NCERT Solutions Class 10 Maths Chapter 4

Q.1

$\begin{array}{l}\text{Check whether the following are quadratic equations :}\\ \left(\text{i}\right){(\mathrm{x}+1)}^2=2(\mathrm{x}-3)\left(\text{ii}\right){\mathrm{x}}^2-2\mathrm{x}=-2(3-\mathrm{x})\\ \left(\mathrm{iii}\right)(\mathrm{x}-2)(\mathrm{x}+1)=(\mathrm{x}-1)(\mathrm{x}+3)\left(\text{iv}\right)(\mathrm{x}-3)(2\mathrm{x}+1)=\mathrm{x}(\mathrm{x}+5)\\ \left(\mathrm{v}\right)(2\mathrm{x}-1)(\mathrm{x}-3)=(\mathrm{x}+5)(\mathrm{x}-1)\left(\text{vi}\right){\text{x}}^2+3\mathrm{x}+1={(\mathrm{x}-2)}^2\\ \left(\mathrm{vii}\right){(\mathrm{x}+2)}^3=2\mathrm{x}({\mathrm{x}}^2-1)\left(\text{viii}\right){\text{x}}^3-4{\mathrm{x}}^2-\mathrm{x}+1={(\mathrm{x}-2)}^3\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ {(\mathrm{x}+1)}^2=2(\mathrm{x}-3)\\ \Rightarrow {\mathrm{x}}^2+2\mathrm{x}+1=2\mathrm{x}-6\\ \Rightarrow {\mathrm{x}}^2+2\mathrm{x}+1-2\mathrm{x}+6=0\\ \Rightarrow {\mathrm{x}}^2+7=0\\ \text{It is of the form a}{\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}=0.\\ \text{Therefore, the given equation is a quadratic equation.}\\ \text{(ii)}\\ {\mathrm{x}}^2-2\mathrm{x}=(-2)(3-\mathrm{x})\\ \Rightarrow {\mathrm{x}}^2-2\mathrm{x}=2\mathrm{x}-6\\ \Rightarrow {\mathrm{x}}^2-2\mathrm{x}-2\mathrm{x}+6=0\\ \Rightarrow {\mathrm{x}}^2-4\mathrm{x}+6=0\\ \text{It is of the form a}{\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}=0.\\ \text{Therefore, the given equation is a quadratic equation.}\\ \text{(iii)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ (\mathrm{x}-2)(\mathrm{x}+1)=(\mathrm{x}-1)(\mathrm{x}+3)\\ \Rightarrow {\mathrm{x}}^2-\mathrm{x}-2={\mathrm{x}}^2+2\mathrm{x}-3\\ \Rightarrow {\mathrm{x}}^2-\mathrm{x}-2-{\mathrm{x}}^2-2\mathrm{x}+3=0\\ \Rightarrow -3\mathrm{x}+1=0\\ \text{It is not of the form}{\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0.\\ \text{Therefore, the given equation is not a quadratic equation.}\\ \text{(iv)}\\ (\mathrm{x}-3)(2\mathrm{x}+1)=\mathrm{x}(\mathrm{x}+5)\\ \Rightarrow \text{2}{\mathrm{x}}^2-5\mathrm{x}-3={\mathrm{x}}^2+5\mathrm{x}\\ \Rightarrow \text{2}{\mathrm{x}}^2-5\mathrm{x}-3-{\mathrm{x}}^2-5\mathrm{x}=0\\ \Rightarrow {\mathrm{x}}^2-10\mathrm{x}-3=0\\ \text{It is of the form}{\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0.\\ \text{Therefore, the given equation is a quadratic equation.}\\ \text{(v)}\\ (2\mathrm{x}-1)(\mathrm{x}-3)=(\mathrm{x}+5)(\mathrm{x}-1)\\ \Rightarrow \text{2}{\mathrm{x}}^2-7\mathrm{x}+3={\mathrm{x}}^2+4\mathrm{x}-5\\ \Rightarrow \text{2}{\mathrm{x}}^2-7\mathrm{x}+3-{\mathrm{x}}^2-4\mathrm{x}+5=0\\ \Rightarrow {\mathrm{x}}^2-11\mathrm{x}+8=0\\ \text{It is of the form}{\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0.\\ \text{Therefore, the given equation is a quadratic equation.}\\ \text{(vi)}\\ {\mathrm{x}}^2+3\mathrm{x}+1={(\mathrm{x}-2)}^2\\ \Rightarrow {\mathrm{x}}^2+3\mathrm{x}+1={\mathrm{x}}^2-4\mathrm{x}+4\\ \Rightarrow {\mathrm{x}}^2+3\mathrm{x}+1-{\mathrm{x}}^2+4\mathrm{x}-4=0\\ \Rightarrow 7\mathrm{x}-3=0\\ \text{It is not of the form}{\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0.\\ \text{Therefore, the given equation is not a quadratic equation.}\\ \text{(vii)}\\ {(\mathrm{x}+2)}^3=2\mathrm{x}({\mathrm{x}}^2-1)\\ \Rightarrow {\mathrm{x}}^3+6{\mathrm{x}}^2+12\mathrm{x}+8=2{\mathrm{x}}^3-2\mathrm{x}\\ \Rightarrow {\mathrm{x}}^3+6{\mathrm{x}}^2+12\mathrm{x}+8-2{\mathrm{x}}^3+2\mathrm{x}=0\\ \Rightarrow -{\mathrm{x}}^3+6{\mathrm{x}}^2+14\mathrm{x}+8=0\\ \text{It is not of the form}{\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0.\\ \text{Therefore, the given equation is not a quadratic equation.}\\ \text{(viii)}\\ {\mathrm{x}}^3-4{\mathrm{x}}^2-\mathrm{x}+1={(\mathrm{x}-2)}^3\\ \Rightarrow {\mathrm{x}}^3-4{\mathrm{x}}^2-\mathrm{x}+1={\mathrm{x}}^3-6{\mathrm{x}}^2+12\mathrm{x}-8\\ \Rightarrow {\mathrm{x}}^3-4{\mathrm{x}}^2-\mathrm{x}+1-{\mathrm{x}}^3+6{\mathrm{x}}^2-12\mathrm{x}+8=0\\ \Rightarrow 2{\mathrm{x}}^2-13\mathrm{x}+9=0\\ \text{It is of the form}{\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0.\\ \text{Therefore, the given equation is a quadratic equation.}\end{array}$

Q.2 Represent the following situations in the form of quadratic equations:

1. The area of a rectangular plot is 528 m­­­². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

2. The product of two consecutive positive integers is 306. We need to find the integers.

3. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

4. A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Ans.

$\begin{array}{l}\text{(i) Let the breadth of the plot be}\mathrm{x}\mathrm{m}\text{. Then, as per given}\\ \text{information, the length of the plot is}(2\mathrm{x}+1)\mathrm{m}\text{.}\\ \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Also, given that area of the rectangular plot is}528{\mathrm{m}}^2.\\ \text{Therefore,}\\ (2\mathrm{x}+1)\mathrm{x}=528[\text{Area of a rectangle}=\text{Length}\times \text{Breadth}]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2{\mathrm{x}}^2+\mathrm{x}-528=0.\\ \text{(ii) Let two consecutive positive integers are}\mathrm{x}\text{and}\mathrm{x}+1.\\ \text{Then we have,}\\ \mathrm{x}(\mathrm{x}+1)=306\\ \Rightarrow {\mathrm{x}}^2+\mathrm{x}-306=0\\ \text{(iii) Let Rohan’s present age is}\mathrm{x}\text{years. Then Rohan’s mother}\\ \text{is}\mathrm{x}+26\text{years old.}\\ \text{According to question,}\\ (\mathrm{x}+3)(\mathrm{x}+26+3)=360\\ \Rightarrow (\mathrm{x}+3)(\mathrm{x}+29)=360\\ \Rightarrow {\mathrm{x}}^2+32\mathrm{x}+87-360=0\\ \Rightarrow {\mathrm{x}}^2+32\mathrm{x}-273=0\\ \text{(iv) Let uniform speed of the train is}\mathrm{x}\text{km/h. Then the time}\\ \text{\hspace{0.17em} taken to travel the distance of 480 km is}\frac{480}{\mathrm{x}}\text{hours.}\\ \text{Again, when the speed is (}\mathrm{x}-8)\text{km/h then the time taken}\\ \text{to travel the distance of 480 km is}\frac{480}{\mathrm{x}-8}\text{hours.}\\ \text{According to question,}\\ \frac{480}{\mathrm{x}-8}-\frac{480}{\mathrm{x}}=3\\ \Rightarrow 480(\mathrm{x}-\mathrm{x}+8)=3\mathrm{x}(\mathrm{x}-8)\\ \Rightarrow \frac{480\times 8}{3}={\mathrm{x}}^2-8\mathrm{x}\\ \Rightarrow {\mathrm{x}}^2-8\mathrm{x}-1280=0\end{array}$

Q.3

$\begin{array}{l}\text{Find the roots of the following quadratic equations}\\ \text{by factorisation:}\\ \text{(i)}{\mathrm{x}}^2-3\mathrm{x}-10=0\text{(ii)}2{\mathrm{x}}^2+\mathrm{x}-6=0\\ \text{(iii)}\sqrt{2}{\mathrm{x}}^2+7\mathrm{x}+5\sqrt{2}=0{\text{(iv) 2x}}^2-\mathrm{x}+\frac{1}{8}=0\\ {\text{(v) 100x}}^2-20\mathrm{x}+1=0\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ {\mathrm{x}}^2-3\mathrm{x}-10=0\\ \Rightarrow {\mathrm{x}}^2-5\mathrm{x}+2\mathrm{x}-10=0\\ \Rightarrow \text{\hspace{0.17em} x(}\mathrm{x}-5)+2(\mathrm{x}-5)=0\\ \Rightarrow \text{\hspace{0.17em} (}\mathrm{x}-5\left)\right(\mathrm{x}+2)=0\\ \Rightarrow \mathrm{x}-5=0\text{or\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+2=0\\ \Rightarrow \mathrm{x}=5\text{or\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=-2\\ \text{Therefore, roots of the equation}{\mathrm{x}}^2-3\mathrm{x}-10=0\text{are 5 and –2.}\\ \text{(ii)}\\ 2{\mathrm{x}}^2+\mathrm{x}-6=0\\ \Rightarrow \text{\hspace{0.17em} 2}{\mathrm{x}}^2+4\mathrm{x}-3\mathrm{x}-6=0\\ \Rightarrow \text{\hspace{0.17em} 2}\mathrm{x}\text{(}\mathrm{x}+2)-3(\mathrm{x}+2)=0\\ \Rightarrow \text{\hspace{0.17em} (2}\mathrm{x}-3\left)\right(\mathrm{x}+2)=0\\ \Rightarrow \text{\hspace{0.17em} 2}\mathrm{x}-3=0\text{or\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+2=0\\ \Rightarrow \mathrm{x}=\frac{3}{2}\text{or\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=-2\\ \text{Therefore, roots of the equation}2{\mathrm{x}}^2+\mathrm{x}-6=0\text{are}\frac{3}{2}\text{and –2.}\\ \text{(iii)}\\ \sqrt{2}{\mathrm{x}}^2+7\mathrm{x}+5\sqrt{2}=0\\ \Rightarrow \sqrt{2}{\mathrm{x}}^2+2\mathrm{x}+5\mathrm{x}+5\sqrt{2}=0\\ \Rightarrow \sqrt{2}\mathrm{x}\text{(}\mathrm{x}+\sqrt{2})+5(\mathrm{x}+\sqrt{2})=0\\ \Rightarrow \text{\hspace{0.17em} (}\sqrt{2}\mathrm{x}+5\left)\right(\mathrm{x}+\sqrt{2})=0\\ \Rightarrow \sqrt{2}\mathrm{x}+5=0\text{or\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+\sqrt{2}=0\\ \Rightarrow \mathrm{x}=-\frac{5}{\sqrt{2}}\text{or\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=-\sqrt{2}\\ \text{Therefore, roots of the equation}\sqrt{2}{\mathrm{x}}^2+7\mathrm{x}+5\sqrt{2}=0\\ \text{are}-\frac{5}{\sqrt{2}}\text{and –}\sqrt{2}\text{.}\\ \text{(iv)}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2x}}^2-\mathrm{x}+\frac{1}{8}=0\\ \Rightarrow {\text{\hspace{0.17em} 16x}}^2-8\mathrm{x}+1=0\\ \Rightarrow {\text{\hspace{0.17em} 16x}}^2-4\mathrm{x}-4\mathrm{x}+1=0\\ \Rightarrow \text{\hspace{0.17em} 4x(4x}-1)-1(4\mathrm{x}-1)=0\\ \Rightarrow \text{\hspace{0.17em} (4x}-1\left)\right(4\mathrm{x}-1)=0\\ \Rightarrow \text{\hspace{0.17em} 4x}-1=0\text{or\hspace{0.17em}\hspace{0.17em}}4\mathrm{x}-1=0\\ \Rightarrow \mathrm{x}=\frac{1}{4}\text{or\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{1}{4}\\ {\text{Therefore, roots of the equation 2x}}^2-\mathrm{x}+\frac{1}{8}=0\text{are}\frac{1}{4}\text{and}\frac{1}{4}\text{.}\\ \text{(v)}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}100x}}^2-20\mathrm{x}+1=0\\ \Rightarrow {\text{\hspace{0.17em} 100x}}^2-10\mathrm{x}-10\mathrm{x}+1=0\\ \Rightarrow \text{\hspace{0.17em} 10x(10x}-1)-1(10\mathrm{x}-1)=0\\ \Rightarrow \text{\hspace{0.17em} (10x}-1)\text{(10x}-1)=0\\ \Rightarrow \text{\hspace{0.17em} 10x}-1=0\text{or\hspace{0.17em}\hspace{0.17em}}10\mathrm{x}-1=0\\ \Rightarrow \mathrm{x}=\frac{1}{10}\text{or\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{1}{10}\\ {\text{Therefore, roots of the equation 100x}}^2-20\mathrm{x}+\; 1=0\text{are}\frac{1}{10}\text{and}\frac{1}{10}\text{.}\end{array}$

Q.4 Solve the problems given in Example 1.
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day.

Ans.

$\begin{array}{l}\text{(i)}\\ \text{John and Jivanti together have 45 marbles.}\\ \text{So, let the number of marbles John had be x.}\\ \text{Then, number of marbles Jivanti had}=45-\mathrm{x}\\ \text{Both of them lost 5 marbles each, then}\\ \text{number of marbles left with John}=\mathrm{x}-5\\ \text{and}\\ \text{number of marbles left with Jivanti}=45-\mathrm{x}-5=40-\mathrm{x}\\ \text{The product of the number of marbles they now have is 124}.\\ \text{Therefore,}\\ (\mathrm{x}-5)(40-\mathrm{x})=\text{124}\\ \Rightarrow 40\mathrm{x}-{\mathrm{x}}^2-200+5\mathrm{x}-124=0\\ \Rightarrow {\mathrm{x}}^2-45\mathrm{x}+324=0\\ \Rightarrow {\mathrm{x}}^2-36\mathrm{x}-9\mathrm{x}+324=0\\ \Rightarrow \mathrm{x}(\mathrm{x}-36)-9(\mathrm{x}-36)=0\\ \Rightarrow (\mathrm{x}-9)(\mathrm{x}-36)=0\\ \Rightarrow \mathrm{x}-9=0\text{or\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-36=0\\ \Rightarrow \mathrm{x}=9\text{or\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=36\\ \text{So, if John had 9 marbles then Jivanti had 36 marbles}\\ \text{and if John had 36 marbles then Jivanti had 9 marbles.}\\ \text{(ii)}\\ \text{Let the number of toys produced be x.}\\ \text{Given that cost of production of each toy is}\\ \text{55 minus the number of toys produced in a day.}\\ \therefore \text{Cost of production of each toy =}₹\text{(55}-\text{x)}\\ \text{Given that total cost of production of the toys in a day}=₹\text{750}\\ \text{Therefore,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(55}-\text{x)}\mathrm{x}=\text{750}\\ \Rightarrow -{\mathrm{x}}^2+55\mathrm{x}-750=0\\ \Rightarrow {\mathrm{x}}^2-55\mathrm{x}+750=0\\ \Rightarrow {\mathrm{x}}^2-25\mathrm{x}-30\mathrm{x}+750=0\\ \Rightarrow \mathrm{x}(\mathrm{x}-25)-30(\mathrm{x}-25)=0\\ \Rightarrow (\mathrm{x}-30)(\mathrm{x}-25)=0\\ \Rightarrow \mathrm{x}-30=0\text{or\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-25=0\\ \Rightarrow \mathrm{x}=30\text{or\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=25\\ \text{So, number of toys is either 25 or 30.}\end{array}$

Q.5 Find two numbers whose sum is 27 and product is 182.

Ans.

$\begin{array}{l}\text{Let the numbers are x and y, then we have}\\ \mathrm{x}+\mathrm{y}=27...\left(1\right)\\ \text{and}\\ \mathrm{xy}=182\\ \Rightarrow \mathrm{x}(27-\mathrm{x})=182[\text{Putting the value of y from equation\hspace{0.17em}\hspace{0.17em}(1)]}\\ \Rightarrow 27\mathrm{x}-{\mathrm{x}}^2-182=0\\ \Rightarrow {\mathrm{x}}^2-27\mathrm{x}+182=0\\ \Rightarrow {\mathrm{x}}^2-13\mathrm{x}-14\mathrm{x}+182=0\\ \Rightarrow \mathrm{x}(\mathrm{x}-13)-14(\mathrm{x}-13)=0\\ \Rightarrow (\mathrm{x}-13)(\mathrm{x}-14)=0\\ \Rightarrow \mathrm{x}-13=0\text{or}\mathrm{x}-14=0\\ \Rightarrow \mathrm{x}=13\text{or}\mathrm{x}=14\\ \text{Putting}\mathrm{x}=13\text{in equation (1), we get}\mathrm{y}=14\\ \text{and}\\ \text{Putting}\mathrm{x}=14\text{in equation (1), we get}\mathrm{y}=13\\ \text{So, the two numbers are 13 and 14.}\end{array}$

Q.6 Find two consecutive positive integers, sum of whose squares is 365.

Ans.

$\begin{array}{l}\text{Let the two consecutive positive integers are}\mathrm{x}\text{and}\mathrm{x}+1\text{and}\\ \text{sum of their square, is 365.}\\ \text{Therefore,}\\ {\mathrm{x}}^2+{\left(\mathrm{x}+1\right)}^2=365\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{x}}^2+2\mathrm{x}+1-365=0\\ \Rightarrow 2{\mathrm{x}}^2+2\mathrm{x}-364=0\\ \Rightarrow {\mathrm{x}}^2+\mathrm{x}-182=0\\ \Rightarrow {\mathrm{x}}^2+14\mathrm{x}-13\mathrm{x}-182=0\\ \Rightarrow \mathrm{x}(\mathrm{x}+14)-13(\mathrm{x}+14)=0\\ \Rightarrow (\mathrm{x}+14)(\mathrm{x}-13)=0\\ \Rightarrow \mathrm{x}+14=0\text{or}\mathrm{x}-13=0\\ \Rightarrow \mathrm{x}=-14\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or}\mathrm{x}=13\\ \text{Integers are positive. So, we take}\mathrm{x}=13\\ \text{Thus the consecutive positive integers are 13 and 14.}\end{array}$

Q.7 The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Ans.

$\begin{array}{l}\text{Let the base of a right triangle is}\mathrm{x}\text{cm. Then its altitude is}\\ (\mathrm{x}-7)\text{cm. It is given that hypotenuse of the triangle is 13 cm.}\\ \text{Using Pythagoras theorem, we have}\\ {\mathrm{x}}^2+{\left(\mathrm{x}-7\right)}^2={13}^2\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{x}}^2-14\mathrm{x}+49=169\\ \Rightarrow \text{2}{\mathrm{x}}^2-14\mathrm{x}+49-169=0\\ \Rightarrow \text{2}{\mathrm{x}}^2-14\mathrm{x}-120=0\\ \Rightarrow {\mathrm{x}}^2-7\mathrm{x}-60=0\\ \Rightarrow {\mathrm{x}}^2-12\mathrm{x}+5\mathrm{x}-60=0\\ \Rightarrow \mathrm{x}(\mathrm{x}-12)+5(\mathrm{x}-12)=0\\ \Rightarrow (\mathrm{x}-12)(\mathrm{x}+5)=0\\ \Rightarrow \mathrm{x}-12=0\text{or}\mathrm{x}+5=0\\ \Rightarrow \mathrm{x}=12\text{or}\mathrm{x}=-5\\ \text{Length can not be negative. Thus,}\\ \text{base}=\mathrm{x}=12\text{cm}\\ \text{and}\\ \text{altitude}=\mathrm{x}-7=12-7=5\text{cm.}\end{array}$

Q.8 A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.

Ans.

$\begin{array}{l}\text{Let the number of pottery articles produced be x.}\\ \text{Cost of production of each article is 3 more than twice}\\ \text{the number of articles produced in a day.}\\ \therefore \text{Cost of production of each article}=₹\text{(}2\mathrm{x}+3\text{)}\\ \text{Total cost of production of the rticles in a day}=₹9\text{0}\\ \text{Therefore,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(}2\mathrm{x}+3\text{)}\mathrm{x}=\text{90}\\ \Rightarrow \text{\hspace{0.17em} 2}{\mathrm{x}}^2+3\mathrm{x}-90=0\\ \Rightarrow \text{\hspace{0.17em} 2}{\mathrm{x}}^2+15\mathrm{x}-12\mathrm{x}+90=0\\ \Rightarrow \mathrm{x}(2\mathrm{x}+15)-6(2\mathrm{x}+15)=0\\ \Rightarrow (2\mathrm{x}+15)(\mathrm{x}-6)=0\\ \Rightarrow 2\mathrm{x}+15=0\text{or\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-6=0\\ \Rightarrow \mathrm{x}=-\frac{15}{2}\text{or\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=6\\ \text{So, number of articles produced in a day is 6.}\\ \text{Cost of each article}=2\times 6+3=₹15\end{array}$

Q.9 

$\begin{array}{l}\text{Find the roots of the following quadratic equations,}\\ \text{if they exist, by the method of completing the square:}\\ \text{(i)}2{\mathrm{x}}^2-7\mathrm{x}+3=0\text{(ii)\hspace{0.17em}\hspace{0.17em}}2{\mathrm{x}}^2+\mathrm{x}-4=0\\ \text{\hspace{0.17em}\hspace{0.17em}(iii) 4}{\mathrm{x}}^2+4\sqrt{3}\mathrm{x}+3=0\text{(iv)}2{\mathrm{x}}^2+\mathrm{x}+4=0\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ 2{\mathrm{x}}^2-7\mathrm{x}+3=0\\ \Rightarrow {\mathrm{x}}^2-\frac{7}{2}\mathrm{x}+\frac{3}{2}=0\\ \Rightarrow {\mathrm{x}}^2-2\times \frac{7}{4}\mathrm{x}+{\left(\frac{7}{4}\right)}^2-{\left(\frac{7}{4}\right)}^2+\frac{3}{2}=0\\ \Rightarrow {(\mathrm{x}-\frac{7}{4})}^2-\frac{49}{16}+\frac{3}{2}=0\\ \Rightarrow {(\mathrm{x}-\frac{7}{4})}^2-\frac{25}{16}=0\\ \Rightarrow {(\mathrm{x}-\frac{7}{4})}^2-{\left(\frac{5}{4}\right)}^2=0\\ \Rightarrow {(\mathrm{x}-\frac{7}{4})}^2={\left(\frac{5}{4}\right)}^2\\ \Rightarrow \mathrm{x}-\frac{7}{4}=\pm \frac{5}{4}\\ \Rightarrow \mathrm{x}=\frac{7}{4}\pm \frac{5}{4}\\ \Rightarrow \mathrm{x}=\frac{12}{4}=3\text{or \hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{2}{4}=\frac{1}{2}\\ \text{(ii)}\\ 2{\mathrm{x}}^2+\mathrm{x}-4=0\\ \Rightarrow {\mathrm{x}}^2+\frac{1}{2}\mathrm{x}-2=0\\ \Rightarrow {\mathrm{x}}^2+2\times \frac{1}{4}\mathrm{x}+{\left(\frac{1}{4}\right)}^2={\left(\frac{1}{4}\right)}^2+2\\ \Rightarrow {(\mathrm{x}+\frac{1}{4})}^2=\frac{1}{16}+2=\frac{33}{16}\\ \Rightarrow {(\mathrm{x}+\frac{1}{4})}^2={\left(\frac{\sqrt{33}}{4}\right)}^2\\ \Rightarrow \mathrm{x}+\frac{1}{4}=\pm \frac{\sqrt{33}}{4}\\ \Rightarrow \mathrm{x}=-\frac{1}{4}\pm \frac{\sqrt{33}}{4}\\ \Rightarrow \mathrm{x}=\frac{-1+\sqrt{33}}{4}\text{or \hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{-1-\sqrt{33}}{4}\\ \\ \text{(iii)}\\ \text{4}{\mathrm{x}}^2+4\sqrt{3}\mathrm{x}+3=0\\ \Rightarrow {\left(2\mathrm{x}\right)}^2+2(2\mathrm{x})\; (\sqrt{3})\; +{\left(\sqrt{3}\right)}^2=0\\ \Rightarrow {(2\mathrm{x}+\sqrt{3})}^2=0\\ \Rightarrow (2\mathrm{x}+\sqrt{3})(2\mathrm{x}+\sqrt{3})=0\\ \Rightarrow 2\mathrm{x}+\sqrt{3}=0\text{or}2\mathrm{x}+\sqrt{3}=0\\ \Rightarrow \mathrm{x}=-\frac{\sqrt{3}}{2}\text{or}\mathrm{x}=-\frac{\sqrt{3}}{2}\\ \\ \text{(iv)}\\ 2{\mathrm{x}}^2+\mathrm{x}+4=0\\ \Rightarrow {\mathrm{x}}^2+\frac{1}{2}\mathrm{x}+2=0\\ \Rightarrow {\mathrm{x}}^2+2\times \frac{1}{4}\mathrm{x}+{\left(\frac{1}{4}\right)}^2-{\left(\frac{1}{4}\right)}^2+2=0\\ \Rightarrow {(\mathrm{x}+\frac{1}{4})}^2=\frac{1}{16}-2=-\frac{31}{16}\\ \text{Square of a number can not be negative.}\\ \text{Therefore, there is no real root for the given equation.}\end{array}$

Q.10

$\begin{array}{l}\text{Find the roots of the following quadratic equations,}\\ \text{by applying quadratic formula:}\\ \text{(i)}2{\mathrm{x}}^2-7\mathrm{x}+3=0\text{(ii)\hspace{0.17em}\hspace{0.17em}}2{\mathrm{x}}^2+\mathrm{x}-4=0\\ \text{\hspace{0.17em}\hspace{0.17em}(iii) 4}{\mathrm{x}}^2+4\sqrt{3}\mathrm{x}+3=0\text{(iv)}2{\mathrm{x}}^2+\mathrm{x}+4=0\end{array}$

Ans.

$\begin{array}{l}\text{(i)}2{\mathrm{x}}^2-7\mathrm{x}+3=0\\ \text{On comparing the given equation with a}{\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}=0,\text{we get}\\ \mathrm{a}=2,\mathrm{b}=-7,\mathrm{c}=3\\ \text{By using quadratic formula, we get}\\ \mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}\\ \Rightarrow \mathrm{x}=\frac{-(-7)\pm \sqrt{{(-7)}^2-4\times 2\times 3}}{2\times 2}\\ \Rightarrow \mathrm{x}=\frac{7\pm \sqrt{49-24}}{4}\\ \Rightarrow \mathrm{x}=\frac{7\pm \sqrt{25}}{4}=\frac{7\pm 5}{4}\\ \Rightarrow \mathrm{x}=\frac{7+5}{4}\text{or}\frac{7-5}{4}\\ \Rightarrow \mathrm{x}=\frac{12}{4}=3\text{or \hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{2}{4}=\frac{1}{2}\\ \text{(ii)}2{\mathrm{x}}^2+\mathrm{x}-4=0\\ \text{On comparing the given equation with a}{\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}=0,\text{\hspace{0.17em}\hspace{0.17em}we get}\\ \mathrm{a}=2,\mathrm{b}=1,\mathrm{c}=-4\\ \text{By using quadratic formula, we get}\\ \mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}\\ \Rightarrow \mathrm{x}=\frac{-1\pm \sqrt{1^2-4\times 2\times (-4)}}{2\times 2}\\ \Rightarrow \mathrm{x}=\frac{-1\pm \sqrt{1^2+32}}{4}\\ \Rightarrow \mathrm{x}=\frac{-1\pm \sqrt{33}}{4}\\ \Rightarrow \mathrm{x}=\frac{-1+\sqrt{33}}{4}\text{or \hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{-1-\sqrt{33}}{4}\\ \text{(iii) 4}{\mathrm{x}}^2+4\sqrt{3}\mathrm{x}+3=0\\ \text{On comparing the given equation with a}{\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}=0,\text{\hspace{0.17em}\hspace{0.17em}we get}\\ \mathrm{a}=4,\mathrm{b}=4\sqrt{3},\mathrm{c}=3\\ \text{By using quadratic formula, we get}\\ \mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}\\ \Rightarrow \mathrm{x}=\frac{-4\sqrt{3}\pm \sqrt{{\left(4\sqrt{3}\right)}^2-4\times 4\times 3}}{2\times 4}\\ \Rightarrow \mathrm{x}=\frac{-4\sqrt{3}\pm \sqrt{48-48}}{8}\\ \Rightarrow \mathrm{x}=\frac{-4\sqrt{3}\pm 0}{8}\\ \Rightarrow \mathrm{x}=-\frac{\sqrt{3}}{2}\text{or}\mathrm{x}=-\frac{\sqrt{3}}{2}\\ \text{(iv)}2{\mathrm{x}}^2+\mathrm{x}+4=0\\ \text{On comparing the given equation with a}{\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}=0,\text{\hspace{0.17em}\hspace{0.17em}we get}\\ \mathrm{a}=2,\mathrm{b}=1,\mathrm{c}=4\\ \text{By using quadratic formula, we get}\\ \mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}\\ \Rightarrow \mathrm{x}=\frac{-1\pm \sqrt{1^2-4\times 2\times 4}}{2\times 2}\\ \Rightarrow \mathrm{x}=\frac{-1\pm \sqrt{-31}}{4}\\ \text{Square of a number can not be negative.}\\ \text{Therefore,}\sqrt{-31}\text{is not real and so there is no real root for}\\ \text{the given equation.}\end{array}$

Q.11

$\begin{array}{l}\text{Find the roots of the following equations:}\\ \text{(i)}\mathrm{x}-\frac{1}{\mathrm{x}}=3,\mathrm{x}\ne 0\text{(ii)\hspace{0.17em}\hspace{0.17em}}\frac{1}{\mathrm{x}+4}-\frac{1}{\mathrm{x}-7}=\frac{11}{30},\mathrm{x}\ne -4,7\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \mathrm{x}-\frac{1}{\mathrm{x}}=3,\mathrm{x}\ne 0\\ \Rightarrow \frac{{\mathrm{x}}^2-1}{\mathrm{x}}=3\\ \Rightarrow {\mathrm{x}}^2-1=3\mathrm{x}\\ \Rightarrow {\mathrm{x}}^2-3\mathrm{x}-1=0\\ \text{On comparing this equation with a}{\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}=0,\text{\hspace{0.17em}\hspace{0.17em}we get}\\ \mathrm{a}=1,\mathrm{b}=-3,\mathrm{c}=-1\\ \text{By using quadratic formula, we get}\\ \mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}\\ \Rightarrow \mathrm{x}=\frac{-(-3)\pm \sqrt{{(-3)}^2-4\times 1\times (-1)}}{2\times 1}\\ \Rightarrow \mathrm{x}=\frac{3\pm \sqrt{9+4}}{2}\\ \Rightarrow \mathrm{x}=\frac{3+\sqrt{13}}{2}\text{or}\mathrm{x}=\frac{3-\sqrt{13}}{2}\\ \text{(ii)\hspace{0.17em}\hspace{0.17em}}\frac{1}{\mathrm{x}+4}-\frac{1}{\mathrm{x}-7}=\frac{11}{30},\mathrm{x}\ne -4,7\\ \Rightarrow \frac{\mathrm{x}-7-\mathrm{x}-4}{(\mathrm{x}+4)(\mathrm{x}-7)}=\frac{11}{30}\\ \Rightarrow \frac{-11}{{\mathrm{x}}^2-3\mathrm{x}-28}=\frac{11}{30}\\ \Rightarrow {\mathrm{x}}^2-3\mathrm{x}-28=-30\\ \Rightarrow {\mathrm{x}}^2-3\mathrm{x}+2=0\\ \text{On comparing this equation with a}{\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}=0,\text{\hspace{0.17em}\hspace{0.17em}we get}\\ \mathrm{a}=1,\mathrm{b}=-3,\mathrm{c}=2\\ \text{By using quadratic formula, we get}\\ \mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}\\ \Rightarrow \mathrm{x}=\frac{-(-3)\pm \sqrt{{(-3)}^2-4\times 1\times 2}}{2\times 1}\\ \Rightarrow \mathrm{x}=\frac{3\pm \sqrt{9-8}}{2}\\ \Rightarrow \mathrm{x}=\frac{3+1}{2}\text{or}\mathrm{x}=\frac{3-1}{2}\\ \Rightarrow \mathrm{x}=2\text{or}\mathrm{x}=1\end{array}$

Q.12

$\begin{array}{l}\text{The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago}\\ \text{and 5 years from now is}\frac{1}{3}.\text{Find his present age.}\end{array}$

Ans.

$\begin{array}{l}\text{Let the present age of Rehman is}\mathrm{x}\text{years. Then 3 years ago,}\\ \text{Rehman’s age}=(\mathrm{x}-3)\text{years}\\ \text{5 years from now,}\\ \text{Rehman’s age}=(\mathrm{x}+5)\text{years}\\ \text{According to question,}\\ \frac{1}{\mathrm{x}-3}+\frac{1}{\mathrm{x}+5}=\frac{1}{3}\\ \Rightarrow \frac{\mathrm{x}+5+\mathrm{x}-3}{(\mathrm{x}-3)(\mathrm{x}+5)}=\frac{1}{3}\\ \Rightarrow \frac{2\mathrm{x}+2}{{\mathrm{x}}^2+2\mathrm{x}-15}=\frac{1}{3}\\ \Rightarrow 6\mathrm{x}+6={\mathrm{x}}^2+2\mathrm{x}-15\\ \Rightarrow {\mathrm{x}}^2+2\mathrm{x}-15-6\mathrm{x}-6=0\\ \Rightarrow {\mathrm{x}}^2-4\mathrm{x}-21=0\\ \text{On comparing this equation with a}{\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}=0,\text{\hspace{0.17em}\hspace{0.17em}we get}\\ \mathrm{a}=1,\mathrm{b}=-4,\mathrm{c}=-21\\ \text{By using quadratic formula, we get}\\ \mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}\\ \Rightarrow \mathrm{x}=\frac{-(-4)\pm \sqrt{{(-4)}^2-4\times 1\times (-21)}}{2\times 1}\\ \Rightarrow \mathrm{x}=\frac{4\pm \sqrt{16+84}}{2}\\ \Rightarrow \mathrm{x}=\frac{4+10}{2}\text{or}\mathrm{x}=\frac{4-10}{2}\\ \Rightarrow \mathrm{x}=7\text{or}\mathrm{x}=-3\\ \text{Age can not be a negative quantity. Therefore, Rehman’s}\\ \text{present age is 7 years.}\end{array}$

Q.13 In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Ans.

$\begin{array}{l}\text{Let Shefali’s marks in Mathematics and English be}\mathrm{x}\text{and}\mathrm{y}\\ \text{respectively.}\\ \text{According to question,}\\ \mathrm{x}+\mathrm{y}=30...\text{(1)}\\ \text{and}\\ (\mathrm{x}+2)(\mathrm{y}-3)=210...\left(2\right)\\ \Rightarrow (\mathrm{x}+2)(30-\mathrm{x}-3)=210\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} [from (1),}\mathrm{y}=30-\mathrm{x}\text{]}\\ \Rightarrow (\mathrm{x}+2)(27-\mathrm{x})=210\\ \Rightarrow 27\mathrm{x}-{\mathrm{x}}^2+54-2\mathrm{x}=210\\ \Rightarrow 25\mathrm{x}-{\mathrm{x}}^2=210-54=156\\ \Rightarrow {\mathrm{x}}^2-25\mathrm{x}+156=0\\ \Rightarrow {\mathrm{x}}^2-12\mathrm{x}-13\mathrm{x}+156=0\\ \Rightarrow \mathrm{x}(\mathrm{x}-12)-13(\mathrm{x}-12)=0\\ \Rightarrow (\mathrm{x}-12)(\mathrm{x}-13)=0\\ \Rightarrow \mathrm{x}=12\text{or}\mathrm{x}=13\\ \text{On putting these values of x in (1), we get}\\ \mathrm{y}=18\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}when}\mathrm{x}=12\\ \text{and}\\ \mathrm{y}=17\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}when}\mathrm{x}=13\\ \text{Therefore, Shefali’s marks in Mathematics and English are}\\ \text{either}12\text{and}18\text{or}13\text{and}17\text{respectively.}\end{array}$

Q.14

$\begin{array}{l}\text{The diagonal of a rectangular field is 60 metres more}\\ \text{than the shorter side. If the longer side is 30 metres}\\ \text{more than the shorter side, find the sides of the field.}\end{array}$

Ans.

$\begin{array}{l}\text{Let length of the shorter side is}\mathrm{x}\text{. Then length of the}\\ \text{longer side is (}\mathrm{x}+30)\text{and length of the diagonal is}\\ (\mathrm{x}+60)\text{.}\\ \text{Using Pythagoras theorem, we have}\\ {\mathrm{x}}^2+{(\mathrm{x}+30)}^2={(\mathrm{x}+60)}^2\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{x}}^2+60\mathrm{x}+900={\mathrm{x}}^2+120\mathrm{x}+3600\\ \Rightarrow {\mathrm{x}}^2-60\mathrm{x}-2700=0\\ \Rightarrow {\mathrm{x}}^2-90\mathrm{x}+30\mathrm{x}-2700=0\\ \Rightarrow \mathrm{x}(\mathrm{x}-90)+30(\mathrm{x}-90)=0\\ \Rightarrow (\mathrm{x}-90)(\mathrm{x}+30)=0\\ \Rightarrow \mathrm{x}-90=0\text{or}\mathrm{x}+30=0\\ \Rightarrow \mathrm{x}=90\text{or}\mathrm{x}=-30\\ \text{Length can not be negative.}\\ \text{Thus,}\mathrm{x}=90\text{m}\\ \text{and}\mathrm{x}+30=90+30=120\text{m}\\ \text{Hence, length of the shorter and longer sides of the rectangular}\\ \text{field are 90 m and 120 m respectively.}\end{array}$

Q.15 The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Ans.

$\begin{array}{l}\text{Let smaller and larger number be x and y respectively.}\\ \text{Then, we have}\\ {\mathrm{y}}^2-{\mathrm{x}}^2=180\\ \text{and}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^2=8\mathrm{y}\\ {\text{Replacing x}}^2\text{with 8y in}{\mathrm{y}}^2-{\mathrm{x}}^2=180,\text{we get}\\ {\mathrm{y}}^2-8\mathrm{y}=180\\ \Rightarrow {\mathrm{y}}^2-8\mathrm{y}-180=0\\ \Rightarrow {\mathrm{y}}^2-18\mathrm{y}+10\mathrm{y}-180=0\\ \Rightarrow \mathrm{y}(\mathrm{y}-18)+10(\mathrm{y}-18)=0\\ \Rightarrow (\mathrm{y}-18)(\mathrm{y}+10)=0\\ \Rightarrow \mathrm{y}-18=0\text{or}\mathrm{y}+10=0\\ \Rightarrow \mathrm{y}=18\text{or}\mathrm{y}=-10\\ {\text{Negative value of y does not satisfy the equation x}}^2=8\mathrm{y}.\\ \text{Therefore,}\mathrm{y}=\text{18}\\ \text{and hence}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^2=8\mathrm{y}\\ \Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^2=8\times \text{18}=144={12}^2\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}=\pm 12\\ \text{Hence larger number is 18 and smaller number is either}\\ \text{12 or}-12.\end{array}$

Q.16 A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Ans.

$\begin{array}{l}\text{Let the speed of the train be}\mathrm{x}\text{km/h.}\\ \text{Time taken to cover 360 km is}\frac{360}{\mathrm{x}}\text{hours.}\\ \text{According to question,}\\ (\mathrm{x}+5)(\frac{360}{\mathrm{x}}-1)=360\\ \Rightarrow (\mathrm{x}+5)(360-\mathrm{x})=360\mathrm{x}\\ \Rightarrow 360\mathrm{x}-{\mathrm{x}}^2+1800-5\mathrm{x}=360\mathrm{x}\\ \Rightarrow -{\mathrm{x}}^2+1800-5\mathrm{x}=0\\ \Rightarrow {\mathrm{x}}^2+5\mathrm{x}-1800=0\\ \Rightarrow {\mathrm{x}}^2+45\mathrm{x}-40\mathrm{x}-1800=0\\ \Rightarrow \mathrm{x}(\mathrm{x}+45)-40(\mathrm{x}+45)=0\\ \Rightarrow (\mathrm{x}+45)(\mathrm{x}-40)=0\\ \Rightarrow \mathrm{x}+45=0\text{or}\mathrm{x}-40=0\\ \Rightarrow \mathrm{x}=-45\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or}\mathrm{x}=40\\ \text{Here, speed can not be negative.}\\ \text{Therefore, speed}=\mathrm{x}=40\text{km/h}\end{array}$

Q.17

$\begin{array}{l}\begin{array}{l}\text{Two water taps together can fill a tank in \hspace{0.33em}}9\frac{3}{8}\text{\hspace{0.33em}hours.}\\ \text{The tap of larger diameter takes 10 hours less than}\end{array}\\ \begin{array}{l}\text{the smaller one to fill the tank separately.}\\ \text{Find the time in which each tap can separately fill the tank.}\end{array}\end{array}$

Ans.

$\begin{array}{l}\text{Let the tap of smaller diameter fills the tank in}\mathrm{x}\text{hours.}\\ \text{So, in 1 hour it fills}\frac{1}{\mathrm{x}}\text{part of the tank.}\\ \text{The tap of larger diameter fills the tank in (}\mathrm{x}-10)\text{hours}\\ \text{and so in 1 hour it fills}\frac{1}{\mathrm{x}-10}\text{part of the tank.}\\ \text{The two water taps together fill the tank in}9\frac{3}{8}\text{hours.}\\ \text{So, we have}\\ 9\frac{3}{8}\left(\frac{1}{\mathrm{x}}\right)+9\frac{3}{8}\left(\frac{1}{\mathrm{x}-10}\right)=1\\ \Rightarrow \frac{75}{8}\left(\frac{1}{\mathrm{x}}\right)+\frac{75}{8}\left(\frac{1}{\mathrm{x}-10}\right)=1\\ \Rightarrow \frac{75}{8}\left(\frac{\mathrm{x}-10+\mathrm{x}}{\mathrm{x}(\mathrm{x}-10)}\right)=1\\ \Rightarrow \frac{75}{8}\left(\frac{2\mathrm{x}-10}{{\mathrm{x}}^2-10\mathrm{x}}\right)=1\\ \Rightarrow 150\mathrm{x}-750=8{\mathrm{x}}^2-80\mathrm{x}\\ \Rightarrow 8{\mathrm{x}}^2-230\mathrm{x}+750=0\\ \Rightarrow 8{\mathrm{x}}^2-200\mathrm{x}-30\mathrm{x}+750=0\\ \Rightarrow 8\mathrm{x}(\mathrm{x}-25)-30(\mathrm{x}-25)=0\\ \Rightarrow (8\mathrm{x}-30)(\mathrm{x}-25)=0\\ \Rightarrow 8\mathrm{x}-30=0\text{or \hspace{0.17em}\hspace{0.17em}}\mathrm{x}-25=0\\ \Rightarrow \mathrm{x}=\frac{30}{8}=\frac{15}{4}\text{or \hspace{0.17em}\hspace{0.17em}}\mathrm{x}=25\\ \text{If}\mathrm{x}=\frac{15}{4}\text{, then}\mathrm{x}-10=\frac{15}{4}-10=\frac{-25}{4}\\ \text{But time can not be negative.}\\ \text{So, here we take}\mathrm{x}=25\text{and so we have}\mathrm{x}-10=25-10=15\\ \text{Therefore, the tap of smaller and larger diameters fill the}\\ \text{tank in}25\text{hours and 15 hours respectively.}\end{array}$

Q.18 An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.

Ans.

$\begin{array}{l}\text{Let average speed of the passenger train be}\mathrm{x}\text{km/h.}\\ \text{Then average speed of the express train is (x + 11) km/h.}\\ \\ \text{It is given that the time taken by the express train}\\ \text{to cover 132 km is 1 hour less than the passenger}\\ \text{train to cover the same distance.}\\ \text{Therefore,}\\ \frac{132}{\mathrm{x}}-\frac{132}{\mathrm{x}+11}=1\\ \text{or 132}(\mathrm{x}+11-\mathrm{x})=\mathrm{x}(\mathrm{x}+11)\\ \text{or 1452}={\mathrm{x}}^2+11\mathrm{x}\\ \text{or}{\mathrm{x}}^2+11\mathrm{x}-\text{1452}=0\\ \text{or}{\mathrm{x}}^2+44\mathrm{x}-33\mathrm{x}-\text{1452}=0\\ \text{or}\mathrm{x}(\mathrm{x}+44)-33(\mathrm{x}+44)=0\\ \text{or (}\mathrm{x}-33\left)\right(\mathrm{x}+44)=0\\ \text{or}\mathrm{x}-33=0\text{or}\mathrm{x}+44=0\\ \text{or}\mathrm{x}=33\text{or}\mathrm{x}=-44\\ \text{Speed can not be negative.}\\ \text{Therefore,}\\ \text{Average speed of the passanger train}=\mathrm{x}\text{km/h}=33\text{km/h}\\ \text{Average speed of the express train}=(\mathrm{x}+11)\text{km/h}=44\text{km/h}\end{array}$

Q.19 Sum of the areas of two squares is 468 m­². If the difference of their perimeters is 24 m, find the sides of the two squares.

Ans.

$\begin{array}{l}\text{Let sides of the two squares are}\mathrm{x}\text{and}\mathrm{y}\text{.}\\ \text{According to question,}\\ 4\mathrm{x}-4\mathrm{y}=24\Rightarrow \mathrm{x}-\mathrm{y}=6\Rightarrow \mathrm{x}=\mathrm{y}+6...\left(1\right)\\ \text{Also, we have}\\ {\mathrm{x}}^2+{\mathrm{y}}^2=468\\ \Rightarrow {(\mathrm{y}+6)}^2+{\mathrm{y}}^2=468\\ \Rightarrow {\mathrm{y}}^2+12\mathrm{y}+36+{\mathrm{y}}^2=468\\ \Rightarrow 2{\mathrm{y}}^2+12\mathrm{y}+36-468=0\\ \Rightarrow 2{\mathrm{y}}^2+12\mathrm{y}-432=0\\ \Rightarrow {\mathrm{y}}^2+6\mathrm{y}-216=0\\ \Rightarrow {\mathrm{y}}^2+18\mathrm{y}-12\mathrm{y}-216=0\\ \Rightarrow \mathrm{y}(\mathrm{y}+18)-12(\mathrm{y}+18)=0\\ \Rightarrow (\mathrm{y}-12)(\mathrm{y}+18)=0\\ \Rightarrow \mathrm{y}-12=0\text{or}\mathrm{y}+18=0\\ \Rightarrow \mathrm{y}=12\text{or}\mathrm{y}=-18\\ \text{Side of a square can not be negative.}\\ \text{So,}\mathrm{y}=12\\ \text{Substituting this value of y in equation (1), we get}\\ \mathrm{x}=12+6=18\\ \text{Hence, the sides of the two squares are 12 m and 18 m.}\end{array}$

Q.20

$\begin{array}{l}\text{Find the nature of the roots of the following quadratic}\\ \text{equations. If the real roots exist, find them.}\\ \text{(i)}2{\mathrm{x}}^2-3\mathrm{x}+5=0\text{(ii)\hspace{0.17em}\hspace{0.17em}}3{\mathrm{x}}^2-4\sqrt{3}\mathrm{x}+4=0\\ \text{(iii)}2{\mathrm{x}}^2-6\mathrm{x}+3=0\end{array}$

Ans.

$\begin{array}{l}\text{(i)}2{\mathrm{x}}^2-3\mathrm{x}+5=0\\ \text{Here,}\mathrm{a}=2,\mathrm{b}=-3,\mathrm{c}=5\\ \text{Therefore, discriminant}{\mathrm{b}}^2-4\mathrm{ac}={\left(-3\right)}^2-4\times 2\times 5\\ =9-40\\ =-31<0\\ \text{So, the given equation has no real roots.}\\ \text{(ii)\hspace{0.17em}\hspace{0.17em}}3{\mathrm{x}}^2-4\sqrt{3}\mathrm{x}+4=0\\ \text{Here,}\mathrm{a}=3,\mathrm{b}=-4\sqrt{3},\mathrm{c}=4\\ \text{Therefore, discriminant}{\mathrm{b}}^2-4\mathrm{ac}={\left(-4\sqrt{3}\right)}^2-4\times 3\times 4\\ =48-48\\ =0\\ \text{So, the given equation has two equal real roots.}\\ \text{The roots are}\frac{-\mathrm{b}}{2\mathrm{a}},\frac{-\mathrm{b}}{2\mathrm{a}},\text{\hspace{0.17em}\hspace{0.17em}i.e.,}\frac{4\sqrt{3}}{6},\frac{4\sqrt{3}}{6},\text{i.e.,}\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}.\\ \text{(iii)}2{\mathrm{x}}^2-6\mathrm{x}+3=0\\ \text{Here,}\mathrm{a}=2,\mathrm{b}=-6,\mathrm{c}=3\\ \text{Therefore, discriminant}{\mathrm{b}}^2-4\mathrm{ac}={\left(-6\right)}^2-4\times 2\times 3\\ =36-24\\ =12>0\\ \text{So, the given equation has two distinct real roots.}\\ \text{The roots are given by}\frac{-\mathrm{b}+\sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}\text{and \hspace{0.17em}}\frac{-\mathrm{b}–\sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}.\\ \frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}=\frac{6\pm \sqrt{12}}{4}=\frac{3\pm \sqrt{3}}{2}\\ \text{Hence, the roots are}\frac{3+\sqrt{3}}{2}\text{and}\frac{3-\sqrt{3}}{2}.\end{array}$

Q.21 Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx +3 = 0
(ii) kx(x – 2) + 6 = 0

Ans.

$\begin{array}{l}\text{(i)}2{\mathrm{x}}^2+\mathrm{kx}+3=0\\ \text{Discriminant must be zero for two roots of the given}\\ \text{equation to be equal}.\\ \text{Therefore,}\\ {\mathrm{b}}^2-4\mathrm{ac}=0\\ \Rightarrow {\mathrm{k}}^2-4\times 2\times 3=0\\ \Rightarrow {\mathrm{k}}^2-24=0\\ \Rightarrow {\mathrm{k}}^2=24\\ \Rightarrow \mathrm{k}=\sqrt{24}=\pm 2\sqrt{6}\\ \text{(ii)\hspace{0.17em}\hspace{0.17em}}\\ \mathrm{kx}(\mathrm{x}-2)+6=0\\ \Rightarrow {\mathrm{kx}}^2-2\mathrm{kx}+6=0\\ \text{Discriminant must be zero for two roots of the given}\\ \text{equation to be equal}.\\ \text{Therefore,}\\ {\mathrm{b}}^2-4\mathrm{ac}=0\\ \Rightarrow 4{\mathrm{k}}^2-4\times \mathrm{k}\times 6=0\\ \Rightarrow {\mathrm{k}}^2-6\mathrm{k}=0\\ \Rightarrow \mathrm{k}(\mathrm{k}-6)=0\\ \Rightarrow \mathrm{k}=0\text{or}\mathrm{k}=6\\ \text{But}\mathrm{k}=0\text{does not satisfy the given equation.}\\ \text{Thus, the given equation has equal roots if the value of}\mathrm{k}\\ \text{is}6\text{only.}\end{array}$

Q.22 Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.

Ans.

$\begin{array}{l}\text{Let breadth of the rectangular mango grove is}\mathrm{x}\text{.}\\ \text{According to question,}\\ \mathrm{x}\times 2\mathrm{x}=\text{800}\\ \Rightarrow {\text{x}}^2=400\\ \Rightarrow {\text{x}}^2-400=0\\ \text{Discriminant}={\mathrm{b}}^2-4\mathrm{ac}=0-4\times 1\times (-400)=1600>0\\ \text{Therefore, the above equation has two distinct real roots.}\\ \text{Now,}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^2-400=0\\ \Rightarrow {\text{x}}^2=400\\ \Rightarrow \text{x}=\pm 20\\ \text{But breadth can not be negative.}\\ \text{So,}\mathrm{x}=20\text{and 2}\mathrm{x}=40\\ \text{Hence, length}=40\text{m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}breadth}=20\text{m}\end{array}$

Q.23 Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Ans.

$\begin{array}{l}\text{Let age of one friend be}\mathrm{x}\text{years.}\\ \text{Age of the other friend}=(20-\mathrm{x})\text{years}\\ \text{According to question,}\\ (\mathrm{x}-4)(20-\mathrm{x}-4)=4\text{8}\\ \Rightarrow (\mathrm{x}-4)(16-\mathrm{x})=4\text{8}\\ \Rightarrow -{\text{x}}^2+16\mathrm{x}-64+4\mathrm{x}-48=0\\ \Rightarrow {\mathrm{x}}^2-20\mathrm{x}+112=0\\ \text{Discriminant}={\mathrm{b}}^2-4\mathrm{ac}={(-20)}^2-4\times 112=400-448=-48<0\\ \text{Therefore, no real root is possible and so the given situation}\\ \text{is not possible.}\end{array}$

Q.24 Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.

Ans.

$\begin{array}{l}\text{Let length and breadth of the rectangular park be}\mathrm{x}\text{.}\\ \text{and}\mathrm{y}\text{respectively.}\\ \text{According to question, we have}\\ 2(\mathrm{x}+\mathrm{y})=\text{80}\\ \Rightarrow \mathrm{x}+\mathrm{y}=40\\ \Rightarrow \mathrm{y}=40-\mathrm{x}...\text{(1)}\\ \text{Also, we have}\\ \mathrm{xy}=400\\ \Rightarrow \mathrm{x}(40-\mathrm{x})=400\\ \Rightarrow 40\mathrm{x}-{\mathrm{x}}^2=400\\ \Rightarrow {\text{x}}^2-40\mathrm{x}+400=0\\ \text{Discriminant}={\mathrm{b}}^2-4\mathrm{ac}={(-40)}^2-4\times 400=0\\ \text{So, the above equation has two equal real roots.}\\ \text{Now,}\\ {\mathrm{x}}^2-40\mathrm{x}+400=0\\ \Rightarrow {(\mathrm{x}-20)}^2=0\\ \Rightarrow \mathrm{x}=20\text{or}\mathrm{x}=20\\ \text{From equation (1), we have}\\ \mathrm{y}=40-\mathrm{x}=40-20=20\\ \text{Hence, length and breadth of the rectangular park are}\\ \text{20 m and 20 m respectively.}\end{array}$