Q&A: practice questions with answers

Concept: Each term is 4 times the previous term.

1. $5\times 4=20$, $20\times 4=80$, $80\times 4=320$, $320\times 4=1280$
2. $1280\times 4=5120$

Trick: Every term ends in 0 after 20 (× 4 of a multiple of 10), so only 5120 and 5130 fit, and $1280\times 4=5120$.

Answer: (A) 5120

Concept: Check the gaps between letter positions.

1. GKH: 7 → 11 → 8 (+4, −3)
2. JNK: 10 → 14 → 11; QUR: 17 → 21 → 18 (+4, −3)
3. INJ: 9 → 14 → 10 (+5, −4), which does not fit

Trick: In all four the 3rd letter = 1st + 1, so check the middle letter: it should be 1st + 4. INJ has +5.

Answer: (A) INJ

Concept: Each letter moves forward by the same number of places (cycling after Z).

1. GDA → WTQ: G(7)+16 = W(23), D+16 = T, A+16 = Q
2. URO → KHE: U(21)+16 = 37 − 26 = 11 = K; also −10, the same thing
3. AXU: A+16 = Q, X(24)+16 = 40 − 26 = 14 = N, U+16 = K, giving QNK

Trick: +16 = −10 in the alphabet cycle. Use whichever is quicker: A+16 = Q and U−10 = K.

Answer: (A) AXU−QNK

Concept: Find the net displacement from A.

1. East 11 → left (North) 8 → left (West) 13 → left (South) 13 → left (East) 2
2. East–West: 11 − 13 + 2 = 0
3. North–South: 8 − 13 = −5, so P is 5 km south of A
4. To return: 5 km to the north

Trick: East–west cancels (11 + 2 = 13), so the answer is only the vertical gap: 13 − 8 = 5, going north.

Answer: (A) 5 km to the north

Concept: Letters common to both words share the common digits.

1. KIDS = 1248, SILK = 2941
2. Common K, I, S ↔ common digits 1, 2, 4
3. Left over: D = 8, L = 9

Trick: The digit that appears only in SILK (9) belongs to the letter that appears only in SILK (L).

Answer: (A) 9

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

1. J = 2; three boxes between J and I, so I = 6
2. K below J, so K = 1; only G between I and L, so L = 4 and G = 5
3. H above I, so H = 7; O takes the last place, 3
4. Order from bottom: K, J, O, L, G, I, H. Between O(3) and K(1): only J

Trick: Start with the definite clues (J = 2, I = 6, K = 1); the rest fall into place one by one.

Answer: (C) One

Concept: Letters: 1st letter −1, 2nd letter +1. Number: −50.

1. FG → EH, 151 → 101; MN → LO, 262 → 212
2. UV → U−1 = T, V+1 = W, so TW
3. 379 − 50 = 329, giving TW 329

Trick: The number rule (−50) leaves only TW 329 and TX 329. V + 1 = W, so the answer is TW.

Answer: (B) TW 329

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

1. Q's neighbours are T and W. If W is on Q's left, R (2nd to W's right) would land on T's seat, so W is on Q's right and T is on Q's left
2. U is 3rd to the left of Q; R is 2nd to the right of W, which makes R next to U
3. Remaining E and Y: E is not next to U, so Y sits next to U on the T side
4. Clockwise: Q, T, Y, U, R, E, W. Going right from Y: T, Q, then W, so 2 people sit between them

Trick: Test the two options for W (left or right of Q); one of them clashes immediately, which fixes the circle.

Answer: (D) 2

Concept: Check the gaps between letter positions.

1. PSQ: 16 → 19 → 17 (+3, −2)
2. DGE: 4 → 7 → 5; JMK: 10 → 13 → 11 (+3, −2)
3. GFT: 7 → 6 → 20, which does not fit

Trick: In the other three the 3rd letter = 1st + 1 (P-Q, D-E, J-K). GFT clearly breaks this.

Answer: (B) GFT

Concept: Check the gaps between letters; the alphabet cycles after Z.

1. DZA: D − 4 = Z, Z + 1 = A
2. CYZ: C − 4 = Y, +1 = Z; EAB: E − 4 = A, +1 = B
3. MJK: M − 3 = J, +1 = K. The gap is −3, not −4

Trick: In all four the last two letters are consecutive, so compare only the first jump (−4 vs −3).

Answer: (C) MJK

Concept: Glass is optically denser than air, and its refractive index differs for each colour.

1. White light splits into VIBGYOR, so (i) is true
2. Light slows down in a denser medium ($v=\frac{c}{n}$, $n\approx 1.5$ for glass), so (iii) is true and (ii) is false

Trick: (ii) and (iii) contradict each other, so only one can be true. Speed falls in glass, so the answer is (i) + (iii).

Answer: (B) Both (i) and (iii)

Concept: $\text{Power}=\frac{\text{Work}}{\text{Time}}=\frac{\text{Weight}\times \text{height}}{\text{time}}$.

1. Work = $420\text{N}\times 6\text{m}=2520\text{J}$
2. Power = $2520÷20=126$ W

Trick: Weight is already in newtons, so do not multiply by g again: $P=\frac{W\times h}{t}$ directly.

Answer: (A) 126 W

Concept: Alcohol + carboxylic acid → ester (sweet smell) + water, in the presence of conc. H₂SO₄.

1. CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ (ethyl ethanoate) + H₂O
2. Conc. H₂SO₄ acts as a catalyst and removes the water formed (dehydrating agent), which pushes the reaction forward

Trick: Sweet smell = ester, so the reaction is esterification. Only option C names it.

Answer: (C) It is an esterification reaction; H₂SO₄ acts as a catalyst and a dehydrating agent.

Concept: Energy released depends on how completely glucose is broken down.

1. Anaerobic: glucose is only partly broken down to ethanol + CO₂ or lactic acid, giving 2 ATP
2. Aerobic: glucose is fully oxidised to CO₂ + H₂O, giving about 36–38 ATP
3. Both use enzymes, so A is wrong

Yaad rakhein: Yeast gives ethanol + CO₂; tired muscles give lactic acid (which causes cramps).

Answer: (B) It partially breaks down glucose.

Concept: Speed of a wave v = frequency × wavelength; time = distance/speed.

1. $v=2000\text{Hz}\times 0.45\text{m}=900\text{m/s}$
2. $t=1800\text{m}÷900\text{m/s}=2\text{s}$

Trick: Convert units first: 2 kHz = 2000 Hz, 45 cm = 0.45 m, 1.8 km = 1800 m.

Answer: (A) 2 seconds

Concept: $F=\frac{G{m}_1{m}_2}{r^2}$, so $G=\frac{F{r}^2}{m_1{m}_2}$.

1. Unit = $\frac{\text{N}\times {\text{m}}^2}{{\text{kg}}^2}$
2. $=\text{N}{\text{m}}^2{\text{kg}}^{-2}$

Yaad rakhein: $G=6.67\times {10}^{-11}\text{N}{\text{m}}^2{\text{kg}}^{-2}$; it is the same everywhere in the universe, unlike g.

Answer: (A) Nm2kg-2

Concept: Epithelial tissue covers body surfaces and lines organs.

1. It forms a protective covering (skin, lining of mouth)
2. Glandular epithelium secretes substances (e.g. sweat, enzymes)
3. Coordination is done by nervous tissue, support by connective tissue, contraction by muscle

Yaad rakhein: Types: squamous (skin, lungs), cuboidal (kidney tubules), columnar (intestine), ciliated, glandular.

Answer: (B) Secretion and protection

Concept: A more reactive element displaces a less reactive one from its compound.

1. A displaces B, so A > B
2. C displaces A, so C > A
3. Order: C > A > B

Trick: The displacer is always on the left: chain the two clues as C > A > B.

Answer: (B) C > A > B

Concept: Groundwater is stored in porous water-bearing layers (aquifers).

1. A dug well draws water that seeps from the water-bearing strata around it
2. Impervious strata (clay, hard rock) do not let water through, so they cannot supply a well

Yaad rakhein: The upper level of groundwater is called the water table; it falls when we take out more water than rain recharges.

Answer: (C) Water-bearing strata

Concept: Only about 10% of energy passes from one trophic level to the next.

1. About 90% is lost as heat and in life processes at each level
2. After 4 levels, too little energy remains to support another level

Trick: 10% law: 1000 J → 100 → 10 → 1 J. By the 4th level almost nothing is left.

Answer: (D) of large loss of energy at each level