Q&A: practice questions with answers

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$.

1. Let onward time $=3t$. Distance $=30t+60(2t)=150t$.
2. Return time $=\frac{150t}{75}=2t$.
3. Average $=\frac{300t}{3t+2t}=60\text{km/h}$.

Trick: Onward average is $\frac{150t}{3t}=50$; for equal distances at 50 and 75, average $=\frac{2\times 50\times 75}{125}=60$.

Answer: (C) 60 km/hr

Concept: Area $=\frac{1}{2}\times \text{base}\times \text{height}$.

1. First area $=\frac{1}{2}\times 18\times 20=180{\text{cm}}^2$; required area $=360{\text{cm}}^2$.
2. $\frac{1}{2}\times 24\times h=360\Rightarrow h=30$ cm.

Trick: $b\times h$ must double: $18\times 20=360\to 720$; $\frac{720}{24}=30$.

Answer: (C) 30 cm

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

1. Decrease $=200-180=20$.
2. $\frac{20}{200}\times 100=10\%$.

Trick: Always divide by the original (₹200), not the new price.

Answer: (A) 10%

Concept: Convert the fraction to a decimal (denominator 16 terminates) and add.

1. $\frac{11}{16}=0.6875$.
2. $0.6875+0.4375=1.125$.

Trick: $0.4375=\frac{7}{16}$, so the sum is $\frac{18}{16}=\frac{9}{8}=1.125$.

Answer: (C) 1.125

Concept: Solve the innermost bracket first.

1. $26-13+5=18$.
2. $-9\times 18=-162$.
3. $30+(-162)=-132$.

Trick: Keep the sign of $-9$ with the product; only one sign change is needed.

Answer: (B) −132

Concept: Draw the Venn diagram: "All jackets are hats" puts the Jacket circle inside Hats; "Some jackets are caps" makes Caps overlap Jackets.

1. The overlap of Caps and Jackets lies inside Hats, so some hats are definitely caps → Conclusion I follows.
2. Caps may extend outside Jackets; "all caps are jackets" is not certain → Conclusion II does not follow.

Trick: All A are B + Some A are C ⇒ Some B are C (always true); the reverse "All C are A" is never guaranteed.

Answer: (A) A

Concept: Check each position of the three-letter groups separately, with wrap-around in the alphabet.

1. 1st letters: Y(25) → Q(17) → I(9) → A(1) → ? : each time $-8$; $1-8=-7\to 26-7=19$ = S.
2. 2nd letters: K(11) → C(3) → U(21) → M(13) → E(5): again $-8$ with wrap-around.
3. 3rd letters: Q(17) → I(9) → A(1) → S(19) → K(11): $-8$.
4. So the next group is SEK.

Trick: When $-8$ goes below 1, add 26: $-7+26=19$ (S).

Answer: (A) ज

Concept: Build the family tree from each statement.

1. G is the father of Y, and Y is the son of J → G and J are the parents (a couple).
2. W is the son of J and M is the brother of W → W and M are also children of G and J.
3. Therefore G is the father of W.

Trick: Same mother (J) ⇒ same father (G): once a couple is fixed, every child of one is a child of the other.

Answer: (B) ए

Concept: Find the position rule for both steps and apply it to each option.

1. DENY → DYNE: positions 1-2-3-4 become 1-4-3-2 (2nd and 4th letters swap). HUGE → HEGU follows the same.
2. DYNE → YNED: positions become 2-3-4-1 (first letter goes to the end). HEGU → EGUH ✓.
3. LOST → LTSO (swap 2nd & 4th ✓) → TSOL (first letter to end ✓).
4. GROW → GRWO swaps 3rd/4th, TURN → TRUN swaps 2nd/3rd, CLOT → LCOT swaps 1st/2nd – all wrong.

Trick: Step 1: swap the 2nd and 4th letters; Step 2: move the first letter to the end.

Answer: (A) च

Concept: Consecutive letters of the alphabet are grouped in threes, continuing cyclically after Z.

1. MNO, PQR, STU, VWX – each group is the next three letters.
2. After X come Y, Z and then the alphabet restarts with A → YZA.

Trick: Z is followed by A when a series wraps around.

Answer: (A) श

Concept: Number the positions 1 (bottom) to 7 (top) and place the fixed boxes first.

1. Only one box above C → C at 6. Only two boxes below D → D at 3.
2. G is below D but above F → G at 2, F at 1.
3. Remaining positions 4, 5, 7. A is not the top box and E is below A → A at 5, E at 4; so B at 7.
4. Stack (top→bottom): B, C, A, E, D, G, F. A and E are adjacent → no box between them.

Trick: Fill "only n above/below" clues first – they fix exact positions.

Answer: (B) "

Concept: Find the relation among the three numbers of each given set.

1. (12, 36, 3): $12\times 3=36$. (15, 75, 5): $15\times 5=75$. Rule: first × third = middle.
2. Check options: $7\times 12=84\ne 72$; $8\times 9=72\ne 56$; $13\times 6=78\ne 188$; $14\times 4=56$ ✓.

Trick: Middle number = product of the two outer numbers.

Answer: (D) ब

Concept: Write the alphabetical positions and compare the gaps (with wrap-around).

1. EWO: 5 → 23 → 15 : $+18,-8$. ASK: 1 → 19 → 11 : $+18,-8$. XPH: 24 → 16 → 8 : $24+18=42-26=16$ ($+18$), then $-8$.
2. MEO: 13 → 5 → 15 : $-8,+10$ – a different pattern.
3. So MEO is the odd one out.

Trick: Pattern of the group: 1st $+18$ = 2nd, 2nd $-8$ = 3rd (equivalently 1st $+10$ = 3rd).

Answer: (B) स

Concept: Arrange the letters of the word alphabetically, then count letters between the required positions in the alphabet.

1. KNIFERS → alphabetical order: E, F, I, K, N, R, S.
2. 3rd from the left = I (9); 3rd from the right = N (14).
3. Letters between I and N in the alphabet: J, K, L, M = $14-9-1=4$.

Trick: Letters between two letters = difference of positions $-1$.

Answer: (D) द

Concept: Compare the differences between consecutive letters in each cluster.

1. NKP: 14 → 11 → 16 : $-3,+5$. VSX: 22 → 19 → 24 : $-3,+5$. ROT: 18 → 15 → 20 : $-3,+5$.
2. KHN: 11 → 8 → 14 : $-3,+6$ – the second gap differs.
3. Hence KHN does not belong to the group.

Trick: Write positions, then gaps; the cluster whose gaps differ is the odd one.

Answer: (A) स

Concept: Vowels (A, E, I, O, U) move one step forward, consonants one step backward.

1. WONDERLANDS → W→V, O→P, N→M, D→C, E→F, R→Q, L→K, A→B, N→M, D→C, S→R.
2. New cluster: V P M C F Q K B M C R.
3. M appears twice (from the two N's); C also appears twice, but among the options only M qualifies. F, Q and R appear once.

Trick: Repeated source letters give repeated coded letters – WONDERLANDS has N twice and D twice.

Answer: (B) म

Concept: Both statements are universal with the same subject "beads", so Beads lies inside both Chairs and Monks.

1. Beads are common to Chairs and Monks, so some chairs are monks → Conclusion I follows.
2. "All monks are beads" reverses the second statement; monks may exist outside beads → Conclusion II does not follow.

Trick: All A are B, All A are C ⇒ Some B are C; a universal statement never reverses to another universal.

Answer: (C) A

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

1. KR 19 → NM 6: K $+3$ = N, R $-5$ = M, $19-13=6$.
2. MV 11 → PQ −2: M $+3$ = P, V $-5$ = Q, $11-13=-2$ ✓ same rule.
3. CS 7 → C $+3$ = F, S $-5$ = N, $7-13=-6$ → FN −6.

Trick: Rule: (1st letter $+3$), (2nd letter $-5$), (number $-13$).

Answer: (A) न

Concept: Try the reverse alphabetical positions (A = 26 … Z = 1) when normal positions do not fit.

1. ZCKU: reverse positions Z = 1, C = 24, K = 16, U = 6; sum $=47$; $47\times 2=94$ ✓.
2. SAPI: S = 8, A = 26, P = 11, I = 18; sum $=63$; $63\times 2=126$ ✓.
3. GXRB: G = 20, X = 3, R = 9, B = 25; sum $=57$; $57\times 2=114$.

Trick: Reverse position $=27-\text{normal position}$. Code = 2 × (sum of reverse positions).

Answer: (C) उ

Concept: Number the seats 1 to 7 from left to right (all face north, so left/right are as we see).

1. No one is to the right of Q → Q at 7; O immediately left of Q → O at 6.
2. Exactly three people between Q and H → H at 3.
3. F is 3rd to the left of L and G is immediately right of L: free seats 1, 2, 4, 5 → L at 4, G at 5, F at 1; M takes seat 2.
4. Row: F, M, H, L, G, O, Q. Between M (2) and O (6) sit H, L, G = three.

Trick: Start with the "end-seat" clue (nobody to the right) – it anchors the whole row.

Answer: (C) "