Class 10: Quadratic Equations

$\displaystyle \textbf{Question 1: }\text{The product of two consecutive positive integers is }306.\text{ Form the}$
$\displaystyle \text{quadratic equation to find the integers, if }x\text{ denotes the smaller integer.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller positive integer be }x.$
$\displaystyle \therefore \text{The next consecutive positive integer is }x+1.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle x(x+1)=306$
$\displaystyle \Rightarrow x^2+x-306=0$
$\displaystyle \therefore \text{The required quadratic equation is }x^2+x-306=0.$
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$\displaystyle \textbf{Question 2: }\text{John and Jivanti together have }45\text{ marbles. Both of them lost }5$
$\displaystyle \text{marbles each, and the product of the number of marbles they now have is }124.$
$\displaystyle \text{Form the quadratic equation to find how many marbles they had to start with, if John}$
$\displaystyle \text{had }x\text{ marbles.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{John initially had }x\text{ marbles.}$
$\displaystyle \therefore \text{Jivanti initially had }(45-x)\text{ marbles.}$
$\displaystyle \text{After losing }5\text{ marbles, John has }(x-5)\text{ marbles.}$
$\displaystyle \text{Jivanti has }(45-x-5)=(40-x)\text{ marbles.}$
$\displaystyle \text{According to the given condition,}$
$\displaystyle (x-5)(40-x)=124$
$\displaystyle \Rightarrow 40x-x^2-200+5x=124$
$\displaystyle \Rightarrow x^2-45x+324=0$
$\displaystyle \therefore \text{The required quadratic equation is }x^2-45x+324=0.$
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$\displaystyle \textbf{Question 3: }\text{The height of a right triangle is }7\text{ cm less than its base. If the}$
$\displaystyle \text{hypotenuse is }13\text{ cm, form the quadratic equation to find the base of the triangle.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the base of the right triangle be }x\text{ cm.}$
$\displaystyle \therefore \text{Height of the right triangle }=(x-7)\text{ cm.}$
$\displaystyle \text{By Pythagoras' theorem,}$
$\displaystyle x^2+(x-7)^2=13^2$
$\displaystyle \Rightarrow x^2+x^2-14x+49=169$
$\displaystyle \Rightarrow 2x^2-14x-120=0$
$\displaystyle \Rightarrow x^2-7x-60=0$
$\displaystyle \therefore \text{The required quadratic equation is }x^2-7x-60=0.$
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$\displaystyle \textbf{Question 4: }\text{A cottage industry produces a certain number of toys in a day. The cost of}$
$\displaystyle \text{production of each toy (in rupees) was found to be }55\text{ minus the number of toys}$
$\displaystyle \text{produced in a day. On a particular day, the total cost of production was Rs. }750.$
$\displaystyle \text{If }x\text{ denotes the number of toys produced that day, form the quadratic equation to find }x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Number of toys produced in a day}=x.$
$\displaystyle \therefore \text{Cost of production of each toy}=\text{Rs. }(55-x).$
$\displaystyle \text{Total cost of production}=\text{Rs. }750.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle x(55-x)=750$
$\displaystyle \Rightarrow 55x-x^2=750$
$\displaystyle \Rightarrow x^2-55x+750=0$
$\displaystyle \therefore \text{The required quadratic equation is }x^2-55x+750=0.$
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$\displaystyle \textbf{Question 5: }\text{An express train takes }1\text{ hour less than a passenger train to travel}$
$\displaystyle 132\text{ km between Mysore and Bangalore. If the average speed of the express train is}$
$\displaystyle 11\text{ km/hr more than that of the passenger train, form the quadratic equation to find the}$
$\displaystyle \text{average speed of the express train.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the average speed of the express train be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Average speed of the passenger train}=(x-11)\text{ km/hr.}$
$\displaystyle \text{Time taken by the express train}=\frac{132}{x}\text{ hours.}$
$\displaystyle \text{Time taken by the passenger train}=\frac{132}{x-11}\text{ hours.}$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{132}{x-11}-\frac{132}{x}=1$
$\displaystyle \Rightarrow \frac{132x-132(x-11)}{x(x-11)}=1$
$\displaystyle \Rightarrow \frac{1452}{x(x-11)}=1$
$\displaystyle \Rightarrow x(x-11)=1452$
$\displaystyle \Rightarrow x^2-11x-1452=0$
$\displaystyle \therefore \text{The required quadratic equation is }x^2-11x-1452=0.$
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$\displaystyle \textbf{Question 6: }\text{A train travels }360\text{ km at a uniform speed. If the speed had been}$
$\displaystyle 5\text{ km/hr more, it would have taken }1\text{ hour less for the same journey. Form the}$
$\displaystyle \text{quadratic equation to find the speed of the train.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the uniform speed of the train be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Increased speed of the train}=(x+5)\text{ km/hr.}$
$\displaystyle \text{Time taken at the original speed}=\frac{360}{x}\text{ hours.}$
$\displaystyle \text{Time taken at the increased speed}=\frac{360}{x+5}\text{ hours.}$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{360}{x}-\frac{360}{x+5}=1$
$\displaystyle \Rightarrow \frac{360(x+5)-360x}{x(x+5)}=1$
$\displaystyle \Rightarrow \frac{1800}{x(x+5)}=1$
$\displaystyle \Rightarrow x(x+5)=1800$
$\displaystyle \Rightarrow x^2+5x-1800=0$
$\displaystyle \therefore \text{The required quadratic equation is }x^2+5x-1800=0.$
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