Q&A: practice questions with answers

Concept: Empirical formula: $\text{Mode}=3\text{Median}-2\text{Mean}$, i.e. $\text{Mode}-\text{Mean}=3(\text{Median}-\text{Mean})$.

1. $16.2=3(\text{Median}-\text{Mean})\Rightarrow \text{Median}-\text{Mean}=5.4$.
2. $\text{Mode}-\text{Median}=16.2-5.4=10.8$.

Trick: Mode − Median $=\frac{2}{3}$ of (Mode − Mean) $=\frac{2}{3}\times 16.2=10.8$.

Answer: (B) 10.8

Concept: Two successive discounts $a\%$ and $b\%$ equal a single discount of $(a+b-\frac{ab}{100})\%$.

1. $a=15,b=20$.
2. Net $=15+20-\frac{15\times 20}{100}=35-3=32\%$.

Trick: On ₹100: after 15% → 85; after 20% → 68; discount $=32$.

Answer: (A) 32%

Concept: Saving % $=100\%-$ expenditure %.

1. Saving $=100\%-65\%=35\%$.
2. $35\%$ of 720 $=\frac{35\times 720}{100}=252$.

Trick: $10\%=72$, $5\%=36$ → $35\%=3\times 72+36=252$.

Answer: (B) ₹252

Concept: $\text{SP}=\text{CP}\times \frac{100\pm \text{gain/loss}\%}{100}$.

1. At 15% loss, SP $=85\%$ of CP $=18700$.
2. CP $=\frac{18700\times 100}{85}=22000$.
3. SP for 15% gain $=22000\times \frac{115}{100}=25300$.

Trick: Direct: new SP $=18700\times \frac{115}{85}=220\times 115=25300$.

Answer: (B) ₹25,300

Concept: Take the youngest person's age as $x$ and express the others through it.

1. Tanu $=x$, Kanu $=2x$, Anu $=2x+2$.
2. $x+2x+2x+2=32\Rightarrow 5x=30\Rightarrow x=6$.
3. Anu $=14$, Tanu $=6$; difference $=8$ years.

Trick: Difference Anu − Tanu $=(2x+2)-x=x+2=8$.

Answer: (B) 8 years

Concept: Angle sum of a triangle is ${180}^{\circ }$; with parallel lines, alternate angles are equal; in an isosceles triangle, angles opposite equal sides are equal.

1. In $△PQR$: $\angle PRQ={180}^{\circ }-{96}^{\circ }-{34}^{\circ }={50}^{\circ }$.
2. $QR\parallel PS$ → $\angle RPS=\angle PRQ={50}^{\circ }$ (alternate).
3. $PR=RS$ → $\angle RSP=\angle RPS={50}^{\circ }$.
4. $\angle PRS={180}^{\circ }-{50}^{\circ }-{50}^{\circ }={80}^{\circ }$.

Trick: Diagonal PR is a transversal of the parallel sides – look for the alternate-angle pair first.

Answer: (C) ${80}^{\circ }$

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

1. WIFE = 1852, FIRE = 7258.
2. Common letters: F, I, E; common digits: 8, 5, 2.
3. Remaining letter R ↔ remaining digit 7.

Trick: Strike out the common letters/digits; what is left in FIRE (R) matches what is left in 7258 (7).

Answer: (A) 7

Concept: Find the shift applied to each letter by alphabet positions.

1. J→N, G→K, L→P, I→M: each $+4$.
2. PMRO→TQVS: also $+4$ each.
3. MJOL: M→Q, J→N, O→S, L→P = QNSP.

Trick: Check only the first and last letters against options: Q…P → QNSP.

Answer: (C) QNSP

Concept: Perfect-square digits: 0, 1, 4, 9. Find odd digits with a perfect square just before and an odd digit just after.

1. Scan pairs: 9-1-1, 1-1-5, 1-5-7.
2. 4-5-9 (positions 8–10).
3. 4-3-5 (near the end).
4. Total $=5$.

Trick: First mark every odd–odd pair, then check only the digit before the first of each pair.

Answer: (A) 5

Concept: Each letter of a term moves $+3$ to give the next term.

1. G→J→M→P→S.
2. I→L→O→R→U.
3. N→Q→T→W→Z.
4. Next term $=$ SUZ.

Trick: Within each term the gaps are $+2,+5$ (G→I→N); S→U→Z keeps this.

Answer: (D) SUZ

Concept: Test each option to find a single rule linking both pairs.

1. Try X $=8$: $8\times 11-1=87$ ✓
2. Same rule: $14\times 11-1=154-1=153$.
3. So X $=8$, Y $=153$.

Trick: Options pair X with Y; fix the rule using X, then test Y – only (B) is consistent.

Answer: (B) X = 8, Y = 153

Concept: Decode each symbol and build a family tree.

1. P + Q: P is Q's wife.
2. Q − R: Q is R's brother.
3. R × S: R is S's father; S ÷ T: S is T's sister → R is T's father.
4. Q is T's father's brother, so P is T's father's brother's wife.

Trick: Read the chain from T backwards: T → father R → his brother Q → Q's wife P.

Answer: (C) Father’s brother’s wife

Concept: Test the options to find one rule that fits both pairs.

1. Try X $=12$: $12\times 6=72$ ✓
2. Same rule: $22\times 6=132$.
3. So X $=12$, Y $=132$.

Trick: Fix the rule from the left pair using X, then apply it to 22 – only (B) matches.

Answer: (B) X = 12, Y = 132

Concept: A course of action follows if it practically addresses the stated problem or reduces its impact.

1. Cause: pesticide use. I restricts pesticides during peak pollination – tackles the cause directly.
2. Effect: falling bee numbers. II boosts bee populations and pollination – reduces the impact.
3. Both are reasonable and practical.

Yaad rakhein: A valid action may attack the cause (I) or soften the effect (II).

Answer: (D) Both I and II follow

Concept: Fix definite positions first (positions 1–6 from the left, all facing north).

1. Only one to Clara's left → Clara at 2.
2. One between Clara and Brad → Brad at 4.
3. Adam not next to Clara (1, 3) and not at an end (6) → Adam at 5.
4. Emily–David with David second to her right: only 1 and 3 → Emily 1, David 3; Farah at 6.

Trick: Place people with fixed positions first; the “second to the right” pair then has only one fit.

Answer: (D) Farah

Concept: Seven seats, 1–7 from the left, all facing north.

1. Five between B and L → they occupy the ends 1 and 7.
2. Z is immediately left of L, so L $=7$, Z $=6$, B $=1$.
3. One between Z and D → D $=4$.
4. Left seats 2, 3, 5 with X before A before Y → X $=2$, A $=3$, Y $=5$.

Trick: “Five between” in a row of seven always means the two ends.

Answer: (D) A

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

1. ‘over the top’ → ‘ak ot yh’.
2. ‘the rock is’ → ‘ot gb tk’.
3. Common word ‘the’ ↔ common code ‘ot’.

Trick: Match common word with common code.

Answer: (B) ot

Concept: Write letter positions and find the pattern: in $ab$-$cd$-$ef$: $b=a+4$, $c=a+2$, $d=b-5$, $e=c+1$, $f=d+6$.

1. TX-VS-WY: T(20)→X(24) $+4$; V $=$ T$+2$; S $=$ X$-5$; W $=$ V$+1$; Y $=$ S$+6$.
2. OS-QN-RT: O(15)→S(19) $+4$; Q $=$ O$+2$; N(14) $=$ S$-5$; R $=$ Q$+1$; T(20) $=$ N$+6$ ✓
3. IM-KH-LO fails: H$+6=$ N, not O.

Trick: Options often differ only in the last letter – check the final step ($+6$) first.

Answer: (A) OS-QN-RT

Concept: A universal negative (“No A is B”) can be reversed: “No B is A”. Two negatives give no definite conclusion about the end terms.

1. “No car is a bus” → “No bus is a car”: I follows.
2. Car and plane are linked only through two negative statements, so they may or may not overlap: II does not follow.

Yaad rakhein: Two negative premises → no conclusion between the end terms.

Answer: (A) Only conclusion I follows

Concept: Check the gaps between consecutive letters by alphabet position.

1. CEG: $+2,+2$.
2. JLN: $+2,+2$.
3. TVX: $+2,+2$.
4. FUT: F(6)→U(21)→T(20): $+15,-1$ – odd.

Trick: Three clusters are three alternate letters in order; FUT breaks the pattern.

Answer: (C) FUT