Q&A: practice questions with answers

Concept: Express both selling prices in terms of the cost price (CP).

1. $\frac{8}{13}$ of actual SP $=80\%$ of CP → actual SP $=0.8\text{CP}\times \frac{13}{8}=1.3\text{CP}$.
2. New SP $=85\%$ of $1.3\text{CP}=1.105\text{CP}$.
3. Profit $\%=(1.105-1)\times 100=10.5\%$.

Trick: Let CP $=100$: $\frac{8}{13}$SP $=80$ → SP $=130$; $0.85\times 130=110.5$ → $10.5\%$ profit.

Answer: (D) $10.5\%$

Concept: Share $=\frac{\text{ratio part}}{\text{sum of parts}}\times \text{total}$.

1. Sum of parts $=5+12+11=28$; one part $=\frac{616}{28}=22$.
2. Largest $-$ smallest $=(12-5)$ parts $=7\times 22=154$.

Trick: Find the value of one part, then multiply by the difference of ratio terms – no need to find each share.

Answer: (A) 154

Concept: Compute both quantities separately and subtract.

1. $65\%$ of $75=\frac{65\times 75}{100}=48.75$.
2. $\frac{4}{5}$ of $15=12$.
3. Difference $=48.75-12=36.75$.

Trick: $65\%$ of 75 $=75\times 0.65$: $75\times 0.6=45$, $75\times 0.05=3.75$ → 48.75.

Answer: (A) 36.75

Concept: Average $=\frac{\text{sum of observations}}{\text{number of observations}}$.

1. Sum $=40+49+56+75+35=255\text{kg}$.
2. Average $=\frac{255}{5}=51\text{kg}$.

Trick: Assume 50 as base: deviations $-10,-1,+6,+25,-15$ sum to $+5$ → $50+\frac{5}{5}=51$.

Answer: (B) 51

Concept: Rate of a pipe $=\frac{\text{part filled}}{\text{time}}$; an emptying pipe has a negative rate.

1. P: $\frac{3}{7}÷12=\frac{1}{28}$ per hour; Q: $\frac{5}{6}÷10=\frac{1}{12}$ per hour.
2. In 2 h, P + Q fill $2(\frac{1}{28}+\frac{1}{12})=2\times \frac{10}{84}=\frac{5}{21}$.
3. R empties $\frac{5}{21}$ in 5 h → R $=\frac{1}{21}$ per hour.
4. P + Q − R $=\frac{1}{28}+\frac{1}{12}-\frac{1}{21}=\frac{3+7-4}{84}=\frac{1}{14}$ → 14 hours.

Trick: LCM of 28, 12, 21 $=84$ units: P = 3, Q = 7, R = −4 → net 6 units/h → $84÷6=14$ h.

Answer: (C) 14 hours

Concept: Form an equation and divide to get the ratio.

1. $\frac{18}{100}a=\frac{2}{5}b$.
2. $\frac{a}{b}=\frac{2}{5}\times \frac{100}{18}=\frac{40}{18}=\frac{20}{9}$.

Trick: If $x\%$ of A $=y\%$ of B then $A:B=y:x$; here $40:18=20:9$.

Answer: (A) 20 : 9

Concept: Perimeter of a square $=4a$, area $=a^2$.

1. Side $a=\frac{340}{4}=85\text{m}$.
2. Area $={85}^2=7225{\text{m}}^2$.

Trick: Square of a number ending in 5: ${85}^2$ → $8\times 9=72$, append 25 → 7225.

Answer: (D) 7225

Concept: For the same time period, profit is shared in the ratio of investments.

1. Pawan's share of profit $=\frac{4800}{20000}=\frac{6}{25}$, so Aman's $=\frac{19}{25}$.
2. Pawan's investment $=\frac{6}{25}\times 76500=18360$; Aman's $=58140$.
3. Difference $=58140-18360=39780$ → ₹39,780.

Trick: Difference $=\frac{19-6}{25}\times 76500=\frac{13}{25}\times 76500=39780$.

Answer: (A) ₹39,780

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

1. In $△ABC$: $\angle BCA={180}^{\circ }-{17}^{\circ }-{125}^{\circ }={38}^{\circ }$.
2. $BC\parallel AD$ with transversal AC → $\angle CAD=\angle BCA={38}^{\circ }$ (alternate angles).
3. $AC=CD$ → $\angle CDA=\angle CAD={38}^{\circ }$.
4. $\angle ACD={180}^{\circ }-{38}^{\circ }-{38}^{\circ }={104}^{\circ }$.

Trick: Isosceles triangle: vertex angle $={180}^{\circ }-2\times \text{base angle}$.

Answer: (A) ${104}^{\circ }$

Concept: Surface area of a sphere $=4\pi r^2$.

1. Radius $r=\frac{14}{2}=7\text{cm}$.
2. $S=4\times \frac{22}{7}\times 7\times 7=4\times 22\times 7=616{\text{cm}}^2$.

Trick: With $r=7$: $\pi r^2=154$; sphere surface $=4\times 154=616$.

Answer: (D) $616{\text{cm}}^2$

Concept: Change each digit by the given rule, then compare the new digits.

1. 925746: odd digits 9, 5, 7 → $-4$ → 5, 1, 3; even digits 2, 4, 6 → $+2$ → 4, 6, 8.
2. New number: 5 4 1 3 6 8.
3. Highest digit $=8$; digits in ascending order 1, 3, 4, … → second lowest $=3$.
4. Difference $=8-3=5$.

Trick: Write the new digits in ascending order once – both 'highest' and 'second lowest' can be read off directly.

Answer: (A) 5

Concept: Look for the pattern number – symbol – number (a symbol with a digit on both sides).

1. Scan every symbol and check its left and right neighbours.
2. '1 % 2' (near the right end): % has 1 before and 2 after ✔.
3. '2 + 2': + has 2 before and 2 after ✔.
4. Others fail, e.g. '5 > ' is followed by a symbol, '6 < +' is followed by a symbol, '7' and '1' are adjacent digits with no symbol between.

Trick: Only check symbols that sit right after a digit – this cuts the work in half.

Answer: (D) Two

Concept: A letter common to two words takes the digit common to their codes.

1. WHAT = 3157 and HORN = 2416.
2. Common letter: H. Common digit: 1.
3. So H = 1.

Trick: Match common letters with common digits – the positions inside the code do not matter.

Answer: (C) 1

Concept: Track each position of the three-letter groups separately.

1. 1st letters: U, C, K, S → each $+8$ (U(21) + 8 = 29 → C) → next S(19) + 8 = 27 → A.
2. 2nd letters: T, B, J, R → $+8$ → R + 8 → Z.
3. 3rd letters: P, X, F, N → $+8$ → N + 8 → V.
4. Next group: AZV.

Trick: Each group is (letter, letter − 1, letter − 5): A, Z, V fits (A − 1 = Z, A − 5 = V cyclically).

Answer: (B) AZV

Concept: Facing the centre, 'left' means clockwise and 'right' means anticlockwise.

1. Fix D. E is immediate right of D; A is third to the left of D.
2. E is second to the left of C → C is two places to the right of E, i.e. opposite side next to A.
3. Two people between G and C counted from G's left → G is immediate left of D.
4. B neighbours A → B sits between G and A; F takes the last seat between C and E.
5. From B going right: G, D, then E → 2 people between.

Trick: Draw the circle and mark left (clockwise) once with an arrow before placing anyone.

Answer: (C) 2

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

1. Swap + ↔ ÷ and − ↔ ×: $24÷3-10\times 2+41$.
2. $24÷3=8$, $10\times 2=20$.
3. $8-20+41=29$.

Trick: Rewrite the full expression with new signs before calculating – never swap in your head.

Answer: (D) 29

Concept: Compare letter positions within and across each pair.

1. TQ − PM: T(20) → Q(17) is $-3$; P(16) → M(13) is $-3$.
2. ZW − VS: $-3$, $-3$. IF − EB: $-3$, $-3$.
3. NK − II: N(14) → K(11) is $-3$, but I(9) → I(9) is $0$ → breaks the pattern.

Trick: In each regular pair both clusters have a gap of 3, and the second cluster starts 4 letters before the first.

Answer: (C) NK − II

Concept: Find the shift for each letter from the example pairs.

1. LEAV → SLHC: L + 7 = S, E + 7 = L, A + 7 = H, V + 7 = C.
2. NGCX → UNJE also $+7$ each.
3. IBXS: I + 7 = P, B + 7 = I, X + 7 = E, S + 7 = Z → PIEZ.

Trick: Check only the first letter (I + 7 = P) and second (B + 7 = I) – option A alone starts with PI.

Answer: (A) PIEZ

Concept: Decode each symbol one by one and build the family tree.

1. D ¥ O: D is the son of O.
2. O ÷ I: O is the father of I. I + N: I is the brother of N → N is also O's child.
3. N @ G: N is the wife of G.
4. D is a son of O, so D is N's brother; N is G's wife → D is G's wife's brother.

Trick: Read the chain from the person asked about (D) towards G and name each link.

Answer: (A) Wife's brother

Concept: Find the shift of each letter position in the example pairs.

1. VALG → YYOE: V + 3 = Y, A − 2 = Y, L + 3 = O, G − 2 = E.
2. BWRC → EUUA: $+3$, $-2$, $+3$, $-2$ ✔.
3. HSXY: H + 3 = K, S − 2 = Q, X + 3 = A, Y − 2 = W → KQAW.

Trick: Alternating pattern $+3,-2,+3,-2$; wrap around Z → A when needed (X + 3 = A).

Answer: (A) KQAW