Q&A: practice questions with answers

Concept: Volume of cone $V=\frac{1}{3}\pi r^{2}h$; water $=\frac{2}{3}$ of $V$.

1. $V=\frac{1}{3}\pi \times {31}^2\times 45=15\times 961\pi =14415\pi {\text{cm}}^3$.
2. Water $=\frac{2}{3}\times 14415\pi =9610\pi {\text{cm}}^3$.

Trick: Combine fractions: $\frac{2}{3}\times \frac{1}{3}\times 45=10$, so answer $=10\times 961\pi$.

Answer: (C) $9610\pi$

Concept: $\text{CP}=\frac{\text{SP}}{1+\text{gain}}$.

1. SP of 1 chocolate $=\frac{1}{4}$ rupee.
2. Gain $100\%$ → CP $=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$ rupee per chocolate.
3. So for ₹1 he bought $8$ chocolates.

Trick: 100% gain means CP is half of SP → he bought twice as many per rupee: $2\times 4=8$.

Answer: (C) $8$

Concept: Subtract the same $x$ years from both ages.

1. $x$ years ago: Arvind $=40-x$, Asha $=25-x$.
2. $40-x=4(25-x)\Rightarrow 40-x=100-4x$.
3. $3x=60\Rightarrow x=20$.

Trick: Check: 20 years ago ages were $20$ and $5$; $20=4\times 5$. ✓

Answer: (B) $20$ years

Concept: Average = sum ÷ number of values.

1. Sum $=20+30+35+40+45=170$.
2. Average $=\frac{170}{5}=34$ years.

Trick: Deviation from 35: $-15,-5,0,+5,+10$ → total $-5$ → $35-1=34$.

Answer: (A) $34$

Concept: Winning margin as a percentage of total votes.

1. Loser $=48\%$, winner $=52\%$ (only two candidates).
2. Margin $=52\%-48\%=4\%$ of total $=756$.
3. Total $=\frac{756\times 100}{4}=18900$.

Trick: Margin % $=100-2\times 48=4\%$; then $756\times 25$.

Answer: (D) $18900$

Concept: The perpendicular from the centre bisects the chord; use Pythagoras.

1. Half chord $AM=\frac{12}{2}=6\text{cm}$, $OM=5\text{cm}$.
2. $r^2=6^2+5^2=36+25=61$.
3. $r=\sqrt{61}\approx 7.81\text{cm}$.

Trick: $\sqrt{61}$ lies between ${7.8}^2=60.84$ and ${7.9}^2=62.41$ → only 7.81 fits.

Answer: (A) $7.81\text{cm}$

Concept: Volume is conserved; surface area $=4\pi r^2$.

1. $\frac{4}{3}\pi (10{)}^3=125\times \frac{4}{3}\pi r^3\Rightarrow r^3=8\Rightarrow r=2\text{cm}$.
2. Big sphere SA $=4\pi (10{)}^2=400\pi$.
3. 6 small spheres SA $=6\times 4\pi (2{)}^2=96\pi$.
4. Ratio $=400:96=25:6$.

Trick: $n$ equal spheres → $r=\frac{R}{\sqrt[3]{n}}$; here $\frac{10}{5}=2$.

Answer: (B) $25:6$

Concept: Equation with sum $S$ and product $P$ of roots: $x^2-Sx+P=0$.

1. $S=4-3\sqrt{2}$, $P=-28$.
2. $x^2-(4-3\sqrt{2})x+(-28)=0$.
3. $x^2-(4-3\sqrt{2})x-28=0$.

Trick: Sign rule: minus sum, plus product — so a negative product appears as $-28$.

Answer: (D) $x^2-(4-3\sqrt{2})x-28=0$

Concept: Check each condition on the given group, then write the codes from the table.

1. Codes: @ = Z, 9 = C, 3 = L, 8 = T, 5 = A.
2. First element @ is a symbol and last element 5 is a number → condition (i) applies: swap the codes of first and last.
3. Condition (iii) fails (3 is not a perfect square); (ii) fails (first is not a number).
4. So @ → A, 9 → C, 3 → L, 8 → T, 5 → Z.

Trick: Test only the first, second, third and last elements – the middle ones are always coded directly.

Answer: (D) A C L T Z

Concept: Look for the pattern Number – Number – Symbol.

1. Series: 2 * 3 & @ @ 6 4 $ @ 3 * % 6 2 1 3 9 7 8 2 €
2. 6 4 $ → 4 qualifies.
3. 8 2 € → 2 qualifies. No other number has a number before and a symbol after.

Trick: Scan only the symbols: check whether the two elements just before each symbol are both numbers.

Answer: (D) 2

Concept: Use the product and sum clues to fix E first.

1. F × E = 15 → {3, 5}. D + E = 7 → if E = 3, D = 4; if E = 5, D = 2.
2. Case E = 5, F = 3, D = 2: free floors 1, 4, 6 – no two consecutive for B, C. Rejected.
3. Case E = 3, F = 5, D = 4: free floors 1, 2, 6 → B = 1, C = 2, A = 6.
4. Between B (1) and D (4): C and E → 2 people.

Trick: Clue with only two possible values (E) → test both cases and reject the one that breaks 'immediately above'.

Answer: (B) 2

Concept: Five people between L and Q in a row of 7 → they sit at the two ends.

1. P is immediately left of Q, so Q cannot be at the left end → Q = 7, P = 6, L = 1.
2. One person between P and I → I = 4.
3. Positions 2, 3, 5 left for J, O, K with J – O – K order → J = 2, O = 3, K = 5.
4. Third from left = O.

Trick: '5 between' in a row of 7 always means extreme ends.

Answer: (C) O

Concept: Compare each letter of the first cluster with the corresponding letter of the second.

1. NM – PO: N + 2 = P, M + 2 = O.
2. SR – UT: +2, +2; UT – WV: +2, +2.
3. NQ – RM: N + 4 = R, Q − 4 = M → different.

Trick: Also inside each cluster: NM, SR, UT are reverse consecutive letters; NQ is not.

Answer: (A) NQ - RM

Concept: Compare the first cluster with the second, letter by letter.

1. PS – QT: P + 1 = Q, S + 1 = T.
2. KN – LO: +1, +1; GJ – HK: +1, +1.
3. DH – FP: D + 2 = F, H + 8 = P → different.

Trick: Inside the first cluster also: PS, KN, GJ have gap +3; DH has gap +4.

Answer: (D) DH - FP

Concept: Find the differences between consecutive terms.

1. $1001-1=1000$, $1000-3=997$, $997-5=992$, $992-7=985$
2. Differences are consecutive odd numbers 1, 3, 5, 7, 9.
3. Next $=985-9=976$

Trick: Odd-number differences → also $1001-n^2$ for $n=0,1,2,\dots$: $1001-25=976$.

Answer: (B) 976

Concept: Facing the centre: a person's left neighbour is clockwise and right neighbour is anticlockwise.

1. From the clues, the anticlockwise chain is P – L – R – Q – M – O – N, so clockwise from L: L, P, N, O, M, Q, R.
2. New clockwise order from L: L, M, N, O, P, Q, R.
3. Compare seat by seat: L, N, O, Q, R stay; P and M swap seats.
4. Unchanged = 5.

Trick: Write the present clockwise order starting at L under the alphabet and count matches.

Answer: (D) 5

Concept: Replace letters by signs, then apply BODMAS.

1. $45\times 14÷21-17+25$
2. $45\times 14=630$, $630÷21=30$
3. $30-17+25=38$

Trick: $\frac{14}{21}=\frac{2}{3}$ → $45\times \frac{2}{3}=30$ quickly.

Answer: (D) 38

Concept: Each letter moves $+4$ in the alphabet (Z wraps to A).

1. 1st letters: S → W → A → E → I
2. 2nd letters: K → O → S → W → A
3. 3rd letters: C → G → K → O → S
4. Next term: IAS

Trick: Positions $+4$: S(19) W(23) A(27 = 1) E(5) I(9).

Answer: (A) IAS

Concept: Start with the fixed day (S = Wednesday) and chain the 'immediately' clues.

1. S Wed → O Thu → U Fri.
2. T after U and before Q → T Sat, Q Sun.
3. P and R take Mon, Tue with R after P → P Mon, R Tue.
4. Between S (Wed) and Q (Sun): O, U, T → three.

Trick: Chain 'immediately after' clues first – they fix several days at once.

Answer: (D) Three

Concept: Vowel → next letter; consonant → previous letter.

1. BAD → A B C (has A); FIG → E J F (has E)
2. ICE → J B F (no vowel)
3. NOT → M P S (no vowel)
4. Clusters without a vowel: 2

Trick: A new vowel can come only from a consonant just after a vowel (B→A, F→E, J→I, P→O, V→U).

Answer: (A) Two