Q&A: practice questions with answers

Concept: Equal SI for equal time means principal × rate is the same for both parts.

1. Let the sum at $14\%$ be $x$: $x\times 14=(21300-x)\times 11$
2. $25x=234300\Rightarrow x=9372$

Trick: Principals are in inverse ratio of rates, $11:14$: $\frac{11}{25}\times 21300=9372$.

Answer: (A) ₹9,372

Concept: Average of the first $n$ even numbers $=n+1$.

1. Sum of first $n$ even numbers $=n(n+1)$
2. Average $=\frac{n(n+1)}{n}=n+1=161+1=162$

Yaad rakhein: First $n$ even numbers: average $n+1$; first $n$ odd numbers: average $n$.

Answer: (D) $162$

Concept: Profit share is proportional to capital × time.

1. M + N $=6915$, M − N $=461\Rightarrow$ M $=3688$, N $=3227$; ratio $=8:7$
2. Let N's capital $=x$, M's $=x+38000$: $\frac{2(x+38000)}{3x}=\frac{8}{7}$
3. $14x+532000=24x\Rightarrow x=53200$
4. M's capital $=53200+38000=91200$

Trick: $3688=8\times 461$ and $3227=7\times 461$ – spotting the common factor gives $8:7$ instantly.

Answer: (C) ₹91,200

Concept: Expand the brackets and combine like terms.

1. $x(2x-8)=2x^2-8x$
2. $5(x^2-3)=5x^2-15$
3. Total: $2x^2-8x+5x^2-15+12=7x^2-8x-3$

Trick: Constant term check: $-15+12=-3$ removes options (A) and (D); $x^2$ coefficient $+7$ removes (B).

Answer: (C) $7x^2-8x-3$

Concept: $SI=\frac{P\times R\times T}{100}$, with $T$ in years $=\frac{\text{days}}{366}$ in a leap year (count the day of deposit out, day of withdrawal in).

1. 2024 is a leap year: Feb has 29 days. Days: Feb 3 (27, 28, 29) + Mar 31 + Apr 27 $=61$
2. $SI=\frac{4000\times 6.75\times 61}{100\times 366}$
3. $=\frac{270\times 61}{366}=\frac{16470}{366}=45$

Trick: $366=6\times 61$, so $\frac{61}{366}=\frac{1}{6}$ year: $SI=\frac{270}{6}=45$.

Answer: (D) ₹45

Concept: Product of two numbers $=\text{HCF}\times \text{LCM}$.

1. Other number $=\frac{83\times 498}{166}$
2. $=\frac{498}{2}=249$

Trick: Both numbers must be multiples of the HCF 83: $249=3\times 83$ ✔, while 230 is not.

Answer: (D) $249$

Concept: $\text{Average speed}=\frac{\text{total distance}}{\text{total time}}$.

1. Times: $\frac{200}{25}=8$ h, $\frac{210}{21}=10$ h, $\frac{204}{51}=4$ h; total $22$ h
2. Distance $=200+210+204=614\text{km}$
3. Average speed $=\frac{614}{22}\approx 27.91\text{km/h}$

Trick: Never average the three speeds directly – that gives 32.33, which is wrong.

Answer: (C) $27.91$

Concept: Test divisibility of odd numbers by primes up to $\sqrt{476}\approx 21.8$.

1. Even numbers 468, 470, 472, 474, 476 are not prime; 475 ends in 5
2. $469=7\times 67$, $471=3\times 157$, $473=11\times 43$
3. No prime lies in this range.

Trick: Check 3 (digit sum), 7 and 11 first – they catch most composite numbers.

Answer: (A) $0$

Concept: Replace letters with signs, then apply BODMAS.

1. $18÷3-8\times 2+12$
2. $=6-16+12$
3. $=2$

Trick: Do ÷ and × first ($6$ and $16$), then + and − left to right.

Answer: (C) $2$

Concept: Decode each symbol from left to right and build a family tree.

1. W $ A: W is the brother of A
2. A = C: A is the mother of C
3. C @ D: C is the wife of D; D * R: D is the father of R, so C is R's mother
4. A is R's mother's mother; W (A's brother) is R's mother's mother's brother

Trick: Climb the chain from R upward: mother (C) → her mother (A) → her brother (W).

Answer: (C) Mother's mother's brother

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

1. DELT → {1, 4, 7, 3}; BLED → {7, 2, 3, 4}
2. Common letters D, E, L ↔ common digits 3, 4, 7
3. DELT has extra T ↔ extra 1; BLED has extra B ↔ extra 2

Trick: Cancel the common part; what remains in BLED's code (2) is B.

Answer: (D) 2

Concept: Fix the definite positions first, then place the pairs.

1. Only S above I $\Rightarrow$ S = 1st, I = 2nd from top
2. Four boxes between I and K $\Rightarrow$ K = 7th
3. M and G (two between) must take 3rd and 6th; H, L (H just above L) take 4th, 5th
4. G not immediately above H $\Rightarrow$ G = 6th, M = 3rd. Order: S, I, M, H, L, G, K

Trick: Only the pair 3–6 has exactly two gaps among the free slots 3, 4, 5, 6.

Answer: (B) H

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

1. Fix S; O is second to the left of S; R is third to the left of O
2. Q is third to the right of L. L must sit immediately to the right of S, which puts Q immediately to the left of O
3. N is a neighbour of Q, M takes the last seat: clockwise S, M, O, Q, N, R, L
4. Third to the right of M (anticlockwise): S, L, R

Trick: The other placement of L–Q leaves no free seat beside Q for N, so it is rejected.

Answer: (C) R

Concept: Track each letter position of the clusters separately.

1. 1st letters: F → A → V → Q, each $-5$ (A $-5$ wraps to V); next Q $-5$ = L
2. 2nd letters: I → D → Y → T, each $-5$; next O
3. 3rd letters: L → G → B → W, each $-5$; next R

Trick: Each cluster is the previous one shifted back by 5 letters: QTW → LOR.

Answer: (B) LOR

Concept: Find the letter-wise shifts in the given pairs.

1. MUV → QPZ: M $+4$ = Q, U $-5$ = P, V $+4$ = Z
2. UKD → YFH: U $+4$ = Y, K $-5$ = F, D $+4$ = H
3. CAL: C $+4$ = G, A $-5$ = V (wrap), L $+4$ = P → GVP ✔

Trick: Check the middle letter ($-5$) first – it rejects TLW (L → Q) and SLU at once.

Answer: (C) CAL : GVP

Concept: A letter common to both words is coded by the digit common to both codes.

1. LOAN → {4, 2, 5, 7}; NIKE → {1, 3, 4, 6}
2. Only common letter: N; only common digit: 4
3. So N = 4

Trick: One common letter ↔ one common digit.

Answer: (D) 4

Concept: Rearrange the digits and read positions carefully from each end.

1. Ascending order: 1 2 3 4 5 6 7 9
2. Third from left = 3
3. Fourth from right (8 digits) = 5th from left = 5
4. Sum $=3+5=8$

Trick: $k$th from right $=(n-k+1)$th from left; here $8-4+1=5$.

Answer: (C) 8

Concept: Use positions: in each pair XY-ZW, compare Y, Z, W with X.

1. NQ-SL: N14 → Q17 ($+3$), S19 ($+5$ from N), L12 ($-7$ from S)
2. JM-OH: J10 → M13 ($+3$), O15 ($+5$), H8 ($-7$)
3. FI-KD: F6 → I9, K11, D4 – same pattern
4. CF-HB: C3 → F6, H8, but B2 is H $-6$ – odd

Trick: Only the last letter breaks the pattern, so compare 4th letter with 3rd first.

Answer: (C) CF-HB

Concept: Look at the differences between consecutive terms.

1. $561\stackrel{+1}{\to }562\stackrel{+2}{\to }564\stackrel{+3}{\to }?$
2. $564+3=567$; check $567+4=571$, $571+5=576$ ✔

Trick: Differences 1, 2, 3, 4, 5 – a simple increasing pattern.

Answer: (C) 567

Concept: Use 'only X above / below' clues to fix the ends first.

1. Only C above D $\Rightarrow$ C = 1st, D = 2nd (from top)
2. Two boxes between D and E $\Rightarrow$ E = 5th
3. Only A below B $\Rightarrow$ A = 7th, B = 6th
4. F not immediately above E $\Rightarrow$ F = 3rd, G = 4th. Between B and F: G, E

Trick: Two 'only' clues fix four boxes at once – start there.

Answer: (B) Two