Ratio and Proportion: Concept, Short Trick with Solved Example
Ratio: The comparative relation between two amounts/quantities of same type is called ratio.
Let an amount be x and another is y, then, the ratio between them is x : y or x/y.
Here ‘x’ is called antecedent and ‘y’ is called consequent
Mixed ratio – Let x:y and P:Q be two ratios, then Px : Qy is called mixed ratio.
Duplicate Ratio–The mixed ratio of two equal ratios is called the duplicate Ratio as
Duplicate ratio of a:b is a2 : b2
Sub duplicate Ratio–The square root of a certain ratio is called its sub duplicate.
The sub duplicate ratio of a: b = $\sqrt{a}:\sqrt{b}$
Triplicate Ratio–The cube of a certain ratio is called triplicate ratio.
The triplicate ratio of a:b = a3 : b3
Sub triplicate Ratio–The cube root of a certain ratio is called sub triplicate ratio as–
The Sub triplicate Ratio of a:b = $\sqrt[3]{a}:\sqrt[3]{b}$
Inverse Ratio–The Reciprocal of quantities of ratio is called its inverse. Reciprocal or inverse ratio of a:b = $\frac{1}{a}:\frac{1}{b}\,\,or\,\,\frac{ab}{a}:\frac{ab}{b}\,\,or\,\,b:a$
Directly Proportional: If x = ky, where k is a constant, then we say that x is directly proportional to y. It is written as x ∝ y.
Inversely Proportional: If x = k/y where k is a constant, then we say that x is inversely proportional to y.
It is written as x ∝ 1/y
Proportion: When two ratios are equal to each other, then they are called proportional as
a:b = c:d, then, a, b, c and d are in proportion.
or,
a:b :: c:d
Example: 2:5 = 6: 15, then we write 2:5 :: 6:15
Example: If 8, 14, 7x + 2, 28 are proportion then find the value of x.
Solution: 8 : 14 = 7x + 2 : 28
⇒ $\frac{8}{14}=\frac{7x+2}{28}$
⇒ $7x+2 =\frac{8\times 28}{14}=16$
⇒ x = (16 – 2)/7 = 2
Fourth Proportional– Let x be the fourth proportional of a, b and c, then a:b::c:x
i.e. $\frac{a}{b}=\frac{c}{x}\,\,or\,\,x\,\,=\frac{bc}{a}$
∴ Fourth proportional of a, b and c = $\frac{bc}{a}$
Example: Find the Fourth proportional of 4, 6 and 8.
Solution: Fourth proportional = $\frac{6\times 8}{4}=12$
Third proportional–Let ‘x’ be the third proportional of a and b then, a:b :: b:x
i.e. $\frac{a}{b}=\frac{b}{x}\,\,or\,\,ax\,\,=b^2\,\,or\,\,x=\frac{b^2}{a}$
∴ Third proportional of a and b = $\frac{b^2}{a}$
Example: Find the third proportional of 4 and 6.
Solution: Third proportional = $\frac{6^2}{4}=9$
Mean Proportion – Let x be the mean proportion between a and b, then a:x::x:b (Real condition)
i.e. $\frac{a}{x}=\frac{x}{b}\,\,or\,\,x^2=ab\,\,or\,\,x\,\,=\,\,\sqrt{ab}$
∴ Mean proportional of a and b = $\sqrt{ab}$
Example: Find the third proportional of 4 and 9.
Solution: Mean proportional = $\sqrt{4\times 9}=6$
Invertendo–The proportion in which antecedent and consequent quantities change their places, is called invertendo.
Invertendo of a:b = c:d is b:a = d:c
Alternendo–If a:b ::c:d is a proportion then its alternendo is a:c::b:d. i.e. alternendo of
$\frac{a}{b}=\frac{c}{d}\,\,is\,\,\frac{a}{c}=\frac{b}{d}$
Componendo–If a:b::c:d is a proportion, then componendo is (a + b) : b :: (c + d) : d
It means, If $\frac{a}{b}=\frac{c}{d}\,\,\Rightarrow \frac{a+b}{b}=\frac{c+d}{d}$
Dividendo–If a:b :: c:d is a proportion, then its dividendo is (a – b): b:: (c – d):d
It means, If $\frac{a}{b}=\frac{c}{d}\,\,\Rightarrow \frac{a-b}{b}=\frac{c-d}{d}$
Componendo and dividendo–If there is a proportion a:b::c:d then its componendo and dividendo is (a + b):(a – b)::(c + d):(c – d)
It means, If $\frac{a}{b}=\frac{c}{d}\,\,\Rightarrow \frac{a+b}{a-b}=\frac{c+d}{c-d}$
Dividendo and Componendo –If there is a proportion a:b::c:d then its componendo and dividendo is (a – b):(a b)::(c – d):(c + d)
It means, If $\frac{a}{b}=\frac{c}{d}\,\,\Rightarrow \frac{a-b}{a+b}=\frac{c-d}{c+d}$
Some Important rule of Ratio and proportion:
RULE 1 : Let ‘x’ be a number which is subtracted from a, b, c and d to make them proportional, then
x = $\frac{ad-bc}{\left( a-b \right) -\left( b-c \right)}$
and
Let ‘x’ be a number which is added to a, b, c and d to make them proportional, then
x = $\frac{bc-ad}{\left( a-b \right) -\left( b-c \right)}$
Example: How much added to each of 6, 8, 10 and 13 to make them proportional?
Solution: Result = $\frac{80-78}{\left( -2 \right) -\left( -3 \right)}=2$
Example-1: How much subtract from each of 13, 18, 19 and 27 to make them proportional?
Solution: Result = $\frac{351-342}{\left( -5 \right) -\left( -8 \right)}=3$
RULE 2 : If A:B = x:y and B:C = y:z then
(i) A:C = x : z
(ii) A:B:C = x : y : z
Example: If A:B = 4 : 5 and B:C = 5 : 6 then (i) A:C = ? and (ii) A:B:C = ?
Solution: (i) A:C = 4 : 6 = 2 : 3
(ii) A:B:C = 4 : 5 : 6
RULE 3 : If A:B = x:y and B:C = p:q then
(i) A:C = product of antecedent : product of consequent = xp : yq
(ii) A:B:C = xp : yp : yq
It is done as follows:
A:B = x:y = xp : yp ( by multiplying p)
B:C = p:q = yp : yq ( by multiplying y)
A:B:C = xp:yp:yq
Remember this trick:
Example: If A:B = 5 : 6 and B:C = 7 : 9 then find (i) A:C and (ii) A:B:C
Solution: (i) A:C = 5 : 9
(ii) Using trick
RULE 4 : If A:B = m : n, B:C = n : o and C : D = o : p then
(i) A:C = m : p
(ii) A:B:C :D= m : n : o : q
Example: If A:B = 5 : 6, B:C = 6 : 7 and C : D = 7 : 8 then (i) A:C = ? and (ii) A:B:C:D = ?
Solution: (i) A : D = 5 : 8
(ii) Using Trick
A : B : C : D = = 5 : 6 : 7 : 8
RULE 5 : If A:B = m : n, B:C = o : p and C : D = q : r then
(i) A:D = moq : npr
(ii) It is done as follows:
A : B = m : n
B : C = o : p
C : D = q : r
A : B : C : D = moq : noq : npq : npr
Remember this trick:
Example: If A:B = 4 : 5, B:C = 6 : 7 and C : D = 8 : 9 then find (i) A : D and (ii) A : B : C : D
Solution: (i) A : D = 192 : 315 = 64 : 105
(ii) Using trick
RULE 7 : If A : x = B : y = C : z then
(i) $\frac{A+B+C}{A}=\frac{x+y+z}{x}$
(ii) $\frac{A+B+C}{B}=\frac{x+y+z}{y}$
(iii) $\frac{A+B+C}{C}=\frac{x+y+z}{z}$
Example: If $\frac{a}{3}=\frac{b}{4}=\frac{c}{7}$ , then $\frac{a+b+c}{c}$ is equal to :
Solution: $\frac{a+b+c}{c}\,\,=\frac{3+4+7}{7}=2$
RULE 8 : If xA = yB = cZ then A: B :C then A: B :C = ?
A : B : C = $\frac{1}{\mathrm{x}}:\frac{1}{\mathrm{y}}:\frac{1}{\mathrm{z}}=\mathrm{yz}:\mathrm{xz}:\mathrm{xy}$
Example: If A of 30% = B of 0.25 = C of 1/5 then A : B : C = ?
Solution: A of 30% = B of 0.25 = C of 1/5 ⇒ 0.3A = 0.25B = 0.2C
A : B : C = $\frac{1}{0.3}:\frac{1}{0.25}:\frac{1}{0.2}=\frac{10}{3}:4:5=10:12:15$
RULE 9 : If an amount S is to be divided between A and B in the ratio x : y then
(i) Part of A = $\frac{x}{x+y}\times S$
(ii) Part of B = $\frac{y}{x+y}\times S$
(iii) Difference of part of A and B = $\frac{x-y}{x+y}\times S$ where x > y
Example: Abdullah and Amaan share their profit in the ratio 6:5. If their total profit is 1661 the what is their profit?
Solution: Abdullah’s Profit = $\frac{6}{11}\times 1661=906$
Amaan’s Profit = $\frac{5}{11}\times 1661=755$
RULE 10 : If the ratio of share of A and B is x : y and the difference in their share is ‘D’ then,
(i) Part of A = $\frac{x}{x-y}\times D$
(ii) Part of B = $\frac{x}{x-y}\times D$
(iii) The sum of parts of A and B = $\frac{x+y}{x-y}\times D$ where x > y
Example: The ratio of share of Rashid and Amit is 13:11 and their share difference is 144. Find the share of Rashid and Amit?
Solution: Rashid’s share = $\frac{13}{13-11}\times 144=936$
Amit’s share = $\frac{11}{13-11}\times 144=792$
RULE 11: If the ratio of A and B is x : y and the part of A is ‘R’, then
(i) Share of B = $\frac{y}{x}\times R$
(ii) Total share of A and B = $\frac{x+y}{x}\times R$
(iii) Difference in share of A and B = $\frac{x\sim y}{x}\times R$
Example: The ratio of daily earning of Madhu and Rahul is 7:9. If Madhu’s daily earning is Rs. 175 then what is the Rahul’s daily earning and what was their total earning?
Solution: Rahul’s daily earning = $\frac{9}{7}\times 175=225$
Their total earning = $\frac{7+9}{7}\times 175=400$
RULE 12 : If Addition of two number ‘S’ and their difference is ‘D’ the ratio of the two number
Ratio = (S +D) : (S –D)
Example: Sum of two number is 145 and their difference is 25. Find the ratio of two numbers?
Solution: ratio of the two number = 145 + 25 : 145 – 25 = 17:12
RULE 13 : If the amount S is divided among A, B and C in the ratio x : y : z, then
(i) The share of A = $\frac{xS}{x+y+z}$
(ii) The share of B =$\frac{yS}{x+y+z}$
(iii) The share of C =$\frac{zS}{x+y+z}$
Example: If Rs. 782 be divided into three part, proportional to $\frac{1}{2}:\frac{2}{3}:\frac{3}{4}$ then the first part is
Solution: $\frac{1}{2}:\frac{2}{3}:\frac{3}{4}\,\,=6:8:9$
First part = $\frac{6\times 782}{23}=204$
Rule 14: The ratio of three number is x : y : z and their product is ‘P’. The numbers are:
1st Number = $x\times \sqrt[3]{\frac{P}{xyz}}$
2nd Number = $y\times \sqrt[3]{\frac{P}{xyz}}$
3rd Number = $z\times \sqrt[3]{\frac{P}{xyz}}$
Example: The ratio of three number is 3 : 4 : 7 and their product is 18144. The numbers are:
Solution: 1st Number = $3\times \sqrt[3]{\frac{18144}{3\times 4\times 7}}=18$
2nd Number = $4\times \sqrt[3]{\frac{18144}{3\times 4\times 7}}=24$
3rd Number = $7\times \sqrt[3]{\frac{18144}{3\times 4\times 7}}=42$
RULE 15: If the ratio of A, B and C is x : y : z and the part of A is ‘D’ then,
(i) Part of B = $\frac{y}{x}\times D$
(ii) Part of C = $\frac{z}{x}\times D$
Example: The income of A, B and C are in the ratio 3 : 7 : 4. If income of A is Rs.2400 then what is income of B and C?
Solution: Income of B = $\frac{7}{3}\times 2400=5600$
Income of C = $\frac{4}{3}\times 2400=3200$
RULE 16: If an amount is to be divided among A, B and C in the ratio x : y : z and the difference between A and B is share is ‘D’, then
(i) Part of C = $\frac{z}{x\sim y}\times D$
Example: The income of A, B and C are in the ratio 7 : 9 : 12 and Difference of income of A and B is Rs.560. Find C income?
Solution: C’s income = $\frac{12}{9-7}\times 560=3360$
RULE 17 : If the ratio of allegation of milk and water in a glass is m:n and in other glass allegation is p:q, then the ratio of milk and water in third glass which contains allegation of both glasses is
Ratio = $\left( \frac{m}{m+n}+\frac{p}{p+q} \right) :\left( \frac{n}{m+n}+\frac{q}{p+q} \right) $
Example: Two equal glasses filled with alcohol and water in the proportions 2 : 1 and 3 : 2 are emptied into a third glass. The proportion of alcohol and water in the third glass will be
Solution: Ratio = $\frac{2}{2+1}+\frac{3}{3+2}:\frac{1}{2+1}+\frac{2}{3+2}$
= $\frac{2}{3}+\frac{3}{5}:\frac{1}{3}+\frac{2}{5}=\frac{19}{15}:\frac{11}{15}=19:11$
RULE 18 : If the ratio of milk and water in the allegation of A liter is m:n then water must be added in it so that ratio of milk and water would be p:q is
Required amount of water = $\frac{A\left( mq-np \right)}{p\left( m+n \right)}$
Example: In a mixture of 60 liters, the ratio of milk and water is 2 : 1. How much more water must be added to make its ratio 1 : 2?
Solution: Required amount of water = $\frac{60\left( 4-1 \right)}{1\left( 3 \right)}=60$
RULE 19: The ratio of income of two persons A and B is m:n. If the ratio of their expenditures is p:q, then the monthly income of A and B, when each one of them saves ‘S’ rupees will be
Monthly income of A = $\frac{Sm\left( q-p \right)}{mq-np}$
Monthly income of B =$\frac{Sn\left( q-p \right)}{mq-np}$
Example: A and B have monthly incomes in the ratio 5 : 6 and monthly expenditures in the ratio 3 : 4. If they save Rs.2550, find the monthly income of B:
Solution: Monthly income of A =$\frac{2550\times 5\times 1}{20-18}=6375$
Monthly income of B = $\frac{2550\times 6\times 1}{20-18}=7650$
RULE 20: Two numbers are in the ratio a:b and if each number is increased by x, the ratio becomes c:d. Then the two numbers will be
$\frac{ax\left( c-d \right)}{ad-bc}\,\,and\,\,\frac{bx\left( c-d \right)}{ad-bc}$
Difference of the Number = $\frac{\mathrm{x}\left( \mathrm{a}-\mathrm{b} \right) \left( \mathrm{c}-\mathrm{d} \right)}{\mathrm{ad}-\mathrm{bc}}$ [Take positive value]
Sum of the Numbers = $\frac{\mathrm{x}\left( \mathrm{a}+\mathrm{b} \right) \left( \mathrm{c}-\mathrm{d} \right)}{\mathrm{ad}-\mathrm{bc}}$
Example: The present ages of A and B are in the ratio 3 : 4. Ten years ago, this ratio was 4 : 7. The present ages of A and B are respectively:
Solution: 10 years ago A’s Age = $\frac{4\times 10\times -1}{16-21}=8$
The present ages of A = 10 + 8 =18
10 years ago B’s Age = $\frac{7\times 10\times -1}{16-21}=14$
The present ages of B = 14 + 10 = 24
Example-2: The present ages of Father and Son are in the ratio 7:2. Eight years later ratio of their age is 5 : 2. What is The present ages of Father and Son?
Solution: Father’s age = $\frac{7\times 8\times 3}{14-10}=42$
Son’s age = $\frac{2\times 8\times 3}{14-10}=12$
Example-3: If two numbers are in the ratio 2 : 3 and the ratio becomes 3 : 4 when 8 is added to both the numbers, then the sum of the two numbers and difference of the numbers is:
Solution: Sum of the Numbers = $\frac{8\left( 2+3 \right) \left( 3-4 \right)}{8-9}=40$
Difference of the numbers = $\frac{8\left( 2-3 \right) \left( 3-4 \right)}{8-9}=8$(since difference, we take positive value)
First number = $\frac{2\times 8\times \left( 3-4 \right)}{8-9}=16$
Second Number = $\frac{3\times 8\times \left( 3-4 \right)}{8-9}=24$
RULE 21: Two numbers are in the ratio a:b and if x is subtracted from each number the ratio becomes c:d. The two numbers will be
$\frac{ax\left( d-c \right)}{ad-bc}\,\,and\,\,\frac{bx\left( d-c \right)}{ad-bc}$
Difference of the number = $\frac{\mathrm{x}\left( \mathrm{a}-\mathrm{b} \right) \left( \mathrm{d}-\mathrm{c} \right)}{\mathrm{ad}-\mathrm{bc}}$ [Take positive value]
Sum of the Numbers = $\frac{\mathrm{x}\left( \mathrm{a}+\mathrm{b} \right) \left( \mathrm{d}-\mathrm{c} \right)}{\mathrm{ad}-\mathrm{bc}}$
Example: Two numbers are in the ratio 5 : 7. On diminishing each of them by 40, they become in the ratio 17 : 27. The difference of the numbers and sum of the numbers is:
Solution: Difference of the number = $\frac{40\left( 5-7 \right) \left( 27-17 \right)}{5\times 27-7\times 17}=50$
Sum of the Numbers = $\frac{40\left( 5+7 \right) \left( 27-17 \right)}{5\times 27-7\times 17}=300$
First number = $\frac{5\times 40\times \left( 27-17 \right)}{5\times 27-7\times 17}=125$
Second Number = $\frac{7\times 40\times \left( 27-17 \right)}{5\times 27-7\times 17}=175$
Rules 22: x numbers of coins consists of 1 rupee, 50 paisa and 25 paisa coins in a bag. The ratio of their rupee values being a: b : c. The number of each type of coins are present in the bag:
Number of 1 rupee coin = $\mathrm{x}\times \frac{\mathrm{a}}{\mathrm{a}+2\mathrm{b}+4\mathrm{c}}$
Number of 50 paisa coin = $\mathrm{x}\times \frac{2\mathrm{b}}{\mathrm{a}+2\mathrm{b}+4\mathrm{c}}$
Number of 25 paisa coin =$\mathrm{x}\times \frac{4\mathrm{c}}{\mathrm{a}+2\mathrm{b}+4\mathrm{c}}$
Example: A box contains 315 coins in rupee, 50 paisa and 25 paisa coins. The ratio of their rupee values being 13 : 11 : 7. The number of each type of coins are present in the box:
Solution:
Number of 1 rupee coin = $315\times \frac{13}{13+2\times 11+4\times 7}=65$
Number of 50 paisa coin = $315\times \frac{22}{13+2\times 11+4\times 7}=110$
Number of 25 paisa coin =$315\times \frac{28}{13+2\times 11+4\times 7}=140$
Rules 23: ‘x’ Rupees consists of 1 rupee, 50 paisa and 25 paisa coins in a bag. Ratio of the coin is a: b : c. The number of each type of coins present in the bag
Number of 1 rupee coin = $\mathrm{x}\times \frac{1\times 4\mathrm{a}}{4\mathrm{a}+2\mathrm{b}+\mathrm{c}}$
Number of 50 paisa coin = $\mathrm{x}\times \frac{2\times 2\mathrm{b}}{4\mathrm{a}+2\mathrm{b}+\mathrm{c}}$
Number of 25 paisa coin = $\mathrm{x}\times \frac{4\times \mathrm{c}}{4\mathrm{a}+2\mathrm{b}+\mathrm{c}}$
Example: In a bag, there are three types of coins — 1-rupee, 50 paisa and 25-paisa in the ratio of 3 : 8 : 20. Their total value is 372. The total number of coins is
Solution:
Number of 1 rupee coin = $372\times \frac{12}{12+16+20}=93$
Number of 50 paisa coin = $372\times \frac{2\times 16}{12+16+20}=248$
Number of 25 paisa coin = $372\times \frac{4\times 20}{12+16+20}=620$
The total number of coins = 93 + 248 + 620 = 961
Practice: Ratio and Proportion MCQ Question