Ratio & Proportion — Exercise 7(B)

If x, 2, 10 and y are in proportion, the values of x and y are respectively :

1. 0.2 and 0.25

2. 0.2 and 50

3. 0.4 and 50

4. 0.4 and 25

Answer

Given,

x, 2, 10 and y are in proportion.

$\therefore \dfrac{x}{2} = \dfrac{2}{10} = \dfrac{10}{y}$ .......(1)

Considering L.H.S. of the equation (1) :

$\Rightarrow \dfrac{x}{2} = \dfrac{2}{10} \\[1em] \Rightarrow x = \dfrac{2 \times 2}{10} \\[1em] \Rightarrow x = \dfrac{4}{10} = 0.4$

Considering R.H.S. of the equation (1) :

$\Rightarrow \dfrac{2}{10} = \dfrac{10}{y} \\[1em] \Rightarrow y = \dfrac{10 \times 10}{2} \\[1em] \Rightarrow y = \dfrac{100}{2} \\[1em] \Rightarrow y = 50.$

Hence, Option 3 is the correct option.

If x : y = y : z, then x² : y² is :

1. 1 : x

2. x : y

3. x : z

4. z : x

Answer

Given,

⇒ x : y = y : z

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

Substituting value of y² in x² : y², we get :

⇒ x² : xz

⇒ x : z.

Hence, Option 3 is the correct option.

The mean proportion between $3 + 2\sqrt{2} \text{ and } 3 - 2\sqrt{2}$ is :

1. 1

2. -1

3. $2\sqrt{2}$

4. 3

Answer

Let mean proportion be x.

$\Rightarrow \dfrac{3 + 2\sqrt{2}}{x} = \dfrac{x}{3 - 2\sqrt{2}} \\[1em] \Rightarrow x^2 = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) \\[1em] \Rightarrow x^2 = 3^2 - (2\sqrt{2})^2 \\[1em] \Rightarrow x^2 = 9 - (4 \times 2) \\[1em] \Rightarrow x^2 = 9 - 8 \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = \sqrt{1} = \pm 1.$

Since, geometrical mean cannot be negative,

∴ x = 1.

Hence, Option 1 is the correct option.

If 2x, 9 and 18 are in continued proportion, the value of x is :

1. $2\dfrac{1}{4}$

2. $\dfrac{4}{9}$

3. 1

4. 9

Answer

Given,

2x, 9 and 18 are in continued proportion.

$\therefore \dfrac{2x}{9} = \dfrac{9}{18} \\[1em] \Rightarrow x = \dfrac{9 \times 9}{18 \times 2} \\[1em] \Rightarrow x = \dfrac{9}{4} = 2\dfrac{1}{4}.$

Hence, Option 1 is the correct option.

Find the fourth proportional to 1.5, 4.5 and 3.5

Answer

Let fourth proportional to 1.5, 4.5 and 3.5 be x

$\Rightarrow 1.5 : 4.5 = 3.5 : x \\[1em] \Rightarrow \dfrac{1.5}{4.5} = \dfrac{3.5}{x} \\[1em] \Rightarrow x = \dfrac{3.5 \times 4.5}{1.5} \\[1em] \Rightarrow x = 3.5 \times 3 \\[1em] \Rightarrow x = 10.5$

Hence, fourth proportional to 1.5, 4.5 and 3.5 is 10.5

Find the fourth proportional to 3a, 6a² and 2ab²

Answer

Let fourth proportional to 3a, 6a² and 2ab² be x

$\Rightarrow 3a : 6a^2 = 2ab^2 : x \\[1em] \Rightarrow \dfrac{3a}{6a^2} = \dfrac{2ab^2}{x} \\[1em] \Rightarrow x = \dfrac{2ab^2 \times 6a^2}{3a} \\[1em] \Rightarrow x = 4a^2b^2$

Hence, fourth proportional to 3a, 6a² and 2ab² is 4a²b².

Find the third proportional to $2\dfrac{2}{3}$ and 4.

Answer

Let the third proportion to $2\dfrac{2}{3}$ and 4 be x,

$\Rightarrow 2\dfrac{2}{3} : 4 = 4 : x \\[1em] \Rightarrow \dfrac{8}{3} : 4 = 4 : x \\[1em] \Rightarrow \dfrac{\dfrac{8}{3}}{4} = \dfrac{4}{x} \\[1em] \Rightarrow \dfrac{8}{12} = \dfrac{4}{x} \\[1em] \Rightarrow x = \dfrac{4 \times 12}{8} \\[1em] \Rightarrow x = 6.$

Hence, third proportional to $2\dfrac{2}{3}$ and 4 is 6.

Find the third proportional to a - b and a² - b².

Answer

Let the third proportion to a - b and a² - b² be x,

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

Hence, third proportional to a - b and a² - b² is (a + b)(a² - b²).

Find the mean proportional between 6 + $3\sqrt{3} \text{ and } 8 - 4\sqrt{3}$

Answer

Let mean proportional between 6 + $3\sqrt{3} \text{ and } 8 - 4\sqrt{3}$ be x

$\therefore 6 + 3\sqrt{3} : x = x : 8 - 4\sqrt{3} \\[1em] \Rightarrow \dfrac{6 + 3\sqrt{3}}{x} = \dfrac{x}{8 - 4\sqrt{3}} \\[1em] \Rightarrow x^2 = (6 + 3\sqrt{3})(8 - 4\sqrt{3}) \\[1em] \Rightarrow x^2 = 48 - 24\sqrt{3} + 24\sqrt{3} - 36 \\[1em] \Rightarrow x^2 = 48 - 36 \\[1em] \Rightarrow x^2 = 12 \\[1em] \Rightarrow x = 2\sqrt{3}.$

Hence, x = $2\sqrt{3}$.

Find the mean proportional between a - b and a³ - a²b

Answer

Let mean proportional between a - b and a³ - a²b be x

$\Rightarrow \dfrac{a - b}{x} = \dfrac{x}{a^3 - a^2b} \\[1em] \Rightarrow x^2 = (a - b)(a^3 - a^2b) \\[1em] \Rightarrow x^2 = a^4 - a^3b - a^3b + a^2b^2 \\[1em] \Rightarrow x^2 = a^4 - 2a^3b + a^2b^2 \\[1em] \Rightarrow x^2 = a^2(a^2 + b^2 - 2ab) \\[1em] \Rightarrow x^2 = a^2(a - b)^2 \\[1em] \Rightarrow x = a(a - b).$

Hence, mean proportional between a - b and a³ - a²b = a(a - b).

If x + 5 is the mean proportion between x + 2 and x + 9; find the value of x.

Answer

Given,

x + 5 is the mean proportion between x + 2 and x + 9

$\therefore x + 2 : x + 5 = x + 5 : x + 9 \\[1em] \Rightarrow \dfrac{x + 2}{x + 5} = \dfrac{x + 5}{x + 9} \\[1em] \Rightarrow (x + 2)(x + 9) = (x + 5)(x + 5) \\[1em] \Rightarrow x^2 + 9x + 2x + 18 = x^2 + 5x + 5x + 25 \\[1em] \Rightarrow x^2 + 11x + 18 = x^2 + 10x + 25 \\[1em] \Rightarrow 11x - 10x = 25 - 18 \\[1em] \Rightarrow x = 7.$

Hence, x = 7.

If x², 4 and 9 are in continued proportion, find x.

Answer

Given,

x², 4 and 9 are in continued proportion.

$\therefore \dfrac{x^2}{4} = \dfrac{4}{9} \\[1em] \Rightarrow x^2 = \dfrac{16}{9} \\[1em] \Rightarrow x = \sqrt{\dfrac{16}{9}} \\[1em] \Rightarrow x = \dfrac{4}{3}.$

Hence, x = $\dfrac{4}{3}.$

If y is the mean proportional between x and z; show that xy + yz is the mean proportional between x² + y² and y² + z².

Answer

Given,

y is the mean proportional between x and z

$\therefore \dfrac{x}{y} = \dfrac{y}{z} \Rightarrow y^2 = xz.$

To prove,

xy + yz is the mean proportional between x² + y² and y² + z²

$\therefore \dfrac{x^2 + y^2}{xy + yz} = \dfrac{xy + yz}{y^2 + z^2} \\[1em] \Rightarrow (xy + yz)(xy + yz) = (x^2 + y^2)(y^2 + z^2) \\[1em] \Rightarrow x^2y^2 + xy^2z + xy^2z + y^2z^2 = x^2y^2 + x^2z^2 + y^4 + y^2z^2 ....[\text{i}]$

Substituting y² = xz in L.H.S. of (i)

⇒ x²(xz) + x(xz)z + x(xz)z + (xz)z²

⇒ x³z + x²z² + x²z² + xz³

⇒ x³z + 2x²z² + xz³ .........(ii)

Substituting y² = xz in R.H.S. of (i)

⇒ x²(xz) + x²z² + (xz)² + (xz)z²

⇒ x³z + x²z² + x²z² + xz³

⇒ x³z + 2x²z² + xz³ .........(iii)

Since, (ii) = (iii)

Hence, proved that xy + yz is the mean proportional between x² + y² and y² + z².

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

pqr(p + q + r)³ = (pq + qr + pr)³.

Answer

Since, q is the mean proportional between p and r

$\therefore \dfrac{p}{q} = \dfrac{q}{r} \Rightarrow q^2 = pr.$

Substituting pr = q² in L.H.S. of pqr(p + q + r)³ = (pq + qr + pr)³

⇒ q.q²(p + q + r)³

⇒ q³(p + q + r)³

⇒ [q(p + q + r)]³

⇒ (pq + q² + qr)³

⇒ (pq + pr + qr)³ = R.H.S. (As q² = pr).

Hence, proved that pqr(p + q + r)³ = (pq + qr + pr)³.

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

Answer

Let x, y and z be in continued proportion.

∴ x : y = y : z

$\Rightarrow \dfrac{x}{y} = \dfrac{y}{z} \Rightarrow y^2 = xz$

To prove :

$\dfrac{x}{z} = \dfrac{x^2}{y^2}$

Substituting y² = xz in R.H.S. of above equation we get,

$\dfrac{x^2}{xz} = \dfrac{x}{z}$ = L.H.S.

Hence, proved that ratio of the first to third is duplicate ratio of first to the second.

Find the third proportional to $\dfrac{x}{y} + \dfrac{y}{x} \text{ and } \sqrt{x^2 + y^2}$

Answer

Let third proportional be p.

$\therefore \dfrac{\dfrac{x}{y} + \dfrac{y}{x}}{\sqrt{x^2 + y^2}} = \dfrac{\sqrt{x^2 + y^2}}{p} \\[1em] \Rightarrow \Big(\sqrt{x^2 + y^2}\Big)^2 = p\Big(\dfrac{x}{y} + \dfrac{y}{x}\Big) \\[1em] \Rightarrow x^2 + y^2 = p \times \dfrac{x^2 + y^2}{xy} \\[1em] \Rightarrow p = \dfrac{(x^2 + y^2)(xy)}{x^2 + y^2} \\[1em] \Rightarrow p = xy.$

Hence, third proportional to $\dfrac{x}{y} + \dfrac{y}{x} \text{ and } \sqrt{x^2 + y^2}$ is xy.

If p : q = r : s; then show that:

mp + nq : q = mr + ns : s

Answer

Given,

$\Rightarrow \dfrac{p}{q} = \dfrac{r}{s}$

Multiplying both sides by m:

$\Rightarrow \dfrac{mp}{q} = \dfrac{mr}{s}$

Adding n on both sides:

$\Rightarrow \dfrac{mp}{q} + n = \dfrac{mr}{s} + n \\[1em] \Rightarrow \dfrac{mp + nq}{q}= \dfrac{mr + ns}{s}$

Hence, proved that mp + nq : q = mr + ns : s.

If p + r = mq and $\dfrac{1}{q} + \dfrac{1}{s} = \dfrac{m}{r};$ then prove that : p : q = r : s.

Answer

Given,

$\Rightarrow \dfrac{1}{q} + \dfrac{1}{s} = \dfrac{m}{r} \\[1em] \Rightarrow \dfrac{s + q}{qs} = \dfrac{m}{r} \\[1em] \Rightarrow \dfrac{s + q}{s} = \dfrac{mq}{r} \\[1em] \Rightarrow \dfrac{s + q}{s} = \dfrac{p + r}{r} \text{ (As mq = p + r)} \\[1em] \Rightarrow 1 + \dfrac{q}{s} = \dfrac{p}{r} + 1 \\[1em] \Rightarrow \dfrac{q}{s} = \dfrac{p}{r} \\[1em] \Rightarrow \dfrac{p}{q} = \dfrac{r}{s}.$

Hence, proved that p : q = r : s.