Q&A: practice questions with answers

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

1. $P=920,R=5,T=3$.
2. SI $=\frac{920\times 5\times 3}{100}=\frac{13800}{100}$.
3. SI $=₹138$.

Trick: $5\%\times 3=15\%$; $10\%$ of $920=92$, $5\%=46$ → $138$.

Answer: (D) ₹138

Concept: A number is divisible by 9 if the sum of its digits is divisible by 9.

1. Digit sum $=1+x+3+6=10+x$.
2. Next multiple of 9 after 10 is 18 → $x=8$.
3. Number $=1836=9\times 204$.

Trick: Digit-sum remainder: $10$ leaves $1$, so add $8$ to reach $18$.

Answer: (B) $8$

Concept: $\%\text{ decrease}=\frac{\text{decrease}}{\text{original}}\times 100$.

1. Decrease $=8000-7200=800$.
2. $\frac{800}{8000}\times 100=10\%$.

Trick: Always divide by the original (old) value, not the new one.

Answer: (A) $10\%$

Concept: Take total work as LCM of the days and find per-day efficiency.

1. Work $=\text{LCM}(20,15)=60$ units; X $=3$/day, Y $=4$/day.
2. Together 5 days: $5\times 7=35$ units.
3. Remaining $=60-35=25$ units.
4. Y alone: $\frac{25}{4}=6.25$ days.

Trick: LCM method avoids fractions – work in whole units.

Answer: (A) $6.25$ days

Concept: $\text{Gain}\%=\frac{\text{true weight}-\text{false weight}}{\text{false weight}}\times 100$.

1. He charges for $60$ g but gives only $50$ g.
2. Gain $=10$ g on $50$ g actually given.
3. Gain $\%=\frac{10}{50}\times 100=20\%$.

Trick: Divide by what the customer actually receives ($50$), not by $60$ – $16.67\%$ is the trap.

Answer: (A) $20\%$

Concept: Compare letters position-wise between the two clusters.

1. DCF → GFI: D+3 = G, C+3 = F, F+3 = I.
2. YXA → BAD: Y+3 = B, X+3 = A, A+3 = D (wrapping after Z).
3. RQT → UTW: R+3 = U, Q+3 = T, T+3 = W ✓.
4. TRT → WVY fails (R+3 = U, not V); TSV → VVX and TRT → VUX fail at the first letter.

Trick: Check only the first letter of each option first – T+3 = W, T+2 = V – it eliminates fast.

Answer: (B) RQT − UTW

Concept: Each term = previous term $\times 2$ minus an increasing number.

1. $8\times 2-1=15$, $15\times 2-2=28$, $28\times 2-3=53$
2. $53\times 2-4=102$
3. Check: $102\times 2-5=199$ ✓

Trick: Terms roughly double each time → test $\times 2$ and look at the small leftover.

Answer: (A) $102$

Concept: Find the alphabet-position gap between consecutive letters (A = 1 … Z = 26, wrapping around).

1. ICW: I(9) − 6 = C(3); C(3) − 6 → W(23).
2. NHB: N(14) − 6 = H(8); H(8) − 6 = B(2).
3. GAU: G(7) − 6 = A(1); A(1) − 6 → U(21).
4. FZO: F(6) − 6 → Z(26), but Z − 6 = T, not O. So FZO is odd.

Trick: Backward −6 with wrap: from A go to U, from C go to W.

Answer: (A) FZO

Concept: Each step moves the first letter of the word to the end.

1. FORK → ORKF (F to end) → RKFO (O to end).
2. TUCK → UCKT → CKTU – same rule.
3. BAKE → AKEB → KEBA ✓.
4. THEM → MTHE moves the last letter to the front – opposite rule; (C) and (D) break the order.

Trick: Read the 2nd word: it should start with the 2nd letter of the 1st word.

Answer: (A) BAKE - AKEB - KEBA

Concept: Look for the pattern Odd – Even – Even and count the middle digit.

1. Scan: 3 8 4 ✓ (3 odd, 4 even).
2. 9 4 6 ✓ (9 odd, 6 even).
3. 7 2 4 ✓ (7 odd, 4 even).
4. Others fail: 9 8 9 (followed by odd), 3 6 3 (followed by odd). Total = 3.

Trick: First mark every odd digit, then check the next two digits are both even.

Answer: (C) Three

Concept: Decode each symbol and build the tree from H.

1. Y ÷ H: Y is the son of H.
2. R # Y: R is the wife of Y (H's son's wife).
3. D & R: D is the father of R; S + D: S is the brother of D.
4. So S is H's son's wife's father's brother.

Trick: Read the chain backwards from H: son → wife → father → brother.

Answer: (C) Son's wife's father's brother

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

1. No box below L → L = 1. Four boxes between M and L → M = 6. H above M → H = 7.
2. Left: positions 2–5 for G, I, N, O. O and N need two boxes between them → 2 and 5.
3. I immediately above G → G = 3, I = 4.
4. N not immediately above I → N ≠ 5, so N = 2, O = 5. Fourth from bottom = I.

Trick: Start with words like 'no box below' – they fix an exact position.

Answer: (D) I

Concept: Take A as origin; east and north positive.

1. North 2 → (0, 2); right = east 5 → (5, 2).
2. Right = south 7 → (5, −5); right = west 8 → (−3, −5).
3. Right = north 5 → (−3, 0) = P.
4. P is 3 km west of A, so she must drive 3 km to the east.

Trick: Net east-west: $5-8=-3$ (3 km west); net north-south: $2-7+5=0$.

Answer: (A) 3 km to the east

Concept: Vowel → next letter; consonant → previous letter.

1. C→B, O→P, N→M, T→S, A→B, C→B, T→S.
2. New group: B P M S B B S.
3. First from the right = S.

Trick: Only the last letter matters: T is a consonant → S.

Answer: (C) S

Concept: Compare the first letters (+7) and the second letters of the two clusters.

1. QM – XJ: Q+7 = X, M−3 = J.
2. HP – OM: H+7 = O, P−3 = M.
3. RD – YA: R+7 = Y, D−3 = A.
4. LG – SC: L+7 = S but G−4 = C (not −3). Odd one.

Trick: When all first letters fit, the odd one hides in the second letters.

Answer: (D) LG – SC

Concept: Apply the letter-wise change: 1st +4, 2nd −6, 3rd −1.

1. DKW : HEV → D+4 = H, K−6 = E, W−1 = V.
2. EGK : IAJ → E+4 = I, G−6 = A, K−1 = J.
3. LRD : PLC → L+4 = P, R−6 = L, D−1 = C ✓.
4. SOT : WJS fails (O−6 = I, not J); FQK : JKL and JGX : NAV fail at the 3rd letter.

Trick: The '−1' on the last letter alone removes (C) and (D).

Answer: (B) LRD : PLC

Concept: Number positions 1 (bottom) to 7 (top).

1. Five boxes above N → N = 2. Three boxes between N and S → S = 6.
2. O below N → O = 1. Only D between S and R → D = 5, R = 4.
3. A above S → A = 7; the remaining box M = 3.
4. Between R (4) and O (1): N and M → two boxes.

Trick: Boxes between = difference of positions − 1 = $4-1-1=2$.

Answer: (D) Two

Concept: In each cluster the second letter is 3 before the first, and the second cluster is +2 of the first.

1. KH – MJ: K−3 = H, M−3 = J, K+2 = M.
2. IF – KH: I−3 = F, K−3 = H, I+2 = K.
3. NK – PM: N−3 = K, P−3 = M, N+2 = P.
4. PM – NM: N−1 = M (not −3) and P−2 = N. Odd one.

Trick: Check the gap inside each cluster first – −3 everywhere except NM.

Answer: (D) PM - NM

Concept: Odd digit +3, even digit −2, then pick positions.

1. 8→6, 2→0, 5→8, 4→2, 3→6, 1→4, 6→4.
2. New number: 6 0 8 2 6 4 4.
3. Third from left = 8; third from right = 6.
4. Sum $=8+6=14$.

Trick: Change only the needed digits: 3rd from left is 5 → 8, 3rd from right is 3 → 6.

Answer: (A) $14$

Concept: Use reverse alphabet values (Z = 1, Y = 2 … A = 26), add them and subtract 6.

1. JCPV: $17+24+11+5=57$, $57-6=51$ ✓.
2. DYSF: $23+2+8+21=54$, $54-6=48$ ✓.
3. HTOB: $19+7+12+25=63$.
4. Code $=63-6=57$.

Trick: Reverse value $=27-\text{forward value}$.

Answer: (D) $57$