Q&A: practice questions with answers

Concept: Total work $=\text{LCM}(12,18)=36$ units; efficiency $=$ work per day.

1. A $=3$ units/day, B $=2$ units/day.
2. In 2 days together: $5\times 2=10$ units; remaining $26$.
3. B alone: $\frac{26}{2}=13$ days.

Trick: Fractions: done $=2(\frac{1}{12}+\frac{1}{18})=\frac{5}{18}$; left $\frac{13}{18}\times 18=13$ days.

Answer: (C) 13

Concept: Convert each ratio to a decimal and arrange.

1. $\frac{4}{11}\approx 0.364$, $\frac{23}{55}\approx 0.418$, $\frac{11}{15}\approx 0.733$, $\frac{26}{35}\approx 0.743$, $\frac{32}{33}\approx 0.970$.
2. Ascending: $4:11<23:55<11:15<26:35<32:33$.

Trick: Only $11:15$ vs $26:35$ is close – cross-multiply: $11\times 35=385<26\times 15=390$, so $11:15$ is smaller.

Answer: (A) 4:11, 23:55, 11:15, 26:35, 32:33

Concept: $A=P{(1+\frac{R}{100})}^T$; CI $=A-P$.

1. $A=10000\times (1.1{)}^2=10000\times 1.21=12100$.
2. CI $=12100-10000=₹2,100$.

Trick: 2-year CI rate $=R+R+\frac{R^2}{100}=10+10+1=21\%$ of 10,000 $=2100$.

Answer: (D) ₹2,100

Concept: If $a+b+c=0$, then $a^3+b^3+c^3=3abc$.

1. $25+(-20)+(-5)=0$ ✓.
2. $3abc=3\times 25\times (-20)\times (-5)=3\times 2500=7500$.

Trick: Whenever three numbers sum to zero, skip the cubes and use $3abc$.

Answer: (C) 7500

Concept: Find the cost of one unit, then multiply.

1. Cost of 1 dress $=\frac{4955}{5}=₹991$.
2. Cost of 9 dresses $=991\times 9=₹8,919$.

Trick: $991\times 9=(1000-9)\times 9=9000-81=8919$.

Answer: (A) ₹8,919

Concept: Apply the rule digit by digit: odd digit $+2$, even digit $-2$, then compare the digits of the new number.

1. 7 2 1 3 5 4 6 → $7+2=9$, $2-2=0$, $1+2=3$, $3+2=5$, $5+2=7$, $4-2=2$, $6-2=4$.
2. New number: 9 0 3 5 7 2 4.
3. Highest digit 9, lowest digit 0 → difference $9-0=9$.

Trick: Write the new digits directly under the old ones – the lone even digit 2 becomes 0, which usually becomes the minimum.

Answer: (A) 9

Concept: Sort the digits in ascending order and compare position by position with the original.

1. Original: 9 3 4 2 1 5 6; ascending: 1 2 3 4 5 6 9.
2. Compare pairs (old, new): (9, 1), (3, 2), (4, 3), (2, 4), (1, 5), (5, 6), (6, 9) – no position has the same digit in both.
3. Hence no digit keeps its place.

Trick: Write both numbers one below the other and tick only columns where the digits match.

Answer: (A) None

Concept: Take successive differences; if they grow by a constant, extend the pattern.

1. $57-37=20$, $81-57=24$, $109-81=28$, $141-109=32$, $177-141=36$.
2. Differences increase by 4 each time, so the next difference is $40$.
3. $177+40=217$.

Trick: Differences 20, 24, 28 … (multiples of 4) – a second-level arithmetic progression.

Answer: (C) 217

Concept: Compare corresponding letters of each pair using alphabet positions.

1. ON → QP: $O+2=Q$, $N+2=P$.
2. SR → UT: $S+2=U$, $R+2=T$; LK → NM: $L+2=N$, $K+2=M$.
3. UT → WX: $U+2=W$ but $T+4=X$ – breaks the $(+2,+2)$ pattern.

Trick: In the three alike pairs the second cluster is also two consecutive letters in reverse (QP, UT, NM); WX is forward order – spot it at a glance.

Answer: (B) UT - WX

Concept: Look for the pattern vowel – consonant – vowel in the series (vowels: A, E, I, O, U).

1. Series: O K I Y C R L U K V I E H L Y V O N Q O S Z.
2. Vowel-consonant-vowel triplets: O-K-I ✓; I-Y-C ✗; U-K-V ✗; I-E-H (E is a vowel, H followed by L) ✗; O-N-Q ✗; O-S-Z ✗.
3. Only K (2nd letter) qualifies.

Trick: Underline the vowels first; then check only the single letters sandwiched between two underlined ones.

Answer: (D) 1

Concept: Arrange digits in ascending order, then pick the required positions.

1. Digits of 3164857: 3, 1, 6, 4, 8, 5, 7 → ascending: 1 3 4 5 6 7 8.
2. Third from left = 4; third from right = 6.
3. Sum $=4+6=10$.

Trick: In a 7-digit sorted number, 3rd from left and 3rd from right are symmetric about the middle digit (5) – here $4+6=2\times 5=10$.

Answer: (C) 10

Concept: Number the seats 1–7 from left to right (all face north, so left/right are the same as ours) and fix the definite positions first.

1. Only one person left of R → R at seat 2. G fourth from right → seat 4. Two between R and E → E at seat 5.
2. D is immediately left of N, so D–N must occupy consecutive free seats: only 6–7 fit → D at 6, N at 7.
3. H and C take seats 1 and 3; H is not to the left of C → C at 1, H at 3.
4. Between C (1) and D (6): R, H, G, E = four people.

Trick: Place the "from the end" and "only one" clues first; the adjacent pair (D–N) then has a single slot.

Answer: (B) Four

Concept: Build the family tree generation by generation, then trace the path from Kunal up to Chetna.

1. Bablu (husband) – Chetna (wife); Anita and Ekam are their children.
2. Manu is Anita's son; Kunal is Manu's brother → Kunal is also Anita's son.
3. Kunal → mother Anita → her mother Chetna. So Chetna is Kunal's mother's mother (maternal grandmother).

Trick: Always start from the person asked about (Kunal) and climb upward; "mother's mother" = nani.

Answer: (A) Mother's mother

Concept: Number positions 1 (bottom) to 7 (top) and use the "only" clues to pin the ends.

1. Only G is above A → G at 7, A at 6.
2. Two boxes between A and C → C at 3. Only D is below E → D at 1, E at 2.
3. Positions 4 and 5 remain for B and F; F is not immediately above C (position 4) → F at 5, B at 4.
4. Between E (2) and F (5): C and B = 2 boxes.

Trick: "Only X is above/below Y" fixes two boxes at an end at once – start there.

Answer: (B) 2

Concept: Track each letter position separately and the number separately.

1. 1st letters H → M → R → W: $+5$ each; next $W+5=B$ (wrap after Z).
2. 2nd letters E → C → A → Y: $-2$ each; next $Y-2=W$. 3rd letters P → N → L → J: $-2$; next $J-2=H$.
3. Numbers 27, 41, 55, 69: $+14$; next $69+14=83$.

Trick: Check the number first – only options A and C end in 83; then the first letter ($W+5=B$) settles it.

Answer: (C) BWH 83

Concept: Number positions 1 (bottom) to 7 (top); fix the stated ends, then fit the "only one between" pairs.

1. B is top-most → B at 7. Only U is below W → U at 1, W at 2.
2. Only C is between B and X → C at 6, X at 5.
3. Only A is between V and X → A at 4, V at 3.
4. Third from the bottom = position 3 = V.

Trick: Each "only one between" clue, once one end is known, fixes two boxes in a row.

Answer: (A) V

Concept: Find the uniform shift of each letter and the number.

1. Z → D → H → L: $+4$ each; next $L+4=P$.
2. Q → U → Y → C: $+4$; next $C+4=G$. F → J → N → R: $+4$; next $R+4=V$.
3. Numbers 19, 28, 37, 46: $+9$; next $55$.

Trick: All three letters move $+4$ and the number $+9$ – only PGV 55 fits both.

Answer: (A) PGV 55

Concept: Draw the Venn diagram: "All Corns are Cauliflowers" puts the Corn circle inside Cauliflowers; "No Cauliflowers are Spinaches" keeps Spinaches completely apart.

1. Since every Corn lies inside Cauliflowers and Cauliflowers never touch Spinaches, no Corn can be a Spinach → Conclusion (I) follows.
2. Conclusion (II) "Some Spinaches are Corns" directly contradicts this → does not follow.

Trick: All A are B + No B are C ⇒ No A are C (and No C are A) – a standard valid pair.

Answer: (B) Only conclusion (I) follows

Concept: Find the gap pattern between consecutive letters of each cluster.

1. GKN: $G+4=K$, $K+3=N$ → $(+4,+3)$.
2. BFI: $B+4=F$, $F+3=I$; PTW: $P+4=T$, $T+3=W$ → same $(+4,+3)$.
3. QNJ: $Q-3=N$, $N-4=J$ → $(-3,-4)$, moving backwards – the odd one.

Trick: Three clusters run forward in the alphabet; QNJ runs backward – visible even before counting gaps.

Answer: (B) QNJ

Concept: Find how each letter and the number change from the first term to the second, then apply the same change.

1. HI 26 → FF 11: $H-2=F$, $I-3=F$, $26-15=11$.
2. MG 22 → KD 7: $M-2=K$, $G-3=D$, $22-15=7$ ✓ same rule.
3. KF 14 → $K-2=I$, $F-3=C$, $14-15=-1$ → IC −1.

Trick: Pattern $(-2,-3,-15)$; among options only IC −1 has both letters and number matching.

Answer: (C) IC −1