RS Aggarwal Solutions Class 10 Ex 10C

Q1 : \( \frac{x – 4}{x – 5} + \frac{x – 6}{x – 7} = 3 \frac{1}{3} , x \neq 5, 7 \)

\( \frac{x – 4}{x – 5} + \frac{x – 6}{x – 7} = 3 \frac{1}{3} , x \neq 5, 7 \)

\( \Rightarrow \frac{(x – 4)(x – 7) + (x – 5)(x – 6)}{(x – 5)(x – 7)} = \frac{10}{3} \)

\( \Rightarrow \frac{x^{2} – 11x + 28 + x^{2} – 11x + 30}{x^{2} – 12x + 35} = \frac{10}{3} \)

\( \Rightarrow \frac{2x^{2} – 22x + 58}{x^{2} – 12x + 35} = \frac{10}{3} \)

\( \Rightarrow \frac{x^{2} – 11x + 29}{x^{2} – 12x + 35} = \frac{5}{3} \)

\( \Rightarrow \) 3x2 – 33x + 87 = 5x2 – 60x + 175

\( \Rightarrow \) 2x2 – 16x – 11x + 88 = 0

\( \Rightarrow 2x^{2} – 16x – 11x + 88 = 0 \)

\( \Rightarrow \) 2x(x – 8) – 11( x – 8) = 0

\( \Rightarrow \) (x – 8)(2x – 11) = 0

\( \Rightarrow \) x – 8 = 0 or 2x – 11 = 0

\( \Rightarrow x = 8 or x = \frac{11}{2} \)

Hence, 8 and (11/2) are the roots of the equation.

Q2: \( \frac{x – 1}{x – 2} + \frac{x – 3}{x – 4} = 3 \frac{1}{3}, x \neq 2, 4 \)

\( \frac{x – 1}{x – 2} + \frac{x – 3}{x – 4} = 3 \frac{1}{3}, x \neq 2, 4 \)

\( \Rightarrow \frac{(x – 1)(x – 4) + (x – 2)(x – 3)}{(x – 2)(x – 4)} = \frac{10}{3} \)

\( \Rightarrow \frac{x^{2} – 5x + 4 + x^{2} – 5x + 6}{x^{2} – 6x + 8} = \frac{10}{3} \)

\( \Rightarrow \frac{2x^{2} – 10x + 10}{x^{2} – 6x + 8} \frac{10}{3} \)

\( \Rightarrow \frac{x^{2} – 5x + 5}{x^{2} – 6x + 8} = \frac{5}{3} \)

\( \Rightarrow \) 3x2 – 15x + 15 = 5x2 – 30x + 40

\( \Rightarrow \) 2x2 – 15x + 25 = 0

\( \Rightarrow \) 2x2 – 10x – 5x + 25 = 0

\( \Rightarrow \) 2x(x – 5) – 5(x – 5) = 0

\( \Rightarrow \) x – 5 = 0 or 2x – 5 = 0

\( \Rightarrow x = 5 \;or \;x = \frac{5}{2} \)

Hence, 5 and (5/2) are the roots of the given equation.

Q3: \( \frac{1}{x – 2} + \frac{2}{x – 1} = \frac{6}{x} \)

Given:

\( \frac{1}{x – 2} + \frac{2}{x – 1} = \frac{6}{x} \)

\( \Rightarrow \frac{(x – 1) + 2(x – 2)}{(x – 1)(x – 2)} = \frac{6}{2} \)

\( \Rightarrow \frac{3x – 5}{x^{2} – 3x + 2} = \frac{6}{x} \)

\( \Rightarrow \) 3x2 – 5x = 6x2 – 18x + 12 [On cross multiplication]

\( \Rightarrow \) 3x2 – 13x + 12 = 0

\( \Rightarrow \) 3x2 – (9 + 4)x + 12 = 0

\( \Rightarrow \) 3x2 – 9x – 4x + 12 = 0

\( \Rightarrow \) 3x(x – 3) – 4(x – 3) = 0

\( \Rightarrow \) (x – 3)(3x – 4) = 0

\( \Rightarrow x= \frac{4}{3} or x = 3 \)

Hence, the roots of the equation are (4/3) and 3.

Q4: \( \frac{1}{x + 1} + \frac{2}{x + 2} = \frac{5}{x + 4} , x \neq -1, -2, -4 \)

\( \frac{1}{x + 1} + \frac{2}{x + 2} = \frac{5}{x + 4} , x \neq -1, -2, -4 \)

\( \Rightarrow \frac{x + 2+ 2x + 2}{(x +1)(x + 2)} = \frac{5}{x + 4} \)

\( \Rightarrow \frac{3x + 4}{x^{2} + 3x + 2} = \frac{5}{x + 4} \)

\( \Rightarrow \) (3x + 4)(x + 4) = 5(x2 + 3x + 2)

\( \Rightarrow \) 3x2 + 16x + 16 = 5x2 + 15x + 10

\( \Rightarrow \) 2x2 – x – 6 = 0

\( \Rightarrow \) 2x2 – 4x + 3x – 6 = 0

\( \Rightarrow \) 2x(x – 2) + 3(x – 2) = 0

\( \Rightarrow \) (x – 2)(2x + 3) = 0

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

Hence, 2, -(3/2) are the roots of the given equation.

Q5: \( 3 \frac{3x – 1}{2x + 3} – 2\frac{2x + 3}{3x – 1} = 5, x\neq \frac{1}{3}, \frac{-3}{2} \)

\( 3 \frac{3x – 1}{2x + 3} – 2\frac{2x + 3}{3x – 1} = 5, x\neq \frac{1}{3}, \frac{-3}{2} \)

\( \Rightarrow \frac{3(3x – 1)^{2} – 2(2x + 3)^{2}}{(2x + 3)(3x – 1)} = 5 \)

\( \Rightarrow \frac{3(9x^{2} – 6x + 1) – 2 (4x^{2} + 12x + 9}{6x^{2} + 7x – 3} = 5 \)

\( \Rightarrow \frac{27x^{2} – 18x + 3 – 8x^{2} – 24x – 18}{6x^{2} + 7x – 3} = 5 \)

\( \Rightarrow \frac{19x^{2} – 42x – 15}{6x^{2} + 7x – 3} = 5 \)

\( \Rightarrow \) 19x2 – 42x – 15 = 30x2 + 35x – 15

\( \Rightarrow \) 11x2 + 77x = 0

\( \Rightarrow \) 11x(x + 7) = 0

\( \Rightarrow \) x = 0 or x + 7 = 0

\( \Rightarrow \) x = 0 or x = -7

Hence, 0 and -7 are the roots of the given equation.

Q6: \( 3 \frac{7x + 1}{5x – 3} – 4 \frac{5x – 3}{7x + 1} = 11, x \neq \frac{3}{5} , \frac{-7}{7} \)

\( 3 \frac{7x + 1}{5x – 3} – 4 \frac{5x – 3}{7x + 1} = 11, x \neq \frac{3}{5} , \frac{-7}{7} \)

\( \Rightarrow \frac{3(7x + 1)^{2} – 4(5x – 3)^{2}}{(5x – 3)(7x + 1)} = 11 \)

\( \Rightarrow \frac{3(49x^{2} + 14x + 1) – 4(25x^{2} – 30x + 0)}{(5x – 3)(7x + 1)} = 11 \)

\( \Rightarrow \frac{147x^{2} + 42x + 3 – 100x^{2} + 120x – 36}{35x^{2} – 16x – 3} = 11 \)

\( \Rightarrow \frac{47x^{2} + 162x – 33}{35x^{2} – 16x – 3} = 11 \)

\( \Rightarrow \) 47x2 + 162x – 33 = 385x2 – 176x – 33

\( \Rightarrow \) 338x2 – 338x = 0

\( \Rightarrow \) 338x(x – 1) = 0

\( \Rightarrow \) x = 0 or x – 1 = 0

\( \Rightarrow \) x = 0 or x = 1

Hence, 0 and 1 are the roots of the given equation.

Q7: \( \frac{4x – 3}{2x + 1} – 10\frac{(2x + 1)}{(4x – 3)} = 3 \)

\( \frac{4x – 3}{2x + 1} – 10\frac{(2x + 1)}{(4x – 3)} = 3 \)

Putting, \( \frac{4x – 3}{2x + 1} = y \), we get

\( y – \frac{10}{y} = 3 \)

\( \Rightarrow \frac{y^{2} – 10}{y} = 3 \)

\( \Rightarrow \) y2 – 10 = 3y [On cross multiplication]

\( \Rightarrow \) y2 – 3y – 10 = 0

\( \Rightarrow \) y2 – (5 – 2)y – 10 = 0

\( \Rightarrow \) y2 – 5y + 2y – 10 = 0

\( \Rightarrow \) y(y – 5) + 2(y – 5) = 0

\( \Rightarrow \) (y – 5)(y + 2) = 0

\( \Rightarrow \) y – 5 = 0 or y + 2 = 0

\( \Rightarrow \) y = 5 or y = -2

Case 1:

If y = 5, we get

\( \frac{4x – 3}{2x + 1} = 5 \)

\( \Rightarrow \) 4x – 3 = 5(2x + 1) [On cross multiplying]

\( \Rightarrow \) 4x – 3 = 10x + 5

\( \Rightarrow \) -6x = 8

\( \Rightarrow x = \frac{-4}{3} \)

Case 2:

If y = -2, we get

\( \frac{4x – 3}{2x + 1} = -2 \)

\( \Rightarrow \) 4x – 3 = -2(2x + 1)

\( \Rightarrow \) 4x – 3 = -4x – 2

\( \Rightarrow \) 8x = 1

\( \Rightarrow x = \frac{1}{8} \)

Hence, the roots of the equation are \( \frac{-4}{3} and \frac{1}{8} \)

Q8: \( (\frac{x}{x + 1})^{2} – 5(\frac{x}{x + 1}) + 6 = 0 \)

Given:

\( (\frac{x}{x + 1})^{2} – 5(\frac{x}{x + 1}) + 6 = 0 \)

Putting \( \frac{x}{x + 1} = y \) , we get

Y2 – 5y + 6 = 0

\( \Rightarrow \) y2 – 5y + 6 = 0

\( \Rightarrow \) y2 – (3 + 2)y + 6 = 0

\( \Rightarrow \) y2 – 3y – 2y + 6 = 0

\( \Rightarrow \) y(y – 3) – 2( y – 3) = 0

\( \Rightarrow \) (y – 3)(y – 2) = 0

\( \Rightarrow \) y – 3 = 0 or y – 2 = 0

\( \Rightarrow \) y = 3 or y = 2

Case 1:

If y = 3, we get

\( \frac{x}{x + 1} = 3 \)

\( \Rightarrow \) x = 3(x + 1) [On cross multiplication]

\( \Rightarrow \) x = 3x + 3

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

Case 2:

If y = 2, we get

\( \frac{x}{x + 1} = 2 \)

\( \Rightarrow \) x = 2(x + 1)

\( \Rightarrow \) x = 2x + 2

\( \Rightarrow \) -x = 2

\( \Rightarrow \) x = -2

Hence, the roots of the equation are (-3/2) and -2.

Q9: \( \frac{a}{x – b} + \frac{b}{x – a} = 2 \)

\( \frac{a}{x – b} + \frac{b}{x – a} = 2 \)

\( \Rightarrow [\frac{a}{x – b} – 1 ] + [ \frac{b}{x – a} – 1] = 0 \)

\( \Rightarrow \frac{a – (x – b)}{x – b} + \frac{b – (x – a)}{x – a} = 0 \)

\( \Rightarrow \frac{a – x + b}{x – b} + \frac{a – x + b}{x – a} = 0 \)

\( \Rightarrow (a – x + b) [\frac{1}{x – b} + \frac{1}{x – a} ] = 0 \)

\( \Rightarrow (a – x + b) [ \frac{(x – a) + (x – b)}{(x – b)(x – a)} ] = 0 \)

\( \Rightarrow (a – x + b) [ \frac{2x – (a + b)}{(x – b)(x – a)} ] = 0 \)

\( \Rightarrow (a – x + b) [2x – (a + b)] = 0 \)

\( \Rightarrow \) a – x + b = 0 or 2x – (a + b) = 0

\( \Rightarrow x = a + b\; or\; x = \frac{a + b}{2} \)

Hence, the roots of the equation are \( (a + b) and \frac{a + b}{2} \)

Q11: \( \frac{a}{ax – 1} + \frac{b}{bx – 1} = (a + b) \)

\( \frac{a}{ax – 1} + \frac{b}{bx – 1} = (a + b) \)

\( \Rightarrow [ \frac{a}{ax – 1} – b] + [\frac{b}{bx – 1} – a] = 0 \)

\( \Rightarrow \frac{a – b(ax – 1)}{ax – 1} + \frac{b – a(bx – 1)}{bx – 1} = 0 \)

\( \Rightarrow \frac{a – abx + b}{ax – 1} + \frac{a – abz + b}{bx – 1} = 0 \)

\( \Rightarrow (a – abx + b) [\frac{1}{ax – 1} + \frac{1}{bx – 1}] = 0 \)

\( \Rightarrow (a – abx + b) [\frac{(bx – 1) + (ax – 1)}{(ax – 1)(bx – 1)}] = 0 \)

\( \Rightarrow (a – abx + b) [\frac{(a + b)x – 2}{(ax – 1)(bx – 1)}] = 0 \)

\( \Rightarrow (a – abx + b) [(a + b)x – 2] = 0 \)

\( \Rightarrow \) a – abx + b = 0 or (a + b)x – 2 = 0

\( \Rightarrow x = \frac{a + b}{ab} or x = \frac{2}{a + b} \)

Hence, the roots of the equation are, \( \frac{a + b}{ab} and \frac{2}{a + b} \)

Q12: 3(x + 2) + 3-x = 10

3(x + 2) + 3-x = 10

3x . 9 + \( \frac{1}{3^{x}} \) = 10

Let, 3x be equal to y

Hence,

\( 9y + \frac{1}{y} = 10 \)

\( \Rightarrow 9y^{2} + 1 = 10 y \)

\( \Rightarrow 9 y^{2} – 10 + 1 = 0 \)

\( \Rightarrow (y – 1) (9y – 1_ = 0 \)

\( \Rightarrow y – 1 = 0 or 9y – 1 = 0 \)

\( \Rightarrow y = 1 or y = \frac{1}{9} \)

\( \Rightarrow 3^{x} = 1 or 3^{x} = \frac{1}{9} \)

\( \Rightarrow 3^{x} = 3^{0} or 3^{x} = 3^{-2} \)

\( \Rightarrow x = 0 or x = -2 \)

Hence, 0 and -2 are the roots of the given equation.

Q13: 4(x + 1) + 4(1 – x) = 10

Given:

4(x + 1) + 4(1 – x) = 10

\( \Rightarrow 4^{x} . 4 + 4^{1} \frac{1}{4^{x}} = 10 \)

Let 4x be y

Hence,

\( 4y + \frac{4}{y} = 10 \)

\( \Rightarrow 4y^{2} – 10y + 4 = 0 \)

\( \Rightarrow 4y^{2} – 8y – 2y + 4 = 0 \)

\( \Rightarrow 4y(y – 2) – 2(y – 2) = 0 \)

\( \Rightarrow y = 2 or y = \frac{2}{4} = \frac{1}{2} \)

\( \Rightarrow 4^{x} = 2^{2x} = 2^{1} or 2^{2x} = 2^{-1} \)

\( \Rightarrow x = \frac{1}{2} or x = \frac{-1}{2} \)

Hence, \( \frac{1}{2} and \frac{-1}{2} \) are roots of the given equation.

Q14: 22x – 3(2(x + 2)) + 32 = 0

Given:

22x – 3(2(x + 2)) + 32 = 0

\( \Rightarrow \) (2x)2 – 3(2x22) + 32 = 0

Let, 2x be y

Hence,

Y2 – 12y + 32 = 0

\( \Rightarrow \) y2 – 8y – 4y + 32 = 0

\( \Rightarrow \) y (y – 8) – 4(y – 8) = 0

\( \Rightarrow \) (y – 8) = 0 or (y – 4) = 0

\( \Rightarrow \) y = 8 or y = 4

Hence,

2x = 8 or 2x = 4

\( \Rightarrow \) 2x = 23 or 2x = 22

\( \Rightarrow \) x = 2 or 3

Hence, 2 and 3 are the roots of the given equation.

15: 4\(\sqrt{6}\)x^{2}  — 13x — 2\(\sqrt{6}\) = 0  

Given:

4\(\sqrt{6}\)x^{2} — 13x — 2\(\sqrt{6}\) = 0

4\(\sqrt{6}\)x^{2}  — 16x + 3x — 2\(\sqrt{6}\) = 0

4\(\sqrt{2}\)x(\(\sqrt{3}\)x — 2\(\sqrt{2}\)) + \(\sqrt{3}\) (\(\sqrt{3}\)x — 2\(\sqrt{2}\)) = 0

(4\(\sqrt{2}\)x + \(\sqrt{3}\)) (\(\sqrt{3}\)x — 2\(\sqrt{2}\)) = 0

4\(\sqrt{2}\)x + \(\sqrt{3}\) = 0 or \(\sqrt{3}\)x — 2\(\sqrt{2}\) = 0

x = \(\frac{-sqrt{3}}{4[\sqrt{2}}\)= \(\frac{-\sqrt{3}x\sqrt{2}}{4\sqrt{2}x\sqrt{2}}\) = \(\frac{-\sqrt{6}}{8}\) or x = \(\frac{2\sqrt{2}}{ \sqrt{3}}\) = \(\frac{2\sqrt{2}x\sqrt{3}}{\sqrt{3}x\sqrt{3}}\) = \(\frac{2\sqrt{6}}{3}\)

Hence, the roots of the equation are  and \(\frac{-\sqrt{6}}{8}\) and \(\frac{2\sqrt{6}}{3}\).


Practise This Question

Circumference of a circle bears a constant ratio with its diameter.