Q&A: practice questions with answers

Concept: Increasing a number by $100\%$ makes it $\frac{200}{100}=2$ times.

1. Let the number be $x$. Then $x\times \frac{200}{100}=2820$.
2. $2x=2820\Rightarrow x=1410$.

Trick: Up by $100\%$ = double, so just halve: $2820÷2=1410$.

Answer: (C) 1410

Concept: Form two linear equations in the cost of a pen ($p$) and a pencil ($q$) and solve.

1. $8p+10q=132$ … (i)
2. $17(p-2)+6(q+7)=136\Rightarrow 17p+6q+8=136\Rightarrow 17p+6q=128$ … (ii)
3. (i)$\times 3$: $24p+30q=396$; (ii)$\times 5$: $85p+30q=640$. Subtract: $61p=244\Rightarrow p=4$.
4. From (i): $10q=132-32=100\Rightarrow q=10$.
5. $12p+9q=48+90=138$.

Trick: Check options quickly: $p=4,q=10$ satisfies both equations ($32+100=132$; $34+102=136$).

Answer: (A) ₹138

Concept: Use the angle-sum property of triangles and the linear pair at E.

1. In right $△BDC$ ($\angle BDC={90}^{\circ }$): $\angle C={180}^{\circ }-{90}^{\circ }-{16}^{\circ }={74}^{\circ }$.
2. In $△AEC$: $\angle AEC={180}^{\circ }-\angle CAE-\angle C={180}^{\circ }-{51}^{\circ }-{74}^{\circ }={55}^{\circ }$.
3. $\angle AEB$ and $\angle AEC$ form a linear pair: $\angle AEB={180}^{\circ }-{55}^{\circ }={125}^{\circ }$.

Trick: Exterior angle of $△AEC$ at E: $\angle AEB=\angle CAE+\angle C={51}^{\circ }+{74}^{\circ }={125}^{\circ }$.

Answer: (D) ${125}^{\circ }$

Concept: If $3A=9B=4C=k$, then $A:B:C=\frac{1}{3}:\frac{1}{9}:\frac{1}{4}$.

1. Multiply by LCM $36$: $A:B:C=12:4:9$; total parts $=25$.
2. Share of A $=\frac{12}{25}\times 75400=12\times 3016=36192$.

Trick: Answer must be a multiple of $12$ (and $75400÷25=3016$): only $36192=12\times 3016$ fits.

Answer: (B) ₹36,192

Concept: Divisibility by 7 (also 11, 13): split the number into groups of three from the right and take the alternating sum.

1. $774|359|327$: $327-359+774=742$.
2. $742=7\times 106$, so the number is divisible by 7 ($774359327=7\times 110622761$).
3. It is odd, so not divisible by 2, 12 or 16.

Trick: Odd number → 2, 12, 16 are out at once; only 7 is left to check.

Answer: (A) 7

Concept: A number is divisible by 3 if the sum of its digits is divisible by 3.

1. Digit sum of $545714151$: $5+4+5+7+1+4+1+5+1=33$, and $33$ is divisible by 3. ✔
2. Last digit is 1 (odd, not 0), so it is not divisible by 10 or by 6 (needs 2 as well).
3. By 7: $151-714+545=-18$, not a multiple of 7.

Trick: Odd number → 6 and 10 are out; digit sum test settles 3.

Answer: (D) 3

Concept: Perimeter of a sector $=2r+l$, where arc length $l=\frac{\theta }{{360}^{\circ }}\times 2\pi r$.

1. $l=\frac{18}{360}\times 2\times 3.14\times 9=\frac{1}{20}\times 56.52=2.826\text{cm}$.
2. Perimeter $=2\times 9+2.826=20.826\approx 20.83\text{cm}$.

Trick: ${18}^{\circ }$ is $\frac{1}{20}$ of a full turn; arc $=\frac{2\pi r}{20}=\frac{\pi r}{10}$.

Answer: (C) $20.83\text{cm}$

Concept: Work step by step from registered voters $R$ to the winning margin.

1. Votes polled $=85\%$ of $R=0.85R$; valid votes $=96\%$ of polled $=0.96\times 0.85R=0.816R$.
2. Winner $65\%$, loser $35\%$ of valid votes, so margin $=30\%$ of valid $=0.3\times 0.816R=0.2448R$.
3. $0.2448R=2142\Rightarrow R=\frac{2142}{0.2448}=8750$.

Trick: $R\times \frac{85}{100}\times \frac{96}{100}\times \frac{30}{100}=2142$; only 8750 makes this exact.

Answer: (B) 8750

Concept: The third proportional $x$ of $a$ and $b$ satisfies $a:b=b:x$, so $x=\frac{b^2}{a}$.

1. $x=\frac{{56}^2}{16}=\frac{3136}{16}=196$.

Trick: $\frac{56}{16}=3.5$, so $x=56\times 3.5=196$.

Answer: (A) 196

Concept: Take total work $=$ LCM of $26,13,10=130$ units and find each person's daily units.

1. Priya $=\frac{130}{26}=5$ units/day; Priya + Varsha $=\frac{130}{13}=10$, so Varsha $=5$.
2. All three $=\frac{130}{10}=13$ units/day, so Nisha $=13-10=3$ units/day.
3. Priya + Nisha $=5+3=8$ units/day; time $=\frac{130}{8}=\frac{65}{4}=16\frac{1}{4}$ days.

Trick: LCM method avoids fractions: total ÷ combined daily units.

Answer: (B) $\frac{65}{4}$

Concept: For the same distance, time is inversely proportional to speed: $T_1{S}_1=T_2{S}_2$.

1. Distance $=138\times 8=1104\text{km}$.
2. Malaika's time $=\frac{1104}{92}=12$ hours.

Trick: Speed ratio $138:92=3:2$, so time ratio $=2:3$. $8\times \frac{3}{2}=12$ hours.

Answer: (B) 12 hours

Concept: With a false weight, the customer pays for 1 kg but gets less. Effective $\text{SP}:\text{CP}=\frac{1+\text{profit}}{\text{actual weight}}$.

1. Let marked weight be 1 kg costing ₹100; actual weight given $=0.92$ kg, costing ₹92.
2. He charges $100\times 1.47=147$.
3. Profit $\%=\frac{147-92}{92}\times 100=\frac{55}{92}\times 100\approx 59.8\%\approx 60\%$.

Trick: $\frac{1.47}{0.92}=1.598$ → about $60\%$ profit.

Answer: (A) $60\%$

Concept: Convert both statements into linear equations and solve.

1. Let Mary $=M$, Diya $=D$. $3M=5D+20$ and $2D=M-2\Rightarrow M=2D+2$.
2. $3(2D+2)=5D+20\Rightarrow 6D+6=5D+20\Rightarrow D=14$.
3. $M=2(14)+2=30$. Difference $=30-14=16$ years.

Trick: Check options: difference 16 → $M=D+16$ and $M=2D+2$ give $D=14$, $M=30$; $3\times 30=90=5\times 14+20$ ✓.

Answer: (C) 16

Concept: Successive discounts multiply the remaining fractions: price left $=(1-d_1)(1-d_2)(1-d_3)$.

1. Remaining $=0.80\times 0.75\times 0.85$.
2. $0.80\times 0.75=0.60$; $0.60\times 0.85=0.51$.
3. Single discount $=1-0.51=0.49=49\%$.

Trick: Take ₹100: $100\to 80\to 60\to 51$. Discount $=49\%$.

Answer: (A) $49\%$

Concept: Test each number for divisibility by primes up to $\sqrt{345}\approx 18.6$ (2, 3, 5, 7, 11, 13, 17).

1. Even numbers and those ending in 5 are out: 332, 334, …, 335, 345.
2. 333 and 339 are divisible by 3; $341=11\times 31$; $343=7^3$.
3. Left: 331 and 337 — both prime. Count $=2$.

Yaad rakhein: 341 looks prime but $=11\times 31$ (alternating digit sum $3-4+1=0$).

Answer: (D) 2

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

1. CP of one eraser $=\frac{150}{12}=12.5$ (₹).
2. SP $=18$, profit $=5.5$.
3. Profit $\%=\frac{5.5}{12.5}\times 100=44\%$.

Trick: Per dozen: SP $=18\times 12=216$ vs CP 150 → $\frac{66}{150}=44\%$.

Answer: (C) 44

Concept: Make decimal places equal, find the LCM of the whole numbers, then put back the decimal places.

1. $15.4=15.400$ and $0.006$ → whole numbers 15400 and 6 (3 decimal places).
2. $15400=2^3\times 5^2\times 7\times 11$, $6=2\times 3$ → LCM $=15400\times 3=46200$.
3. Put back 3 decimal places: $46.200=46.2$.

Trick: LCM must be a multiple of 15.4: $15.4\times 3=46.2$, and $46.2÷0.006=7700$ ✓.

Answer: (C) 46.2

Concept: Divide by $\cos \theta$ to get $\tan \theta$; use ${\sec }^2\theta =1+{\tan }^2\theta$.

1. $\frac{\tan \theta +1}{\tan \theta -4}=11\Rightarrow \tan \theta +1=11\tan \theta -44\Rightarrow \tan \theta =\frac{45}{10}=\frac{9}{2}$.
2. Expression $=13{\tan }^2\theta +3-3{\sec }^2\theta =13{\tan }^2\theta +3-3-3{\tan }^2\theta =10{\tan }^2\theta$.
3. $=10\times \frac{81}{4}=\frac{405}{2}$.

Trick: $\frac{3}{{\cos }^2\theta }=3{\sec }^2\theta$; the constants 3 and $-3$ cancel, leaving $10{\tan }^2\theta$.

Answer: (C) $\frac{405}{2}$

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

1. Shorter side $=14\text{m}$ → diameter $=14$, radius $=7\text{m}$.
2. Circumference $=\pi d=\frac{22}{7}\times 14=44\text{m}$.

Yaad rakhein: The longer side (116 m) does not matter — the circle is limited by the width.

Answer: (B) 44

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

1. Total interest rate $=4\times 5=20\%$.
2. $P\times 1.20=2400\Rightarrow P=\frac{2400}{1.2}=2000$.

Trick: Amount $=120\%$ of principal → $\frac{2400}{120}\times 100=2000$.

Answer: (C) ₹2,000