Class 10: Quadratic Equations – Miscellaneous Problems

$\displaystyle \textbf{Question 1: } \text{Find two natural numbers which differ by }3\text{ and the sum of whose squares is }117.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the natural numbers be }x\text{ and }x+3.$
$\displaystyle x^2+(x+3)^2=117$
$\displaystyle \Rightarrow x^2+x^2+6x+9=117$
$\displaystyle \Rightarrow 2x^2+6x-108=0$
$\displaystyle \Rightarrow x^2+3x-54=0$
$\displaystyle \Rightarrow (x+9)(x-6)=0$
$\displaystyle \Rightarrow x+9=0 \text{ or } x-6=0$
$\displaystyle \Rightarrow x=-9 \text{ or } x=6$
$\displaystyle \text{Since }-9\text{ is not a natural number,}$
$\displaystyle \text{one number }=6\text{ and the other number }=6+3=9$
$\displaystyle \therefore \text{The required numbers are }6\text{ and }9.$
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$\displaystyle \textbf{Question 2: } \text{Five times a certain whole number is equal to three less than twice the square of the number.}$
$\displaystyle \text{Find the number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number be }x.$
$\displaystyle \text{Given, }5x=2x^2-3$
$\displaystyle \Rightarrow 2x^2-5x-3=0$
$\displaystyle \Rightarrow (x-3)(2x+1)=0$
$\displaystyle \Rightarrow x-3=0 \text{ or } 2x+1=0$
$\displaystyle \Rightarrow x=3 \text{ or } x=-\frac{1}{2}$
$\displaystyle \text{Since the number is a whole number, }x=3.$
$\displaystyle \therefore \text{The required whole number is }3.$
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$\displaystyle \textbf{Question 3: } \text{Divide }8\text{ into two parts such that the sum of their reciprocals is }\frac{8}{15}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two parts be }x\text{ and }8-x.$
$\displaystyle \therefore \frac{1}{x}+\frac{1}{8-x}=\frac{8}{15}$
$\displaystyle \Rightarrow \frac{8-x+x}{x(8-x)}=\frac{8}{15}$
$\displaystyle \Rightarrow 120=8(8x-x^2)$
$\displaystyle \Rightarrow x^2-8x+15=0$
$\displaystyle \Rightarrow (x-5)(x-3)=0$
$\displaystyle \Rightarrow x=5\text{ or }x=3$
$\displaystyle \therefore \text{The required parts are }3\text{ and }5.$
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$\displaystyle \textbf{Question 4: } \text{For the same amount of work, A takes }6\text{ hours less than B.}$
$\displaystyle \text{If together they complete the work in }13\text{ hours }20\text{ minutes, find how much time B alone takes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let B alone take }x\text{ hours.}$
$\displaystyle \therefore \text{A alone takes }(x-6)\text{ hours.}$
$\displaystyle 13\text{ hours }20\text{ minutes}=13\frac{1}{3}\text{ hours}=\frac{40}{3}\text{ hours}$
$\displaystyle \therefore \frac{1}{x-6}+\frac{1}{x}=\frac{3}{40}$
$\displaystyle \Rightarrow \frac{x+x-6}{x(x-6)}=\frac{3}{40}$
$\displaystyle \Rightarrow 40(2x-6)=3x(x-6)$
$\displaystyle \Rightarrow 80x-240=3x^2-18x$
$\displaystyle \Rightarrow 3x^2-98x+240=0$
$\displaystyle \Rightarrow 3x^2-90x-8x+240=0$
$\displaystyle \Rightarrow 3x(x-30)-8(x-30)=0$
$\displaystyle \Rightarrow (x-30)(3x-8)=0$
$\displaystyle \Rightarrow x=30\text{ or }x=\frac{8}{3}$
$\displaystyle \text{Since }x>6,\ x=30.$
$\displaystyle \therefore \text{B alone will take }30\text{ hours to complete the work.}$
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$\displaystyle \textbf{Question 5: } \text{The hypotenuse of a right triangle is }13\text{ cm and the difference between the other two sides is }7\text{ cm.}$
$\displaystyle \text{Taking }x\text{ as the shorter side, write an equation and solve it to find the two unknown sides.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the shorter side be }x\text{ cm.}$
$\displaystyle \therefore \text{The longer side }=(x+7)\text{ cm.}$
$\displaystyle \text{Using Pythagoras theorem,}$
$\displaystyle x^2+(x+7)^2=13^2$
$\displaystyle \Rightarrow x^2+x^2+14x+49=169$
$\displaystyle \Rightarrow 2x^2+14x-120=0$
$\displaystyle \Rightarrow x^2+7x-60=0$
$\displaystyle \Rightarrow (x+12)(x-5)=0$
$\displaystyle \Rightarrow x=-12\text{ or }x=5$
$\displaystyle \text{Since the side of a triangle cannot be negative, }x=5.$
$\displaystyle \therefore \text{The two unknown sides are }5\text{ cm and }12\text{ cm.}$
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$\displaystyle \textbf{Question 6: } \text{The length of a verandah is }3\text{ m more than its breadth.}$
$\displaystyle \text{The numerical value of its area is equal to the numerical value of its perimeter.}$
$\displaystyle \text{(i) Taking }x\text{ as the breadth, write an equation in }x.$
$\displaystyle \text{(ii) Solve the equation and find the dimensions of the verandah.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the breadth be }x\text{ m.}$
$\displaystyle \therefore \text{Length }=(x+3)\text{ m.}$
$\displaystyle \text{(i) Given, numerical value of area }=\text{ numerical value of perimeter}$
$\displaystyle x(x+3)=2\{x+(x+3)\}$
$\displaystyle \Rightarrow x^2+3x=2(2x+3)$
$\displaystyle \Rightarrow x^2+3x=4x+6$
$\displaystyle \Rightarrow x^2-x-6=0$
$\displaystyle \text{(ii) }x^2-x-6=0$
$\displaystyle \Rightarrow (x-3)(x+2)=0$
$\displaystyle \Rightarrow x=3\text{ or }x=-2$
$\displaystyle \text{Since breadth cannot be negative, }x=3.$
$\displaystyle \therefore \text{Breadth }=3\text{ m and length }=6\text{ m.}$
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$\displaystyle \textbf{Question 7: } \text{By increasing the speed of a car by }10\text{ km/hr, the time of journey for }72\text{ km is reduced by }36\text{ minutes.}$
$\displaystyle \text{Find the original speed of the car.}\hfill\text{[ICSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original speed of the car be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken to cover }72\text{ km }=\frac{72}{x}\text{ hours}$
$\displaystyle \text{New speed of the car }=(x+10)\text{ km/hr}$
$\displaystyle \therefore \text{New time taken to cover }72\text{ km }=\frac{72}{x+10}\text{ hours}$
$\displaystyle 36\text{ minutes }=\frac{36}{60}\text{ hours}=\frac{3}{5}\text{ hours}$
$\displaystyle \therefore \frac{72}{x}-\frac{72}{x+10}=\frac{3}{5}$
$\displaystyle \Rightarrow \frac{72(x+10)-72x}{x(x+10)}=\frac{3}{5}$
$\displaystyle \Rightarrow 5(720)=3x(x+10)$
$\displaystyle \Rightarrow 3600=3x^2+30x$
$\displaystyle \Rightarrow x^2+10x-1200=0$
$\displaystyle \Rightarrow (x-30)(x+40)=0$
$\displaystyle \Rightarrow x=30\text{ or }x=-40$
$\displaystyle \text{Since speed cannot be negative, }x=30.$
$\displaystyle \therefore \text{The original speed of the car is }30\text{ km/hr.}$
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$\displaystyle \textbf{Question 8: } \text{Car A travels }x\text{ km per litre of petrol, while car B travels }(x+5)\text{ km per litre.}$
$\displaystyle \text{(i) Write the litres of petrol used by A and B for }400\text{ km.}$
$\displaystyle \text{(ii) If A uses }4\text{ litres more than B, form an equation and find petrol used by B.}\hfill\text{[ICSE 1997]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Petrol used by car A }=\frac{400}{x}\text{ litres}$
$\displaystyle \text{Petrol used by car B }=\frac{400}{x+5}\text{ litres}$
$\displaystyle \text{(ii) Given, }\frac{400}{x}-\frac{400}{x+5}=4$
$\displaystyle \Rightarrow \frac{400(x+5)-400x}{x(x+5)}=4$
$\displaystyle \Rightarrow \frac{2000}{x(x+5)}=4$
$\displaystyle \Rightarrow 4x(x+5)=2000$
$\displaystyle \Rightarrow x^2+5x-500=0$
$\displaystyle \Rightarrow (x+25)(x-20)=0$
$\displaystyle \Rightarrow x=-25\text{ or }x=20$
$\displaystyle \text{Since distance per litre cannot be negative, }x=20.$
$\displaystyle \therefore \text{Petrol used by car B }=\frac{400}{20+5}=16\text{ litres.}$
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$\displaystyle \textbf{Question 9: } \text{By selling an article for Rs. }24,\text{ a trader loses as much per cent as the cost price.}$
$\displaystyle \text{Calculate the cost price of the article.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the cost price of the article be Rs. }x.$
$\displaystyle \therefore \text{Loss }=x\%\text{ of Rs. }x$
$\displaystyle =\frac{x}{100}\times x=\frac{x^2}{100}$
$\displaystyle \text{Selling price }=\text{Cost price }-\text{Loss}$
$\displaystyle \Rightarrow x-\frac{x^2}{100}=24$
$\displaystyle \Rightarrow 100x-x^2=2400$
$\displaystyle \Rightarrow x^2-100x+2400=0$
$\displaystyle \Rightarrow (x-60)(x-40)=0$
$\displaystyle \Rightarrow x=60\text{ or }x=40$
$\displaystyle \therefore \text{The cost price of the article is Rs. }40\text{ or Rs. }60.$
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$\displaystyle \textbf{Question 10: } \text{The sum }S\text{ of first }n\text{ natural numbers is }S=\frac{1}{2}n(n+1).$
$\displaystyle \text{Find }n\text{ if the sum is }276.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }S=276$
$\displaystyle \Rightarrow \frac{1}{2}n(n+1)=276$
$\displaystyle \Rightarrow n(n+1)=552$
$\displaystyle \Rightarrow n^2+n-552=0$
$\displaystyle \Rightarrow (n+24)(n-23)=0$
$\displaystyle \Rightarrow n=-24\text{ or }n=23$
$\displaystyle \text{Since }n\text{ is a natural number, }n\neq-24.$
$\displaystyle \therefore n=23$
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$\displaystyle \textbf{Question 11: } \text{A two digit number is such that the product of its digits is }6.$
$\displaystyle \text{When }9\text{ is added to this number, the digits interchange their places. Find the number.}\hfill\text{[ICSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required two digit number be }10x+y.$
$\displaystyle \text{Given, }xy=6\text{ and }10x+y+9=10y+x$
$\displaystyle \Rightarrow 10x+y+9=10y+x$
$\displaystyle \Rightarrow 9x+9=9y$
$\displaystyle \Rightarrow y=x+1$
$\displaystyle \text{Now, }xy=6$
$\displaystyle \Rightarrow x(x+1)=6$
$\displaystyle \Rightarrow x^2+x-6=0$
$\displaystyle \Rightarrow (x+3)(x-2)=0$
$\displaystyle \Rightarrow x=-3\text{ or }x=2$
$\displaystyle \text{Since }-3\text{ is not a digit, }x=2.$
$\displaystyle \Rightarrow y=x+1=3$
$\displaystyle \therefore \text{The required number is }10x+y=10(2)+3=23.$
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$\displaystyle \textbf{Question 12: } \text{Five years ago, a woman's age was the square of her son's age.}$
$\displaystyle \text{Ten years hence, her age will be twice that of her son's age. Find:}\hfill\text{[ICSE 2007]}$
$\displaystyle \text{(i) the age of the son five years ago.}$
$\displaystyle \text{(ii) the present age of the woman.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the age of the son }5\text{ years ago be }x\text{ years.}$
$\displaystyle \therefore \text{Woman's age }5\text{ years ago }=x^2\text{ years}$
$\displaystyle \text{Present age of woman }=(x^2+5)\text{ years}$
$\displaystyle \text{Present age of son }=(x+5)\text{ years}$
$\displaystyle \text{Woman's age }10\text{ years hence }=x^2+15\text{ years}$
$\displaystyle \text{Son's age }10\text{ years hence }=x+15\text{ years}$
$\displaystyle \text{According to the given condition,}$
$\displaystyle x^2+15=2(x+15)$
$\displaystyle \Rightarrow x^2+15=2x+30$
$\displaystyle \Rightarrow x^2-2x-15=0$
$\displaystyle \Rightarrow (x+3)(x-5)=0$
$\displaystyle \Rightarrow x=-3\text{ or }x=5$
$\displaystyle \text{Since age cannot be negative, }x=5.$
$\displaystyle \therefore \text{(i) The age of the son }5\text{ years ago was }5\text{ years.}$
$\displaystyle \text{(ii) Present age of the woman }=x^2+5=25+5=30\text{ years.}$
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$\displaystyle \textbf{Question 13: } \text{A motor-boat, whose speed is }9\text{ km/h in still water, goes }12\text{ km downstream}$
$\displaystyle \text{and comes back in a total time of }3\text{ hours. Find the speed of the stream.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speed of the stream be }x\text{ km/hr.}$
$\displaystyle \text{Speed downstream }=(9+x)\text{ km/hr}$
$\displaystyle \text{Speed upstream }=(9-x)\text{ km/hr}$
$\displaystyle \text{Time taken downstream }=\frac{12}{9+x}\text{ hours}$
$\displaystyle \text{Time taken upstream }=\frac{12}{9-x}\text{ hours}$
$\displaystyle \therefore \frac{12}{9+x}+\frac{12}{9-x}=3$
$\displaystyle \Rightarrow \frac{12(9-x)+12(9+x)}{(9+x)(9-x)}=3$
$\displaystyle \Rightarrow \frac{216}{81-x^2}=3$
$\displaystyle \Rightarrow 216=243-3x^2$
$\displaystyle \Rightarrow x^2=9$
$\displaystyle \Rightarrow x=\pm3$
$\displaystyle \text{Since speed cannot be negative, }x=3.$
$\displaystyle \therefore \text{The speed of the stream is }3\text{ km/hr.}$
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$\displaystyle \textbf{Question 14: } \text{A piece of cloth costs Rs. }200.\text{ If it was }5\text{ m longer and each metre cost Rs. }2\text{ less,}$
$\displaystyle \text{the cost would have remained unchanged. Find its length and original rate per metre.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the length of the piece be }x\text{ m.}$
$\displaystyle \text{Original rate per metre }=\text{Rs. }\frac{200}{x}$
$\displaystyle \text{New length }=(x+5)\text{ m}$
$\displaystyle \text{New rate per metre }=\text{Rs. }\frac{200}{x+5}$
$\displaystyle \therefore \frac{200}{x}-\frac{200}{x+5}=2$
$\displaystyle \Rightarrow \frac{200(x+5)-200x}{x(x+5)}=2$
$\displaystyle \Rightarrow \frac{1000}{x(x+5)}=2$
$\displaystyle \Rightarrow 2x(x+5)=1000$
$\displaystyle \Rightarrow x^2+5x-500=0$
$\displaystyle \Rightarrow (x+25)(x-20)=0$
$\displaystyle \Rightarrow x=-25\text{ or }x=20$
$\displaystyle \text{Since length cannot be negative, }x=20.$
$\displaystyle \therefore \text{Length of the piece }=20\text{ m}$
$\displaystyle \text{Original rate per metre }=\frac{200}{20}=\text{Rs. }10$
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$\displaystyle \textbf{Question 15: } \text{A shopkeeper buys a certain number of books for Rs. }960.$
$\displaystyle \text{If each book cost Rs. }8\text{ less, }4\text{ more books could be bought. Find }x.\hfill\text{[ICSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original cost of each book be Rs. }x.$
$\displaystyle \text{Number of books bought for Rs. }960=\frac{960}{x}$
$\displaystyle \text{If the cost of each book is Rs. }(x-8),$
$\displaystyle \text{number of books bought for Rs. }960=\frac{960}{x-8}$
$\displaystyle \therefore \frac{960}{x-8}-\frac{960}{x}=4$
$\displaystyle \Rightarrow \frac{960x-960(x-8)}{x(x-8)}=4$
$\displaystyle \Rightarrow \frac{7680}{x(x-8)}=4$
$\displaystyle \Rightarrow 7680=4(x^2-8x)$
$\displaystyle \Rightarrow x^2-8x-1920=0$
$\displaystyle \Rightarrow (x-48)(x+40)=0$
$\displaystyle \Rightarrow x=48\text{ or }x=-40$
$\displaystyle \text{Since cost cannot be negative, }x=48.$
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$\displaystyle \textbf{Question 16: } \text{Some students planned a picnic. The budget for food was Rs. }480.$
$\displaystyle \text{As }8\text{ failed to join, the cost for each member increased by Rs. }10.\text{ Find how many went.}\hfill\text{[ICSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of students who planned the picnic be }x.$
$\displaystyle \text{Original share of each student }=\text{Rs. }\frac{480}{x}$
$\displaystyle \text{Number of students who went for the picnic }=x-8$
$\displaystyle \text{New share of each student }=\text{Rs. }\frac{480}{x-8}$
$\displaystyle \therefore \frac{480}{x-8}-\frac{480}{x}=10$
$\displaystyle \Rightarrow \frac{480x-480(x-8)}{x(x-8)}=10$
$\displaystyle \Rightarrow \frac{3840}{x(x-8)}=10$
$\displaystyle \Rightarrow 3840=10(x^2-8x)$
$\displaystyle \Rightarrow x^2-8x-384=0$
$\displaystyle \Rightarrow (x-24)(x+16)=0$
$\displaystyle \Rightarrow x=24\text{ or }x=-16$
$\displaystyle \text{Since the number of students cannot be negative, }x=24.$
$\displaystyle \therefore \text{Number of students who went for the picnic }=24-8=16.$
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