In this section, we will study the Ratio and Proportion topic

[RATIO]

• Ratio is used to show comparison between two similar quantities. For example, Suppose Aman bought 4 apples and 5 mangos then the ratio of Apple: Mango= 4:5 (read as 4 is to 5 or 4 ratio 5). Here we have compared number of items (similar quantity). But if we want to compare number of item and price also then it’s not possible to do so in a ratio.

Let’s solve some questions on ratio  

E.g. 1: In a class, number of students who like geography is 42 and those who like mathematics is 36. Find the ratio of students who like geography to those who like mathematics.

 Required ratio = No. of students who like geography: No. of students who like mathematics

                           = 42:36 = 7:6 (Ans).

(This means for every 7 students who like geography we have 6 students who like mathematics) 

E.g. 2: A’s height is 5/8th of B’s height. What is the ratio of B’s height to A’s height?

A/Q,

A’s height = 5/8 of B’s Height

     ⇒Ah = 5/8 of Bh

     ⇒Ah/Bh = 5/8 or Ah:Bh = 5:8

∴Bh:Ah = 8:5 (Ans).

If we have mA = nB = pC, then A:B:C = 1/m:1/n:1/p

Let’s take an example to have more clarity on this formula. 

E.g : If 3A = 2B = C, then A:B:C=?

Let’s divide the equality in two parts

3A= 2B                                                                     b. 2B=C

           ⇒ A/B = 2/3                                                             ⇒ B/C= ½

In both the equations we have different values of B so we cannot compare A and C for different values of B. We will convert both the values of B to a single value by multiplying in numerator and denominator.

In a) value of B = 3 and in b) value of B = 1, so we can get single value of B by multiplying by 3 in numerator and denominator of b)

A/B = 2/3                                                         b) B/C = 1 × 3/2 × 3 = 3/6

Therefore, we have A:B:C = 2:3:6.

By formula:

A:B:C = 1/m:1/n:1/p = 1/3:1/2:1/1 or 2:3:6 (Multiplying by the LCM of denominators 2 and 3 i.e. 6) (Ans). 

If the X is divided in a:b:c, then 1st part = {a/(a + b + c)} × X or aX/(a + b + c)

                                                           2nd part = {b/(a + b + c)} × X or bX/(a + b + c)

                                                            3rd part = {c/(a + b + c)} × X or cX/(a + b + c)

Questions of this type are most common questions from ratio topic in Exams. Let’s solve some questions to understand it in a better way. 

E.g. 1: The ratio of two numbers is 3:1 and their sum is 72. Find the difference between the numbers.

Ratio of numbers= 3:1

Total sum, X= 72

By division,

Step 1: Take the sum of ratio

Sum of ratio = 3 + 1 = 4

Step 2: Divide total sum by sum of ratio to get value of one part of ratio

Value of one part = Total sum/Sum of ratio

                            = 72/4 = 18

Step 3: Multiply in value of one part by share of each part.

First number= Share of part 1 × Value of one part

                   = 3 × 18 = 54

Second part = 1 × 18 = 18.

∴Required difference = 54 – 18 = 36 (Ans).

By formula,

1st number = {a/(a + b)} × X

                     = {3/(3 + 1)} × 72

                     = ¾ × 72 = 3 × 18 = 54

2nd Number = {b/(a + b)} × X

                       = {1/(3 + 1)} × 72

                       = ¼ × 72 = 1 × 18 = 18.

∴ Difference between the numbers = 54 – 18 = 36 (Ans).

Short trick

Here, Ratio of two numbers = 3:1

Therefore, we can clearly say that first number is two parts more than the second number.

So required difference = (Difference in ratios/ Total sum of parts)×Total sum

                                        =(2/4) × 72

                                        = 72/2 = 36 (Ans).

E.g. 2: Divide Rs. 162 in the ratio 2:3:6:7. The rupees in the respective ratios are given by:

By formula,

First part = {a/(a + b + c + d)} × X

                 = {2/(2 + 3 + 6 + 7)} × 162

                 = 2/18 × 162

                 = 2 × 9 = Rs. 18

Similarly,

Second part = (3/18) × 162 = 3 × 9 = 27

Third part = (6/18) × 162 = 6 × 9 = 54

Fourth part = (7/18) × 162 = 7 × 9 = 63 (Ans). 

[PROPORTION]

Proportion:

Proportion is the equality of two ratios. For example, if we have a/b = c/d or a:b = c:d then a, b, c and d are in proportion and we can write them as a:b::c:d.

Here, a and d are called extremes and b and c are called means and the relation between them is-

Product of extremes = Product of means

ad = bc

Note:

1. When proportion is of a:b::c:d type then d is the fourth proportional.

Fourth proportional, d = bc/a

2. When proportion is of a:b::b:c type then c is the third proportional.

Third proportional, c = b²/a

We will use these relations to find the missing proportional. Let’s practice this formula through some questions.

E.g. 1: If N:38::3:57, find N.

By relation, ad = bc

We have

N × 57 = 38 × 3

⇒N = 38 × 3/57

 ⇒N = 2 × 3/3

⇒N = 2 (Ans). 

E.g. 2: Find the fourth proportional: 3:27::5:?

In this problem,

a = 3, b = 27, c = 5, d = ?

Fourth proportional, d= bc/a

d = 27 × 5/3

   = 27 × 5/3

   =9 × 5 = 45

∴ Fourth Proportional = 45 (Ans). 

E.g. 3: Find the third proportional to 4 and 8.

Here,  a=4, b=8

Third proportional, c= b2/a

c= 82/4

  = 64/4 = 16 (Ans). 

[WORD PROBLEMS]

• If two numbers are in the ratio a:b and m is added to the numbers, then the ratio becomes c:d. Two numbers will be ^am(c-d)/_(ad-bc) and ^bm(c-d)/_(ad-bc).

Let’s take an example on this topic. 

E.g.: Two numbers are in the ratio 5:6. When 6 is added to each, the ratio becomes 7:8 then the numbers are

Let the number be 5x and 6x.

A/Q,

5x + 6/6x + 6= 7/8

Doing cross multiplication,

⇒ 8 × (5x + 6)=7 × (6x + 6)

⇒ 40x + 48 = 42x + 42

⇒ 48 – 42 = 42x – 40x

⇒ 6 = 2x

⇒ x = 6/2 = 3

∵Since numbers are 5x and 6x.

∴Required numbers = 5 × 3 and 6 × 3

                                    = 15 and 18 (Ans).

By formula,

Here, a:b = 5:6, c:d = 7/8, m = 6

First number = am (c – d)/(ad – bc) = 5 × 6(7 – 8)/(8 × 5 – 6 × 7) = 30 × (-1)/(40 – 42) = 30 × (-1)/(-2)= 15.

Second Number= bm (c – d)/(ad – bc)= 6 × 6(7 – 8)/(8 × 5 – 6 × 7)= 36 × (-1)/(40 – 42) = 36 × (-1)/(-2) = 18

∴Required Number = 15 and 18 (Ans). 

• Two numbers are in ratio a:b and m is subtracted from the numbers, then the ratio becomes c:d. The two numbers will be ^{am(d – c)}/_{(ad – bc)} and ^{bm(d – c)}/_{(ad – bc),}respectively.

Let’s solve an example to see the application of this formula. 

E.g.: Two numbers are in the ratio 3:4. When 3 is subtracted from both the numbers, the ratio becomes 2:3. Find the sum of the numbers.

Let the number be 3x and 4x.

A/Q,

3x – 3/4x – 3 = 2/3

Doing cross multiplication,

3 × (3x – 3)=2 × (4x – 3)

⇒ 9x – 9 = 8x – 6

⇒ 9x – 8x = 9 – 6

⇒ x = 3

∴ Required numbers= 3 × 3 and 4 × 3

                                      = 9 and 12

Hence, sum of the numbers= 9 + 12 = 21 (Ans).

By formula,

Here, a:b = 3:4, c:d = 2:3 and m=3

First number= am(d – c)/(ad – bc)= 3 × 3(3 – 2)/(3 × 3 – 4 × 2)= 9/(9 – 8) =9

Second Number= bm(d – c)/(ad – bc)= 4 × 3(3 – 2)/(3 × 3 – 4 × 2)= 12 × 1/(9 – 8) =12

∴Required Number= 9 and 12

Hence, Sum of numbers= 21 (Ans).

Increment Problems

In increment problems suppose the numbers in multiple of 10 and do the increment and take the final ratio after the increment. Let’s take some increment problems and try to solve it.

E.g. 1: Three categories of employees get wages in the ratio of 1:2:3. If they get increments of 5%, 10% and 15% respectively, what will be the new ratio of their wages?

Step 1: Suppose the numbers in multiple of 10.

Ratio = 1:2:3

Let the number be 10, 20 and 30

Step 2: Do the increment

Increment in first number = 10 + 5% of 10 = 10 + 0.5 = 10.5

Increment in second number = 20 + 10% of 20 = 20 + 2 = 22

Increment in third number = 30 + 15% of 30 = 30 + 4.5 = 34.5

Step 3: Take the final ratio

Final ratio = 10.5:22:34.5 0r 21:44:69 (Ans).

E.g. 2: The ratio of marks obtained by a student in 3 subjects was 1:2:3. The school decided to allow grace marks of 5% for each subject. Find the student’s new ratio.

•  

Ratio = 1:2:3

Let the numbers be 10, 20 and 30

Increment in first number = 10 + 5% of 10= 10 + 0.5= 10.5

Increment in second number = 20 + 5% of 20= 20 + 1= 21

Increment in third number = 30 + 5% of 30= 30 + 1.5= 31.5

Therefore,

Required ratio = 10.5:21:31.5

                          = 21:42:63 or 1:2:3 (Ans).

Note: If there is an increment in the all the parts of a ratio by same percentage then the ratio remains the same.

Miscellaneous Problem

Let’s understand it with an example.

• Problem: If three numbers are in the ratio 2:3:5 and twice their sum is 100. Find the square of the largest of three numbers.

Ratio = 2:3:5

Let the numbers be 2x, 3x and 5x.

Twice their sum = 100

⇒ Their Sum = 100/2 = 50

⇒ 2x + 3x + 5x = 50

⇒ 10x = 50

⇒ x = 50/10 = 5

Required numbers are 2x, 3x and 5x i.e. 10, 15 and 25 respectively.

∴Square of the largest number = 252 = 625 (Ans). 

[PARTNERSHIP]

We all are aware of term ‘Partnership’. When two or more people enter into a business or association and invest some money, then they are called partners in business. After some time when the profit is earned, it is split into partners according to the investment they made.

Ratio of profit = Ratio of (Investment × Time period)

We have two cases in partnership:

1. When capital is same but time period is different. 

E.g.: A, B and C start a business with investment of Rs. 50000 each. A remain in Partnership for 9 months, B for 6 months and C for 12 months. Then, find the ratio of their profits.

In such problem the ratio of investment is obtained from the ratio of time period for which they invested the money.

Ratio of investment = Ratio of time period

A:B:C = 9:6:12 or 3:2:4

Therefore, Ratio of profit = 3:2:4 (Ans).

• When capital is different and time period is also different.

E.g.: A starts a business with Rs. 2000 and B joins him after3 months with Rs. 8000. Find the ratio of their profit at the end of the year.

Profit = Amount Invested × Time period

Time period of A = 1 year = 12 months

Time period of B = (12 – 3) months = 9 months 

A’s Profit: B’s Profit = Amount invested by A × Time period: Amount invested by B × Time period

                                 = 2000 × 12 : 8000 × 9           [simplify it like a fraction]

                                 = 1 × 4:4 × 3 = 1:3 (Ans). 

Let’s see some more types of Partnership problem.

• If x₁😡₂😡₃… is the ratio of capital invested and P₁😛₂😛₃… is the ratio of profits, then the ratio of time period is given by

P₁/x₁: P₂/x₂😛₃/x₃…

 Let’s see some examples

E.g.1: Mohan invested Rs. 100,000 in a garment business. After few months, Sohan joined him with Rs. 40000. At the end of the year, total profit was divided between them in ratio 3:1. After how many months did Sohan join the business?

Investment by Mohan = 100,000 × 12

Let Sohan invested for x months.

Investment by Sohan = 40,000 × x

Ratio of profit = Ratio of (investment × Time period)

⇒ 3:1 = 100,000 × 12 : 40,000 × x or

⇒ 3/1 = 100,000 × 12/40,000 × x

Doing cross multiplication,

⇒ 3 × (40,000 × x) = 1 × (100,000 × 12)

⇒ 120,000x = 1,200,000

⇒ x = 1,200,000/120,000 = 10

∴ Sohan invested for 10 months.

Hence, he joined after 2 months (Ans).

By formula,

Ratio of capital invested, x1:x2 = 100,000:40000 = 5:2

Ratio of Profits, P1:P2 = 3:1

∴Ratio of time period, t1:t2 = P1/x1: P2/x2 = 3/5:1/2

To convert ratio from fraction to integer multiply by LCM in numerator and denominator.

t1:t2 = 3 × 10/5:1 × 10/2 = 3 × 2:1 × 5 = 6:5

Time period of Mohan = 12 months

Hence multiplying factor = 2

∴ Time period of Sohan = 5 × 2 = 10 months.

So, he joined after 2 months (Ans). 

E.g. 2: Ramdas invested R. 90000 in a cosmetics business. After few months, Shyamdas joined him with Rs. 30,000. At the end of the year, the total profit was divided between them in the ratio 4:1. After how many months did Shyamdas joined business?

Ratio of Capital invested= 90000:30000 = 3:1

Ratio of profit = 4:1

By formula,

Ratio of time period, t1:t2= P1/x1: P2/x2= 4/3:1/1 = 4/3 : 1

Multiplying by 3 in numerator and denominator,

t1:t2 = 4 × 3/3 : 1 × 3 = 4:3

We know that, t1 = 12 months.

So, multiplying factor = 3

∴Time period for which Shyamdas invested money, t2 = 3 × 3 = 9 months.

Hence, Shyamdas joined after 3 months had passed in the year (Ans).

• If P₁😛₂😛₃… is the ratio of profits and t₁:t₂:t₃… is the ratio of time period of investments, then the ratio of capital invested is given by

P₁/t₁: P₂/t₂😛₃/t₃…

Let’s see an example for more understanding of the formula. 

E.g: If A, B and C each does investments for time periods 5 months, 6 months and 8 months respectively. At the end of the business terms, they received the profits in the ratio of 5:3:12. Find the ratio of investments of A, B and C.

Ratio of time period = 5:6:8

Ratio of Profit= Ratio of (Capital invested ×Time period)

⇒5:3:12 = 5×1:6×2:8×3

Taking first two terms of the ratio,

5×1/6×2 = 5/3

⇒x1/x2 = 2/1

Taking last two terms of the ratio,

6×2/8×3 = 3/12

⇒x2/x3 = 1/3

∴x1:x2:x3 = 2:1:3 (Ans).

By formula,

Ratio of time period = 5:6:8

Ratio of Profit = 5:3:12

x1:x2:x3 = P1/t1: P2/t2:P3/t3 = 5/5 : 3/6 : 12/8

 = 1: ½ : 3/2    [Multiplying by 2 in numerator]

= 2:1:3 (Ans). 

[AGE BASED PROBLEMS (ON RATIO)]

Problems based on ages generally consist of information of ages of two or more person and their correlation with each other in the past or present or future. Problems are very easy to solve if we understand them.

Basic Rules for problem based on ages:

1. If the ratio of present ages of A and B is x:y and n years ago, the ratio of their ages was p:q, then (kx-n)/(ky-n) = p/q, where k is a constant 

E.g.: The ratio of the ages of A and B at present is 3:1. Four years earlier, the ratio was 4:1. Find the present age of A.

Let the present ages of A and B be 3x and x years respectively.

Now, Four years ago, age of A and B = (3x – 4) and (x – 4) years respectively.

A/q,

(3x – 4)/(x – 4) = 4/1

Doing cross multiplication,

(3x – 4) = 4 × (x – 4)

⇒3x – 4 = 4x – 16

⇒16 – 4 = 4x – 3x

⇒x = 12

∴Present age of A = 3x = 3 × 12 = 36 years (Ans).

• If the ratio of present ages of A and B is x:y and after n years, the ratio of their ages will be p:q, then (kx + n)/(ky + n) =p/q, where k is a constant.

E.g.: At present, the ratio of the ages of Maya and Chhaya is 6:5 and fifteen years from now, the ratio will get changed to 9:8. Find the present age of Maya.

Let the present ages of Maya and Chhaya be 6x and 5x years, respectively.

A/q,

After 15 years,

(6x + 15)/(5x + 15) = 9/8

⇒8 × (6x + 15) = 9 × (5x + 15)

⇒48x + 120 = 45x + 135

⇒48x – 45x = 135 -120

⇒3x = 15

⇒x = 5

∴Present age of Maya= 6x = 6 × 5 = 30 years (Ans).

The problems that we get in our exams from this topic are combination of both the rule that we have just studied. They also involve some concept and knowledge of how to solve linear equation (that we will discuss in algebra). Now that we have understood the rules let’s solve some example. 

E.g. 1: 4 years ago, the ratio of Vikah’s to Rahul’s age was 3:5. After 6 years, this ratio would become 4:5. Find present age of Rahul.

Let the age of Vikah and Rahul four years ago be 3x and 5x.

So, their present age will be (3x + 4) and (5x + 4) years respectively.

Age of Vikah and Rahul after 6 years from present= (3x + 4 + 6) and (5x + 4 + 6) years respectively

  = (3x + 10) and (5x + 10) years respectively.

A/q,

Ratio of Vikah’s age to Rahul’s age after 6 years= 4:5

⇒ (3x + 10):(5x + 10) = 4:5 or

⇒ (3x + 10)/(5x + 10) = 4/5

Doing cross multiplication,

5 × (3x + 10) = 4 × (5x + 10)

⇒15x + 50 = 20x + 40

⇒50 – 40 = 20x – 15x

⇒10 = 5x

⇒x = 2

∴Present age of Rahul= (5x + 4) = 5 × 2 + 4 = 14 years (Ans). 

E.g. 2: Six years ago, the ratio of ages of the two persons P and Q was 3:2. Four years hence the ratio of their ages will be 8:7. What is P’s age?

 Let the age of P and Q six years ago be 3x and 2x.

So, their present age will be (3x + 6) and (2x + 6) years respectively.

A/q,

Ratio of P’s age to Q’s age after 4 years= 8:7

⇒ (3x + 6 + 4):(2x + 6 + 4) = 8:7

⇒ (3x + 10):(2x + 10) = 8:7 or

⇒ (3x + 10)/(2x + 10) = 8/7

Doing cross multiplication,

7 × (3x + 10) = 8 × (2x + 10)

⇒21x + 70 = 16x + 80

⇒21x – 16x = 80 – 70

⇒5x = 10

⇒x = 2

∴Present age of P = (3x + 6) = 3 × 2 + 6 = 12 years (Ans).