Q&A: practice questions with answers

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

1. $23+(-15)+(-8)=0$ ✔
2. Value $=3\times 23\times (-15)\times (-8)$
3. $=3\times 23\times 120=8280$

Trick: Always test $a+b+c=0$ first in sums of three cubes.

Answer: (D) $8280$

Concept: Take total goods $=100$ units costing $1$ each; add up profit/loss of each part.

1. $50$ units at $+12\%$ → $+6$
2. $40\%$ of remaining $50=20$ units at $+16\%$ → $+3.2$
3. Remaining $30$ units at $-19\%$ → $-5.7$
4. Net $=6+3.2-5.7=3.5$ on $100$ → $x=3.5$

Trick: Weighted average: $0.5(12)+0.2(16)+0.3(-19)=3.5$.

Answer: (D) $3.5$

Concept: Work done $\propto$ (workforce $\times$ days). Taking one man's daily work equal to one woman's, the group equals $25$ workers.

1. 10 men + 15 women $=25$ workers; total work $=25\times 12=300$ worker-days
2. Remaining work $=\frac{1}{3}\times 300=100$ worker-days
3. To finish in 2 days: $\frac{100}{2}=50$ workers needed
4. Extra women $=50-25=25$

Trick: Remaining time is halved (4 days → 2 days) → workforce must double → add another 25.

Answer: (B) $25$

Concept: Rule for 7: double the last digit and subtract it from the rest; repeat until small.

1. 24017: $2401-14=2387$ → $238-14=224$ → $22-8=14$ ✔ divisible
2. Check: $7\times 3431=24017$
3. 21234 → $2123-8=2115$ → $211-10=201$ → $20-2=18$ ✘

Trick: Alternatively $24017-21000=3017$, and $3017=7\times 431$.

Answer: (D) $24017$

Concept: $\tan \theta =\frac{\text{height}}{\text{base}}$; the point lies between the poles, so $d=$ sum of the two bases.

1. Taller pole: $\tan {60}^{\circ }=\frac{30}{x}\Rightarrow x=\frac{30}{\sqrt{3}}=10\sqrt{3}$
2. Shorter pole: $\tan {30}^{\circ }=\frac{10}{y}\Rightarrow y=10\sqrt{3}$
3. $d=x+y=20\sqrt{3}\text{m}$

Trick: $\tan {60}^{\circ }=\sqrt{3}$, $\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$; here both bases come out equal.

Answer: (C) $20\sqrt{3}$ meters

Concept: Facing the centre, a person’s right is anticlockwise and left is clockwise (as seen from above).

1. Fix P3. P1 is immediately right of P3.
2. P2 sits between P3 and P7, so P2 is on P3’s left and P7 next to P2.
3. P6 is third to the left of P3; P5 is immediately left of P6. The only seat left goes to P4 (between P1 and P5).
4. From P1, the next seat to the right is P4.

Trick: Fill the fixed clues first (P1, P2–P7, P6); the last empty seat is the answer person.

Answer: (A) Immediate right

Concept: Use alphabet positions and find the change for each letter: FGJ : HAE.

1. $F(6)\to H(8)=+2,G(7)\to A(1)=-6,J(10)\to E(5)=-5$.
2. Check KLO : MFJ: $K\to M=+2,L\to F=-6,O\to J=-5$ ✓.
3. BMV: $B(2)+2=D,M(13)-6=G(7),V(22)-5=Q(17)$ → DGQ ✓.

Trick: Test only the first letter (+2) and then the second (−6) to eliminate options fast.

Answer: (D) BMV : DGQ

Concept: Find the differences between consecutive terms.

1. Differences: $314-277=37,277-244=33,244-215=29,215-190=25,190-169=21$.
2. The differences fall by 4 each time, so the next difference is $21-4=17$.
3. $169-17=152$.

Trick: Differences 37, 33, 29, 25, 21, 17 — an AP with $d=-4$.

Answer: (C) $152$

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

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

Trick: “Only X above Y” fixes Y at position 2; “only X below Y” fixes Y at position 6.

Answer: (B) 2

Concept: Read each letter position of the clusters as a separate series.

1. 1st letters: T, S, R, Q → next P (−1).
2. 2nd letters: A, Z, Y, X → next W (−1, Z comes after A going backwards).
3. 3rd letters: V, U, T, S → next R (−1).
4. So the answer is PWR.

Trick: Every letter moves one step back each time.

Answer: (C) PWR

Concept: Add the alphabet positions of the letters and look at the relation with the code.

1. XEQA: $24+5+17+1=47$, and $47+2=49$ ✓.
2. KLCH: $11+12+3+8=34$, and $34+2=36$ ✓.
3. NJDX: $14+10+4+24=52$, so code $=52+2=54$.

Trick: Code = (sum of positions) $+2$.

Answer: (D) $54$

Concept: Replace the letters by signs and solve using BODMAS.

1. $19C6B5D72A8=19+6\times 5-72÷8$.
2. Division and multiplication first: $72÷8=9$, $6\times 5=30$.
3. $19+30-9=40$.

Trick: Always do $÷$ and $\times$ before $+$ and $-$.

Answer: (B) $40$

Concept: Look for a square relation: the second number is close to the square of the first.

1. ${18}^2=324$ and $324-1=323$ ✓; ${19}^2=361$ and $361-1=360$ ✓.
2. Rule: second number $=n^2-1$.
3. Check options: ${14}^2-1=195$, ${20}^2-1=399$, ${22}^2-1=483$, ${21}^2-1=440$ ✓.

Trick: If the second number is just below a perfect square, try $n^2-1$.

Answer: (D) $21,440$

Concept: Decode each symbol from left to right and build the family tree.

1. S > C: S is the sister of C (S is female).
2. C − O: C is the son of O.
3. O = P: O is the father of P, so S, C and P are siblings.
4. P ! E: P is the wife of E.
5. S is the sister of P, i.e. E’s wife’s sister.

Trick: Sister of a sibling (C) is also sister of the other child (P) of the same father.

Answer: (A) Wife's sister

Concept: Find the differences between consecutive terms.

1. Differences: $149-118=31,211-149=62,242-211=31,304-242=62,335-304=31$.
2. The pattern is $+31,+62$ alternately, so the next step is $+62$.
3. $335+62=397$.

Trick: Alternate differences $31$ and $62=2\times 31$.

Answer: (D) $397$

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

1. Fix P1. P7 is second to the right of P1.
2. Two people between P1 and P6 counting from P1’s right → P6 is third to the right of P1 (P7 is between them).
3. P2 is between P5 and P6 → P2 next to P6 (on the far side), then P5.
4. The two empty seats go to P4 and P3 with P3 second to the right of P4 → P4 next to P1 on its left, P3 immediately right of P1.
5. Seat between P6 and P3 is P7.

Trick: Place the person with the most clues (P1) first and count seats on the drawing.

Answer: (D) P7

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

1. No box above E → E at 1.
2. Only B between E and Q → B at 2, Q at 3.
3. Only R below C → R at 7, C at 6.
4. One box between Q and D → D at 5 (position 1 is taken).
5. The remaining box S is at 4. Between S (4) and R (7): D and C → two boxes.

Trick: “No box above E” = E on top; “only R below C” = C sixth, R last.

Answer: (C) Two

Concept: Track positions of letters 1, 2, 3, 4 in each step.

1. DISK → DKSI: letters taken in order 1, 4, 3, 2.
2. DKSI → KSID: first letter moves to the end (order 2, 3, 4, 1).
3. Check LEAF → LFAE → FAEL ✓.
4. POEM → PMEO (1, 4, 3, 2) → MEOP (first letter to the end) ✓.

Trick: Step 1 swaps the 2nd and 4th letters; step 2 sends the first letter to the end.

Answer: (A) POEM - PMEO - MEOP

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

1. ZY−VU: $Z\to Y=-1$, $V\to U=-1$, and $Z\to V=-4$.
2. JI−FE and ML−IH follow the same: each cluster is two consecutive letters in reverse ($-1$), and the second cluster starts 4 letters back.
3. UT−PR: $P(16)\to R(18)=+2$, not $-1$; also $U\to P=-5$. So UT−PR is odd.

Trick: Three pairs are “reverse-consecutive” twins; spot the cluster that breaks it.

Answer: (A) UT−PR

Concept: Compare each letter of a cluster with the same position in the next cluster.

1. POR: $P=16,O=15,R=18$; VUX: $22,21,24$ → each letter $+6$.
2. VUX → BAD ($V+6=B$ after Z) → HGJ, all $+6$.
3. HGJ $+6$: $H\to N,G\to M,J\to P$ → NMP.

Trick: Inside every cluster the pattern is $-1,+3$ (P→O→R); NMP also follows N→M→P.

Answer: (B) NMP