Q&A: practice questions with answers

Concept: Jain tradition records the later life of Chandragupta Maurya.

1. Chandragupta gave up his throne, became a disciple of the Jain monk Bhadrabahu and went to Shravanabelagola (Karnataka).
2. He is said to have died there by sallekhana (fasting unto death); Chandragiri hill is named after him.
3. Buddhism is linked with his grandson Ashoka, not with Chandragupta.

Yaad rakhein: Chandragupta – Jainism (Bhadrabahu); Ashoka – Buddhism (Upagupta); Bindusara – Ajivika.

Answer: (B) Jainism

Concept: If $a:b=b:c$, then $c$ is the third proportional: $c=\frac{b^2}{a}$.

1. First three even natural numbers: $2+4+6=12$, so $B=12$.
2. $C=\frac{B^2}{44}=\frac{144}{44}$.
3. $\frac{144}{44}=\frac{36}{11}=3.2727\dots \approx 3.27$.

Trick: Third proportional $=\frac{(\text{middle}{)}^2}{\text{first}}$.

Answer: (A) $3.27$

Concept: For $a{y}^2+by+c=0$: sum of roots $\alpha +\beta =-\frac{b}{a}$ and product $\alpha \beta =\frac{c}{a}$.

1. Here $a=1,b=-9,c=8$.
2. $\alpha +\beta =9$ and $\alpha \beta =8$.
3. $\alpha +\beta -\alpha \beta =9-8=1$.

Trick: Roots are actually $1$ and $8$ (since $1+8=9,1\times 8=8$): $9-8=1$.

Answer: (B) $1$

Concept: Product of two numbers $=\text{LCM}\times \text{HCF}$.

1. Let the numbers be $18k$ and $5k$; product $=90k^2=360$, so $k^2=4,k=2$.
2. Numbers: $36$ and $10$; HCF $=2$, LCM $=180$ (check: $180\times 2=360$).
3. $\frac{1}{180}+\frac{1}{2}=\frac{1+90}{180}=\frac{91}{180}$.

Trick: Since 18 and 5 are co-prime, HCF $=k$ and LCM $=90k$.

Answer: (A) $\frac{91}{180}$

Concept: After increases of $a\%$ and $b\%$: present $=\text{original}\times (1+\frac{a}{100})(1+\frac{b}{100})$.

1. $\text{original}\times 1.02\times 1.20=183600$.
2. $1.02\times 1.20=1.224$.
3. Original $=\frac{183600}{1.224}=150000$.

Trick: Check options: $150000\times 1.02=153000$; $153000\times 1.2=183600$ ✓.

Answer: (A) ₹1,50,000

Concept: Number of small cubes $={\left(\frac{\text{big side}}{\text{small side}}\right)}^3$; surface area of one cube $=6a^2$.

1. Number of cubes $={\left(\frac{10}{2}\right)}^3=125$.
2. TSA of one small cube $=6\times 2^2=24{\text{cm}}^2$.
3. Total $=125\times 24=3000{\text{cm}}^2$.

Trick: Total TSA $=\text{original TSA}\times \frac{10}{2}=600\times 5=3000$.

Answer: (A) $3000$

Concept: BODMAS: Brackets → Of → Division/Multiplication → Addition/Subtraction.

1. Bracket: $10+20=30$.
2. Multiplication: $5\times 30=150$.
3. $118-150+55=23$.

Trick: Do $118+55=173$ first, then $173-150=23$.

Answer: (B) $23$

Concept: If two numbers are $a\%$ and $b\%$ more than the same number, their ratio is $(100+a):(100+b)$.

1. Let the third number $=100$.
2. First $=130$, second $=118$.
3. Ratio $=130:118=65:59$.

Trick: Take the base as 100 and simplify.

Answer: (C) $65:59$

Concept: Arrange data in ascending order; for odd $n$, median $={\left(\frac{n+1}{2}\right)}^{\text{th}}$ term.

1. Ascending: 14, 20, 21, 24, 32, 46, 49, 66, 90, 91, 92.
2. $n=11$, so median $={\left(\frac{11+1}{2}\right)}^{\text{th}}=6^{\text{th}}$ term.
3. $6^{\text{th}}$ term $=46$.

Trick: Five values below and five above the median.

Answer: (A) $46$

Concept: Half-yearly compounding: rate $=\frac{R}{2}$ per half-year and time $=2n$ half-years; $A=P{(1+\frac{r}{100})}^t$.

1. Rate $=\frac{20}{2}=10\%$ per half-year, periods $=2$.
2. $A=2400\times (1.1{)}^2=2400\times 1.21$.
3. $A=2904$.

Trick: 10% twice: $2400\to 2640\to 2904$.

Answer: (B) ₹2,904

Concept: The smallest cuboid holding a hemisphere of radius $r$ is $2r\times 2r\times r$. Water $=$ cuboid volume $-$ hemisphere volume $\left(\frac{2}{3}\pi r^3\right)$.

1. Cuboid $=6\times 6\times 3=108{\text{cm}}^3$.
2. Hemisphere $=\frac{2}{3}\times 3.14\times 27=56.52{\text{cm}}^3$.
3. Water $=108-56.52=51.48{\text{cm}}^3$.

Trick: Water $=r^3(4-\frac{2\pi }{3})=27\times 1.9067\approx 51.48$.

Answer: (C) $51.48$

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$.

1. Total distance $=100+408+1460=1968\text{km}$; total time $=\frac{1968}{20}=98.4\text{h}$.
2. Time for 1st and 3rd parts $=\frac{100}{20}+\frac{1460}{20}=5+73=78\text{h}$.
3. Time for 2nd part $=98.4-78=20.4\text{h}$, so $x=\frac{408}{20.4}=20$.

Trick: If the average equals the speed of the other parts (20), the middle speed must also be 20.

Answer: (C) $20$

Concept: Net rate = filling rate − emptying rate; time $=\frac{\text{work}}{\text{net rate}}$.

1. Net rate $=\frac{1}{8}-\frac{1}{11}=\frac{11-8}{88}=\frac{3}{88}$ tank per minute.
2. Time for $\frac{3}{4}$ tank $=\frac{3}{4}÷\frac{3}{88}=\frac{3}{4}\times \frac{88}{3}=22$ min.

Trick: Full tank needs $\frac{88}{3}$ min; three-fourth of it $=22$ min.

Answer: (A) $22$

Concept: Substitute the value of $a$ and simplify inside the root first.

1. $8a^2-4a+1=8\times 4-8+1=25$.
2. $3\sqrt{25}=3\times 5=15$.
3. $17a=34$; total $=15+34=49$.

Trick: A perfect square under the root (25) confirms the substitution is right.

Answer: (B) $49$

Concept: A tangent is perpendicular to the radius at the point of contact, so OTP is a right triangle with OP as hypotenuse.

1. $P{T}^2=O{P}^2-O{T}^2={25}^2-7^2$.
2. $P{T}^2=625-49=576$.
3. $PT=\sqrt{576}=24\text{cm}$.

Trick: Triplet $(7,24,25)$.

Answer: (A) $24$ cm

Concept: Form equations from the given conditions; the age difference between father and son never changes.

1. Let present ages be father $F$ and son $S$. Three years ago: $F-3=2(S-3)+14\Rightarrow F=2S+11$.
2. After $x$ years: $F+x=2(S+x)\Rightarrow x=F-2S$.
3. $x=(2S+11)-2S=11$ years.

Trick: When $F=2S+k$ today, the father becomes twice the son's age after exactly $k$ years.

Answer: (D) 11

Concept: Maximum selling price comes from the scheme whose net multiplier (product of the remaining percentages) is the largest.

1. I: $0.73\times 0.73=0.5329$
2. II: $0.64\times 0.55=0.352$
3. III: $0.81\times 0.68=0.5508$
4. IV: $0.66\times 0.93=0.6138$ ← largest, i.e. least net discount $=38.62\%$.

Trick: Net discount $=a+b-\frac{ab}{100}$: IV gives $34+7-2.38=38.62\%$, the smallest of all.

Answer: (C) IV

Concept: Sum of numbers = average × count.

1. Sum of three numbers $=23\times 3=69$.
2. Third number $=69-(11+26)=69-37=32$.

Trick: Deviations from 23: $11\to -12$, $26\to +3$; third must be $+9$ → $23+9=32$.

Answer: (C) 32

Concept: Find each pen's actual selling price, then total profit = total SP − total CP.

1. Pen 1: marked at 15% profit $=100\times 1.15=$ ₹115; after 29% discount: $115\times 0.71=$ ₹81.65 → loss ₹18.35.
2. Pen 2: sold at 58% profit $=160\times 1.58=$ ₹252.80 → profit ₹92.80.
3. Net profit $=92.80-18.35=$ ₹74.45.

Trick: Total SP $=81.65+252.80=334.45$; total CP $=260$ → profit $=74.45$.

Answer: (B) ₹74.45

Concept: A pure recurring decimal $0.\bar{a}=\frac{a}{9}$.

1. $2.777\dots =2+0.\bar{7}=2+\frac{7}{9}$.
2. $=\frac{18+7}{9}=\frac{25}{9}$.
3. Check: $25÷9=2.777\dots$

Trick: $\frac{\text{whole number without bar}-\text{non-recurring part}}{9}=\frac{27-2}{9}=\frac{25}{9}$.

Answer: (D) $\frac{25}{9}$