Q&A: practice questions with answers

Concept: Track the position of the single 'c' in each 8-letter term.

1. c is at position 3 → 2 → 1: moving one place left.
2. After position 1 it wraps round to position 8.
3. Next term: bbbbbbbc.

Trick: Think of it as a circular rotation – the c 'falls off' the left end and reappears at the right.

Answer: (B) bbbbbbbc

Concept: Handle the letter part and number part separately.

1. Letter positions 1, 2, 4, 7 rise by 1, 2, 3 → next rise 4 gives 11th letter K.
2. Numbers $1,8,27,64=1^3,2^3,3^3,4^3$ → next $5^3=125$.
3. Term = K125.

Trick: Number part fixes 125 (A, B, D); letter K removes H and L.

Answer: (A) K125

Concept: Convert letters to positions and study the gaps.

1. Positions are 3, 6, 10, 15, 21 – the gaps grow 3, 4, 5, 6.
2. Next gap +7: 21 + 7 = 28.
3. 28 − 26 = 2 → B.

Trick: Past Z, subtract 26 to wrap around the alphabet.

Answer: (C) B

Concept: Track the 1st, 2nd and 3rd letters of each cluster separately.

1. 1st: S(19) → V(22) → Y(25) → +3 → $28-26=2$ = B.
2. 2nd: Q → R → S → T (+1).
3. 3rd: N(14) → J(10) → F(6) → −4 → 2 = B. Answer BTB.

Trick: Middle letter T alone leaves A or B; first letter B confirms A.

Answer: (A) BTB

Concept: Each term adds the next alphabet letter at the end.

1. AB, ABC, ABCD … ABCDEFG – lengths 2, 3, …, 7.
2. Next has 8 letters: ABCDEFGH.

Trick: Options B, C, D each swap two letters – just check alphabetical order.

Answer: (A) ABCDEFGH

Concept: Find the repeating block and fill blanks in order.

1. Letters k, l, m, n in a 12-letter string → block 'klmn' × 3.
2. klmn / klmn / klmn.
3. Blanks: l, n, k, n, l.

Trick: Length 12 with 4 distinct letters → try a block of 4 first.

Answer: (A) I, n, k, n, I

Concept: Each letter of a cluster moves the same step to the next cluster.

1. D→F→H, G→I→K, J→L→N: each +2.
2. Add 2 to each letter of HKN → JMP.

Trick: Inside each cluster letters differ by 3 (D-G-J); only JMP keeps that gap.

Answer: (A) JMP

Concept: Each letter moves +1 from one cluster to the next.

1. DGJ → EHK → FIL: every letter +1.
2. Next: G, J, M → GJM.

Trick: Internal gap of 3 (G-J-M) is kept only in option A.

Answer: (A) GJM

Concept: Replace each sign as per the code, then apply BODMAS.

1. Code: + → ÷, × → −, − → ×, ÷ → +.
2. $64+8\times 3-4÷8\to 64÷8-3\times 4+8$.
3. $=8-12+8=4$.

Trick: Pairs of signs usually swap with each other – write the converted expression first, then solve.

Answer: (B) 4

Concept: When differences themselves change steadily, find the next difference.

1. Differences: 12, 16, 20, 24 (increase by 4).
2. Next difference = 28.
3. 85 + 28 = 113.

Trick: Constant second difference (4) → quadratic pattern; just extend the gaps.

Answer: (A) 113

Concept: For identical items, overall % = average of the individual gain/loss %.

1. Gains/losses: +16%, +16%, −12%.
2. Net = $\frac{16+16-12}{3}=\frac{20}{3}\%=6\frac{2}{3}\%\approx 6.67\%$ profit.
3. The total cost ₹5,320 is not needed.

Trick: Equal cost prices → simply average the percentages (loss taken as negative).

Answer: (C) $6.67\%$ Profit

Concept: Successive discounts are applied one after another on the reduced price.

1. After 20%: $12000\times 0.8=$ ₹9600.
2. After 4% more: $9600\times 0.96=$ ₹9216.

Trick: Net discount = $20+4-\frac{20\times 4}{100}=23.2\%$ → $12000\times 0.768=$ ₹9216.

Answer: (B) Rs. 9216

Concept: $\text{SP}=\text{MP}\times \frac{100-\text{discount}\%}{100}$.

1. SP is 92% of MP.
2. $\text{MP}=2668\times \frac{100}{92}=29\times 100=$ ₹2900.

Trick: $2668÷92=29$ exactly – so MP = ₹2900. Check: 8% of 2900 = 232; $2900-232=2668$.

Answer: (A) ₹2900

Concept: Gain of old group above new average = loss of new group below it (balance method).

1. Old students: each 0.5 yr above 29.5 → total excess = $105\times 0.5=52.5$.
2. New students: each 1.5 yr below 29.5.
3. Number = $52.5÷1.5=35$.

Trick: Alligation: $(30-29.5):(29.5-28)=1:3$ = new : old → new = $\frac{105}{3}=35$.

Answer: (B) 35

Concept: Total work fixed → output per day × days = constant.

1. Productive days = 70% of 540 = 378.
2. New : old output = 540 : 378 = 10 : 7.
3. Increase = $\frac{3}{7}\times 100=42\frac{6}{7}\%$.

Trick: Days become $\frac{7}{10}$, so output must become $\frac{10}{7}$ → rise of $\frac{3}{7}=42\frac{6}{7}\%$ (option shown as '4267%').

Answer: (A) $42\frac{6}{7}\%$

Concept: $n(M\cup P)=n(M)+n(P)-n(M\cap P)$.

1. Like Maths or Physics = $60+50-20=90\%$ → remaining 10% = 50 students.
2. Computer Science = 30% of 50 = 15.
3. No preference = $50-15=35$.

Trick: Directly: $500\times 10\%\times 70\%=35$.

Answer: (A) 35

Concept: A perfect square never ends in 2, 3, 7, 8, and a square ending in 0 must end in an even number of zeros.

1. $1600={40}^2,625={25}^2,1225={35}^2$.
2. 840 ends in a single zero → cannot be a square (${28}^2=784,{29}^2=841$).

Trick: Odd number of trailing zeros → not a perfect square; 840 is out instantly.

Answer: (D) 840

Concept: Equal time → profit shared in the ratio of capital.

1. Q's 4 parts = ₹1200 → 1 part = ₹300.
2. Total = 9 parts = $9\times 300=$ ₹2700.

Trick: Total = $1200\times \frac{9}{4}=$ ₹2700.

Answer: (A) 2700

Concept: $\text{Share }\%=\frac{\text{own capital}}{\text{total capital}}\times 100$ (same time).

1. Total capital = $16000+20000+24000=$ ₹60,000.
2. R's share = $\frac{24000}{60000}\times 100=40\%$.

Trick: Ratio 4 : 5 : 6 → $R=\frac{6}{15}=\frac{2}{5}=40\%$.

Answer: (C) $40\%$

Concept: Difference of shares corresponds to difference of ratio parts.

1. Ratio = 25000 : 33000 = 25 : 33; difference = 8 parts = ₹1000.
2. 1 part = ₹125.
3. Total = $58\times 125=$ ₹7250.

Trick: Total = $1000\times \frac{58}{8}=$ ₹7250.

Answer: (A) ₹7250