Chapter 4 – Ratio and Proportion
Practice set 4.1
(1) From the following pairs of numbers, find the reduced form of ratio of first number to second number.
(i) 72, 60
Solution:
Ratio of 72 to 60
= \(\large \frac {72}{60}\)
= \(\large \frac {12\,×\,6}{12\,×\,5}\)
= \(\large \frac {6}{5}\)
= 6 : 5
(ii) 38, 57
Solution:
Ratio of 38 to 57
= \(\large \frac {38}{57}\)
= \(\large \frac {19\,×\,2}{19\,×\,3}\)
= \(\large \frac {2}{3}\)
= 2 : 3
(iii) 52, 78
Solution:
Ratio of 52 to 78
= \(\large \frac {52}{78}\)
= \(\large \frac {26\,×\,2}{26\,×\,3}\)
= \(\large \frac {2}{3}\)
= 2 : 3
(2) Find the reduced form of the ratio of the first quantity to second quantity.
(i) 700 ₹, 308 ₹
Solution:
Ratio of 700 ₹ to 308 ₹
= \(\large \frac {700}{308}\)
= \(\large \frac {28\,×\,25}{28\,×\,11}\)
= \(\large \frac {25}{11}\)
= 25 : 11
(ii) 14 ₹, 12 ₹. 40 paise.
Solution:
₹ 14 = 14 × 100 = 1400 paise
₹ 12, 40 paise = (12 × 100) + 40 = 1240 paise
Ratio of 14 ₹ to 12 ₹, 40 paise =
= \(\large \frac {1400}{1240}\)
= \(\large \frac {4\,×\,35}{4\,×\,31}\)
= \(\large \frac {35}{31}\)
= 35 : 31
(iii) 5 litre, 2500 ml
Solution:
5 litre = 5 × 1000 = 5000 ml
Ratio of 5000 ml to 2500 ml
= \(\large \frac {5000}{2500}\)
= \(\large \frac {2500\,×\,2}{2500\,×\,1}\)
= \(\large \frac {2}{1}\)
= 2 : 1
(iv) 3 years 4 months, 5 years 8 months
Solution:
3 years 4 months = 3 × 12 + 4 = 40 months
5 years 8 months = 5 × 12 + 8 = 68 months
Ratio of 3 years 4 months to 5 years 8 months
= \(\large \frac {40}{68}\)
= \(\large \frac {4\,×\,10}{4\,×\,7}\)
= \(\large \frac {10}{7}\)
= 10 : 7
(v) 3.8 kg, 1900 gm
Solution:
3.8 kg = 3.8 × 1000 = 3800 kg
Ratio of 3.8 kg to 1900 gm
= \(\large \frac {3800}{1900}\)
= \(\large \frac {1900\,×\,2}{1900\,×\,1}\)
= \(\large \frac {2}{1}\)
= 2 : 1
(vi) 7 minutes 20 seconds, 5 minutes 6 seconds.
Solution:
7 minutes 20 seconds = 7 × 60 + 20 = 440 seconds
5 minutes 6 seconds = 5 × 60 + 6 = 306 seconds
Ratio of 7 minutes 20 seconds to 5 minutes 6 seconds
= \(\large \frac {440}{306}\)
= \(\large \frac {2\,×\,220}{2\,×\,103}\)
= \(\large \frac {220}{103}\)
= 220 : 103
(3) Express the following percentages as ratios in the reduced form.
(i) 75 : 100
Solution:
= \(\large \frac {75}{100}\)
= \(\large \frac {25\,×\,3}{25\,×\,4}\)
= \(\large \frac {3}{4}\)
= 3 : 4
(ii) 44 : 100
Solution:
= \(\large \frac {44}{100}\)
= \(\large \frac {4\,×\,11}{4\,×\,25}\)
= \(\large \frac {11}{25}\)
= 11 : 25
(iii) 6.25%
Solution:
= \(\large \frac {6.25}{100}\)
= \(\large \frac {625}{10000}\)
= \(\large \frac {625\,×\,1}{625\,×\,16}\)
= \(\large \frac {1}{16}\)
= 1 : 16
(iv) 52 : 100
Solution:
= \(\large \frac {52}{100}\)
= \(\large \frac {4\,×\,13}{4\,×\,25}\)
= \(\large \frac {13}{25}\)
= 13 : 25
(v) 0.64%
Solution:
= \(\large \frac {0.64}{100}\)
= \(\large \frac {64}{10000}\)
= \(\large \frac {16\,×\,4}{16\,×\,625}\)
= \(\large \frac {4}{625}\)
= 4 : 625
(4) Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required?
Solution:
Suppose x men will construct the house in 6 days.
According to the given condition, it is an inverse proportion
∴ 3 × 8 = x × 6
∴ x = \(\large \frac {24}{6}\)
∴ x = 4
Ans: 4 men will construct the house in 6 days.
(5) Convert the following ratios into percentage.
(i) 15 : 25
Solution:
= \(\large \frac {15}{25}\)
= \(\large \frac {15\,×\,4}{25\,×\,4}\)
= \(\large \frac {60}{100}\)
= 60%
(ii) 47 : 50
Solution:
= \(\large \frac {47}{50}\)
= \(\large \frac {47\,×\,2}{50\,×\,2}\)
= \(\large \frac {94}{100}\)
= 94%
(iii) \(\large \frac {7}{10}\)
Solution:
= \(\large \frac {7}{10}\)
= \(\large \frac {7\,×\,10}{10\,×\,10}\)
= \(\large \frac {70}{100}\)
= 70%
(iv) \(\large \frac {546}{600}\)
Solution:
= \(\large \frac {546}{600}\)
= \(\large \frac {546\,÷\,6}{600\,÷\,6}\)
= \(\large \frac {91}{100}\)
= 91%
(v) \(\large \frac {7}{16}\)
Solution:
= \(\large \frac {7}{16}\)
= \(\large \frac {7\,×\,100}{16\,×\,100}\)
= \(\large \frac {700}{16\,×\,100}\)
= \(\large \frac {43.75}{100}\)
= 43.75%
(6) The ratio of ages of Abha and her mother is 2 : 5. At the time of Abha’s birth her mothers age was 27 year. Find the present ages of Abha and her mother.
Solution:
Let the common multiple of the given ratio be x.
The ratio of ages of Abha and her mother is 2 : 5.
∴ Their ages are 2x years and 5x years respectively
According to the given condition,
5x – 2x = 27
∴ 3x = 27
∴ x = 9
∴ 2x = 2 × 9 = 18
and 5x = 5 × 9 = 45
Ans: Present age of Abha and her mother is 18 years and 45 years respectively.v
(7) Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5 : 4?
Solution:
Let the number of years for which the ratio of their ages will be 5 : 4 be x.
Present ages of Vatsala and Sara are 14 years and 10 years respectively.
After x years, Vatsala’s age = (14 + x) years Sara’s age = (10 + x) years
According to given condition
\(\large \frac {14\,+\,x}{10\,+\,x}\) = \(\large \frac {5}{4}\)
∴ 4 (14 + x) = 5 (10 + x)
∴ 56 + 4x = 50 + 5x
∴ 56 – 50 = 5x – 4x
∴ 6 = x
i.e. x = 6
Ans: After 6 years, ratio of ages of Vatsala and Sara will be 5 : 4
(8) The ratio of present ages of Rehana and her mother is 2 : 7. After 2 years, the ratio of their ages will be 1 : 3. What is Rehana’s present age ?
Solution:
Let the common multiple of the given ratio be x.
The ratio of the present ages of Rehana and her mother is 2 :7.
∴ Their present ages are 2x years and 7x years respectively
After 2 years, Rehana’s age = (2x + 2) years
Mother’s age = (7x + 2) years
According to given condition,
\(\large \frac {2x\,+\,2}{7x\,+\,2}\) = \(\large \frac {1}{3}\)
∴ 3 (2x + 2) = 7x + 2
∴ 6x + 6 = 7x + 2
∴ 6 – 2 = 7x – 6x
∴ 4 = x
∴ x = 4
∴ 2x = 2 × 4 = 8
Ans: Present age of Rehana is 8 years.
Practice set 4.2
(1) Using the property of \(\large \frac {a}{b}\) = \(\large \frac {ak}{bk}\), fill in the blanks substituting proper numbers in the following.
(i) \(\large \frac {5}{7}\) = \(\large \frac {…}{28}\) = \(\large \frac {35}{…}\)= \(\large \frac {…}{3.5}\)
Solution:
\(\large \frac {5}{7}\) = \(\large \frac {…}{28}\)
∴ \(\large \frac {5}{7}\) = \(\large \frac {5\,×\,4}{7\,×\,4}\) = \(\large \frac {20}{28}\)
\(\large \frac {5}{7}\) = \(\large \frac {35}{…}\)
∴ \(\large \frac {5}{7}\) = \(\large \frac {5\,×\,7}{7\,×\,7}\) = \(\large \frac {35}{49}\)
\(\large \frac {5}{7}\) = \(\large \frac {…}{3.5}\)
∴ \(\large \frac {5}{7}\) = \(\large \frac {5\,×\,0.5}{7\,×\,0.5}\) = \(\large \frac {2.5}{3.5}\)
(ii) \(\large \frac {9}{14}\) = \(\large \frac {4.5}{…}\) = \(\large \frac {…}{42}\) = \(\large \frac {…}{3.5}\)
Solution:
\(\large \frac {9}{14}\) = \(\large \frac {4.5}{…}\)
∴ \(\large \frac {9}{14}\) = \(\large \frac {9\,×\,0.5}{\,×\,14}\) = \(\large \frac {20}{28}\)
\(\large \frac {9}{14}\) = \(\large \frac {…}{42}\)
∴ \(\large \frac {9}{14}\) = \(\large \frac {9\,×\,3}{14\,×\,3}\) = \(\large \frac {27}{42}\)
\(\large \frac {9}{14}\) = \(\large \frac {…}{3.5}\)
∴ \(\large \frac {9}{14}\) = \(\large \frac {9\,×\,0.25}{14\,×\,0.25}\) = \(\large \frac {2.25}{3.5}\)
(2) Find the following ratios.
(i) The ratio of radius to circumference of the circle.
Solution:
Ratio of radius of circle to its circumference
= \(\large \frac {Radius}{Circumference}\)
= \(\large \frac {r}{2πr}\)
= \(\large \frac {1}{2π}\)
= 1 : 2π
(ii) The ratio of circumference of circle with radius r to its area.
Solution:
Ratio of circumference of circle to area of circle
= \(\large \frac {Circumference}{Area}\)
= \(\large \frac {2πr}{πr^{2}}\)
= \(\large \frac {2}{r}\)
= 2 : r
(iii) The ratio of diagonal of a square to its side, if the length of side is 7 cm.
Solution:
Side of a square = 7 cm.
Ratio of diagonal of a square to it’s side
= \(\large \frac {Diagonal}{Side}\)
= \(\large \frac {\sqrt{2}\,×\, Side}{Side}\)
= \(\large \frac {\sqrt{2}}{1}\)
= \(\sqrt{2}\) : 1
(iv) The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of its perimeter to area.
Solution:
For rectangle,
l = 5 cm
b = 3.5 cm
Perimeter of rectangle = 2 [l + b]
∴ Perimeter of rectangle = 2 [5 + 3.5]
∴ Perimeter of rectangle = 2 × 8.5
∴ Perimeter of rectangle = 17 cm …(i)
Area of rectangle = l × b
∴ Area of rectangle = 5 × 3.5
∴ Area of rectangle = 17.5 cm² …(ii)
Ratio of perimeter to area of rectangle
= \(\large \frac {Perimeter}{Area}\)
= \(\large \frac {17}{17.5}\)
= \(\large \frac {170}{175}\)
= \(\large \frac {5\,×\,34}{5\,×\,35}\)
= \(\large \frac {34}{35}\)
= 34 : 35
(3) Compare the following pairs of ratios.
(i) \(\large \frac {\sqrt{5}}{3}\), \(\large \frac {3}{\sqrt{7}}\)
Solution:
\(\large \frac {\sqrt{5}}{3}\), \(\large \frac {3}{\sqrt{7}}\)
∴ \(\sqrt{7}\) × \(\sqrt{5}\) , 3 × 3
∴ \(\sqrt{35}\) , 9
∴ \(\sqrt{35}\) , \(\sqrt{81}\)
Now,
\(\sqrt{35}\) < \(\sqrt{81}\)
∴ \(\large \frac {\sqrt{5}}{3}\) < \(\large \frac {3}{\sqrt{7}}\)
(ii) \(\large \frac {3\sqrt{5}}{5\sqrt{7}}\), \(\large \frac {\sqrt{63}}{\sqrt{125}}\)
Solution:
\(\large \frac {3\sqrt{5}}{5\sqrt{7}}\), \(\large \frac {\sqrt{63}}{\sqrt{125}}\)
∴ \(3\sqrt{5}\) × \(\sqrt{125}\) , \(5\sqrt{7}\) × \(\sqrt{63}\)
∴ \(\sqrt{5\,×\,9}\) × \(\sqrt{125}\) , \(\sqrt{7\,×\,25}\) × \(\sqrt{63}\)
∴ \(\sqrt{45\,×\,125}\) , \(\sqrt{175\,×\,63}\)
∴ \(\sqrt{5625}\) , \(\sqrt{11025}\)
Now,
\(\sqrt{5625}\) < \(\sqrt{11025}\)
∴ \(\large \frac {3\sqrt{5}}{5\sqrt{7}}\) < \(\large \frac {\sqrt{63}}{\sqrt{125}}\)
(iii) \(\large \frac {5}{18}\), \(\large \frac {17}{121}\)
Solution:
\(\large \frac {5}{18}\), \(\large \frac {17}{121}\)
∴ 5 × 121 , 17 × 18
∴ 605 > 306
∴ \(\large \frac {5}{18}\) > \(\large \frac {17}{121}\)
(iv) \(\large \frac {\sqrt{80}}{\sqrt{48}}\), \(\large \frac {\sqrt{45}}{\sqrt{27}}\)
Solution:
\(\large \frac {\sqrt{80}}{\sqrt{48}}\), \(\large \frac {\sqrt{45}}{\sqrt{27}}\)
∴ \(\sqrt{80}\) × \(\sqrt{27}\) , \(\sqrt{45}\) × \(\sqrt{48}\)
∴ \(\sqrt{80\,×\,27}\) , \(\sqrt{45\,×\,48}\)
∴ \(\sqrt{2160}\) = \(\sqrt{2160}\)
∴ \(\large \frac {\sqrt{80}}{\sqrt{48}}\), \(\large \frac {\sqrt{45}}{\sqrt{27}}\)
(v) \(\large \frac {9.2}{5.1}\), \(\large \frac {3.4}{7.1}\)
Solution:
\(\large \frac {9.2}{5.1}\), \(\large \frac {3.4}{7.1}\)
∴ 9.2 × 7.1 , 3.4 × 5.1
∴ 65.32 > 17.34
∴ \(\large \frac {9.2}{5.1}\), \(\large \frac {3.4}{7.1}\)
(4) (i) □ABCD is a parallelogram. The ratio of ∠A and ∠B of this parallelogram is 5 : 4. Find the measure of ∠B.
Solution:
Let the common multiple of the given ratio be x.
∴ m∠A = 5x° and m∠B = 4x°
□ ABCD is a parallelogram
∴ m∠A + m∠B = 180° …[Adjacent angles of a parallelogram are supplementary]
∴ 5x + 4x = 180
∴ 9x = 180
∴ x = \(\large \frac {180}{9}\)
∴ x = 20
m∠B = 4x = 4 × 20 = 80
Ans: The measure of B is 80°.
(ii) The ratio of present ages of Albert and Salim is 5 : 9. Five years hence ratio of their ages will be 3 : 5. Find their present ages.
Solution:
Let the common multiple of the given ratio be x.
The ratio of present ages of Albert and Salim is 5 : 9.
Their present ages are 5x years and 9x years respectively.
After five years,
Albert’s age = (5x + 5) years
Salim’s age = (9x + 5) years
According to given condition
\(\large \frac {5x\,+\,5}{9x\,+\,5}\) = \(\large \frac {3}{5}\)
∴ 5 (5x + 5) = 3 (9x + 5)
∴ 25x + 25 = 27x + 15
∴ 25 – 15 = 27x – 25x
∴ 10 = 2x
∴ x = \(\large \frac {10}{2}\)
∴ x = 5
Ans: The present ages of Albert and Salim are 25 years and 45 years respectively.
(iii) The ratio of length and breadth of a rectangle is 3 : 1, and its perimeter is 36 cm. Find the length and breadth of the rectangle.
Solution:
Let the common multiple of the given ratio be x.
The ratio of the length and breadth of a rectangle is 3 : 1.
Length = 3x cm
Breadth = x cm.
Perimeter of rectangle
= 2 [length + breadth]
= 2 [3x + x]
= 2 × 4x
= 8x cm
According to given condition,
Perimeter of rectangle = 36 cm
∴ 8x = 36
∴ x = \(\large \frac {36}{8}\)
∴ x = 4.5
Length = 3x cm = 3 × 4.5 = 13.5
Breadth = x cm = 4.5 cm
Ans: The length and breadth of the rectangle are 13.5 cm and 4.5 cm respectively.
(iv) The ratio of two numbers is 31 : 23 and their sum is 216. Find these numbers.
Solution:
Let the common multiple of the given ratio be x.
The ratio of two numbers is 31 : 23.
The two numbers are 31x and 23 x.
According to given condition
31x + 23x = 216
∴ 54x = 216
∴ x = \(\large \frac {216}{54}\)
∴ x = 4
The two numbers are,
31x = 31 × 4 = 124
23x = 23 × 4 = 92
Ans: The two numbers are 124 and 92.
(v) If the product of two numbers is 360 and their ratio is 10 : 9, then find the numbers.
Solution:
Let the common multiple of the given ratio be x.
The ratio of two numbers is 10 : 9.
The two numbers are 10x and 9x.
According to given condition
10x × 9x = 360
∴ 90x² = 360
∴ x² = \(\large \frac {360}{90}\)
∴ x² = 4
∴ x = 2 …[Taking square root on both sides]
The two numbers are,
10x = 10 × 2 = 20
9x = 9 × 2 = 18
Ans: The two numbers are 20 and 18.
(5*) If a : b = 3 : 1 and b : c = 5 : 1 then find the value of
Solution:
\(\large \frac {a}{b}\) = \(\large \frac {3}{1}\)
∴ a = 3b
\(\large \frac {b}{c}\) = \(\large \frac {5}{1}\)
∴ b = 5c …(i)
Now,
a = 3b
∴ a = 3(5c)
∴ a = 15c …(ii)
(i) \(\large (\frac {a^3}{15b^{2}c})\small ^2\)
= \(\large (\frac {(15c)^3}{15(c)^{2}c})\small ^2\)
= \(\large (\frac {15\,×\,15\,×\,15\,×\,c\,×\,c\,×\,c}{15\,×\,5\,×\,5\,×\,c\,×\,c\,×\,c})\)
= 9 × 9 × 9
= 729
(ii) \(\large \frac {a^2}{7bc}\)
\(\large \frac {a^2}{7bc}\)
= \(\large \frac {(15c)^2}{7(5c)c}\)
= \(\large \frac {15\,×\,15\,×\,c\,×\,c}{7\,×\,5\,×\,c\,×\,c}\)
= \(\large \frac {45}{7}\)
(6*) If \(\sqrt{0.04\,×\,0.4\,×\,a}\) = 0.04 × 0.4 × \(\sqrt{b}\) then find the ratio of \(\large \frac {a}{b}\).
Solution:
\(\sqrt{0.04\,×\,0.4\,×\,a}\) = 0.04 × 0.4 × \(\sqrt{b}\)
Squaring both sides, we get,
\((\sqrt{0.04\,×\,0.4\,×\,a})^2\) = \((\)0.04 × 0.4 × \(\sqrt{b})^2\)
∴ 0.04 × 0.4 × a = 0.16 × 0.0016 × b
∴ \(\large \frac {a}{b}\) = \(\large \frac {0.16\,×\,0.0016}{0.4\,×\,0.04}\)
∴ \(\large \frac {a}{b}\) = \(\large \frac {0.16\,×\,0.0016}{0.016}\)
∴ \(\large \frac {a}{b}\) = 0.16 × 0.1
∴ \(\large \frac {a}{b}\) = 0.016
∴ \(\large \frac {a}{b}\) = \(\large \frac {16}{1000}\)
∴ \(\large \frac {a}{b}\) = \(\large \frac {8\,×\,2}{8\,×\,125}\)
∴ \(\large \frac {a}{b}\) = \(\large \frac {2}{125}\)
∴ \(\large \frac {a}{b}\) = 2 : 125
(7) (x + 3) : (x + 11) = (x – 2) : (x + 1) then find the value of x.
Solution:
(x + 3) : (x + 11) = (x – 2) : (x + 1)
∴ \(\large \frac {(x\,+\,3)}{(x\,+\,11)}\) = \(\large \frac {(x\,–\,2)}{(x\,+\,1)}\)
∴ (x + 3) (x + 1) = (x – 2) (x + 11)
∴ x² + x + 3x + 3 = x² + 11x – 2x – 22
∴ 4x + 3 = 9x – 22
∴ 3 + 22 = 9x – 4x
∴ 25 = 5x
∴ 5x = 25
∴ x = \(\large \frac {25}{5}\)
∴ x = 5
Ans: The value of x is 5
Practice set 4.3
(1) If \(\large \frac {a}{b}\) = \(\large \frac {7}{3}\), then find the values of the following ratios.
(i) \(\large \frac {5a\,+\,3b}{5a\,–\,3b}\)
Solution:
\(\large \frac {a}{b}\) = \(\large \frac {7}{3}\)
∴ \(\large \frac {5}{3}\) × \(\large \frac {a}{b}\) = \(\large \frac {5}{3}\) × \(\large \frac {7}{3}\) …[Multiplying both sides by \(\large \frac {5}{3}\)]
∴ \(\large \frac {5a}{3b}\) = \(\large \frac {35}{9}\)
∴ \(\large \frac {5a\,+\,3b}{5a\,–\,3b}\) = \(\large \frac {35\,+\,9}{35\,–\,9}\) …[By Componendo-Dividendo]
∴ \(\large \frac {5a\,+\,3b}{5a\,–\,3b}\) = \(\large \frac {44}{26}\)
∴ \(\large \frac {5a\,+\,3b}{5a\,–\,3b}\) = \(\large \frac {22}{13}\)
(ii) \(\large \frac {2a²\,+\,3b²}{2a²\,–\,3b²}\)
Solution:
\(\large \frac {a}{b}\) = \(\large \frac {7}{3}\)
∴ \(\large \frac {a²}{b²}\) = \(\large \frac {7²}{3²}\) …[Squaring both sides]
∴ \(\large \frac {a²}{b²}\) = \(\large \frac {49}{9}\)
∴ \(\large \frac {2}{3}\) × \(\large \frac {a²}{b²}\) = \(\large \frac {2}{3}\) × \(\large \frac {49}{9}\) …[Multiplying both sides by \(\large \frac {2}{3}\)]
∴ \(\large \frac {2a²}{3b²}\) = \(\large \frac {98}{27}\)
∴ \(\large \frac {2a²\,+\,3b²}{2a²\,–\,3b²}\) = \(\large \frac {98\,+\,27}{98\,–\,27}\) …[By Componendo-Dividendo]
∴ \(\large \frac {2a²\,+\,3b²}{2a²\,–\,3b²}\) = \(\large \frac {125}{71}\)
(iii) \(\large \frac {a³\,–\,b³}{b³}\)
Solution:
\(\large \frac {a}{b}\) = \(\large \frac {7}{3}\)
∴ \(\large \frac {a³}{b³}\) = \(\large \frac {7³}{3³}\) …[Cubing both sides]
∴ \(\large \frac {a³}{b³}\) = \(\large \frac {343}{27}\)
∴ \(\large \frac {a³\,–\,b³}{b³}\) = \(\large \frac {343\,–\,27}{27}\) …[By Dividendo]
∴ \(\large \frac {a³\,–\,b³}{b³}\) = \(\large \frac {316}{27}\)
(iv) \(\large \frac {7a\,+\,9b}{7a\,–\,9b}\)
Solution:
\(\large \frac {a}{b}\) = \(\large \frac {7}{3}\)
∴ \(\large \frac {7}{9}\) × \(\large \frac {a}{b}\) = \(\large \frac {7}{9}\) × \(\large \frac {7}{3}\) …[Multiplying both sides by \(\large \frac {7}{9}\)]
∴ \(\large \frac {7a}{9b}\) = \(\large \frac {49}{27}\)
∴ \(\large \frac {7a\,+\,9b}{7a\,–\,9b}\) = \(\large \frac {49\,+\,27}{49\,–\,27}\) …[By Componendo-Dividendo]
∴ \(\large \frac {7a\,+\,9b}{7a\,–\,9b}\) = \(\large \frac {76}{22}\)
∴ \(\large \frac {7a\,+\,9b}{7a\,–\,9b}\) = \(\large \frac {38}{11}\)
(2) If \(\large \frac {15a²\,+\,4b²}{15a²\,–\,4b²}\) = \(\large \frac {47}{7}\), then find the values of the following ratios.
(i) \(\large \frac {a}{b}\)
Solution:
\(\large \frac {15a²\,+\,4b²}{15a²\,–\,4b²}\) = \(\large \frac {47}{7}\)
∴ \(\large \frac {(15a²\,+\,4b²)\,+\,(15a²\,–\,4b²)}{(15a²\,+\,4b²)\,–\,(15a²\,–\,4b²)}\) = \(\large \frac {47\,+\,7}{47\,–\,7}\) …[By Componendo-Dividendo]
∴ \(\large \frac {15a²\,+\,4b²\,+\,15a²\,–\,4b²}{15a²\,+\,4b²\,–\,15a²\,+\,4b²}\) = \(\large \frac {54}{40}\)
∴ \(\large \frac {30a²}{8b²}\) = \(\large \frac {27}{20}\)
∴ \(\large \frac {a²}{b²}\) = \(\large \frac {27\,×\,8}{20\,×\,30}\)
∴ \(\large \frac {a²}{b²}\) = \(\large \frac {9}{25}\)
∴ \(\large \frac {a}{b}\) = \(\large \frac {3}{5}\) …[Taking positive square root]
(ii) \(\large \frac {7a\,+\,3b}{7a\,–\,3b}\)
Solution:
\(\large \frac {a}{b}\) = \(\large \frac {3}{5}\) …[From (i)]
∴ \(\large \frac {7}{3}\) × \(\large \frac {a}{b}\) = \(\large \frac {7}{3}\) × \(\large \frac {3}{5}\) …[Multiplying both sides by \(\large \frac {7}{3}\)]
∴ \(\large \frac {7a}{3b}\) = \(\large \frac {7}{5}\)
∴ \(\large \frac {7a\,+\,3b}{7a\,–\,3b}\) = \(\large \frac {7\,+\,5}{7\,–\,5}\) …[By Componendo-Dividendo]
∴ \(\large \frac {7a\,+\,3b}{7a\,–\,3b}\) = \(\large \frac {12}{2}\)
∴ \(\large \frac {7a\,+\,3b}{7a\,–\,3b}\) = \(\large \frac {6}{1}\)
∴ \(\large \frac {7a\,–\,3b}{7a\,+\,3b}\) = \(\large \frac {1}{6}\) …[By Invertendo]
(iii) \(\large \frac {b²\,–\,2a²}{b²\,+\,2a²}\)
Solution:
\(\large \frac {a}{b}\) = \(\large \frac {3}{5}\) …[From (i)]
∴ \(\large \frac {a²}{b²}\) = \(\large \frac {3²}{5²}\) …[Squaring both sides]
∴ \(\large \frac {a²}{b²}\) = \(\large \frac {9}{25}\)
∴ \(\large \frac {b²}{a²}\) = \(\large \frac {25}{9}\) …[By Invertendo]
∴ \(\large \frac {1}{2}\) × \(\large \frac {b²}{a²}\) = \(\large \frac {1}{2}\) × \(\large \frac {25}{9}\) …[Multiplying both sides by \(\large \frac {1}{2}\)]
∴ \(\large \frac {b²}{2a²}\) = \(\large \frac {25}{18}\)
∴ \(\large \frac {b²\,+\,2a²}{b²\,–\,2a²}\) = \(\large \frac {25\,+\,18}{25\,–\,18}\) …[By Componendo-Dividendo]
∴ \(\large \frac {b²\,+\,2a²}{b²\,–\,2a²}\) = \(\large \frac {43}{7}\)
∴ \(\large \frac {b²\,–\,2a²}{b²\,+\,2a²}\) = \(\large \frac {7}{43}\) …[By Invertendo]
(iv) \(\large \frac {b³\,–\,2a³}{b³\,–\,2a³}\)
Solution:
\(\large \frac {a}{b}\) = \(\large \frac {3}{5}\) …[From (i)]
∴ \(\large \frac {a³}{b³}\) = \(\large \frac {3³}{5³}\) …[Cubing both sides]
∴ \(\large \frac {a³}{b³}\) = \(\large \frac {27}{125}\)
∴ \(\large \frac {b³}{a³}\) = \(\large \frac {125}{27}\) …[By Invertendo]
∴ \(\large \frac {1}{2}\) × \(\large \frac {b³}{a³}\) = \(\large \frac {1}{2}\) × \(\large \frac {125}{27}\) …[Multiplying both sides by \(\large \frac {1}{2}\)]
∴ \(\large \frac {b²}{2a³}\) = \(\large \frac {125}{54}\)
∴ \(\large \frac {b³\,+\,2a³}{b³\,–\,2a³}\) = \(\large \frac {125\,+\,54}{125\,–\,54}\) …[By Componendo-Dividendo]
∴ \(\large \frac {b³\,+\,2a³}{b³\,–\,2a³}\) = \(\large \frac {179}{71}\)
∴ \(\large \frac {b³\,–\,2a³}{b³\,+\,2a³}\) = \(\large \frac {71}{179}\) …[By Invertendo]
(3) If \(\large \frac {3a\,+\,7b}{3a\,–\,7b}\) = \(\large \frac {4}{3}\), then find the values of the following ratio \(\large \frac {3a²\,+\,7b²}{3a²\,–\,7b²}\)
Solution:
\(\large \frac {3a\,+\,7b}{3a\,–\,7b}\) = \(\large \frac {4}{3}\)
∴ \(\large \frac {(3a\,+\,7b)\,+\,(3a\,–\,7b)}{(3a\,+\,7b)\,–\,(3a\,–\,7b)}\) = \(\large \frac {4\,+\,3}{4\,–\,3}\)
∴ \(\large \frac {3a\,+\,7b\,+\,3a\,–\,7b}{3a\,+\,7b\,–\,3a\,+\,7b}\) = \(\large \frac {7}{1}\)
∴ \(\large \frac {6a}{14b}\) = \(\large \frac {7}{1}\)
∴ \(\large \frac {a}{b}\) = \(\large \frac {7\,×\,14}{1\,×\,6}\)
∴ \(\large \frac {a}{b}\) = \(\large \frac {49}{3}\)
∴ \(\large \frac {a²}{b²}\) = \(\large \frac {2401}{9}\) …[Squaring both sides]
∴ \(\large \frac {3}{7}\) × \(\large \frac {a²}{b²}\) = \(\large \frac {3}{7}\) × \(\large \frac {2401}{9}\) …[Multiplying both sides by \(\large \frac {3}{7}\)]
∴ \(\large \frac {3a²}{7b²}\) = \(\large \frac {343}{3}\)
∴ \(\large \frac {3a²\,+\,7b²}{3a²\,–\,7b²}\) = \(\large \frac {343\,+\,3}{343\,–\,3}\) …[By Componendo-Dividendo]
∴ \(\large \frac {3a²\,+\,7b²}{3a²\,–\,7b²}\) = \(\large \frac {346}{340}\)
∴ \(\large \frac {3a²\,+\,7b²}{3a²\,–\,7b²}\) = \(\large \frac {173}{170}\)
∴ \(\large \frac {3a²\,+\,7b²}{3a²\,–\,7b²}\) = \(\large \frac {170}{173}\) …[By Invertendo]
(4) Solve the following equations
(i) \(\large \frac {x²\,+\,12x\,–\,20}{3x\,–\,5}\) = \(\large \frac {x²\,+\,8x\,+\,12}{2x\,+\,3}\)
Solution:
\(\large \frac {x²\,+\,12x\,–\,20}{3x\,–\,5}\) = \(\large \frac {x²\,+\,8x\,+\,12}{2x\,+\,3}\)
∴ \(\large \frac {1}{4}\) × \(\large \frac {x²\,+\,12x\,–\,20}{3x\,–\,5}\) = \(\large \frac {1}{4}\) × \(\large \frac {x²\,+\,8x\,+\,12}{2x\,+\,3}\) …[Multiplying by \(\large \frac {1}{4}\) on both sides]
∴ \(\large \frac {x²\,+\,12x\,–\,20}{12x\,–\,20}\) = \(\large \frac {x²\,+\,8x\,+\,12}{8x\,+\,12}\)
∴ \(\large \frac {(x²\,+\,12x\,–\,20)\,–\,(12x\,–\,20)}{12x\,–\,20}\) = \(\large \frac {(x²\,+\,8x\,+\,12)\,–\,(8x\,+\,12)}{8x\,+\,12}\) …[By Dividendo]
∴ \(\large \frac {x²\,+\,12x\,–\,20\,–\,12x\,+\,20}{12x\,–\,20}\) = \(\large \frac {x²\,+\,8x\,+\,12\,–\,8x\,–\,12}{8x\,+\,12}\)
∴ \(\large \frac {x²}{12x\,–\,20}\) = \(\large \frac {x²}{8x\,+\,12}\)
Now for x = 0, the equation is evidently satisfied.
∴ x = 0 is one of the solution
When x ≠ 0, then x² ≠ 0,
∴ \(\large \frac {x²}{12x\,–\,20}\) = \(\large \frac {x²}{8x\,+\,12}\)
∴ \(\large \frac {1}{12x\,–\,20}\) = \(\large \frac {1}{8x\,+\,12}\)
∴ 12x – 20 = 8x + 12
∴ 12x – 8x = 12 + 20
∴ 4x = 32
∴ x = \(\large \frac {32}{4}\)
∴ x = 8
Ans: x = 0 or x = 8 is the solution of given equation.
(ii) \(\large \frac {10x²\,+\,15x\,+\,63}{5x²\,–\,25x\,+\,12}\) = \(\large \frac {2x\,+\,3}{x\,–\,5}\)
Solution:
\(\large \frac {10x²\,+\,15x\,+\,63}{5x²\,–\,25x\,+\,12}\) = \(\large \frac {2x\,+\,3}{x\,–\,5}\)
If we substitute x = 0, then the equation becomes,
\(\large \frac {63}{12}\) = \(\large \frac {3}{–\,5}\), which is not possible.
∴ x ≠ 0
∴ \(\large \frac {10x²\,+\,15x\,+\,63}{2x\,+\,3}\) = \(\large \frac {5x²\,–\,25x\,+\,12}{x\,–\,5}\) …[By Alternendo]
∴ \(\large \frac {1}{5x}\) × \(\large \frac {10x²\,+\,15x\,+\,63}{2x\,+\,3}\) = \(\large \frac {1}{5x}\) × \(\large \frac {5x²\,–\,25x\,+\,12}{x\,–\,5}\) …[Multiplying by \(\large \frac {1}{5x}\) on both sides]
∴ \(\large \frac {10x²\,+\,15x\,+\,63}{5x(2x\,+\,3)}\) = \(\large \frac {5x²\,–\,25x\,+\,12}{5x(x\,–\,5)}\)
∴ \(\large \frac {10x²\,+\,15x\,+\,63}{10x²\,+\,15x}\) = \(\large \frac {5x²\,–\,25x\,+\,12}{5x²\,–\,25x}\)
∴ \(\large \frac {(10x²\,+\,15x\,+\,63)\,–\,(10x²\,+\,15x)}{10x²\,+\,15x}\) = \(\large \frac {(5x²\,–\,25x\,+\,12)\,–\,(5x²\,–\,25x)}{5x²\,–\,25x}\) …[By Dividendo]
∴ \(\large \frac {10x²\,+\,15x\,+\,63\,–\,10x²\,–\,15x}{10x²\,+\,15x}\) = \(\large \frac {5x²\,–\,25x\,+\,12\,–\,5x²\,+\,25x}{5x²\,–\,25x}\)
∴ \(\large \frac {63}{10x²\,+\,15x}\) = \(\large \frac {12}{5x²\,–\,25x}\)
∴ \(\large \frac {3(21)}{5x(2x\,+\,3)}\) = \(\large \frac {3(4)}{5x(x\,–\,5)}\)
∴ \(\large \frac {21}{2x\,+\,3}\) = \(\large \frac {4}{x\,–\,5}\)
∴ 21(x – 5) = 4(2x + 3)
∴ 21x – 105 = 8x + 12
∴ 21x – 8x = 12 + 105
∴ 13x = 117
∴ x = \(\large \frac {117}{13}\)
∴ x = 9
Ans: x = 9 is the solution of given equation.
(iii) \(\large \frac {(2x\,+\,1)²\,+\,(2x\,–\,1)²}{(2x\,+\,1)²\,–\,(2x\,–\,1)²}\) = \(\large \frac {17}{8}\)
Solution:
\(\large \frac {(2x\,+\,1)²\,+\,(2x\,–\,1)²}{(2x\,+\,1)²\,–\,(2x\,–\,1)²}\) = \(\large \frac {17}{8}\)
∴ \(\large \frac {[(2x\,+\,1)²\,+\,(2x\,–\,1)²]\,+\,[(2x\,+\,1)²\,–\,(2x\,–\,1)²]}{[(2x\,+\,1)²\,+\,(2x\,–\,1)²]\,–\,[(2x\,+\,1)²\,–\,(2x\,–\,1)²]}\) = \(\large \frac {17\,+\,8}{17\,–\,8}\) …[By Componendo-Dividendo]
∴ \(\large \frac {(2x\,+\,1)²\,+\,(2x\,–\,1)²\,+\,(2x\,+\,1)²\,–\,(2x\,–\,1)²}{(2x\,+\,1)²\,+\,(2x\,–\,1)²\,–\,(2x\,+\,1)²\,+\,(2x\,–\,1)²}\) = \(\large \frac {25}{9}\)
∴ \(\large \frac {2(2x\,+\,1)²}{2(2x\,–\,1)²}\) = \(\large \frac {25}{9}\)
∴ \(\large \frac {(2x\,+\,1)²}{(2x\,–\,1)²}\) = \(\large \frac {25}{9}\)
∴ \(\large \frac {2x\,+\,1}{2x\,–\,1}\) = \(\large \frac {5}{3}\) …[Taking positive square root]
∴ 3(2x + 1) = 5 (2x – 1)
∴ 6x + 3 = 10x – 5
∴ 3 + 5 = 10x – 6x
∴ 8 = 4x
∴ x = \(\large \frac {8}{4}\)
∴ x = 2
Ans: x = 2 is the solution of given equation.
(iv) \(\large \frac {\sqrt{4x\,+\,1}\,+\,\sqrt{x\,+\,3}}{\sqrt{4x\,+\,1}\,–\,\sqrt{x\,+\,3}}\) = \(\large \frac {4}{1}\)
Solution:
\(\large \frac {\sqrt{4x\,+\,1}\,+\,\sqrt{x\,+\,3}}{\sqrt{4x\,+\,1}\,–\,\sqrt{x\,+\,3}}\) = \(\large \frac {4}{1}\)
∴ \(\large \frac {[\sqrt{4x\,+\,1}\,+\,\sqrt{x\,+\,3}]\,+\,[\sqrt{4x\,+\,1}\,+\,\sqrt{x\,+\,3}]}{[\sqrt{4x\,+\,1}\,–\,\sqrt{x\,+\,3}]\,–\,[\sqrt{4x\,+\,1}\,+\,\sqrt{x\,+\,3}]}\) = \(\large \frac {4\,+\,1}{4\,–\,1}\) …[By Componendo-Dividendo]
∴ \(\large \frac {\sqrt{4x\,+\,1}\,+\,\sqrt{x\,+\,3}\,+\,\sqrt{4x\,+\,1}\,+\,\sqrt{x\,+\,3}}{\sqrt{4x\,+\,1}\,–\,\sqrt{x\,+\,3}\,–\,\sqrt{4x\,+\,1}\,–\,\sqrt{x\,+\,3}}\) = \(\large \frac {5}{3}\)
∴ \(\large \frac {2\sqrt{4x\,+\,1}}{2\sqrt{x\,+\,3}}\) = \(\large \frac {5}{3}\)
∴ \(\large \frac {\sqrt{4x\,+\,1}}{\sqrt{x\,+\,3}}\) = \(\large \frac {5}{3}\)
∴ \(\large (\frac {\sqrt{4x\,+\,1}}{\sqrt{x\,+\,3}})\small ^2\) = \(\large (\frac {5}{3})\small ^2\) …[Taking positive square root]
∴ \(\large \frac {4x\,+\,1}{x\,+\,3}\) = \(\large {\frac {25}{9}}\)
∴ 9(4x + 1) = 25(x + 3)
∴ 36x + 9 = 25x + 75
∴ 36x – 25x = 75 – 9
∴ 11x = 66
∴ x = \(\large \frac {66}{11}\)
∴ x = 6
Ans: x = 6 is the solution of the given equation.
(v) \(\large \frac {(4x\,+\,1)²\,+\,(2x\,+\,3)²}{4x²\,+\,12x\,+\,9}\) = \(\large \frac {61}{36}\)
Solution:
\(\large \frac {(4x\,+\,1)²\,+\,(2x\,+\,3)²}{4x²\,+\,12x\,+\,9}\) = \(\large \frac {61}{36}\)
∴ \(\large \frac {(4x\,+\,1)²\,+\,(2x\,+\,3)²}{(2x\,+\,3)²}\) = \(\large \frac {61}{36}\)
∴ \(\large \frac {[(4x\,+\,1)²\,+\,(2x\,+\,3)²]\,–\,(2x\,+\,3)²}{(2x\,+\,3)²}\) = \(\large \frac {61\,–\,36}{36}\) …[By Dividendo]
∴ \(\large \frac {(4x\,+\,1)²\,+\,(2x\,+\,3)²\,–\,(2x\,+\,3)²}{(2x\,+\,3)²}\) = \(\large \frac {25}{36}\)
∴ \(\large \frac {(4x\,+\,1)²}{(2x\,+\,3)²}\) = \(\large \frac {25}{36}\)
∴ \(\large \frac {4x\,+\,1}{2x\,+\,3}\) = \(\large \frac {5}{6}\) …[Taking positive square root]
∴ 6(4x + 1) = 5 (2x + 3)
∴ 24x + 6 = 10x + 15
∴ 24x – 10x = 15 – 6
∴ 14x = 9
∴ x = \(\large \frac {9}{14}\)
Ans: x = \(\large \frac {9}{14}\) is the solution of the given equation.
(vi) \(\large \frac {(3x\,–\,4)³\,–\,(x\,+\,1)³}{(3x\,–\,4)³\,+\,(x\,+\,1)³}\) = \(\large \frac {61}{189}\)
Solution:
\(\large \frac {(3x\,–\,4)³\,–\,(x\,+\,1)³}{(3x\,–\,4)³\,+\,(x\,+\,1)³}\) = \(\large \frac {61}{189}\)
∴ \(\large \frac {(3x\,–\,4)³\,+\,(x\,+\,1)³}{(3x\,–\,4)³\,–\,(x\,+\,1)³}\) = \(\large \frac {189}{61}\) …[By Invertendo]
∴ \(\large \frac {[(3x\,–\,4)³\,+\,(x\,+\,1)³]\,+\,[(3x\,–\,4)³\,–\,(x\,+\,1)³]}{[(3x\,–\,4)³\,+\,(x\,+\,1)³]\,–\,[(3x\,–\,4)³\,–\,(x\,+\,1)³]}\) = \(\large \frac {189\,+\,61}{189\,–\,61}\) …[By Componendo-Dividendo]
∴ \(\large \frac {(3x\,–\,4)³\,+\,(x\,+\,1)³\,+\,(3x\,–\,4)³\,–\,(x\,+\,1)³}{(3x\,–\,4)³\,+\,(x\,+\,1)³\,–\,(3x\,–\,4)³\,+\,(x\,+\,1)³}\) = \(\large \frac {250}{128}\)
∴ \(\large \frac {2(3x\,–\,4)³}{2(x\,+\,1)³}\) = \(\large \frac {125}{64}\)
∴ \(\large \frac {(3x\,–\,4)³}{(x\,+\,1)³}\) = \(\large \frac {125}{64}\)
∴ \(\large \frac {3x\,–\,4}{x\,+\,1}\) = \(\large \frac {5}{4}\) …[Taking cube root]
∴ 4(3x – 4) = 5 (x + 1)
∴ 12x – 16 = 5x + 5
∴ 12x – 5x = 5 + 16
∴ 7x = 21
∴ x = \(\large \frac {21}{7}\)
∴ x = 3
Ans: x = 3 is the solution of the given equation.
Practice set 4.4
(1) Fill in the blanks of the following
(i) \(\large \frac {x}{7}\) = \(\large \frac {y}{3}\) = \(\large \frac {3x\,+\,5y}{…}\) = \(\large \frac {7x\,–\,9y}{…}\)
Solution:
\(\large \frac {x}{7}\) = \(\large \frac {y}{3}\) = \(\large \frac {3x\,+\,5y}{…}\) = \(\large \frac {7x\,–\,9y}{…}\)
∴ \(\large \frac {x}{7}\) = \(\large \frac {y}{3}\) = \(\large \frac {3x\,+\,5y}{3\,×\,7\,+\,5\,×\,3}\) = \(\large \frac {7x\,–\,9y}{7\,×\,7\,–\,9\,×\,3}\) …[By Theorem on equal ratios]
∴ \(\large \frac {x}{7}\) = \(\large \frac {y}{3}\) = \(\large \frac {3x\,+\,5y}{21\,+\,15}\) = \(\large \frac {7x\,–\,9y}{49\,–\,27}\)
∴ \(\large \frac {x}{7}\) = \(\large \frac {y}{3}\) = \(\large \frac {3x\,+\,5y}{36}\) = \(\large \frac {7x\,–\,9y}{22}\)
(ii) \(\large \frac {a}{3}\) = \(\large \frac {b}{4}\) = \(\large \frac {c}{7}\) = \(\large \frac {a\,–\,2b\,+\,3c}{…}\) = \(\large \frac {…}{6\,–\,8\,+\,14}\)
Solution:
\(\large \frac {a}{3}\) = \(\large \frac {b}{4}\) = \(\large \frac {c}{7}\) = \(\large \frac {a\,–\,2b\,+\,3c}{…}\) = \(\large \frac {…}{6\,–\,8\,+\,14}\)
∴ \(\large \frac {a}{3}\) = \(\large \frac {b}{4}\) = \(\large \frac {c}{7}\) = \(\large \frac {a\,–\,2b\,+\,3c}{3\,–\,2\,×\,4\,+\,3\,×\,7}\) = \(\large \frac {2a\,–\,2b\,+\,2c}{2\,×\,3\,–\,2\,×\,4\,+\,2\,×\,7}\)
∴ \(\large \frac {a}{3}\) = \(\large \frac {b}{4}\) = \(\large \frac {c}{7}\) = \(\large \frac {a\,–\,2b\,+\,3c}{3\,–\,8\,+\,21}\) = \(\large \frac {2a\,–\,2b\,+\,2c}{6\,–\,8\,+\,14}\)
∴ \(\large \frac {a}{3}\) = \(\large \frac {b}{4}\) = \(\large \frac {c}{7}\) = \(\large \frac {a\,–\,2b\,+\,3c}{16}\) = \(\large \frac {2a\,–\,2b\,+\,2c}{6\,–\,8\,+\,14}\)
(2) 5m – n = 3m + 4n then find the values of the following expressions.
Solution:
5m – n = 3m + 4n
∴ 5m – 3m = 4n + n
∴ 2m = 5n
∴ \(\large \frac {m}{n}\) = \(\large \frac {5}{2}\) …(i)
(i) \(\large \frac {m²\,+\,n²}{m²\,–\,n²}\)
Solution:
\(\large \frac {m}{n}\) = \(\large \frac {5}{2}\) …[From (i)]
∴ \(\large (\frac {m}{n})^2\) = \(\large (\frac {5}{2})^2\) …[Taking square on both sides]
∴ \(\large (\frac {m²}{n²})\) = \(\large \frac {25}{4}\)
∴ \(\large (\frac {m²\,+\,n²}{m²\,–\,n²})\) = \(\large \frac {25\,+\,4}{25\,–\,4}\) …[By Componendo-Dividendo]
∴ \(\large (\frac {m²\,+\,n²}{m²\,–\,n²})\) = \(\large \frac {29}{21}\)
(ii) \(\large \frac {3m\,+\,4n}{3m\,–\,4n}\)
Solution:
\(\large \frac {m}{n}\) = \(\large \frac {5}{2}\) …[From (i)]
∴ \(\large \frac {3}{4}\) × \(\large \frac {m}{n}\) = \(\large \frac {3}{4}\) × \(\large \frac {5}{2}\) …[Multiplying by \(\large \frac {3}{4}\) on both sides]
∴ \(\large \frac {3m}{4n}\) = \(\large \frac {15}{8}\)
∴ \(\large \frac {3m\,+\,4n}{3m\,–\,4n}\) = \(\large \frac {15\,+\,8}{15\,–\,8}\) …[By Componendo-Dividendo]
∴ \(\large \frac {3m\,+\,4n}{3m\,–\,4n}\) = \(\large \frac {23}{7}\)
(3) (i) If a(y + z) = b(z + x) = c(x + y) and out of a, b, c no two of them are equal then show that, \(\large \frac {y\,–\,z}{a(b\,–\,c)}\) = \(\large \frac {z\,–\,x}{b(c\,–\,a)}\) = \(\large \frac {x\,–\,y}{c(a\,–\,b)}\)
Solution:
a(y + z) = b(z + x) = c(x + y)
∴ \(\large \frac {a(y\,+\,z)}{abc}\) = \(\large \frac {b(z\,+\,x)}{abc}\) = \(\large \frac {c(x\,+\,y)}{abc}\) …[Dividing each ratio by abc]
∴ \(\large \frac {y\,+\,z}{bc}\) = \(\large \frac {z\,+\,x}{ac}\) = \(\large \frac {x\,+\,y}{ab}\)
Let,
\(\large \frac {y\,+\,z}{bc}\) = \(\large \frac {z\,+\,x}{ac}\) = \(\large \frac {x\,+\,y}{ab}\) = k
By theorem on equal ratios, we get,
k = \(\large \frac {(x\,+\,y)\,–\,(z\,+\,x)}{ab\,–\,ac}\)
∴ k = \(\large \frac {x\,+\,y\,–\,z\,–\,x}{a(b\,–\,c)}\)
∴ k = \(\large \frac {y\,–\,z}{a(b\,–\,c)}\) …(i)
Similarly,
k = \(\large \frac {(y\,+\,z)\,–\,(x\,+\,y)}{bc\,–\,ab}\)
∴ k = \(\large \frac {y\,+\,z\,–\,x\,–\,y}{b(c\,–\,a)}\)
∴ k = \(\large \frac {z\,–\,x}{b(c\,–\,a)}\) …(ii)
Similarly,
k = \(\large \frac {(z\,+\,x)\,–\,(y\,+\,z)}{ac\,–\,bc}\)
∴ k = \(\large \frac {z\,+\,x\,–\,y\,–\,z}{c(a\,–\,b)}\)
∴ k = \(\large \frac {x\,–\,y}{c(a\,–\,b)}\) …(iii)
From (i), (ii) and (iii), we get,
\(\large \frac {y\,–\,z}{a(b\,–\,c)}\) = \(\large \frac {z\,–\,x}{b(c\,–\,a)}\) = \(\large \frac {x\,–\,y}{c(a\,–\,b)}\)
Hence proved.
(ii) If \(\large \frac {x}{3x\,–\,y\,–\,z}\) = \(\large \frac {y}{3y\,–\,z\,–\,x}\) = \(\large \frac {z}{3z\,–\,x\,–\,y}\) and x + y + z ≠ 0 then show that the value of each ratio is equal to 1.
Solution:
Let,
\(\large \frac {x}{3x\,–\,y\,–\,z}\) = \(\large \frac {y}{3y\,–\,z\,–\,x}\) = \(\large \frac {z}{3z\,–\,x\,–\,y}\) = k
By theorem on equal ratios, we get,
k = \(\large \frac {x\,+\,y\,+\,z}{3x\,–\,y\,–\,z\,+\,3y\,–\,z\,–\,x\,+\,3z\,–\,x\,–\,y}\)
∴ k = \(\large \frac {x\,+\,y\,+\,z}{x\,+\,y\,+\,z}\)
∴ k = 1 …[∵ x + y + z ≠ 0]
∴ Each ratio is equal to 1
(iii) If \(\large \frac {ax\,+\,by}{x\,+\,y}\) = \(\large \frac {bx\,+\,az}{x\,+\,z}\) = \(\large \frac {ay\,+\,bz}{y\,+\,z}\) and x + y + z ≠ 0 then show that \(\large \frac {a\,+\,b}{2}\)
Solution:
Let,
\(\large \frac {ax\,+\,by}{x\,+\,y}\) = \(\large \frac {bx\,+\,az}{x\,+\,z}\) = \(\large \frac {ay\,+\,bz}{y\,+\,z}\) = k
By theorem on equal ratios, we get,
k = \(\large \frac {ax\,+\,by\,+\,bx\,+\,az\,+\,ay\,+\,bz}{x\,+\,y\,+\,x\,+\,z\,+\,y\,+\,z}\)
∴ k = \(\large \frac {ax\,+\,by\,+\,bx\,+\,az\,+\,ay\,+\,bz}{2x\,+\,2y\,+\,2z}\)
∴ k = \(\large \frac {a(x\,+\,y\,+\,z)\,+\,b(x\,+\,y\,+\,z)}{2(x\,+\,y\,+\,z)}\)
∴ k = \(\large \frac {(x\,+\,y\,+\,z)(a\,+\,b)}{2(x\,+\,y\,+\,z)}\)
∴ k = \(\large \frac {(a\,+\,b)}{2}\) …[x + y + z ≠ 0]
∴ Each ratio is equal to \(\large \frac {(a\,+\,b)}{2}\)
(iv) If \(\large \frac {y\,+\,z}{a}\) = \(\large \frac {z\,+\,x}{b}\) = \(\large \frac {x\,+\,y}{c}\) then show that \(\large \frac {x}{b\,+\,c\,–\,a}\) = \(\large \frac {y}{c\,+\,a\,–\,b}\) = \(\large \frac {z}{a\,+\,b\,–\,c}\)
Solution:
\(\large \frac {y\,+\,z}{a}\) = \(\large \frac {z\,+\,x}{b}\) = \(\large \frac {x\,+\,y}{c}\)
By theorem on equal ratios we get,
k = \(\large \frac {(z\,+\,x)\,+\,(x\,+\,y)\,–\,(y\,+\,z)}{b\,+\,c\,–\,a}\)
k = \(\large \frac {z\,+\,x\,+\,x\,+\,y\,–\,y\,–\,z}{b\,+\,c\,–\,a}\)
k = \(\large \frac {2x}{b\,+\,c\,–\,a}\) …(i)
Similarly, by theorem on equal ratios,
k = \(\large \frac {(x\,+\,y)\,+\,(y\,+\,z)\,–\,(z\,+\,x)}{c\,+\,a\,–\,b}\)
k = \(\large \frac {x\,+\,y\,+\,y\,+\,z\,–\,z\,–\,x}{c\,+\,a\,–\,b}\)
k = \(\large \frac {2y}{c\,+\,a\,–\,b}\) …(ii)
Similarly, by theorem on equal ratios,
k = \(\large \frac {(x\,+\,y)\,+\,(y\,+\,z)\,–\,(z\,+\,x)}{c\,+\,a\,–\,b}\)
k = \(\large \frac {(y\,+\,z)\,+\,(z\,+\,x)\,–\,(x\,+\,y)}{a\,+\,b\,–\,c}\)
k = \(\large \frac {2z}{a\,+\,b\,–\,c}\) …(iii)
From (i), (ii) and (iii), we get,
\(\large \frac {2x}{b\,+\,c\,–\,a}\) = \(\large \frac {2y}{c\,+\,a\,–\,b}\) = \(\large \frac {2z}{a\,+\,b\,–\,c}\)
∴ \(\large \frac {x}{b\,+\,c\,–\,a}\) = \(\large \frac {y}{c\,+\,a\,–\,b}\) = \(\large \frac {z}{a\,+\,b\,–\,c}\)
(v) If \(\large \frac {3x\,–\,5y}{5z\,+\,3y}\) = \(\large \frac {x\,+\,5z}{y\,–\,5x}\) = \(\large \frac {y\,–\,z}{x\,–\,z}\) then show that every ratio = \(\large \frac {x}{y}\).
Solution:
Let,
\(\large \frac {3x\,–\,5y}{5z\,+\,3y}\) = \(\large \frac {x\,+\,5z}{y\,–\,5x}\) = \(\large \frac {y\,–\,z}{x\,–\,z}\) = k
∴ k = \(\large \frac {3x\,–\,5y}{5z\,+\,3y}\) = \(\large \frac {x\,+\,5z}{y\,–\,5x}\) = \(\large \frac {5(y\,–\,z)}{5(x\,–\,z)}\)
By theorem on equal ratio, we get,
k = \(\large \frac {3x\,–\,5y\,+\,x\,+\,5z\,+\,5(y\,–\,z)}{5z\,+\,3y\,+\,y\,–\,5x\,+\,5x\,–\,5z}\)
∴ k = \(\large \frac {4x}{4y}\)
∴ k = \(\large \frac {x}{y}\)
∴ Each ratio is equal to \(\large \frac {x}{y}\)
(4) Solve:
(i) \(\large \frac {16x²\,–\,20x\,+\,9}{8x²\,–\,12x\,+\,21}\) = \(\large \frac {4x\,–\,5}{2x\,+\,3}\)
Solution:
\(\large \frac {16x²\,–\,20x\,+\,9}{8x²\,+\,12x\,+\,21}\) = \(\large \frac {4x\,–\,5}{2x\,+\,3}\)
If x = 0, then, {9}{21}\) = \(\large \frac {–\,5}{3}\), which is not possible
∴ x ≠ 0
Let,
\(\large \frac {16x²\,–\,20x\,+\,9}{8x²\,+\,12x\,+\,21}\) = \(\large \frac {4x\,–\,5}{2x\,+\,3}\) = k
∴ k = \(\large \frac {16x²\,–\,20x\,+\,9}{8x²\,+\,12x\,+\,21}\) = \(\large \frac {–\,4(4x\,–\,5)}{–\,4(2x\,+\,3)}\)
By theorem on equal ratios,
k = \(\large \frac {16x²\,–\,20x\,+\,9\,–\,4(4x\,–\,5)}{8x²\,+\,12x\,+\,21\,–\,4(2x\,+\,3)}\)
∴ k = \(\large \frac {16x²\,–\,20x\,+\,9\,–\,16x²\,+\,20x}{8x²\,+\,12x\,+\,21\,–\,8x²\,–\,12x}\)
∴ k = \(\large \frac {9}{21}\)
∴ \(\large \frac {4x\,–\,5}{2x\,+\,3}\) = \(\large \frac {9}{21}\)
∴ 21(4x – 5) = 9 (2x + 3)
∴ 84x – 105 = 18x + 27
∴ 84x – 18x = 27 + 105
∴ 66x = 132
∴ x = \(\large \frac {132}{66}\)
∴ x = 2
Ans: x = 2 is the solution of the given equation.
(ii) \(\large \frac {5y²\,+\,40y\,–\,12}{5y²\,+\,10y\,–\,4}\) = \(\large \frac {y\,+\,8}{1\,+\,2y}\)
Solution:
Let,
\(\large \frac {5y²\,+\,40y\,–\,12}{5y²\,+\,10y\,–\,4}\) = \(\large \frac {y\,+\,8}{1\,+\,2y}\) = k
∴ k = \(\large \frac {5y²\,+\,40y\,–\,12}{5y²\,+\,10y\,–\,4}\) = \(\large \frac {–\,5(y\,+\,8)}{–\,5(1\,+\,2y)}\)
By theorem of equal ratios, we get,
∴ k = \(\large \frac {5y²\,+\,40y\,–\,12\,–\,5(y\,+\,8)}{5y²\,+\,10y\,–\,4\,–\,5(1\,+\,2y)}\)
∴ k = \(\large \frac {5y²\,+\,40y\,–\,12\,–\,5y²\,–\,40y}{5y²\,+\,10y\,–\,4\,–\,5y\,–\,10y²}\)
∴ k = \(\large \frac {–\,12}{–\,4}\)
∴ k = 3
∴ \(\large \frac {y\,+\,8}{1\,+\,2y}\) = 3
∴ y + 8 = 3(1 + 2y)
∴ y + 8 = 3 + 6y
∴ 8 – 3 = 6y – y
∴ 5 = 5y
∴ y = \(\large \frac {5}{5}\)
∴ y = 1
Ans: y = 1 is the solution of the given equation.
Practice set 4.5
(1) Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?
Solution:
Let the number to be subtracted be x.
∴ (12 + x), (16 + x) and (21 + x) are in continued proportion
∴ \(\large \frac {12\,+\,x}{16\,+\,x}\) = \(\large \frac {16\,+\,x}{21\,+\,x}\)
∴ (12 + x) (21 + x) = (16 + x) (16 + x)
∴ 252 + 12x + 21x + x² = 256 + 16x + 16x + x²
∴ 252 + 33x = 256 + 32x
∴ 33x – 32x = 256 – 252
∴ x = 4
Ans: The number to be added is 4.
(2) If (28 – x) is the mean proportional of (23 – x) and (19 – x) then find the value of x.
Solution:
∴ (28 – x)² = (23 – x) (19 – x)
∴ 784 – 56x + x² = 437 – 23x – 19x + x²
∴ 784 – 56x = 437 – 42x
∴ – 56x + 42x = 437 – 784
∴ – 14x = – 347
∴ x = \(\large \frac {–\,347}{–\,14}\)
x = \(\large \frac {347}{14}\)
Ans: The value of x is \(\large \frac {347}{14}\)
(3) Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.
Solution:
Let a, b, c, are in continued proportion.
∴ \(\large \frac {a}{b}\) = \(\large \frac {b}{c}\)
i.e. b² = ac … (i)
Here b = 12 and a + c = 26 …[Given]
∴ 12² = ac …[From (i)]
∴ 12² = (26 – c) × c …[∵ a + c = 26]
∴ 144 = 26c – c²
∴ c² – 26c + 144 = 0
∴ c² – 18c – 8c + 144 = 0
∴ c (c – 18) – 8 (c – 18) = 0
∴ (c – 18) (c – 8) = 0
∴ c – 18 = 0 or c – 8 = 0
∴ c = 18 or c = 8
If c = 18 then a = 26 – 18 = 8
And if c = 8 then a = 26 – 8 = 18
Ans: Three numbers in continued proportion are 18, 12, 8 or 8, 12, 18.
(4) If (a + b + c) (a – b + c) = a² + b² + c² show that a, b, c are in continued proportion.
Solution:
(a + b + c) (a – b + c) = a² + b² + c²
∴ a² – ab + ac + ab – b² + bc + ac – bc + c² = a² + b² + c²
∴ ac + ac – b² = b²
∴ 2ac = b² + b²
∴ 2ac = 2b²
∴ ac = b²
∴ b² = ac
∴ b is the mean proportional of a and c.
∴ a, b, c are in continued proportion.
(5) If \(\large \frac {a}{b}\) = \(\large \frac {b}{c}\) and a, b, c > 0 then show that,
Solution:
\(\large \frac {a}{b}\) = \(\large \frac {b}{c}\)
Let \(\large \frac {a}{b}\) = \(\large \frac {b}{c}\) = k
∴ b = ck and a = ck² … (i)
(i) (a + b + c) (b – c) = ab – c²
LHS
= (a + b + c) (b – c)
= (ck2 + ck + c) (ck – c)
= c (k²+ k + 1) c (k – 1)
= c²(k – 1) (k² + k + 1)
RHS
= ab – c²
= ck² × ck – c²
= c²k³ – c²
= c²(k³– 1)
= c²(k – 1) (k²+ k + 1) ….[Using a³ – b³ = (a – b)(a² + ab + b²)]
∵ LHS = RHS
∴ (a + b + c) (b – c) = ab – c²
(ii) (a² + b²) (b² + c²) = (ab + bc)
LHS
= (a² + b²) (b² + c²)
= [(ck²)² + (ck)²] [(ck)² + c²]
= (c²k⁴ + c²k²) (c²k² + c²)
= c²k²(k² + 1) c² (k² + 1)
= c⁴k²(k² + 1)(k² + 1)
= c⁴k²(k² + 1)²
= c⁴k²(k⁴ + 2k² + 1) …[Using (a + b)²= (a² + 2ab + b²)]
RHS
= (ab + bc)²
= [b(a + c)]²
= b²(a + c)²
= b²(a² + 2ac + c²)
= (ck)²[(ck²)² + 2 × ck² × c + c²]
= c²k²(c²k⁴ + 2c²k² + c²)
= c²k² × c²(k⁴ + 2k² + 1)
= c⁴k² (k⁴ + 2k² + 1)
∵ LHS = RHS
∴ (a² + b²) (b² + c²) = (ab + bc)²
(iii) \(\large \frac {a²\,+\,b²}{ab}\) = \(\large \frac {a\,+\,b}{c}\)
Solution:
LHS
= \(\large \frac {a²\,+\,b²}{ab}\)
= \(\large \frac {(ck²)²\,+\,(ck)²}{ck²\,×\,ck}\)
= \(\large \frac {c²k⁴\,+\,c²k²}{c²k³}\)
= \(\large \frac {c²k²(k²\,+\,1)}{c²k²(k)}\)
= \(\large \frac {k²\,+\,1}{k}\)
RHS
= \(\large \frac {a\,+\,b}{c}\)
= \(\large \frac {ck²\,+\,c}{ck}\)
= \(\large \frac {c(k²\,+\,1)}{c(k)}\)
= \(\large \frac {k²\,+\,1}{k}\)
∵ LHS = RHS
∴ \(\large \frac {a²\,+\,b²}{ab}\) = \(\large \frac {a\,+\,b}{c}\)
(6) Find mean proportional of \(\large \frac {x\,+\,y}{x\,–\,y}\), \(\large \frac {x²\,–\,y²}{x²y²}\).
Solution:
Let mean proportion of \(\large \frac {x\,+\,y}{x\,–\,y}\) and \(\large \frac {x²\,–\,y²}{x²y²}\) be b
∴ b² = \(\large \frac {x\,+\,y}{x\,–\,y}\) × \(\large \frac {x²\,–\,y²}{x²y²}\)
∴ b² = \(\large \frac {x\,+\,y}{x\,–\,y}\) × \(\large \frac {(x\,–\,y)(x\,+\,y)}{x²y²}\) …[Using a² – b² = (a – b)(a + b)]
∴ b² = \(\large \frac {(x\,+\,y)(x\,+\,y)}{x²y²}\)
∴ b² = \(\large \frac {(x\,+\,y)²}{(xy)²}\)
∴ b = \(\large \frac {x\,+\,y}{xy}\) …[Taking square root]
Ans: \(\large \frac {x\,+\,y}{xy}\) is the mean proportional of \(\large \frac {x\,+\,y}{x\,–\,y}\) and \(\large \frac {x²\,–\,y²}{x²y²}\).
Problem Set 4
(1) Select the appropriate alternative answer for the following questions.
(i) If 6 : 5 = y : 20 then what will be the value of y?
(A) 15
(B) 24
(C) 18
(D) 22.5
Ans: Option (B) : 24
(ii) What is the ratio of 1 mm to 1 cm ?
(A) 1 : 100
(B) 10 : 1
(C) 1 : 10
(D) 100 : 1
Ans: Option (A) : 1 : 100
(iii*) The ages of Jatin, Nitin and Mohasin are 16, 24 and 36 years respectively. What is the ratio of Nitin’s age to Mohasin’s age?
(A) 3 : 2
(B) 2 : 3
(C) 4 : 3
(D) 3 : 4
Ans: Option (B) : 2 : 3
(iv) 24 Bananas were distributed between Shubham and Anil in the ratio 3 : 5, then how many bananas did Shubham get?
(A) 8
(B) 15
(C) 12
(D) 9
Ans: Option (D) : 9
(v) What is the mean proportional of 4 and 25?
(A) 6
(B) 8
(C) 10
(D) 12
Ans: Option (C) : 10
(2) For the following numbers write the ratio of first number to second number in the reduced form.
(i) 21, 48
Solution:
Ratio of 21 to 48
= \(\large \frac {21}{48}\)
= \(\large \frac {3\,×\,7}{3\,×\,16}\)
= \(\large \frac {7}{16}\)
= 7 : 16
(ii) 36, 90
Solution:
Ratio of 36 to 90
= \(\large \frac {36}{90}\)
= \(\large \frac {18\,×\,2}{18\,×\,5}\)
= \(\large \frac {2}{5}\)
= 2 : 5
(iii) 65, 117
Solution:
Ratio of 65 to 117
= \(\large \frac {65}{117}\)
= \(\large \frac {13\,×\,5}{13\,×\,9}\)
= \(\large \frac {5}{9}\)
= 5 : 9
(iv) 138, 161
Solution:
Ratio of 138 to 161
= \(\large \frac {138}{161}\)
= \(\large \frac {23\,×\,6}{23\,×\,7}\)
= \(\large \frac {6}{7}\)
= 6 : 7
(v) 114, 133
Solution:
Ratio of 114 to 133
= \(\large \frac {114}{133}\)
= \(\large \frac {19\,×\,6}{19\,×\,7}\)
= \(\large \frac {6}{7}\)
= 6 : 7
(3) Write the following ratios in the reduced form.
(i) Radius to the diameter of a circle.
Solution:
Ratio of radius to diameter of a circle
= \(\large \frac {r}{d}\)
= \(\large \frac {r}{2r}\)
= \(\large \frac {1}{2}\)
= 1 : 2
(ii) The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm.
Solution:
Diagonal of a rectangle is equivalent to a hypotenuse of the right angled triangle
∴ (Diagonal)² = (Length)² + (Breadth)² …[Pythagoras theorem]
∴ (Diagonal)² = 4² + 3²
∴ (Diagonal)² = 16 + 9
∴ (Diagonal)² = 25
∴ Diagonal = 5 …[Taking positive square root]
Ratio of diagonal to length of a rectangle
= \(\large \frac {5}{4}\)
= 5 : 4
(iii) The ratio of perimeter to area of a square, having side 4 cm.
Solution:
Side of the square = 4cm
Perimeter of a square
= 4 × side
= 4 × 4
= 16 cm²
Area of a square
= side²
= 4²
= 16 cm
Ratio of perimeter to area of a square
= \(\large \frac {16}{16}\)
= \(\large \frac {1}{1}\)
= 1 : 1
(4) Check whether the following numbers are in continued proportion.
(i) 2, 4, 8
Solution:
If b² = ac then a, b, c are in continued proportion.
Let a = 2, b = 4 and c = 8
b² = 4² = 16
ac = 2 × 8 = 16
∵ b² = ac
∴ 2, 4, 8 are in continued proportion.
(ii) 1, 2, 3
Solution:
If b² = ac then a, b, c are in continued proportion.
Let a = 1, b = 2 and c = 3
b² = 2² = 4
ac = 1 × 3 = 3
∵ b² ≠ ac
∴ 1, 2, 3 are not in continued proportion.
(iii) 9, 12, 16
Solution:
If b² = ac then a, b, c are in continued proportion.
Let a = 9, b = 12 and c = 16
b² = 12² = 144
ac = 9 × 16 = 144
∵ b² = ac
∴ 9, 12, 16 are in continued proportion.
(iv) 3, 5, 8
Solution:
If b² = ac then a, b, c are in continued proportion.
Let a = 3, b = 5 and c = 8
b² = 5² = 25
ac = 3 × 8 = 24
∵ b² ≠ ac
∴ 3, 5, 8 are not in continued proportion.
(5) a, b, c are in continued proportion. If a = 3 and c = 27 then find b.
Solution:
a, b, c are in continued proportion.
∴ b² = ac
a = 2 and c = 8
b² = 3 × 27
b² = 81
∴ b = 9 …[Taking square root]
Ans: The value of b is 9.
(6) Convert the following ratios into percentages.
(i) 37 : 500
Solution:
37 : 500
= \(\large \frac {37}{100\,×\,5}\)
= \(\large \frac {7.4}{100}\)
= 7.4 %
(ii) \(\large \frac {5}{8}\)
Solution:
\(\large \frac {5}{8}\)
= \(\large \frac {5\,×\,125}{8\,×\,124}\)
= \(\large \frac {625}{1000}\)
= \(\large \frac {62.5}{100}\)
= 62.5 %
(iii) \(\large \frac {22}{30}\)
Solution:
\(\large \frac {22}{30}\)
= \(\large \frac {22\,×\,10}{30\,×\,10}\)
= \(\large \frac {7.333\,×\,10}{100}\)
= \(\large \frac {73.33}{100}\)
= 73.33 %
(iv) \(\large \frac {5}{16}\)
Solution:
\(\large \frac {5}{16}\)
= \(\large \frac {5\,×\,100}{16\,×\,100}\)
= \(\large \frac {5\,×\,25}{8\,×\,100}\)
= \(\large \frac {125}{8\,×\,100}\)
= \(\large \frac {31.25}{100}\)
= 31.25 %
(v) \(\large \frac {144}{1200}\)
Solution:
\(\large \frac {144}{1200}\)
= \(\large \frac {144}{12\,×\,100}\)
= \(\large \frac {12}{100}\)
= 12 %
(7) Write the ratio of first quantity to second quantity in the reduced form.
(i) 1024 MB, 1.2 GB [(1024 MB = 1 GB)]
Solution:
Ratio of 1 G.B. to 1.2 G.B.
= \(\large \frac {1}{1.2}\)
= \(\large \frac {10}{12}\)
= \(\large \frac {5}{6}\)
= 5 : 6
(ii) 17 Rupees, 25 Rupees 60 paise
Solution:
₹ 17
= (17 × 100) paise
= 1700 paise
₹ 25 and 60 paise
= (25 × 100 + 60) paise
= (2500 + 60) paise
= 2560 paise
∴ Ratio of 1700 paise to 2560 paise
= \(\large \frac {1700}{2560}\)
= \(\large \frac {85\,×\,20}{128\,×\,20}\)
= \(\large \frac {85}{128}\)
= 85 : 128
(iii) 5 dozen, 120 units
Solution:
5 dozen
= (5 × 12) units
= 60 units
∴ Ratio of 60 units to 120 units
= \(\large \frac {60}{120}\)
= \(\large \frac {1}{2}\)
= 1: 2
(iv) 4 sq.m, 800 sq.cm
Solution:
4m² = (4 × 10000) cm² …[1 m² = 10000 cm²]
∴ 4m² = 40000 cm²
Ratio of 40000 cm² to 800 cm²
= \(\large \frac {40000}{800}\)
= \(\large \frac {50\,×\,800}{800}\)
= \(\large \frac {50}{1}\)
= 50 : 1
(v) 1.5 kg, 2500 gm
Solution:
1.5 kg
= (1.5 × 1000) gm …[1 kg = 1000 gm]
= 1500 gm
Ratio of 1500 gm to 2500 gm
= \(\large \frac {1500}{2500}\)
= \(\large \frac {15}{25}\)
= \(\large \frac {3}{5}\)
= 3 : 5
(8) If \(\large \frac {a}{b}\) = \(\large \frac {2}{3}\) then find the values of the following expressions.
(i) \(\large \frac {4a\,+\,3b}{3b}\)
Solution:
\(\large \frac {a}{b}\) = \(\large \frac {2}{3}\)
∴ \(\large \frac {4}{3}\) × \(\large \frac {a}{b}\) = \(\large \frac {4}{3}\) × \(\large \frac {2}{3}\) …[Multiplying both sides by \(\large \frac {4}{3}\)]
∴ \(\large \frac {4a}{3b}\) = \(\large \frac {8}{9}\)
∴ \(\large \frac {4a\,+\,3b}{3b}\) = \(\large \frac {8\,+\,9}{9}\) …[By Componendo]
∴ \(\large \frac {4a\,+\,3b}{3b}\) = \(\large \frac {17}{9}\)
(ii) \(\large \frac {5a²\,+\,2b²}{5a²\,–\,2b²}\)
Solution:
\(\large \frac {a}{b}\) = \(\large \frac {7}{3}\)
∴ \(\large \frac {a²}{b²}\) = \(\large \frac {7²}{3²}\) …[Squaring both sides]
∴ \(\large \frac {a²}{b²}\) = \(\large \frac {49}{9}\)
∴ \(\large \frac {5}{2}\) × \(\large \frac {a²}{b²}\) = \(\large \frac {5}{2}\) × \(\large \frac {49}{9}\) …[Multiplying both sides by \(\large \frac {5}{2}\)]
∴ \(\large \frac {5a²}{2b²}\) = \(\large \frac {10}{9}\)
∴ \(\large \frac {5a²\,+\,2b²}{5a²\,–\,2b²}\) = \(\large \frac {10\,+\,9}{10\,–\,9}\) …[By Componendo-Dividendo]
∴ \(\large \frac {5a²\,+\,2b²}{5a²\,–\,2b²}\) = \(\large \frac {19}{1}\)
(iii) \(\large \frac {a³\,+\,b³}{b³}\)
Solution:
\(\large \frac {a}{b}\) = \(\large \frac {2}{3}\)
∴ \(\large \frac {a³}{b³}\) = \(\large \frac {2³}{3³}\) …[Cubing both sides]
∴ \(\large \frac {a³}{b³}\) = \(\large \frac {8}{27}\)
∴ \(\large \frac {a³\,+\,b³}{b³}\) = \(\large \frac {8\,+\,27}{27}\) …[By Componendo]
∴ \(\large \frac {a³\,+\,b³}{b³}\) = \(\large \frac {35}{27}\)
(iv) \(\large \frac {7b\,–\,4a}{7b\,+\,4a}\)
Solution:
\(\large \frac {a}{b}\) = \(\large \frac {2}{3}\)
∴ \(\large \frac {b}{a}\) = \(\large \frac {3}{2}\) …[By Invertendo]
∴ \(\large \frac {7}{4}\) × \(\large \frac {b}{a}\) = \(\large \frac {7}{4}\) × \(\large \frac {3}{2}\) …[Multiplying both sides by \(\large \frac {7}{4}\)]
∴ \(\large \frac {7b}{4a}\) = \(\large \frac {21}{8}\)
∴ \(\large \frac {7b\,+\,4a}{7b\,–\,4a}\) = \(\large \frac {21\,+\,8}{21\,–\,8}\) …[By Componendo-Dividendo]
∴ \(\large \frac {7b\,+\,4a}{7b\,–\,4a}\) = \(\large \frac {29}{13}\)
∴ \(\large \frac {7b\,–\,4a}{7b\,+\,4a}\) = \(\large \frac {29}{13}\) …[By Invertendo]
(9) If a, b, c, d are in proportion, then prove that,
(i) \(\large \frac {11a²\,+\,9ac}{11b²\,+\,9bd}\) = \(\large \frac {a²\,+\,3ac}{b²\,+\,3bd}\)
Solution:
a, b, c, d are in proportion
∴ \(\large \frac {a}{b}\) = \(\large \frac {c}{d}\)
Let,
\(\large \frac {a}{b}\) = \(\large \frac {c}{d}\) = k … (k ≠ 0)
∴ a = bk, c = dk
LHS
\(\large \frac {11a²\,+\,9ac}{11b²\,+\,9bd}\)
= \(\large \frac {11(bk)²\,+\,9(bk)(bk)}{11b²\,+\,9bd}\)
= \(\large \frac {11b²k²\,+\,9bdk²}{11b²\,+\,9bd}\)
= \(\large \frac {bk²(11b\,+\,3d)}{b(11b\,+\,6bd)}\)
= k²
RHS
= \(\large \frac {a²\,+\,3ac}{b²\,+\,3bd}\)
= \(\large \frac {(bk)²\,+\,3(bk)(dk)}{b²\,+\,3bd}\)
= \(\large \frac {b²k²\,+\,3bdk²)}{b²\,+\,3bd}\)
= \(\large \frac {k²(b²\,+\,3bd))}{(b²\,+\,3bd)}\)
= k²
∵ LHS = RHS
∴ \(\large \frac {11a²\,+\,9ac}{11b²\,+\,9bd}\) = \(\large \frac {a²\,+\,3ac}{b²\,+\,3bd}\)
(ii) \(\large \sqrt {\frac {a²\,+\,5c²}{b²\,+\,5d²}}\) = \(\large \frac {a}{b}\)
Solution:
\(\large \sqrt {\frac {a²\,+\,5c²}{b²\,+\,5d²}}\) = \(\large \frac {a}{b}\)
LHS
= \(\large \sqrt {\frac {(bk)²\,+\,5(dk)²}{b²\,+\,5d²}}\)
= \(\large \sqrt {\frac {b²k²\,+\,5d²k²}{b²\,+\,5d²}}\)
= \(\large \sqrt {\frac {k²(b²\,+\,5d²)}{b²\,+\,5d²}}\)
= \(\sqrt {k²}\)
= k
RHS
= \(\large \frac {a}{b}\)
= \(\large \frac {bk}{b}\)
= k
∵ LHS = RHS
∴ \(\large \sqrt {\frac {a²\,+\,5c²}{b²\,+\,5d²}}\) = \(\large \frac {a}{b}\)
(iii) \(\large \frac {a²\,+\,ab\,+\,b²}{a²\,–\,ab\,+\,b²}\) = \(\large \frac {c²\,+\,cd\,+\,d²}{c²\,–\,cd\,+\,d²}\)
Solution:
\(\large \frac {a²\,+\,ab\,+\,b²}{a²\,–\,ab\,+\,b²}\) = \(\large \frac {c²\,+\,cd\,+\,d²}{c²\,–\,cd\,+\,d²}\)
LHS
= \(\large \frac {(bk)²\,+\,(bk)b\,+\,b²}{(bk)²\,–\,(bk)b\,+\,b²}\)
= \(\large \frac {b²k²\,+\,b²k\,+\,b²}{b²k²\,–\,b²k\,+\,b²}\)
= \(\large \frac {b²(k²\,+\,k\,+\,1)}{b²(k²\,–\,k\,+\,1)}\)
= \(\large \frac {k²\,+\,k\,+\,1}{k²\,–\,k\,+\,1}\)
RHS
\(\large \frac {c²\,+\,cd\,+\,d²}{c²\,–\,cd\,+\,d²}\)
= \(\large \frac {(dk)²\,+\,(dk)d\,+\,d²}{(dk)²\,–\,(dk)d\,+\,d²}\)
= \(\large \frac {d²k²\,+\,d²k\,+\,d²}{d²k²\,–\,d²k\,+\,d²}\)
= \(\large \frac {d²(k²\,+\,k\,+\,1)}{d²(k²\,–\,k\,+\,1)}\)
= \(\large \frac {k²\,+\,k\,+\,1}{k²\,–\,k\,+\,1}\)
∵ LHS = RHS
∴ \(\large \frac {a²\,+\,ab\,+\,b²}{a²\,–\,ab\,+\,b²}\) = \(\large \frac {c²\,+\,cd\,+\,d²}{c²\,–\,cd\,+\,d²}\)
(10) If a, b, c are in continued proportion, then prove that,
(i) \(\large \frac {a}{a\,+\,2b}\) = \(\large \frac {a\,–\,2b}{a\,–\,4c}\)
Solution:
a, b, c are in continued proportion
∴ \(\large \frac {a}{b}\) = \(\large \frac {b}{c}\)
Let,
\(\large \frac {a}{b}\) = \(\large \frac {b}{c}\) = k …[k ≠ 0]
∴ b = ck, a = bk = (ck)k = ck²
LHS
= \(\large \frac {a}{a\,+\,2b}\)
= \(\large \frac {ck²}{ck²\,+\,2ck}\)
= \(\large \frac {ck(k)}{ck(k\,+\,2)}\)
= \(\large \frac {k}{k\,+\,2}\)
RHS
= \(\large \frac {a\,–\,2b}{a\,–\,4c}\)
= \(\large \frac {ck²\,–\,2ck}{ck²\,–\,4c}\)
= \(\large \frac {ck(k\,–\,2)}{c(k²\,–\,4)}\)
= \(\large \frac {k\,–\,2}{k²\,–\,2²}\)
= \(\large \frac {k\,–\,2}{(k\,–\,2)(k\,+\,2)}\)
= \(\large \frac {k}{k\,+\,2}\)
∵ LHS = RHS
∴ \(\large \frac {a}{a\,+\,2b}\) = \(\large \frac {a\,–\,2b}{a\,–\,4c}\)
(ii) \(\large \frac {b}{b\,+\,c}\) = \(\large \frac {a\,–\,b}{a\,–\,c}\)
Solution:
\(\large \frac {b}{b\,+\,c}\) = \(\large \frac {a\,–\,b}{a\,–\,c}\)
LHS
= \(\large \frac {b}{b\,+\,c}\)
= \(\large \frac {ck}{ck\,+\,c}\)
= \(\large \frac {c(k)}{c(k\,+\,1)}\)
= \(\large \frac {k}{k\,+\,1}\)
RHS
= \(\large \frac {a\,–\,b}{a\,–\,c}\)
= \(\large \frac {ck²\,–\,ck}{ck²\,–\,c}\)
= \(\large \frac {ck(k\,–\,1)}{c(k²\,–\,1)}\)
= \(\large \frac {k(k\,–\,1)}{(k²\,–\,1²)}\)
= \(\large \frac {k(k\,–\,1)}{(k\,–\,1)(k\,+\,1)}\)
= \(\large \frac {k}{k\,+\,1}\)
∵ LHS = RHS
∴ \(\large \frac {b}{b\,+\,c}\) = \(\large \frac {a\,–\,b}{a\,–\,c}\)
(11) Solve : \(\large \frac {12x²\,+\,18x\,+\,42}{18x²\,+\,12x\,+\,58}\) = \(\large \frac {2x\,+\,3}{3x\,+\,2}\)
Solution:
\(\large \frac {12x²\,+\,18x\,+\,42}{18x²\,+\,12x\,+\,58}\) = \(\large \frac {2x\,+\,3}{3x\,+\,2}\)
Let,
\(\large \frac {12x²\,+\,18x\,+\,42}{18x²\,+\,12x\,+\,58}\) = \(\large \frac {2x\,+\,3}{3x\,+\,2}\) = k
∴ \(\large \frac {12x²\,+\,18x\,+\,42}{18x²\,+\,12x\,+\,58}\) = \(\large \frac {–\,6x(2x\,+\,3)}{–\,6x(3x\,+\,2)}\)
By theorem on equal ratios, we get,
k = \(\large \frac {12x²\,+\,18x\,+\,42\,–\,6x(2x\,+\,3)}{18x²\,+\,12x\,+\,58\,–\,6x(3x\,+\,2)}\)
k = \(\large \frac {12x²\,+\,18x\,+\,42\,–\,12x²\,–\,18x}{18x²\,+\,12x\,+\,58\,–\,18x²\,–\,12x}\)
k = \(\large \frac {42}{58}\)
∴ \(\large \frac {2x\,+\,3}{3x\,+\,2}\) = \(\large \frac {42}{58}\)
∴ 29 (2x + 3) = 21(3x + 2)
∴ 58x + 87 = 63x + 42
∴ 87 – 42 = 63x – 58x
∴ 45 = 5x
∴ 5x = 45
∴ x = \(\large \frac {45}{5}\)
∴ x = 9
Ans: x = 9 is the solution of given equation.
(12) If \(\large \frac {2x\,–\,3y}{3z\,+\,y}\) = \(\large \frac {z\,–\,y}{z\,–\,x}\) = \(\large \frac {x\,+\,3z}{2y\,–\,3x}\) then prove that every ratio = \(\large \frac {x}{y}\)
Solution:
If \(\large \frac {2x\,–\,3y}{3z\,+\,y}\) = \(\large \frac {z\,–\,y}{z\,–\,x}\) = \(\large \frac {x\,+\,3z}{2y\,–\,3x}\)
Let,
If \(\large \frac {2x\,–\,3y}{3z\,+\,y}\) = \(\large \frac {z\,–\,y}{z\,–\,x}\) = \(\large \frac {x\,+\,3z}{2y\,–\,3x}\) = k
∴ k = \(\large \frac {2x\,–\,3y}{3z\,+\,y}\) = \(\large \frac {–\,3(z\,–\,y)}{–\,3(z\,–\,x)}\) = \(\large \frac {x\,+\,3z}{2y\,–\,3x}\)
By theorem on equal ratios, we get,
∴ k = \(\large \frac {2x\,–\,3y\,–\,3(z\,–\,y)\,x\,+\,3z}{3z\,+\,y\,–\,3(z\,–\,x)\,2y\,–\,3x}\)
∴ k = \(\large \frac {2x\,–\,3y\,–\,3z\,+\,3y\,x\,+\,3z}{3z\,+\,y\,–\,3z\,+\,3x\,+\,2y\,–\,3x}\)
∴ k = \(\large \frac {3x}{3y}\)
∴ k = \(\large \frac {x}{y}\)
Hence each ratio is equal to \(\large \frac {x}{y}\)
(13*) If \(\large \frac {by\,+\,cz}{b²\,+\,c²}\) = \(\large \frac {cz\,+\,ax}{c²\,+\,a²}\) = \(\large \frac {ax\,+\,by}{a²\,+\,b²}\) then prove that \(\large \frac {x}{a}\) = \(\large \frac {y}{b}\) = \(\large \frac {z}{c}\)
Solution:
\(\large \frac {by\,+\,cz}{b²\,+\,c²}\) = \(\large \frac {cz\,+\,ax}{c²\,+\,a²}\) = \(\large \frac {ax\,+\,by}{a²\,+\,b²}\)
Let,
\(\large \frac {by\,+\,cz}{b²\,+\,c²}\) = \(\large \frac {cz\,+\,ax}{c²\,+\,a²}\) = \(\large \frac {ax\,+\,by}{a²\,+\,b²}\) = k
By theorem on equal ratios, we get,
k = \(\large \frac {by\,+\,cz\,+\,cz\,+\,ax\,+\,ax\,+\,by}{b²\,+\,c²\,+\,c²\,+\,a²\,+\,a²\,+\,b²}\)
∴ k = \(\large \frac {2ax\,+\,2by\,+\,2cz}{2a²\,+\,2b²\,+\,2c²}\)
∴ k = \(\large \frac {2(ax\,+\,by\,+\,cz)}{2(a²\,+\,b²\,+\,c²)}\)
∴ k = \(\large \frac {ax\,+\,by\,+\,cz}{a²\,+\,b²\,+\,c²}\)
Now,
\(\large \frac {by\,+\,cz}{b²\,+\,c²}\) = k
∴ \(\large \frac {by\,+\,cz}{b²\,+\,c²}\) = \(\large \frac {ax\,+\,by\,+\,cz}{a²\,+\,b²\,+\,c²}\)
∴ \(\large \frac {a²\,+\,b²\,+\,c²}{b²\,+\,c²}\) = \(\large \frac {ax\,+\,by\,+\,cz}{by\,+\,cz}\) …[By Alternendo]
∴ \(\large \frac {a²\,+\,b²\,+\,c²\,–\,(b²\,+\,c²)}{b²\,+\,c²}\) = \(\large \frac {ax\,+\,by\,+\,cz\,–\,(by\,+\,cz)}{by\,+\,cz}\) …[By Componendo-Dividendo]
∴ \(\large \frac {a²\,+\,b²\,+\,c²\,–\,b²\,–\,c²}{b²\,+\,c²}\) = \(\large \frac {ax\,+\,by\,+\,cz\,–\,by\,–\,cz}{by\,+\,cz}\)
∴ \(\large \frac {a²}{b²\,+\,c²}\) = \(\large \frac {ax}{by\,+\,cz}\)
∴ \(\large \frac {by\,+\,cz}{b²\,+\,c²}\) = \(\large \frac {ax}{a²}\)
∴ \(\large \frac {x}{a}\) = \(\large \frac {by\,+\,cz}{b²\,+\,c²}\) …(i)
Now,
\(\large \frac {cz\,+\,ax}{c²\,+\,a²}\) = k
∴ \(\large \frac {cz\,+\,ax}{c²\,+\,a²}\) = \(\large \frac {ax\,+\,by\,+\,cz}{a²\,+\,b²\,+\,c²}\)
∴ \(\large \frac {a²\,+\,b²\,+\,c²}{c²\,+\,a²}\) = \(\large \frac {ax\,+\,by\,+\,cz}{cz\,+\,ax}\) …[By Alternendo]
∴ \(\large \frac {a²\,+\,b²\,+\,c²\,–\,(c²\,+\,a²)}{c²\,+\,a²}\) = \(\large \frac {ax\,+\,by\,+\,cz\,–\,(cz\,+\,ax)}{cz\,+\,ax}\) …[By Componendo-Dividendo]
∴ \(\large \frac {a²\,+\,b²\,+\,c²\,–\,c²\,–\,a²}{c²\,+\,a²}\) = \(\large \frac {ax\,+\,by\,+\,cz\,–\,cz\,–\,ax}{cz\,+\,ax}\)
∴ \(\large \frac {b²}{c²\,+\,a²}\) = \(\large \frac {by}{cz\,+\,ax}\)
∴ \(\large \frac {cz\,+\,ax}{c²\,+\,a²}\) = \(\large \frac {by}{b²}\) …[By Alternendo]
∴ \(\large \frac {y}{b}\) = \(\large \frac {cz\,+\,ax}{c²\,+\,a²}\) …(ii)
And,
\(\large \frac {ax\,+\,by}{a²\,+\,b²}\) = k
∴ \(\large \frac {ax\,+\,by}{a²\,+\,b²}\) = \(\large \frac {ax\,+\,by\,+\,cz}{a²\,+\,b²\,+\,c²}\)
∴ \(\large \frac {a²\,+\,b²\,+\,c²}{a²\,+\,b²}\) = \(\large \frac {ax\,+\,by\,+\,cz}{ax\,+\,by}\) …[By Alternendo]
∴ \(\large \frac {a²\,+\,b²\,+\,c²\,–\,(a²\,+\,b²)}{a²\,+\,b²}\) = \(\large \frac {ax\,+\,by\,+\,cz\,–\,(ax\,+\,by)}{ax\,+\,by}\) …[By Componendo-Dividendo]
∴ \(\large \frac {a²\,+\,b²\,+\,c²\,–\,a²\,–\,b²}{a²\,+\,b²}\) = \(\large \frac {ax\,+\,by\,+\,cz\,–\,ax\,–\,by}{ax\,+\,by}\)
∴ \(\large \frac {c²}{a²\,+\,b²}\) = \(\large \frac {cz}{ax\,+\,by}\)
∴ \(\large \frac {ax\,+\,by}{a²\,+\,b²}\) = \(\large \frac {cz}{c²}\)
∴ \(\large \frac {z}{c}\) = \(\large \frac {ax\,+\,by}{a²\,+\,b²}\) …(iii)
from (i), (ii) and (iii), we get,
\(\large \frac {x}{a}\) = \(\large \frac {y}{b}\) = \(\large \frac {z}{c}\)
Hence proved.