# Compound Proportion

“**The proportion involving two or more quantities is called Compound Proportion.**”

__Rules for Solving Compound Proportions__

\[\begin{array}{*{20}{c}} {{\text{Quantity 1}}}&{{\text{Quantity 2}}}&{{\text{Quantity 3}}} \\ {\text{a}}&{\text{b}}&{\text{c}} \\ {\text{d}}&{\text{e}}&x \end{array}\]

**CASE-1**

If quantity 1 and quantity 2 are directly related and quantity 2 and quantity 3 are also directly related, then we use the following rule:

\[\boxed{\frac{{{\text{a x b}}}}{{\text{c}}} = {\text{ }}\frac{{{\text{d x e}}}}{x}}\]

**CASE-2**

If quantity 1 and quantity 2 are directly related and quantity 2 and quantity 3 are inversely related, then we use the following rule:

\[\boxed{\frac{{{\text{b x c}}}}{{\text{a}}}{\text{ = }}\frac{{{\text{e x }}x}}{{\text{d}}}}\]

**CASE-3**

If quantity 1 and quantity 2 are inversely related and quantity 2 and quantity 3 are directly related, then we use the following rule:

\[\boxed{\frac{{{\text{a x b}}}}{{\text{c}}}{\text{ = }}\frac{{{\text{d x e}}}}{x}}\]

**CASE-4**

If quantity 1 and quantity 2 are inversely related and quantity 2 and quantity 3 are also inversely related, then we use the following rule:

\[\boxed{{\text{a x b x c = d x e x }}x}\]

**Example:**

195 men working 10 hours a day can finish a job in 20 days. How many men are employed to finish the job in 15 days if they work 13 hours a day?

**Solution:**

Let $$x$$ be the no. of men required

\[\begin{array}{*{20}{c}} {{\text{Days}}}&{{\text{Hours}}}&{{\text{Men}}} \\ {{\text{20}}}&{{\text{10}}}&{{\text{195}}} \\ {{\text{15}}}&{{\text{13}}}&x \end{array}\]

\[\begin{gathered} 20 \times 10 \times 195 = 15 \times 13{\text{ }} \times x \\ x = \frac{{{\text{20 x 10 x 195}}}}{{{\text{15 x 13}}}} = 200\,men \\ \end{gathered} \]

**Example:**

A soap factory makes 600 units in 9 days with the help of 20 machines. How many units can be made in 12 days with the help of 18 machines?

**Solution:**

\[\begin{array}{*{20}{c}} {{\text{Machines}}}&{{\text{Days}}}&{{\text{Units}}} \\ {{\text{20}}}&{\text{9}}&{{\text{600}}} \\ {{\text{18}}}&{{\text{12}}}&x \end{array}\]

\[\begin{gathered} \frac{{{\text{20 x 9}}}}{{{\text{600}}}}{\text{ = }}\frac{{{\text{18 x 12}}}}{x} \\ 20 \times 9 \times x = {\text{ }}600 \times 18 \times 12 \\ x = \frac{{{\text{600 x 18 x 12}}}}{{{\text{20 x 9}}}} = 720\,units \\ \end{gathered} \]

Qaiser zaman

August 2@ 11:20 pmWhat amout the formula for which the uper side of division is the tails of arrows and the lower side will be the heads of arrows

mohammad siddiq

April 14@ 2:48 pmhow the quantities will be determined as directly related or inversely related.

Innocent

October 7@ 8:28 pmHow to use the method of lining to this questions? Just Like inverse proportion.

Abdur Raheem

March 5@ 9:52 amHow we know the 3rd quantity

This is directly or inversely related to 1st value while third quantity is unknown