Question

Find the perpendicu
A.M. of two numbers is 7 and their product is 45. Find the numbers.​

Answer 1

Appropriate Question :-

• A.M. of two numbers is 7 and their product is 45. Find the numbers.

Solution :-

$\begin{gathered}\begin{gathered}\bf\: Let-\begin{cases} &\sf{first \: number \: be \: x} \\ &\sf{seond \: number \: be \: y} \end{cases}\end{gathered}\end{gathered}$

According to statement,

• Arithmetic mean between 2 numbers is 7

$\rm :\longmapsto\:\dfrac{x + y}{2} = 7$

$\rm :\longmapsto\:x + y = 14$

$\bf\implies \:y = 14 - x - - - (1)$

Again,

According to statement,

• Product of 2 numbers is 45.

$\rm :\longmapsto\:xy = 45$

$\rm :\longmapsto\:x(14 - x) = 45$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: using \: (1)\bigg \}}$

$\rm :\longmapsto\:14x - {x}^{2} = 45$

$\rm :\longmapsto\: {x}^{2} - 14x + 45 = 0$

$\rm :\longmapsto\: {x}^{2} - 9x - 5x + 45 = 0$

$\rm :\longmapsto\:x(x - 9) - 5(x - 9) = 0$

$\rm :\longmapsto\:(x - 5)(x - 9) = 0$

$\bf :\implies\:x = 5 \: \: \: \: or \: \: \: \: x = 9$

Hence,

Numbers are

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y = 14 - x \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 5 & \sf 9 \\ \\ \sf 9 & \sf 5 \end{array}} \\ \end{gathered}$

Basic Concept Used :-

Writing Systems of Equation from Word Problem.

1. Understand the problem.

• Understand all the words used in stating the problem.

• Understand what you are asked to find.

2. Translate the problem to an equation.

• Assign a variable (or variables) to represent the unknown.

• Clearly state what the variable represents.

3. Carry out the plan and solve the problem.