Q&A: practice questions with answers

Concept: Count the day of withdrawal but not the day of deposit; 2024 is a leap year (366 days).

1. Days: February $26$–$29$ = 4, March = 31, April = 26 → total 61 days.
2. Time $=\frac{61}{366}=\frac{1}{6}$ year.
3. SI $=\frac{4000\times 6.75\times 1}{100\times 6}=\frac{270}{6}=45$.

Yaad rakhein: In a leap year, 61 days $=\frac{1}{6}$ year exactly.

Answer: (C) ₹45

Concept: For the same time period, profit is shared in the ratio of investments.

1. Total investment $=1050+1320+1730=4100$.
2. Deepa's share $=\frac{1730}{4100}\times 1230$.
3. $\frac{1230}{4100}=0.3$, so share $=1730\times 0.3=519$.

Trick: Profit is $30\%$ of total capital, so each gets $30\%$ of own investment.

Answer: (B) ₹519

Concept: One-day work of C $=$ (A+B+C) one-day work $-$ A's $-$ B's.

1. A $=\frac{1}{22}$, B $=\frac{1}{11}=\frac{2}{22}$ → A+B $=\frac{3}{22}$.
2. C $=\frac{1}{5}-\frac{3}{22}=\frac{22-15}{110}=\frac{7}{110}$.
3. C alone $=\frac{110}{7}=15\frac{5}{7}$ days.

Trick: Total work = LCM$(22,11,5)=110$ units: A = 5, B = 10, all = 22 → C $=7$ units/day → $\frac{110}{7}$ days.

Answer: (A) $15\frac{5}{7}$ days

Concept: Expand each bracket, then combine like terms.

1. $3(\frac{7}{3}{x}^2-27x+14)=7x^2-81x+42$.
2. $-7(x^2+8x-11)=-7x^2-56x+77$.
3. Add: $(7x^2-7x^2)+(-81x-56x)+(42+77)=-137x+119$.

Trick: $x^2$ terms cancel; constant $42+77=119$ is positive, $x$ term negative → only (C).

Answer: (C) $-137x+119$

Concept: Arrange data in order; for $n$ odd, median is the ${\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Ascending: 17, 20, 21, 27, 35, 41, 42, 49, 68, 76, 95.
2. $n=11$ → $\frac{11+1}{2}=6^{\text{th}}$ value.
3. 6th value $=41$.

Yaad rakhein: Always sort first — 42 is the middle of the unsorted list, a trap.

Answer: (C) 41

Concept: Apply the rule to every digit separately, then count digits that occur more than once.

1. Odd digit $+2$, even digit $-1$.
2. 8354671 → 7 5 7 3 5 9 3.
3. 7, 5 and 3 each appear twice → three digits repeat.

Trick: Write the new digits under the old ones to avoid mistakes.

Answer: (D) Three

Concept: Find the position pattern: word 1 (1234) → word 2 → word 3.

1. NEST → ESNT: letters 2, 3, 1, 4. GONE → ONGE: same.
2. Third word is the first word reversed: NEST → TSEN, GONE → ENOG.
3. LIES → IELS (2, 3, 1, 4) → SEIL (LIES reversed) ✓.

Trick: Check the easy part first — third word $=$ reverse of first. Only LIES–SEIL and COPE–… fit; COPE reversed is EPOC, not OPEC.

Answer: (A) LIES - IELS - SEIL

Concept: Track each letter position and the number separately.

1. 1st letters: I, L, O, R, U ($+3$).
2. 2nd letters: C, G, K, O, S ($+4$).
3. 3rd letters: V, S, P, M, J ($-3$).
4. Numbers: 16, 5, $-6$, $-17$, $-28$ ($-11$).

Trick: Only (C) starts with U and has $-28$.

Answer: (C) USJ $-28$

Concept: Check every number: the element just before must be a symbol and the element just after must be a number.

1. $ 6 7 → 6 is preceded by $ and followed by 7 ✓.
2. @ 1 4 → 1 is preceded by @ and followed by 4 ✓.
3. \ 1 & fails (followed by symbol); % 9 has nothing after it; 7, 4, 2, 2 are preceded by numbers.

Trick: Look for the pattern Symbol–Number–Number only.

Answer: (B) Two

Concept: Draw the least-overlap Venn diagram; a conclusion follows only if it is true in every possible case.

1. Teas lie inside Bins; Socks lie outside Teas.
2. Socks may or may not overlap the Bins outside Teas → 'Some socks are bins' is only possible (I fails).
3. 'All bins are socks' is impossible, since teas are bins but not socks (II fails).

Trick: 'No A is C' + 'All A are B' gives no definite link between B and C.

Answer: (D) Neither conclusion (I) nor (II) follows.

Concept: Trace the path on axes; opposite moves cancel.

1. 19 km West, then right (North) 10 km, right (East) 11 km, right (South) 10 km.
2. North 10 and South 10 cancel → Z lies on the same line as Y.
3. Net West $=19-11=8$ km, so Y is 8 km East of Z.

Trick: Net displacement: West–East $=19-11=8$; North–South $=0$.

Answer: (D) 8 km towards east

Concept: Use alphabet positions and compare the gaps between letters.

1. HLF: $8\to 12\to 6$ ($+4$, $-6$). MQK: $13\to 17\to 11$ ($+4$, $-6$).
2. KOI: $11\to 15\to 9$ ($+4$, $-6$).
3. FJC: $6\to 10\to 3$ ($+4$, $-7$) ✗.

Trick: First-to-last gap is $-2$ in HLF, MQK, KOI (H–F, M–K, K–I) but $-3$ in FJC.

Answer: (D) FJC

Concept: Fix genders from words like son/father, then draw the family tree.

1. M is a son (male) and is married to A, so A is female; T is their daughter → M is T's father.
2. J is a father (male) and married to K, so K is female; M is K's son → K is M's mother.
3. K is T's father's mother (paternal grandmother).

Yaad rakhein: D, N and G are distractors — they do not affect the K–T link.

Answer: (A) Father's mother

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

1. 'like his attitude' → nt yg fr.
2. 'his first salary' → kt lo yg.
3. Common word: his; common code: yg.

Trick: Match one common word with one common code — no further work needed.

Answer: (A) yg

Concept: Total $=$ position from left $+$ position from right $-1$.

1. Rikshit 31st and Palak 32nd from right → they are adjacent, 0 people between.
2. So 0 people between Palak and Huma → Huma is next to Palak on the other side: 33rd from right.
3. Huma is 10th from left → total $=10+33-1=42$.

Trick: Huma cannot be 31st (Rikshit's place), so she must be 33rd from right.

Answer: (D) 42

Concept: Replace letters by signs, then follow BODMAS.

1. $46÷2+21-8\times 3$.
2. $=23+21-24$.
3. $=20$.

Trick: Do $÷$ and $\times$ first: 23 and 24, then $23+21-24=20$.

Answer: (B) 20

Concept: Change only the first digit, then test divisibility by the second digit.

1. 324 → first 6, second 2: $6÷2=3$ ✓.
2. 523 → 8, 2: $8÷2=4$ ✓.
3. 643 → 9, 4 ✗; 136 → 4, 3 ✗; 441 → 7, 4 ✗.
4. Count $=2$.

Trick: Second digit 2 always divides an even first digit — only 324 and 523 have it.

Answer: (D) 2

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

1. No box above B → B = 7. Three between B and N → N = 3; A just above N → A = 4.
2. Only C below O → C = 1, O = 2.
3. One between D and A → D = 2 or 6; 2 is taken → D = 6 (four between D and C ✓).
4. M = 5 → boxes below M: A, N, O, C = 4.

Trick: 'Only C is below O' fixes the bottom two boxes at once.

Answer: (B) Four

Concept: Codes are jumbled; compare common letters with common digits.

1. GALE → {7, 5, 1, 3}; FLEA → {5, 1, 9, 7}.
2. Common letters A, L, E ↔ common digits 7, 5, 1.
3. Leftover: G → 3 and F → 9.

Trick: The only letter unique to FLEA gets the only digit unique to its code: 9.

Answer: (B) 9

Concept: Number seats 1–7 from the left (all face north) and place definite clues first.

1. Two to the right of C → C = 5. V immediately left of C → V = 4.
2. Two between C and U → U = 2. Two between A and V → A = 1 or 7.
3. W, T must be adjacent (T right of W): if A = 7, free seats 1, 3, 6 have no adjacent pair ✗. So A = 1, W = 6, T = 7, B = 3.

Trick: Use the W–T pair to decide where A goes.

Answer: (D) B