CBSE Class 10 Maths Important Questions - Chapter 4 Quadratic Equations

1. Solve by factorization

a. $4x^2-4a^{2}x+(a^4-b^4)=0$

Ans: 

$4x^2-[2(a^2+b^2)+2(a^2-b^2)]x+(a^2-b^2)(a^2+b^2)=0$ 

$\Rightarrow 2x[2x-(a^2+b^2)]-(a^2-b^2)[2x-(a^2+b^2)]=0$ 

$\Rightarrow x={\displaystyle \frac{a^2+b^2}{2}},x={\displaystyle \frac{a^2-b^2}{2}}$

Therefore, $x={\displaystyle \frac{a^2+b^2}{2}}\left(or\right)x={\displaystyle \frac{a^2-b^2}{2}}$

b. $x^2+({\displaystyle \frac{a}{a+b}}x+{\displaystyle \frac{a+b}{a}})x+1=0$

Ans:

            $x^2+({\displaystyle \frac{a}{a+b}}x+{\displaystyle \frac{a+b}{a}})x+1=0$ 

            $\Rightarrow x^2+({\displaystyle \frac{a}{a+b}}x+{\displaystyle \frac{a+b}{a}}x+{\displaystyle \frac{a}{a+b}}.{\displaystyle \frac{a+b}{a}})=0$ 

            $\Rightarrow [x+{\displaystyle \frac{a}{a+b}}]+{\displaystyle \frac{a+b}{a}}[x+{\displaystyle \frac{a}{a+b}}]=0$ 

            $\Rightarrow x=-{\displaystyle \frac{a}{a+b}},x={\displaystyle \frac{(-a+b)}{a}},a+b=0$ 

c. ${\displaystyle \frac{1}{a+b+x}}={\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{x}},(a+b\ne 0)$

Ans:

${\displaystyle \frac{1}{a+b+x}}={\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{x}}$ 

$\Rightarrow {\displaystyle \frac{1}{a+b+x}}-{\displaystyle \frac{1}{x}}={\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}$ 

$\Rightarrow {\displaystyle \frac{x-(a+b+x)}{x(a+b+x)}}={\displaystyle \frac{a+b}{ab}}$ 

$\Rightarrow (a+b)\{x(a+b+x)+ab\}=0$ 

$\Rightarrow x(a+b+x)+ab=0$ 

$\Rightarrow x^2+ax+bx+ab=0$ 

$\Rightarrow (x+a)(x+b)=0$ 

$\Rightarrow x=-a\left(or\right)x=-b$ 

d. $(x-3)(x-4){\displaystyle \frac{34}{{33}^2}}$

$(x-3)(x-4)={\displaystyle \frac{34}{{33}^2}}$ 

$\Rightarrow x^2-7x+12={\displaystyle \frac{34}{{33}^2}}$ 

$\Rightarrow x^2-7x+{\displaystyle \frac{13034}{{33}^2}}=0$ 

Ans:

$\Rightarrow x^2-7x+{\displaystyle \frac{98}{33}}\times {\displaystyle \frac{133}{33}}=0$ 

$\Rightarrow x^2-{\displaystyle \frac{231}{33}}x+{\displaystyle \frac{98}{33}}\times {\displaystyle \frac{133}{33}}=0$ 

$\Rightarrow x^{2}x-{\displaystyle \frac{231}{33}}x+{\displaystyle \frac{98}{33}}\times {\displaystyle \frac{133}{33}}=0$ 

 $\Rightarrow x^2-({\displaystyle \frac{98}{33}}+{\displaystyle \frac{133}{33}})x+{\displaystyle \frac{98}{33}}\times {\displaystyle \frac{133}{33}}=0$ 

 $\Rightarrow x^2-{\displaystyle \frac{98}{33}}x-{\displaystyle \frac{133}{33}}x+{\displaystyle \frac{98}{33}}\times {\displaystyle \frac{133}{33}}=0$ 

 $\Rightarrow (x-{\displaystyle \frac{98}{33}})x-{\displaystyle \frac{133}{33}}(x-{\displaystyle \frac{98}{33}})=0$ 

 $\Rightarrow (x-{\displaystyle \frac{98}{33}})(x-{\displaystyle \frac{133}{33}})=0$ 

 $\Rightarrow x={\displaystyle \frac{98}{33}}\left(or\right)x={\displaystyle \frac{133}{33}}$ 

e. $\text{x=}{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-x}}}}}}x\ne 2$

Ans:

$\text{x=}{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-x}}}}}}x\ne 2$ 

 $\Rightarrow \text{ x=}{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-x}}}}}}$ 

 $\Rightarrow \text{x=}{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-{\displaystyle \frac{(2-x)}{4-2x-1}}}}}}$ 

 $\Rightarrow \text{x=}{\displaystyle \frac{1}{2-{\displaystyle \frac{2-x}{3-2x}}}}$ 

 $\Rightarrow x={\displaystyle \frac{3-2x}{2(3-2x)-(2-x)}}$ 

 $\Rightarrow x={\displaystyle \frac{3-2x}{4-3x}}$ 

 $\Rightarrow 4x-3x^2=3-2x$ 

 $\Rightarrow 3x^2-6x+3=0$ 

 $\Rightarrow {(x-1)}^2=0$ 

 $\Rightarrow x=1,1.$ 

2. By the method of completion of squares show that the equation $\mathbf{4}{\mathbf{x}}^{\mathbf{2}}+\mathbf{3}\mathbf{x}+\mathbf{5}=\mathbf{0}$ has no real roots.

Ans:

 $4x^2+3x+5=0$ 

 $\Rightarrow x^2+{\displaystyle \frac{3}{4}}x+{\displaystyle \frac{5}{4}}=0$ 

 $\Rightarrow x^2+{\displaystyle \frac{3}{4}}x+{\left({\displaystyle \frac{3}{8}}\right)}^2=-{\displaystyle \frac{5}{4}}+{\displaystyle \frac{9}{64}}$ 

 $\Rightarrow {(x+{\displaystyle \frac{3}{8}})}^2=-{\displaystyle \frac{71}{64}}$ 

 $\Rightarrow x+{\displaystyle \frac{3}{8}}=\sqrt{-{\displaystyle \frac{71}{64}}}$ 

 Which is not a real number. Hence the equation has no real roots.

3. The sum of areas of two squares is 468m2. If the difference of their perimeters is 24cm, find the sides of the two squares. 

Ans:

Let, the side of the larger square be x. 

Let, the side of the smaller square be y. 

$x^2+y^2=468$ 

Cond. II 4x-4y = 24 

 $\Rightarrow xy=6$ 

 $\Rightarrow x=6+y$ 

 $\Rightarrow x^2+y^2=468$ 

 $\Rightarrow {(6+y)}^2+y^2=468$ 

on solving we get 

y = 12 

⇒ x = (12+6) = 18 m 

∴ The length of the sides of the two squares are 18m and 12m. 

4. A dealer sells a toy for Rs.24 and gains as much percent as the cost price of the toy. Find the cost price of the toy.

Ans:

Let the C.P be x 

∴Gain = x% 

$\Rightarrow Gain=x{\displaystyle \frac{x}{100}}$

S.P = C.P +Gain 

SP = 24 

$\Rightarrow x+{\displaystyle \frac{x^2}{100}}=24$

On solving we get x = 20 or x = -120 (reject this as cost cannot be negative) 

∴ C.P of toy = Rs.20 

5. A fox and an eagle lived at the top of a cliff of height 6m, whose base was at a distance of 10m from a point A on the ground. The fox descends the cliff and went straight to point A. The eagle flew vertically up to height x meters and then flew in a straight line to a point A, the distance traveled by each being the same. 3 Find the value of x.

Ans: 

Distance traveled by the fox = distance traveled by the eagle 

            $(6+x{)}^2+(10{)}^2$ = $(16–x{)}^2$  

            on solving we get x = 2.72m. 

6. A lotus is 2m above the water in a pond. Due to wind, the lotus slides on the side and only the stem completely submerges in the water at a distance of 10m from the original position. Find the depth of water in the pond. 

Ans: 

From,above figure,We can write as,

${(x+2)}^2=x^2+{10}^2$ 

$\Rightarrow x^2+4x+4$ = $x^2+100$ 

$\Rightarrow 4x+4=100$

$\Rightarrow x=24$

Therefore, the depth of water in the pond is 24m. 

7. Solve $x=\sqrt{6+\sqrt{6+\sqrt{6.....}}}$

Ans: 

 $x=\sqrt{6+\sqrt{6+\sqrt{6.....}}}$ 

 $\Rightarrow x=\sqrt{6+x}$ 

 $\Rightarrow x^2=6+x$ 

 $\Rightarrow x^2-x-6=0$ 

 $\Rightarrow (x-3)(x+2)=0$ 

 $\Rightarrow x=3$ 

As x cannot be negative x is not equal to 2.

8. The hypotenuse of a right triangle is 20m. If the difference between the length of the 4 other sides is 4m. Find the sides. 

Ans: 

From above figure,

$x^2+y^2={20}^2$ 

 $x^2+y^2=400$ 

 $alsox-y=4$ 

 $\Rightarrow x=404+y$ 

 $\Rightarrow {(4+y)}^2+y^2=400$ 

 $\Rightarrow 2y^2+8y-384=0$ 

 $\Rightarrow (y+16)\left(y12\right)=0$ 

 $\Rightarrow y=12\left(or\right)y=16\left(notpossible\right)$ 

∴sides are 12cm and 16cm 

9. The positive value of k for which ${\mathbf{x}}^{\mathbf{2}}+\mathbf{Kx}+\mathbf{64}=\mathbf{0}\mathrm{\&}{\mathbf{x}}^{\mathbf{2}}-\mathbf{8}\mathbf{x}+\mathbf{k}=\mathbf{0}$will have real roots.

Ans:

$\begin{array}{l}{x}^2+Kx+64=0\\ \Rightarrow b^2-4ac\ge 0\\ \Rightarrow K^2-256\ge 0\\ \Rightarrow K\ge 16orK\le -16\dots \dots \dots \dots \dots \left(1\right)\\ x^2-8x+K=0\\ 644K\ge 0\\ \Rightarrow 4K\le 64\\ \Rightarrow K\le 16\dots \dots \dots \dots \dots \left(2\right)\\ From\left(1\right)\mathrm{\&}\left(2\right)K=16\end{array}$

10. A teacher attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left over. When he increased the size of the square by one student he found he was short of 25 students. Find the number of students. 

Ans: 

Let the side of the square be $x.$

No. of students = $x^2+24$

New side = $x+1$

No. of students = ${(x+1)}^2-25$

$\begin{array}{l}\Rightarrow x^2+24={(x+1)}^2-25\\ \Rightarrow x^2+24=x^2+2x+1-25\\ \Rightarrow 2x=48\\ \Rightarrow x=24\end{array}$

∴ side of square = 24 

No. of students = $576+24=600$

11. A pole has to be erected at a point on the boundary of a circular park of diameter 13m in such a way that the differences of its distances from two diametrically opposite fixed gates A $ B on the boundary in 7m. Is it possible to do so? If the answer is yes at what distances from the two gates should the pole be erected? 

Ans: 

AB = 13 m 

BP = x 

$\begin{array}{l}\Rightarrow AP-BP=7\\ \Rightarrow AP=x+7APQ\\ \Rightarrow {\left(13\right)}^2={(x+7)}^2+x^2\\ \Rightarrow x^2+7x-60=0\\ \Rightarrow (x+12)(x-5)=0\\ \Rightarrow x=-12\left(notpossible\right)orx=5\end{array}$

∴Pole has to be erected at a distance of 5m from gate B & 12m from gate A. 

12. If the roots of the equation $(\mathbf{a}-\mathbf{b}){\mathbf{x}}^{\mathbf{2}}+\mathbf{b}-\mathbf{c})\mathbf{x}+(\mathbf{c}-\mathbf{a})=\mathbf{0}$ are equal. Prove that$\mathbf{2}\mathbf{a}=\mathbf{b}+\mathbf{c}$. 

Ans: 

$\begin{array}{l}\left(ab\right)x^2+\left(bc\right)x+\left(ca\right)=0\\ Given:2a=b+c\\ B^2-4AC=0\\ {\left(bc\right)}^2\left[4\left(ab\right)\left(ca\right)\right]=0\\ \Rightarrow b^2-2bc+c^2\left[4(ac{a}^{2}bc+ab)\right]=0\\ \Rightarrow b^2-2bc+c^{2}4ac+4a^2+4bc-\text{4}ab=0\\ \Rightarrow b^2+2bc+c^2+4a^{2}4ac4ab=0\\ \Rightarrow {(\text{b}+c-2a)}^2=0\\ \Rightarrow b+c=2a\end{array}$

Hence proved.

13. X and Y are centers of circles of radius 9cm and 2cm and XY = 17cm. Z is the center of a circle of radius 4 cm, which touches the above circles externally. Given that $\mathrm{\angle }XYZ=90{}^{\circ }$, write an equation in r and solve it for r. 

Ans: 

Let r be the radius of the third circle 

XY = 17cm 

⇒ XZ = 9 + r 

    YZ = 2 + r

$\begin{array}{l}{(r+9)}^2+{(r+2)}^2={\left(17\right)}^2\\ \Rightarrow r^2+18r+81+r^2+4r+4=289\\ \Rightarrow r^2+22r-204=0\\ \Rightarrow r^2+11r-102=0\\ \Rightarrow (r+17)(r-6)=0\\ \Rightarrow r=-17\left(notpossible\right)orr=6cm\\ \therefore radius=6cm.\end{array}$

Level - 01 (01 Marks) 

1. Check whether the following are quadratic equation or not 

i. $(x-3)(2x+1)=x(x+5)$

Ans:

Yes, this is a quadratic equation as the highest power of x is 2.

ii. ${(\mathbf{x}+\mathbf{2})}^{\mathbf{2}}=\mathbf{2}\mathbf{x}({\mathbf{x}}^{\mathbf{2}}-\mathbf{1})$

Ans: No, this is not a quadratic equation as the highest power of x is 1.

2. Solve by factorization method ${\mathbf{x}}^{\mathbf{2}}-\mathbf{7}\mathbf{x}+\mathbf{12}=\mathbf{0}$

Ans: $x=3;x=4$

3. Find the discriminant ${\mathbf{x}}^{\mathbf{2}}-\mathbf{3}\mathbf{x}-\mathbf{10}=\mathbf{0}$

Ans: $D=49$ (D = discriminant)

4. Find the nature of root $\mathbf{2}{\mathbf{x}}^{\mathbf{2}}+\mathbf{3}\mathbf{x}-\mathbf{4}=\mathbf{0}$

Ans: root are real and unequal. 

5. Find the value k so that quadratic equation $\mathbf{3}{\mathbf{x}}^{\mathbf{2}}-\mathbf{kx}+\mathbf{38}=\mathbf{0}$ has equal root

Ans: $5\pm 18$

6. Determine whether given value of x is a solution or not 

${\mathbf{x}}^{\mathbf{2}}-\mathbf{3}\mathbf{x}-\mathbf{1}=\mathbf{0}:\mathbf{x}=\mathbf{1}$

Ans: not a solution 

Level 2 (02 Marks)

7. Solve by quadratic equation $\mathbf{16}{\mathbf{x}}^{\mathbf{2}}-\mathbf{24}\mathbf{x}-\mathbf{1}=\mathbf{0}$ by using quadratic formula. 

Ans: $1$

8. Determine the value of for which the quadratic equation $\mathbf{2}{\mathbf{x}}^{\mathbf{2}}+\mathbf{3}\mathbf{x}+\mathbf{k}=\mathbf{0}$ have both roots real.

Ans: 

$k\le {\displaystyle \frac{9}{8}}x,x={\displaystyle \frac{3+\sqrt{10}}{4}},{\displaystyle \frac{3-\sqrt{10}}{4}}$

9. Find the roots of equation $\mathbf{2}{\mathbf{x}}^{\mathbf{2}}+\mathbf{x}-\mathbf{6}=\mathbf{0}$

Ans: $x=2,x={\displaystyle \frac{3}{2}}$

10. Find the roots of equation $x-{\displaystyle \frac{1}{x}}=3;x\ne 0$

Ans:$x={\displaystyle \frac{3}{2}}$ 

Level 3 (03 Marks)

1. The sum of the squares of two consecutive positive integers is 265. Find the integers. 

Ans: number are 11, 12 

2. Divide 39 into two parts such that their product is 324. 

Ans: 27, 12

3. The sum of the number and its reciprocals is. Find the number. 

Ans: $4{\displaystyle \frac{1}{4}}$ 

4. The length of a rectangle is 5cm more than its breadth if its area is 150 Sq. cm. 

Ans: 10cm, 15cm 

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

Ans: 12cm and 5cm 

1 Marks Questions 

1. Which of the following is a quadratic equation? 

$a)x^3-2x-\sqrt{5}-x=0$ 

$b)3x^2-5x+9=x^2-7x+3$ 

$c){(x+{\displaystyle \frac{1}{x}})}^2=3(x+{\displaystyle \frac{1}{x}})+4$ 

$d)x^3+x+3=0$ 

Ans: $b)3x^2-5x+9=x^2-7x+3$

2. Factor of $a^2{x}^2-3abx+2b^2=0$ is 

$a){\displaystyle \frac{2b}{a}},{\displaystyle \frac{b}{a}}$ 

$b){\displaystyle \frac{3b}{a}},{\displaystyle \frac{a}{b}}$ 

$c){\displaystyle \frac{b}{a}},{\displaystyle \frac{a}{b}}$ 

$d){\displaystyle \frac{a}{b}},{\displaystyle \frac{a}{b}}$ 

Ans: $a){\displaystyle \frac{2b}{a}},{\displaystyle \frac{b}{a}}$ 

3. Which of the following have real roots? 

$\text{a) 2}{\text{x}}^2+x-1=0$ 

$b){\text{x}}^2+x+1=0$ 

$c){\text{x}}^2-6x+6=0$ 

$d)\text{ 2}{\text{x}}^2+15x+30=0$ 

Ans: $c){\text{x}}^2-6x+6=0$

4. Solve for x:

$x={\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2-x}}}}}}$

 $a)x=2$ 

 $b)x=-1$ 

 $c)x=1$ 

 $d)x=3$ 

Ans: $b)x=-1$ 

5. Solve by factorization $\sqrt{3}{x}^2+10x+7\sqrt{3}=0$

 $a)x=-\sqrt{3},-{\displaystyle \frac{7}{\sqrt{3}}}$ 

 $b)x=-\sqrt{3},{\displaystyle \frac{7}{\sqrt{3}}}$ 

 $c)x=2,{\displaystyle \frac{1}{2}}$ 

 $d)\pm 3$ 

Ans: $a)x=-\sqrt{3},-{\displaystyle \frac{7}{\sqrt{3}}}$ 

6. The quadratic equation whose roots are 3 and -3 is 

 $a)x^2-9=0$ 

 $b)x^2-3x-3=0$ 

 $c)x^2-2x+2=0$ 

 $d)x^2+9=0$ 

Ans: $a)x^2-9=0$

7. Discriminant of $-x^2+{\displaystyle \frac{1}{2}}x+{\displaystyle \frac{1}{2}}=0$ is

a) $-{\displaystyle \frac{1}{2}},1$ 

b) ${\displaystyle \frac{1}{2}},1$ 

c) ${\displaystyle \frac{-1}{2}},1$ 

d) ${\displaystyle \frac{1}{2}},{\displaystyle \frac{-1}{2}}$ 

Ans: 

(a) $-{\displaystyle \frac{1}{2}},1$

8. For equal root,$kx(x-2)+6=0$ , value of k is

a). $k=6$ 

b). $k=3$ 

c). $k=2$ 

d. $k=8$ 

Ans:

(a) $k=6$

9. Quadratic equation whose roots are $2+\sqrt{s},2-\sqrt{s}$ is

a). $x^2-4x-1=0$ 

b). $x^2+4x+1=0$ 

c). $x^2+(x+\sqrt{5})x-\left(2\sqrt{5}\right)=0$ 

d). $x^2-4x+2=0$ 

Ans: 

(a) $x^2-4x-1=0$

10. If $\alpha$ and $\beta$ are roots of the equation $3x^2+5x-7=0$ then $\alpha \beta$ equal to

a). ${\displaystyle \frac{7}{3}}$ 

b). ${\displaystyle \frac{-7}{3}}$ 

c). ${\displaystyle \frac{-5}{3}}$ 

d). $21$ 

Ans: 

(b) ${\displaystyle \frac{-7}{3}}$

2 Marks Questions

1. Solve the following problems given-

i. $x{}^2-45x+324=0$ 

Ans:

$x{}^2-45x+324=0$

                    $\Rightarrow x^2-36x-9x+324=0$ 

                    $\Rightarrow x(x-36)-9(x-36)=0$ 

    $\Rightarrow (x-9)(x-36)=0$ 

    $\therefore x=36,9$ 

ii. $x{}^2-55x+750=0$ 

Ans: 

                    $\Rightarrow x{}^2-25x-30x+750=0$ 

                    $\Rightarrow x(x-25)30(x-25)=0$ 

                    $\Rightarrow (x-30)(x-25)=0$ 

                    $\therefore x=30,25$ 

2. Find two numbers whose sum is $27$ and the product is $182$ 

Ans: 

Let first number be x and let second number be $(27-x)$ 

According to given condition, the product of two numbers is $182$.

Therefore,

      $x(27-x)=182$

      $\Rightarrow 27x-x{}^2=182$

      $\Rightarrow x{}^2-27x+182=0$ 

              $\Rightarrow x{}^2-27x+182=0$ 

              $\Rightarrow x(x-14)13(x-14)=0$ 

      $\Rightarrow (x-14)(x-13)=0$ 

              $\therefore (x-14)(x-13)=0$ 

       Therefore, the first number is equal to $14\text{ or 13}$ 

       And, second number is $=27x=27-14=13$ or Second number $=27-13=14$ 

       Therefore, two numbers are $13\text{ and }14$ 

3. Find two consecutive positive integers, the sum of whose squares is $365$.

Ans: 

Let first number be x and let second number be $(x+1)$ 

According to given condition

$x^2+{(x+1)}^2=365\{{(a+b)}^2=a^2+b{}^2+2ab\}$ 

$\Rightarrow x^2+x^2+1+2x=365$ 

$\Rightarrow 2x^2+2x-364=0$ 

Dividing equation by $2$ 

$\Rightarrow x^2+x-182=0$ 

$\Rightarrow x^2+14x-13x-182=0$ 

     $\Rightarrow x(x+14)-13(x+14)=0$ 

   $\Rightarrow (x+14)(x-13)=0$ 

    $\therefore x=13,-14$ 

Therefore, first number $=13$ {We discard $-14$ because it is negative number)
Second number $=x+1=13+1=14$ 

Therefore, two consecutive positive integers are $13\text{and 14}$ whose sum of squares is equal to $365$.

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

Ans: 

Let base of triangle be x cm and let altitude of triangle be (x−7) cm

It is given that hypotenuse of triangle is 13 cm

According to Pythagoras Theorem,

$132=x^2+{(x-7)}^2{(a+b)}^2=a^2+b^2+2ab$ 

$\Rightarrow 169=x^2+x^2+49-14x$ 

$\Rightarrow 169=2x^2-14x+49$ 

$\Rightarrow 2x^2-14x120=0$ 

Dividing equation by $2$ 

$\Rightarrow x^2-7x60=0$ 

$\Rightarrow x^2-12x+5x60=0$ 

$\Rightarrow x(x-12)+5(x-12)=0$ 

$\Rightarrow (x-12)(x+5)$ 

$\therefore x=-5,12$ 

We discard $x=-5$ because the length of the side of the triangle cannot be negative.

Therefore, the base of triangle $=12$cm

Altitude of triangle $=(x-7)=12-7=5$ cm

5. 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 Rs.$90$ , find the number of articles produced and the cost of each article.

Ans:  

Let cost of production of each article be Rs $x$ 

We are given total cost of production on that particular day $=Rs90$ 

Therefore, total number of articles produced that day $={\displaystyle \frac{90}{x}}$ 

According to the given conditions,

$x=2\left({\displaystyle \frac{90}{x}}\right)+3$ 

$\Rightarrow x={\displaystyle \frac{180}{x}}+3$ 

$\Rightarrow x={\displaystyle \frac{180+3x}{x}}$ 

$\Rightarrow x^2=180+3x$ 

$\Rightarrow x^2-3x180=0$ 

$\Rightarrow x^2-15x+12x180=0$ 

$\Rightarrow x(x-15)+12(x-15)=0$ 

$\Rightarrow (x-15)(x+12)=0$     

    $\therefore x=15,-12$      

Cost cannot be in negative; therefore, we discard $x=-12$ 

Therefore$,x=Rs15$which is the cost of production of each article.

Number of articles produced on that particular day      $={\displaystyle \frac{90}{15}}=6$ 

6. 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: 

Let Shefali's marks in Mathematics $=x$ 

Let Shefali's marks in English $=30-x$ 

If, she had got 2 marks more in Mathematics, her marks would be $=x+2$ 

If, she had got 3 marks less in English, her marks in English would be $=30x-3=27-x$ 

According to given condition:

$(x+2)(27-x)=210$ 

$\Rightarrow 27x-x^2+54-2x=210$ 

$\Rightarrow x^2-25x+156=0$ 

Comparing quadratic equation $x^2-25x+156=0$with general form$a{x}^2+bx+c=0$,

We get $a=1,b=-25andc=156$ 

Applying Quadratic Formula$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ 

$x={\displaystyle \frac{25\pm \sqrt{{\left(25\right)}^2-4\left(1\right)\left(156\right)}}{2\times 1}}$ 

$\Rightarrow x={\displaystyle \frac{25\pm \sqrt{625-624}}{2}}$ 

$x={\displaystyle \frac{25\pm \sqrt{1}}{2}}$ 

$\Rightarrow x={\displaystyle \frac{25+1}{2}},{\displaystyle \frac{25-1}{2}}$ 

$\therefore x=13,12$ 

Therefore, Shefali's marks in Mathematics $=13or12$ 

Shefali's marks in English $=30x=3013=17$ 

Or Shefali's marks in English $=30x=3012=18$ 

Therefore, her marks in Mathematics and English are $(13,17)or(12,18).$ 

7. The diagonal of a rectangular field is $60$ meters more than the shorter side. If, the longer side is$30$ meters more than the shorter side, find the sides of the field.

Ans:

Let shorter side of rectangle $=x$meters

Let diagonal of rectangle $=(x+60)$meters 

Let longer side of rectangle $=(x+30)$meters 

According to Pythagoras theorem,

${(x+60)}^2={(x+30)}^2+x^2$ 

$\Rightarrow x^2+3600+120x=x^2+900+60x+x^2$ 

$\Rightarrow x^2-60x2700=0$ 

$x^2-60x2700=0$ 

Comparing equation $x^2-60x2700=0$with standard form$a{x}^2+bx+c=0$, 

We get $a=1,b=-60andc=-2700$ 

Applying quadratic formula$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ 

$x={\displaystyle \frac{60\pm \sqrt{{\left(60\right)}^2-4\left(1\right)(-2700)}}{2\times 1}}$ 

$\Rightarrow x={\displaystyle \frac{60\pm \sqrt{3600+10800}}{2}}$ 

$\Rightarrow x={\displaystyle \frac{60\pm \sqrt{14400}}{2}}={\displaystyle \frac{60\pm 120}{2}}$ 

$\Rightarrow x={\displaystyle \frac{60+120}{2}},{\displaystyle \frac{60-120}{2}}$ 

$\therefore x=90,-30$ 

We ignore $-30$ . Since length cannot be negative. 

Therefore, $x=90$ which means length of shorter side $=90$ meters 

And length of longer side $=x+30=90+30=120$meters 

Therefore, length of sides is $90\text{ and }120$ in meters.

8. 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: 

Let smaller number $=x$and let larger number $=y$ 

According to condition: 

$y^2-x^2=180\dots \left(1\right)$ 

Also, we are given that square of smaller number is $8$ times the larger number. 

$\Rightarrow x^2=8y\dots \left(2\right)$ 

Putting equation (2) in (1), we get

 $y^2-8y=180$ 

$\Rightarrow y^2-8y180=0$ Comparing equation $y^2-8y180=0$with general form $a{y}^2+by+c=0$, 

We get $a=1,b=-8\text{ and }c=-180$ 

Using quadratic formula $y={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ 

$y={\displaystyle \frac{8\pm \sqrt{{(-8)}^2-4\left(1\right)(-180)}}{2\times 1}}$ 

$\Rightarrow y={\displaystyle \frac{8\pm \sqrt{64+720}}{2}}$ 

$\Rightarrow y={\displaystyle \frac{8\pm \sqrt{784}}{2}}={\displaystyle \frac{8\pm 28}{2}}$ 

$\Rightarrow y={\displaystyle \frac{8+28}{2}},{\displaystyle \frac{8-28}{2}}$ 

$\therefore y=18,-10$ 

Using equation (2) to find smaller number:

$x{}^2=8y$ 

$\Rightarrow x^2=8y=8\times 18=144$ 

     $\Rightarrow x=\pm 12$ 

     And,$x^2=8y=8\times -10=-80$  {No real solution for $x$} 

  Therefore, two numbers are $(12,18)or(-12,18)$ 

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

Ans: 

Let the speed of the train = x km/hr 

If, speed had been 5km/hr more, train would have taken 1 hour less. 

So, according to this condition

${\displaystyle \frac{360}{x}}={\displaystyle \frac{360}{x+5}}+1$ 

$\Rightarrow 360({\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{x+5}})=1$ 

$\Rightarrow 360\left({\displaystyle \frac{x+5-x}{x(x+5)}}\right)=1$ 

$\Rightarrow 360\times 5=x^2+5x$ 

$\Rightarrow x^2+5x1800=0$