Q&A: practice questions with answers

Concept: Opposite directions: first meeting time $=\frac{L}{v_1+v_2}$; both at the start together after LCM of their lap times.

1. Speeds: $\frac{400}{100}=4$ m/s and $\frac{400}{80}=5$ m/s; relative speed $=9$ m/s.
2. They meet every $\frac{400}{9}$ s; both return to the start at LCM(100, 80) $=400$ s.
3. Meetings in 400 s $=\frac{400}{400/9}=9$, but the 9th meeting is at the starting point itself, so 8 meetings occur before it.

Trick: Opposite directions, lap ratio $4:5$ ⇒ meetings per cycle $=4+5=9$; subtract 1 for the meeting at the start.

Answer: (B) 8

Concept: “X beats Y by $d$ m in an $L$ m race” ⇒ when X runs $L$, Y runs $L-d$; speed ratio $=L:(L-d)$.

1. X : Y $=750:740=75:74$.
2. Y : Z $=444:438=74:73$.
3. X : Z $=75:73$. When X runs 150 m, Z runs $150\times \frac{73}{75}=146$ m ⇒ X beats Z by 4 m.

Trick: The ratios chain neatly through 74, giving X : Z $=75:73$ – a 2 m gap per 75 m, so 4 m per 150 m.

Answer: (B) 4

Concept: B's one-day work $=\frac{1}{\text{together}}-\frac{1}{\text{A}}$.

1. $\frac{1}{35}-\frac{1}{60}=\frac{12-7}{420}=\frac{5}{420}=\frac{1}{84}$ ⇒ B alone takes 84 days.
2. One-third of the work: $\frac{84}{3}=28$ days.

Trick: B alone $=\frac{35\times 60}{60-35}=\frac{2100}{25}=84$; then divide by 3.

Answer: (A) 28

Concept: If the ratio is $7:2$, the ages are $7k$ and $2k$.

1. $7k=49\Rightarrow k=7$.
2. Son's age $=2k=14$ years.

Trick: $49÷7=7$ per ratio unit; son $=2\times 7=14$.

Answer: (A) 14 years

Concept: Cutting squares of side $a$ from the corners and folding gives a box of $(l-2a)\times (b-2a)\times a$; number of small boxes $=\frac{\text{volume of big box}}{\text{volume of one small box}}$.

1. Box dimensions: $(56-20)\times (40-20)\times 10=36\times 20\times 10$ ⇒ volume $=7200$ cm³.
2. Match box volume $=4\times 2\times 1.5=12$ cm³.
3. Number $=\frac{7200}{12}=600$.

Trick: Subtract twice the cut side from both length and breadth; the cut side becomes the height.

Answer: (B) 600 boxes

Concept: Track the position of each letter through the three steps.

1. BARS → ARBS: the first three letters shift one place left (B A R → A R B), the 4th letter stays.
2. ARBS → SRAB: the last letter jumps to the front, the 2nd stays, and the old 1st and 3rd swap (A R B S → S R A B). FLAM → LAFM → MALF follows the same rule.
3. Apply to KEAT: K E A → E A K gives EAKT ✓; then E A K T → T A E K ✓.
4. BALK → ABLK fails at step 1 (should be ALBK); MIND and WAST also break the rule.

Trick: Write positions as numbers: step 1 = 2-3-1-4, step 2 = 4-2-1-3. Test only step 1 first – it kills most options.

Answer: (A) KEAT - EAKT - TAEK

Concept: Facing the centre, "immediate left" is the next seat clockwise and "immediate right" the next seat anticlockwise. Fix one person and place the rest.

1. Statement I: Gauri is on Lovely's right, and Ragini–Kajal sit together. The remaining three seats take Ragini, Kajal and Navya, so Lovely's left neighbour could be Kajal or Navya – not sufficient.
2. Statement II: put Kajal at a seat; Navya is 2nd to her left. Gauri must touch both Navya and Lovely: the only free seat next to Navya that also has a free neighbour fixes Gauri, then Lovely. Ragini takes the last seat.
3. This gives the order (clockwise) Navya, Gauri, Lovely, Kajal, Ragini – the immediate left of Lovely is Kajal. Statement II alone is sufficient.

Trick: In data sufficiency, as soon as one statement leaves two possibilities, mark it "not sufficient" and move on; do not try to combine unless both fail.

Answer: (C) Data in Statement II alone is sufficient to answer the question while data in statement I is not

Concept: Check each number's left neighbour (must be a letter) and right neighbour (must be a symbol).

1. Numbers in the series: 2, 7, 7, 4, 5, 4.
2. 2 → S, U (letter, letter) ✗; 7 → B, N ✗; 7 → M, 4 ✗; 4 → 7, C ✗.
3. 5 → Z (letter) and @ (symbol) ✓; last 4 → U, D ✗.
4. Only one such number: 5.

Trick: Underline only the numbers first, then glance at both sides of each – never scan the whole string blindly.

Answer: (A) One

Concept: Compare first letters across the triad and second letters across the triad separately.

1. CF – GI – KL: first letters C → G → K (+4, +4); second letters F → I → L (+3, +3).
2. MP – QS – UV: M → Q → U (+4, +4); P → S → V (+3, +3). Same pattern.
3. OR – SU – WX: O → S → W (+4, +4) and R → U → X (+3, +3) ✓.
4. RM–QT–PR goes backwards, BF–EI–HL uses +3/+3, PL–RN–TP uses +2/+2.

Trick: In two-letter triads, always separate "first-letter chain" and "second-letter chain" – the gaps are usually different for the two.

Answer: (D) OR – SU – WX

Concept: Elder = born earlier. Arrange the 7 months as positions 1 (Jan) to 7 (Dec) and fill the fixed people first.

1. "Only two elder than M" → M is 3rd (May). "Only two between L and M" → L is 6th (Sep) (position 0 does not exist).
2. "Only U is elder than O" → U 1st (Jan), O 2nd (Mar).
3. Remaining V, T, N in positions 4, 5, 7 with V elder than T elder than N → V Jun, T Jul, N Dec.
4. Between T (5th) and U (1st) sit O, M, V → three people.

Trick: "Only two are elder than X" pins X to the 3rd slot instantly; start every ranking puzzle with such absolute clues.

Answer: (B) Three

Concept: Number the 8 seats 1–8 clockwise (corners odd, mid-sides even). Facing the centre: right = anticlockwise, left = clockwise.

1. H at 1. B 2nd to the right → seat 7 (a corner, as required). D 3rd to the left of H → seat 4 (mid-side). G is at a corner → 3 or 5.
2. F touches both E and B: F at 6 with E at 5 (F at 8 would need E at 7 or 1, occupied). So G = 3.
3. A and C fill seats 2 and 8 with C 2nd to the right of A → A = 2, C = 8.
4. From D (4) going left (clockwise) to A (2): E, F, B, C, H → five people.

Trick: "Not at the centre of a side" tells you the parity of all corner seats at once – fix corners/mid-sides before placing anyone else.

Answer: (A) Five

Concept: Replace the signs as told, then apply BODMAS (÷ and × before + and −).

1. 35 + 10 × 2 − 15 ÷ 3 becomes $35-10÷2+15\times 3$.
2. $10÷2=5$, $15\times 3=45$.
3. $35-5+45=75$.

Trick: Rewrite the whole expression with new signs first, then solve – never swap "in your head" midway.

Answer: (A) 75

Concept: Draw the tree generation by generation; a woman's husband's children are her own children unless stated otherwise.

1. L → M → O; O is the sister of H, so H is also M's daughter.
2. H is the wife of P, and P is the father of F → F is the child of H and P. F is a mother, so F is female.
3. M is the mother of H, and H is the mother of F → M is F's mother's mother (maternal grandmother).

Trick: Always travel through the female line when the question says "mother": M → H → F gives the answer in two hops.

Answer: (C) Mother’s mother

Concept: Swap + with − and × with ÷, then solve by BODMAS (brackets first).

1. New expression: $81÷9+14\times 2-(60÷6)\times 3+49-15$.
2. Bracket: $60÷6=10$; then $81÷9=9$, $14\times 2=28$, $10\times 3=30$.
3. $9+28-30+49-15=41$.

Trick: Add all positives ($9+28+49=86$) and all negatives ($30+15=45$) separately, then subtract: $86-45=41$.

Answer: (B) 41

Concept: To name the tallest, every one of the five must be shown shorter than one person.

1. Statement I: K > J > H, I. L is not mentioned, so L could be taller than K – not sufficient.
2. Statement II: K > L > J. H and I are not mentioned – not sufficient.
3. Together: K > J > H, I and K > L → K is taller than all four → K is the tallest. Both statements are needed.

Trick: For "who is tallest/shortest" count the unplaced names; one missing name means "not sufficient".

Answer: (C) Both statements I and II together (and not statement I alone or statement II alone) are sufficient to answer the question

Concept: Convert letters to their alphabet numbers and look at the gaps between consecutive letters.

1. RUP: 18, 21, 16 → +3, −5.
2. ORM: 15, 18, 13 → +3, −5. GJE: 7, 10, 5 → +3, −5.
3. LMK: 12, 13, 11 → +1, −2 – different pattern.

Trick: In three-letter clusters, write the two gaps (with sign) under each option; the odd one shows a different pair.

Answer: (D) LMK

Concept: Number seats from the left end (position 1) and place people one clue at a time; the last position fixed by "extreme end" gives the total.

1. E = 1. Seven between E and F → F = 9. A is 3rd to the right of F → A = 12.
2. C is exactly between E and F → C = 5. Four between C and B → B = 10 (B = 0 impossible).
3. Four between B and D → D = 15 (D = 5 is taken by C). D is at an extreme end, so the row ends at 15.

Trick: When the total is unknown, never guess it – anchor at the given end and let the farthest person define the length.

Answer: (C) 15

Concept: Number seats 1–8 clockwise. Facing the centre: left = clockwise, right = anticlockwise. "Only three between J and W" means they sit opposite each other.

1. Put J at a mid-side seat (seat 2) → W opposite at seat 6. I is 3rd to the left of W → seat 1. H is 2nd to the left of J → seat 4.
2. V touches both W (6) and H (4) → seat 5. U touches J → seat 3 (seat 1 is I). X is 2nd to the left of W → seat 8. K gets seat 7.
3. From K (7) counting leftwards (clockwise) to H (4): X, I, J, U → four people.

Trick: "Only three between A and B" in an 8-seat circle always means A and B are face to face – place that pair first.

Answer: (C) Four

Concept: Substitute the letters by their operators, then apply BODMAS.

1. 24 C 16 D 8 A 4 B 2 → $24+16-8÷4\times 2$.
2. $8÷4=2$, then $2\times 2=4$ (÷ and × left to right).
3. $24+16-4=36$.

Trick: Write the decoded expression on paper before computing; most errors come from mixing D (−) and A (÷).

Answer: (C) 36

Concept: Rank the six from highest (1) to lowest (6) using the absolute clues first.

1. "Only two scored more than Y" → Y is 3rd. "Only S scored less than P" → P 5th, S 6th.
2. Z, T, Q remain for ranks 1, 2, 4 with Z > T > Q → Z 1st, T 2nd, Q 4th.
3. Only Z scored more than T → one person.

Trick: Count the slots: once Y, P, S are fixed, only three slots remain and the chain Z > T > Q drops straight into them.

Answer: (C) One