RATIO AND PROPORTTON

Ratio and Proportion is the portion from where a wide variety of questions are asked in various competitive exams. You must workout all the types in detail because in all exams, questions of almost all types are asked. 

RATIO

Ratio is a comparison between two quantities by division. The number of times one quantity contains another quantity of the same kind is called the ratio of the two quantities.The numbers forming the ratio are called "Terms" of the ratio. The ratio between
a and b is denoted as a:b or a/b Here,and b are the terms of the ratio. The first term "a" is called the "Antecedent" and the second tem "b" is called the "Consequent"

TYPE - I

1.

If P:Q =2:3 and
Q:R = 4:5 what is P:R 
P : Q = 2 : 3 __ (i)
Q : R =4:5___(ii)
(i) x4 :=>
P : Q = 8 : 12 
(ii) x3 =>
Q: R= 12:15
ie, P:Q:R= 8:12:15 
.'. P:R=8:15
2.If P:Q:R=2:3:4,then P/Q:Q/R:R/P is equal to
Let P = 2X, Q = 3x and R = 4x. 
Then,P/Q=2x/3x, Q/R=3x/4x and
R/P=4x/2x=2/1
P/Q:Q/R:R/P=2/3:3/4:2/1
=8:9:24

3.

If 40% of A = 88% of B, then A:B = ? 
40/100A=88/100b
=>A/B=88/40
=11/5
.'.A : B = 11.5 

4.

If 6a = 5b and 3b=2c find a:b:c
6a = 5b =>a/b=5/6;a:b=5:6__(i)
3b=2c=>b/c=2/3;b:c=2:3___(ii)
(ii)*3=>b:c=6:9
So a:b:c=5:6:9

5.

If (2x  3y) : (2x-3y) = 4:1, find x:y
2x3y/2x-3y=4/1
2x  3y = 8x - 12y 
6x =15y
x/y =15/6=5/2
.'.x : y = 5 : 2

TYPE - II

1.

The sum of two numbers is 30 and difference is 2. What is the ratio of these numbers.
Let one number be x and the other number be y.
Given,
xy=30
x-y=2/2x=32
x=16
.'.The numbers are 16 and 14 
.'.Ratio= 16/14 = 7 =8:7

2.

Two numbers are in the ratio 9:5. If 5 is subtracted from each, the ratio becomes 2:1. Find the numbers
Let the numbers be 9x and 5x.
If 5 is substracted =>
9x-5/5x-5=2/1
9x-5=10x-10
x=5
.'.Required number are 42 and 25

3.

The ratio of the number of gents and ladies attended in a party is 2:5. If a total of 350 people came to the party, find the number of ladies in the party?
Ratio = 2:5 
2x  5x = 350 
7x = 350 
x = 50
.'.No. of ladies = 5 x 50 = 250

4.

If the ratio of the areas of two squares is 1:9, what is the ratio of their perimeters
A1^2/A2^2=1/9
.'.A1/A2=1/3
Perimeter of square =4A
ie,4A1/4A2=A1/A2=1/3
.'.Ratio of perimeters=1.3

5.

Two numbers are in the ratio 3:4.If the sum of these two numbers is 112, find the smaller number.
Let the number be 3x and 4x 
Then, 3x  4x= 112 
7x = 112
x = 16
.'.Smaller number = 16 x 3 = 48

POINTS TO REMEMBER

*Duplicate ratio of a:b is a2:b2 
*Triplicate ratio of a:b is a3:b3
*Sub-duplicate ratio of a:b is /a:/b
*Sub-triplicate ratio of a:b is 1/a3:1/b3
*If a:b and c:d are two ratios, then ac;bd is called the compounded ratio of the given ratios.

TYPE-III

1.

Athul added 4 litres of water to 10 litres of milk and another 6 litres of water to 8 litres of milk. What is the ratio of the strengths of milk in the two mixtures.
Strength of milk in the 1st mixture
=10/10  4= 10/14=5/7
Strength of milk in the 2nd mixture
=8/86=8/14=4/7
.'.Ratio of their strengths =5/7/4/7= 5:4

2.

The ratio of water and spirit in the mixture is 3:5. If the volume of the solution is increased by 50% by adding spirit only, what is the resultant ratio of water and spirit.
Let the spirit to be added = k
Then, 3x (5x  k) =150/100*8x
8x  k =12x
k =4x
Required ratio = 3x : (5x k)
= 3x : 9x = 1:3

3.

36 litre of a mixture contains milk and
water in the ratio 5:4. If 4 litres of this mixture be replaced by 4 litres of milk, the ratio of milk to water in the new mixture would be_____
Quantity of milk in 36 litres of mix
=36x5/9= 20 litres
Quantity of milk in 40 litres of new mix
=(20  4) = 24 litres
Quantity of water in it = (40 - 24) = 16 litres
.'.Ratio of milk and water in the new mix = 24 : 16 = 3 : 2

TYPE - IV

1.

The salaries P,Q,R are in the ratio 2:3:5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries,then what will be the new ratio of their salaries?
Let P = 2x, Q = 3x and R = 5x 
Then,
P's new salary =115/100of 2x
=115/100*2x=23x/10
Q's new salary =110/100of 3x
110/100 = 3x=33x/10
R's new salary = 120/100 of 5x
=120/100*5x=6x
.'.New ratio =23x/10: 33x/ 10:6x 
= 23 :33 : 60

2.

The ratio of male and female workers in a company is 2:3. The daily wages of a male and female worker are Rs. 100 and Rs. 180 respectively. If the daily wages of all workers be Rs. 1480, find the number of both male and female workers.
Let the no. of male workers be 2x and that of female workers be 3x.
(2x x 100) (3x x 180) = 1480
200x 540x = 1480
x=1480/740
x=2
.'.No. of male workers =2x2 = 4 
No. of female workers = 2x3 = 6

3.

A company employs clerks, sweepers and part-time labourers in the ratio 8:5:1 and
their wages are in the ratio 5:2:3. When 20 sweepers are employed, the total daily wages of all amount to Rs. 318. Find the daily wages paid to each category of employees.
Clerks : Sweepers : Part-time labourers =8:5:1
Ratio of the respective wages = 5:2:3 
Amount must be paid in the ratio 8*5:5*2:3*l = 40 : 10 : 3
Sum of the ratios = 40  10  3 = 53
If the total amount is Rs. 53, derks get Rs.40
If the total amount is Rs. 318, clerks get 
Rs. 40/53x318 = Rs.240
Sweepers get Rs. 10/53x318 = Rs. 60
Part time labourers get Rs. 3/53*318 = Rs. 18

PROPORTION

The equality of two ratios is called Proportion.
If a : b = c : d, then a,b,c and d are said to be in proportion. It is denoted as:
a : b : : c : d
Here, a and d are called extremes and b and c are means.

CONTINUED PROPORTION

Three quantities of the same kind (and in same unit) are said to be in continued Proportion if the ratios of first to second and second to third are the same.
Eg:- If a, b and c are in continued proportion, then a : b :: b : c
'b' is called the Mean proportion between a and c
(Mean)2 =Product of extremes
ie, b2 =ac
b =/ac
Here, a is the first proportion,b is the
second or mean proportion and c is the third proportion.

EXAMPLES

1.

Find the third proportional to 6 and 18.
Third Proportional means 4th term of the Proportion.
.'.6 : 18 : : 18 : x
6/18=18/x
6x=18*18
x = 18*18/6 =54
.'.Third Proportional is 54

2.

Find the fourth proportional to the numbers 3,5 and 12
3 : 5 : : 12 : x 
3/5 = 12/x 
3x = 12x5 
X = 20
.'.Fourth Proportional is = 20

3.

Find the mean proportional between 9 and 81.
b2 = ac
b = /9x81 = 27 
.'.Mean proportional = 27

POINTS TO REMEMBER

*If a:b::c:d, then
(1)a/b=c/d
(2)b/a=d/c
(3)a/c=b/d
(4)c/a=d/b (inverse ratios)
(5)ab/b=cd/d
(6)a-b/b=c-d/d
(7)ab/a-b=cd/d
(8)a/a-b=c/c-d

PERCENTAGE

'percentage' is the most familiar word that you must have heard since your school days. It may be percentage of marks. Here, we deal with percentage of marks, population, salary, etc. Two or three questions are asked from topic in all exams. Studying formulae and shortcut keys help you to save time.
The word 'percent' came from the Latin word meaning 'per hundred'. Percentage is expressed by the symbol '%'
50 percentage is written as 50% and
solved as=50/100=1/2=0.5

TYPE-I

a.To convert fraction to rate percent.
To convert a . fraction to rate percent, multiply it by 100, Thus,
a/b=a/bxl00  percent

EXAMPLES

1.Express 6/4 as rate percent=6/4*100 =150% 2.Find rate percent of 3*1/2
=3*1/2xl00
=7/2*100=350%
b.To convert a rate percent to a fraction
To convert a rate percent to a fraction, divide it by 100.

EXAMPLES

1.20%
=20/100=1/5
2.70%
=70/100=7/10
C.To find the rate percent of a number
Rate percent of a number is the product of the number and equivalent fraction of the rate percent.

EXAMPLES

1.Find 30% of 500
30% of 500 = 30/100x500 = 150
2.Find 33 1/3%of 60
1/3x60 = 20

TYPE II

If any number is decreased by x%, then
the new number = Original number x [100-x/100]
If any number is decreased by x%,then the new number=Original number x [100-x/100]

EXAMPLES

1.

If 1500 is increased by 20%, find the new number.
New number = 1500x [10020/100]
=1500* 120/100=1800

2.

If 2600 is decreased by 60%, find the new number.
New number = 2600*[100-60/100]
=2600*40/100 = 1040

3.

Subtracting 60% of a number from the number, we get the result as 80. Find the number.
x -60x/100=80
x=80*100/40=200
.'.The number is 200

TYPE-III

*If a number is first increased by x%, and then decreased by y%, then
total changes=[xyxy/100]%
*If a number is first increased by x%, and then decreased by y%, then
total changes=[x-y-xy/100]%
If a number is first decreased by x%, and then decreased by y%, then
total change=[-x-yxy/100]%

EXAMPLES

1.

The salary of an employee was first increased by 10% and thereafter decreased by 5%, what was the total effect in his salary.
=[x-y-xy/100]%
=[10-5-10*5/100]%
=[5-50/100]%
=[5-1/2]%=[9/2]%=4.5%

2.

A number was first decreased by 4% and then again decreased by 8%. What is the change.
=[-x-y xy/100]%
=[-4-84*8/100]%
=[-1232/100]%
=[-120032/100]%
=[-1168/100]%=-11.68%
.'.Net change is decrease of 11.68%

SHORTCUT KEYS

1%=1/100
2%=1/50
4%=1/25
5%=1/20
8%=2/25
10%=1/10
16%=4/25
20%=1/5
40%=2/5
60%=3/5
64%=16/25
80%=4/5
6 1/4%=1/16
12 1/2%=1/8
25%=1/4
37 1/2%=3//8
50%=1/2
87 1/2%=7/8
8 1/3 %=1/12
16 2/3%=1/6
33 1/3%=1/3
66 2/3%=2/3
83 1/3%=5/6
133 1/3%=4/3

TYPE-IV

*If the population of town is P and the annual increase is R%, then population of N years
P[100R/100]^N
The population N years ago is
P[100/100R]^N

EXAMPLES

1.

The population of a town increases 5% annually. If it is 15,435 now, what was it 2 years ago.
Let the initial population be x.
Population after 1st year
=x5x/100
=1.05x
1.1025X = 15,435 
.'.x = 14000
Population 2 years before = 14,000

2.

The population of a town is 20,000. It increased by 20% during first year. During second year it decreased 10% and increased by 20% again during the third year. What is the population after 3 years.
=[20000*120/100*90/100*120/100]
= 25920
.'.Population after 3 years = 25920

TYPE-V

*If the present value of a machine is P which depreciates at R% per annum,
then the value of the machine after N years=p[100-R/100]N
The value of the machine N years ago
p[100/100-R]N

EXAMPLES

1.

The value of a machine depreciates at the rate of 50% per annum. If its present value is Rs. 20,000. What will be its worth after 2 years.
The value of the machine after 2 years
=20000 x [1-50/100]2
= 20,000 x [1-1/2]2
=20,000 x 1/2 x 1/2 = 5000

2.

If the present value of a machine is ?60,000 and if it depreciates at 10% per annum, what will be its value after 3 years.
The value of the machine after 3 years
= 60,000 *[1-10/100]3
=60,000*[9/10]3
=60000*9/10*9/10*9/10
= 43740
.'.Value of machine after 3 years = Rs. 43740

TYPE -VI

*If a number first increases by x% and then decreases by x%, then
total decrease=x2/100%
*If two commodities are sold at the same price, one with a profit of x% and the other with a loss of x%, then
net loss =x2/100%
*If the sides of any two dimensional figure, (square, triangle, rectangle, circle, etc) are increased by x%, then its area is increased by:

EXAMPLES

1.

If the radius of a circle is increased by 3%, what will be the percentage increase in area.
% increase = 2xx2/100 
=2*33^2/100
=69/100=6.09
.'.% increase in area = 6.09%

2.

If the sides of a square are increased by 6%, what will be the net increase in the area of the square.
Net increase in area =2x66^2/100
=1236/100
=148/100=1.48
.'.% increase in area = 1.48%

TYPE -VII

When a number 'A' exceeds another number B by x%, then B is
[x/100x] *100% shorter than A, ie B is
less than A byx[x/100-x]00%, 
*When a number A is shorter than B by x
x%, then B  [x/100-x ]x 100% excess than A.

EXAMPLES

1.

If A is taller than B by 5% then B is
shorter by how much percent than A
A=[x/100x]*100%
=[5/1005]*100%
= 5/105*l00% =4.76%
.'.B is shorter than A by 4.76%

2.

If Priya is shorter than Preethi by 2%, then Preethi is taller by how much percent than Priya
=[x/100-x]*100%
[2/100-2]*100%
=2/98*100%=2.04%
.'.Preethi is taller than Priya by 2.04%