Ganit Prakash - Class-X - Ratio and Proportion
Let us work out 5.1 Step-by-Step Solutions
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Ratio Basics(i) 4 months and 1 year 6 months
(ii) 75 paise and 1 rupee 25 paise
(iii) 60 cm. and 0.6 metre
(iv) 1.2 kg and 60 gram.
(i) 4 months and 1 year 6 months
1 year 6 months = (12 + 6) months = 18 months.
Ratio = \(4 : 18 = 2 : 9\). Since \(2 < 9\), it is a ratio of less inequality.
(ii) 75 paise and 1 rupee 25 paise
1 rupee 25 paise = 125 paise.
Ratio = \(75 : 125 = 3 : 5\). Since \(3 < 5\), it is a ratio of less inequality.
(iii) 60 cm. and 0.6 metre
0.6 metre = \(0.6 \times 100\) cm = 60 cm.
Ratio = \(60 : 60 = 1 : 1\). This is a ratio of equality.
(iv) 1.2 kg and 60 gram.
1.2 kg = \(1.2 \times 1000\) g = 1200 g.
Ratio = \(1200 : 60 = 20 : 1\). Since \(20 > 1\), it is a ratio of greater inequality.
(ii) Let us write when it is possible to find ratio of x days and z months.
(iii) Let us write what type of mixed ratio of a ratio and its inverse ratio.
(iv) Let us find mixed ratio of \(\frac{a}{b}:c\), \(\frac{b}{c}:a\), \(\frac{c}{a}:b\).
(v) Let us write by calculating what ratio and \(x^2:yz\) will form the mixed ratio \(xy:z^2\)
(vi) Let us calculate compound ratio of inverse ratio of \(x^2:\frac{yz}{x}\), \(y^2:\frac{zx}{y}\), \(z^2:\frac{yx}{z}\)
(i) \(p \text{ kg} = 1000p \text{ gram}\).
The required Ratio = \(1000p : q\).
(ii) \(z\) months = \(30z\) days.
So the required ratio is \(x : 30z\)
(iii) Let a ratio be \(x:y\) and its inverse ratio will be \(y:x\).
So, mixed ratio \(= (x \times y) : (y \times x) = xy : xy = 1 : 1\).
Answer: Ratio of equality.
(iv) Mixed ratio = \((\frac{a}{b} \times \frac{b}{c} \times \frac{c}{a}) : (c \times a \times b) = 1 : abc\).
(v) Let the required ratio be \(a:b\).
Then, the mixed ratio of \(a:b\) and \(x^2:yz\) is \((a \times x^2) : (b \times yz)\).
ATP, \(\frac{ax^2}{byz} = \frac{xy}{z^2} \implies \frac{a}{b} = \frac{xy \cdot yz}{z^2 \cdot x^2}\)
\(= \frac{xy^2z}{x^2z^2} = \frac{y^2}{xz}\).
Answer: \(y^2 : xz\).
(vi) Inverse ratios: \(\frac{yz}{x}:x^2\), \(\frac{zx}{y}:y^2\), \(\frac{yx}{z}:z^2\).
Compound ratio \(= (\frac{yz}{x} \cdot \frac{zx}{y} \cdot \frac{yx}{z}) : (x^2 \cdot y^2 \cdot z^2)\)
\(= \frac{x^2y^2z^2}{xyz} : x^2y^2z^2 = xyz : x^2y^2z^2 = 1 : xyz\).
(i) \(4:5\), \(5:7\) and \(9:11\)
(ii) \((x+y):(x-y)\), \((x^2+y^2):(x+y)^2\) and \((x^2-y^2)^2:(x^4-y^4)\)
(i) Mixed ratio \(= (4 \times 5 \times 9) : (5 \times 7 \times 11) = 180 : 385 = 36 : 77\).
(ii) Product of antecedents: \((x+y) \cdot (x^2+y^2) \cdot (x^2-y^2)^2\)
Product of consequents: \((x-y) \cdot (x+y)^2 \cdot (x^4-y^4)\)
\(= (x-y) \cdot (x+y)^2 \cdot (x^2-y^2)(x^2+y^2) = (x-y)(x+y) \cdot (x+y) \cdot (x^2-y^2)(x^2+y^2) = (x^2-y^2) \cdot (x+y) \cdot (x^2-y^2)(x^2+y^2) = ((x+y) \cdot (x^2+y^2) \cdot (x^2-y^2)^2\)
So, the required ratio \(= 1 : 1\).
Answer: Ratio of equality (1:1).
(ii) If \(A:B = 2:3\), \(B:C = 4:5\) and \(C:D = 6:7\), let us find \(A:D\).
(iii) If \(A:B = 3:4\) and \(B:C = 2:3\), let us find \(A:B:C\)
(iv) If \(x:y = 2:3\) and \(y:z = 4:7\) let us find \(x:y:z\)
(i) \(\frac{A}{B} \times \frac{B}{C} = \frac{6}{7} \times \frac{8}{7} \implies \frac{A}{C} = \frac{48}{49}\).
Answer: \(48 : 49\).
(ii) \(\frac{A}{B} \times \frac{B}{C} \times \frac{C}{D} = \frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} = \frac{48}{105} = \frac{16}{35}\).
Answer: \(16 : 35\).
(iii) \(A:B = 3:4\),
\(B:C = 2:3 = 4:6\) (Multiplying by 2).
So, \(A:B:C = 3:4:6\)
Answer: \(3 : 4 : 6\).
(iv) \(x:y = 2:3 = 8:12\) (Multiplying by 4),
\(y:z = 4:7 = 12:21\) (Multiplying by 3).
So, \(x:y:z = 8:12:21\)
Answer: \(8 : 12 : 21\).
(ii) If \(a:b = 8:7\), let us show that \((7a-3b):(11a-9b) = 7:5\)
(iii) If \(p:q = 5:7\) and \(p-q = -4\), let us find the value of \(3p+4q\).
(i) Let \(x = 3k\) and \(y = 4k\).
\((3y-x) = 3(4k) - 3k = 12k - 3k = 9k\)
\((2x+y) = 2(3k) + 4k = 6k + 4k = 10k\)
Ratio \(= 9k : 10k = 9:10\).
(ii) Let \(a = 8k\) and \(b = 7k\).
LHS: \(\frac{7(8k)-3(7k)}{11(8k)-9(7k)} = \frac{56k-21k}{88k-63k} = \frac{35k}{25k} = \frac{7}{5}\). [Showed]
(iii) Let \(p=5k, q=7k\).
Given \(p-q = -4 \)
\(\implies 5k-7k = -4 \)
\(\implies -2k = -4 \)
\(\implies k=2\).
Now, \(3p+4q = 3(5 \times 2) + 4(7 \times 2) = 30 + 56 = 86\).
(ii) If \((3a+7b):(5a-3b) = 5:3\), let us find \(a:b\).
(i) \(\frac{5x-3y}{2x+4y} = \frac{11}{12}\)
\(\implies 60x - 36y = 22x + 44y\)
\(\implies 60x - 22x = 44y + 36y\)
\(\implies 38x = 80y \implies \frac{x}{y} = \frac{80}{38} = \frac{40}{19}\).
Answer: \(40:19\).
(ii) \(\frac{3a+7b}{5a-3b} = \frac{5}{3} \)
\(\implies 9a + 21b = 25a - 15b\)
\(\implies 15b + 21b = 25a - 9a\)
\(\implies 36b = 16a \implies \frac{a}{b} = \frac{36}{16} = \frac{9}{4}\).
Answer: \(9:4\).
(ii) If \((10x+3y):(5x+2y) = 9:5\), let us show that, \((2x+y):(x+2y) = 11:13\)
(i) Given: \(\frac{7x-5y}{3x+4y} = \frac{7}{11}\)
\(\implies 11(7x - 5y) = 7(3x + 4y)\)
\(\implies 77x - 55y = 21x + 28y\)
\(\implies 77x - 21x = 28y + 55y\)
\(\implies 56x = 83y\)
\(\implies x = \frac{83}{56}y\)
LHS: \(\frac{3x - 2y}{3x + 4y}\)
\(= \frac{3\left(\frac{83}{56}y\right) - 2y}{3\left(\frac{83}{56}y\right) + 4y}\)
\(= \frac{\frac{249y - 112y}{56}}{\frac{249y + 224y}{56}}\)
\(= \frac{249y - 112y}{249y + 224y}\)
\(= \frac{137y}{473y} = 137:473\). [Showed]
(ii) Given: \(\frac{10x+3y}{5x+2y} = \frac{9}{5}\)
By cross-multiplication:
\(5(10x + 3y) = 9(5x + 2y)\)
\(\implies 50x + 15y = 45x + 18y\)
Arranging terms with \(x\) on the left and \(y\) on the right:
\(\implies 50x - 45x = 18y - 15y\)
\(\implies 5x = 3y\)
\(\implies x = \frac{3}{5}y\)
LHS: \(\frac{2x + y}{x + 2y}\)
Substituting the value of \(x = \frac{3}{5}y\):
\(= \frac{2\left(\frac{3}{5}y\right) + y}{\left(\frac{3}{5}y\right) + 2y}\)
\(= \frac{\frac{6y}{5} + y}{\frac{3y}{5} + 2y}\)
\(= \frac{\frac{6y + 5y}{5}}{\frac{3y + 10y}{5}}\)
\(= \frac{6y + 5y}{3y + 10y}\)
\(= \frac{11y}{13y}\)
\(= \frac{11}{13} = 11:13\). [Showed]
(ii) Let us calculate what term should be subtracted from each term of the ratio \(a:b\) to make the ratio \(m:n\).
(iii) What term should be added to antecedent and subtracted from consequent of ratio \(4:7\) to make a compound ratio of \(2:3\) and \(5:4\).
(i) Let \(x\) be added to both terms of the ratio \(2 : 5\).
According to the condition:
\(\frac{2 + x}{5 + x} = \frac{6}{11}\)
By cross-multiplication:
\(11(2 + x) = 6(5 + x)\)
\(\implies 22 + 11x = 30 + 6x\)
\(\implies 11x - 6x = 30 - 22\)
\(\implies 5x = 8\)
\(\implies x = \frac{8}{5} = 1.6\)
Answer: 1.6
(ii) Let \(x\) be subtracted from each term of the ratio \(a : b\).
According to the condition:
\(\frac{a - x}{b - x} = \frac{m}{n}\)
By cross-multiplication:
\(n(a - x) = m(b - x)\)
\(\implies an - nx = bm - mx\)
Arranging terms with \(x\) on the left:
\(\implies mx - nx = bm - an\)
\(\implies x(m - n) = bm - an\)
\(\implies x = \frac{bm - an}{m - n}\)
Answer: \(\frac{bm - an}{m - n}\)
(iii) First, let us find the compound ratio of \(2:3\) and \(5:4\).
Compound Ratio \(= (2 \times 5) : (3 \times 4) = 10 : 12 = 5 : 6\).
Let \(x\) be added to the antecedent and subtracted from the consequent of \(4 : 7\).
According to the condition:
\(\frac{4 + x}{7 - x} = \frac{5}{6}\)
By cross-multiplication:
\(6(4 + x) = 5(7 - x)\)
\(\implies 24 + 6x = 35 - 5x\)
\(\implies 6x + 5x = 35 - 24\)
\(\implies 11x = 11\)
\(\implies x = \frac{11}{11} = 1\)
Answer: 1
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