1.
Introduction
2.
Tricks to remember
3.
Properties to Remember
4.
Chain Ratio
4.1.
Methods to Solve Chain Ratio Problems
4.1.1.
Bypass Method:
5.
Variation and its 3 types
5.1.
Direct Variation
5.2.
Inverse Variation
6.
Key Takeaways
Last Updated: Mar 27, 2024

# Ratio and Proportions

Deeksha
Roadmap to SDE career at Amazon
Speaker
Anubhav Sinha
SDE-2 @
25 Jun, 2024 @ 01:30 PM

## Introduction

An ordered pair of numbers x and y where y is not equal to infinity is known as a ratio. x/y is the ratio of x to y; y≠∞. Now let’s discuss the term proportion. When two ratios are equal, they are said to be in proportion. Let’s discuss an example for it: Suppose a/b = c/d, then a/b is in proportion with c/d and can be written as a:b:: c:d. Here ‘a’ and ‘d’ are called extremes, and ‘c’ and ‘b’ are the means. Let me share a rule of proportion here that is:

If you want a, b, c and d in proportion then, Product of the extremes = Product of means, i.e., ad = bc.

The above point is important. Let’s discuss an example on it:

Example 1:Let us say 4:2::a:22. What is the value of a?

Solution: The  Product of the extremes = Product of means

4*22 = 2*a

a = 44.

So till now, we understood the terms ratio and proportions. Let’s move to discuss some cool tricks that will help you solve questions in just a few seconds.

## Tricks to remember

Trick #1

Suppose you are given a question where you have to divide a sum of money S into the ratio a: b: c then the three parts will be:

• (a*S/(a+b+c))
• (b*S/(a+b+c))
• (c*S/(a+b+c))

Nice shortcut, isn’t it?

Trick#2

Suppose you are given a question that goes like this: If X and Y are two persons investing in the ratio of m: n for the same period of

time, then the ratio of profits will be?

Now one method is that you solve it by doing all the long calculations but who will give you this much time in your aptitude paper. So here is the trick:

The ratio of their profits will be m:n. Yes! That’s simple.

Trick#3

Suppose you come across a question like this in your apti test that goes like If m kg of one kind costing ‘a’ rupees/kg is mixed with ‘n’ kg of another kind

costing Rs.b/kg, then the price of the mixture is:

As usual, the candidate who has missed this super cool blog of coding ninjas will invest his time in solving this from scratch and might miss the remaining coding questions of his paper. But you are smart you gonna solve it in just 2 seconds:

The trick is that the price of the mixture will be (ma+nb)/(m+n).

Okay so till now we read some shortcuts now let’s see some properties of ratio and proportions.

Get the tech career you deserve, faster!
Connect with our expert counsellors to understand how to hack your way to success
User rating 4.7/5
1:1 doubt support
95% placement record
Akash Pal
Senior Software Engineer
326% Hike After Job Bootcamp
Himanshu Gusain
Programmer Analyst
32 LPA After Job Bootcamp
After Job
Bootcamp

## Properties to Remember

Property#1:

If we multiply the numerator and the denominator of the ratio by the same number, the ratio does not change.
Example: We have a ratio of 3: 4. Now on multiplying denominator and numerator by 2 we will get 6:8. Now you can see that the lowest basic form of 6:8 is again 3:4.

Property#2:

If we divide the numerator and the denominator of a ratio by the same number, then the ratio does not change i.e Dividing ‘d’, by both numerator and denominator or ratio a/b gives, a/b = (𝑎 ÷ 𝑑)/𝑏 ÷ 𝑑)

Example: We have a ratio of 6: 4. Now on dividing denominator and numerator by 2 we will get 3:2. Now you can see that the lowest basic form of 6:4 is again 3:2.

Property#3:

If some ratio is in fractional form, then to convert it into an integral ratio, multiply all fractions by LCM of their denominators

Example: Suppose we are given a ratio 1/2: 3/5: 7/6

Now, to convert this ratio into an integral ratio, multiply all the fractions by LCM of their denominators (2,5&6). LCM(2,5,6) = 30. i.e 30/2 : (3×30)/5 : (7×30)/6 = 15:18:35.

Property#4:

If a/b = c/d = e/f = k then;

(a+c+e)/(b+d+f) = k. Let’s see an example on it:

For example : 2/3 = 4/6 = 10/15 = 200/300 = k then,

(2+4+10+200) / (3+6+15+300) = 216/324 = 2/3.

Property#5

Suppose you added a number to the numerator and denominator to maintain the equality, then the numbers should have the same ratio as that of the original ratio.

Property#6:

This property says that if you add two numbers such that their ratio is larger than the original ratio, then the final ratio becomes larger.

Let’s understand this property with an example:

Let's say a ratio = 400/800. (400+5)/(800+7).Here, ratio 5/7 is larger than the original ratio(400/800 =1/2). i.e c/d > a/b then (a + c)/(b + d) > a/b i.e. (400+5)/(800+7) > 400/800

Now let’s see a new concept in ratio and proportions:

## Chain Ratio

A ratio in which one to next, next to the next, and next to next ratios are given is known as the Chain ratio. Let’s understand it with the help of an example:

Suppose you are given a question like this:A: B = 3:5, B: C = 7:8 then, convert chain ratios into a single ratio A:B: C

Then in such questions, we have to apply the concept of chain ratio. Here B is a common element in both the ratios. To equate 5 & 7, take LCM of 5 & 7. LCM(5,7) = 35. To make common element 35. Multiply the ratios A: B and B: C by 7 and 5 respectively. Thus, A: B will become 21:35, and B: C will become 35:40. B is the same in both cases. Hence A: B: C is 21:35:40. Isn’t this very simple?

### Methods to Solve Chain Ratio Problems

Taking LCM is a tedious task so here I’ll be sharing a bypass of taking LCM.

#### Bypass Method:

Suppose you encounter a question that goes like this: Let us say A: B is N1:D1, B: C is N2:D2, C:D is N3:D3, and D: E is N4:D4. Find A:B: C:D: E

• The value of A would correspond to the multiplication of all numerators. So, A would be N1N2N3N4.
• The value of B would be D1N2N3N4.
• The value of C would be D1D2N3N4.
• The value of D would be D1D2D3N4.
• And the value of E would be D1D2D3D4.

A                         B                 C                    D                   E

N1N2N3N4 : D1N2N3N4 : D1D2N3N4 : D1D2D3N4 : D1D2D3ND

Now there are few terms in ratio and proportions that you should remember for solving problems:

• Invertendo: If x/y = p/q then y/x = q/p
• Alternando: If x/y = p/q, then x/p = y/q
• Componendo: If a/b = c/d, then (a+b)/b = (c+d)/d.
• Dividendo: If a/b = c/d, then (a-b)/b = (c-d)/d.
• Componendo and Dividendo: If x/y = p/q, then (x + y)/(x – y) = (p + q)/(p – q)

Now let’s see some problems based on ratio and proportions.

Ques 1. A bag contains 50 P, 25 P, and 10 P coins in the ratio 5: 9: 4, amounting to Rs. 206. Find the number of coins of each type respectively.

Ans 1. Let the ratio be x.

Hence no. of coins be 5x ,9x , 4x respectively Now given total amount = Rs.206

=> (.50)(5x) + (.25)(9x) + (.10)(4x) = 206

we get x = 40

=> No. of 50p coins = 200

=> No. of 25p coins = 360

=> No. of 10p coins = 160

Ques 2. Divide rupees 252 amongst A, B, and C such that 1/3rd of what A gets is equal to 1/5th of what B gets is equal to 1/4th of what C gets. How much A, B, and C will get individually?

Ans2. Here 2 equations are formed; A + B + C = 252 ………(1)

A/3 = B/5 = C/4 ……….(2)

Let’s say, eq (2) is equal to k. A/3 = B/5 = C/4 = k , A = 3k , B = 5k , C = 4k

Put value of A,B and C in eq(1) we get;

3k+5k+4k = 252

12k = 252 and k = 252/12 = 21.

Hence the numbers are; A = 3 × 21= 63, B = 5 × 21= 110 and C =4 × 21= 84

Ques 3. If 2A = 3B = 4C, find A : B : C

Ans 3. Let 2A = 3B = 4C = x

So, A = x/2 B = x/3 C = x/4

The L.C.M of 2, 3 and 4 is 12

Therefore, A : B : C = x/2 × 12 : x/3 × 12 : x/4 = 12

= 6x : 4x : 3x

= 6 : 4 : 3

Therefore, A : B : C = 6 : 4 : 3

Ques 4. The income of P & Q is in ratio 1:2 and expenditure of P & Q is in ratio 1:3. If each saves 500 of their income. Find the P’s income.

Ans 4. Let,

P’s income = x and Q’s income = 2x.

P’s expenditure = y and Q’s expenditure = 3y.

And we know:

Saving = Income - Expenditure

Saving for P; x - y = 500 …………(1)

Saving for Q; 2x - 3y = 500 ………..(2)

Solving eq(1) and (2) we get; x = 1000 and y = 500.

Hence P’s income = 1000.

Now let’s discuss one more concept

## Variation and its 3 types

Variation is quite an important concept while solving problems related to ratio and proportions. Let’s understand three types of variation

### Direct Variation

Direct variation implies that x varies directly as y or x is directly proportional to y, or mathematically, we can say xαy

Now you must be thinking what's the use of this in questions? So now see its logical meaning:

Direct variation is saying that when x increases, y increases, and if x decreasing y also decreases. In the form of a ratio, you can say that If x increases by 1/10, then y will also increase by 1/10.

### Inverse Variation

Inverse variation implies that x varies inversely as y or product of x and y is constant, or mathematically, we can say xα1/y.Logically, you can say that suppose we have two values, a and b, which are inversely proportional to each other than if x increases by 75%, then you can say that y will decrease by 75%.

Now is the time to solve some questions on these concepts. So be ready with your pen and notebook:

Ques 5. Given that, x directly varies with y and x is 18 when y is 7. Find x when y is 21?

Ans 5. It is given that, x directly varies with y i.e xαy or x = ky ……….(1) Now Replace x and y with their respective values.

So; eq (1) becomes 18 = k×7 ⇒k = 18/7.

When y =21 the value of x is;

from(1); x = 18/7×21 = 54.

Ques 6. 3 monkeys have bananas in a ratio of 1/3:1/5:1/7. The total number of bananas with 3 monkeys is 284. How many bananas does each monkey have?

Ans 6. Now this question is quite easy. Just apply trick 1 discussed in this blog. Now check your answer with my solution:

Banana with first monkey will be: ((1/3) / (1/3+1/5+1/7+))*284) = 140

Banana with second monkey one will be: ((1/5) / (1/3+1/5+1/7+))*284) = 84

Banana with third monkey one will be: ((1/7) / (1/3+1/5+1/7+))*284) = 60

## Key Takeaways

In this blog, we covered all the tricks and concepts for the problems on ratio and proportions. Now you are ready to rock in your online assessments. Remember, the more you practice, the better your speed will be. If you want to be quick at solving aptitude problems, just click here and practice as much as possible.

If you feel that this blog has added some value to your knowledge, then do share it with your friends.

Live masterclass