Question

13. A sum of 800 is in the form of denominations of 10 and 20. If the total number of notes is 50, find the number of notes of each type.​

Answer 1

let there be x notes of rs 10.total number of notes is= 50 so, there are 20 notes of rs 10 and 30 notes of rd 20. hope it is helpful you

Answer 2

$\large\underline{\bold{Given \:Question - }}$

• A sum of 800 is in the form of denominations of 10 and 20. If the total number of notes is 50, find the number of notes of each type.

$\large\underline{\bold{Solution-}}$

Concept Used :-

Writing Systems of Linear Equations from Word Problems :

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.

Let's solve the problem now!!

$\begin{gathered}\begin{gathered}\bf \:Let - \begin{cases} &\sf{denominations \: of \: 10 = x} \\ &\sf{denominations \: of \: 20 = y} \end{cases}\end{gathered}\end{gathered}$

$\bf \: According \: to \: Ist \: condition \: -$

• A sum of 800 is in the form of denominations of 10 and 20.

So,

• Total value of 10 denomination = 10x

and

• Total value of 20 denomination = 20y

Hence,

$\sf \: 10x + 20y = 800$

$\bf \: \therefore \: x + 2y = 80 - - - (1)$

$\bf \: According \: to \: 2nd \: condition \: -$

• Total number of notes is 50.

So,

$\bf \: x + y = 50 - - - (2)$

• On Subtracting equation (2) from equation (1), we get

$\bf \: y \: = \: 30 \: - - - (3)$

• On substituting value of y = 30 in equation (2), we get

$\sf \: x + 30 = 50$

$\bf \therefore \: x \: = \: 20$

$\begin{gathered}\begin{gathered}\bf \: Hence - \begin{cases} &\sf{denominations \: of \: 10 \: = \: 20} \\ &\sf{denominations \: of \: 20 \: = \: 30} \end{cases}\end{gathered}\end{gathered}$