Q&A: practice questions with answers

Concept: The outer and inner shapes exchange positions.

1. Given third figure: six-pointed star outside, five-pointed star inside.
2. After the swap: five-pointed star outside, six-pointed star inside.
3. Only option C shows this.

Trick: Focus only on the outer boundary of each option – it must be the five-pointed star.

Answer: (C) Option C

Concept: Each element turns 90° clockwise in its own place; the bottom-right element also gets shaded.

1. In A→B: arrow right → down, triangle tips onto its side, star tilts, right triangle turns and is filled.
2. Apply to C: up arrow → right arrow; every other shape turns 90° clockwise.
3. Bottom-right shape becomes shaded – matches option D.

Trick: Track one easy element (the arrow) first – it shows the direction of rotation and removes wrong options quickly.

Answer: (D) Option D

Concept: Number of symbols in the 2nd figure = total sides of the shapes inside the frame of the 1st.

1. First: square (4) + pentagon (5) = 9 → 9 symbols (3 × 3).
2. Third: pentagon (5) + square (4) = 9.
3. So the answer needs 9 crosses: option B (3 rows of 3).

Trick: Just count the crosses: A = 5, C = 7, D = 6, B = 9.

Answer: (B) Option B

Concept: Track each mark separately in the 5-sector circle.

1. Shaded dot: moves 2 sectors clockwise each step.
2. Hollow circle: moves 1 sector anticlockwise each step.
3. From figure 4: shaded dot upper-right → bottom; hollow circle right → upper-right.
4. Option B shows this.

Trick: Follow one mark at a time; when two marks move, they usually move in opposite directions or by different steps.

Answer: (B) Option B

Concept: Right number = $(\text{left number}+2{)}^2$.

1. 14 → ${16}^2=256$ ✓; 12 → ${14}^2=196$ ✓; 16 → ${18}^2=324$ ✓.
2. 18 → ${20}^2=400$, but given 420 ✗.

Trick: Perfect squares end in 0, 1, 4, 5, 6, 9; 420 is not a square at all – it stands out instantly.

Answer: (D) 18-420

Concept: Find the shift of each letter position.

1. SPPU → NJKO: S−5, P−6, P−5, U−6.
2. GBOT → BVJN: same −5, −6, −5, −6 (B−6 wraps to V).
3. Apply to KBNT: K→F (back 5), B→V (back 6), N→I (back 5), T→N (back 6) → FVIN.

Trick: Second letter B−6 = V; only option D has V in second place.

Answer: (D) FVIN

Concept: A word appearing in two sentences has the code common to both.

1. (ii) & (iii): common word 'happiness', common code 'mj'.
2. 'bring' is in (i) and (ii); the only code present in both is 'eb' → bring = eb.
3. 'bring happiness' = eb + mj.

Trick: Find 'happiness' = mj first; then only options with both mj and eb survive – B.

Answer: (B) mj eb

Concept: Compare pairs of sentences to fix codes one by one.

1. 'very' occurs in (i), (iii), (iv); 'st' is in all three → very = st.
2. (i) & (iii) share apple, very → kp, st → apple = kp. (i) & (iv) share is → lm.
3. Left in (i): sweet = qr.
4. 'fruit' in (ii), (iii), (iv); common code 'mn' → fruit = mn.
5. very sweet fruit = st qr mn → order-free match: qr mn st.

Trick: Code order is not fixed; just check the set {st, qr, mn} in the options.

Answer: (C) qr mn st

Concept: All letters of LIGHT (L, I, G, H, T) must be present in the option.

1. LITHOTOMY: no G. HOLLOWING: no T. HOLOGRAPH: no I, no T.
2. LIGHTER contains L, I, G, H, T.

Trick: Check the rare letters first (G and T) – they eliminate three options in seconds.

Answer: (C) LIGHTER

Concept: Reverse the cluster, then move each letter one step forward.

1. QPSU → reverse USPQ → +1 → VTQR ✓.
2. GXOE → reverse EOXG → +1 → FPYH ✓.
3. HCSJ → reverse JSCH → +1 → KTDI.

Trick: Last letter of answer = first letter + 1: H + 1 = I; only options B and C end in I, and second letter S+1 = T leaves B.

Answer: (B) KTDI

Concept: For n5, $(n5{)}^2=n\times (n+1)$ followed by 25.

1. 175 → $n=17$, $n+1=18$.
2. $17\times 18=306$.
3. Write 25 after it → ${175}^2=30625$.

Trick: Units digit of the square must be 25 and ${175}^2<{180}^2=32400$ → only 30625 fits.

Answer: (A) 30625

Concept: Substitute standard trigonometric values.

1. $\cos {45}^{\circ }\times \sin {45}^{\circ }=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)=\frac{1}{2}$.
2. $\cot {60}^{\circ }\times \csc {30}^{\circ }=\frac{1}{\sqrt{3}}\times 2=\frac{2}{\sqrt{3}}$.
3. $\sec {30}^{\circ }\times \tan {45}^{\circ }=\frac{2}{\sqrt{3}}\times 1=\frac{2}{\sqrt{3}}$.
4. Sum = $\frac{1}{2}+\frac{4}{\sqrt{3}}$.

Trick: Notice the last two terms are both $\frac{2}{\sqrt{3}}$ – just double it.

Answer: (B) $\left(\frac{1}{2}+\frac{4}{\sqrt{3}}\right)$

Concept: Use tan in the right triangles formed from the hilltop.

1. Foot of pole at 60°: $\tan {60}^{\circ }=\frac{300}{d}\to d=\frac{300}{\sqrt{3}}=100\sqrt{3}$ m.
2. Top of pole at 30°: $\tan {30}^{\circ }=\frac{300-h}{d}\to 300-h=100\sqrt{3}\times \frac{1}{\sqrt{3}}=100\to h=200$ m.
3. $h-d=200-100\sqrt{3}=100(2-\sqrt{3})$ m.

Trick: With 60° and 30° from the same point, the vertical drops are in ratio 3 : 1 → $300-h=100$ directly.

Answer: (D) $100\left(2-\sqrt{3}\right)$

Concept: Water overflow = volume of the solid immersed.

1. Sphere diameter = 12 cm = cube side, so it fits fully.
2. Volume = $\frac{4}{3}\pi r^3=\frac{4}{3}\pi \times 216=288\pi {\text{cm}}^3$.

Trick: Don't subtract from the cube volume – overflow equals the sphere's volume itself.

Answer: (A) $288\pi$

Concept: In a right triangle the median to the hypotenuse = half the hypotenuse.

1. Check: ${85}^2+{120}^2=7225+14400=21625={145}^2$ → right angle at B.
2. M is the midpoint of hypotenuse CA.
3. $BM=\frac{145}{2}=72.5$ cm.

Trick: 85 : 120 : 145 = 17 : 24 : 29 (×5) – a known triplet, so spot the right angle instantly.

Answer: (B) 72.5 cm

Concept: $\text{Degrees}=\text{radians}\times \frac{{180}^{\circ }}{\pi }$.

1. $-\frac{4\pi }{3}\times \frac{{180}^{\circ }}{\pi }=-4\times {60}^{\circ }=-{240}^{\circ }$.
2. Written as $-{240}^{\circ }{0}^{\prime }{0}^{\prime \prime }$.

Trick: $\frac{\pi }{3}={60}^{\circ }$; so $\frac{4\pi }{3}=4\times {60}^{\circ }$ and keep the minus sign.

Answer: (A) -240°0'0"

Concept: Sum of the two smaller sides must exceed the largest side.

1. A: 2 + 4 = 6 > 5 ✓; B: 6 + 6 > 6 ✓; D: 6 + 8 = 14 > 11 ✓.
2. C: 3 + 2 = 5 < 7 ✗ – cannot form a triangle.

Trick: Only test the two smallest sides against the largest – one check per option.

Answer: (C) $7\mathrm{cm},3\mathrm{cm},2\mathrm{cm}$

Concept: Diagonals of an n-gon = $\frac{n(n-3)}{2}$.

1. Heptagon: $n=7$.
2. $\frac{7\times (7-3)}{2}=\frac{28}{2}=14$.

Trick: Alternative: ${}^n{C}_2-n=21-7=14$ (all joining lines minus sides).

Answer: (B) 14

Concept: Under SI, the amount grows by a fixed interest each year.

1. 2 years' SI = $21800-20500=$ ₹1300 → 1 year = ₹650.
2. 4 years' SI = ₹2600 → $P=20500-2600=$ ₹17,900.
3. SI at 10% for 5 years = $\frac{17900\times 10\times 5}{100}=$ ₹8,950.

Trick: 10% for 5 years = 50% of P → half of 17,900 = ₹8,950.

Answer: (C) ₹8,950

Concept: Total SI% = sum of (rate × years) for each period.

1. Periods: 2 yrs @7%, 4 yrs @10%, 3 yrs @12%.
2. Total % = 14 + 40 + 36 = 90% of P.
3. 90% of P = ₹8100 → P = ₹9,000.

Trick: Add 'rate × time' blocks first; one division gives the principal.

Answer: (C) ₹9,000