Ratio and Proportion — Chapter Test

Find the compounded ratio of

(a + b)² : (a - b)², (a² - b²) : (a² + b²) and (a⁴ - b⁴) : (a + b)⁴.

Answer

The compounded ratio is :

$\dfrac{(a + b)^2}{(a - b)^2} \times \dfrac{(a^2 - b^2)}{(a^2 + b^2)} \times \dfrac{(a^4 - b^4)}{(a + b)^4} \\[1em] = \dfrac{(a + b)^2}{(a - b)^2} \times \dfrac{(a - b)(a + b)}{(a^2 + b^2)} \times \dfrac{(a^2 - b^2)(a^2 + b^2)}{(a + b)^4} \\[1em] = \dfrac{(a + b)^2}{(a - b)^2} \times \dfrac{(a - b)(a + b)}{(a^2 + b^2)} \times \dfrac{(a - b)(a + b)(a^2 + b^2)}{(a + b)^4} \\[1em] = \dfrac{(a + b)^4(a - b)^2(a^2 + b^2)}{(a - b)^2(a^2 + b^2)(a + b)^4} \\[1em] = \dfrac{1}{1} = 1 : 1 .$

Hence, the compounded ratio is 1 : 1.

If (7p + 3q) : (3p - 2q) = 43 : 2, find p : q.

Answer

Given, (7p + 3q) : (3p - 2q) = 43 : 2.

$\Rightarrow \dfrac{7p + 3q}{3p - 2q} = \dfrac{43}{2} \\[1em] \Rightarrow 2(7p + 3q) = 43(3p - 2q) \\[1em] \Rightarrow 14p + 6q = 129p - 86q \\[1em] \Rightarrow 92q = 115p \\[1em] \Rightarrow \dfrac{p}{q} = \dfrac{92}{115} = \dfrac{4}{5}. \\[1em]$

Hence, the value of ratio p : q is 4 : 5.

If a : b = 3 : 5, find (3a + 5b) : (7a - 2b).

Answer

Given,

$\dfrac{a}{b} = \dfrac{3}{5} \\[1em] \therefore a = \dfrac{3b}{5}$

Putting value of a in the ratio (3a + 5b) : (7a - 2b),

$= \dfrac{3 \times \dfrac{3b}{5} + 5b}{7 \times \dfrac{3b}{5} - 2b} \\[1em] = \dfrac{\dfrac{9b}{5} + 5b}{\dfrac{21b}{5} - 2b} \\[1em] = \dfrac{\dfrac{9b + 25b}{5}}{\dfrac{21b - 10b}{5}} \\[1em] = \dfrac{34b}{11b} \\[1em] = \dfrac{34}{11} = 34 : 11.$

Hence, the value of ratio (3a + 5b) : (7a - 2b) is 34 : 11.

The ratio of the shorter sides of a right-angled triangle is 5 : 12. If the perimeter of the triangle is 360 cm, find the length of the longest side.

Answer

Let the length of the shorter side be 5x cm and 12x cm.

Length of hypotenuse = $\sqrt{(5x)^2 + (12x)^2} \text{ cm}$

$= \sqrt{25x^2 + 144x^2} \text{ cm} \\[0.75em] = \sqrt{169x^2} \text{ cm} \\[0.75em] = 13x \text{ cm} \\[0.75em]$

According to given,

$5x + 12x + 13x = 360 \\[0.5em] 30x = 360 \\[0.5em] x = 12.$

∴ x = 12, 13x = 13 $\times$ 12 = 156.

Hence, the length of longest side is 156 cm.

The ratio of the pocket money saved by Lokesh and his sister is 5 : 6. If the sister saves ₹ 30 more, how much more the brother should save in order to keep the ratio of their savings unchanged.

Answer

Let the savings of Lokesh and his sister are 5x and 6x.

Let Lokesh save ₹ y more then according to question,

$\Rightarrow \dfrac{5x + y}{6x + 30} = \dfrac{5}{6} \\[0.5em] \Rightarrow 6(5x + y) = 5(6x + 30) \\[0.5em] \Rightarrow 30x + 6y = 30x + 150 \\[0.5em] \Rightarrow 6y = 150 \\[0.5em] \Rightarrow y = 25.$

Hence, Lokesh should save ₹ 25 more.

In an examination, the number of those who passed and the number of those who failed were in the ratio 3 : 1. Had 8 more appeared, and 6 less passed, the ratio of passed to failures would have been 2 : 1. Find the number of candidates who appeared.

Answer

In first case,
Let the number of students passed = 3x and failed = x.
Total candidates appeared = 3x + x = 4x.

In second case,
Total candidates appeared = 4x + 8
Number of passed students = 3x - 6
Number of failed students = (4x + 8) - (3x - 6)
= 4x + 8 - 3x + 6
= x + 14.

According to question, ratio of number of passed to failed student in this case = 2 : 1.

$\therefore \dfrac{3x - 6}{x + 14} = \dfrac{2}{1} \\[0.5em] \Rightarrow (3x - 6) = 2(x + 14) \\[0.5em] \Rightarrow 3x - 6 = 2x + 28 \\[0.5em] \Rightarrow 3x - 2x = 28 + 6 \\[0.5em] \Rightarrow x = 34.$

∴ x = 34, 4x = 136.

Hence, the number of students who appeared were 136.

What number must be added to each of the numbers 4, 6, 8, 11 to make them proportional ?

Answer

Let x be added to each number.

So, 4 + x, 6 + x, 8 + x and 11 + x must be in proportion.

$\therefore \dfrac{4 + x}{6 + x} = \dfrac{8 + x}{11 + x} \\[1em] \Rightarrow (4 + x) \times (11 + x) = (8 + x) \times (6 + x) \\[1em] \Rightarrow 44 + 4x + 11x + x^2 = 48 + 8x + 6x + x^2\\[1em] \Rightarrow 44 + 15x + x^2 = 48 + 14x + x^2 \\[1em] \Rightarrow 15x - 14x = 48 - 44 \\[1em] \Rightarrow x = 4.$

Hence, required number is 4.

If (a + 2b + c), (a - c) and (a - 2b + c) are in continued proportion, prove that b is the mean proportional between a and c.

Answer

Since, the numbers are in continued proportion,

$\therefore \dfrac{(a + 2b + c)}{(a - c)} = \dfrac{(a - c)}{(a - 2b + c)} \\[1em] \Rightarrow (a + 2b + c)(a - 2b + c) = (a - c)^2 \\[1em] \Rightarrow (a^2 - 2ab + ac + 2ab - 4b^2 + 2bc + ac - 2bc + c^2) = (a^2 + c^2 - 2ac) \\[1em] \Rightarrow a^2 + c^2 + 2ac - 4b^2 = a^2 + c^2 - 2ac \\[1em] \Rightarrow 4b^2 = 4ac \\[1em] \Rightarrow b^2 = ac.$

Since, b² = 4ac, hence proved that b is the mean proportional between a and c.

If 2, 6, p, 54 and q are in continued proportion, find the values of p and q.

Answer

2, 6, p, 54 and q are in continued proportion then,

$\Rightarrow \dfrac{2}{6} = \dfrac{6}{p} = \dfrac{p}{54} = \dfrac{54}{q}.\\[1em] \text{Solving, } \dfrac{2}{6} = \dfrac{6}{p} \text{ for p,} \\[1em] \Rightarrow p = \dfrac{6}{2} \times 6 \\[1em] \Rightarrow p = 18. \\[1em] \text{Now solving, } \dfrac{p}{54} = \dfrac{54}{q} \text{ for q,} \\[1em] \Rightarrow q = \dfrac{54}{p} \times 54 \\[1em] \Rightarrow q = \dfrac{54}{18} \times 54 \\[1em] \Rightarrow q = 3 \times 54 \\[1em] \Rightarrow q = 162.$

Hence, the value of p = 18 and q = 162.

If a, b, c, d, e are in continued proportion, prove that a : e = a⁴ : b⁴.

Answer

Since, a, b, c, d, e are in continued proportion.

Let, $\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = \dfrac{d}{e} = k.$

∴ d = ek, c = ek², b = ek³, a = ek⁴.

Now,

$\text{L.H.S. } = \dfrac{a}{e} \\[1em] = \dfrac{ek^4}{e} = k^4 \\[1em] \text{R.H.S. } = \dfrac{(a^4)}{(b^4)} \\[1em] = \dfrac{(ek^4)^4}{(ek^3)^4} \\[1em] = \dfrac{e^4k^{16}}{e^4k^{12}} \\[1em] = k^4$

Since, L.H.S. = R.H.S. hence, proved that a : e = a⁴ : b⁴.

Find two numbers whose mean proportional is 16 and the third proportional is 128.

Answer

Let the two numbers be a and b.

Given, mean proportional between a and b is 16.

$\therefore \sqrt{ab} = 16 \qquad \text{[....Eq 1]}$

Given, third proportional is 128.

$\therefore \dfrac{a}{b} = \dfrac{b}{128} \\[0.5em] \Rightarrow a = \dfrac{b^2}{128} \qquad \text{[....Eq 2]}$

Putting this value of a in Eq 1,

$\Rightarrow \sqrt{\Big(\dfrac{b^2}{128}\Big)b} = 16 \\[0.5em] \Rightarrow \sqrt{\Big(\dfrac{b^3}{128}\Big)} = 16$

Squaring both sides,

$\Rightarrow \dfrac{b^3}{128} = 256 \\[1em] \Rightarrow b^3 = 256 \times 128 \\[1em] \Rightarrow b^3 = 32768 \\[1em] \Rightarrow b = \sqrt[3]{32768} \\[1em] \Rightarrow b = 32$

Putting value of b in Eq 2, $a = \dfrac{b^2}{128} \qquad \text{[....Eq 2]} \\[1em] \Rightarrow a = \dfrac{(32)^2}{128} \\[1em] \Rightarrow a = \dfrac{1024}{128} \\[1em] \Rightarrow a = 8$

Hence, the value of a = 8 and b = 32.

If q is the mean proportional between p and r, prove that :

$p^2 - 3q^2 + r^2 = q^4\Big(\dfrac{1}{p^2} - \dfrac{3}{q^2} + \dfrac{1}{r^2}\Big).$

Answer

Since, q is the mean proportional between p and r,

$\therefore q^2 = pr$

Given,

$\Rightarrow p^2 - 3q^2 + r^2 = q^4\Big(\dfrac{1}{p^2} - \dfrac{3}{q^2} + \dfrac{1}{r^2}\Big). \\[1em] \text{L.H.S.} = p^2 - 3q^2 + r^2 \\[1em] = p^2 + r^2 - 3pr. \\[1em] \text{R.H.S.} = q^4\Big(\dfrac{1}{p^2} - \dfrac{3}{q^2} + \dfrac{1}{r^2}\Big) \\[1em] \Rightarrow p^2r^2\Big(\dfrac{1}{p^2} - \dfrac{3}{pr} + \dfrac{1}{r^2}\Big) \\[1em] \Rightarrow p^2r^2\Big(\dfrac{r^2 - 3pr + p^2}{p^2r^2}\Big) \\[1em] \Rightarrow p^2 + r^2 - 3pr.$

Since, L.H.S. = R.H.S. hence proved that,

$p^2 - 3q^2 + r^2 = q^4\big(\dfrac{1}{p^2} - \dfrac{3}{q^2} + \dfrac{1}{r^2}\big).$

If $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f}$, prove that each ratio is equal to

(i) $\sqrt{\dfrac{3a^2 - 5c^2 + 7e^2}{3b^2 - 5d^2 + 7f^2}}$

(ii) $\Big(\dfrac{2a^3 + 5c^3 + 7e^3}{2b^3 + 5d^3 + 7f^3}\Big)^{1/3}.$

Answer

(i) Let, $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k,$ then,

a = bk, c = dk, e = fk.

Given,

$\sqrt{\dfrac{3a^2 - 5c^2 + 7e^2}{3b^2 - 5d^2 + 7f^2}}$

Putting values of a, c, e in above equation,

$= \sqrt{\dfrac{3(bk)^2 - 5(dk)^2 + 7(fk)^2}{3b^2 - 5d^2 + 7f^2}} \\[1em] = \sqrt{\dfrac{3b^2k^2 - 5d^2k^2 + 7f^2k^2}{3b^2 - 5d^2 + 7f^2}} \\[1em] = \sqrt{k^2\Big(\dfrac{3b^2 - 5d^2 + 7f^2}{3b^2 - 5d^2 + 7f^2}\Big)} \\[1em] = k.$

Since, the values of all ratios = k. Hence, proved.

(ii) Let, $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = k,$ then,

a = bk, c = dk, e = fk.

Given,

$\Big(\dfrac{2a^3 + 5c^3 + 7e^3}{2b^3 + 5d^3 + 7f^3}\Big)^{1/3}$

Putting values of a, c, e in above equation, $= \Big(\dfrac{2(bk)^3 + 5(dk)^3 + 7(fk)^3}{2b^3 + 5d^3 + 7f^3}\Big)^{1/3} \\[1em] = \Big(\dfrac{2b^3k^3 + 5d^3k^3 + 7f^3k^3}{2b^3 + 5d^3 + 7f^3}\Big)^{1/3} \\[1em] = \Big[k^3\Big(\dfrac{2b^3 + 5d^3 + 7f^3}{2b^3 + 5d^3 + 7f^3}\Big)\Big]^{1/3} \\[1em] = k^{3 \times 1/3} \\[1em] = k.$

Since, the values of all ratios = k. Hence, proved.

If $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$, prove that

$\dfrac{3x^3 - 5y^3 + 4z^3}{3a^3 - 5b^3 + 4c^3} = \Big(\dfrac{3x - 5y + 4z}{3a - 5b + 4c}\Big)^3.$

Answer

Let $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} = k,$ then,

x = ak, y = bk, z = ck.

Given,

$\dfrac{3x^3 - 5y^3 + 4z^3}{3a^3 - 5b^3 + 4c^3} = \Big(\dfrac{3x - 5y + 4z}{3a - 5b + 4c}\Big)^3$

Putting values of x, y, z in above equation and solving L.H.S.,

$= \dfrac{3(ak)^3 - 5(bk)^3 + 4(ck)^3}{3a^3 - 5b^3 + 4c^3} \\[1em] = \dfrac{3a^3k^3 - 5b^3k^3 + 4c^3k^3}{3a^3 - 5b^3 + 4c^3} \\[1em] = k^3\Big(\dfrac{3a^3 - 5b^3 + 4c^3}{3a^3 - 5b^3 + 4c^3}\Big) \\[1em] = k^3.$

Solving, R.H.S. now, $= \Big(\dfrac{3(ak) - 5(bk) + 4(ck)}{3a - 5b + 4c}\Big)^3 \\[1em] = \Big[k\Big(\dfrac{3a - 5b + 4c}{3a - 5b + 4c}\Big)\Big]^3 \\[1em] = k^3.$

Since, L.H.S. = R.H.S. Hence, proved that,

$\dfrac{3x^3 - 5y^3 + 4z^3}{3a^3 - 5b^3 + 4c^3} = \Big(\dfrac{3x - 5y + 4z}{3a - 5b + 4c}\Big)^3$.

If x : a = y : b, prove that,

$\dfrac{x^4 + a^4}{x^3 + a^3} + \dfrac{y^4 + b^4}{y^3 + b^3} = \dfrac{(x + y)^4 + (a + b)^4}{(x + y)^3 + (a + b)^3}.$

Answer

Given, x : a = y : b,

Let, $\dfrac{x}{a}$ = $\dfrac{y}{b}$ = k

∴ x = ak, y = bk

Putting values of x and y in L.H.S. first,

$= \dfrac{(ak)^4 + a^4}{(ak)^3 + a^3} + \dfrac{(bk)^4 + b^4}{(bk)^3 + b^3} \\[1em] = \dfrac{a^4(k^4 + 1)}{a^3(k^3 + 1)} + \dfrac{b^4(k^4 + 1)}{b^3(k^3 + 1)} \\[1em] = \dfrac{a(k^4 + 1) + b(k^4 + 1)}{k^3 + 1} \\[1em] = \dfrac{(a + b)(k^4 + 1)}{k^3 + 1}$

Now, putting values in R.H.S., $= \dfrac{(x + y)^4 + (a + b)^4}{(x + y)^3 + (a + b)^3} \\[1em] = \dfrac{(ak + bk)^4 + (a + b)^4}{(ak + bk)^3 + (a + b)^3} \\[1em] = \dfrac{k^4(a + b)^4 + (a + b)^4}{k^3(a + b)^3 + (a + b)^3} \\[1em] = \dfrac{(a + b)^4(k^4 + 1)}{(a + b)^3(k^3 + 1)} \\[1em] = \dfrac{(a + b)(k^4 + 1)}{k^3 + 1}.$

Since, L.H.S. = R.H.S. Hence, proved that,

$\dfrac{x^4 + a^4}{x^3 + a^3} + \dfrac{y^4 + b^4}{y^3 + b^3} = \dfrac{(x + y)^4 + (a + b)^4}{(x + y)^3 + (a + b)^3}.$

If $\dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c},$ prove that each ratio is equal to

$\dfrac{x + y + z}{a + b + c}.$

Answer

Let, $\dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c} = k.$

∴ x = k(b + c - a), y = k(c + a - b), z = k(a + b - c).

Putting values of x, y and z in $\dfrac{x + y + z}{a + b + c}$ we get,

$\dfrac{k(b + c - a) + k(c + a - b) + k(a + b - c)}{a + b + c} \\[1em] = \dfrac{kb + kc - \cancel{ak} + \cancel{kc} + \cancel{ak} - \cancel{bk} + ak + \cancel{bk} - \cancel{kc}}{a + b + c} \\[1em] = \dfrac{k(a + b + c)}{(a + b + c)} \\[1em] = k.$

Since, the value of all ratios = k, hence, each ratio = $\dfrac{x + y + z}{a + b + c}.$

If a : b = 9 : 10, find the value of

(i) $\dfrac{5a + 3b}{5a - 3b}$

(ii) $\dfrac{2a^2 - 3b^2}{2a^2 + 3b^2}.$

Answer

(i) Given,

$\dfrac{a}{b} = \dfrac{9}{10} \\[0.5em] \Rightarrow a = \dfrac{9b}{10}$

Putting $a = \dfrac{9b}{10}$ in $\dfrac{5a + 3b}{5a - 3b}$, we get,

$\dfrac{5 \times \dfrac{9b}{10} + 3b}{5 \times \dfrac{9b}{10} - 3b} \\[1em] = \dfrac{\dfrac{9b}{2} + 3b}{\dfrac{9b}{2} - 3b} \\[1em] = \dfrac{\dfrac{9b + 6b}{2}}{\dfrac{9b - 6b}{2}} \\[1em] = \dfrac{15b}{3b} \\[1em] = 5.$

Hence, the value of $\dfrac{5a + 3b}{5a - 3b}$ = 5.

(ii) Given,

$\dfrac{a}{b} = \dfrac{9}{10} \\[0.5em] \Rightarrow a = \dfrac{9b}{10}$

Putting $a = \dfrac{9b}{10}$ in $\dfrac{2a^2 - 3b^2}{2a^2 + 3b^2}$, we get,

$\dfrac{2 \times \Big(\dfrac{9b}{10}\Big)^2 - 3b^2}{2 \times \Big(\dfrac{9b}{10}\Big)^2 + 3b^2} \\[1em] = \dfrac{2 \times \dfrac{81b^2}{100} - 3b^2}{2 \times \dfrac{81b^2}{100} + 3b^2} \\[1em] = \dfrac{\dfrac{81b^2 - 150b^2}{50}}{\dfrac{81b^2 + 150b^2}{50}} \\[1em] = -\dfrac{69b^2}{231b^2} \\[1em] = -\dfrac{23}{77}.$

Hence, the value of $\dfrac{2a^2 - 3b^2}{2a^2 + 3b^2} = -\dfrac{23}{77}$.

If (3x² + 2y²) : (3x² - 2y²) = 11 : 9, find the value of $\dfrac{3x^4 + 25y^4}{3x^4 - 25y^4}.$

Answer

Given,

$\dfrac{3x^2 + 2y^2}{3x^2 - 2y^2} = \dfrac{11}{9}$

Applying componendo and dividendo to above equation,

$\Rightarrow \dfrac{3x^2 + 2y^2 + 3x^2 - 2y^2}{3x^2 + 2y^2 - 3x^2 + 2y^2} = \dfrac{11 + 9}{11 - 9} \\[1em] \Rightarrow \dfrac{6x^2}{4y^2} = \dfrac{20}{2} \\[1em] \Rightarrow \dfrac{3x^2}{2y^2} = 10 \\[1em] \Rightarrow \dfrac{x^2}{y^2} = \dfrac{20}{3}$

Putting value of $\dfrac{x^2}{y^2}$ = $\dfrac{20}{3}$ in $\dfrac{3x^4 + 25y^4}{3x^4 - 25y^4}$,

$\Rightarrow \dfrac{3\big(\dfrac{x^2}{y^2}\big)^2 + 25}{3\big(\dfrac{x^2}{y^2}\big)^2 - 25} \\[1em] \Rightarrow \dfrac{3\big(\dfrac{400}{9}\big) + 25}{3\big(\dfrac{400}{9}\big) - 25} \\[1em] \Rightarrow \dfrac{\dfrac{400}{3} + 25}{\dfrac{400}{3} - 25} \\[1em] \Rightarrow \dfrac{\dfrac{400 + 75}{3}}{\dfrac{400 - 75}{3}} \\[1em] \Rightarrow \dfrac{475}{325} \\[1em] \Rightarrow \dfrac{19}{13}.$

Hence, the value of $\dfrac{3x^4 + 25y^4}{3x^4 - 25y^4} \text{ is } \dfrac{19}{13}.$

If x = $\dfrac{2mab}{a + b},$ find the value of

$\dfrac{x + ma}{x - ma} + \dfrac{x + mb}{x - mb}.$

Answer

$x = \dfrac{2mab}{a + b}$

Putting this value of x in

$\phantom{= }\dfrac{x + ma}{x - ma} + \dfrac{x + mb}{x - mb} \\[1.5em] = \dfrac{\dfrac{2mab}{a + b} + ma}{\dfrac{2mab}{a + b} - ma} + \dfrac{\dfrac{2mab}{a + b} + mb}{\dfrac{2mab}{a + b} - mb} \\[1em] = \dfrac{\dfrac{2mab + ma^2 + mab}{a + b}}{\dfrac{2mab - ma^2 - mab}{a + b}} + \dfrac{\dfrac{2mab + mb^2 + mab}{a + b}}{\dfrac{2mab - mb^2 - mab}{a + b}} \\[1em] = \dfrac{2mab + ma^2 + mab}{2mab - ma^2 - mab} + \dfrac{2mab + mb^2 + mab}{2mab - mb^2 - mab} \\[1em] = \dfrac{ma(2b + a + b)}{ma(2b - a - b)} + \dfrac{mb(2a + b + a)}{mb(2a - b - a)} \\[1em] = \dfrac{3b + a}{b - a} + \dfrac{3a + b}{a - b} \\[1em] = \dfrac{3b + a}{b - a} - \dfrac{3a + b}{b - a} \\[1em] = \dfrac{3b + a - 3a - b}{b - a} \\[1em] = \dfrac{2b - 2a}{b - a} \\[1em] = \dfrac{2(b - a)}{(b - a)} \\[1em] = 2.$

Hence, the value of $\dfrac{x + ma}{x - ma} + \dfrac{x + mb}{x - mb} = 2.$

If x = $\dfrac{pab}{a + b}$, prove that

$\dfrac{x + pa}{x - pa} - \dfrac{x + pb}{x - pb} = \dfrac{2(a^2 - b^2)}{ab}.$

Answer

Given,

$x = \dfrac{pab}{a + b} \\[1em] \Rightarrow \dfrac{x}{pa} = \dfrac{b}{a + b} \text{ and } \dfrac{x}{pb} = \dfrac{a}{a + b}$

Applying componendo and dividendo on both equations,

$\Rightarrow \dfrac{x + pa}{x - pa} = \dfrac{b + a + b}{b - a - b} \text{ and } \dfrac{x + pb}{x - pb} = \dfrac{a + a + b}{a - a - b} \\[1em] \Rightarrow \dfrac{x + pa}{x - pa} = -\dfrac{2b + a}{a} \text{ and } \dfrac{x + pb}{x - pb} = -\dfrac{2a + b}{b}$

Subtracting both the equations,

$\Rightarrow \dfrac{x + pa}{x - pa} - \dfrac{x + pb}{x - pb} = -\dfrac{2b + a}{a} - \Big(-\dfrac{2a + b}{b}\Big) \\[1em] = -\dfrac{2b + a}{a} + \dfrac{2a + b}{b} \\[1em] = \dfrac{-2b^2 - \cancel{ab} + 2a^2 + \cancel{ab}}{ab} \\[1em] = \dfrac{2(a^2 - b^2)}{ab} = \text{R.H.S.} \\[1em]$

Since, L.H.S. = R.H.S. , hence, proved that,

$\dfrac{x + pa}{x - pa} - \dfrac{x + pb}{x - pb} = \dfrac{2(a^2 - b^2)}{ab}.$

Find x from the equation :

$\dfrac{a + x + \sqrt{a^2 - x^2}}{a + x - \sqrt{a^2 - x^2}} = \dfrac{b}{x}.$

Answer

Given,

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

Applying componendo and dividendo,

$\Rightarrow \dfrac{a + x + \sqrt{a^2 - x^2} + a + x - \sqrt{a^2 - x^2}}{a + x + \sqrt{a^2 - x^2} - a - x + \sqrt{a^2 - x^2}} = \dfrac{b + x}{b - x} \\[1em] \Rightarrow \dfrac{2(a + x)}{2\sqrt{a^2 - x^2}} = \dfrac{b + x}{b - x} \\[1em] \Rightarrow \dfrac{(a + x)}{\sqrt{a^2 - x^2}} = \dfrac{b + x}{b - x} \\[1em]$

Squaring both sides,

$\Rightarrow \dfrac{(a + x)^2}{a^2 - x^2} = \dfrac{(b + x)^2}{(b - x)^2} \\[1em] \Rightarrow \dfrac{(a + x)^2}{(a + x)(a - x)} = \dfrac{(b + x)^2}{(b - x)^2} \\[1em] \Rightarrow \dfrac{(a + x)}{(a - x)} = \dfrac{(b + x)^2}{(b - x)^2} \\[1em]$

Again applying componendo and dividendo,

$\Rightarrow \dfrac{a + x + a - x}{a + x - a + x} = \dfrac{(b + x)^2 + (b - x)^2}{(b + x)^2 - (b - x)^2} \\[1em] \Rightarrow \dfrac{2a}{2x} = \dfrac{b^2 + x^2 + 2bx + b^2 + x^2 - 2bx}{b^2 + x^2 + 2bx - (b^2 + x^2 - 2bx)} \\[1em] \Rightarrow \dfrac{2a}{2x} = \dfrac{b^2 + b^2 + x^2 + x^2 + 2bx - 2bx}{b^2 - b^2 + x^2 - x^2 + 2bx - (-2bx)} \\[1em] \Rightarrow \dfrac{2a}{2x} = \dfrac{2(b^2 + x^2)}{4bx} \\[1em] \Rightarrow \dfrac{a}{x} = \dfrac{b^2 + x^2}{2bx}$

Multiplying both sides by x we get,

$\Rightarrow a = \dfrac{b^2 + x^2}{2b}$

On cross-multiplication,

$\Rightarrow 2ab = b^2 + x^2 \\[1em] \Rightarrow x^2 = 2ab - b^2 \\[1em] \Rightarrow x = \sqrt{2ab - b^2}.$

Hence, the value of x is $\sqrt{2ab - b^2}.$

If x = $\dfrac{\sqrt[3]{a + 1} + \sqrt[3]{a - 1}}{\sqrt[3]{a + 1} - \sqrt[3]{a - 1}},$

prove that x³ - 3ax² + 3x - a = 0.

Answer

Given,

$\dfrac{x}{1} = \dfrac{\sqrt[3]{a + 1} + \sqrt[3]{a - 1}}{\sqrt[3]{a + 1} - \sqrt[3]{a - 1}}$

Applying componendo and dividendo,

$\Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt[3]{a + 1} + \sqrt[3]{a - 1} + \sqrt[3]{a + 1} - \sqrt[3]{a - 1}}{\sqrt[3]{a + 1} + \sqrt[3]{a - 1} - \sqrt[3]{a + 1} + \sqrt[3]{a - 1}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt[3]{a + 1}}{2\sqrt[3]{a - 1}}$

Cubing both the sides,

$\Rightarrow \dfrac{(x + 1)^3}{(x - 1)^3} = \dfrac{a + 1}{a - 1}$

Again applying componendo and dividendo,

$\Rightarrow \dfrac{(x + 1)^3 + (x - 1)^3}{(x + 1)^3 - (x - 1)^3} = \dfrac{a + \cancel{1} + a - \cancel{1}}{\cancel{a} + 1 - \cancel{a} + 1} \\[1em] \Rightarrow \dfrac{x^3 + 1 + 3x(x + 1) + x^3 - 1 - 3x(x - 1)}{x^3 + 1 + 3x(x + 1) - (x^3 - 1 - 3x(x - 1))} = \dfrac{2a}{2} \\[1em] \Rightarrow \dfrac{x^3 + x^3 + 1 - 1 + 3x^2 - 3x^2 + 3x + 3x}{x^3 - x^3 + 1 + 1 + 3x^2 + 3x^2 + 3x - 3x} = \dfrac{2a}{2} \\[1em] \Rightarrow \dfrac{2x^3 + 6x}{2 + 6x^2} = a \\[1em] \Rightarrow \dfrac{2(x^3 + 3x)}{2(1 + 3x^2)} = a \\[1em] \Rightarrow \dfrac{x^3 + 3x}{1 + 3x^2} = a$

On cross-multiplication,

$\Rightarrow x^3 + 3x = a(1 + 3x^2) \\[1em] \Rightarrow x^3 + 3x = a + 3ax^2 \\[1em] \Rightarrow x^3 - 3ax^2 + 3x - a = 0.$

Hence, proved that x³ - 3ax² + 3x - a = 0.

If $\dfrac{(a + b)^3}{(a - b)^3} = \dfrac{64}{27}$

(a) Find $\dfrac{a + b}{a - b}$

(b) Hence using properties of proportion, find a : b.

Answer

(a) Solving,

$\Rightarrow \dfrac{(a + b)^3}{(a - b)^3} = \dfrac{64}{27} \\[1em] \Rightarrow \dfrac{(a + b)^3}{(a - b)^3} = \dfrac{4^3}{3^3} \\[1em] \Rightarrow \Big(\dfrac{a + b}{a - b}\Big)^3 = \Big(\dfrac{4}{3}\Big)^3 \\[1em] \Rightarrow \dfrac{a + b}{a - b} = \dfrac{4}{3} \\[1em]$

Hence, $\dfrac{a + b}{a - b} = \dfrac{4}{3}.$

(b) Solving further,

$\Rightarrow 3(a + b) = 4(a - b) \\[1em] \Rightarrow 3a + 3b = 4a - 4b \\[1em] \Rightarrow 4a - 3a = 3b + 4b \\[1em] \Rightarrow a = 7b \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{7}{1} \\[1em] \Rightarrow a : b = 7 : 1.$

Hence, a : b = 7 : 1.

If x, y and z are in continued proportion, prove that :

$\dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} = \dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3}$

Answer

Given,

x, y and z are in continued proportion.

$\therefore \dfrac{x}{y} = \dfrac{y}{z} \\[1em] \Rightarrow y^2 = xz$

To prove :

$\dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} = \dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3}$

Solving L.H.S.,

$\Rightarrow \dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} \\[1em] \Rightarrow \dfrac{x^3 + y^3 + z^3}{x^2.y^2.z^2} \\[1em] \Rightarrow \dfrac{x^3 + y^3 + z^3}{x^2.xz.z^2} \\[1em] \Rightarrow \dfrac{x^3 + y^3 + z^3}{x^3.z^3} \\[1em] \Rightarrow \dfrac{x^3}{x^3.z^3} + \dfrac{y^3}{x^3z^3} + \dfrac{z^3}{x^3.z^3} \\[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{y^3}{(xz)^3} + \dfrac{1}{x^3} \\[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{y^3}{(y^2)^3} + \dfrac{1}{x^3} \\[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{y^3}{y^6} + \dfrac{1}{x^3} \\[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{1}{y^3} + \dfrac{1}{x^3}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} = \dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3}$.