Class 10: Ratio and Proportion – ICSE Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: If $x, y \ and \ z$ are in continued proportion, prove that: $\frac{(x+y)^2}{(y+z)^2}$ $=$ $\frac{x}{y}$ . [2010]

Answer:

If $x, y \ and \ z$ are in continued proportion, then

$\frac{x}{y}$ $=$ $\frac{y}{z}$ $\Rightarrow x =$ $\frac{y^2}{z}$

Applying componendo and dividendo

$\frac{x+y}{x-y}$ $=$ $\frac{y+z}{y-z}$

$\Rightarrow$ $\frac{x+y}{y+z}$ $=$ $\frac{x-y}{y-z}$

Squaring both sides

$\Rightarrow$ $\frac{(x+y)^2}{(y+z)^2}$ $=$ $(\frac{x-y}{y-z})^2$

Substituting

$\Rightarrow$ $\frac{(x+y)^2}{(y+z)^2}$ $=$ $(\frac{\frac{y^2}{z}-y}{y-z})^2$

$\Rightarrow$ $\frac{(x+y)^2}{(y+z)^2}$ $=$ $(\frac{y^2-yz}{z(y-z)})^2$ $=$ $\frac{y^2}{z^2}$ $=$ $\frac{zx}{z^2}$ $=$ $\frac{x}{z}$

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Question 2: Given $x=$ $\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 }}$ . Use componendo and dividendo to prove that: $x^2=$ $\frac{2a^2 x}{x^2+1}$ . [2010]

Answer:

Given $x=$ $\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 }}$

Applying componendo and dividendo

$\frac{x+1}{x-1}$ $=$ $\frac{(\sqrt{a^2+b^2}+\sqrt{a^2-b^2})+(\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 })}{(\sqrt{a^2+b^2}+\sqrt{a^2-b^2})-(\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 })}$

Simplifying

$\frac{x+1}{x-1}$ $=$ $\frac{\sqrt{a^2+b^2}}{\sqrt{a^2-b^2 }}$

Square both sides

$\frac{x^2+1+2x}{x^2-2x+1}$ $=$ $\frac{a^2+b^2}{a^2-b^2}$

Applying componendo and dividendo

$\frac{x^2+1+2x+x^2-2x+1}{x^2+1+2x-x^2+2x-1}$ $=$ $\frac{a^2+b^2+a^2-b^2}{a^2+b^2-a^2+b^2}$

$\frac{2(x^2+1)}{4x}$ $=$ $\frac{2a^2}{2b^2}$

$\frac{x^2+1}{2x}$ $=$ $\frac{a^2}{b^2}$

Simplifying

$b^2 =$ $\frac{2a^2x}{x^2+1}$

$\\$

Question 3: If $\frac{x^2+y^2}{x^2-y^2}$ $=2$ $\frac{1}{8}$ , find:

i) $\frac{x}{y}$ ii) $\frac{x^3+y^3}{x^3-y^3 }$ [2014]

Answer:

i) Given $\frac{x^2+y^2}{x^2-y^2}$ $=2$ $\frac{1}{8}$ $=$ $\frac{17}{8}$

Applying componendo and dividendo

$\frac{x^2+y^2+x^2-y^2}{x^2+y^2-x^2+y^2}$ $=$ $\frac{17+8}{17-8}$

$\frac{2x^2}{2y^2}$ $=$ $\frac{25}{9}$

Simplifying, we get

$\frac{x}{y}$ $=$ $\frac{5}{3}$

ii) $\frac{x^3+y^3}{x^3-y^3 }$

Applying componendo and dividendo

$\frac{x^3+y^3+x^3-y^3 }{x^3+y^3-x^3+y^3 }$

$\frac{2x^3}{2y^3}$ $= \Big($ $\frac{x}{y}$ $\Big)^3 = \Big($ $\frac{5}{3}$ $\Big)^3 =$ $\frac{125}{9}$

$\\$

Question 4: Using componendo and dividendo, find the value of $x$ : $\frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}$ $=9$ [2011]

Answer:

Given $\frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}$ $=9$

Applying componendo and dividendo

$\frac{(\sqrt{3x+4}+\sqrt{3x-5})+(\sqrt{3x+4}-\sqrt{3x-5})}{(\sqrt{3x+4}+\sqrt{3x-5})-(\sqrt{3x+4}-\sqrt{3x-5})}$ $=$ $\frac{9+1}{9-1}$

$\frac{2\sqrt{3x+4}}{2\sqrt{3x-5}}$ $=$ $\frac{10}{8}$

Simplifying

$\frac{\sqrt{3x+4}}{\sqrt{3x-5}}$ $=$ $\frac{5}{4}$

Square both sides

$\frac{3x+4}{3x-5}$ $=$ $\frac{25}{14}$

$42x+56 = 75x-125$

Simplifying we get $x = 7$

$\\$

Question 5: If $x=$ $\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}$ , using properties of proportion show that: $x^2-2ax+1$ [2012]

Answer:

Given $x=$ $\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}$

Applying componendo and dividendo

$\frac{x+1}{x-1}$ $=$ $\frac{(\sqrt{a+1}+\sqrt{a-1})+(\sqrt{a+1}-\sqrt{a-1})}{(\sqrt{a+1}+\sqrt{a-1})-(\sqrt{a+1}-\sqrt{a-1})}$

Simplify

$\frac{x+1}{x-1}$ $=$ $\frac{\sqrt{a+1}}{\sqrt{a-1}}$

Now square both sides

$\frac{x^2+1+2x}{x^2-2x+1}$ $=$ $\frac{a+1}{a-1}$

Simplifying

$x^2+1 = 2ax$

or $x^2-2ax+1 = 0$

$\\$

Question 6: Given, $\frac{a}{b}$ $=$ $\frac{c}{d}$ , prove that: $\frac{3a-5b}{3a+5b}$ $=$ $\frac{3c-5d}{3c+5d}$ [2000]

Answer:

Given $\frac{a}{b}$ $=$ $\frac{c}{d}$

$\Rightarrow \frac{3a}{5b}$ $=$ $\frac{3c}{5d}$

By componendo and dividendo

$\frac{3a+5b}{3a-5b}$ $=$ $\frac{3c+5d}{3c-5d}$

By Alternendo

$\frac{3a-5b}{3a+5b}$ $=$ $\frac{3c-5d}{3c+5d}$

$\\$

Question 7: If $x=\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}}$ , prove that: $3bx^2-2ax+3b=0$ . [2007]

Answer:

Given $x=$ $\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}}$

Applying componendo and dividendo

$\frac{x+1}{x-1}$ $=$ $\frac{(\sqrt{a+3b}+\sqrt{a-3b})-(\sqrt{a+3b}-\sqrt{a-3b})}{(\sqrt{a+3b}+\sqrt{a-3b})-(\sqrt{a+3b}-\sqrt{a-3b})}$

$\frac{x+1}{x-1}$ $=$ $\frac{2\sqrt{a+3b}}{-2\sqrt{a-3b}}$

Squaring both sides

$\frac{x^2+2x+1}{x^2-2x+1}$ $=$ $\frac{a+3b}{a-3b}$

Applying componendo and dividendo once again

$\frac{(x^2+2x+1)-(x^2-2x+1)}{(x^2+2x+1)-(x^2-2x+1)}$ $=$ $\frac{(a+3b)-(a-3b)}{(a+3b)-(a-3b)}$

simplifying

$\frac{x^2+1}{2x}$ $=$ $\frac{a}{3b}$

$3b(x^2+1) = 2ax$

$3bx^2-2ax+3b=0$ Hence proved.

$\\$

Question 8: Using the properties of proportion, solve for $x$ Given: $\frac{(x^4+1)}{2x^2}$ $=$ $\frac{17}{8}$ . [2013]

Answer:

Given $\frac{(x^4+1)}{2x^2}$ $=$ $\frac{17}{8}$

Applying componendo and dividendo

$\frac{(x^4+1)+2x^2}{(x^4+1)-2x^2}$ $=$ $\frac{17+8}{17-8}$

$\frac{(x^2+1)^2}{(x^2-1)^2}$ $=$ $\frac{25}{9}$

Taking the square root of both sides

$\frac{x^2+1}{x^2-1}$ $=$ $\frac{5}{3}$

$3x^2+3=5x^2-5$

$x^2 = 4 \ or \ x = \pm 2$

$\\$

Question 9: What least number must be added to each of the numbers $6, 15, 20, \ and \ 43$ to make them proportional. [2005, 2013]

Answer:

Let the number added be $x$

Therefore $(6+x): (15+x) = (20+x): (43+x)$

$\Rightarrow (6+x) \times (43+x) = (20+x) \times (15+x)$

$\Rightarrow x^2+49x+258 = x^2+ 35x +300$

$\Rightarrow x = 3$

$\\$

Question 10: The monthly pocket money of Ravi and Sanjeev are in the ratio of 5:7 Their expenditures are in the ratio of 3:5. If each saves Rs. 80 per month, find their monthly pocket money. [2012]

Answer:

Let monthly pocket of Rave and Sanjeev by $x and y$ respectively.

$\frac{x}{y}$ $=$ $\frac{5}{7}$ $\Rightarrow x =$ $\frac{5}{7}$ $y$

$\frac{x-80}{y-80}$ $=$ $\frac{3}{5}$

Substituting

$\frac{ \frac{5}{7} y-80}{y-80}$ $=$ $\frac{3}{5}$

$\frac{25}{7}$ $x-400=3x-240 \Rightarrow x=280$

Substituting

$y =$ $\frac{5}{7}$ $\times 280 = 200$

$\\$

Question 11: If $(x-9):(3x+6)$ is the triplicate ratio of $4:9$ , find $x$ . [2014]

Answer:

$\frac{x-9}{3x+6}$ $=$ $\frac{4^2}{9^2}$ $=$ $\frac{16}{81}$

$81x-729=48x+96$

$x=25$

$\\$

Question 12: If $a:b=5:3$ , find $(5a+8b):(6a-7b)$ . [2002]

Answer:

Given $a:b=5:3$

or $\frac{a}{b}$ $=$ $\frac{5}{3}$ $\Rightarrow a = b$ $\frac{5}{3}$

Now substituting

$\frac{5a+8b}{6a-7b}$ $=$ $\frac{5 \times b\frac{5}{3}+8b}{6 \times b\frac{5}{3}-7b}$ $=$ $\frac{25+24}{30-21}$ $=$ $\frac{49}{9}$

Hence $(5a+8b):(6a-7b) =$ $\frac{49}{9}$

$\\$

Question 13: The work done by $(x-3)$ men in $(2x+1)$ days and the work done by $(2x+1)$ men in $(x+4)$ days are in the ratio $3:10$ . Find the value of $x$ . [2003]