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

Question 1: Find the fourth proportional to:

$\displaystyle \text{i) } 1.5, 4.5 \text{ and } 3.5$ $\displaystyle \text{ii) } 3a, 6a^2 \text{ and } 2ab^2$

Answer:

i) Let the $\displaystyle 4^{th}$ proportion be $\displaystyle x$

Therefore

$\displaystyle \frac{1.5}{4.5} = \frac{3.5}{x} \Rightarrow x = \frac{3.5 \times 4.5}{1.5} = 10.5$

$\displaystyle \text{Hence } x = 10.5$

ii) Let the $\displaystyle 4^{th}$ proportion be $\displaystyle x$

Therefore

$\displaystyle \frac{3a}{6a^2} = \frac{2ab^2}{x} \Rightarrow x = \frac{2ab^2 \times 6a^2}{3a} = 4a^2b^2$

$\displaystyle \text{Hence } x = 4a^2b^2$

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Question 2: Find the third proportional to:

$\displaystyle \text{i) } 2 \frac{2}{3} \text{ and } 4$ $\displaystyle \text{ii) } a-b \text{ and } a^2-b^2$

Answer:

i) Let the $\displaystyle 3^{rd}$ proportion be $\displaystyle x$

Therefore

$\displaystyle 2 \frac{2}{3} : 4 = 4 : x \Rightarrow x = \frac{4 \times 4}{2 \frac{2}{3}} = 6$

$\displaystyle \text{Hence } x = 6$

ii) Let the $\displaystyle 3^{rd}$ proportion be $\displaystyle x$

Therefore

$\displaystyle a-b : a^2-b^2 = a^2-b^2: x \Rightarrow x = \frac{(a^2-b^2) \times (a^2-b^2)}{(a-b)} = (a+b)(a^2-b^2)$

$\displaystyle \text{Hence } x = (a+b)(a^2-b^2)$

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Question 3: Find the mean proportional between:

$\displaystyle \text{i) } 17.5 \text{ and } 0.007$ $\displaystyle \text{ii) } 6+ 3\sqrt{3} \text{ and } 8-4\sqrt{3}$ $\displaystyle \text{iii) } a-b \text{ and } a^3-a^2b$

Answer:

i) Let the mean proportional be $\displaystyle x$

$\displaystyle \text{Therefore } 17.5 : x = x:0.007 \Rightarrow x^2= 17.5 \times 0.007$

$\displaystyle \Rightarrow x = 0.35$

ii) Let the mean proportional be $\displaystyle x$

$\displaystyle \text{Therefore } 6+ 3\sqrt{3} : x = x : 8-4\sqrt{3} \Rightarrow x^2 = (6+ 3\sqrt{3}) \times (8-4\sqrt{3})$

$\displaystyle \Rightarrow x = 2\sqrt{3}$

iii) Let the mean proportional be $\displaystyle x$

$\displaystyle \text{Therefore } a-b : x= x: a^3-a^2b \Rightarrow x^2 = (a-b) \times (a^3-a^2b )$

$\displaystyle \Rightarrow x = a(a-b)$

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Question 4: If $\displaystyle x+5$ is the means proportion between $\displaystyle x+2 \text{ and } x+9$ ; find the value of $\displaystyle x$ .

Answer:

Given $\displaystyle x+5$ is the means proportion between $\displaystyle x+2 \text{ and } x+9$

Therefore

$\displaystyle (x+2) :(x+5)= (x+5):(x+9)$

$\displaystyle (x+5)^2 = (x+2)(x+9)$

$\displaystyle \Rightarrow x^2+25+10x = x^2+11x+18$

$\displaystyle \Rightarrow x = 7$

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Question 5: what least number must be added to each of the numbers $\displaystyle 16, 7, 79 \text{ and } 43$ so that the resulting numbers are in proportion?

Answer:

Let the number added be $\displaystyle x$

$\displaystyle \text{Therefore } (16+x): (7+x) = (79+x): (43+x)$

$\displaystyle \Rightarrow (16+x) \times (43+x) = (79+x) \times (7+x)$

$\displaystyle \Rightarrow x^2+59x+688 = x^2+ 86x +553$

$\displaystyle \Rightarrow x = 5$

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

Answer:

Let the number added be $\displaystyle x$

$\displaystyle \text{Therefore } (6+x): (15+x) = (20+x): (43+x)$

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

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

$\displaystyle \Rightarrow x = 3$

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Question 7: What number must be added to each of the numbers $\displaystyle 16, 26 \text{ and } 40$ so that the resulting numbers may be in continued proportion?

Answer:

Let the number added be $\displaystyle x$

$\displaystyle \text{Therefore } (16+x): (26+x) = (26+x): (40+x)$

$\displaystyle \Rightarrow (16+x) \times (40+x) = (26+x) \times (26+x)$

$\displaystyle \Rightarrow x^2+56x+640 = x^2+ 52x +676$

$\displaystyle \Rightarrow x = 9$

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Question 8: What least number must be subtracted from each of the numbers $\displaystyle 7, 17 \text{ and } 47$ so that the remainders are in continued proportion?

Answer:

Let the number subtracted be $\displaystyle x$

$\displaystyle \text{Therefore } (7-x): (17-x) = (17-x): (47-x)$

$\displaystyle \Rightarrow (7-x) \times (47-x) = (17-x) \times (17-x)$

$\displaystyle \Rightarrow x^2-54x+329 = x^2-34x +289$

$\displaystyle \Rightarrow x = 2$

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Question 9: If $\displaystyle y$ is the mean proportional between $\displaystyle x \text{ and } z$ ; show that $\displaystyle xy+yz$ is the mean proportional between $\displaystyle x^2+y^2 \text{ and } y^2+z^2$ .

Answer:

Since $\displaystyle y$ is the mean proportional between $\displaystyle x \text{ and } z$

$\displaystyle x:y = y : z \Rightarrow y^2 = xz$

Let the mean proportional between $\displaystyle x^2+y^2 \text{ and } y^2+z^2$ by $\displaystyle p$

$\displaystyle \text{Therefore } (x^2+y^2 ): p = p : (x^2+z^2)$

$\displaystyle \Rightarrow p^2 = (x^2+y^2 ) \times (y^2+z^2)$

$\displaystyle \Rightarrow p^2= (x^2+xz ) \times (xz+z^2)$

$\displaystyle = xz(x+z)(x+z) = y^2(x+z)^2$

$\displaystyle \Rightarrow p = y(x+z) = yx+yz$ . Hence proved.

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Question 10: If $\displaystyle q$ is the mean proportional between $\displaystyle p \text{ and } r$ . show that: $\displaystyle pqr(p+q+r)^3=(pq+qr+pr)^3$ .

Answer:

Given $\displaystyle q$ is the mean proportional between $\displaystyle p \text{ and } r$

$\displaystyle \text{Therefore } q^2 = pr$

$\displaystyle LHS = pqr(p+q+r)^3 = q^3(p+q+r)^3$

$\displaystyle = [q(p+q+r)]^3 = (pq+qr+pr)^3 = RHS$ . Hence proved.

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Question 11: If three quantities are in continued proportion; show that the ratio of the first to the third is the duplicate ratio of the first to the second.

Answer:

Let the three quantities by $\displaystyle x, y \text{ and } z$

If they are in proportion, then we have $\displaystyle x:y =y: z \Rightarrow y^2 = xz$

Now we have to prove that $\displaystyle x: z = x^2 : y^2$

$\displaystyle \Rightarrow x \times y^2 = x^2 \times z$

Substituting $\displaystyle y^2 = xz \Rightarrow LHS = x^2z = RHS$ . Hence proved.

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Question 12: If $\displaystyle y$ is the mean proportional between $\displaystyle x \text{ and } z$ , prove that: $\displaystyle \frac{(x^2-y^2+z^2)}{(x^{-2}-y^{-2}+z^{-2} )} =y^4$ .

Answer:

Given $\displaystyle y$ is the mean proportional between $\displaystyle x \text{ and } z$

$\displaystyle \text{Therefore } y^2 = xz$

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

$\displaystyle = \frac{(x^2-y^2+z^2)}{\frac{1}{x^2}-\frac{1}{y^2} + \frac{1}{z^2}}$

$\displaystyle = \frac{(x^2-xz+z^2)}{\frac{1}{x^2}-\frac{1}{xz} + \frac{1}{z^2}}$

$\displaystyle = \frac{(x^2-xz+z^2)}{\frac{z^2-xz+z^2}{x^2z^2}}$

$\displaystyle =x^2z^2 = (xy)^2 = (y^2)^2 = y^4 = RHS$ . Hence proved.

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Question 13: Give four quantities $\displaystyle a, b, c \text{ and } d$ are in proportion. Show that: $\displaystyle (a-c)b^2:(b-d)cd=(a^2 -b^2-ab) : (c^2-d^2-cd)$

Answer:

Given $\displaystyle a, b, c \text{ and } d$ are in proportion

$\displaystyle \Rightarrow \frac{a}{b} = \frac{c}{d} = k$

$\displaystyle \Rightarrow a = bk \text{ and } a = bk \text{ and } c = dk$

To prove $\displaystyle (a-c)b^2:(b-d)cd=(c^2-d^2-cd)$

$\displaystyle \text{LHS } = \frac{(a-c)b^2}{(b-d)cd} = \frac{(bk-dk)b^2}{(b-d)d^2k} = \frac{b^2}{d^2}$

$\displaystyle \text{RHS } = \frac{a^2 -b^2-ab}{c^2-d^2-cd} = \frac{b^2k^2 -b^2-bkd}{d^2k^2-d^2-kd^2}$

$\displaystyle = \frac{b^2(k^2-1-k)}{d^2(k^2-1-k)} = \frac{b^2}{d^2}$

$\displaystyle \text{Hence } LHS = RHS$ . Hence proved.

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Question 14: Find two numbers such that the mean proportional between them is $\displaystyle 12$ and the third proportional to them is $\displaystyle 96$ .

Answer:

Let the two numbers be $\displaystyle a \text{ and } b$

$\displaystyle \text{Therefore } a: 12 = 12: b \Rightarrow ab = 144$

If $\displaystyle 96$ is the third proportion

$\displaystyle \Rightarrow a : b = b : 96$

$\displaystyle \Rightarrow b^2 = 96a$

$\displaystyle \Rightarrow \Big( \frac{144}{a} \Big)^2 = 96a$

$\displaystyle \Rightarrow a ^3 = 216 \text{ or } a = 6 \text{ and } b = \frac{144}{6} = 24.$

$\displaystyle \text{Hence } a = 6 \text{ and } b = 24.$

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$\displaystyle \text{Question 15: Find the third proportional to } \frac{x}{y} + \frac{y}{x} \text{ and } \sqrt{x^2+y^2 }$

Answer:

Let the third proportion by $\displaystyle p$

$\displaystyle \text{Therefore } \frac{x}{y} + \frac{y}{x} : \sqrt{x^2+y^2 } = \sqrt{x^2+y^2 } : p$

$\displaystyle \Rightarrow p \Big( \frac{x}{y} + \frac{y}{x} \Big) = (\sqrt{x^2+y^2 })^2$

$\displaystyle \Rightarrow p \Big( \frac{x^2+y^2}{xy} \Big) = x^2+y^2$

$\displaystyle \Rightarrow p = xy$

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$\displaystyle \text{Question 16: If } p:q=r:s \text{ ; then show that: } mp+nq:q=mr+ns:s$

Answer:

Given $\displaystyle p:q=r:s$

$\displaystyle \text{Therefore } \frac{p}{q} = \frac{r}{s}$

Multiplying both sides by $\displaystyle m$

$\displaystyle \frac{mp}{q} = \frac{mr}{s}$

Adding $\displaystyle n$ to both sides

$\displaystyle \frac{mp}{q} +n = \frac{mr}{s} +n$

$\displaystyle \Rightarrow \frac{mp+nq}{q} = \frac{mr+sn}{s}$

or $\displaystyle mp+nq:q=mr+ns:s$

Hence proved.

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$\displaystyle \text{Question 17: If } p+r=mq \text{ and } \frac{1}{q} + \frac{1}{s} = \frac{m}{r} \text{ ; then prove that: } p:q=r:s$ .

Answer:

$\displaystyle \text{Given } \frac{1}{q} + \frac{1}{s} = \frac{m}{r}$

$\displaystyle \frac{1}{q} + \frac{1}{s} = \frac{m}{r}$

$\displaystyle \frac{s+q}{qs} = \frac{m}{r}$

$\displaystyle \Rightarrow \frac{s+q}{s} = \frac{mq}{r}$

Given $\displaystyle p+r=mq$

$\displaystyle \Rightarrow \frac{s+q}{s} = \frac{p+r}{r}$

$\displaystyle \Rightarrow 1+ \frac{q}{s} = 1+ \frac{p}{r}$

or $\displaystyle p:q=r:s$

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$\displaystyle \text{Question 18: If } \frac{a}{b} = \frac{c}{d} \text{ , prove that each of the given ratio is equal to: }$

$\displaystyle \text{i) } \frac{5a+4c}{5b+4d}$ $\displaystyle \text{ii) } \frac{13a-8c}{13b-8d}$ $\displaystyle \text{iii) } \sqrt{\frac{3a^2-10c^2}{3b^2-10d^2}}$ $\displaystyle \text{iv) } \Big( \frac{8a ^3+ 15c ^3}{ 8b ^3+ 15d ^3 } \Big)^\frac{1}{3}$

Answer:

$\displaystyle \text{Given } \frac{a}{b} = \frac{c}{d} = k$

$\displaystyle a = bk \text{ and } c = dk$

$\displaystyle \text{i) } \frac{5a+4c}{5b+4d} = \frac{5(bk) +4(dk)}{5b+4d} = k \Big( \frac{5b +4d}{5b+4d} \Big) = k$

$\displaystyle \text{ii) } \frac{13a-8c}{13b-8d} = \frac{13(bk)-8(dk)}{13b-8d} =k \Big( \frac{13b-8d}{13b-8d} \Big) = k$

$\displaystyle \text{iii) } \sqrt{\frac{3a^2-10c^2}{3b^2-10d^2}} = \sqrt{\frac{3(bk)^2-10(dk)^2}{3b^2-10d^2}} = k \sqrt{\frac{3b^2-10d^2}{3b^2-10d^2}} = k$

$\displaystyle \text{iv) } \Big( \frac{8a ^3+ 15c ^3}{ 8b ^3+ 15d ^3 } \Big)^\frac{1}{3} = \Big( \frac{8(bk) ^3+ 15(dk) ^3}{ 8b ^3+ 15d ^3 } \Big)^\frac{1}{3} = \Big[ k^3 \Big( \frac{8b^3+ 15d^3}{ 8b ^3+ 15d ^3 } \Big) \Big]^\frac{1}{3} = k$

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Question 19: If $\displaystyle a, b, c \text{ and } d$ are in proportion, prove that:

$\displaystyle \text{i) } \frac{13a+17b}{13c+17d} = \sqrt{\frac{2ma ^2- 3nb ^2}{ 2mc ^2- 3nd ^2 }}$

$\displaystyle \text{ii) } \sqrt {\frac{ 4a ^2+ 9b ^2}{ 4c ^2+ 9d ^2 }} = \Big( \frac{ xa ^3-5yb^3}{ xc ^3-5yd^3 } \Big)^\frac{1}{3}$

Answer:

$\displaystyle \text{Given } \frac{a}{b} = \frac{c}{d} = k$

$\displaystyle a = bk \text{ and } c = dk$

$\displaystyle \text{i) } \frac{13a+17b}{13c+17d} = \sqrt{\frac{2ma ^2- 3nb ^2}{ 2mc ^2- 3nd ^3 }}$

$\displaystyle \text{LHS } = \frac{13a+17b}{13c+17d} = \frac{13bk+17b}{13dk+17d} = \frac{b(13k+17)}{d(13dk+17)} = \frac{b}{d}$

$\displaystyle \text{RHS } = \sqrt{\frac{2ma ^2- 3nb ^2}{ 2mc ^2- 3nd ^3 }} = \sqrt{\frac{2m(bk)^2- 3nb ^2}{ 2m(dk)^2- 3nd ^2 }} = \sqrt{\frac{b^2(2mk^2- 3n)}{d^2( 2mk^2- 3n)}} = \frac{b}{d}$

Hence LHS = RHS.

$\displaystyle \text{ii) } \sqrt {\frac{ 4a ^2+ 9b ^2}{ 4c ^2+ 9d ^2 }} = \Big( \frac{ xa ^3-5yb^3}{ xc ^3-5yd^3 } \Big)^\frac{1}{3}$

$\displaystyle \text{LHS } = \sqrt {\frac{4a^2+ 9b^2}{ 4c^2+ 9d^2}} = \sqrt {\frac{ 4(bk)^2+ 9b ^2}{4(dk)^2+ 9d ^2}} = \sqrt {\frac{ b^2(4k^2+ 9)}{d^2(4k^2+ 9)}} = \frac{b}{d}$

$\displaystyle \text{RHS } = \Big( \frac{ xa ^3-5yb^3}{ xc ^3-5yd^3 } \Big)^\frac{1}{3}= \Big( \frac{ x(bk)^3-5yb^3}{ x(dk)^3-5yd^3 } \Big)^\frac{1}{3} = \Big( \frac{ b^3(xk^3-5y)}{ d^3(xk^3-5y)} \Big)^\frac{1}{3} = \frac{b}{d}$

Hence LHS = RHS.

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$\displaystyle \text{Question 20: If } \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \text{ , prove that: } \frac{ 2x ^3- 3y ^3+ 4z ^3}{ 2a ^3- 3b ^3+ 4c ^3 } = \Big( \frac{2x-3y+4z}{2a-3b+4c} \Big)^3$