Ratio and Proportion
Class9Mathematics1Chapter4Maharashtra Board
Notes
Topics to be learn : Part 1

Ratio and proportion
Ratio : A ratio is the relation which one quantity bears to another of the same kind and having the same units.
Direct and Inverse Proportion :
(i) Direct Proportion : If the ratio of the two quantities is constant, then it is said that the two quantities are in direct proportion.
Example :
The information about the distance covered by a car using certain quantity of petrol is given in the table.
Petrol : x litres  1  3  4  6 
Distance : y km  12  36  48  72 
\(\frac{x}{y}\)  \(\frac{1}{12}\)  \(\frac{3}{36}=\frac{1}{12}\)  \(\frac{4}{48}=\frac{1}{12}\)  \(\frac{6}{72}=\frac{1}{12}\) 
The ratio of the consumption of petrol (in litres) and the distance covered by the car (in km) is constant. In this case, it is said that the two quantities are in direct
proportion.
(ii) Inverse Proportion : If the product of the two quantities is constant, then it is said that the two quantities are in inverse proportion.
Example :
The information about the speed of the vehicle (in km/h) and the time taken (in hours) to cover 120 km is given in the following table.
Speed (km/h) : x km/h  15  30  45  60 
Time (hrs) : y  6  3  2  1.5 
x × y km  90  90  90  90 
Here, the product of the speed of the vehicle and the time taken to cover 90 km is constant. In such a case, it is said that the quantities are in inverse proportion.
Properties of ratio :
(i) Ratio of numbers a and b is written as a : b or \(\frac{a}{b}\). In this ratio a is called the predecessor (first term) and b is called successor (Second term).
(ii) In the ratio of two numbers, if the second term is 100 then it is known as a percentage.
(iii) The ratio remains unchanged, if its terms are multiplied or divided by nonzero number.
 Example : 3 : 4 = 6 : 8 = 9 : 12, Similarly 2 : 3 : 5 = 8 : 12 : 20. If k is a nonzero number, then a : b = ak : bk a : b : c = ak : bk : ck
(iv) The quantities taken in the ratio must be expressed in the same unit.
(v) The ratio of two quantities is unitless.
 Example : The ratio of 2 kg and 300 g is not 2 : 300 , but it is 2000 : 300 as (2 kg = 2000 gm) i.e. 20 : 3
Comparison of ratios :
The numbers a, b, c, d being positive, comparison of ratios \(\frac{a}{b}\), \(\frac{c}{d}\) can be done
using following rules :
(i) If ad > bc then \(\frac{a}{b}>\frac{c}{d}\)
(ii) If ad < bc then \(\frac{a}{b}<\frac{c}{d}\)
(iii) If ad = bc then \(\frac{a}{b}=\frac{c}{d}\)
Operations on equal ratios :
Using the properties of equality, we can perform some operations on ratios.
a, b, c, d are positive integers.
(1) Invertendo : If \(\frac{a}{b}=\frac{c}{d}\) then \(\frac{b}{a}=\frac{d}{c}\) This result is known as invertendo.
(2) Alternando : If \(\frac{a}{b}=\frac{c}{d}\) then \(\frac{a}{c}=\frac{b}{d}\) This result is known as alternando.
(3) Componendo : If \(\frac{a}{b}=\frac{c}{d}\) then \(\frac{a+b}{b}=\frac{c+d}{d}\)
\(\frac{a}{b}=\frac{c}{d}\) ...add 1 to both side
\(\frac{a}{b}+1=\frac{c}{d}+1\)
∴ \(\frac{a+b}{b}=\frac{c+d}{d}\)
This result is known as componendo
(4) Dividendo : If \(\frac{a}{b}=\frac{c}{d}\) then \(\frac{ab}{b}=\frac{cd}{d}\)
\(\frac{a}{b}=\frac{c}{d}\) ...subtracting 1 from both side
\(\frac{a}{b}1=\frac{c}{d}1\)
∴ \(\frac{ab}{b}=\frac{cd}{d}\)
This result is known as dividendo.
(5) Componendodividendo : If \(\frac{a}{b}=\frac{c}{d}\) then \(\frac{a+b}{ab}=\frac{c+d}{cd}\)
By componendo, \(\frac{a+b}{b}=\frac{c+d}{d}\) …..(1)
By dividend, \(\frac{ab}{b}=\frac{cd}{d}\) …..(2)
Dividing (1) by (2),
\((\frac{a+b}{b})÷(\frac{ab}{b})=(\frac{c+d}{d})÷(\frac{cd}{d})\)
\((\frac{a+b}{b})×(\frac{b}{ab})=(\frac{c+d}{d})×(\frac{d}{cd})\)
∴ \(\frac{a+b}{ab}=\frac{c+d}{cd}\)
General form of Componendo and Dividendo :
If \(\frac{a}{b}=\frac{c}{d}\) then \(\frac{a+b}{b}=\frac{c+d}{d}\) …..... (Performing componendo once)
\(\frac{a+2b}{b}=\frac{c+2d}{d}\) …..... (Performing componendo twice)
Generally \(\frac{a+mb}{b}=\frac{c+md}{d}\) …..... (Performing componendo m times) …(1)
Similarly If \(\frac{a}{b}=\frac{c}{d}\) then \(\frac{amb}{b}=\frac{cmd}{d}\) ... (Performing dividendo m times) ... (2)
And If \(\frac{a}{b}=\frac{c}{d}\) then \\(\frac{a+mb}{amb}=\frac{c+md}{cmd}\) ... [Dividing (1) by (2)]
Q. Which of the above statements are true or false ? If false, explain why.
(i) \(\frac{a+b}{b}=\frac{c+d}{d}\), (ii) \(\frac{a}{c}=\frac{b}{d}\), (iii) \(\frac{a}{b}=\frac{ac}{bd}\), (iv) \(\frac{c}{d}=\frac{ca}{db}\), (v) \(\frac{a}{b}=\frac{rc}{rd}\)
Solution :
True : (i), (ii), (v).
False :
(iii). Here, the numerator and denominator on the right hand side are multiplied by two different variables c and d.
(iv) Here, different variables a and b are subtracted from the numerator and denominator on the right hand side.
Application of properties of equal ratios :
To solve some types of equations, it is convenient to use properties of equal ratios rather than using other methods.
Q. If \(\frac{a}{b}=\frac{3}{4}\) then find (i) \(\frac{a}{2b}\), (ii) \(\frac{3a}{4b}\), (iii) \(\frac{2ab}{2a+b}\), (iv) \(\frac{a^2+b^2}{b^2}\), (v) \(\frac{5a^2+2b^2}{5a^22b^2}\)
Solution :
(i) \(\frac{a}{b}=\frac{3}{4}\)
∴ \(\frac{a}{2b}=\frac{3}{2×4}=\frac{3}{8}\) … (multiplying both side by \(\frac{1}{2}\) )
(ii) \(\frac{a}{b}=\frac{3}{4}\)
∴ \(\frac{3}{4}×\frac{a}{b}=\frac{3}{4}×\frac{3}{4}=\frac{9}{16}\) … (multiplying both side by \(\frac{3}{4}\))
∴ \(\frac{3a}{4b}\) = \(\frac{9}{16}\)
(iii) \(\frac{a}{b}=\frac{3}{4}\)
∴ \(\frac{2a}{b}=\frac{2×3}{4}=\frac{6}{4}=\frac{3}{2}\)
∴ \(\frac{2ab}{2a+b}\) = \(\frac{32}{3+2}=\frac{1}{5}\) ….(using componendo and componendodividendo)
(iv) \(\frac{a}{b}=\frac{3}{4}\)
∴ \(\frac{a^2}{b^2}=\frac{3^2}{4^2}=\frac{9}{16}\)
∴ \(\frac{a^2+b^2}{b^2}\) = \(\frac{9+16}{16}=\frac{25}{16}\) ….(using componendo and componendodividendo)
(v) \(\frac{a^2}{b^2}=\frac{9}{16}\)
∴ \(\frac{5}{2}×\frac{a^2}{b^2}=\frac{5}{2}×\frac{9}{16}\)
∴ \(\frac{5a^2}{2b^2}=\frac{45}{32}\)
using componendo and componendodividendo
∴ \(\frac{5a^2+2b^2}{5a^22b^2}\) = \(\frac{45+32}{4532}\) = \(\frac{77}{13}\)
Theorem on equal ratios :
(i) If \(\frac{a}{b}=\frac{c}{d}\) then (\frac{a}{b}=\frac{a+c}{b+d}=\frac{c}{d}\) . This property is called the theorem of equal ratios.
(ii) If \(\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=......\) (finite terms) and if l, m, n, ... are nonzero numbers such that lb + md + nf + ... # 0,
then each ratio = \(\frac{la+mc+ne+...}{lb+md+nf+...}\)
This is the general form of the above theorem.
Q.1 : Fill in the blanks in the following statements.
\(\frac{x}{3}=\frac{y}{5}=\frac{z}{4}=\frac{5x3y+4z}{}\)
Solution :
\(\frac{x}{3}=\frac{y}{5}=\frac{z}{4}=\frac{5×x}{5×3}=\frac{3×y}{3×5}=\frac{4×z}{4×4}\)
∴ \(\frac{5x}{15}=\frac{3y}{15}=\frac{4z}{16}\)
= \(\frac{5x3y+4z}{1515+16}\) = \(\frac{5x3y+4z}{16}\) (by the theorem of equal ratio)
Q.2 : In a certain gymnasium, there are 35 girls and 42 boys in the kid's section, 30 girls and 36 boys in the children's section and 20 girls and 24 boys in the teen's section.
(i) What is the ratio of the number of boys to the number of girls in every section?
(ii) For physical exercises, all three groups gathered on the ground. Now what is the ratio of number of boys to the number of girls ?
(iii) From the answers of the above questions, did you verify the theorem of equal ratios ?
Solution :
(i) The ratio of number of boys to the number of girls in
Kid's section = 42 : 35 = 6 : 5 ... (1)
Children's section = 36 : 30 = 6 : 5 ... (2)
Teen's section = 24 : 20 = 6 : 5 ... (3)
The ratio of the number of boys to the number of girls in each group is the same. i.e. \(\frac{6}{5}\)
(ii) Total number of boys = 42 + 36 + 24 = 102,
Total number of girls = 35 + 30 + 20 = 85
The ratio of total number of boys to the total number of girls = 102 : 85 = 6 : 5 ... (4)
(iii) From (1), (2), (3) and (4), We verify the theorem on equal ratios.
Continued Proportion :
(i) If \(\frac{a}{b}=\frac{b}{c}\) then, a, b, c are in continued proportion.
(ii) If ac = b^{2}, then dividing both the sides by bc, we get, \(\frac{a}{b}=\frac{b}{c}\)
(iii) If \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}=\frac{e}{f}....\) then a, b, c, d, e, and f.... are said to be in continued proportion.
k – method :
(1) Examples on equal ratios can be easily solved by equating each ratio to k. This method is popularly known as the kmethod.
(2) If four positive real numbers a, b, c and d are in proportion, we have, \(\frac{a}{b}=\frac{b}{c}\)
If we equate each of these ratios to k, we get, \(\frac{a}{b}=\frac{b}{c}\) = k or a = bk and c = dk
Q. If \(\frac{a}{b}=\frac{b}{c}\) then show that \(\frac{5a3c}{5b3d}=\frac{7a2c}{7b2d}\)
Solution :
Let \(\frac{a}{b}=\frac{b}{c}\) = k ∴ a = bk, c = dk
Substituting values of a and c in both sides,
LHS = \(\frac{5a3c}{5b3d}\) = \(\frac{5(bk)3(dk)}{5b3d}\) = \(\frac{k(5b3d)}{5b3d}\) = k
RHS = \(\frac{7a2c}{7b2d}\) = \(\frac{7(bk)2(dk)}{7b2d}\) = \(\frac{k(7b2d)}{7b2d}\) = k
∴ LHS = RHS
∴ \(\frac{5a3c}{5b3d}=\frac{7a2c}{7b2d}\)
(3) If a, b, c are in continued proportion, we have, \(\frac{a}{b}=\frac{b}{c}\) . Equating each of these ratios to k, we get, \(\frac{a}{b}=\frac{b}{c}\) = k. Thus, b = ck and a = bk = ck.k. Hence, we have, b = ck and a = ck^{2}.
(4) If four numbers a, b, c and d are in continued proportion, we have, \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
In this case, c = dk, b = dk^{2} and a = dk^{3}.
(5) If five numbers a, b, c, d and e are in continued proportion, we have, \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}\) = k. In this case, d = ek, c = ek^{2}, b = ek^{3} and a = ek^{4}.
Q. Five numbers are in continued proportion. The first term is 5 and the last term is 80. Find these numbers.
Solution :
Let the numbers in continued proportion be a, ak, ak^{2}, ak^{3}, ak^{4}.
Here a = 5 and ak^{4} = 80
∴ 5 × k^{4} = 80
∴ k^{4} = 16
∴ k = 2 ….(∵ 2^{4} = 16)
ak = 5 × 2 = 10, ak^{2} = 5 × 4 = 20, ak^{3} = 5 × 8 = 40 , ak^{4} = 5 × 16 = 80
∴ the numbers are 5, 10, 20, 40, 80.
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