Q&A: practice questions with answers

Concept: Translate both statements into two linear equations and solve.

1. $A+4B=37$ and $7A+6B=61$.
2. From the first: $A=37-4B$. Substitute: $7(37-4B)+6B=61\Rightarrow 259-22B=61$.
3. $22B=198\Rightarrow B=9$, so $A=37-36=1$.
4. $A+B=1+9=10$.

Trick: Check options: $A+B=10$ with $B=9$ fits both equations instantly.

Answer: (C) $10$

Concept: Find CP from the loss case, then the gain % at the new SP.

1. $\text{CP}=\frac{950}{0.95}=$ ₹$1000$.
2. New SP = ₹$1340$, gain = ₹$340$.
3. $x\%=\frac{340}{1000}\times 100=34\%$.

Trick: A 5% loss means SP is 95% of CP: $950\to 1000$ at a glance.

Answer: (D) $34$

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

1. So Median divides Mean–Mode in ratio $1:2$.
2. $\text{Mode}-\text{Mean}=17.7=3$ parts ⇒ 1 part $=5.9$.
3. $\text{Mode}-\text{Median}=2\times 5.9=11.8$.

Trick: $\text{Mode}-\text{Median}=\frac{2}{3}(\text{Mode}-\text{Mean})=\frac{2}{3}\times 17.7$.

Answer: (D) $11.8$

Concept: $\text{Average}=\frac{\text{Sum}}{\text{Number of items}}$.

1. Sum $=288$, number $=6$.
2. Average $=\frac{288}{6}=48$.

Trick: $6\times 48=288$ – quick check by multiplying back.

Answer: (A) $48$

Concept: Ladder, wall and ground form a right triangle; the ladder is the hypotenuse.

1. Distance of foot from wall $=x$ (base), angle with ground $={60}^{\circ }$.
2. $\cos {60}^{\circ }=\frac{x}{10}\Rightarrow x=10\times \frac{1}{2}=5\text{m}$.

Trick: At ${60}^{\circ }$ the base is always half the ladder (30-60-90 triangle).

Answer: (C) $5$ metres

Concept: For runners on a circular track, distinct meeting points = sum of their speed ratio (opposite directions) or difference (same direction), ratio in lowest terms.

1. Speeds: Amit $=\frac{2100}{42}=50\text{m/s}$, Sumit $=\frac{2100}{75}=28\text{m/s}$.
2. Ratio $50:28=25:14$ (lowest terms).
3. Opposite directions ⇒ meeting points $=25+14=39$.

Trick: Speed ratio = inverse of time ratio: $75:42=25:14$ directly.

Answer: (B) $39$

Concept: Total = average × count; then share the total in the given ratio.

1. Total $=3\times 15255=$ ₹$45765$.
2. Ratio parts $=3+5+7=15$, 1 part $=\frac{45765}{15}=$ ₹$3051$.
3. Cheapest $=3\times 3051=$ ₹$9153$.

Trick: Cheapest $=\frac{3}{15}\times 3\times 15255=\frac{3}{5}\times 15255$.

Answer: (A) ₹$9153$

Concept: Amount $A=P(1+\frac{RT}{100})$ for simple interest.

1. $\frac{RT}{100}=\frac{4\times 3}{100}=0.12$.
2. $840=P\times 1.12\Rightarrow P=\frac{840}{1.12}=750$.

Trick: 12% of 750 = 90 and $750+90=840$ ✔.

Answer: (A) ₹$750$

Concept: A root satisfies the equation: substitute $x=3$ in both.

1. $9-24+P=0\Rightarrow P=15$.
2. $9+12+Q=0\Rightarrow Q=-21$.
3. $P+Q=15-21=-6$.

Trick: Adding directly: $(9-24)+(9+12)+P+Q=0\Rightarrow P+Q=-6$.

Answer: (B) $-6$

Concept: Heron's formula: $\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$, $s=\frac{a+b+c}{2}$.

1. $s=\frac{8+9+13}{2}=15$.
2. $s-a=7,s-b=6,s-c=2$.
3. Area $=\sqrt{15\times 7\times 6\times 2}=\sqrt{1260}\approx 35.50{\text{cm}}^2$.

Trick: ${35}^2=1225$, ${36}^2=1296$; $1260$ is almost midway ⇒ about $35.5$.

Answer: (D) $35.50{\text{cm}}^2$

Concept: Mode = the observation that occurs most often.

1. Count repeats: 78 occurs 5 times; 90 and 66 occur 3 times each; 86 and 83 twice; the rest once.
2. Highest frequency = 5 for 78.

Trick: Tick only the option values in the list – no need to tally every number.

Answer: (C) 78

Concept: In a proportion $a:b::c:d$, $\frac{b}{a}=\frac{d}{c}$.

1. $6:42$ ⇒ multiplier 7, so $x=7.6\times 7=53.2$.
2. $3:6$ ⇒ multiplier 2, so $y=6\times 2=12$.
3. $x:y=53.2:12=532:120=133:30$.

Trick: Multiply by 10 to clear the decimal, then cancel by 4.

Answer: (A) $133:30$

Concept: A works for the full 54 minutes; B fills only the remaining part.

1. Work by A in 54 min $=\frac{54}{72}=\frac{3}{4}$.
2. Remaining $\frac{1}{4}$ is filled by B: time $=\frac{1}{4}\times 76=19$ min.
3. So B is turned off after 19 minutes.

Trick: Pipe that runs throughout – compute its share first; the other pipe covers the rest.

Answer: (C) 19 minutes

Concept: For 2 years: $\text{CI}-\text{SI}=P{\left(\frac{R}{100}\right)}^2$.

1. $967=P\times {\left(\frac{17}{100}\right)}^2=P\times 0.0289$.
2. $P=\frac{967}{0.0289}\approx 33460.2$.
3. Rounded: ₹$33460$.

Trick: $\frac{967\times 10000}{289}$: $289\times 33460=9669940\approx 9670000$.

Answer: (B) ₹$33460$

Concept: A letter common to two words gets the digit common to their two codes (digits may be jumbled).

1. HIRE → 7, 5, 3, 9 and INKS → 2, 4, 9, 6.
2. Only common letter: I.
3. Only common digit: 9.
4. So I = 9.

Trick: Common letter ↔ common digit.

Answer: (B) 9

Concept: Check the alphabet-position gap inside each cluster.

1. GC: $7\to 3$ ($-4$); ZV: $26\to 22$ ($-4$)
2. CY: $3\to 25$ ($-4$, wrapping); VR: $22\to 18$ ($-4$)
3. RN: $18\to 14$ ($-4$); KG: $11\to 7$ ($-4$)
4. LH: $12\to 8$ ($-4$) but DC: $4\to 3$ ($-1$) – breaks the pattern.

Trick: Both clusters of each pair must show the same backward jump of 4.

Answer: (B) LH−DC

Concept: Read only the region that is inside Surgeon and Dietician but outside Barber.

1. Surgeon ∩ Dietician has two parts: 2 (also inside Barber) and 9 (outside Barber).
2. 'Not barbers' removes the centre 2.
3. Required region = 9.

Trick: Common to two, not the third = the two-circle overlap minus the centre.

Answer: (C) 9

Concept: Trace each turn on a grid (left of West = South, right of South = West, etc.).

1. F → 2 km West; left (South) 4 km; right (West) 10 km; left (South) 4 km.
2. Left (East) 12 km brings him back to F's vertical line, 8 km south of F.
3. Left (North) 2 km → O is 6 km south of F.
4. To return he drives 6 km towards the North.

Trick: West moves: 2 + 10 = 12 = East 12, so only the North–South gap remains: 8 − 2 = 6.

Answer: (C) 6 km towards the north

Concept: Swap the signs first, then apply BODMAS.

1. After interchange: $48\times 14+168÷3-47$
2. $48\times 14=672$, $168÷3=56$
3. $672+56-47=681$

Trick: Rewrite the whole expression with new signs before calculating anything.

Answer: (A) 681

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

1. Series: 6 £ & 8 4 * % 7 2 & 2 2 9 £ 1 & € 1 2 6 6 2
2. & 8 4 – followed by a number, no. % 7 2 – no. & 2 2 – no. € 1 2 – no.
3. £ 1 & – yes, the only one.

Trick: Scan only the numbers that follow a symbol, then check the next element.

Answer: (B) 1