Question

In a purse there are 20 rupee notes, 10 rupee notes and 50 rupee notes. The number of 50 rupee notes exceeds 2 times the number of 10 rupee notes by 1 and the number of 20 rupee notes is 5 less than the number of 10 rupee notes. If the total value of money in the purse is 860. Find the number of each variety of notes.

Answer 1

Hint: For solving this question you should know about the making equations by statements. In this problem here many statements are given to us. We will collect them all and first of all we make the required equations from that. And then by solving these equations from that. And then by solving these equations with the help of each other we find the values of all assumed variables. And then put the value again in the equation and find the value for that specified condition.

Complete step by step answer:
According to the question it is asked to find the number of each type note if the total value of each type notes if the total value of money is 860 and the statements (situations) are given to us.
So, according to the given situations if we make the equations for solving this problem then the equations according to situational statements are: -
Let the number of 10 rupees note be \[=X\]
Let the number of 20 rupees note \[=X-5\]
Let the number of 50 rupees note \[=2X+1\]
Given that the total value is 860.
According to the data equation can be formed as
\[\left( X\times 10 \right)+\left[ \left( X-5 \right)\times 20 \right]+\left[ \left( 2X+1 \right)\times 50 \right]=860\]
\[10X+20X-100+100X+50=860\]
\[130X-50=860\]
\[130X=860+50=910\]
\[X=\dfrac{910}{130}=7\]
Hence,
The number of 10 rupee note \[=X=7\]
The number of 20 rupee note \[=X-5=7-5=2\]
The number of 50 rupee note \[=\left[ 2X+1 \right]=2\times 7+1\]
                                                      \[=14+1=15\]

Note: While solving these types of questions you have to be very careful about the statements and the equations made from these. And the equations must be correct with digits and signs also. And the equations are solved with each other's help.