# Ratio and Proportion: Concepts and Tricks The ratio is defined as the quantitative relation between two values showing the number of times one value contains or is contained within the other.The ratio in the mathematical term used to compare two similar quantities expressed in the same units. The ratio of two numbers ‘x’ and ‘y’ is denoted as x:y Fractions and ratio are same but the only difference is that ratio is unit less quantity but fraction is not.

Question
Ratios, rates, proportions
Ratio and Proportion: Concepts and Tricks The ratio is defined as the quantitative relation between two values showing the number of times one value contains or is contained within the other.The ratio in the mathematical term used to compare two similar quantities expressed in the same units. The ratio of two numbers ‘x’ and ‘y’ is denoted as x:y Fractions and ratio are same but the only difference is that ratio is unit less quantity but fraction is not.

2021-04-16

### Basics Properties of Ratio

• A:B = mA : mB where m is constant
• a:b:c = A:B:C is equivalent to $$\frac { a }{ A } = \frac { b }{ B } = \frac { c }{ C }$$
• This property has to be used in the ratio of three things
• The inverse ratios of two equal ratios are equal. This property is called invertendo.
• If $$\frac { a }{ b } = \frac { c }{ d } then \frac { b }{ a } = \frac { d }{ c }$$
• The ratio of antecedents and consequents of two equal ratio are equal. This property is called Alternendo.
• If $$\frac { a }{ b } = \frac { c }{ d } then \frac { a }{ c } = \frac { b }{ d }$$

Componendo:

If $$\frac { a }{ b } = \frac { c }{ d } then \frac { (a+b) }{ b } = \frac { (c+d) }{ d }$$

Dividendo:

If $$\frac { a }{ b } = \frac { c }{ d } then \frac { (a-b) }{ b } = \frac { (c-d) }{ d }$$

Componendo-Dividendo:

If $$\frac { a }{ b } = \frac { c }{ d } then \frac { (a+b) }{ (a-b) } = \frac { (c+d) }{ (c-d) }$$

## RATIO TRICKS

### Trick 1: Splitting number in the given ratio

1. Divide 900 in the ratio 4:5
We cannot use x for solving

Total parts = 4 + 5 = 9
First number = $$(\frac { 4 }{ 9 })$$900 = 400

Second number = $$(\frac { 5 }{ 9 })$$900 = 500

2. Divide 1500 in the ratio 5:7:3

Total parts = 5 + 7 + 3 = 15
First number = $$(\frac { 5 }{ 15 })$$1500 = 500

Second number = $$(\frac { 7 }{ 15 })$$1500 = 700

Third number =$$(\frac { 3 }{ 15 })$$ 1500 = 300

### TRICK 2: Directly Proportional

When two parameters are directly proportional if one parameter increases the other one also increases and if one decreases another one also decreases.
If parameters A and B are directly proportional then they will satisfy the following equation

$$\frac { A1 }{ A2 } = \frac { B1 }{ B2 }$$

Example:

Price of diamond is directly proportional to its weight. If 3 grams of diamond costs Rs 45000. Then what will be the price of 8 grams of the diamond.
Solution:
As two parameters which are weight and cost are directly proportional.
Hence,$$\frac { W1 }{ W2 } = \frac { C1 }{ C2 }$$

3/8 = 45000/C2
C2 = 120000

TRICK 3: Inverse Proportional

When two parameters are inversely proportional if one parameter increases then other one decreases and if the one decreases then the other increases.
If parameter A and B are inversely proportional then they will satisfy the following equation

$$\frac { A1 }{ A2 } = \frac { B1 }{ B2 }$$

TRICK 4:

If the ratio’s A:B and B:C are given then we can find A:C without solving any equation

 A B C x y y m m n

A: B: C → X m : Y m : Y n

Examples:

If the ratio of A:B is 4:5 and the ratio of B:C is 3:4. Then find the ratio A:C

 A B C 4 5 5 3 4 4

3 : 5 3 : 5 4
A:B:C → 12:15:20
If the ratio of a salary of A and B is 2:3, B and C is 4:5, C and D is 6:7. If the salary of A is 48000 then Find the salary of D.

 A B C D 2 3 3 3 4 4 5 5 6 6 6 7

A:B:C:D à 48: 72: 90: 105
A:D = 48:105

D’s salary = 105000

### TRICK 5: USAGE OF COMMON FACTOR ‘X’

Common factor “X” should be used only when there is no direct relation with ratio either directly or inversely

Example:

Two numbers are in the ratio 3:5. Sum of their squares is 30600. Find the numbers.
Ratio ⟶ 3: 5
Assume the actual number 3x and 5x
According to the question,
(3x)^2+(5x)^2 = 30600
34x^2= 30600
x^2= 900
x = 30
Substitute the value of x in 3x and 5x to find numbers
Hence the numbers are 90 and 150

### TRICK 6: COIN PROBLEMS

Quantity Ratio ⟶ Q1 : Q2 : Q3
Value Ratio⟶ V1 : V2 : V3
(Q1 V1) + (Q2 V2) + (Q3 V3) = T
Common factor X = Total Amount/ T
Quantity of each coin = Q1X , Q2X, Q3X

Example:

1. A bag contains 5 paisa, 10 paisa and 20 paisa coins in the ratio 2:4:5. Total amount in the bag is Rs. 4.50 How many coins are there of 20 paisa?
Solutions:
T = (5 2) + (10 4) + (20 5)
T = 150 paisa = Rs. 1.50
X = 4.50/ 1.50 = 3

Hence the quantity of 20 paisa coins = 5 x3 = 15 coins

2. The bag contains 440 coins of 1 rupee, 50 paisa and 25 paisa and their value are in the ratio 6:7:9. Find the number of 50 paisa coins.
Solution:
Value Ratio ⟶ 6:7:9
Coins Ratio ⟶ 6 1: 7 2: 9 4
3:7:18
Number of 50 paisa coins = $$(\frac { 7 }{ 28 }) \times$$440 = 110

## Examples of Ratio

We have explained some of the tricks of ratio in the first part of Ratio. Now we will discuss some more examples of the previous tricks of Ratio.

### Example 1.

If A:B is 2:5 and B:C is 7:3 then find A:B:C

 A B C 2 5 5 7 3 3

A:B:C = 7 2 : 7 5 : 5 3
A:B:C = 14:21:15

### Example 2.

A wooden stick is broken into two parts one bigger and one smaller part. The ratio of the bigger part and the smaller part is proportional to the ratio of lengths of the full stick and bigger part. Find the ratio.
Solution:
Let us consider the length of bigger part be 1 m and the smaller part be ‘x’ m.
So length of full stick = (1+x) m
Acc. to the ques.
=$$\frac { 1 }{ x }=$$$$\frac { 1 + x }{ x }$$

x^2+ x = 1
x =$$\frac { (-1+\sqrt { 5 } ) }{ 2 }$$ (As the length will be positive)

### Example 3.

Three persons A,B,C whose salary together amount to Rs. 96000. But the expenses of all three person is 75% , 85% and 80% of their salaries. If their savings is in the ratio of 8: 9:20. Find the salary of B
Solution:
Let us suppose the salary of A be Rs. x, B be Rs. y and C be Rs. Z
Then the savings of them is
$$\frac { (x\times 25)}{100}$$, $$\frac { (y\times 15)}{100}$$,$$\frac { (Z\times 20)}{100}$$
$$\frac { x }{ 4 }$$, $$\frac { 3y }{ 20 }$$,$$\frac { z }{ 5 }$$ = 8:9:20
x:y:z = 32:60:100 = 8:15:25
B’s salary will be $$(\frac { 15 }{ 48 })\times$$96000 = Rs. 30000

### Example 4.

Seats in medical, commerce and art department in the college is in the ratio of 5:7:8. But due to the popularity of the college, management decided to increase the number of seats by 40%, 50% and 75% respectively. Find the new ratio of a number of seats available.
Solution:
Initial ratio of seats = 5:7:8
There is increase of 40%, 50% and 75%
New ratio of seats:
5 x140% : 7 x150% : 8 x175%
7: $$\frac { 21 }{ 2 }$$:14
2:3:4
Hence the new ratio of seats is 2:3:4

### Example 5.

The ratio of the first and second class fares between two railway stations in 4:1 and that of the number of passengers is 2:17. If on a day Rs. 2500 is collected as a fare. Find the amount collected from the second class passengers.
Solution:
Given è Ratio of fares = 4:1
Ratio of number of passengers = 2:17
Ratio of amount collection = 4 x2 : 1 x17 = 8:17
Hence amount collected from second-class passengers is
$$(\frac {17 }{ 25 } )\times$$2500 = 1700
Required amount collection is Rs. 1700

### Example 6.

Price of the diamond is directly proportional to the square of its weight. If the diamond is dropped and broken into three pieces of the ratio 1:2:3.If the price of original diamond is Rs. 1296 Find the loss after breakage of diamond
Solution:
Let us consider the total weight be ‘x’
Acc. to ques. Weight is proportional to square of weight
x^2= 1296
X = 36
After breakage ratio of weight = 1:2:3
Actual Weight = 6, 12, 18
Total value = 6^2+12^2 +18^2
= 36+144+324 = 504
Loss after breakage = 1296 — 504 = 792

### Some Examples:

• A person covers a certain distance by train, bus and car in ratio 4:3:2. The ratio of the fare is 1:2:4 per km. The total expenditure as a fare is Rs.720. Find the total expenditure as fare on car
• A sum of money is distributed between A, B, C and D in the ratio 1:3:6:7. If the share of D is Rs.3744 more than A. Find the total amount of money of B&C together?
• A bag contains 1 rupee, 50 paisa and 25 paisa coins in the ratio 2:3:5. Their value is Rs.288. Find the value of 50 paisa coins.
• The salary of A,B, C are in the ratio of 2:3:5. If the increment of 10%, 15& and 20% are allowed respectively on new year. Find the ratio of new incremented salaries of A,B,C.
• In a business, A and C invested in the ratio 2:1 whereas the ratio between amount invested by A and B was 3:2. If Rs.1,57,300 was their profit. Find the amount invested by B.

## Proportion

The numbers or quantities are said to be in proportion when the two ratio between them are equal.
For two ratio to be equal there is requirement of four variables.
If $$\frac { a }{ b } = \frac {c }{ d }$$then a, b, c, d are said to be in proportion
This is expressed as ‘a’ is to ‘b’ is to ‘c’ is to ‘d’
And written as a:b :: c:d
( product of means = product of extremes)

### Mean or Third Proportion

When there are only three variables or quantities like a, b, c. Then the middle number is to be repeated
a:b = b:c
Middle number b is called mean proportion and a and c are called extreme numbers.
B^2= Ax C

### Example 1.

Find the third proportion of 6 and 12 Given it is third proportion where a = 6 and b = 12
B^2= Ax C
12^2= 6 xC
C = 24

### Some Examples:

• The fourth proportion to 75, 192 and 200 is equal to fourth proportion to 90, 384 and Q. Find the value of Q.
• What is the value of ‘x’ if it is 45% of the fourth proportion of 5, 8 and 25
• By how much the fourth proportion to 26, 143 and 68 will be greater than third proportion to 49 and 63?
• Find the mean proportion between (3+√2) and (12-√32)
• What is the least number must be subtracted from the numbers 14, 17, 34 and 42 so that the numbers may be in proportion?

### Relevant Questions

Mixture or Alligation method of Quantitative Aptitude Alligation means linking. Today I'm going to discuss alligation method with you all. This is one of an important chapters of Quantitative exams. Basically, alligation is a rule that enables us to find the ratio of the quantities of the ingredients at given prices and are mixed to produce a mixture of given price.
The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
Suspect was Armed:
Black - 543
White - 1176
Hispanic - 378
Total - 2097
Suspect was unarmed:
Black - 60
White - 67
Hispanic - 38
Total - 165
Total:
Black - 603
White - 1243
Hispanic - 416
Total - 2262
Give your answer as a decimal to at least three decimal places.
a) What percent are Black?
b) What percent are Unarmed?
c) In order for two variables to be Independent of each other, the P $$(A and B) = P(A) \cdot P(B) P(A and B) = P(A) \cdot P(B).$$
This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
d) Now let's get the real percent that are Black and Unarmed by using the table?
If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
e) What percent are White and Unarmed?
f) What percent are Hispanic and Unarmed?
If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
Why is that?
This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
g) What percent of blacks shot and killed by police were unarmed?
h) What percent of whites shot and killed by police were unarmed?
i) What percent of Hispanics shot and killed by police were unarmed?
You can see by the answers to part g and h, that the percentage of blacks that were unarmed and killed by police is approximately twice that of whites that were unarmed and killed by police.
j) Why do you believe this is happening?
Do a search on the internet for reasons why blacks are more likely to be killed by police. Read a few articles on the topic. Write your response using the articles as references. Give the websites used in your response. Your answer should be several sentences long with at least one website listed. This part of this problem will be graded after the due date.
Tricks to solve Ratio and Proportion Problems Today I'm going to discuss a very helpful trick of ratio and proportion of Quantitative Aptitude section. I'm sure this would be very helpful for you and this trick will save your more than half time, you generally take to solve the question.
Finance bonds/dividends/loans exercises, need help or formulas
Some of the exercises, calculating the Ri is clear, but then i got stuck:
A security pays a yearly dividend of 7€ during 5 years, and on the 5th year we could sell it at a price of 75€, market rate is 19%, risk free rate 2%, beta 1,8. What would be its price today? 2.1 And if its dividend growths 1,7% each year along these 5 years-what would be its price?
A security pays a constant dividend of 0,90€ during 5 years and thereafter will be sold at 10 €, market rate 18%, risk free rate 2,5%, beta 1,55, what would be its price today?
At what price have i purchased a security if i already made a 5€ profit, and this security pays dividends as follows: first year 1,50 €, second year 2,25€, third year 3,10€ and on the 3d year i will sell it for 18€. Market rate is 8%, risk free rate 0,90%, beta=2,3.
What is the original maturity (in months) for a ZCB, face value 2500€, required rate of return 16% EAR if we paid 700€ and we bought it 6 month after the issuance, and actually we made an instant profit of 58,97€
You'll need 10 Vespas for your Parcel Delivery Business. Each Vespa has a price of 2850€ fully equipped. Your bank is going to fund this operation with a 5 year loan, 12% nominal rate at the beginning, and after increasing 1% every year. You'll have 5 years to fully amortize this loan. You want tot make monthly installments. At what price should you sell it after 3 1/2 years to lose only 10% of the remaining debt.
In there a relationship between confidence intervals and two-tailed hypothesis tests? The answer is yes. Let c be the level of confidence used to construct a confidence interval from sample data. Let * be the level of significance for a two-tailed hypothesis test. The following statement applies to hypothesis tests of the mean: For a two-tailed hypothesis test with level of significance a and null hypothesis H_0 : mu = k we reject Ho whenever k falls outside the c = 1 — alpha confidence interval for mu based on the sample data. When A falls within the c = 1 — alpha confidence interval. we do reject H_0. For a one-tailed hypothesis test with level of significance Ho : mu = k and null hypothesiswe reject Ho whenever A falls outsidethe c = 1 — 2alpha confidence interval for p based on the sample data. When A falls within thec = 1 — 2alpha confidence interval, we do not reject H_0. A corresponding relationship between confidence intervals and two-tailed hypothesis tests is also valid for other parameters, such as p,mu1 — mu_2, and p_1, - p_2. (b) Consider the hypotheses H_0 : p_1 — p_2 = O and H_1 : p_1 — p_2 != Suppose a 98% confidence interval for p_1 — p_2 contains only positive numbers. Should you reject the null hypothesis when alpha = 0.05? Why or why not?
In there a relationship between confidence intervals and two-tailed hypothesis tests? The answer is yes. Let c be the level of confidence used to construct a confidence interval from sample data. Let * be the level of significance for a two-tailed hypothesis test. The following statement applies to hypothesis tests of the mean:
For a two-tailed hypothesis test with level of significance a and null hypothesis $$H_{0} : \mu = k$$ we reject Ho whenever k falls outside the $$c = 1 — \alpha$$ confidence interval for $$\mu$$ based on the sample data. When A falls within the $$c = 1 — \alpha$$ confidence interval. we do reject $$H_{0}$$.
For a one-tailed hypothesis test with level of significance Ho : $$\mu = k$$ and null hypothesiswe reject Ho whenever A falls outsidethe $$c = 1 — 2\alpha$$ confidence interval for p based on the sample data. When A falls within the $$c = 1 — 2\alpha$$ confidence interval, we do not reject $$H_{0}$$.
A corresponding relationship between confidence intervals and two-tailed hypothesis tests is also valid for other parameters, such as p, $$\mu1 — \mu_2,\ and\ p_{1}, - p_{2}$$.
(a) Consider the hypotheses $$H_{0} : \mu_{1} — \mu_{2} = O\ and\ H_{1} : \mu_{1} — \mu_{2} \neq$$ Suppose a 95% confidence interval for $$\mu_{1} — \mu_{2}$$ contains only positive numbers. Should you reject the null hypothesis when $$\alpha = 0.05$$? Why or why not?