RATIO, PROPORTION AND VARIATION

Ratio - Anupat:

Ratio is the relation which one quantity bears to another of the same kind, the comparison
being made by considering what multiple, part or parts, one quantity is of the other. The ratio of
two quantities "a" and "b" is represented as a : b and read as "a is to b". "a" is called antecedent,
"b" is the consequent. Since the ratio expresses the number of times one quantity contains the
other, it's an abstract quantity.

A ratio a : b can also be expressed as a/b. So if two items are in the ratio 2 : 3, we can say
that their ratio is 2/3. If two terms are in the ratio 2, it means that they are in the ratio of 2/1, i.e., 2 : 1 .

"A ratio is said to be a ratio of greater inequality or less inequality or of equality according
as antecedent is greater than, less than or equal to consequent". From this we find that a ratio of
greater inequality is diminished and a ratio of less inequality is increased by adding same quantity to
both terms, i.e., in a : b

if a < b then (a + x) : (b + x) > a:b and if a > b then (a + x) : {b + x) < a:b

𝑎 𝑐 𝑒 𝑎+𝑐+𝑒....
If = = . . . . . . . . ..then each of these ratios is equal to
𝑏 𝑑 𝑓 𝑏+𝑑+𝑓.....

Proportion:
When two ratios are equal the four quantities involved in the two rations are said to be
proportional i.e., if a/b = c/d, then a, b, c and d are proportional. This is represented as a: b :: c : d
and is read as "a is to b (is) as c is to d".
When a, b, c and d are in proportion, then the items a and d are called the EXTREMES and
the items b and c are called the MEANS. We also have the relationship;
Product of the MEANS = Product of the EXTREMES i.e., bc = ad
If three quantities a, b and c are such that

a: b :: b : c, then we say that they are in CONTINUED PROPORTION.

We also get b2 = ac
If a:b is a ratio, then:

• 𝑎2 : 𝑏 2 is a duplicate ratio;

• √𝑎: √𝑏 is the sub-duplicate ratio;

• 𝑎3 : 𝑏 3 is a triplicate ratio;

If a : b = c : d, then:

• d is called the fourth proportional to a, b, c.

• c is called the third proportion to a and b.

• Mean proportional between a and b is √𝑎𝑏.

S.No Ratio Proportion

1 The ratio is used to compare the size of two The proportion is used to express the relation
things with the same unit of two ratios
2 It is expressed using a colon (:), slash (/) It is expressed using the double colon (::) or
equal to the symbol (=)
3 It is an expression It is an equation
4 Keyword to identify ratio in a problem is “to Keyword to identify proportion in a problem is
every” “out of”

Variation
Two quantities A and B may be such that as one quantity changes in value, the other quantity
also changes in value depending on the change in the value of the first quantity.

Direct variation:
One quantity A is said to vary directly as another quantity B if the two quantities depend upon each
other in such a manner that if B is increased in a certain ratio, A is increased in the same ratio
and if B is decreased, A is decreased in the same ratio. This is denoted as A α B (A varies
directly as B).
If A α B then A = kB, where k is a constant. It is called a constant of proportionality.
Inverse variation:
A quantity A is said to vary inversely as another quantity B if the two quantities depend upon
each ether in such a manner that if B is increased in a certain ratio, A is decreased in the same
ration and if B is decreased then A is increased in the same ratio. It is the same as saying that A
varies directly with 1/B.
It is denoted as If A α 1/B i.e., A = k/B where is k is a constant of proportionality.

Joint variation:
If there are three quantities A, B and C such that A varies with B when C is constant and
varies with C when B is constant, then A is said to vary jointly with B and C when both B and C are
varying. Then A α BC or A = kBC where k is the constant of proportionality.

Worked out examples

Ex: If (x/y) = (3/4), find (5x - 3y)/(7x + 2y)

Sol.: (x/y) = (3/4) => x = (3y/4).
Substituting in (5x - 3y)/(7x + 2y)
(5 × 3𝑦)/4 − (3𝑦) (15𝑦 − 12𝑦)/4 3
( )⇒ = .
(7 × 3𝑦)/4 − (2𝑦) (21𝑦 + 8𝑦)/4 29
Alternatively,
Since x/y = 3/4; we can solve the question by assuming x value as 3 and y value as 4 and substituting in
the equation.

Ex: Two numbers are in the ratio 5: 8. If 9 be added to each, they are in the ratio 8:11. Find the
numbers.
Sol.: Let the numbers be x and y;
𝑥 5 5𝑦
= ; 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒; 𝑥 = and
𝑦 8 8
5𝑦
(𝑥 + 9) 8 [ + 9] 8 5𝑦 + 72 8
= ⇒ 8 = ⇒ =
(𝑦 + 9) 11 𝑦+9 11 8(𝑦 + 9) 11
Cross multiplying,
55y + 72*11 = 64y + 72*8
Thus, 9y = 72*3 = 216
y = 24; Thus, x = 15

Ex. 3x - 1: 6x- 7 = 7x- 10: 9x + 10

Sol.: (3x - 1)/(6x - 7) - (7x - 10)/(9x+10)
=> (9x + 10) (3x - 1) = (7x - 10) (6x - 7) =27x z -9x+30x-10
= 42x2 - 49x - 60x + 70
 => 15x 2 -130x + 80 = 0
=> (5x - 40) (3x - 2) = 0 => x = 8 or 2/3

Ex: If x varies inversely as y 2 - 1 and is equal to 24 when y is 10, find x when y=5
Sol.: x α(1/y 2 -1)
x = m/(y 2 -1); where m is the constant
Given, x = 24 when y = 10
𝑚
24 = => m = 24 X 99
𝑦 2 −1
𝑚
Now, x = ; When y = 5,
𝑦 2 −1
x = (24 X 99)/(25 - 1)
x = 99

Ex: The volume of a pyramid varies jointly as its height and the area of its base; when the area
of the base is 60 square feet and the height 14 feet the volume is 280cu.ft. What is the area
of the base of a pyramid whose volume is 390cu.ft and whose height is 26ft.?
Sol.: Let v be volume; a= area of base; h the height;
Therefore, 𝑣 ∝ 𝑎 × ℎ; 𝑣 = 𝑚 × 𝑎 × ℎ, where m is a constant.
Therefore, 280 = m x 60 x 14 => m = 1/3;
v = 1/3ah.
When v = 390, h = 26.
Therefore,
390 = (1/3)a x 26; a = 45.
Hence, the area of the base is 45 Sq.ft.
 CONCEPT REVISED PROBLEMS

1. Rs.432 is divided amongst three workers A, B and C such that 8 times A’s share is equal to 12
times B’s share which is equal to 6 times C’s share. How much did A get?

Solved in the class

2. The monthly incomes of A and B are in the ratio 4 : 5, their expenses are in the ratio 5 : 6. If
'A' saves Rs.25 per month and 'B' saves Rs.50 per month, what are their respective incomes?

Solved in the class

3. The proportion of milk and water in 3 samples is 2:1, 3:2 and 5:3. A mixture comprising of
equal quantities of all 3 samples is made. The proportion of milk and water in the mixture is?

Solved in the class

4. A, B and C play cricket. A's runs are to B's runs and B's runs are to C's as 3:2. They get
altogether 342 runs. How many runs did A make?
Solution: Ans 162

Given, a: b = 3: 2 and b : c = 3: 2
Thus, We shall first find, a: b: c
a: b: c = 9: 6: 4
Given, 9x + 6x + 4x = 19x = 342
x = 342/19 = 18
Thus, A made 9x = 9 * 18 = 162
 5. The monthly salaries of two persons are in the ratio of 4:7. If each receives an increase of
Rs.25 in the salary, the ratio is altered to 3: 5. Find their respective salaries.

Solution: Ans (200 and 350)

Let monthly salaries be 4x and 7x
Given,
4𝑥 + 25 3
=
7𝑥 + 25 5

5(4x + 25) = 3(7x + 25)

20x + 125 = 21x + 75
x = 50
Therefore, salaries are 200 and 350
 6. The present ages of A and B are as 6 : 4. Five years ago their ages were in the ratio 5 : 3. Find
their present ages.

Solution: Ans (30 and 20)

Let ages be 6x and 4x
Five years ago the ages will be 6x – 5 and 4x – 5;
Given,
6𝑥 − 5 5
=
4𝑥 − 5 3

18x – 15 = 20x – 25

2x = 10; x = 5

Thus, ages are 30 and 20

 7. The marks scored by a student in three subjects are in the ratio of 4 : 5 : 6. If the candidate
scored an overall aggregate of 60% of the sum of the maximum marks and the maximum
marks in all three subjects is the same, in how many subjects did he score more than 60%?

Solution:

Ans: 1 subject

Let the marks scored by the student be 4x, 5x and 6x,

Given, maximum marks in all subjects is the same, so let us assume maximum marks of each subject
is 100.

Thus, total maximum marks of 3 subjects will be 300.

Given, the student scores 60% of the total maximum marks. Thus, the student scored

60% of 300 = 180 marks

Thus, 4x + 5x + 6x = 180

15x = 180; x = 12

Therefore, the marks scored by the student in each subject is 48, 60 and 72

Thus, only in 1 subject the student has scored more than 60%.
 8. If a: b = 3: 4, then find ( 1/a + 1/b) / (1/a – 1/b).

Solution: Ans 7

The easiest way of solving these questions is by considering a and b value as 3 and 4 and
substituting in the equation

1 1 4+3
(3 + 4) ( 12 ) 7
= =
1 1 4−3
(3 − 4) ( 12 ) 1
 9. I have one-rupee coins, fifty paise coins and twenty five paise coins. The number of coins is
in the ratio of 2.5: 3: 4. If the total amount with me is Rs. 210, find the number of one-rupee
coins.
Solution: Ans 105

Let the number of 1 Re, 50 p and 25p coins be 2.5x, 3x and 4x

Then the total amount will be
(2.5x) * 1 + (3x)*(0.5) + (4x) * (0.25) = 210
(NOTE: You have to multiple the number of coins with the value of each coin)
2.5x + 1.5x + 1x = 210
5x = 210
x = 42
Thus, the number of one rupee coins = 2.5*42 = 105
 3𝑥 2 −4𝑦 2 8 5𝑥+𝑦
10. If = , 𝑓𝑖𝑛𝑑 x and y are positive.
2𝑥 2 −𝑦 2 7 3𝑥−𝑦

Solution: Ans 11/5

3𝑥 2 − 4𝑦 2 8
=
2𝑥 2 − 𝑦 2 7
There are multiple ways of solving this problem, but one of the easiest way is considering equations
Thus, Take
3𝑥 2 − 4𝑦 2 = 8
2𝑥 2 − 𝑦 2 = 7
as two equations. Multiplying the second equation by 4, we get
3𝑥 2 − 4𝑦 2 = 8
8𝑥 2 − 4𝑦 2 = 28
Subtracting the two, we get
5𝑥 2 = 20; 𝑥 2 = 4; 𝑥 = 2
Substituting
2(2)2 − 𝑦 2 = 7; 8 − 𝑦 2 = 7; 𝑦 2 = 1; 𝑦 = 1
Thus,
5𝑥 + 𝑦 5(2) + 1 11
= =
3𝑥 − 𝑦 3(2) − 1 5
 𝑎 2 𝑥 1 3𝑎𝑥+𝑏𝑦
11. If = 𝑎𝑛𝑑 = 2 ; 𝑓𝑖𝑛𝑑
𝑏 3 𝑦 2 𝑎𝑥−𝑏𝑦

Solution: Ans 9

𝑎 2 𝑥 5
Given, = 𝑎𝑛𝑑 =
𝑏 3 𝑦 2

Consider, a = 2, x = 5, b = 3 and y = 2 and substitute

3 × 2 × 5 + 3 × 2 36
= =9
2×5−3×2 4
 3𝑥+2𝑦 8 6𝑥−𝑦
12. If = ; 𝑓𝑖𝑛𝑑
2𝑥−𝑦 3 𝑥+2𝑦

Solution: Ans 11/4

3x + 2y =8
2x – y = 3
Solving the two equation, x =2 and y = 1
Substituting in the given equation
We get, 11/4
13. The ratio of marks obtained by X and Y is 3: 2. If X gets 30 marks more than Y, find their
individual marks.

Solution: Ans 90 & 60

Let marks be 3a and 2a,

Thus, X got 3a – 2a = a marks more than Y
Given, a = 30
Thus, individual marks is 90 and 60
14. The area of a circle varies as the square of its radius. Given that the area of circle is 180
sq.ft when radius is 6 ft, find the radius of circle with area of 245 sq.ft.

Solved in the class

15. The weight of a cylinder varies directly with the square of the radius when the height is
constant and with the height when the radius is constant. What is the ratio of the radii of
two cylinders of the same weights whose heights are in the ratio 9: 25?

Solution: Ans 5/3

Given, 𝑤 ∝ 𝑟 2 × ℎ
𝑤 = 𝑘 × 𝑟 2 × ℎ; 𝑤ℎ𝑒𝑟𝑒 𝑘 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑜𝑓 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙𝑖𝑡𝑦
Assume the heights of the cylinders is 9 and 25.
Given, the weights are same
Let the weights of the two cylinders be w, and radii of the two cylinders be r1 and r2.
Therefore,

𝑤 = 𝑘 × 𝑟1 2 × 9 𝑎𝑛𝑑 𝑎𝑙𝑠𝑜
𝑤 = 𝑘 × 𝑟2 2 × 25
Equating the two; 𝑘 × 𝑟1 2 × 9 = 𝑘 × 𝑟2 2 × 25
𝑟1 2 × 9 = 𝑟2 2 × 25

𝑟1 2 25 𝑟1 5
= ; =
𝑟2 2 9 𝑟2 3
 16. Twice the first number added to the second number gives 20 whereas doubles the first
number exceeds the second number by 12. Find the two numbers.

Solution: Ans (8 and 4)

Let the numbers be x and y

Given, 2x + y = 20 and 2x – y = 12

Solving the two, we get x = 8 and y = 4

 17. The ratio of ages of a mother and her daughter is 2: 1. Mother was 24 years old when she
gave birth to her only son who is 6 years younger to her only daughter. Find the present ages
of daughter and mother (in years).

Solution: Ans (36 and 18)

Given, Son is 6 years younger to daughter and when son was born, mother’s age was 24. Thus, when
the son was born, the daughter should have been 6 years old. Which means, when daughter was born,
mother’s age was 18 years old.

Thus, If daughters age is x years as of today, then mothers age will be 18 + x

Given,

(18 + x)/x = 2/1; therefore, x = 18

So daughter is 18 years old and mother is 18+18 = 36 years old

18. If a: b is 2: 5 and c: b is 3: 5, find a: b: c.

Solution:
Rewriting the ratio as
a: b and b: c
we get, 2: 5 and 5: 3
Thus,
a: b: c = 2: 5: 3
19. X varies directly as the square of Y (when Z is constant) and inversely with Z (when Y is
constant). Given X is 12 when Y is 2 and Z is 3, find the value of X given Y is 4 and Z is 6.

Solved in the class

 1
20. A father divides his property among his three sons in the ratio 1 : 2/3: 3/4. Find the share of each
2
son if the total property of the father is Rs.70, 000.

Solution: Ans 36000, 16000, and 18000

Given the ratio is

1 2 3
1 : :
2 3 4

3 2 3
: :
2 3 4
Converting the fractional ratios into integers. We multiple each ratio by LCM of denominators, i.e.
12
Thus, the ratio will become 18: 8: 9
Total of this is 18 + 8 + 9 = 35
Given, the total is 70000 which is 2000 times of 35. Thus every value will increase by 2000 times,
So the shares that each son will get is
36000, 16000 and 18000