ML Aggarwal Solution Class 10 Chapter 5 Quadratic Equations in One Variable Test

Test

Question 1

(i) x² + 6x – 16 = 0

(ii) 3x² + 11x + 10 = 0

Sol :

(i)

x² + 6x – 16 = 0

⇒ x² + 8x – 2x – 16 = 0

⇒ x (x + 8) – 2 (x + 8) = 0

⇒(x+8)(x-2)=0

Either x+8=0, then x=-8

or x-2=0, then x=2

Hence x=-8,2

(ii) $3 x^{2}+11 x+10=0$

$3 x^{2}+11 x+10=0$

$\Rightarrow 3 x^{2}+6 x+5 x+10=0$

⇒3x(x+2)+5(x+2)=0

⇒ (x+2)(3 x+5)=0

Either x+2=0, then x=-2 

or 3x+5=0, then

3 x=-5 

⇒$x=\frac{-5}{3}$

Hence x=-2,$\frac{-5}{3}$

Question 2

(i) $2 x^{2}+a x-a^{2}=0$

(ii) $\sqrt{3 x^{2}+10 x+7 \sqrt{3}=0}$

Sol :

(i) $2 x^{2}+a x-a^{2}=0$

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

⇒2x(x+a)-a(x+a)=0

⇒(x+a)(2 x-a)=0

Either $x+a=0,$ then $x=-a$

or 2x-a=0, then

2x=a 

$\Rightarrow x=\frac{a}{2}$

Hence x=-a, $\frac{a}{2}$

(ii) $\sqrt{3} x^{2}+10 x+7 \sqrt{3}=0$

$\sqrt{3} x^{2}+10 x+7 \sqrt{3}=0$

$\Rightarrow \quad \sqrt{3} x^{2}+3 x+7 x+7 \sqrt{3}=0$

$\Rightarrow \quad \sqrt{3} x(x+\sqrt{3})+7(x+\sqrt{3})=0$

$\Rightarrow \quad(x+\sqrt{3})(\sqrt{3} x+7)=0$

Either $x+\sqrt{3}=0,$ then $x=-\sqrt{3}$

or $\sqrt{3} x+7=0,$ then $\sqrt{3} x=-7$

$\Rightarrow x=\frac{-7}{\sqrt{3}}$

$x=\frac{-7 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{-7 \sqrt{3}}{3}$

Hence $x=-\sqrt{3}, \frac{-7 \sqrt{3}}{3}$

Question 3

(i) x(x + 1) + (x + 2)(x + 3) = 42

(ii) $\frac{6}{x}-\frac{2}{x-1}=\frac{1}{x-2}$

Sol :

$\Rightarrow 2 x^{2}+6 x+6-42=0$

$\Rightarrow 2 x^{2}+6 x-36=0$

$\Rightarrow x^{2}+3 x-18=0$

$\Rightarrow x^{2}+6 x-3 x-18=0$

⇒x(x+6)-3(x+6)=0

⇒(x+6)(x-3)=0

Either x+6=0, then x=-6

or x-3=0, then x=3 

Hence  x=-6,3 

(ii) $\frac{6}{x}-\frac{2}{x-1}=\frac{1}{x-2}$

$\Rightarrow \quad \frac{6 x-6-2 x}{x(x-1)}=\frac{1}{x-2}$

$\Rightarrow \quad \frac{4 x-6}{x^{2}-x}=\frac{1}{x-2}$

$\Rightarrow \quad(4 x-6)(x-2)=x^{2}-x$

$\Rightarrow \quad 4 x^{2}-8 x-6 x+12=x^{2}-x$

$\Rightarrow \quad 4 x^{2}-14 x+12-x^{2}+x=0$

$\Rightarrow \quad 3 x^{2}-13 x+12=0$

$\Rightarrow \quad 3 x^{2}-9 x-4 x+12=0$

⇒3x(x-3)-4(x-3)=0

⇒(x-3)(3 x-4)=0

Either x-3=0, then x=3

or 3x-4=0, then

3x=4 

$\Rightarrow x=\frac{4}{3}$

Hence x=3, $\frac{4}{3}$

Question 4

(i) $\sqrt{x+15}=x+3$

(ii) $\sqrt{3 x^{2}-2 x-1}=2 x-2$

Sol :

(i) $\sqrt{x+15}=x+3$

Squaring on both sides

$x+15=(x+3)^{2}$

$\Rightarrow x+15=x^{2}+6 x+9$

$\Rightarrow \quad x^{2}+6 x+9-x-15=0$

$\Rightarrow \quad x^{2}+5 x-6=0$

$\Rightarrow \quad x^{2}+6 x-x-6=0$

⇒ x(x+6)-1(x+6)=0

⇒ (x+6)(x-1)=0

Either x+6=0 then x=-6

or x-1=0, then x=1

∴x=-6,1

Check :

(i) If x=-6 then

L.H.S. $=\sqrt{x+15}$

$=\sqrt{-6+15}=\sqrt{9}=3$

R.H.S. =x+3=-6+3=-3

∵ L.H.S≠R.H.S.

∴ x=-6 is not a root

(ii) If x=1, then

L.H.S. $=\sqrt{x+15}$

$=\sqrt{1+15}=\sqrt{16}=4$

R.H.S. =x+3=1+3=4

∵ L.H.S=R.H.S

∴ x=1 is a root of this equation 

Hence x=1

(ii) $\sqrt{3 x^{2}-2 x-1}=2 x-2$

$\sqrt{3 x^{2}-2 x-1}=2 x-2$

Squaring both sides 

$3x^{2}-2 x-1=(2 x-2)^{2}$

$\Rightarrow \quad 3x^{2}-2 x-1=4 x^{2}-8 x+4$

$\Rightarrow \quad 4 x^{2}-8 x+4-3 x^{2}+2 x+1=0$

$\Rightarrow \quad x^{2}-6 x+5=0$

$\Rightarrow \quad x^{2}-5 x-x+5=0$

⇒x(x-5)-1(x-5)=0

⇒(x-5)(x-1)=0

Either x-5=0, then x=5 or x-1=0, then x=1

Check. (i) If x=5,  then

L.H.S. $=\sqrt{3 x^{2}-2 x-1}$

$=\sqrt{3 \times(5)^{2}-2 \times 5-1}$

$=\sqrt{3 \times 25-10-1}=\sqrt{75-10-1}$

$=\sqrt{64}=8$

R.H.S=2x-2=2×5-2

=10-2=8

∵L.H.S=R.H.S

∴x=5 is a root

L.H.S. $=\sqrt{3 x^{2}-2 x-1}$

$=\sqrt{3(1)^{2}-2(1)-1}$

$=\sqrt{3 \times 1-2-1}=\sqrt{3-2-1}=0$

R.H.S. =2x-2=2×1-2=2-2=0

∴ L.H.S.=R.H.S.

Hence x=5,1

Question 5

(i) 2x² – 3x – 1 = 0

(ii) $x\left(3 x+\frac{1}{2}\right)=6$

Sol :

(i) $2x^{2}-3 x-1=0$

Here a=2, b=-3, c=-1

$\mathrm{D}=b^{2}-4 a c=(-3)^{2}-4 \times 2 \times(-1)$

=9+8=17

$x=\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}$

$=\frac{-(-3) \pm \sqrt{17}}{2 \times 2}$

$=\frac{3 \pm \sqrt{17}}{4}$

$\therefore x=\frac{3+\sqrt{17}}{4}, \frac{3-\sqrt{17}}{4}$

(ii) $x\left(3 x+\frac{1}{2}\right)=6$

$3 x^{2}+\frac{x}{2}=6$

$\Rightarrow 6 x^{2}+x=12$

$\Rightarrow 6 x^{2}+x-12=0$

Here a=6, b=1, c=-12

$D=b^{2}-4 a c=(1)^{2}-4 \times 6 \times(-12)$

=1+288=289

$\therefore x=\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}$

$=\frac{-1 \pm \sqrt{289}}{2 \times 6}$

$=\frac{-1 \pm 17}{12}$

∴$x_{1}=\frac{-1+17}{12}=\frac{16}{12}=\frac{4}{3}$

$x_{2}=\frac{-1-17}{12}=\frac{-18}{12}=-\frac{3}{2}$

∴$x=\frac{4}{3},-\frac{3}{2}$

Question 6

(i) $\frac{2 x+5}{3 x+4}=\frac{x+1}{x+3}$

(ii) $\frac{2}{x+2}-\frac{1}{x+1}=\frac{4}{x+4}-\frac{3}{x+3}$

Sol :

(i) $\frac{2 x+5}{3 x+4}=\frac{x+1}{x+3}$

$(2 x+5)(x+3)=(x+1)(3 x+4)$

$2 x^{2}+6 x+5 x+15=3 x^{2}+4 x+3 x+4$

$\Rightarrow \quad 3 x^{2}+7 x+4-2 x^{2}-11 x-15=0$

$\Rightarrow \quad x^{2}-4 x-11=0$

Here a=1, b=-4, c=-11

$\mathrm{D}=b^{2}-4 a c=(-4)^{2}-4 \times 1 \times(-11)$

=16+44=60

∵$x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-(-4) \pm \sqrt{60}}{2 \times 1}=\frac{4 \pm \sqrt{4 \times 15}}{2}$

$=\frac{4 \pm 2 \sqrt{15}}{2}=2 \pm \sqrt{15}$

∴ $x=2+\sqrt{15}, 2 \div \sqrt{15}$

(ii)$\frac{2}{x+2}-\frac{1}{x+1}=\frac{4}{x+4}-\frac{3}{x+3}$

$\frac{2}{x+2}-\frac{1}{x+1}=\frac{4}{x+4}-\frac{3}{x+3}$

$\frac{2 x+2-x-2}{(x+2)(x+1)}=\frac{4 x+12-3 x-12}{(x+4)(x+3)}$

$\Rightarrow \quad \frac{x}{(x+2)(x+1)}=\frac{x}{(x+4)(x+3)}$

$\Rightarrow \quad \frac{1}{(x+2)(x+1)}=\frac{1}{(x+4)(x+3)}$ [dividing by x if x≠0]

$\Rightarrow \quad \frac{1}{x^{2}+3 x+2}=\frac{1}{x^{2}+7 x+12}$

$\Rightarrow \quad x^{2}+7 x+12-x^{2}-3 x-2=0$

$\Rightarrow \quad 4 x+10 \Rightarrow 2 x+5=0$

$\Rightarrow \quad 2 x=-5 \Rightarrow x=\frac{-5}{2}$

If x=0 , then

$\frac{0}{(x+2)(x+1)}=\frac{0}{(x+4)(x+3)}$

Which is correct

Hence $x=0, \frac{-5}{2}$

Question 7

(i) $\frac{3 x-4}{7}+\frac{7}{3 x-4}=\frac{5}{2}, x \neq \frac{4}{3}$

(ii) $\frac{4}{x}-3=\frac{5}{2 x+3}, x \neq 0,-\frac{3}{2}$

Sol :

(i) $\frac{3 x-4}{7}+\frac{7}{3 x-4}=\frac{5}{2}, x \neq \frac{4}{3}$

let $\frac{3 x-4}{7}=y,$ then

$y+\frac{1}{y}=\frac{5}{2}$

$\Rightarrow 2 y^{2}+2=5 y$

$\Rightarrow 2 y^{2}-5 y+2=0$

$\Rightarrow 2 y^{2}-y-4 y+2=0$

⇒ y(2y-1)-2(2y-1)=0

⇒ (2y-1)(y-2)=0

Either 2y-1=0, then

2y=1 

$\Rightarrow y=\frac{1}{2}$

or y-2=0, then y=2

When $y=\frac{1}{2},$ then

$\frac{3 x-4}{7}=\frac{1}{2}$

⇒ 6x-8=7

⇒  6x=7+8 

⇒ 6x=15

⇒ $x=\frac{15}{6}=\frac{5}{2}$

y=2 then

$\frac{3 x-4}{7}=\frac{2}{1} \Rightarrow 3 x-4=14$

⇒ 3x=14+4=18

$\Rightarrow x=\frac{18}{3}=6$

$\therefore x=6, \frac{5}{2}$

(ii) $\frac{4}{x}-3=\frac{5}{2 x+3}, x \neq 0,-\frac{3}{2}$

$\Rightarrow(4-3 x)(2 x+3)=5 x$

$\Rightarrow 8x+12-6 x^{2}-9 x-5 x=0$

$\Rightarrow-6 x^{2}-6 x+12=0$

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

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

⇒ x(x+2)-1(x+2)=0

⇒(x+2)(x-1)=0

Either x+2=0, then x=-2

or x-1=0, then x=1

∴x=1,-2

Question 8

(i)x² + (4 – 3a)x – 12a = 0

(ii)10ax² – 6x + 15ax – 9 = 0,a≠0

Sol :

(i)x² + (4 – 3a)x – 12a = 0

Here a = 1, b = 4 – 3a, c = -12a

$\therefore d=b^{2}-4 a c$

$\quad=(4-3 a)^{2}-4 \times 1 \times(-12 a)$

$\quad=16-24 a+9 a^{2}+48 a$

$\quad=16+24 a+9 a^{2}=(4+3 a)^{2}$

$\therefore \quad x=\frac{-b \pm \sqrt{d}}{2 \mathrm{~A}}$

$=\frac{-(4-3 a) \pm \sqrt{(4+3 a)^{2}}}{2 \times 1}$

$=\frac{3 a-4 \pm 3 a+4}{2}$

$\therefore x_{1}=\frac{3 a-4+3 a+4}{2}$

$=\frac{6 a}{2}=3 a$

and $x_{2}=\frac{3 a-4-3 a-4}{2}$

$=\frac{-8}{2}=-4$

∴Roots are 3a, -4

(ii) $10 a x^{2}-(6-15 a) x-9=0$

Here $a=10 a, b=-(6-15 a), c=-9$

$d=b^{2}-4 a c$

$=[-(6-15 a)]^{2}-4 \times 10 a(-9)$

$=36-180 a+225 a^{2}+360 a$

$=36+180 a+225 a^{2}=(6+15 a)^{2}$

∵$x=\frac{-b \pm \sqrt{d}}{2 a}$

$=\frac{-[-(6-15 a)] \pm \sqrt{(6+15 a)^{2}}}{2 \times 10 a}$

$=\frac{(6-15 a) \pm(6+15 a)}{20 a}$

∵$x_{1}=\frac{6-15 a+6+15 a}{20 a}$

$=\frac{12}{20 a}=\frac{3}{5 a}$

$x_{2}=\frac{6-15 a-6-15 a}{20 a}$

$=\frac{-30 a}{20 a}=\frac{-3}{2}$

Hence $x=\frac{3}{5 a}, \frac{-3}{2}$

Question 9

Solve for x using the quadratic formula. Write your answer correct to two significant figures: (x – 1)² – 3x + 4 = 0. (2014)

Sol :

(x – 1)² – 3x + 4 = 0

x² + 1 – 2x – 3x + 4 = 0

$x^{2}-5 x+5=0$

Here a=1, b=-5 and c=5

$x=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(5)}}{2}$

$=\frac{5 \pm \sqrt{25-20}}{2}=\frac{5 \pm \sqrt{5}}{2}$

$=\frac{5+2.236}{2}$ or $\frac{5-2.236}{2}$

$=\frac{7.236}{2}$ or $\frac{2.764}{2}$

=3.618 or 1.382

∴x=3.6, 1.4

Question 10

Discuss the nature of the roots of the following equations:

(i) 3x² – 7x + 8 = 0

(ii) $x^{2}-\frac{1}{2} x-4=0$

(iii) $5 x^{2}-6 \sqrt{5} x+9=0$

(iv) $\sqrt{3} x^{2}-2 x-\sqrt{3}=0$

Sol :

(i) $3 x^{2}-7 x+8=0$

Here a=3, b=-7, c=8

$\therefore D=b^{2}-4 a c=(-7)^{2}-4 \times 3 \times 8$

=49-96=-47

∵ D<0

∴ Roots are not real

(ii) $x^{2}-\frac{1}{2} x-4=0$

Here  $a=1, b=\frac{1}{2}, c=-4$

$\therefore \mathrm{D}=b^{2}-4 a c=\left(\frac{1}{2}\right)^{2}=4 \times 1 \times(-4)$

$=\frac{1}{4}+16=\frac{65}{4}$

∵D>0

∴Roots are real and distinct

(iii) $5 x^{2}-6 \sqrt{5} x+9=0$

Here a=5, $b=-6 \sqrt{5}$, c=9

∴$\mathrm{D}=b^{2}-4 a c$

$=(-6 \sqrt{5})^{2}-4 \times 5 \times 9=180-180=0$

∴D=0

∴Roots are real and equal

(iv) $\sqrt{3} x^{2}-2 x-\sqrt{3}=0$

Here $a=\sqrt{3}, b=-2, c=-\sqrt{3}$

∴$D=b^{2}-4 a c$

$=(-2)^{2}-4 \times \sqrt{3} \times(-\sqrt{3})=4+12=16$

∵D>0

∴Roots are real and distinct

Question 11

Find the values of k so that the quadratic equation (4 – k) x² + 2 (k + 2) x + (8k + 1) = 0 has equal roots.

Sol :

(4 – k) x² + 2 (k + 2) x + (8k + 1) = 0

Here a = (4 – k), b = 2 (k + 2), c = 8k + 1

$\therefore \mathrm{D}=b^{2}-4 a c$

$=[2(k+2)]^{2}-4 \times(4-k)(8 k+1)=0$

$=4(k+2)^{2}-4\left(32 k+4-8 k^{2}-k\right)$

$=4\left(k^{2}+4 k+4\right)-4\left(32 k+4-8 k^{2}-k\right)$

$=4 k^{2}+16 k+16-128 k-16+32 k^{2}+4 k$

$=36 k^{2}-108 k=36 k(k-3)$

∵Roots are equal

∴D=0

⇒36k(k-3)=0

⇒k(k-3)=0

Either k=0

or k – 3 = 0, then k= 3

k = 0, 3

Question 12

Find the values of m so that the quadratic equation 3x² – 5x – 2m = 0 has two distinct real roots.

Sol :

3x² – 5x – 2m = 0

Here a = 3, b = -5, c = -2m

∴$\mathrm{D}=b^{2}-4 a c$

$=(-5)^{2}-4 \times 3 \times(-2 m)=25+24 m$

∵D>0

∴25+24m>0

25m>-25

$m>-\frac{25}{24}$

Question 13

Find the value(s) of k for which each of the following quadratic equation has equal roots:

(i)3kx² = 4(kx – 1)

(ii)(k + 4)x² + (k + 1)x + 1 =0

Also, find the roots for that value (s) of k in each case.

Sol :

(i)3kx² = 4(kx – 1)

⇒ 3kx² = 4kx – 4

⇒ 3kx² – 4kx + 4 = 0

Here a=3 k, b=-4 k, c=4

∴$\mathrm{D}=b^{2}-4 a c=(-4 k)^{2}-4 \times 3 k \times 4$

$=16 k^{2}-48 k$

∴Roots are equal

∴D=0

⇒ $\Rightarrow 16 k^{2}-48 k=0$

$\Rightarrow k^{2}-3 k=0$

⇒ k(k-3)=0$

Either k=0

or k-3=0 then k=3

∴$x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{-b}{2 a}$ (∵D=0)

$\frac{4 k}{2 \times 3 k}=\frac{4 \times 3}{2 \times 3 \times 3}=\frac{12}{18}=\frac{2}{3}$

∴$x=\frac{2}{3}, \frac{2}{3}$

(ii)

$(k+4) x^{2}+(k+1) x+1=0$

Here a=k+4, b=k+1, c=1

∴$\mathrm{D}=b^{2}-4 a c=(k+1)^{2}-4 \times(k+4) \times 1$

$=k^{2}+2 k+1-4 k-16=k^{2}-2 k-15$

∵Root are equal

∵$k^{2}-2 k-15=0$

$\Rightarrow k^{2}-5 k+3 k-15=0$

⇒k(k-5)+3(k-5)=0

⇒(k-5)(k+3)=0

Either k-5=0, then k=5

or k+3=0, then k=-3

(a) When k=5, then

$x=\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}=\frac{-b}{2 a}=\frac{-k-1}{2(k+4)}=\frac{-5-1}{2(5+4)}$

$=\frac{-6}{18}=\frac{-1}{3}$

∴$x=\frac{-1}{3}, \frac{-1}{3}$

(b) When k=-3, then

$x=\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}=\frac{-b}{2 a}=\frac{-k-1}{2(k+4)}$

$=\frac{(-3)-1}{2(-3+4)}=\frac{2}{2 \times 1}=1$

∴ x=1, 1

Question 14

Find two natural numbers which differ by 3 and whose squares have the sum 117.

Sol :

Let first natural number = x

then second natural number = x + 3

According to the condition :

x² + (x + 3)² = 117

$\Rightarrow \quad x^{2}+x^{2}+6 x+9=117$

$\Rightarrow \quad 2 x^{2}+6 x+9-117=0$

$\Rightarrow \quad 2 x^{2}+6 x-108=0$

$\Rightarrow \quad x^{2}+3 x-54=0$ (dividing by 2)

$\Rightarrow \quad x^{2}+9 x-6 x-54=0$

⇒ x(x+9)-6(x+9)=0

⇒ (x+9)(x-6)=0

Either x+9=0, then x=-9, but it is not a natural number.

or x-6=0, then x=6

∴First natural number=6

and second number=6+3=9

Question 15

Divide 16 into two parts such that the twice the square of the larger part exceeds the square of the smaller part by 164.

Sol :

Let larger part = x

then smaller part = 16 – x

(∵ sum = 16)

According to the condition

$2 x^{2}-(16-x)^{2}=164$

$\Rightarrow 2 x^{2}-\left(256-32 x+x^{2}\right)=164$

$\Rightarrow 2 x^{2}-256+32 x-x^{2}=164$

$\Rightarrow x^{2}+32 x-256-164=0$

$\Rightarrow x^{2}+32 x-420=0$

$\Rightarrow x^{2}+42 x-10 x-420=0$

⇒ x(x+42)-10(x+42)=0

⇒ (x+42)(x-10)=0

Either x+42=0, then x=-42, but it is not possible

or x-10=0, then x=10

∴Largest part=10

and smaller part=16-10=6

Question 16

Two natural numbers are in the ratio 3 : 4. Find the numbers if the difference between their squares is 175.

Sol :

Ratio in two natural numbers = 3 : 4

Let the numbers be 3x and 4x

According to the condition,

$(4 x)^{2}-(3 x)^{2}=175$

$\Rightarrow 16 x^{2}-9 x^{2}=175$

$\Rightarrow 7 x^{2}=175$

$\Rightarrow x^{2}=\frac{175}{7}=25=(\pm 5)^{2}$

∴x=5, -5

But x=-5 is not a natural number

∴x=5

∴Natural numbers are 3x, 4x

=3×5, 4×5=15 ,20

Question 17

Two squares have sides A cm and (x + 4) cm. The sum of their areas is 656 sq. cm.Express this as an algebraic equation and solve it to find the sides of the squares.

Sol :

Side of first square = x cm .

and side of second square = (x + 4) cm

Now according to the condition,

$(x)^{2}+(x+4)^{2}=656$

$\Rightarrow x^{2}+ x^{2}+8 x+16-656$

$\Rightarrow 2 x^{2}+8 x+16-656=0$

$\Rightarrow 2 x^{2}+8 x-640=0$

$\Rightarrow x^{2}+4 x-320=0$ (Dividing by 2)

$\Rightarrow x^{2}+20 x-16 x-320=0$

$\Rightarrow x(x+20)-16(x+20)=0$

$\Rightarrow(x+20)(x-16)=0$

Either x+20=0, then x=-20 , but it not possible as it is negative

or x – 16 = 0 then x = 16

Side of first square = 16 cm

and side of second square = 16 + 4 – 4 = 20 cm

Question 18

The length of a rectangular garden is 12 m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.

Sol :

Let breadth = x m

then length = (x + 12) m

Area = l × b = x (x + 12) m²

and perimeter = 2 (l + b) = 2(x + 12 + x) = 2 (2x + 12) m

According to the condition.

x(x+12)=4 \times 2(2 x+12)

$\Rightarrow x^{2}+12 x=16 x+96$

$\Rightarrow x^{2}+12 x-16 x-96=0$

$\Rightarrow x^{2}-4 x-96=0$

$\Rightarrow x^{2}-12 x+8 x-96=0$

⇒x(x-12)+8(x-12)=0 

⇒(x-12)(x+8)=0

Either x-12=0, then x=12 or x+8=0, then x=-8, but it is not possible as it is in negative.

∴ Breadth =12 m

and length =12+12=24 m

Question 19

A farmer wishes to grow a 100 m² rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side fence. Find the dimensions of his garden.

Sol :

Area of rectangular garden = 100 cm²

Length of barbed wire = 30 m

Let the length of the side opposite to wall = x

and length of other each side $=\frac{30-x}{2}$

According to the condition, 

$\frac{x(30-x)}{2}=100$

$\Rightarrow x(30-x)=200$

$\Rightarrow 30 x-x^{2}=200$

$\Rightarrow x^{2}-30 x+200=0$

$\Rightarrow x^{2}-20 x-10 x+200=0$

⇒x(x-20)-10(x-20)=0

⇒(x-20)(x-10)=0

Either x-20=0, then x=20

or x-10=0, then x=10

(i) If x=20 then side opposite to the wall =20m

and other side $\frac{30-20}{2}=\frac{10}{2}=5 \mathrm{~m}$

(ii) If x=10, then side opposite to wall

and other side $e=\frac{30-10}{2}=\frac{20}{2}$

=10 m

∴Sides are 20m, 5, or 10m, 10m

Question 20

The hypotenuse of a right-angled triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.

Sol :

Let the length of shortest side = x m

Length of hypotenuse = 2x – 1

and third side = x + 1

Now according to the condition,

$(2 x-1)^{2}=(x)^{2}+(x+1)^{2}$

(By Pythagoras Theorem)

$\Rightarrow 4 x^{2}- 4 x+1=x^{2}+x^{2}+2 x+1$

$\Rightarrow 4 x^{2}-4 x+1-2 x^{2}-2 x-1=0$

$\Rightarrow 2 x^{2}-6 x=0$

$\Rightarrow x^{2}-3 x=0$  (Dividing by 2)

$\Rightarrow x(x-3)=0$

Either x=0, but it is not possible

or x-3=0, then x=3

∴Shortest side=3m

Hypotenuse=2×3-1=6-1=5

Third side =x+1=3+1=4

Hence sides are 3.4,5 (in m)

Question 21

A wire ; 112 cm long is bent to form a right angled triangle. If the hypotenuse is 50 cm long, find the area of the triangle.

Sol :

Perimeter of a right angled triangle = 112 cm

Hypotenuse = 50 cm

∴ Sum of other two sides = 112 – 50 = 62 cm

Let the length of first side = x

and length of other side = 62 – x

$(x)^{2}+(62-x)^{2}=(50)^{2}$

(By Pythagoras Theorem)

$\Rightarrow x^{2}+3844-124 x+x^{2}=2500$

$\Rightarrow 2 x^{2}-124 x+3844-2500=0$

$\Rightarrow 2 x^{2}-124 x+1344=0$

$\Rightarrow x^{2}-62 x+672=0$

$\Rightarrow x^{2}-48 x-14 x+672=0$ (Dividing by 2)

⇒x(x-48) -14(x-48)=0

⇒(x-48)(x-14)=0

Either x-48=0, then x=48

or x-14=0, then x=14

(i) If x=48, then one side=48 cm

and other side=62-48=14 cm

(ii) If x=14, then one side =14 cm

and other side=62-14=48

Hence sides are 14 cm, 48 cm

Question 22

Car A travels x km for every litre of petrol, while car B travels (x + 5) km for every litre of petrol.

(i) Write down the number of litres of petrol used by car A and car B in covering a distance of 400 km.

(ii) If car A uses 4 litres of petrol more than car B in covering 400 km. write down an equation, in A and solve it to determine the number of litres of petrol used by car B for the journey.

Sol :

Distance traveled by car A in one litre = x km

and distance traveled by car B in one litre = (x + 5) km

(i) Consumption of car A in covering 400 km

$=\frac{400}{x}$ litres and consumption of car B $=\frac{400}{x+5}$ litres.

(ii) According to the condition. 

$\frac{400}{x}-\frac{400}{x+5}=4$

$\frac{400(x+5-x)}{x(x+5)}=4$

$\Rightarrow \frac{400 \times 5}{x^{2}+5 x}=4$

$\Rightarrow \quad 2000=4 x^{2}+20 x$

$\Rightarrow \quad 4 x^{2}+20 x-2000=0$

$\Rightarrow \quad x^{2}+5 x-500=0$ (Dividing by 4)$

$\Rightarrow \quad x^{2}+25 x-20 x-500=0$

⇒ x(x+25)-20(x+25)=0

⇒ (x+25)(x-20)=0

Either x+25=0, then x=-25, but it is not possible as it is in negative. or x-20=0, then x=20

∴Petrol used by car B=20-4=16 litres

Question 23

The speed of a boat in still water is 11 km/ hr. It can go 12 km up-stream and return downstream to the original point in 2 hours 45 minutes. Find the speed of the stream

Sol :

Speed of a boat in still water = 11 km/hr

Let the speed of stream = x km/hr.

Distance covered = 12 km.

Time taken = 2 hours 45 minutes

$=2 \frac{3}{4}=\frac{11}{4}$ hours

Now according to the condition

$\frac{12}{11-x}+\frac{12}{11+x}=\frac{11}{4}$

$\Rightarrow \frac{12(11+x+11-x)}{(11-x)(11+x)}=\frac{11}{4}$

$\Rightarrow \frac{12 \times 22}{121-x^{2}}=\frac{11}{4}$

$\Rightarrow 1331-11 x^{2}=4 \times 12 \times 22=1056$

$\Rightarrow 1331-11 x^{2}=1056$

$\Rightarrow 1331-1056-11 x^{2}=0$

$\Rightarrow-11 x^{2}+275=0$

$\Rightarrow x^{2}-25=0$ (Dividing by -11)$

$\Rightarrow(x+5)(x-5)=0$

Either x+5=0, then x=-5, but it is not possible as it is negative 

or x-5=0, then x=5

Hence,speed of stream=5 km/hr

Question 24

By selling an article for Rs. 21, a trader loses as much per cent as the cost price of the article. Find the cost price.

Sol :

S.P. of an article = Rs. 21

Let cost price = Rs. x

Then loss = x%

$\therefore$ S.P. $=\frac{\text { C.P. }(100-\text { Loss } \%)}{100}$

$21=\frac{x(100-x)}{100}$

$2100=100 x-x^{2}$

$\Rightarrow x^{2}-100 x+2100=0$

$\Rightarrow x^{2}-30 x-70 x+2100=0$

⇒ x(x-30)-70(x-30)=0

⇒ (x-30)(x-70)=0

Either x-30=0, then x=30

or x-70=0, then x=70

∴Cost price=30 or 70

Question 25

A man spent Rs. 2800 on buying a number of plants priced at Rs x each. Because of the number involved, the supplier reduced the price of each plant by Rupee 1.The man finally paid Rs. 2730 and received 10 more plants. Find x.

Sol :

Amount spent = Rs. 2800

Price of each plant = Rs. x

Reduced price = Rs. (x – 1)

No. of plants in first case $=\frac{2800}{x}$

No. of plants received in second case $=\frac{2800}{x}+10$

Amount paid=Rs 2730

According to the condition

$\left(\frac{2800}{x}+10\right)(x-1)=2730$

$\Rightarrow \quad \frac{(2800+10 x)(x-1)}{x}=2730$

$\Rightarrow \quad(2800+10 x)(x-1)=2730 x$

$\Rightarrow \quad 2800 x-2800+10 x^{2}-10 x=2730 x$

$\Rightarrow 10 x^{2}+2800 x-10 x-2730 x-2800=0$

$\Rightarrow \quad 10 x^{2}+60 x-2800=0$

$\Rightarrow \quad x^{2}+6 x-280=0$ (Dividing by $\left.10\right)$

$\Rightarrow \quad x^{2}+20 x-14 x-280=0$

⇒x(x+20)-14(x+20)=0

⇒(x+20)(x-14)=0

Either x+20=0, then x=-20

, but it is not possible as it is negative

or x-14=0, then x=14

Question 26

Forty years hence, Mr. Pratap’s age will be the square of what it was 32 years ago. Find his present age.

Sol :

Let Partap’s present age = x years

40 years hence his age = x + 40

and 32 years ago his age = x – 32

According to the condition

$x+40=(x-32)^{2}$

$\Rightarrow x+40=x^{2}-64 x+1024$

$\Rightarrow x^{2}-64 x+1024-x-40=0$

$\Rightarrow x^{2}-65 x+984=0$

$\Rightarrow x^{2}-24 x-41 x+984=0$

⇒ x(x-24)-41(x-24)=0

⇒ (x-24)(x-41)=0

Either x-24=0, then x=24 but it is not possible as it is less than 32

or x-41=0, then x=41

Hence present age=41 year