Q&A: practice questions with answers

Concept: For two numbers: HCF × LCM = product of the numbers.

1. LCM = $98304÷128$
2. $=768$

Trick: The LCM must be a multiple of the HCF (128): $768=128\times 6$ and no other option is. Pick 768 without dividing.

Answer: (B) 768

Concept: Increased by 110% means the new value is 210% of the original.

1. $2.1\times x=3570$
2. $x=3570÷2.1=1700$

Trick: Check: 1700 + 110% of 1700 = $1700+1870=3570$ ✓

Answer: (D) 1700

Concept: Gain factor = (1 + profit%) × (claimed weight/actual weight).

1. Gives 86 g for every 100 g; charges 115% of the cost of 100 g
2. Profit factor = $\frac{115}{86}=1.3372$
3. Profit ≈ $33.72\%\approx 34\%$

Trick: Think of it as a sale of 86 at a price of 115: profit = $\frac{29}{86}\times 100\approx 33.7\%$.

Answer: (C) 34%

Concept: Each spend of a percentage of the 'remaining' amount multiplies the income by $(1-p\%)$.

1. After food + education (40%): $0.6x$
2. After shopping/transport (40% of rest): $0.6\times 0.6x=0.36x$
3. After medical (20% of rest): $0.36\times 0.8x=0.288x=23,040$
4. $x=23040÷0.288=$ ₹80,000

Trick: Check the option: 80,000 → 48,000 → 28,800 → 23,040 ✓

Answer: (C) ₹80,000

Concept: $P=\frac{SI\times 100}{R\times T}$.

1. $P=\frac{720\times 100}{8\times 3}$
2. $=\frac{72000}{24}=$ ₹3000

Trick: SI for 3 years at 8% = 24% of P. $24\%=720$, so $1\%=30$ and $P=3000$.

Answer: (C) 3000

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

1. Fix D. F is 2nd to the left of D, and E is 3rd to the left of F, which puts E just to the right of D
2. Only J sits between F and G, so J and G go on the side of F away from D; K takes the seat between D and F (K is then not next to E ✓)
3. Clockwise order: D, K, F, J, G, E
4. Going right (anticlockwise) from G: J, F, then K, so 2 people sit between them

Trick: In a 6-seat circle, '3rd to the left' = '3rd to the right' (directly opposite), so E is opposite F.

Answer: (A) Two

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

1. Total = 22 + 10 − 1
2. = 31

Yaad rakhein: Subtract 1 because Nehal is counted twice, once from each end.

Answer: (D) 31

Concept: Track the net East–West and North–South movement.

1. West 32 → left (South) 51 → right (West) 66 → right (North) 14 → right (East) 45 → left (North) 37
2. West–East: −32 − 66 + 45 = −53, so Z is 53 km west of Y
3. South–North: −51 + 14 + 37 = 0, so no vertical gap
4. To return to Y: drive 53 km towards the east

Trick: North–south cancels out (51 = 14 + 37), so just add up the east–west parts.

Answer: (C) 53 km towards east

Concept: Vowels → next letter, consonants → previous letter.

1. L→K, A→B, R→Q, K→J, I→J, S→R, H→G
2. New group: K B Q J J R G
3. Only J appears exactly twice, so the count is 1

Trick: A repeat can come only when a consonant's −1 lands on a vowel's +1: K−1 = J = I+1.

Answer: (C) 1

Concept: Shift each letter by +1, then sort the new letters alphabetically.

1. BESWARM → C F T X B S N
2. Sorted: B C F N S T X
3. 4th from right: X, T, S, N, so the answer is N

Trick: With 7 letters, 4th from the right = 4th from the left (the middle one) = N.

Answer: (D) N

Concept: Each term is double the previous term.

1. $13\times 2=26$, $26\times 2=52$, … $208\times 2=416$

Trick: The answer must be even (it is double of something), so 415 and 417 are out right away.

Answer: (B) 416

Concept: Find the net displacement from A.

1. North 5 → left (West) 6 → left (South) 7 → left (East) 8 → left (North) 2
2. East–West: −6 + 8 = +2, so P is 2 km east of A
3. North–South: 5 − 7 + 2 = 0
4. To return to A: 2 km to the west

Trick: Four left turns bring him back to facing north; just net the opposite pairs (W 6 vs E 8; N 5 + 2 vs S 7).

Answer: (A) 2 km to the west

Concept: For each digit, check: previous digit even AND next digit odd.

1. Scan for an even digit followed by X and then an odd digit
2. 2 5 1, 6 8 3, 4 9 7, 2 6 1, 6 1 7
3. Total = 5

Trick: First mark every even digit, then check only the two digits that come after it. This is faster than checking every position.

Answer: (C) Five

Concept: Check the gaps between letter positions.

1. KNL: 11 → 14 → 12 (+3, −2)
2. NQO: 14 → 17 → 15 (+3, −2); RUS: 18 → 21 → 19 (+3, −2)
3. MHG: 13 → 8 → 7 (−5, −1), which does not fit

Trick: In three clusters the 3rd letter is just after the 1st (K-L, N-O, R-S). MHG breaks this.

Answer: (A) MHG

Concept: Decode each symbol and link the relations step by step.

1. A + B: A is the daughter of B
2. B @ H: B is the father of H, so A is H's sister
3. H # K: H is the mother of K; K ÷ T: K is the wife of T
4. H is T's wife's mother, so A is T's wife's mother's sister

Trick: Read from T backwards: T → wife K → her mother H → H's sister A.

Answer: (B) Wife's mother's sister

Concept: Letters common to both words share the common digits.

1. CURL = 4386, LURE = 5846
2. Common letters U, R, L ↔ common digits 4, 8, 6
3. Left over: C = 3 and E = 5

Trick: Cancel the common digits. The extra digit in LURE (5) belongs to its extra letter E.

Answer: (A) 5

Concept: Replace the letters with signs, then apply BODMAS.

1. $4\times 5-20÷2+17=?-3$
2. $20-10+17=27$
3. $?-3=27\Rightarrow ?=30$

Trick: Do ÷ and × first: $20÷2=10$ and $4\times 5=20$. Then remember to add 3 back on the right side.

Answer: (D) 30

Concept: Find the change in each letter's position from the first to the second cluster.

1. ZRD → WPC: Z−3 = W, R−2 = P, D−1 = C
2. WET → TCS: −3, −2, −1
3. YDG → VBF: Y−3 = V, D−2 = B, G−1 = F ✓

Trick: Check the last letter first (−1): only P→N (−2) and G→F (−1) are candidates, and YDG→VBF fits fully.

Answer: (D) YDG : VBF

Concept: The digit must be odd, the one before it odd, and the one after it even.

1. Look for the pattern odd → odd → even
2. 7 1 4, 9 5 2, 1 3 2
3. Total = 3

Trick: Look for two odd digits in a row; then just check whether the next digit is even (e.g. 3 9 5 counts only 5, not 9).

Answer: (D) Three

Concept: Replace the letters with signs, then apply BODMAS.

1. $5\times 2-10÷2+12=?-3$
2. $10-5+12=17$
3. $?=17+3=20$

Trick: Solve the left side, then add the 3 that is subtracted on the right.

Answer: (B) 20