Q&A: practice questions with answers

Concept: Empirical relation: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $\text{Mode}=3\times 100-2\times 76$.
2. $=300-152=148$.

Trick: Remember '3 Median minus 2 Mean'.

Answer: (A) 148

Concept: Form two linear equations from the statements and solve.

1. $9A+8B=29$ and $2B-2A=20\Rightarrow B=A+10$.
2. $9A+8(A+10)=29\Rightarrow 17A=-51\Rightarrow A=-3,B=7$.
3. Present sum $A+B=4$.
4. Sum after 6 years $=4+6+6=16$ years.

Trick: Sum of two ages after n years = present sum + 2n.

Answer: (C) 16

Concept: $SI=\frac{P\times R\times T}{100}\Rightarrow R=\frac{SI\times 100}{P\times T}$.

1. $R=\frac{500\times 100}{1250\times 10}$.
2. $=\frac{50000}{12500}=4\%$.

Trick: Interest is $\frac{500}{1250}=40\%$ in 10 years → $4\%$ per year.

Answer: (B) $4\%$

Concept: Overtaking in the same direction: time = (sum of lengths) ÷ (difference of speeds).

1. Relative speed $=110-100=10\text{km/h}=10\times \frac{5}{18}=\frac{25}{9}\text{m/s}$.
2. Distance $=100+500=600\text{m}$.
3. Time $=600\times \frac{9}{25}=216\text{s}=\frac{216}{60}=3.6$ minutes.

Trick: In km and hours: $\frac{0.6}{10}\text{h}=0.06\times 60=3.6$ min.

Answer: (C) 3.6

Concept: Express both conditions in terms of Burhan's time and subtract.

1. Let Anirudh's speed $=a$, Burhan's time $=t$. Then $\frac{90}{a}=t+3$ and $\frac{90}{2a}=t-2$.
2. Subtract: $\frac{90}{a}-\frac{90}{2a}=5\Rightarrow \frac{45}{a}=5$.
3. $a=9\text{km/h}$ (check: 10 h normally, 5 h at double speed; Burhan 7 h ✓).

Trick: Doubling speed halves the time: half of Anirudh's time = 3 + 2 = 5 h, so time = 10 h, speed = 9 km/h.

Answer: (C) 9 km/hr

Concept: ${(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2$ and $x^3-\frac{1}{x^3}={(x-\frac{1}{x})}^3+3(x-\frac{1}{x})$.

1. ${(x-\frac{1}{x})}^2=83-2=81\Rightarrow x-\frac{1}{x}=9$.
2. $x^3-\frac{1}{x^3}=9^3+3\times 9=729+27=756$.

Trick: If $x-\frac{1}{x}=k$, then $x^3-\frac{1}{x^3}=k^3+3k$.

Answer: (D) 756

Concept: Track the position with East–West and North–South totals; a left turn while facing West means facing South, a right turn means facing North.

1. West 18, left (South) 6, right (West) 21 → he is 39 km West and 6 km South of A.
2. Right (North) 18 → 6 km South becomes 12 km North; right (East) 39 → East–West offset becomes 0.
3. So P is exactly 12 km North of A; to return he must drive 12 km towards the South.

Trick: Net E–W: $-18-21+39=0$; net N–S: $-6+18=12$ North → go back South.

Answer: (A) 12 km to the South

Concept: A conclusion follows only if it is true in every possible Venn diagram of the statements.

1. Induction lies inside Stove; Oven meets Stove only in “some” part – that part may lie outside Induction.
2. Tandoor meets Oven only in “some” part, so it need not touch Induction → (I) Some tandoor are induction: not certain.
3. Oven may overlap only the non-induction part of Stove → (II) Some oven are induction: not certain.

Trick: “Some + Some” or “All + Some” never gives a definite “some” link to the first term.

Answer: (A) Neither conclusion (I) nor (II) follows

Concept: Check the differences between consecutive terms.

1. Differences: $13-6=7$, $25-13=12$, $42-25=17$, $64-42=22$
2. The differences rise by 5: next difference $=27$
3. $64+27=91$

Trick: Second differences constant (5) → keep adding 5 to the last gap.

Answer: (C) $91$

Concept: Decode each symbol step by step and draw the family tree.

1. P + Q: P is the father of Q. Q × R: Q is the son of R, so R is Q’s mother (wife of P).
2. R – S: R is the daughter of S. S ÷ T: S is the wife of T, so T is R’s father.
3. R is T’s daughter and P is R’s husband → P is T’s daughter’s husband (son-in-law).

Trick: Go from T downwards: T → daughter R → her husband P.

Answer: (A) Daughter's husband

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

1. ‘talk to him’ → ‘yd hp mu’; ‘when you talk’ → ‘nc mu bt’
2. Common word: talk; common code: mu
3. So talk = mu.

Trick: Match the common word with the common code – order of codes does not matter.

Answer: (D) mu

Concept: Treat the first letter, second letter and number as three separate series.

1. First letters: J, I, H, G → each −1 → F
2. Second letters: N, M, L, K → each −1 → J
3. Numbers: 28, 40, 52, 64 → each $+12$ → $76$

Trick: Both letters move back 1; number $+12$ → FJ76.

Answer: (D) FJ76

Concept: Swap the signs first, then solve using BODMAS.

1. Swap ÷ ↔ − and × ↔ +: $4320÷96+37\times 21-372$
2. $4320÷96=45$, $37\times 21=777$
3. $45+777-372=450$

Trick: Write the new expression fully before calculating – most errors come from half-swapped signs.

Answer: (B) $450$

Concept: Find the letter-shift that turns the left term into the right term in each pair.

1. GLI → %: try +5 on each letter: G+5 = L, L+5 = Q, I+5 = N → % = LQN.
2. Then # → QVS must also be +5: L+5 = Q, Q+5 = V, N+5 = S → # = LQN.
3. Only option (B) gives # = LQN and % = LQN.

Trick: Check one option at a time with the same +5 rule on both sides.

Answer: (B) # = LQN, % = LQN

Concept: Look for the pattern symbol – number – symbol.

1. Series: 9 # 2 6 7 © 6 8 & © $ & 3 9 1 7 $ 5 # @ * 4
2. # 2 6 (6 is a number), & 3 9 (9 is a number) – not valid.
3. Only $ 5 # fits → count = 1.

Trick: Scan only the numbers that come right after a symbol, then check the next element.

Answer: (D) 1

Concept: Add up East–West and North–South movements separately.

1. West 5, right (North) 11, right (East) 3, left (North) 4, right (East) 3, right (South) 15.
2. East–West: $-5+3+3=+1$ → 1 km East. North–South: $11+4-15=0$.
3. B is 1 km East of A, so he must drive 1 km towards the West.

Trick: North–South cancels ($15=11+4$) → only the East–West gap of 1 km remains.

Answer: (A) 1 km towards the West

Concept: Find the shift applied to every letter of the first pair.

1. F→J, B→F, H→L, P→T: each letter $+4$. Check VRXF → ZVBJ: also $+4$ (X+4 = B wraps round).
2. HDJR: H+4 = L, D+4 = H, J+4 = N, R+4 = V → LHNV.

Trick: Same +4 to each letter, keep the order – LHNV, not a jumbled form.

Answer: (C) LHNV

Concept: Add 1 to every digit (both odd and even), then pick the required digits.

1. 57831426 → 6 8 9 4 2 5 3 7
2. First two: $6+8=14$; last two: $3+7=10$
3. Sum $=14+10=24$

Trick: Only the four needed digits matter: $(5+7+3+6)+4=24$.

Answer: (C) $24$

Concept: Code each element from the table, then apply whichever condition fits.

1. Elements: 5, 3, 6, +, 8 → F, C, P, J, K.
2. Condition I: first element 5 is not a symbol – no. Condition III: 3 and 6 are not perfect squares – no.
3. Condition II: first 5 is odd, last 8 is even → both coded as @ → @CPJ@.

Trick: Check first and last elements first – they decide conditions I and II.

Answer: (A) @CPJ@

Concept: The letter common to both words takes the digit common to both codes.

1. KITE → 3641 and LACK → 1295
2. Common letter: K; common digit: 1
3. So K = 1.

Trick: Common letter ↔ common digit; positions may be jumbled.

Answer: (B) 1