Ganit Prakash - Class-X - Ratio and Proportion
Let us work out 5.2 Proportions & Variables
📘 Exercise 5.2 Solutions
Proportion Laws(i) \(10 : 35 :: x : 42\)
(ii) \(x : 50 :: 3 : 2\)
(i) \(10 : 35 :: x : 42\)
By property of proportion: \(\frac{10}{35} = \frac{x}{42}\)
\(\implies 35x = 10 \times 42\)
\(\implies x = \frac{420}{35} = 12\)
Answer: \(x = 12\)
(ii) \(x : 50 :: 3 : 2\)
\(\frac{x}{50} = \frac{3}{2}\)
\(\implies 2x = 150\)
\(\implies x = \frac{150}{2} = 75\)
Answer: \(x = 75\)
(i) \(\frac{1}{3}, \frac{1}{4}, \frac{1}{5}\)
(ii) 9.6 kg, 7.6 kg, 28.8 kg
(iii) \(x^2y, y^2z, z^2x\)
(iv) \((p-q), (p^2-q^2), p^2-pq+q^2\)
(i) Let fourth proportional be \(x\).
\(\frac{1/3}{1/4} = \frac{1/5}{x} \)
\(\implies \frac{4}{3} = \frac{1}{5x}\)
\( \implies 20x = 3 \)
\(\implies x = \frac{3}{20}\).
(ii) Let fourth proportional be \(x\).
\(\frac{9.6}{7.6} = \frac{28.8}{x} \)
\(\implies x = \frac{7.6 \times 28.8}{9.6} \)
\(= 7.6 \times 3 = 22.8\).
Answer: 22.8 kg
(iii) Let fourth proportional be \(A\).
\(\frac{x^2y}{y^2z} = \frac{z^2x}{A}\)
\( \implies A \cdot x^2y = y^2z \cdot z^2x \)
\(\implies A = \frac{xy^2z^3}{x^2y} = \frac{yz^3}{x}\).
(iv) Let fourth proportional be \(x\).
\(\frac{p-q}{p^2-q^2} = \frac{p^2-pq+q^2}{x} \)
\(\implies \frac{p-q}{(p-q)(p+q)} = \frac{p^2-pq+q^2}{x}\)
\(\implies \frac{1}{p+q} = \frac{p^2-pq+q^2}{x}\)
\( \implies x = (p+q)(p^2-pq+q^2) = p^3+q^3\).
(i) 5, 10 (ii) 0.24, 0.6 (iii) \(p^3q^2, q^2r\) (iv) \((x-y)^2, (x^2-y^2)^2\)
(i) Let 3rd proportional be \(x\).
\(5:10 :: 10:x \)
\(\implies \frac{5}{10} = \frac{10}{x} \)
\(\implies 5x = 100 \)
\(\implies x = 20\).
(ii) Let 3rd proportional be \(x\).
\(\frac{0.24}{0.6} = \frac{0.6}{x}\)
\(\implies x = \frac{0.6 \times 0.6}{0.24} = \frac{0.36}{0.24} = 1.5\).
(iii) Let 3rd proportional be \(x\).
\(\frac{p^3q^2}{q^2r} = \frac{q^2r}{x}\)
\( \implies x \cdot p^3q^2 = (q^2r)^2 \)
\(\implies x = \frac{q^4r^2}{p^3q^2} = \frac{q^2r^2}{p^3}\).
(iv) Let 3rd proportional be \(A\).
\(\frac{(x-y)^2}{(x^2-y^2)^2} = \frac{(x^2-y^2)^2}{A} \)
\(\implies A = \frac{((x-y)(x+y))^4}{(x-y)^2} = \frac{(x-y)^4(x+y)^4}{(x-y)^2} = (x-y)^2(x+y)^4\).
(i) 5 and 80 (ii) 8.1 and 2.5 (iii) \(x^3y\) and \(xy^3\) (iv) \((x-y)^2, (x+y)^2\)
(i) Let the mean proportional be \(x\).
\(5:x :: x:80\)
\(\implies x^2 = 5 \times 80\)
\(\implies x^2 = 400\)
\(\implies x = \sqrt{400} = 20\).
(ii) Let the mean proportional be \(x\).
\(8.1:x :: x:2.5\)
\(\implies x^2 = 8.1 \times 2.5\)
\(\implies x^2 = 20.25\)
\(\implies x = \sqrt{20.25} = 4.5\).
(iii) Let the mean proportional be \(A\).
\(x^3y:A :: A:xy^3\)
\(\implies A^2 = x^3y \cdot xy^3\)
\(\implies A^2 = x^4y^4\)
\(\implies A = \sqrt{x^4y^4} = x^2y^2\).
(iv) Let the mean proportional be \(M\).
\((x-y)^2:M :: M:(x+y)^2\)
\(\implies M^2 = (x-y)^2 \cdot (x+y)^2\)
\(\implies M^2 = \{(x-y)(x+y)\}^2\)
\(\implies M = (x-y)(x+y) = x^2-y^2\).
If the ratios \(a:b\) and \(c:d\) express mutually opposite relations, then their inverse ratios \(b:a\) and \(d:c\) will also express mutually opposite relations.
For three numbers \(a, b, c\) in continued proportion, the relation is \(\frac{a}{b} = \frac{b}{c}\).
This can be arranged in the following ways:
(1) \(a:b :: b:c\)
(2) \(c:b :: b:a\) (Inverse relation)
Thus, 2 continued proportions can be constructed.
Let the five numbers in continued proportion be \(a, b, c, d\), and \(e\).
Given, first number \(a = 2\) and second number \(b = 6\).
According to the condition of continued proportion:
\(\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{d}{e}\)
Taking the first two ratios:
\(\frac{2}{6} = \frac{6}{c}\)
\(\implies 2c = 6 \times 6\)
\(\implies 2c = 36\)
\(\implies c = \frac{36}{2} = 18\)
Taking the next ratios:
\(\frac{b}{c} = \frac{c}{d}\)
\(\implies \frac{6}{18} = \frac{18}{d}\)
\(\implies 6d = 18 \times 18\)
\(\implies 6d = 324\)
\(\implies d = \frac{324}{6} = 54\)
Taking the final ratios:
\(\frac{c}{d} = \frac{d}{e}\)
\(\implies \frac{18}{54} = \frac{54}{e}\)
\(\implies 18e = 54 \times 54\)
\(\implies 18e = 2916\)
\(\implies e = \frac{2916}{18} = 162\)
Answer: 162
Let \(x\) be added to each of 6, 15, 20, and 43 to make them proportional.
According to the problem: \((6+x) : (15+x) :: (20+x) : (43+x)\)
\(\implies \frac{6+x}{15+x} = \frac{20+x}{43+x}\)
By cross-multiplication:
\(\implies (6+x)(43+x) = (15+x)(20+x)\)
\(\implies 258 + 6x + 43x + x^2 = 300 + 15x + 20x + x^2\)
\(\implies 258 + 49x = 300 + 35x\)
\(\implies 49x - 35x = 300 - 258\)
\(\implies 14x = 42\)
\(\implies x = \frac{42}{14} = 3\).
Answer: 3
Let \(x\) be subtracted from each of 23, 30, 57, and 78 to make them proportional.
According to the problem: \((23-x) : (30-x) :: (57-x) : (78-x)\)
\(\implies \frac{23-x}{30-x} = \frac{57-x}{78-x}\)
By cross-multiplication:
\(\implies (23-x)(78-x) = (30-x)(57-x)\)
\(\implies 1794 - 23x - 78x + x^2 = 1710 - 30x - 57x + x^2\)
\(\implies 1794 - 101x = 1710 - 87x\)
\(\implies 1794 - 1710 = 101x - 87x\)
\(\implies 84 = 14x\)
\(\implies x = \frac{84}{14} = 6\).
Answer: 6
Let \(x\) be subtracted from each of \(p, q, r, s\).
According to the problem: \((p-x) : (q-x) :: (r-x) : (s-x)\)
\(\implies \frac{p-x}{q-x} = \frac{r-x}{s-x}\)
By cross-multiplication:
\(\implies (p-x)(s-x) = (q-x)(r-x)\)
\(\implies ps - px - sx + x^2 = qr - qx - rx + x^2\)
\(\implies ps - x(p+s) = qr - x(q+r)\)
\(\implies x(q+r) - x(p+s) = qr - ps\)
\(\implies x(q + r - p - s) = qr - ps\)
\(\implies x = \frac{qr - ps}{q + r - p - s}\).
Answer: \(\frac{qr - ps}{q + r - p - s}\)
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