Q&A: practice questions with answers

Concept: Innermost brackets first, then ÷/× before +/−.

Solution:

1. $24÷6=4\to (120-4)\times 2=232$; $50÷5+6=16\to 232-16=216$.
2. $18+24-8=34$; $9÷3+5=8\to 34-8=26$.
3. Total = $216+26=242$.

Yaad rakhein: Inside each bracket solve ÷ first — in $24÷(4+2)$, compute $(4+2)$ first.

Answer: (B) 242

Concept: Fourth proportional of a, b, c = $\frac{bc}{a}$.

Solution:

1. $x=7.5\times \frac{6.3}{3.5}$.
2. $\frac{6.3}{3.5}=1.8$.
3. $x=7.5\times 1.8=13.5$.

Trick: Spot $\frac{6.3}{3.5}=\frac{9}{5}$ first, then $7.5\times \frac{9}{5}=13.5$.

Answer: (A) 13.5

Concept: $\text{CSA}=2\pi rh,V=\pi r^{2}h$, so $V=\frac{\text{CSA}\times r}{2}$.

Solution:

1. $r=\frac{775.8}{2\times 3.14\times 24}=\frac{775.8}{150.72}\approx 5.1473$ cm.
2. $V=\frac{775.8\times 5.1473}{2}$.
3. $V\approx 1996.63{\text{cm}}^3$.

Trick: $V=\frac{\text{CSA}\times r}{2}$ skips recomputing π and h.

Answer: (D) 1996.63 cm3

Concept: Savings % = 100 − total spending %.

Solution:

1. Spending = $25+20+15=60\%$.
2. Savings = 40% of 84,000 = ₹33,600.

Trick: 10% of 84,000 = 8,400 → $40\%=4\times 8,400=33,600$.

Answer: (C) 33,600

Concept: Opposite direction: relative speed = sum; distance = sum of lengths.

Solution:

1. Train speed = $\frac{200}{10}=20$ m/s.
2. Other train = $60\times \frac{5}{18}=\frac{50}{3}$ m/s; relative = $20+\frac{50}{3}=\frac{110}{3}$ m/s.
3. Distance = $200+350=550$ m.
4. Time = $550\times \frac{3}{110}=15$ s.

Trick: 20 m/s = 72 km/h; $72+60=132$ km/h $=\frac{110}{3}$ m/s → $550÷\frac{110}{3}=15$ s.

Answer: (A) 15 seconds

Concept: Build the family tree step by step.

Solution:

1. Rajat, Garima and Harish are siblings → Rajat is Harish's brother.
2. Harish is Jyoti's father; Jyoti is Karan's wife.
3. So Rajat = Karan's wife's father's brother.

Trick: Start from Karan: wife (Jyoti) → her father (Harish) → his brother (Rajat).

Answer: (B) Wife's father's brother

Concept: Fix definite positions first (left-right counts).

Solution:

1. Positions 1–7 from left: E = 5 (4 on left), L = 3 (4 on right).
2. One between K and L → K = 1 (5 is taken).
3. V immediately left of A → only free pair 6–7: V = 6, A = 7.
4. C not left of N → N = 2, C = 4; third right of C = 7 = A.

Trick: Seat counts like 'four to the left' give exact positions — place them first.

Answer: (A) A

Concept: Replace letters with signs, then apply BODMAS.

Solution:

1. $3\times 2-12÷4+3$.
2. $=6-3+3=6$.

Yaad rakhein: After substitution always do ÷ and × before + and −.

Answer: (A) 6

Concept: Rearrange letters alphabetically, then count positions.

Solution:

1. LANDMARK → A A D K L M N R.
2. 4th from left = K; 2nd from right = N.
3. Letters between K and N in the alphabet: L, M = 2.

Yaad rakhein: K = 11, N = 14 → letters between = 14 − 11 − 1 = 2.

Answer: (A) 2

Concept: Track each letter position separately.

Solution:

1. 1st: D, X, R, L → −6 each → F.
2. 2nd: G, K, O, S → +4 → W.
3. 3rd: Y, T, O, J → −5 → E.
4. 4th: L, Q, V, A → +5 → F. Answer FWEF.

Trick: First letter alone (F) leaves FVDF/FWEF; 2nd letter +4 → W picks FWEF.

Answer: (D) FWEF

Concept: Swap the signs first, then apply BODMAS.

Solution:

1. $3÷5+16\times 8-8$ becomes $3\times 5-16÷8+8$.
2. $=15-2+8=21$.

Yaad rakhein: Do not solve before swapping — rewrite the full expression first.

Answer: (A) 21

Concept: Find the rule linking the numbers in given sets.

Solution:

1. (5, 18, 24): 5 + 13 = 18, 18 + 6 = 24.
2. (9, 22, 28): 9 + 13 = 22, 22 + 6 = 28.
3. (8, 21, 27): 8 + 13 = 21, 21 + 6 = 27 ✓.

Trick: Check differences: given sets have +13, +6; only (8, 21, 27) shows +13, +6.

Answer: (D) (8, 21, 27)

Concept: Decode each symbol and chain the relations.

Solution:

1. H + S: H is mother of S.
2. S − T: S is sister of T → H is also T's mother.
3. T × G: T is father of G.
4. So H is G's father's mother. (G ÷ P only tells G is P's daughter.)

Trick: Read from G backwards: G's father = T, T's mother = H.

Answer: (B) Father’s mother

Concept: Total = rank from top + rank from bottom − 1.

Solution:

1. Total = 64 + 23 − 1 = 86.

Yaad rakhein: Subtract 1 because the person is counted in both ranks.

Answer: (C) 86

Concept: Compare positional shifts of letters.

Solution:

1. WO–QI: −6, −6; TL–NF: −6, −6; YQ–SK: −6, −6.
2. ME–HA: M→H = −5, E→A = −4.
3. ME–HA breaks the −6 pattern.

Yaad rakhein: Learn EJOTY (5, 10, 15, 20, 25) to get letter positions fast.

Answer: (A) ME – HA

Concept: Common letter ↔ common digit.

Solution:

1. KEPT = 7523, OKAY = 4136.
2. Common letter: K; common digit: 3.
3. So K = 3.

Trick: Ignore order; just match the shared letter with the shared digit.

Answer: (C) 3

Concept: Facing centre: left = clockwise, right = anticlockwise.

Solution:

1. Number seats clockwise; Ay = 0, Cv (immediate right) = 6, Du (third left) = 3.
2. Bk next to Cv and Fo → Bk = 5, Fo = 4.
3. Gw not next to Du → Gw = 1, Es = 2.
4. Es is two seats anticlockwise from Fo → second to the right.

Trick: Fix the person named most often (Ay) first, then place others around.

Answer: (D) Second to the right

Concept: Use coordinates: East +x, North +y.

Solution:

1. E 58 → (58, 0); left N 28 → (58, 28); left W 40 → (18, 28).
2. Left S 71 → (18, −43); right (W) 18 → (0, −43) = Z.
3. Y is at (0, 0) → 43 km towards north.

Trick: Net x: 58 − 40 − 18 = 0; net y: 28 − 71 = −43 → go 43 km north.

Answer: (A) 43 km towards north

Concept: Find letter-wise shift.

Solution:

1. CVT → DXW: +1, +2, +3.
2. OBL → PDO: +1, +2, +3.
3. NAU → OCX: +1, +2, +3 ✓.

Trick: Check the third letter first (+3): only U→X fits.

Answer: (C) NAU : OCX

Concept: Use coordinates: East +x, North +y.

Solution:

1. W 11 → (−11, 0); right N 8 → (−11, 8); right E 14 → (3, 8).
2. Right S 11 → (3, −3); right W 3 → (0, −3) = P.
3. A is at (0, 0) → 3 km north.

Trick: Net x: −11 + 14 − 3 = 0; net y: 8 − 11 = −3 → 3 km north.

Answer: (B) 3 km to the north