Q&A: practice questions with answers

Concept: One right turn = ${90}^{\circ }$ clockwise; ${360}^{\circ }$ brings back the start.

1. East → South → West → North → East.
2. That is 4 turns = $4\times {90}^{\circ }={360}^{\circ }$.

Trick: To face the starting direction again with one-way turns: 4 turns (or a multiple of 4).

Answer: (A) 4

Concept: The square's quarters are mirror images about the middle lines.

1. Bottom-left = top-left flipped vertically; bottom-right = bottom-left flipped sideways.
2. So top-right = top-left flipped sideways: long line to the corner, small cut line at the top centre, circle on the outer side.
3. Option (A) shows exactly this.

Trick: Read the missing quarter from its neighbour across the vertical axis (left ↔ right).

Answer: (A)

Concept: Fig 2 is Fig 1 flipped top-to-bottom (water image).

1. Dot and diamond swap top/bottom; left-right unchanged; shading moves to the other half.
2. Flip Fig 3 the same way: triangle points down, stalk with dot goes up, stripes move to the lower part.
3. Option (B) matches.

Trick: Water image: only up–down changes, left–right stays put.

Answer: (B)

Concept: First pair: top and bottom elements interchange – a water image.

1. Circle (top) ↔ pentagon (bottom); triangle flips; sides stay.
2. In Fig 3: '=' goes to the top, '+' stays left, 'p' at top becomes 'b' at bottom, 'g' on right becomes '6'.
3. Option (D) shows this.

Trick: In water image p ↔ b, g ↔ 6 type flips happen – look for them.

Answer: (D)

Concept: Each pair swaps inner and outer shapes and moves to the diagonally opposite cell.

1. A → B: triangle-in-square becomes square-in-triangle, shifted diagonally.
2. Apply to C: top-left gets pentagon-in-diamond, top-right hexagon-in-triangle.
3. Bottom-left: circle-in-square; bottom-right: diamond-in-circle → option (A).

Trick: Check one cell only (e.g. top-left) – it usually leaves just one option.

Answer: (A)

Concept: Frame stays; the end-marks shift one corner clockwise each step.

1. Ringed dot: bottom-left → top-left → top-right → bottom-right → next bottom-left.
2. Plus: top-left → top-right → next bottom-right.
3. Only (D) has ringed dot at bottom-left and plus at bottom-right.

Trick: Track one symbol at a time; two symbols are enough to fix the answer.

Answer: (D)

Concept: Each letter shifts by the same value.

1. C→G, S→W, J→N, P→T: +4 each (E→I, B→F confirm).
2. BUPQ, each moved four ahead: B→F, U→Y, P→T, Q→U.

Trick: Check just the first letter: B+4 = F leaves (C) and (D); U+4 = Y → (D).

Answer: (D) FYTU

Concept: Same letters in the same positions get the same code.

1. HONEY → CYIIT; MONEY differs only in the first letter, so 'ONEY' → 'YIIT'.
2. First letter rule: H(8) → C(3) and D(4) → Y(25): −5.
3. M(13) − 5 = H(8) → HYIIT.

Trick: Compare words with common parts – copy that part of the code directly.

Answer: (B) HYIIT

Concept: Reverse the word, then add +5, +4, +3, +5, +4, +3.

1. GHTMIK reversed = KIMTHG → +5,+4,+3,+5,+4,+3 → PMPYLJ ✓
2. DACBYP reversed = PYBCAD.
3. P→U (+5), Y→C (+4), B→E (+3), C→H (+5), A→E (+4), D→G (+3) → UCEHEG.

Trick: When no direct pattern appears, try reversing the word first.

Answer: (C) UCEHEG

Concept: Every letter of SHOES, with its count, must be present.

1. SHOES needs S×2, H, O, E.
2. SCHOOL – one S, no E; STATEMENT and BATSMAN – no H, no O.
3. SCHOOLERS has S×2, H, O, E ✓

Trick: Check the repeated letter (S twice) first – it removes most options.

Answer: (B) SCHOOLERS

Concept: Mixed recurring: some decimal digits do not repeat, then a block repeats.

1. 0.75 – terminating.
2. 0.333… and 0.2727… – repetition starts right after the point (pure recurring).
3. 2.1454545… – '1' does not repeat, then '45' repeats → mixed recurring.

Trick: Pure = bar starts at first decimal place; Mixed = bar starts later.

Answer: (D) $2.1454545\dots$

Concept: After d% discount, SP = $(100-d)\%$ of MP.

1. SP = 80% of MP = 6500.
2. MP = $6500\times \frac{100}{80}=6500\times \frac{5}{4}=8125$.

Trick: 20% off → MP = $\text{SP}\times \frac{5}{4}$ (add one-fourth of SP: $6500+1625$).

Answer: (A) 8125

Concept: $\angle BIC={90}^{\circ }+\frac{A}{2}$ (incentre); $\angle BOC={180}^{\circ }-A$ (orthocentre).

1. $({90}^{\circ }+\frac{A}{2})-({180}^{\circ }-A)={60}^{\circ }$.
2. $\frac{3A}{2}-{90}^{\circ }={60}^{\circ }\to \frac{3A}{2}={150}^{\circ }$.
3. $A={100}^{\circ }$.

Trick: Remember the pair: Incentre $90+\frac{A}{2}$, Excentre $90-\frac{A}{2}$, Orthocentre $180-A$, Circumcentre 2A.

Answer: (A) 100°

Concept: $\cot \theta =\frac{5}{12}\Rightarrow \tan \theta =\frac{12}{5}$. The two brackets have the same denominator $\cos \theta$.

1. $\frac{1+\sin \theta }{\cos \theta }-\frac{1-\sin \theta }{\cos \theta }=\frac{(1+\sin \theta )-(1-\sin \theta )}{\cos \theta }=\frac{2\sin \theta }{\cos \theta }=2\tan \theta$.
2. $2\tan \theta =2\times \frac{12}{5}=\frac{24}{5}$.

Trick: Same denominators → subtract numerators: $(1+s)-(1-s)=2s$, so the answer is just $2\tan \theta$.

Answer: (B) $\frac{24}{5}$

Concept: Line from external point to centre bisects the angle between tangents; radius ⊥ tangent.

1. Half angle = ${45}^{\circ }$; in right △: $\sin {45}^{\circ }=\frac{r}{R}$.
2. $R=\frac{7}{\sin {45}^{\circ }}=7\sqrt{2}$.
3. $7\times 1.414\approx 9.89$ cm.

Trick: Tangents at ${90}^{\circ }$ → the figure is a square: $R=\text{diagonal}=r\sqrt{2}$.

Answer: (C) 9.89 cm

Concept: Perpendicular from centre bisects the chord.

1. AB = $\frac{30}{2}=15$ cm.
2. $O{B}^2={17}^2-{15}^2=289-225=64$.
3. OB = 8 cm.

Trick: Triplet 8-15-17: see 15 and 17 → other side 8.

Answer: (B) 8 cm

Concept: $(l-b{)}^2=(l+b{)}^2-4lb$.

1. $l+b=\frac{14\sqrt{2}}{2}=7\sqrt{2}\to (l+b{)}^2=98$.
2. $4lb=4\times 18=72$.
3. $(l-b{)}^2=98-72=26\to l-b=\sqrt{26}$ cm.

Trick: Never find l and b separately – use the identity directly.

Answer: (C) $\sqrt{26}\mathrm{cm}$

Concept: Area of rhombus = $\frac{1}{2}\times d_1\times d_2$; $1{\text{m}}^2=10000{\text{cm}}^2$.

1. One tile = $\frac{1}{2}\times 15\times 18=135{\text{cm}}^2$.
2. 500 tiles = $67500{\text{cm}}^2=6.75{\text{m}}^2$.
3. Cost = $6.75\times 20=$ ₹135.

Trick: $\frac{500\times 20}{10000}=1$, so cost in ₹ = area of one tile in cm² = 135.

Answer: (A) Rs. 135

Concept: Diagonals bisect at ${90}^{\circ }$: ${\text{side}}^2={\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2$.

1. Half-diagonals $\frac{7k}{2}$ and $\frac{24k}{2}$ → 7-24-25 triplet scaled: $\frac{25k}{2}=25\to k=2$.
2. Diagonals = 14 cm and 48 cm.
3. Area = $\frac{1}{2}\times 14\times 48=336{\text{cm}}^2$.

Trick: 7 : 24 : 25 triplet – half-diagonals are 7 and 24, so diagonals 14 and 48.

Answer: (B) 14 cm and $48\mathrm{cm},336{\mathrm{cm}}^2$

Concept: Equal time → profit in ratio of investment.

1. Total parts = $5+7=12$.
2. 1 part = $1,20,000÷12=10,000$.
3. P = $5\times 10,000=$ ₹50,000.

Trick: Share must be a multiple of 10,000 here – only ₹50,000 fits.

Answer: (B) Rs. 50,000