Q&A: practice questions with answers

Concept: The acid quantity does not change; only water is increased.

1. Acid = $\frac{2}{3}\times 30=20$ L; water = 10 L.
2. For 1 : 1, water must equal acid = 20 L.
3. Water to add = $20-10=10$ L.

Trick: Ratio 2 : 1 → 1 : 1 means water goes from 1 part to 2 parts; 1 part = 10 L, so add 10 L.

Answer: (B) 10 L

Concept: Assume CP = 100, find old SP, raise it, then get new CP from the new profit %.

1. Old CP = 100, profit 15% → SP = 115.
2. New SP = $115\times 1.1=126.5$; this gives 20% profit.
3. New CP = $126.5÷1.2=105.41$ → CP rises by about 5.41%.

Trick: $\frac{\text{New CP}}{\text{old CP}}$ = $\frac{1.15\times 1.10}{1.20}=\frac{1.265}{1.2}\approx 1.054$ → about 5.4% – only (D) fits.

Answer: (D) 5.41%

Concept: Average = total of the group ÷ number of members in it.

1. Spent by the other 4 = $1000-400=$ ₹600.
2. Their average = $600÷4=$ ₹150.

Trick: Options above ₹150 would make the 4 spend more than ₹600 – impossible, so (A).

Answer: (A) ₹150

Concept: SP = MP $\times \frac{100-\text{discount}\%}{100}$.

1. $6800=\text{MP}\times \frac{85}{100}$.
2. $\text{MP}=6800\times \frac{100}{85}=80\times 100=$ ₹8000.

Trick: $6800÷85=80$ (since $85\times 8=680$) → MP = ₹8000.

Answer: (C) ₹8,000

Concept: Convert everyone into one kind of worker using the given equality.

1. 4 men = 6 women → 2 men = 3 women.
2. So the team 2M + 3W equals 2M + 2M, i.e. a team of 4 men.
3. 4 men take 18 days → answer 18 days.

Trick: 2M + 3W is exactly half of (4M) + half of (6W) = one full 'team' → same 18 days.

Answer: (C) 18 days

Concept: Side = perimeter ÷ 4; area = side${}^2$.

1. Side = $80÷4=20$ cm.
2. Area = ${20}^2=400{\text{cm}}^2$.

Trick: Area $={\left(\frac{P}{4}\right)}^2=\frac{P^2}{16}=\frac{6400}{16}=400{\text{cm}}^2$.

Answer: (C) 400 cm²

Concept: New score = new total − old total.

1. Old total (6 tests) = $6\times 52=312$.
2. New total (7 tests) = $7\times 63=441$.
3. 7th test = $441-312=129$.

Trick: New score = new avg + (old count × rise) = $63+6\times 11=129$.

Answer: (C) 129

Concept: Apply the offers in the given order and check each condition.

1. ₹40,000 $>$ ₹15,000 → 20% off: $40,000\times 0.8=$ ₹32,000.
2. ₹32,000 $>$ ₹20,000 → extra ₹2,000 off.
3. Final = $32,000-2,000=$ ₹30,000.

Trick: Options below ₹20,000 are impossible – after only 20% off the bill is already ₹32,000.

Answer: (A) ₹30,000

Concept: Diluting keeps salt fixed; adding salt keeps water fixed.

1. Salt = 40% of 150 = 60 L, water = 90 L.
2. 20% salt: total = $60÷0.2=300$ L → add $300-150=150$ L water.
3. Now water = 240 L. For 50% salt, salt must also be 240 L → add $240-60=180$ L salt.

Trick: When adding pure salt, fix the water: at 50% salt = water, so salt needed = water amount.

Answer: (A) Add 150 L of water, then 180 L of salt

Concept: $(1+\sin \theta )(1-\sin \theta )=1-{\sin }^2\theta ={\cos }^2\theta$.

1. Given $\cos \theta =\frac{5}{13}$ (the 2nd-quadrant sign does not matter after squaring).
2. ${\cos }^2\theta =\frac{25}{169}$.

Trick: Quadrant info is a distractor here: squaring removes the sign; no need to find $\sin \theta$.

Answer: (D) $\frac{25}{169}$

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

1. $0.6\times \frac{180}{\pi }=\frac{108}{\pi }$.
2. Taking $\pi \approx 3.14$: $108÷3.14\approx {34.39}^{\circ }$.

Trick: 1 radian $\approx {57.3}^{\circ }$, so 0.6 rad $\approx 0.6\times 57.3\approx {34.4}^{\circ }$ – picks (B) instantly.

Answer: (B) 34.39°

Concept: If $\sin A=\cos B$ (acute angles), then $A+B={90}^{\circ }$.

1. $x+(3x-{10}^{\circ })={90}^{\circ }$.
2. $4x={100}^{\circ }\to x={25}^{\circ }$.

Trick: Check: $3\times 25-10={65}^{\circ }$, and ${25}^{\circ }+{65}^{\circ }={90}^{\circ }$ ✓.

Answer: (A) 25°

Concept: Orthocentre = meeting point of the three altitudes.

1. Acute triangle: every altitude falls inside → orthocentre inside.
2. Right triangle: at the vertex with the right angle.
3. Obtuse triangle: outside the triangle.

Yaad rakhein: Orthocentre and circumcentre behave alike (inside / on / outside for acute / right / obtuse); incentre and centroid are always inside.

Answer: (C) Inside the triangle

Concept: Perpendicular from centre bisects the chord: half-chord = $\sqrt{r^2-d^2}$.

1. Half of chord = $\sqrt{r^2-d^2}$ with r = 17, d = 15: $\sqrt{{17}^2-{15}^2}=\sqrt{32\times 2}=\sqrt{64}$.
2. So half-chord is 8 cm and full chord = 16 cm.

Trick: 8–15–17 is a Pythagorean triplet – the half-chord is 8 at once, so chord = 16.

Answer: (C) 16 cm

Concept: Radius ⟂ tangent at the point of contact, so $(\text{distance}{)}^2=r^2+(\text{tangent}{)}^2$.

1. ${15}^2=r^2+9^2$.
2. $r^2=225-81=144$.
3. r = 12 cm.

Trick: 9–12–15 is $3\times$ (3–4–5) – the radius is 12 without calculation.

Answer: (A) 12 cm

Concept: Antonym = word of opposite meaning.

1. Joyous = full of joy, very happy.
2. Lugubrious = gloomy, mournful, sad in a heavy way – the opposite.
3. Happy, Cheerful, Delighted all mean the same as Joyous (synonyms).

Trick: Three options that mean the same thing are synonyms of the word – the odd one out is the antonym.

Answer: (C) Lugubrious

Concept: A cricket idiom: a 'duck' is a score of zero.

1. To break one's duck = to get the first run, i.e. to achieve a first success after failing.
2. In the sentence, after repeated failures she finally passed → first success.

Yaad rakhein: 'Duck' (0 runs) comes from the egg-like shape of zero – 'duck's egg'.

Answer: (C) Achieve a first success after previous failures

Concept: To cut someone dead = to deliberately ignore a person you know.

1. She passed him in the corridor and acted as if he did not exist.
2. No words were spoken, so it is not an open insult or sarcastic praise – it is deliberate ignoring.

Yaad rakhein: Related: 'give someone the cold shoulder' – also means to ignore or treat coldly.

Answer: (A) Ignored me deliberately

Concept: Break the word into known parts: phenomeno + logical.

1. Correct: PHE-NO-ME-NO-LO-GI-CAL.
2. (A) misses 'a' in -cal; (C) has 'Pha' and double 'l'; (D) changes 'meno' to 'minono'.

Trick: Write 'phenomenon', drop the final 'n', add '-logical' → phenomeno-logical.

Answer: (B) Phenomenological

Concept: Circum (around) + locution (way of speaking) = talking around the point.

1. Correct: CIR-CUM-LO-CU-TION.
2. (B) drops the 'u' before -tion; (C) drops the 'c' in 'Cir'; (D) writes 'lu' instead of 'lo'.

Trick: Same root as circumference, circumstance (circum) and elocution (locution).

Answer: (A) Circumlocution