Q&A: practice questions with answers

Concept: Average per article $=\frac{\text{total cost}}{\text{total articles}}$.

1. Shirts $8\times 550=4400$; pants $10\times 1000=10000$; shoes $12\times 850=10200$.
2. Total $=24600$ for $8+10+12=30$ articles.
3. Average $=\frac{24600}{30}=820$.

Trick: Count articles carefully: pairs of shoes are counted as 12 articles.

Answer: (A) ₹820

Concept: Convert averages to sums: the overlapping member B is counted twice.

1. $A+B+C=3\times 45=135$.
2. $A+B=80$, $B+C=84$.
3. $(A+B)+(B+C)-(A+B+C)=B=80+84-135=29$.

Trick: Middle person = sum of two pair-totals − grand total.

Answer: (C) 29

Concept: $SI=\frac{P\times R\times T}{100}$; the interest difference comes from the principal difference.

1. Principal difference $=9000-5000=4000$.
2. $\frac{4000\times R\times 3}{100}=360$.
3. $120R=360\Rightarrow R=3\%$.

Trick: Difference in SI depends only on difference in principal.

Answer: (C) 3%

Concept: Successive changes multiply: new price = original × (1 − decrease) × (1 + increase).

1. $8000\times 0.88=7040$.
2. $7040\times 1.25=8800$.

Trick: Net change $=-12+25-\frac{12\times 25}{100}=+10\%$ → $8000\times 1.1=8800$.

Answer: (C) ₹8,800

Concept: Mean of a frequency distribution $=\frac{\sum fx}{\sum f}$.

1. $\sum fx=1274+2028+4180+2196+8232=17910$.
2. $\sum f=91+78+95+36+98=398$.
3. Mean $=\frac{17910}{398}=45$.

Trick: Check: $398\times 45=17910$ exactly.

Answer: (A) 45

Concept: Take total work = LCM of days and compare daily work.

1. LCM$(8,24,3)=24$ units.
2. A $=\frac{24}{8}=3$, B $=\frac{24}{24}=1$, A+B+C $=\frac{24}{3}=8$ units/day.
3. C $=8-3-1=4$ units/day.
4. C alone $=\frac{24}{4}=6$ days.

Trick: $\frac{1}{3}-\frac{1}{8}-\frac{1}{24}=\frac{8-3-1}{24}=\frac{1}{6}$.

Answer: (C) 6 days

Concept: Evaluate powers and brackets first (BODMAS).

1. ${17}^3=4913$, ${10}^2=100$, ${\left(\frac{26}{13}\right)}^2=2^2=4$, $20\times 12=240$.
2. $4913-100+4-12+240$.
3. $=5045$.

Trick: Unit digit check: $3-0+4-2+0=5$ → only 5055 or 5045; size check gives 5045.

Answer: (D) 5045

Concept: Arithmetic mean $=\frac{\text{sum of observations}}{\text{number of observations}}$.

1. Sum $=79+92+39+57+91+28+74+90+44=594$.
2. Number of observations $=9$.
3. Mean $=\frac{594}{9}=66$.

Trick: Pair to round numbers: $(79+91)+(92+28)+(57+44)+(39+74)+90$.

Answer: (C) 66

Concept: Find the shift from the first cluster to the second on the right side, then apply it to the left side.

1. CHD → %: try option (A) HMI: C+5=H, H+5=M, D+5=I. Each letter moves $+5$.
2. Left side: # → QVR, so # = QVR moved back by 5.
3. Q−5=L, V−5=Q, R−5=M → # = LQM.
4. Option (A) gives # = LQM and % = HMI – both fit.

Trick: Write positions: L(12)→Q(17), C(3)→H(8): same $+5$ on every letter.

Answer: (A) # = LQM, % = HMI

Concept: Draw the smallest possible Venn diagram that satisfies all statements.

1. All jams are bands and all bands are sacks → Jams ⊂ Bands ⊂ Sacks.
2. No sack is a can → the Cans circle lies fully outside Sacks, so it is outside Jams too → (I) No jam is a can follows.
3. All jams are sacks, so some sacks are certainly jams → (II) follows.

Trick: "All A are B" always gives "Some B are A" (conversion).

Answer: (D) Both conclusions (I) and (II) follow.

Concept: Look for one operation linking the three numbers of each set.

1. $(33,429,13)$: $33\times 13=429$.
2. $(44,220,5)$: $44\times 5=220$.
3. Rule: first × third = second.
4. Check (B): $43\times 21=903$ ✓. Others: $20\times 18=360$, $30\times 12=360$, $4\times 5=20$ – fail.

Trick: Middle number much bigger than both? Test product of the outer two first.

Answer: (B) (43, 903, 21)

Concept: Change each digit by the rule (odd → $+1$, even → $-2$) and then read the asked positions.

1. 5673128 → 6 4 8 4 2 0 6, i.e. new number 6484206.
2. Third from left = 8.
3. Second from right = 0.
4. Difference $=8-0=8$.

Trick: Write the new digits in boxes under the old ones – positions can then be read without error.

Answer: (D) 8

Concept: Track each letter position separately.

1. 1st letters: N, L, J, H → each $-2$ → F.
2. 2nd letters: T, R, P, N → each $-2$ → L.
3. 3rd letters: Z, X, V, T → each $-2$ → R.
4. Next term = FLR.

Trick: All three positions move back by 2 together – just shift the last term HNT by $-2$.

Answer: (C) FLR

Concept: Use the product clue first – it allows only a few floor pairs.

1. $F\times D=15$ → floors $\{3,5\}$.
2. If $D=5$: $A=2$, left floors 1, 4, 6 – no two consecutive for C and E. Rejected.
3. So $D=3$, $F=5$, $A=7-3=4$. Left floors 1, 2, 6 → C = 1, E = 2, B = 6.
4. $A+E=4+2=6$.

Trick: Product clues ($15=3\times 5$) fix two people at once – always start with them.

Answer: (A) 6

Concept: Apply odd → $+1$, even → $-2$ to every digit, then count repeated digits.

1. 8271634 → 6 0 8 2 4 4 2, i.e. 6082442.
2. Digit 2 appears twice, digit 4 appears twice.
3. 6, 0, 8 appear once each.
4. So two digits appear more than once.

Trick: Count digits, not occurrences: 2 and 4 → answer "Two".

Answer: (A) Two

Concept: Each part of the term follows its own rule.

1. 1st letter: N, M, L, K → $-1$ → J.
2. 2nd letter: Q, P, O, N → $-1$ → M.
3. Number: 31, 40, 49, 58 → $+9$ → 67.
4. Next term = JM67.

Trick: Options differ only in M or N and 66 or 67 – check just those two parts.

Answer: (C) JM67

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

1. "how are you" → jp zl rc.
2. "are they coming" → zl qn bt.
3. Common word: "are"; common code: zl.
4. So are = zl.

Trick: Common word ↔ common code.

Answer: (A) zl

Concept: Facing the centre: moving to a person's right = anticlockwise, to the left = clockwise.

1. Fix P. One person between P and V on P's right → V is 2 seats anticlockwise from P.
2. Combining "one between V and U", "one between P and Q", "one between T and Q" and "two between R and U", the only arrangement (clockwise from P) is P, S, Q, U, T, V, R.
3. From S going left (clockwise): Q, U, T, V, R, then P.
4. So 5 people sit between S and P.

Trick: With 7 seats, "2 between" one way = "3 between" the other way – use it to test both sides quickly.

Answer: (D) 5

Concept: "Only X lives above Y" puts Y on the second-highest floor; "only P lives below O" puts O on floor 2.

1. Only M above N → M = 6, N = 5.
2. Z immediately below N → Z = 4.
3. Only P below O → P = 1, O = 2.
4. Remaining Y → floor 3.

Trick: "Only one above" = 5th floor; "only one below" = 2nd floor.

Answer: (B) Y

Concept: Compare letter gaps inside each cluster and between the two clusters.

1. MO – KM: M $+2$ O; K $+2$ M; and each letter of the 2nd cluster is 2 behind the 1st (M→K, O→M).
2. SU – QS and QS – OQ follow the same: $+2$ inside, $-2$ from the 1st to the 2nd cluster.
3. CG – LP: C $+4$ G and L $+4$ P; C → L is $+9$. Different.

Trick: Here the second cluster equals the first shifted by $-2$ – spot the one that breaks it.

Answer: (B) CG - LP