Q&A: practice questions with answers

Concept: The given figure (a flat-topped shape with a V-notch in its base) must appear exactly as drawn, same size and same orientation, inside one option figure.

1. Trace the outline: two vertical sides, a horizontal top and a base that dips into a V in the middle.
2. In figure (B) the lower body below the black top band has exactly this outline: straight sides, top edge and the V-shaped notch at the bottom.
3. In (A), (C) and (D) the lines around the V are at different places or extra lines cut the outline, so the shape cannot be traced without rotation.

Trick: Look first for the most special part of the figure (here the V-notch) and check only those options that contain it in the same direction.

Answer: (B) the second figure

Concept: In a cube net, two faces separated by exactly one face in a straight line are opposite; opposite faces can never be seen together.

1. Row 7-8-9: 7 and 9 are opposite.
2. Column 8-X-Y-Z: 8 is opposite Y and X is opposite Z.
3. (A) shows 7 and 9 together, (C) shows 7 and 9 together and (D) shows X and Z together, so all are impossible.
4. (B) shows Z, 9 and Y, no two of which are opposite, so it is possible.

Trick: Mark the three opposite pairs first; then reject every option showing a pair together.

Answer: (B) the cube with Z, 9 and Y

Concept: Check each position of the clusters separately; each position follows its own fixed jump.

1. 1st letters: A, G, M, S, Y (each $+6$).
2. 2nd letters: X, C, H, M, R (each $+5$).
3. 3rd letters: F, J, N, R, V (each $+4$).
4. 4th letters: N, Q, T, W, Z (each $+3$). So the missing cluster is SMRW.

Trick: Jumps $+6,+5,+4,+3$ for the four positions; finding the first two letters (S, M) already leaves only option (D).

Answer: (D) SMRW

Concept: Decode each symbol one by one and draw the family tree.

1. J @ C: J is the husband of C.
2. C $ B: C is the mother of B, so J is the father of B.
3. B # R: B is the father of R.
4. Hence J is the father of R's father, i.e. R's father's father (grandfather).

Trick: Chain the generations: J → B → R is two steps down through males, so J = father's father.

Answer: (D) Father's father

Concept: Code = sum of alphabet positions of the letters + 3 for every letter of the word.

1. SUGAR: $19+21+7+1+18=66$, and $66+5\times 3=81$ ✓.
2. FILL: $6+9+12+12=39$, and $39+4\times 3=51$ ✓.
3. CREAM: $3+18+5+1+13=40$.
4. Code $=40+5\times 3=55$.

Trick: Find (code − letter sum) for both words: 15 for 5 letters, 12 for 4 letters, i.e. 3 per letter.

Answer: (D) 55

Concept: Third number = product of the first two numbers $\times 5$.

1. $(7,2,70)$: $7\times 2\times 5=70$ ✓.
2. $(6,4,120)$: $6\times 4\times 5=120$ ✓.
3. $(3,8,120)$: $3\times 8\times 5=120$ ✓.
4. Others: $13\times 4\times 5=260$, $5\times 9\times 5=225$, $11\times 6\times 5=330$, so they fail.

Trick: Divide the third number by the product of the first two: both give 5; check which option also gives 5.

Answer: (A) $(3,8,120)$

Concept: Second term $=(\text{first term}-6{)}^2+6$.

1. $16$: $(16-6{)}^2+6=100+6=106$ ✓.
2. $10$: $(10-6{)}^2+6=16+6=22$ ✓.
3. $11$: $(11-6{)}^2+6=25+6=31$.

Trick: 106 and 22 are both 6 more than a perfect square ($100$ and $16$), which are the squares of $16-6$ and $10-6$.

Answer: (D) 31

Concept: The four corner symbols follow two alternating steps: (i) every symbol turns ${90}^{\circ }$ clockwise in its own corner, (ii) every symbol moves one corner clockwise without turning.

1. Figure 1 → 2: each symbol turns ${90}^{\circ }$ clockwise in place (U becomes ⊂, T becomes ⊣, ↓ becomes ←).
2. Figure 2 → 3: each symbol shifts one corner clockwise.
3. Figure 3 → 4: turning again (← becomes ↑, ⊂ becomes ∩, ⊣ becomes ⊥).
4. Figure 4 → 5: shifting: ↑ goes top-right, ∩ goes bottom-right, ⊥ goes bottom-left, the D-shape goes top-left. This is figure (D).

Trick: In alternating series, the fifth figure repeats the 'shift' step of figure 2 → 3.

Answer: (D) the fourth option figure

Concept: Check each position of the clusters separately.

1. 1st letters: P, V, B, H, N (each $+6$).
2. 2nd letters: J, B, T, L, D (each $-8$).
3. 3rd letters: V, Z, D, H, L (each $+4$).
4. 4th letters: G, B, W, R, M (each $-5$). Missing cluster: NDLM.

Trick: Find the 4th letter last: options NDLN and NDLM differ only there, and $R-5=M$.

Answer: (D) NDLM

Concept: The series alternates between two figures: figure 1 = figure 3 and figure 2 = figure 4.

1. Figures 1 and 3: two bent arrows (one up, one down) followed by ↑ ↓.
2. Figures 2 and 4: the first two arrows change shape, the last two ↑ ↓ stay the same.
3. So the 5th figure must be the same as figure 1 (and 3), which is option (A).

Trick: Pattern A, B, A, B, ? → the next one is A.

Answer: (A) the first option figure

Concept: The whole T-like figure (a long bar, a short bar and the black joint) turns ${90}^{\circ }$ anticlockwise at every step.

1. Figure 1: long bar vertical, short bar on the right.
2. Figure 2: long bar horizontal, short bar pointing up.
3. Figure 3: long bar vertical, short bar on the left.
4. Next: long bar horizontal with the short bar pointing down, i.e. figure (A).

Trick: Follow only the short bar: right → up → left → down.

Answer: (A) the first option figure

Concept: Compare words letter by letter; when one word is the start of another (Doctor, Doctorate), the shorter word comes first.

1. All start with DOC: next letters are I (Docile), K (Dock), T (Doctor, Doctorate, Doctrine), U (Documentary).
2. Among DOCT…: Doctor, Doctorate (DOCTO) come before Doctrine (DOCTR); Doctor before Doctorate.
3. Order: Docile, Dock, Doctor, Doctorate, Doctrine, Documentary = 3, 5, 1, 2, 6, 4.

Trick: Group by the 4th letter first (I < K < T < U), then sort only inside the T-group.

Answer: (A) 3, 5, 1, 2, 6, 4

Concept: When two positions of a die share one common face, the four faces seen next to it are its neighbours in a fixed cyclic order; the face not seen is opposite the common face.

1. Common face in both positions = 3, so A, 2, B, C are all adjacent to 3 and 1 is opposite 3.
2. Reading the side faces in the same direction: 2 is followed by A, and C is followed by B; so the order around 3 is A, 2, C, B.
3. In this ring, faces two places apart are opposite: A–B and 2–C.
4. Hence B is opposite A.

Trick: Face not visible with the common face (1) is opposite it; the four around it pair up alternately.

Answer: (C) B

Concept: Each term is 11 times the previous term (geometric progression).

1. $\frac{1}{121}\times 11=\frac{1}{11}$.
2. $\frac{1}{11}\times 11=1$.
3. $1\times 11=11$, and $11\times 11=121$ ✓.

Trick: $121={11}^2$; the series is ${11}^{-2},{11}^{-1},{11}^0,{11}^1,{11}^2$.

Answer: (A) 11

Concept: Two 'Some' statements linked through a middle term give no definite conclusion between the end terms.

1. Some movies are dramas; some dramas are fiction. The dramas that are movies may or may not be fiction.
2. I: 'Some movies are fiction' is only possible, not certain, so it does not follow.
3. II: 'All fiction are movies' is not certain either; fiction may lie entirely outside movies.

Trick: Some + Some = no conclusion.

Answer: (C) Neither conclusion I nor II follows

Concept: Replace the letters by signs and solve using BODMAS (÷ and × first, then + and −).

1. $22\times 7-288÷12+27$.
2. $288÷12=24$ and $22\times 7=154$.
3. $154-24+27=157$.

Trick: Do the division first; it removes the big number 288 quickly.

Answer: (D) 157

Concept: Second number $=(\text{first number}+4)\times 4$.

1. $12$: $(12+4)\times 4=64$ ✓.
2. $32$: $(32+4)\times 4=144$ ✓.
3. For 116: $\frac{116}{4}=29$, and $29-4=25$.

Trick: Work backwards: divide by 4, then subtract 4.

Answer: (A) 25

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

1. $21\times 14-154÷14+42$.
2. $154÷14=11$ and $21\times 14=294$.
3. $294-11+42=325$.

Trick: $154=14\times 11$, so the division is exact and quick.

Answer: (B) 325

Concept: Convert the letters into signs and equate both sides.

1. Left side: $56÷8\times 7=7\times 7=49$.
2. Right side: $?\times 7$.
3. $?\times 7=49\Rightarrow ?=7$.

Trick: Both sides have '$\times 7$', so ? $=56÷8=7$ directly.

Answer: (B) 7

Concept: Find the net east–west and north–south displacements, then use Pythagoras.

1. 65 m east, left turn → 28 m north, left turn → 110 m west.
2. Net west $=110-65=45\text{m}$; net north $=28\text{m}$.
3. Distance $=\sqrt{{45}^2+{28}^2}=\sqrt{2025+784}=\sqrt{2809}=53\text{m}$.
4. The library is to the north and west of the house: North-West.

Trick: $(28,45,53)$ is a Pythagorean triplet.

Answer: (D) 53 m, North-West