Class 10: Quadratic Equations

$\displaystyle \textbf{Question 1: }\text{Person }A\text{ requires }10\text{ fewer days than Person }B$
$\displaystyle \text{ to complete a task independently.} \text{Working together, they finish it in }12\text{ days.}$
$\displaystyle \text{Calculate the time Person } B\text{ takes to complete} \ \text{the work on their own.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let Person }B\text{ take }x\text{ days to complete the work alone.}$
$\displaystyle \therefore \text{Person }A\text{ takes }(x-10)\text{ days to complete the work alone.}$
$\displaystyle \text{Person }B\text{'s one day's work}=\frac{1}{x}$
$\displaystyle \text{Person }A\text{'s one day's work}=\frac{1}{x-10}$
$\displaystyle \text{Together, they complete the work in }12\text{ days.}$
$\displaystyle \therefore \frac{1}{x}+\frac{1}{x-10}=\frac{1}{12}$
$\displaystyle \Rightarrow \frac{2x-10}{x(x-10)}=\frac{1}{12}$
$\displaystyle \Rightarrow 12(2x-10)=x(x-10)$
$\displaystyle \Rightarrow 24x-120=x^2-10x$
$\displaystyle \Rightarrow x^2-34x+120=0$
$\displaystyle \Rightarrow x^2-30x-4x+120=0$
$\displaystyle \Rightarrow x(x-30)-4(x-30)=0$
$\displaystyle \Rightarrow (x-30)(x-4)=0$
$\displaystyle \Rightarrow x=30\text{ or }x=4$
$\displaystyle \text{But }x>10,\text{ therefore }x=4\text{ is rejected.}$
$\displaystyle \therefore \text{Person }B\text{ takes }30\text{ days to complete the work alone.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Two inlet pipes operating simultaneously fill a reservoir in }$
$\displaystyle 12\text{ hours. If one pipe fills} \ \text{the reservoir }10\text{ hours faster than the other,}$
$\displaystyle \text{determine the time required by the second pipe to fill the reservoir alone.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the second pipe fill the reservoir alone in }x\text{ hours.}$
$\displaystyle \therefore \text{The faster pipe fills the reservoir alone in }(x-10)\text{ hours.}$
$\displaystyle \text{Second pipe's one hour work}=\frac{1}{x}$
$\displaystyle \text{Faster pipe's one hour work}=\frac{1}{x-10}$
$\displaystyle \text{Together, the two pipes fill the reservoir in }12\text{ hours.}$
$\displaystyle \therefore \frac{1}{x}+\frac{1}{x-10}=\frac{1}{12}$
$\displaystyle \Rightarrow \frac{2x-10}{x(x-10)}=\frac{1}{12}$
$\displaystyle \Rightarrow 12(2x-10)=x(x-10)$
$\displaystyle \Rightarrow 24x-120=x^2-10x$
$\displaystyle \Rightarrow x^2-34x+120=0$
$\displaystyle \Rightarrow x^2-30x-4x+120=0$
$\displaystyle \Rightarrow x(x-30)-4(x-30)=0$
$\displaystyle \Rightarrow (x-30)(x-4)=0$
$\displaystyle \Rightarrow x=30\text{ or }x=4$
$\displaystyle \text{But }x>10,\text{ therefore }x=4\text{ is rejected.}$
$\displaystyle \therefore \text{The second pipe takes }30\text{ hours to fill the reservoir alone.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Two water taps together can fill a tank in }9\frac{3}{8}\text{ hours.}$
$\displaystyle \text{The tap of larger diameter takes }10\text{ hours less than the smaller one to fill the tank}$
$\displaystyle \text{separately. Find the time in which each tap can separately fill the tank.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller tap fill the tank alone in }x\text{ hours.}$
$\displaystyle \therefore \text{The larger tap fills the tank alone in }(x-10)\text{ hours.}$
$\displaystyle \text{Smaller tap's one hour work}=\frac{1}{x}$
$\displaystyle \text{Larger tap's one hour work}=\frac{1}{x-10}$
$\displaystyle \text{Together, the two taps fill the tank in }9\frac{3}{8}=\frac{75}{8}\text{ hours.}$
$\displaystyle \therefore \frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}$
$\displaystyle \Rightarrow \frac{2x-10}{x(x-10)}=\frac{8}{75}$
$\displaystyle \Rightarrow 75(2x-10)=8x(x-10)$
$\displaystyle \Rightarrow 150x-750=8x^2-80x$
$\displaystyle \Rightarrow 8x^2-230x+750=0$
$\displaystyle \Rightarrow 4x^2-115x+375=0$
$\displaystyle \Rightarrow 4x^2-100x-15x+375=0$
$\displaystyle \Rightarrow 4x(x-25)-15(x-25)=0$
$\displaystyle \Rightarrow (x-25)(4x-15)=0$
$\displaystyle \Rightarrow x=25\text{ or }x=\frac{15}{4}$
$\displaystyle \text{But }x>10,\text{ therefore }x=\frac{15}{4}\text{ is rejected.}$
$\displaystyle \therefore \text{The smaller tap takes }25\text{ hours to fill the tank alone.}$
$\displaystyle \therefore \text{The larger tap takes }25-10=15\text{ hours to fill the tank alone.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Two pipes running together can fill a tank in }11\frac{1}{9}\text{ minutes. If one pipe takes}$
$\displaystyle 5\text{ minutes more than the other to fill the tank separately, find the time in which each pipe}$
$\displaystyle \text{would fill the tank separately.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the faster pipe fill the tank alone in }x\text{ minutes.}$
$\displaystyle \therefore \text{The slower pipe fills the tank alone in }(x+5)\text{ minutes.}$
$\displaystyle \text{Faster pipe's one minute work}=\frac{1}{x}$
$\displaystyle \text{Slower pipe's one minute work}=\frac{1}{x+5}$
$\displaystyle \text{Together, the two pipes fill the tank in }11\frac{1}{9}=\frac{100}{9}\text{ minutes.}$
$\displaystyle \therefore \frac{1}{x}+\frac{1}{x+5}=\frac{9}{100}$
$\displaystyle \Rightarrow \frac{2x+5}{x(x+5)}=\frac{9}{100}$
$\displaystyle \Rightarrow 100(2x+5)=9x(x+5)$
$\displaystyle \Rightarrow 200x+500=9x^2+45x$
$\displaystyle \Rightarrow 9x^2-155x-500=0$
$\displaystyle \Rightarrow 9x^2-180x+25x-500=0$
$\displaystyle \Rightarrow 9x(x-20)+25(x-20)=0$
$\displaystyle \Rightarrow (x-20)(9x+25)=0$
$\displaystyle \Rightarrow x=20\text{ or }x=-\frac{25}{9}$
$\displaystyle \text{Since time cannot be negative, }x=-\frac{25}{9}\text{ is rejected.}$
$\displaystyle \therefore \text{The faster pipe takes }20\text{ minutes.}$
$\displaystyle \therefore \text{The slower pipe takes }20+5=25\text{ minutes.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{To fill a swimming pool two pipes are used. If the pipe of larger}$
$\displaystyle \text{diameter is used for 4 hours and the pipe of smaller diameter for }9\text{ hours, only half of the}$
$\displaystyle \text{pool can be filled. Find how long each pipe would take to fill the pool separately, if the}$
$\displaystyle \text{smaller pipe takes 10 hours more than the larger pipe to fill the pool.}\hfill\text{[CBSE 2010, 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the larger-diameter pipe fill the pool alone in }x\text{ hours.}$
$\displaystyle \therefore \text{The smaller-diameter pipe fills the pool alone in }(x+10)\text{ hours.}$
$\displaystyle \text{Larger pipe's one hour work}=\frac{1}{x}$
$\displaystyle \text{Smaller pipe's one hour work}=\frac{1}{x+10}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{4}{x}+\frac{9}{x+10}=\frac{1}{2}$
$\displaystyle \Rightarrow 8(x+10)+18x=x(x+10)$
$\displaystyle \Rightarrow 8x+80+18x=x^2+10x$
$\displaystyle \Rightarrow x^2-16x-80=0$
$\displaystyle \Rightarrow x^2-20x+4x-80=0$
$\displaystyle \Rightarrow x(x-20)+4(x-20)=0$
$\displaystyle \Rightarrow (x-20)(x+4)=0$
$\displaystyle \Rightarrow x=20\text{ or }x=-4$
$\displaystyle \text{Since time cannot be negative, }x=-4\text{ is rejected.}$
$\displaystyle \therefore \text{The larger-diameter pipe takes }20\text{ hours.}$
$\displaystyle \therefore \text{The smaller-diameter pipe takes }20+10=30\text{ hours.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Two water taps together can fill a tank in }1\frac{7}{8}\text{ hours. The tap with }$
$\displaystyle \text{larger diameter takes }2\text{ hours less than the tap with the smaller one to fill the tank}$
$\displaystyle \text{separately. Find the time in which each tap can fill the tank separately.}\hfill\text{[CBSE 2019, 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller-diameter tap fill the tank alone in }x\text{ hours.}$
$\displaystyle \therefore \text{The larger-diameter tap fills the tank alone in }(x-2)\text{ hours.}$
$\displaystyle \text{Smaller tap's one hour work}=\frac{1}{x}$
$\displaystyle \text{Larger tap's one hour work}=\frac{1}{x-2}$
$\displaystyle \text{Together, the two taps fill the tank in }1\frac{7}{8}=\frac{15}{8}\text{ hours.}$
$\displaystyle \therefore \frac{1}{x}+\frac{1}{x-2}=\frac{8}{15}$
$\displaystyle \Rightarrow \frac{2x-2}{x(x-2)}=\frac{8}{15}$
$\displaystyle \Rightarrow 15(2x-2)=8x(x-2)$
$\displaystyle \Rightarrow 30x-30=8x^2-16x$
$\displaystyle \Rightarrow 8x^2-46x+30=0$
$\displaystyle \Rightarrow 4x^2-23x+15=0$
$\displaystyle \Rightarrow 4x^2-20x-3x+15=0$
$\displaystyle \Rightarrow 4x(x-5)-3(x-5)=0$
$\displaystyle \Rightarrow (x-5)(4x-3)=0$
$\displaystyle \Rightarrow x=5\text{ or }x=\frac{3}{4}$
$\displaystyle \text{But }x>2,\text{ therefore }x=\frac{3}{4}\text{ is rejected.}$
$\displaystyle \therefore \text{The smaller-diameter tap takes }5\text{ hours to fill the tank alone.}$
$\displaystyle \therefore \text{The larger-diameter tap takes }5-2=3\text{ hours to fill the tank alone.}$
$\displaystyle \\$