Class 10: Quadratic Equations

$\displaystyle \textbf{Question 1: } \text{The distance by road between two towns }A\text{ and }B\text{ is }216\text{ km,}$
$\displaystyle \text{and by rail it is }208\text{ km. A car travels at a speed of }x\text{ km/hr}$
$\displaystyle \text{and the train travels }16\text{ km/hr faster than the car. Calculate:}$
$\displaystyle \text{(i) the time taken by the car to reach town }B\text{ from }A\text{, in terms of }x;$
$\displaystyle \text{(ii) the time taken by the train to reach town }B\text{ from }A\text{, in terms of }x;$
$\displaystyle \text{(iii) if the train takes }2\text{ hours less than the car, obtain an equation}$
$\displaystyle \text{in }x\text{ and solve it. Hence, find the speed of the train.}\hfill \text{[ICSE 1998]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Time taken by the car}=\frac{216}{x}\text{ hours}$
$\displaystyle \text{Speed of the train}=(x+16)\text{ km/hr}$
$\displaystyle \text{Time taken by the train}=\frac{208}{x+16}\text{ hours}$
$\displaystyle \text{Given, }\frac{216}{x}-\frac{208}{x+16}=2$
$\displaystyle \frac{216(x+16)-208x}{x(x+16)}=2$
$\displaystyle 216x+3456-208x=2x(x+16)$
$\displaystyle 8x+3456=2x^2+32x$
$\displaystyle 2x^2+24x-3456=0$
$\displaystyle x^2+12x-1728=0$
$\displaystyle (x-36)(x+48)=0$
$\displaystyle \Rightarrow x=36\text{ or }x=-48$
$\displaystyle \text{Since speed cannot be negative, }x\neq -48.$
$\displaystyle \therefore x=36$
$\displaystyle \text{Speed of the train}=x+16=36+16=52\text{ km/hr}$
$\displaystyle \therefore \text{The speed of the train is }52\text{ km/hr}.$
$\\$

$\displaystyle \textbf{Question 2: } \text{A trader buys }x\text{ articles for a total cost of Rs. }600.$
$\displaystyle \text{Write down the cost of one article in terms of }x.\text{ If the cost per article}$
$\displaystyle \text{were Rs. }5\text{ more, the number of articles that can be bought for Rs. }600$
$\displaystyle \text{would be four less. Write down the equation in terms of }x\text{ and solve it.}$
$\displaystyle \hfill \text{[ICSE 1999]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Cost of one article}=\text{Rs. }\frac{600}{x}$
$\displaystyle \text{New cost of one article}=\text{Rs. }\left(\frac{600}{x}+5\right)$
$\displaystyle \text{New number of articles}=x-4$
$\displaystyle \therefore \left(\frac{600}{x}+5\right)(x-4)=600$
$\displaystyle \left(\frac{600+5x}{x}\right)(x-4)=600$
$\displaystyle (600+5x)(x-4)=600x$
$\displaystyle 600x-2400+5x^2-20x=600x$
$\displaystyle 5x^2-20x-2400=0$
$\displaystyle x^2-4x-480=0$
$\displaystyle (x-24)(x+20)=0$
$\displaystyle \Rightarrow x=24\text{ or }x=-20$
$\displaystyle \text{Since the number of articles cannot be negative, }x\neq -20.$
$\displaystyle \therefore x=24$
$\displaystyle \therefore \text{The number of articles bought is }24.$
$\\$

$\displaystyle \textbf{Question 3: } \text{A hotel bill for a number of people for overnight stay is Rs. }4800.$
$\displaystyle \text{If there were }4\text{ people more, the bill each person had to pay would}$
$\displaystyle \text{have reduced by Rs. }200.\text{ Find the number of people staying overnight.}$
$\displaystyle \hfill \text{[ICSE 2000]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of people be }x.$
$\displaystyle \therefore \text{Amount paid by each person}=\text{Rs. }\frac{4800}{x}$
$\displaystyle \text{If there were }4\text{ people more, amount paid by each person}=\text{Rs. }\frac{4800}{x+4}$
$\displaystyle \therefore \frac{4800}{x}-\frac{4800}{x+4}=200$
$\displaystyle \frac{4800(x+4)-4800x}{x(x+4)}=200$
$\displaystyle \frac{19200}{x(x+4)}=200$
$\displaystyle x(x+4)=96$
$\displaystyle x^2+4x-96=0$
$\displaystyle (x-8)(x+12)=0$
$\displaystyle \Rightarrow x=8\text{ or }x=-12$
$\displaystyle \text{Since the number of people cannot be negative, }x\neq -12.$
$\displaystyle \therefore x=8$
$\displaystyle \therefore \text{The number of people staying overnight is }8.$
$\\$

$\displaystyle \textbf{Question 4: } \text{An airplane travelled a distance of }400\text{ km at an average}$
$\displaystyle \text{speed of }x\text{ km/hr. On the return journey, the speed was increased by }40\text{ km/hr.}$
$\displaystyle \text{Write down an expression for the time taken for the onward journey}$
$\displaystyle \text{and the return journey. If the airplane takes }2\text{ hours less in returning,}$
$\displaystyle \text{calculate the speed of the airplane.}\hfill \text{[ICSE 2002]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Time taken for onward journey}=\frac{400}{x}\text{ hours}$
$\displaystyle \text{Time taken for return journey}=\frac{400}{x+40}\text{ hours}$
$\displaystyle \text{Given, }\frac{400}{x}-\frac{400}{x+40}=2$
$\displaystyle \frac{400(x+40)-400x}{x(x+40)}=2$
$\displaystyle \frac{16000}{x(x+40)}=2$
$\displaystyle x(x+40)=8000$
$\displaystyle x^2+40x-8000=0$
$\displaystyle x=\frac{-40\pm\sqrt{40^2-4(1)(-8000)}}{2}$
$\displaystyle x=\frac{-40\pm\sqrt{33600}}{2}$
$\displaystyle x=\frac{-40\pm40\sqrt{21}}{2}$
$\displaystyle x=-20\pm20\sqrt{21}$
$\displaystyle \text{Since speed cannot be negative, }x=20(\sqrt{21}-1).$
$\displaystyle \therefore \text{The speed of the airplane is }20(\sqrt{21}-1)\text{ km/hr, i.e., }71.65\text{ km/hr.}$
$\\$

$\displaystyle \textbf{Question 5: } \text{Rs. }6500\text{ was divided equally among a certain number}$
$\displaystyle \text{of persons. Had there been }15\text{ persons more, each would have got Rs. }30$
$\displaystyle \text{less. Find the original number of persons.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original number of persons be }x.$
$\displaystyle \therefore \text{Amount received by each person}=\text{Rs. }\frac{6500}{x}$
$\displaystyle \text{If there were }15\text{ persons more, amount received by each person}=\text{Rs. }\frac{6500}{x+15}$
$\displaystyle \therefore \frac{6500}{x}-\frac{6500}{x+15}=30$
$\displaystyle \frac{6500(x+15)-6500x}{x(x+15)}=30$
$\displaystyle \frac{97500}{x(x+15)}=30$
$\displaystyle 30x(x+15)=97500$
$\displaystyle x^2+15x-3250=0$
$\displaystyle (x-50)(x+65)=0$
$\displaystyle \Rightarrow x=50\text{ or }x=-65$
$\displaystyle \text{Since the number of persons cannot be negative, }x\neq -65.$
$\displaystyle \therefore x=50$
$\displaystyle \therefore \text{The original number of persons is }50.$
$\\$

$\displaystyle \textbf{Question 6: } \text{A plane left }30\text{ minutes later than the scheduled time and,}$
$\displaystyle \text{in order to reach its destination }1500\text{ km away in time, it had to increase}$
$\displaystyle \text{its speed by }250\text{ km/hr from its usual speed. Find its usual speed.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the usual speed of the plane be }x\text{ km/hr.}$
$\displaystyle \text{Increased speed}=(x+250)\text{ km/hr}$
$\displaystyle 30\text{ minutes}=\frac{1}{2}\text{ hour}$
$\displaystyle \therefore \frac{1500}{x}-\frac{1500}{x+250}=\frac{1}{2}$
$\displaystyle \frac{1500(x+250)-1500x}{x(x+250)}=\frac{1}{2}$
$\displaystyle \frac{375000}{x(x+250)}=\frac{1}{2}$
$\displaystyle x(x+250)=750000$
$\displaystyle x^2+250x-750000=0$
$\displaystyle (x-750)(x+1000)=0$
$\displaystyle \Rightarrow x=750\text{ or }x=-1000$
$\displaystyle \text{Since speed cannot be negative, }x\neq -1000.$
$\displaystyle \therefore x=750$
$\displaystyle \therefore \text{The usual speed of the plane is }750\text{ km/hr}.$
$\\$

$\displaystyle \textbf{Question 7: } \text{Two trains leave a railway station at the same time.}$
$\displaystyle \text{The first train travels due west and the second train due north.}$
$\displaystyle \text{The first train travels }5\text{ km/hr faster than the second train.}$
$\displaystyle \text{If after }2\text{ hours, they are }50\text{ km apart, find the speed of each train.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speed of the second train be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Speed of the first train}=(x+5)\text{ km/hr}$
$\displaystyle \text{Distance travelled by second train in }2\text{ hours}=2x\text{ km}$
$\displaystyle \text{Distance travelled by first train in }2\text{ hours}=2(x+5)\text{ km}$
$\displaystyle \text{Since the trains travel in perpendicular directions,}$
$\displaystyle [2(x+5)]^2+(2x)^2=50^2$
$\displaystyle 4(x+5)^2+4x^2=2500$
$\displaystyle 4(x^2+10x+25)+4x^2=2500$
$\displaystyle 8x^2+40x+100=2500$
$\displaystyle 8x^2+40x-2400=0$
$\displaystyle x^2+5x-300=0$
$\displaystyle (x-15)(x+20)=0$
$\displaystyle \Rightarrow x=15\text{ or }x=-20$
$\displaystyle \text{Since speed cannot be negative, }x\neq -20.$
$\displaystyle \therefore x=15$
$\displaystyle \text{Speed of first train}=x+5=20\text{ km/hr}$
$\displaystyle \therefore \text{The speeds are }20\text{ km/hr and }15\text{ km/hr}.$
$\\$

$\displaystyle \textbf{Question 8: } \text{The sum }S\text{ of first }n\text{ even natural numbers is given}$
$\displaystyle \text{by the relation }S=n(n+1).\text{ Find }n,\text{ if the sum is }420.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }S=n(n+1)$
$\displaystyle 420=n(n+1)$
$\displaystyle n^2+n-420=0$
$\displaystyle (n-20)(n+21)=0$
$\displaystyle \Rightarrow n=20\text{ or }n=-21$
$\displaystyle \text{Since }n\text{ cannot be negative, }n\neq -21.$
$\displaystyle \therefore n=20$
$\\$

$\displaystyle \textbf{Question 9: } \text{The sum of the ages of a father and his son is }45\text{ years.}$
$\displaystyle \text{Five years ago, the product of their ages was }124.\text{ Determine their present ages.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the present age of the father be }x\text{ years.}$
$\displaystyle \therefore \text{Present age of the son}=(45-x)\text{ years}$
$\displaystyle \text{Father's age }5\text{ years ago}=(x-5)\text{ years}$
$\displaystyle \text{Son's age }5\text{ years ago}=(45-x-5)=(40-x)\text{ years}$
$\displaystyle \text{Given, }(x-5)(40-x)=124$
$\displaystyle 40x-x^2-200+5x=124$
$\displaystyle -x^2+45x-200=124$
$\displaystyle x^2-45x+324=0$
$\displaystyle (x-36)(x-9)=0$
$\displaystyle \Rightarrow x=36\text{ or }x=9$
$\displaystyle \text{Since father is older than son, }x=36.$
$\displaystyle \text{Son's age}=45-36=9\text{ years}$
$\displaystyle \therefore \text{The present ages of the father and son are }36\text{ years and }9\text{ years.}$
$\\$

$\displaystyle \textbf{Question 10: } \text{In an auditorium, seats were arranged in rows and columns.}$
$\displaystyle \text{The number of rows was equal to the number of seats in each row.}$
$\displaystyle \text{When the number of rows was doubled and the number of seats in each row}$
$\displaystyle \text{was reduced by }10,\text{ the total number of seats increased by }300.\text{ Find:}$
$\displaystyle \text{(i) the number of rows in the original arrangement;}$
$\displaystyle \text{(ii) the number of seats in the auditorium after re-arrangement.}\hfill \text{[ICSE 2003]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of rows in the original arrangement be }x.$
$\displaystyle \therefore \text{Number of seats in each row}=x$
$\displaystyle \text{Original number of seats}=x^2$
$\displaystyle \text{New number of rows}=2x$
$\displaystyle \text{New number of seats in each row}=x-10$
$\displaystyle \text{Given, }2x(x-10)-x^2=300$
$\displaystyle 2x^2-20x-x^2=300$
$\displaystyle x^2-20x-300=0$
$\displaystyle (x-30)(x+10)=0$
$\displaystyle \Rightarrow x=30\text{ or }x=-10$
$\displaystyle \text{Since the number of rows cannot be negative, }x\neq -10.$
$\displaystyle \therefore x=30$
$\displaystyle \text{Number of rows in original arrangement}=30$
$\displaystyle \text{Number of seats after re-arrangement}=2x(x-10)$
$\displaystyle =2(30)(30-10)=1200$
$\displaystyle \therefore \text{The original number of rows is }30\text{ and the number of seats}$
$\displaystyle \text{after re-arrangement is }1200.$
$\\$

$\displaystyle \textbf{Question 11: } \text{Mohan takes }16\text{ days less than Manoj to do a piece}$
$\displaystyle \text{of work. If both working together can do it in }15\text{ days, in how many}$
$\displaystyle \text{days will Mohan alone complete the work?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let Manoj take }x\text{ days to complete the work.}$
$\displaystyle \therefore \text{Mohan takes }(x-16)\text{ days to complete the work.}$
$\displaystyle \text{Given, }\frac{1}{x}+\frac{1}{x-16}=\frac{1}{15}$
$\displaystyle \frac{x-16+x}{x(x-16)}=\frac{1}{15}$
$\displaystyle 15(2x-16)=x(x-16)$
$\displaystyle 30x-240=x^2-16x$
$\displaystyle x^2-46x+240=0$
$\displaystyle (x-40)(x-6)=0$
$\displaystyle \Rightarrow x=40\text{ or }x=6$
$\displaystyle \text{Since }x>16,\ x\neq 6.$
$\displaystyle \therefore x=40$
$\displaystyle \text{Mohan's time}=x-16=24\text{ days}$
$\displaystyle \therefore \text{Mohan alone will complete the work in }24\text{ days.}$
$\\$

$\displaystyle \textbf{Question 12: } \text{Two years ago, a man's age was three times the square}$
$\displaystyle \text{of his son's age. In three years' time, his age will be four times}$
$\displaystyle \text{his son's age. Find their present ages.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the man's present age be }x\text{ years.}$
$\displaystyle \text{Let the son's present age be }y\text{ years.}$
$\displaystyle \text{Given, }x-2=3(y-2)^2$
$\displaystyle \text{Also, }x+3=4(y+3)$
$\displaystyle \therefore x=4(y+3)-3=4y+9$
$\displaystyle \text{Substituting in }x-2=3(y-2)^2,$
$\displaystyle 4y+9-2=3(y^2-4y+4)$
$\displaystyle 4y+7=3y^2-12y+12$
$\displaystyle 3y^2-16y+5=0$
$\displaystyle (3y-1)(y-5)=0$
$\displaystyle \Rightarrow y=\frac{1}{3}\text{ or }y=5$
$\displaystyle \text{Since son's age is taken in whole years, }y\neq \frac{1}{3}.$
$\displaystyle \therefore y=5$
$\displaystyle x=4y+9=4(5)+9=29$
$\displaystyle \therefore \text{The present ages of the man and his son are }29\text{ years and }5\text{ years.}$
$\\$

$\displaystyle \textbf{Question 13: } \text{In a certain positive fraction, the denominator is greater}$
$\displaystyle \text{than the numerator by }3.\text{ If }1\text{ is subtracted from the numerator}$
$\displaystyle \text{and the denominator both, the fraction reduces by }\frac{1}{14}.\text{ Find the fraction.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the fraction be }\frac{x}{x+3}.$
$\displaystyle \text{Given, }\frac{x}{x+3}-\frac{x-1}{x+2}=\frac{1}{14}$
$\displaystyle \frac{x(x+2)-(x-1)(x+3)}{(x+3)(x+2)}=\frac{1}{14}$
$\displaystyle \frac{x^2+2x-(x^2+2x-3)}{(x+3)(x+2)}=\frac{1}{14}$
$\displaystyle \frac{3}{(x+3)(x+2)}=\frac{1}{14}$
$\displaystyle (x+3)(x+2)=42$
$\displaystyle x^2+5x+6=42$
$\displaystyle x^2+5x-36=0$
$\displaystyle (x+9)(x-4)=0$
$\displaystyle \Rightarrow x=-9\text{ or }x=4$
$\displaystyle \text{Since the numerator is positive, }x\neq -9.$
$\displaystyle \therefore x=4$
$\displaystyle \therefore \text{The required fraction is }\frac{4}{7}.$
$\\$

$\displaystyle \textbf{Question 14: } \text{In a two-digit number, the ten's digit is bigger.}$
$\displaystyle \text{The product of the digits is }27\text{ and the difference between}$
$\displaystyle \text{the two digits is }6.\text{ Find the number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the unit's digit be }x.$
$\displaystyle \therefore \text{Ten's digit}=x+6$
$\displaystyle \text{Given, }x(x+6)=27$
$\displaystyle x^2+6x-27=0$
$\displaystyle (x+9)(x-3)=0$
$\displaystyle \Rightarrow x=-9\text{ or }x=3$
$\displaystyle \text{Since }x\text{ is a digit, }x\neq -9.$
$\displaystyle \therefore x=3$
$\displaystyle \text{Ten's digit}=3+6=9$
$\displaystyle \therefore \text{The required number is }93.$
$\\$

$\displaystyle \textbf{Question 15: } \text{Some school children went on an excursion by bus to a picnic spot}$
$\displaystyle \text{at a distance of }300\text{ km. While returning, it was raining and the bus had}$
$\displaystyle \text{to reduce its speed by }5\text{ km/hr and it took }2\text{ hours longer for returning.}$
$\displaystyle \text{Find the time taken to return.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the usual speed of the bus be }x\text{ km/hr.}$
$\displaystyle \text{Returning speed}=(x-5)\text{ km/hr}$
$\displaystyle \text{Given, }\frac{300}{x-5}-\frac{300}{x}=2$
$\displaystyle \frac{300x-300(x-5)}{x(x-5)}=2$
$\displaystyle \frac{1500}{x(x-5)}=2$
$\displaystyle 2x(x-5)=1500$
$\displaystyle 2x^2-10x-1500=0$
$\displaystyle x^2-5x-750=0$
$\displaystyle (x-30)(x+25)=0$
$\displaystyle \Rightarrow x=30\text{ or }x=-25$
$\displaystyle \text{Since speed cannot be negative, }x\neq -25.$
$\displaystyle \therefore x=30$
$\displaystyle \text{Returning speed}=30-5=25\text{ km/hr}$
$\displaystyle \text{Time taken to return}=\frac{300}{25}=12\text{ hours}$
$\displaystyle \therefore \text{The time taken to return is }12\text{ hours}.$
$\\$

$\displaystyle \textbf{Question 16: } \text{Rs. }480\text{ is divided equally among }x\text{ children.}$
$\displaystyle \text{If the number of children were }20\text{ more, then each would have got Rs. }12$
$\displaystyle \text{less. Find }x.\hfill \text{[ICSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Amount received by each child}=\text{Rs. }\frac{480}{x}$
$\displaystyle \text{If there were }20\text{ children more, amount received by each child}=\text{Rs. }\frac{480}{x+20}$
$\displaystyle \text{Given, }\frac{480}{x}-\frac{480}{x+20}=12$
$\displaystyle \frac{480(x+20)-480x}{x(x+20)}=12$
$\displaystyle \frac{9600}{x(x+20)}=12$
$\displaystyle 12x(x+20)=9600$
$\displaystyle x^2+20x-800=0$
$\displaystyle (x-20)(x+40)=0$
$\displaystyle \Rightarrow x=20\text{ or }x=-40$
$\displaystyle \text{Since the number of children cannot be negative, }x\neq -40.$
$\displaystyle \therefore x=20$
$\displaystyle \therefore \text{The value of }x\text{ is }20.$
$\\$