Q&A: practice questions with answers

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

1. $\text{SP}=546\times 1.6=873.6$
2. $\text{MP}\times 0.78=873.6$
3. $\text{MP}=\frac{873.6}{0.78}=1120$

Trick: $\frac{\text{MP}}{\text{CP}}=\frac{100+P}{100-D}=\frac{160}{78}$, and $546=78\times 7$, so MP $=160\times 7=1120$.

Answer: (B) 1,120

Concept: Income = Expenditure + Savings. Savings 30% ⇒ expenditure is 70% of income.

1. $70\%\text{ of income}=840$
2. $\text{Income}=\frac{840\times 100}{70}=1200$

Trick: $70\%\to 840$ ⇒ $10\%\to 120$ ⇒ $100\%\to 1200$.

Answer: (D) 1,200

Concept: For $a{x}^2+bx+c=0$, discriminant $D=b^2-4ac$ decides the nature of roots.

1. $D>0$: real and distinct; $D=0$: real and equal
2. $D<0$: $\sqrt{D}$ is imaginary, so roots are $\frac{-b\pm i\sqrt{|D|}}{2a}$ – a pair of complex conjugates (for real coefficients).

Trick: Negative D ⇒ no real roots ⇒ complex pair $p\pm iq$.

Answer: (B) complex conjugates

Concept: When two words share a letter, the digit common to both codes is the code of that letter.

1. CART → 4268 and RISE → 3527.
2. The only common letter is R.
3. The only common digit in the two codes is 2.

Trick: Common letter ↔ common digit; ignore order of digits.

Answer: (A) 2

Concept: Fix genders from marriages and 'brother' clues, then trace the path from B to A.

1. D is a brother (male), so his wife R is female – B's mother.
2. E is R's sister, so E is B's mother's sister (maasi).
3. A is married to E, so A is the husband of B's mother's sister.

Trick: Mother's sister's husband = maasa (मौसा).

Answer: (A) Mother's sister's husband

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

1. HUNT → 6438 and USER → 5723.
2. Common letter: U.
3. Common digit: 3.

Trick: Match common letter with common digit.

Answer: (A) 3

Concept: Check each number: the element just before it and just after it must both be numbers.

1. Run 4 3 5 6 7: inside numbers 3, 5, 6 → 3 qualify.
2. Run 9 2 7 5 9 (after %): inside numbers 2, 7, 5 → 3 qualify.
3. 1 3 (between * and %) and lone 7 have a symbol on one side.

Trick: In a run of k consecutive numbers, exactly $k-2$ qualify: $(5-2)+(5-2)=6$.

Answer: (B) 6

Concept: Look only at symbols; each must have a number immediately on both sides.

1. 3 * 2 → * qualifies.
2. 9 & 7 → & qualifies.
3. 9 @ 4 → last @ qualifies; other @, © and # touch another symbol or the end.

Trick: Scan for the pattern number–symbol–number.

Answer: (D) 3

Concept: Find the step pattern inside each set.

1. (22, 28, 31): $22+6=28,28+3=31$
2. (39, 45, 48): $39+6=45,45+3=48$
3. (27, 33, 36): $27+6=33,33+3=36$ ✓; others fail (e.g. 53, 59, 65 is +6, +6).

Trick: Pattern +6, +3 → also first and third differ by 9.

Answer: (D) (27, 33, 36)

Concept: Draw the minimum Venn diagram consistent with all statements.

1. Sparrows ⊂ Parrots ⊂ Macau, so every sparrow is a macau → I follows.
2. Some sparrows are doves; those sparrows are parrots, so some doves are parrots → II follows.

Trick: 'All A are B' + 'Some A are C' → 'Some C are B' is always definite.

Answer: (A) Both conclusions (I) and (II) follow

Concept: Find the shift from the right pair's rule and test the options.

1. Take # = UZV: U−3 = R, Z−3 = W, V−3 = S → RWS ✓
2. Apply the same −3 to INJ: I−3 = F, N−3 = K, J−3 = G → FKG
3. So # = UZV and % = FKG.

Trick: Positions: U(21)→R(18), I(9)→F(6): a uniform −3.

Answer: (C) # = UZV, % = FKG

Concept: Track coordinates (east +x, north +y) after every turn.

1. Y(0, 0) → E 22 → (22, 0); right = S 33 → (22, −33); left = E 25 → (47, −33)
2. Left = N 51 → (47, 18); left = W 47 → (0, 18); left = S 23 → Z(0, −5)
3. Z is 5 km directly south of Y, so he must go 5 km north.

Trick: Net E–W: $22+25-47=0$; net N–S: $-33+51-23=-5$ → 5 km north to return.

Answer: (A) 5 km towards the north

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

1. Two below T → T = 3; one above U → U = 6; one box between U and D → D = 4.
2. Free: 1, 2, 5, 7. G just above S → only S = 1, G = 2 works.
3. F below E → F = 5, E = 7.
4. Between E (7) and G (2): U, F, D, T → four boxes.

Trick: Boxes between positions p and q $=|p-q|-1=7-2-1=4$.

Answer: (D) Four

Concept: Facing the centre, a person's left is clockwise and right is anticlockwise.

1. Fix T; D is just clockwise of T (T is D's immediate right).
2. S: three people between T and S going to T's left → S is 4th seat clockwise of T.
3. C must be 4 seats from D → the seat after S; A just right of R gives order T, D, A, R, S, C, B clockwise.
4. From R going left (clockwise): S, C, then B → two people.

Trick: In a 7-seat circle, '3 between' one way means 2 between the other way.

Answer: (B) Two

Concept: Change each digit by the rule, then read the required positions.

1. 7648135 → 7+1, 6−2, 4−2, 8−2, 1+1, 3+1, 5+1
2. New number: 8 4 2 6 2 4 6
3. 2nd from left = 4, 2nd from right = 4 → $4-4=0$

Trick: Only the 2nd and 6th digits matter: 6→4 and 3→4.

Answer: (C) 0

Concept: Check the differences between consecutive terms.

1. Differences: 7, 14, 28, 56 – each doubles.
2. Next difference $=112$
3. $107+112=219$

Trick: Also $a_{n+1}=2a_n+5$: $2\times 107+5=219$.

Answer: (D) 219

Concept: Compare positions within and across the two clusters.

1. RP – AU: R−2 = P; A−6 = U (cyclic); R+9 = A.
2. IG – RL: I−2 = G; R−6 = L; I+9 = R. NL – WQ: N−2 = L; W−6 = Q; N+9 = W.
3. BZ – KF: B−2 = Z, B+9 = K, but K−6 = E, not F → odd one.

Trick: Pattern: (x, x−2) – (x+9, x+3); BZ–KF breaks the last step.

Answer: (C) BZ – KF

Concept: Apply the rule digit by digit.

1. 7526318 → 7+2, 5+2, 2−1, 6−1, 3+2, 1+2, 8−1
2. New number: 9 7 1 5 5 3 7
3. Highest 9, lowest 1 → $9-1=8$

Trick: Odd digits rise, even fall – the old 2 becomes the new minimum 1.

Answer: (B) 8

Concept: Only the second digit from the left is needed.

1. 7235614 → 8 0 4 6 4 2 2
2. Second digit 2 is even → $2-2=0$

Trick: Change only the digit asked for – saves time.

Answer: (C) 0

Concept: Find separate rules for each letter and for the number.

1. HT 17 → GQ 6: H−1 = G, T−3 = Q, $17-11=6$
2. LX 9 → KU −2: L−1 = K, X−3 = U, $9-11=-2$
3. QR 6 → Q−1 = P, R−3 = O, $6-11=-5$ → PO −5

Trick: Rule: (−1, −3, −11).

Answer: (C) PO −5