Class 10 – Quadratic Equations – ICSE Board Problems (2)

$\displaystyle \textbf{Question 1: }\text{The hotel bill for a number of people for overnight stay is Rs }4800.$
$\displaystyle \text{If there were }4\text{ more, then the bill each had to pay would have reduced Rs }200.$
$\displaystyle \text{Find the number of people staying overnight. (ICSE 2000)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of people staying overnight be }x$
$\displaystyle \text{Total amount for overnight stay }=\text{Rs }4800$
$\displaystyle \text{Amount to be paid by each person }=\frac{4800}{x}$
$\displaystyle \text{If there are }4\text{ more persons i.e. }x+4,\text{ then amount paid by each person }=\frac{4800}{x+4}$
$\displaystyle \text{According to the question, }\frac{4800}{x}-\frac{4800}{x+4}=200$
$\displaystyle 4800\left(\frac{1}{x}-\frac{1}{x+4}\right)=200$
$\displaystyle 4800\left(\frac{x+4-x}{x(x+4)}\right)=200$
$\displaystyle \frac{4800\times 4}{x(x+4)}=200$
$\displaystyle \frac{24(4)}{x(x+4)}=1\ \ \text{[divide both sides by 200]}$
$\displaystyle x(x+4)=96$
$\displaystyle x^2+4x-96=0$
$\displaystyle x^2+12x-8x-96=0\ \ \text{[splitting the middle term]}$
$\displaystyle x(x+12)-8(x+12)=0$
$\displaystyle (x-8)(x+12)=0$
$\displaystyle x=8\text{ or }x=-12$
$\displaystyle \text{Since number of people cannot be negative, }x=8$
$\displaystyle \text{Hence, the number of people staying overnight is }8$

$\displaystyle \textbf{Question 2: }\text{An aeroplane travelled a distance of }400\text{ km at an average} \\ \text{speed of }x\text{ km/h. On the return journey, the speed was increased} \\ \text{by }40\text{ km/h.} \text{Write down an expression for the time taken for}$
$\displaystyle \text{(i) the onward journey.}$
$\displaystyle \text{(ii) the return journey.}$
$\displaystyle \text{If the return journey took }30\text{ min less than the onward journey, then write down an} \\ \text{equation in }x\text{ and find its value. (ICSE 2002)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Average speed }=x\text{ km/h}$
$\displaystyle \text{Time for onward journey }=\frac{400}{x}$
$\displaystyle \text{Return speed }=(x+40)\text{ km/h}$
$\displaystyle \text{Return time }=\frac{400}{x+40}$
$\displaystyle \text{Given return time is }30\text{ min}=\frac{1}{2}\text{ h less}$
$\displaystyle \frac{400}{x}-\frac{400}{x+40}=\frac{1}{2}$
$\displaystyle 400\left(\frac{1}{x}-\frac{1}{x+40}\right)=\frac{1}{2}$
$\displaystyle 400\left(\frac{40}{x(x+40)}\right)=\frac{1}{2}$
$\displaystyle \frac{16000}{x(x+40)}=\frac{1}{2}$
$\displaystyle 32000=x(x+40)$
$\displaystyle x^2+40x-32000=0$
$\displaystyle x^2+200x-160x-32000=0$
$\displaystyle x(x+200)-160(x+200)=0$
$\displaystyle (x+200)(x-160)=0$
$\displaystyle x=160\text{ or }x=-200$
$\displaystyle \text{Since speed cannot be negative, }x=160$
$\displaystyle \text{Average speed of aeroplane }=160\text{ km/h}$
$\displaystyle \text{(i) The speed of the onward journey is }160\text{ km/h.}$
$\displaystyle \text{(ii) The speed of the return journey is }160+40=200\text{ km/h.}$

$\displaystyle \textbf{Question 3: }\text{In an auditorium, seats were arranged in rows and columns. The number of} \\ \text{rows was equal to the number of seats in each row.} \text{When the number of rows was doubled} \\ \text{and the number of seats in each row was reduced by }10\text{, the total number of seats increased} \\ \text{by }300. \text{Find }$
$\displaystyle \text{(i) the number of rows in the original arrangement.}$
$\displaystyle \text{(ii) the number of seats in the auditorium after re-arrangement. (ICSE 2003)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let number of rows }=x$
$\displaystyle \text{Seats in each row }=x$
$\displaystyle \text{Total seats }=x^2$
$\displaystyle \text{After change, rows }=2x,\ \text{seats per row }=x-10$
$\displaystyle \text{Total seats }=2x(x-10)$
$\displaystyle 2x(x-10)=x^2+300$
$\displaystyle 2x^2-20x=x^2+300$
$\displaystyle x^2-20x-300=0$
$\displaystyle x^2-30x+10x-300=0$
$\displaystyle x(x-30)+10(x-30)=0$
$\displaystyle (x-30)(x+10)=0$
$\displaystyle x=30\text{ or }x=-10$
$\displaystyle \text{Since rows cannot be negative, }x=30$
$\displaystyle \text{(i) Original rows }=30$
$\displaystyle \text{(ii) Seats after rearrangement }=30^2+300=1200$

$\displaystyle \textbf{Question 4: }\text{By increasing the speed of a car by }10\text{ km/h, the time of journey for a} \\ \text{distance of }72\text{ km is reduced by }36\text{ min.} \text{Find the original speed of the car. (ICSE 2005)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let speed of car }=x\text{ km/h}$
$\displaystyle \text{New speed }=(x+10)\text{ km/h}$
$\displaystyle \text{Distance }=72\text{ km}$
$\displaystyle t_1=\frac{72}{x},\quad t_2=\frac{72}{x+10}$
$\displaystyle t_1-t_2=36\text{ min}=\frac{36}{60}=\frac{3}{5}\text{ h}$
$\displaystyle \frac{72}{x}-\frac{72}{x+10}=\frac{3}{5}$
$\displaystyle 72\left(\frac{1}{x}-\frac{1}{x+10}\right)=\frac{3}{5}$
$\displaystyle 72\left(\frac{10}{x(x+10)}\right)=\frac{3}{5}$
$\displaystyle \frac{720}{x(x+10)}=\frac{3}{5}$
$\displaystyle 720\times5=3x(x+10)$
$\displaystyle 1200=x(x+10)$
$\displaystyle x^2+10x-1200=0$
$\displaystyle x^2+40x-30x-1200=0$
$\displaystyle x(x+40)-30(x+40)=0$
$\displaystyle (x+40)(x-30)=0$
$\displaystyle x=30\text{ or }x=-40$
$\displaystyle \text{Since speed cannot be negative, }x=30$
$\displaystyle \text{Hence, speed of car is }30\text{ km/h}$

$\displaystyle \textbf{Question 5: }\text{A shopkeeper buys a certain number of books for Rs }720. \text{If the} \\ \text{cost per book was Rs }5\text{ less, then the number of books that could be bought for} \\ \text{Rs }720\text{ would be }2\text{ more.} \text{Taking the original cost of each book to be} \\ \text{Rs }x,\text{ write an equation in }x\text{ and solve it. (ICSE 2006)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Original cost per book = Rs }x$
$\displaystyle \text{Number of books purchased = }\frac{720}{x}$
$\displaystyle \text{If cost per book is reduced by Rs }5,\text{ new cost = Rs }(x-5)$
$\displaystyle \text{Number of books then purchased = }\frac{720}{x-5}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{720}{x-5}=\frac{720}{x}+2$
$\displaystyle \frac{720}{x-5}-\frac{720}{x}=2$
$\displaystyle \frac{720x-720(x-5)}{x(x-5)}=2$
$\displaystyle \frac{720x-720x+3600}{x(x-5)}=2$
$\displaystyle \frac{3600}{x(x-5)}=2$
$\displaystyle 3600=2x(x-5)$
$\displaystyle 3600=2x^2-10x$
$\displaystyle 2x^2-10x-3600=0$
$\displaystyle x^2-5x-1800=0$
$\displaystyle x^2-45x+40x-1800=0$
$\displaystyle x(x-45)+40(x-45)=0$
$\displaystyle (x+40)(x-45)=0$
$\displaystyle x=-40\text{ or }x=45$
$\displaystyle \text{Since cost cannot be negative, }x=45$
$\displaystyle \text{Original cost of each book = Rs }45$

$\displaystyle \textbf{Question 6: }\text{5 yr ago, a woman's age was the square of her son's age.} \text{After }10\text{ yr,} \\ \text{her age will be twice that of her son's age. Find }$
$\displaystyle \text{(i) the age of the son }5\text{ yr ago.}$
$\displaystyle \text{(ii) the present age of the woman. (ICSE 2007)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let present age of woman }=x\text{ yr, son }=y\text{ yr}$
$\displaystyle 5\text{ yr ago, }x-5=(y-5)^2$
$\displaystyle 10\text{ yr hence, }x+10=2(y+10)$
$\displaystyle x+10=2y+20$
$\displaystyle x=2y+10$
$\displaystyle \text{Substitute in first equation:}$
$\displaystyle 2y+10-5=(y-5)^2$
$\displaystyle 2y+5=y^2-10y+25$
$\displaystyle y^2-12y+20=0$
$\displaystyle y^2-10y-2y+20=0\ \ \text{[splitting the middle term]}$
$\displaystyle y(y-10)-2(y-10)=0$
$\displaystyle (y-2)(y-10)=0$
$\displaystyle y=2\text{ or }y=10$
$\displaystyle \text{Since }y=2\text{ not possible, }y=10$
$\displaystyle x=2(10)+10=30$
$\displaystyle \text{(i) Age of son }5\text{ yr ago }=10-5=5\text{ yr}$
$\displaystyle \text{(ii) Present age of woman }=30\text{ yr}$

$\displaystyle \textbf{Question 7: }\text{Some students planned a picnic. The budget for food was Rs }480. \text{As} \\ \text{eight of them failed to join, the cost per member increased by Rs }10. \text{Find how many} \\ \text{students went for the picnic. (ICSE 2008)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let number of students who planned picnic }=x$
$\displaystyle \text{Number who went }=x-8$
$\displaystyle \text{Total budget }=\text{Rs }480$
$\displaystyle \text{Budget per student planned }=\frac{480}{x}$
$\displaystyle \text{Budget per student went }=\frac{480}{x-8}$
$\displaystyle \text{According to question, }\frac{480}{x-8}-\frac{480}{x}=10$
$\displaystyle \frac{480x-480(x-8)}{x(x-8)}=10$
$\displaystyle \frac{3840}{x(x-8)}=10$
$\displaystyle 3840=10x(x-8)$
$\displaystyle 384=x(x-8)$
$\displaystyle x^2-8x-384=0$
$\displaystyle x^2-24x+16x-384=0\ \ \text{[splitting the middle term]}$
$\displaystyle x(x-24)+16(x-24)=0$
$\displaystyle (x-24)(x+16)=0$
$\displaystyle x=24\text{ or }x=-16$
$\displaystyle \text{Since }x\text{ cannot be negative, }x=24$
$\displaystyle \text{Hence, number of students is }24$

$\displaystyle \textbf{Question 8: }\text{The speed of an express train is }x\text{ km/h and the speed of an ordinary} \\ \text{train is }12\text{ km/h less.} \text{If the ordinary train takes }1\text{ h more than the express train to} \\ \text{cover }240\text{ km, find the speed of the express train. (ICSE 2009)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Speed of express train }=x\text{ km/h}$
$\displaystyle \text{Speed of ordinary train }=(x-12)\text{ km/h}$
$\displaystyle \text{Distance covered by both }=240\text{ km}$
$\displaystyle t_1=\frac{240}{x},\quad t_2=\frac{240}{x-12}$
$\displaystyle \text{According to the question, }\frac{240}{x}=\frac{240}{x-12}-1$
$\displaystyle 240(x-12)=x(252-x)$
$\displaystyle 240x-2880=252x-x^2$
$\displaystyle x^2-12x-2880=0$
$\displaystyle x^2-60x+48x-2880=0\ \ \text{[splitting the middle term]}$
$\displaystyle x(x-60)+48(x-60)=0$
$\displaystyle (x-60)(x+48)=0$
$\displaystyle x=60\text{ or }x=-48$
$\displaystyle \text{Since speed cannot be negative, }x=60$
$\displaystyle \text{Hence, speed of express train is }60\text{ km/h}$

$\displaystyle \textbf{Question 9: }\text{A positive number is divided into two parts, such that the sum of the} \\ \text{squares is }208. \text{The square of the larger part is }18\text{ times the smaller part. Taking }x\text{ as the} \\ \text{smaller part, find the number. (ICSE 2010)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller and larger parts of a positive number be }x\text{ and }y,\text{ respectively.}$
$\displaystyle \text{According to the question, }(\text{Larger part})^2=18(\text{Smaller part})$
$\displaystyle \Rightarrow y^2=18x$
$\displaystyle \Rightarrow y=\sqrt{18x}$
$\displaystyle \text{Also, given sum of the squares of the two parts is }208$
$\displaystyle \therefore x^2+y^2=208$
$\displaystyle x^2+18x=208$
$\displaystyle x^2+18x-208=0$
$\displaystyle x^2+26x-8x-208=0\ \ \text{[splitting the middle term]}$
$\displaystyle x(x+26)-8(x+26)=0$
$\displaystyle (x-8)(x+26)=0$
$\displaystyle x=8\text{ or }x=-26$
$\displaystyle \text{Since }x\text{ cannot be negative, }x=8$
$\displaystyle \text{Smaller part }=8$
$\displaystyle y=\sqrt{18x}=\sqrt{18\times8}=\sqrt{144}=12$
$\displaystyle \text{Required number }=8+12=20$

$\displaystyle \textbf{Question 10: }\text{A car covers a distance of }400\text{ km at a certain speed.} \text{Had the} \\ \text{speed been }12\text{ km/h more, the time taken would have been }1\text{ h }40\text{ min less.} \text{Find the} \\ \text{original speed of the car. (ICSE 2012)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original speed of the car = }x\text{ km/h}$
$\displaystyle \text{Distance cover by a car }=400\text{ km}$
$\displaystyle \text{Original time taken, }T=\frac{400}{x}\text{ h}\ \ \text{[time = distance/speed]}$
$\displaystyle \text{Further, speed is increased by }12\text{ km/h}$
$\displaystyle \text{Increased speed }=(x+12)\text{ km/h}$
$\displaystyle \text{Now, new time taken }=\frac{400}{x+12}\text{ h}$
$\displaystyle \text{According to the question, }\frac{400}{x}-\frac{400}{x+12}=\frac{5}{3}$
$\displaystyle \text{[}\because 1\text{ h }40\text{ min}=1\text{ h}+\frac{40}{60}\text{ h}=1+\frac{2}{3}=\frac{5}{3}\text{ h]}$
$\displaystyle 400\left(\frac{1}{x}-\frac{1}{x+12}\right)=\frac{5}{3}$
$\displaystyle \Rightarrow 400\left(\frac{x+12-x}{x(x+12)}\right)=\frac{5}{3}$
$\displaystyle \Rightarrow \frac{400\times 12}{x(x+12)}=\frac{5}{3}$
$\displaystyle \Rightarrow 80(12)\times 3=x(x+12)\times 1\ \ \text{[divide both sides by 5]}$
$\displaystyle \Rightarrow 2880=x^2+12x$
$\displaystyle \Rightarrow x^2+12x-2880=0$
$\displaystyle \Rightarrow x^2+60x-48x-2880=0\ \ \text{[splitting the middle term]}$
$\displaystyle \Rightarrow x(x+60)-48(x+60)=0$
$\displaystyle \Rightarrow (x+60)(x-48)=0$
$\displaystyle \Rightarrow x=48\text{ or }x=-60$
$\displaystyle \text{Since speed cannot be negative, }x=48$
$\displaystyle \text{Hence, the original speed of the car is }48\text{ km/h}$

$\displaystyle \textbf{Question 11: }\text{Rs }480\text{ is divided equally among }x\text{ children. If the number of children} \\ \text{was }20\text{ more,} \text{then each would have got Rs }12\text{ less. Find the value of }x.\text{ (ICSE 2011)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Total money }=\text{Rs }480$
$\displaystyle \text{Let number of children }=x$
$\displaystyle \text{Money received by each child }=\frac{480}{x}$
$\displaystyle \text{If children were }20\text{ more, total children }=x+20$
$\displaystyle \text{Money received then }=\frac{480}{x+20}$
$\displaystyle \frac{480}{x}-\frac{480}{x+20}=12$
$\displaystyle \frac{480(x+20)-480x}{x(x+20)}=12$
$\displaystyle \frac{9600}{x(x+20)}=12$
$\displaystyle 9600=12x(x+20)$
$\displaystyle 800=x(x+20)$
$\displaystyle x^2+20x-800=0$
$\displaystyle x^2+40x-20x-800=0$
$\displaystyle x(x+40)-20(x+40)=0$
$\displaystyle (x-20)(x+40)=0$
$\displaystyle x=20\text{ or }x=-40$
$\displaystyle \text{Since }x\text{ cannot be negative, }x=20$
$\displaystyle \text{Hence, number of children is }20$

$\displaystyle \textbf{Question 12: }\text{Without solving, find the value of }m\text{ for which the equation has real} \\ \text{and equal roots.} x^2+2(m-1)x+(m+5)=0 \text{ (ICSE 2012)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given quadratic equation }x^2+2(m-1)x+(m+5)=0$
$\displaystyle \text{On comparing with }ax^2+bx+c=0,\text{ we get }a=1,b=2(m-1),c=m+5$
$\displaystyle D=b^2-4ac=0$
$\displaystyle [2(m-1)]^2-4\times1\times(m+5)=0$
$\displaystyle 4(m-1)^2-4(m+5)=0$
$\displaystyle 4[m^2-2m+1-m-5]=0$
$\displaystyle 4[m^2-3m-4]=0$
$\displaystyle m^2-3m-4=0$
$\displaystyle m^2-4m+m-4=0$
$\displaystyle m(m-4)+1(m-4)=0$
$\displaystyle (m+1)(m-4)=0$
$\displaystyle m=-1,4$

$\displaystyle \textbf{Question 13: }\text{Without solving, find the value of }p\text{ for which the equation has real} \\ \text{and equal roots.} x^2+(p-3)x+p=0 \text{ (ICSE 2013)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given quadratic equation is}$
$\displaystyle x^2+(p-3)x+p=0$
$\displaystyle \text{On comparing above equation with }ax^2+bx+c=0,\text{ we get}$
$\displaystyle a=1,b=p-3\text{ and }c=p$
$\displaystyle \text{For real and equal roots, discriminant should be zero.}$
$\displaystyle D=b^2-4ac=0$
$\displaystyle (p-3)^2-4\times1\times p=0$
$\displaystyle p^2+9-6p-4p=0$
$\displaystyle p^2-10p+9=0$
$\displaystyle p^2-9p-p+9=0$
$\displaystyle p(p-9)-1(p-9)=0$
$\displaystyle (p-1)(p-9)=0$
$\displaystyle p=1\text{ or }p=9$

$\displaystyle \textbf{Question 14: }\text{A two-digit positive number, such that the product of its digits is }6. \text{If }9\text{ is} \\ \text{added to the number, then the digits interchange their places. Find the number.} \\ \text{ (ICSE 2014)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let required two-digit number be }10x+y$
$\displaystyle \text{Product of digits }=xy=6$
$\displaystyle 10x+y+9=10y+x$
$\displaystyle 9x-9y+9=0$
$\displaystyle 9(x-y+1)=0$
$\displaystyle x-y+1=0$
$\displaystyle y=x+1$
$\displaystyle x(x+1)=6$
$\displaystyle x^2+x-6=0$
$\displaystyle x^2+3x-2x-6=0$
$\displaystyle x(x+3)-2(x+3)=0$
$\displaystyle (x-2)(x+3)=0$
$\displaystyle x=2\text{ or }x=-3$
$\displaystyle \text{Since digit cannot be negative, }x=2,\ y=3$
$\displaystyle \text{Required number is }23$

$\displaystyle \textbf{Question 15: }\text{Sum of two natural numbers is }8\text{ and the difference of their reciprocals} \\ \text{is }\frac{2}{15}. \text{ Find the numbers. (ICSE 2015)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let required natural numbers be }x\text{ and }8-x$
$\displaystyle \frac{1}{x}-\frac{1}{8-x}=\frac{2}{15}$
$\displaystyle \frac{8-x-x}{x(8-x)}=\frac{2}{15}$
$\displaystyle \frac{8-2x}{x(8-x)}=\frac{2}{15}$
$\displaystyle 15(8-2x)=2x(8-x)$
$\displaystyle 120-30x=16x-2x^2$
$\displaystyle 2x^2-46x+120=0$
$\displaystyle x^2-23x+60=0$
$\displaystyle x^2-20x-3x+60=0$
$\displaystyle x(x-20)-3(x-20)=0$
$\displaystyle (x-20)(x-3)=0$
$\displaystyle x=20\text{ or }x=3$
$\displaystyle \text{Since }x=20\text{ not possible, }x=3$
$\displaystyle \text{Required numbers are }3\text{ and }5$

$\displaystyle \textbf{Question 16: }\text{A bus covers a distance of }240\text{ km at a uniform speed.} \text{Due to} \\ \text{heavy rain its speed gets reduced by }10\text{ km/h and it takes }2\text{ hours longer.} \text{Assuming the} \\ \text{uniform speed to be }x\text{ km/h, form an equation and solve to evaluate }x.\text{ (ICSE 2016)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let uniform speed be }x\text{ km/h}$
$\displaystyle \text{Time taken to cover }240\text{ km }=\frac{240}{x}\text{ h}$
$\displaystyle \text{Reduced speed }=(x-10)\text{ km/h}$
$\displaystyle \text{Time taken at reduced speed }=\frac{240}{x-10}\text{ h}$
$\displaystyle \text{According to question, }\frac{240}{x-10}-\frac{240}{x}=2$
$\displaystyle \frac{240x-240(x-10)}{x(x-10)}=2$
$\displaystyle \frac{240x-240x+2400}{x(x-10)}=2$
$\displaystyle \frac{2400}{x(x-10)}=2$
$\displaystyle 2400=2x(x-10)$
$\displaystyle 1200=x^2-10x$
$\displaystyle x^2-10x-1200=0$
$\displaystyle x^2-40x+30x-1200=0$
$\displaystyle x(x-40)+30(x-40)=0$
$\displaystyle (x-40)(x+30)=0$
$\displaystyle x=40\text{ or }x=-30$
$\displaystyle \text{Since speed cannot be negative, }x=40$
$\displaystyle \text{Hence original speed is }40\text{ km/h}$

$\displaystyle \textbf{Question 17: }\text{Bosco wishes to start a }200\text{ m}^2\text{ rectangular vegetable garden.} \text{Since,} \\ \text{he has only }50\text{ m barbed wire, he fences three sides of the rectangle} \text{letting his} \\ \text{house compound wall act as the fourth side. Find the dimensions. (ICSE 2017)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let length of garden be }l\text{ and width be }b$
$\displaystyle \text{Area }=l\times b=200$
$\displaystyle b=\frac{200}{l}$
$\displaystyle \text{Perimeter of three sides }=2l+b=50$
$\displaystyle 2l+\frac{200}{l}=50$
$\displaystyle 2l^2+200=50l$
$\displaystyle 2l^2-50l+200=0$
$\displaystyle l^2-25l+100=0$
$\displaystyle l^2-20l-5l+100=0$
$\displaystyle l(l-20)-5(l-20)=0$
$\displaystyle (l-20)(l-5)=0$
$\displaystyle l=20\text{ or }l=5$
$\displaystyle \text{If }l=20,\ b=10\quad\text{If }l=5,\ b=40$
$\displaystyle \text{Hence dimensions are }(20\text{ m and }10\text{ m})\text{ or }(40\text{ m and }5\text{ m})$

$\displaystyle \textbf{Question 18: }\text{The sum of the ages of Vivek and his younger brother Amit is }47\text{ yr.}$
$\displaystyle \text{The product of their ages in years is }550.\text{ Find their ages. (ICSE 2017)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let age of Vivek }=x,\text{ age of Amit }=47-x$
$\displaystyle x(47-x)=550$
$\displaystyle 47x-x^2=550$
$\displaystyle x^2-47x+550=0$
$\displaystyle x^2-25x-22x+550=0$
$\displaystyle x(x-25)-22(x-25)=0$
$\displaystyle (x-25)(x-22)=0$
$\displaystyle x=25\text{ or }x=22$
$\displaystyle \text{Age of Vivek }=25\text{ yr, age of Amit }=22\text{ yr}$

$\displaystyle \textbf{Question 19: }\text{Find the value of }k\text{ for which the equation has equal roots. (ICSE 2018)}$
$\displaystyle x^2+4kx+(k^2-k+2)=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given quadratic equation }x^2+4kx+(k^2-k+2)=0$
$\displaystyle \text{On comparing with }ax^2+bx+c=0,\text{ we get }a=1,b=4k,c=k^2-k+2$
$\displaystyle b^2-4ac=0$
$\displaystyle (4k)^2-4\times1\times(k^2-k+2)=0$
$\displaystyle 16k^2-4k^2+4k-8=0$
$\displaystyle 12k^2+4k-8=0$
$\displaystyle 3k^2+k-2=0$
$\displaystyle 3k^2+3k-2k-2=0$
$\displaystyle 3k(k+1)-2(k+1)=0$
$\displaystyle (3k-2)(k+1)=0$
$\displaystyle k=\frac{2}{3},-1$
$\displaystyle \text{Answer: }k=\frac{2}{3},-1$

$\displaystyle \textbf{Question 20: }\text{The product of two consecutive natural numbers which are multiples of }3$
$\displaystyle \text{is equal to }810.\text{ Find the two numbers. (ICSE 2019)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let two consecutive natural numbers having multiples of }3\text{ be }3x\text{ and }3x+3$
$\displaystyle \text{According to given condition, }3x(3x+3)=810$
$\displaystyle 9(x^2+x)=810$
$\displaystyle x^2+x=90$
$\displaystyle x^2+x-90=0$
$\displaystyle x^2+10x-9x-90=0$
$\displaystyle x(x+10)-9(x+10)=0$
$\displaystyle (x-9)(x+10)=0$
$\displaystyle x=9\text{ or }x=-10$
$\displaystyle \text{Since numbers are natural, discard negative value}$
$\displaystyle \text{Required numbers are }27\text{ and }30$

$\displaystyle \textbf{Question 21: }\text{For what value of }k\text{ will the quadratic equation have real} \\ \text{and equal roots?} (k+1)x^2-4kx+9=0 \text{(ICSE 2020)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given quadratic equation }(k+1)x^2-4kx+9=0$
$\displaystyle \text{On comparing with }ax^2+bx+c=0,\text{ we get }a=k+1,b=-4k,c=9$
$\displaystyle D=b^2-4ac=0$
$\displaystyle (-4k)^2-4(k+1)(9)=0$
$\displaystyle 16k^2-36(k+1)=0$
$\displaystyle 16k^2-36k-36=0$
$\displaystyle 4k^2-9k-9=0$
$\displaystyle 4k^2-12k+3k-9=0$
$\displaystyle 4k(k-3)+3(k-3)=0$
$\displaystyle (4k+3)(k-3)=0$
$\displaystyle k=3,-\frac{3}{4}$
$\displaystyle \Rightarrow 4x^2-6x-6x+9=0$
$\displaystyle \Rightarrow 2x(2x-3)-3(2x-3)=0$
$\displaystyle \Rightarrow (2x-3)(2x-3)=0$
$\displaystyle \Rightarrow x=\frac{3}{2},\frac{3}{2}$
$\displaystyle \text{Again, if }k=-\frac{3}{4},\text{ then equation becomes}$
$\displaystyle \frac{1}{4}x^2+3x+9=0$
$\displaystyle \Rightarrow x^2+12x+36=0$
$\displaystyle \Rightarrow (x+6)^2=0$
$\displaystyle \Rightarrow x=-6,-6$

$\displaystyle \textbf{Question 22: }\text{The roots of the quadratic equation }x^2+2x+1=0\text{ are}$
$\displaystyle \text{(a) real and distinct \ \ (b) real and equal \ \ (c) distinct \ \ (d) not real/imaginary}$
$\displaystyle \text{ (ICSE Semester I 2022)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b): }\text{Given quadratic equation }x^2+2x+1=0$
$\displaystyle \text{On comparing with }ax^2+bx+c=0,\text{ we get }a=1,b=2,c=1$
$\displaystyle D=b^2-4ac=(2)^2-4(1)(1)=4-4=0$
$\displaystyle D=0$
$\displaystyle \text{Hence, the roots are real and equal.}$
$\displaystyle \text{Answer: Real and equal}$

$\displaystyle \textbf{Question 23: }\text{Two cars }X\text{ and }Y\text{ use }1\text{ L of diesel to travel }x\text{ km and }(x+3)\text{ km respectively.}$
$\displaystyle \text{If both the cars covered a distance of }72\text{ km, then}$

$\displaystyle \text{(i) The number of litres of diesel used by car }X\text{ is}$
$\displaystyle \text{(a) }\frac{72}{x-3}\text{ L \ \ (b) }\frac{72}{x+3}\text{ L \ \ (c) }\frac{72}{x}\text{ L \ \ (d) }\frac{12}{x}\text{ L}$

$\displaystyle \text{(ii) The number of litres of diesel used by car }Y\text{ is}$
$\displaystyle \text{(a) }\frac{72}{x-3}\text{ L \ \ (b) }\frac{72}{x+3}\text{ L \ \ (c) }\frac{72}{x}\text{ L \ \ (d) }\frac{12}{x+3}\text{ L}$

$\displaystyle \text{(iii) If car }X\text{ used }4\text{ L of diesel more than car }Y\text{ in the journey, then}$
$\displaystyle \text{(a) }\frac{72}{x-3}-\frac{12}{x}=4$
$\displaystyle \text{(b) }\frac{72}{x+3}-\frac{72}{x}=4$
$\displaystyle \text{(c) }\frac{72}{x}-\frac{72}{x+3}=4$
$\displaystyle \text{(d) }\frac{72}{x-3}-\frac{12}{x+3}=4$

$\displaystyle \text{(iv) The amount of diesel used by the car }X\text{ is}$
$\displaystyle \text{(a) }6\text{ L \ \ (b) }12\text{ L \ \ (c) }18\text{ L \ \ (d) }24\text{ L \ \ (ICSE Semester I 2022)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) To travel }x\text{ km, diesel used by car }X=1\text{ L}$
$\displaystyle \text{To travel }1\text{ km, diesel used by car }X=\frac{1}{x}\text{ L}$
$\displaystyle \text{To travel }72\text{ km, diesel used by car }X=\frac{72}{x}\text{ L}$

$\displaystyle \text{(ii) Similarly, to travel }72\text{ km, diesel used by car }Y=\frac{72}{x+3}\text{ L}$

$\displaystyle \text{(iii) Given car }X\text{ used }4\text{ L more than car }Y$
$\displaystyle \frac{72}{x}=\frac{72}{x+3}+4$
$\displaystyle \Rightarrow \frac{72}{x}-\frac{72}{x+3}=4\ \ \ ...(i)$

$\displaystyle \text{(iv) From Eq. (i),}$
$\displaystyle \frac{72}{x}-\frac{72}{x+3}=4$
$\displaystyle 72\left(\frac{x+3-x}{x(x+3)}\right)=4$
$\displaystyle \frac{216}{x(x+3)}=4$
$\displaystyle x(x+3)=54$
$\displaystyle x^2+3x-54=0$
$\displaystyle x^2+9x-6x-54=0$
$\displaystyle x(x+9)-6(x+9)=0$
$\displaystyle (x-6)(x+9)=0$
$\displaystyle x=6,\ x=-9\text{ (neglected)}$
$\displaystyle \text{Required amount of diesel }=\frac{72}{x}=\frac{72}{6}=12\text{ L}$

$\displaystyle \textbf{Question 24: }\text{When the roots of a quadratic equation are real and equal, then} \\ \text{the discriminant is}$
$\displaystyle \text{(a) infinite \ \ (b) positive \ \ (c) zero \ \ (d) negative \ \ (ICSE 2023)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c): }\text{Let }\alpha\text{ and }\beta\text{ be the roots and }\alpha,\beta\text{ are real.}$
$\displaystyle \text{Since, }\alpha=\beta\ \ \ ...(i)$
$\displaystyle \text{Then, quadratic equation is }x^2-(\alpha+\beta)x+\alpha\beta=0$
$\displaystyle \Rightarrow x^2-(\alpha+\alpha)x+\alpha\alpha=0\ \ \text{[from Eq. (i)]}$
$\displaystyle \Rightarrow x^2-2\alpha x+\alpha^2=0$
$\displaystyle \text{On comparing }ax^2+bx+c=0,\text{ we get }a=1,b=-2\alpha,c=\alpha^2$
$\displaystyle D=b^2-4ac=(-2\alpha)^2-4\times1\times\alpha^2$
$\displaystyle =4\alpha^2-4\alpha^2=0$
$\displaystyle \text{Hence, discriminant is zero.}$

$\displaystyle \textbf{Question 25: }\text{A man covers a distance of }100\text{ km, travelling with a uniform} \\ \text{speed of }x\text{ km/h.} \text{Had the speed been }5\text{ km/h more it would have taken }1\text{ h less.} \\ \text{Find }x\text{ the original speed. (ICSE 2023)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The increased speed of the train }=(x+5)\text{ km/h}$
$\displaystyle \text{Time taken under uniform speed to cover }100\text{ km }=\frac{100}{x}\text{ h}$
$\displaystyle \text{Time taken under increased speed to cover }100\text{ km }=\frac{100}{x+5}\text{ h}$
$\displaystyle \text{According to question, }\frac{100}{x}-\frac{100}{x+5}=1$
$\displaystyle \frac{100(x+5)-100x}{x(x+5)}=1$
$\displaystyle 100x+500-100x=x(x+5)$
$\displaystyle x^2+5x=500$
$\displaystyle x^2+5x-500=0$
$\displaystyle x^2+25x-20x-500=0$
$\displaystyle x(x+25)-20(x+25)=0$
$\displaystyle (x+25)(x-20)=0$
$\displaystyle x=-25\text{ or }x=20$
$\displaystyle \text{Since speed cannot be negative, }x=20$
$\displaystyle \text{Hence, original speed is }20\text{ km/h}$

$\displaystyle \textbf{Question 26: }\text{The roots of the quadratic equation }px^2-qx+r=0\text{ are real and equal, if }$
$\displaystyle \text{(a) }p^2=4qr\ \ \text{(b) }q^2=4pr\ \ \text{(c) }-q^2=4pr\ \ \text{(d) }p^2>4qr\ \text{ (ICSE 2024)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b): }\text{Given, }px^2-qx+r=0$
$\displaystyle \text{The roots are real and equal.}$
$\displaystyle b^2-4ac=0$
$\displaystyle (-q)^2-4\times p\times r=0$
$\displaystyle q^2=4pr$
$\displaystyle \text{Answer: }q^2=4pr$