Q&A: practice questions with answers

Concept: Form two equations from the two statements and solve.

1. $4P-Q=19$ ...(i)
2. After 5 years: $3(Q+5)=9(P+5)-27\Rightarrow 3Q+15=9P+18\Rightarrow Q=3P+1$ ...(ii)
3. Put (ii) in (i): $4P-(3P+1)=19\Rightarrow P=20$
4. $Q=3\times 20+1=61$

Trick: Check options in (i): $Q=61\Rightarrow P=\frac{61+19}{4}=20$; then $3\times 66=198$ and $9\times 25-27=198$ ✓.

Answer: (D) $61$ years

Concept: Use $1+{\cot }^{2}A={\csc }^{2}A$ and $1+{\tan }^{2}A={\sec }^{2}A$.

1. $\frac{1}{{\cot }^{2}A}={\tan }^{2}A$, so the expression $=(1+{\cot }^{2}A)+(1+{\tan }^{2}A)={\csc }^{2}A+{\sec }^{2}A$
2. $=\frac{1}{{\sin }^{2}A}+\frac{1}{{\cos }^{2}A}=\frac{{\cos }^{2}A+{\sin }^{2}A}{{\sin }^{2}A{\cos }^{2}A}=\frac{1}{{\sin }^{2}A{\cos }^{2}A}$
3. $={\csc }^{2}A{\sec }^{2}A$

Trick: Put $A={45}^{\circ }$: expression $=2+2=4$ and ${\csc }^{2}A{\sec }^{2}A=2\times 2=4$ ✓.

Answer: (C) ${\csc }^{2}A{\sec }^{2}A$

Concept: Profit is shared in the ratio of investment.

1. Total parts $=59+14+72=145$
2. Difference of R and Q $=72-14=58$ parts
3. Difference $=\frac{58}{145}\times 4790=\frac{2}{5}\times 4790=1916$

Trick: $\frac{58}{145}=\frac{2}{5}$ – so just take $40\%$ of ₹4,790.

Answer: (D) ₹$1,916$

Concept: Take square roots of perfect squares, counting decimal places in pairs.

1. $\sqrt{121}=11$
2. $\sqrt{0.0196}=0.14$ (since ${14}^2=196$, 4 decimal places → 2)
3. $\sqrt{8.41}=2.9$ (since ${29}^2=841$)
4. $11+0.14-2.9=8.24$

Trick: Decimal places in the root = half of those under the root.

Answer: (C) $8.24$

Concept: Perimeter of a semicircle = curved arc + diameter $=\pi r+2r$.

1. Arc $=\frac{22}{7}\times 7=22\text{cm}$
2. Diameter $=2\times 7=14\text{cm}$
3. Perimeter $=22+14=36\text{cm}$

Trick: Perimeter $=\frac{36}{7}r$; with $r=7$ it is directly $36$.

Answer: (C) $36\text{cm}$

Concept: Take total work = LCM of days and find each person's daily units.

1. LCM$(30,10,9)=90$ units. Anjani $=\frac{90}{30}=3$, Anjani + Khushbu $=9$, all three $=10$ units/day
2. Sushmita $=10-9=1$ unit/day
3. Anjani + Sushmita $=3+1=4$ units/day
4. Time $=\frac{90}{4}=\frac{45}{2}$ days

Trick: Sushmita = (A+K+S) − (A+K): $\frac{1}{9}-\frac{1}{10}=\frac{1}{90}$.

Answer: (C) $\frac{45}{2}$ days

Concept: Test each number for divisibility by primes up to $\sqrt{494}\approx 22$.

1. 488, 490, 492, 494 are even – not prime.
2. $489=3\times 163$ (digit sum 21) and $493=17\times 29$ – not prime.
3. 491 is not divisible by 2, 3, 5, 7, 11, 13, 17, 19 – prime.
4. So only one prime.

Trick: Remove evens first; only 489, 491, 493 need checking.

Answer: (C) $1$

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

1. $r=\frac{54}{2}=27\text{cm}$
2. $4\pi r^2=4\pi \times {27}^2=4\times 729\pi$
3. $=2916\pi {\text{cm}}^2$

Trick: $4\pi r^2=\pi d^2={54}^2\pi =2916\pi$ directly.

Answer: (D) $2916\pi {\text{cm}}^2$

Concept: $\text{Savings}=\text{Income}-\text{Expenditure}$.

1. Savings $=22.5\%$ of 86400 $=19440$; Expenditure $=86400-19440=66960$
2. New income $=86400\times 1.22=105408$
3. New expenditure $=66960\times 1.25=83700$
4. New savings $=105408-83700=21708$; change $=21708-19440=+2268$

Trick: Change $=22\%\text{ of }86400-25\%\text{ of }66960=19008-16740=2268$.

Answer: (B) increase by ₹$2,268$

Concept: Required speed $=\frac{\text{remaining distance}}{\text{remaining time}}$.

1. First 5 h: $5\times 91=455\text{km}$
2. Next 84 km at $84\text{km/h}$ takes $1$ h
3. Covered $=455+84=539\text{km}$ in $6$ h; left $=811-539=272\text{km}$ in $14-6=8$ h
4. Speed $=\frac{272}{8}=34\text{km/h}$

Trick: Same number distance and speed (84 km at 84 km/h) = exactly 1 hour.

Answer: (D) $34\text{km/h}$

Concept: For a period in days, $\text{SI}=\frac{P\times R\times \text{days}}{100\times \text{days in year}}$. Count the day of deposit OR withdrawal, not both. 2024 is a leap year (366 days).

1. Days: February $29-5=24$, March $31$, April $6$ → $24+31+6=61$ days.
2. $\text{SI}=\frac{4000\times 7.5\times 61}{100\times 366}$
3. $=\frac{1830000}{36600}=50$

Trick: $61$ days of $366$ = $\frac{1}{6}$ year → $\text{SI}=4000\times 7.5\%\times \frac{1}{6}=300÷6=50$.

Answer: (C) ₹$50$

Concept: Increasing a number by $60\%$ means multiplying it by $1.6$.

1. Let the number be $x$. Then $x\times \frac{160}{100}=3600$.
2. $x=\frac{3600\times 100}{160}=2250$.

Trick: $160\%\to 3600$, so $1\%\to 22.5$ and $100\%\to 2250$.

Answer: (C) $2250$

Concept: Sum of first $n$ even numbers $=n(n+1)$, so their average $=n+1$.

1. Here $n=165$.
2. Average $=165+1=166$.
3. Check: first and last even numbers are $2$ and $330$; average $=\frac{2+330}{2}=166$.

Yaad rakhein: Average of first $n$ even numbers $=n+1$; of first $n$ odd numbers $=n$.

Answer: (A) $166$

Concept: Successive discounts are applied one after another: $\text{SP}=\text{MP}\times (1-d_1)(1-d_2)$.

1. After $10\%$: $720\times 0.9=648$.
2. After $15\%$: $648\times 0.85=550.80$.

Trick: Net discount $=10+15-\frac{10\times 15}{100}=23.5\%$ → $720\times 0.765=550.80$.

Answer: (B) ₹$550.80$

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

1. Total interest rate $=5\times 4=20\%$.
2. $P\times \frac{120}{100}=2400$ → $P=2000$.

Trick: $120\%\to 2400$, so $100\%\to 2000$.

Answer: (B) ₹$2,000$

Concept: Total $=\text{average}\times \text{number of items}$; then divide the total in the given ratio.

1. Total price $=16005\times 3=48015$.
2. Ratio parts $=3+5+7=15$ → 1 part $=48015÷15=3201$.
3. Cheapest item $=3\times 3201=9603$.

Trick: Average of 3 items = middle share when ratio is in AP: $5$ parts $=16005$ → 1 part $=3201$ → $3$ parts $=9603$.

Answer: (A) $9,603$

Concept: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$, compared per pen.

1. CP of 1 pen $=150÷12=12.5$.
2. SP of 1 pen $=45$; profit $=45-12.5=32.5$.
3. $\text{Profit}\%=\frac{32.5}{12.5}\times 100=260\%$.

Trick: Compare a dozen: CP $=150$, SP $=45\times 12=540$; profit $=390$ → $\frac{390}{150}\times 100=260\%$.

Answer: (B) $260$

Concept: $a^m÷a^n=a^{m-n}$ and $a^m\times a^n=a^{m+n}$; work left to right.

1. ${83}^{-2}÷{83}^7={83}^{-2-7}={83}^{-9}$.
2. ${83}^{-9}\times {83}^{-10}={83}^{-9-10}={83}^{-19}$.

Trick: Only the powers matter: $-2-7+(-10)=-19$.

Answer: (A) ${83}^{-19}$

Concept: For $n(n+1)=930$, $n$ is just below $\sqrt{930}$.

1. $\sqrt{930}\approx 30.5$, so try $30\times 31$.
2. $30\times 31=930$ ✓.
3. Greater number $=31$.

Trick: Unit digit $0$ in $930$ → one number ends in $0$ or $5$; $30\times 31$ fits at once.

Answer: (D) $31$

Concept: Expand the brackets and combine like terms.

1. $x(7x-2)=7x^2-2x$.
2. $5(x^2-3)=5x^2-15$.
3. Sum $+13$: $7x^2+5x^2-2x-15+13=12x^2-2x-2$.

Trick: Put $x=0$: expression $=-15+13=-2$ → constant term must be $-2$ with positive $x^2$; only (C) fits.

Answer: (C) $12x^2-2x-2$