Q&A: practice questions with answers

Concept: Selling price $=\text{MP}\times \frac{100-d}{100}$.

1. Discount $=25\%$ of $800=200$.
2. SP $=800-200=600$.

Trick: $25\%$ off means pay $\frac{3}{4}$: $\frac{3}{4}\times 800=600$.

Answer: (B) ₹$600$

Concept: One day's work $=\frac{1}{\text{days}}$; work done together = sum of rates × days.

1. $A+B$ per day $=\frac{1}{13}+\frac{1}{24}=\frac{24+13}{312}=\frac{37}{312}$.
2. In 5 days: $\frac{185}{312}$.
3. Remaining $=1-\frac{185}{312}=\frac{127}{312}$.

Trick: LCM $13\times 24=312$ units: A does 24, B does 13 per day → $5\times 37=185$, left $127$.

Answer: (C) $\frac{127}{312}$

Concept: Successive discounts are multiplied: price $\times (1-d_1)(1-d_2)(1-d_3)$.

1. After $20\%$: $10000\times 0.8=8000$.
2. After $10\%$: $8000\times 0.9=7200$.
3. After $5\%$: $7200\times 0.95=6840$.

Trick: Never add discounts ($35\%$ would give ₹$6500$, a trap).

Answer: (D) ₹$6,840$

Concept: Literates among women = total literates − literate men.

1. Total literates $=22\%$ of $359000=78980$.
2. Literate men $=22\%$ of $179000=39380$.
3. Women $=180000$; literate women $=39600$; rate $=\frac{39600}{180000}\times 100=22\%$.

Trick: If the whole and one part both have $22\%$, the other part must also be $22\%$.

Answer: (C) $22\%$

Concept: The largest circle inside a rectangle has diameter = the shorter side.

1. Diameter $=217\text{m}$.
2. Circumference $=\pi d=\frac{22}{7}\times 217=22\times 31=682\text{m}$.

Trick: $217=7\times 31$, so $\frac{22}{7}$ cancels neatly.

Answer: (C) $682$

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

1. P: $\frac{3}{4}$ in 15 h → full in $20$ h. Q: $\frac{5}{6}$ in 25 h → full in $30$ h.
2. In 2 h, P+Q fill $2(\frac{1}{20}+\frac{1}{30})=\frac{1}{6}$.
3. R empties $\frac{1}{6}$ in 5 h → R empties full tank in $30$ h.
4. P+Q−R: $\frac{1}{20}+\frac{1}{30}-\frac{1}{30}=\frac{1}{20}$ → $20$ hours.

Trick: Q and R have equal rates and cancel, so only P works: $20$ hours.

Answer: (D) 20 hours

Concept: Form two linear equations and solve.

1. $5s+f=79$ and $2.5f+s=94$.
2. From the first, $f=79-5s$; so $2.5(79-5s)+s=94\Rightarrow 197.5-11.5s=94$.
3. $s=9$, $f=79-45=34$.

Trick: Option check: $f=34\Rightarrow s=9$; $2.5\times 34+9=94$ ✓.

Answer: (A) $34$

Concept: Total = average × number; split total in the given ratio.

1. Total $=3\times 15990=47970$.
2. Cheapest $=\frac{3}{15}\times 47970=9594$.

Trick: $\frac{3}{15}\times 3\times \text{avg}=\frac{3}{5}\times 15990=9594$.

Answer: (B) $9,594$

Concept: $SI=\frac{Prt}{100}$; amount $=P+SI$.

1. $P\times \frac{7.7}{100}\times \frac{20}{7}=1188\Rightarrow 0.22P=1188\Rightarrow P=5400$.
2. New SI $=5400\times \frac{8.5}{100}\times \frac{41}{9}=51\times 41=2091$.
3. Amount $=5400+2091=7491$.

Trick: $7.7\times \frac{20}{7}=22\%$ total interest – so $P=\frac{1188}{0.22}$.

Answer: (A) ₹$7,491$

Concept: Distance = speed × time (convert minutes to hours).

1. $5.4\text{min}=\frac{5.4}{60}=0.09\text{h}$.
2. Length $=42\times 0.09=3.78\text{km}$.

Trick: km/h × minutes ÷ 60: $\frac{42\times 5.4}{60}=0.7\times 5.4$.

Answer: (D) $3.78\text{km}$

Concept: Replace letters by signs, then apply BODMAS.

1. $9\times 5-12÷2+11$.
2. $=45-6+11=50$.

Trick: Write the converted expression fully before calculating.

Answer: (A) $50$

Concept: Track each letter position separately.

1. 1st letters: X, U, R, O → each $-3$ → L.
2. 2nd letters: Z, W, T, Q → $-3$ → N.
3. 3rd letters: R, O, L, I → $-3$ → F.

Trick: All three positions move $-3$; just shift OQI back by 3.

Answer: (C) LNF

Concept: Total = rank from top + rank from bottom − 1.

1. $25+35-1=59$.

Trick: Subtract 1 because the person is counted twice.

Answer: (A) $59$

Concept: Find the gap within each group and between groups.

1. Within groups: E→H ($+3$), H→R ($+10$); L→O, O→Y; S→V, V→F; Z→C, C→M.
2. First letters: E, L, S, Z → $+7$ → G.
3. G $+3$ = J, J $+10$ = T → GJT.

Trick: Get the first letter (+7), then apply the fixed +3, +10 inside.

Answer: (A) GJT

Concept: Write the positions and see how they move.

1. SKIP → KSPI: swap letters 1–2 and 3–4.
2. KSPI → PIKS: swap the two halves (KS)(PI) → (PI)(KS).
3. MEND → EMDN → DNEM follows the same rule.
4. KITE → IKET → ETIK ✓.

Trick: Step 1 = pair swap, Step 2 = half swap.

Answer: (A) KITE - IKET - ETIK

Concept: Look at the differences.

1. Differences: $+18,+9,+36,+18,+54,\dots$
2. Pattern: $+18k$ then half of it: $18,9$; $36,18$; $54,27$.
3. $148+27=175$.

Trick: Alternate differences are multiples of 18 and their halves.

Answer: (C) $175$

Concept: Fix the most definite clue first (positions counted from the left).

1. Only two to the right of N → N at 5; Q immediately left → Q at 4.
2. Two between N and R → R at 2. Two between L and Q → L at 1 or 7.
3. P–O must be adjacent (O right of P): only possible if L = 1 and P, O = 6, 7; M = 3.
4. Row: L R M Q N P O → 3rd from left is M.

Trick: Start with “only two to the right” – it pins a person to a fixed seat.

Answer: (B) M

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

1. “row” is in both sentences.
2. Common code in “eg kv oc” and “jl vf eg” is “eg”.

Trick: Common word ↔ common code.

Answer: (D) eg

Concept: Check the differences.

1. Differences: $11,13,15,17$ (odd numbers increasing by 2).
2. Next difference $=19$: $100+19=119$.

Trick: Differences rising by 2 → second difference is constant.

Answer: (C) $119$

Concept: Draw the least-overlap Venn diagram allowed by the statements.

1. Watches lie fully inside Belts; Ties lie completely outside Belts.
2. So no watch is a tie → I (Some watches are ties) is false.
3. No tie is a belt → II (All ties are belts) is false.

Trick: “All A are B” + “No B is C” ⇒ “No A is C”.

Answer: (D) Neither conclusion (I) nor (II) follows.