (1st) and H (6th): boxes between = L, I, J, G/F = 4.

Trick: “Only one above L” pins L at 2 – always start from such absolute clues.

Answer: (D) Four

Concept: Vowels (A, E, I, O, U) move +1, consonants move −1 in the alphabet.

1. R A D U L A E → R−1 = Q, A+1 = B, D−1 = C, U+1 = V, L−1 = K, A+1 = B, E+1 = F.
2. New group: Q B C V K B F.
3. Only B appears exactly twice (from the two A’s) → one letter.

Trick: Repeated letters in the original word (A, A) stay repeated after the shift – count them first.

Answer: (C) One

Concept: Find a rule linking the three numbers (a, b, c) that works for both given sets.

1. (2, 5, 7): $2^2+1=5$ and $5+2=7$.
2. (3, 10, 13): $3^2+1=10$ and $10+3=13$. Rule: $b=a^2+1,c=b+a$.
3. (6, 37, 43): $6^2+1=37$ ✓, $37+6=43$ ✓. Others fail: (7, 49, 51) needs 50; (5, 26, 36) needs 31; (8, 63, 75) needs 65.

Trick: Check the first pair rule ($a^2+1$) on options – only (6, 37) survives.

Answer: (B) (6, 37, 43)

Concept: A word common to two sentences has the code common to both.

1. ‘stage pig add’ → ‘yv se kk’; ‘third add bee’ → ‘se ec wm’.
2. Common word = add; common code = se.
3. So add = se.

Trick: Match common words with common codes – no need to decode the rest.

Answer: (D) se

Concept: Vowels are A, E, I, O, U; Y is a consonant. Check each vowel’s left and right neighbours.

1. Series: F M I H Y F Y E G K Y O N V K H Q T E K N – vowels at positions 3, 8, 12, 19.
2. I: M–I–H (followed by consonant ✗); E: Y–E–G ✗; O: Y–O–N ✗; E: T–E–K ✗.
3. No vowel is followed by another vowel → none.

Trick: “Followed by a vowel” means two vowels together – scan for any adjacent vowel pair; there is none.

Answer: (A) None

Concept: Facing the centre: left = clockwise, right = anticlockwise. Number the 7 seats clockwise with Q at seat 1.

1. Two between Q and L from Q’s right (anticlockwise) → L at seat 5.
2. Two between L and O from O’s left (clockwise) → O is 3 seats anticlockwise of L → O at seat 2.
3. Three between Q and F from Q’s right → F at seat 4.
4. Seats 3, 6, 7 left for G, H, M. M immediately left (clockwise) of H → H = 6, M = 7; G = 3.
5. Third to the left of G (clockwise): F, L, H.

Trick: Write “right = anticlockwise” on the diagram once; every clue then becomes a simple count.

Answer: (A) H

Concept: Compare the alphabetical gap between corresponding letters of each pair.

1. ST – YZ: S→Y = +6, T→Z = +6.
2. CD – IJ: +6, +6. IJ – OP: +6, +6.
3. MJ – LP: M→L = −1, J→P = +6 – breaks the pattern.

Trick: Use place values (A=1 … Z=26); three pairs differ by +6 for both letters.

Answer: (D) MJ - LP

Concept: Track each position of the cluster separately across the series.

1. 1st letters: P → W → D → K → ? : each +7 → K+7 = R.
2. 2nd letters: R → Y → F → M → ? : +7 → T.
3. 3rd letters: Y → F → M → T → ? : +7 → A.

Trick: +7 wraps around Z (K=11, 11+7=18=R; T=20, 20+7=27→1=A).

Answer: (A) RTA

Concept: Number positions 1 (top) to 7 (bottom); “before” = above, “after” = below.

1. Y is 2nd from bottom → position 6. Two boxes between Y and V → V at 3.
2. Only Z between Y and U → Z at 5, U at 4 (going upward; downward is impossible).
3. A is above V but below W → W = 1, A = 2.
4. Only X is left → X at 7, the bottom.

Trick: With 7 boxes, “second from bottom” = 6th from top; convert all clues to one direction.

Answer: (C) X

Concept: Use place values and check the differences between consecutive letters in each cluster.

1. OGY: O(15)→G(7) = −8, G(7)→Y(25) = +18.
2. KCU: K(11)→C(3) = −8, C→U(21) = +18. PHZ: P(16)→H(8) = −8, H→Z(26) = +18.
3. QAI: Q(17)→A(1) = −16, A→I(9) = +8 – different.

Trick: Pattern “−8, +18” in three clusters; QAI breaks it.

Answer: (B) QAI

Concept: Vowels move +1, consonants move −1 in the alphabet.

1. N O T F E R N → N−1 = M, O+1 = P, T−1 = S, F−1 = E, E+1 = F, R−1 = Q, N−1 = M.
2. New group: M P S E F Q M.
3. Only M appears exactly twice → one letter.

Trick: The two N’s both become M; no other letter repeats.

Answer: (B) One

Concept: Letters common to both words get the common digits; the leftover letter gets the leftover digit.

1. LOTS → 8 4 3 9; JOLS → 4 8 5 3.
2. Common letters O, L, S ↔ common digits {8, 4, 3}.
3. Leftover in JOLS is J ↔ leftover digit 5.

Trick: Digits need not be in letter order – just compare the sets.

Answer: (C) 5

Concept: Compare the shift of each letter from the first cluster to the second.

1. TR − OM: T→O = −5, R→M = −5.
2. PN − KI: −5, −5. XV − SQ: −5, −5.
3. US − OP: U→O = −6, S→P = −3 – odd one.

Trick: Three pairs have a uniform −5 shift; check the first option last if it looks different.

Answer: (A) US−OP

Concept: Find the letter-wise shift in the given pair and apply it to the new cluster.

1. LOSM → QTXR: L+5 = Q, O+5 = T, S+5 = X, M+5 = R.
2. ILPJ → NQUO: each letter +5 (confirms rule).
3. FIMG: F+5 = K, I+5 = N, M+5 = R, G+5 = L → KNRL.

Trick: Uniform +5; options with the same letters in a different order (KNLR, NKRL) are traps.

Answer: (C) KNRL

Concept: Decode left to right and draw the family tree.

1. A ÷ B: A is sister of B. B × C: B is husband of C.
2. C + D: C is mother of D; so B and C are parents of D.
3. D – E: D is brother of E; E ÷ F: E is sister of F → D, E, F are siblings, children of B and C.
4. A is the sister of B, F’s father → A is F’s father’s sister.

Trick: Siblings share parents – once B is D’s father, he is F’s father too.

Answer: (A) Father’s sister