Meena went to a bank to withdraw Rs.2000. She asked the cashier to give her Rs.50 and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received.
Hint: The question is related to the linear equation in two variables. Try to make two equations using the information given in the problem statement and solve them simultaneously.
Complete step-by-step answer: Complete step-by-step answer: In the question, it is given that Meena withdrew \[Rs.2000\] from a bank in the form of notes of $Rs.50$ and $Rs.100$. It is also given that she got $25$ notes in all. So, we will consider $x$ as the number of $Rs.50$ notes received by Meena and $y$ as the number of $Rs.100$ notes received by Meena. Now, in the first case, it is given that Meena withdrew \[Rs.2000\] from a bank in the form of notes of $Rs.50$ and $Rs.100$. So, the amount in the form of $Rs.50$ notes is equal to $50\times x=Rs.50x$. Also, the amount in the form of $Rs.100$ notes is equal to $100\times y=Rs.100y$. So, the total amount will be $Rs.\left( 50x+100y \right)$. But it is given that the total amount withdrawn is \[Rs.2000\]. So, $50x+100y=2000..........(i)$ Now, we have considered $x$ as the number of $Rs.50$ notes received by Meena and $y$ as the number of $Rs.100$ notes received by Meena. So, the total number of notes will be equal to $x+y$. But it is given that the total number of notes is equal to $25$. So, $x+y=25.....(ii)$ Now, we will solve the linear equations to find the values of $x$ and $y$. From equation$(ii)$, we have $x+y=25$ $\Rightarrow y=25-x$ On substituting $y=25-x$ in equation$(i)$, we get $50x+100\left( 25-x \right)=2000$ $\Rightarrow 50x+2500-100x=2000$ $\Rightarrow 500-50x=0$ $\Rightarrow 50x=500$ $\Rightarrow x=10$ Now, substituting \[x=10\] in equation$(ii)$, we get $10+y=25$ \[\Rightarrow y=15\] Hence, the numbers of $Rs.50$ notes and \[Rs.100\] notes that are received by Meena from the cashier are $10$ and $15$ respectively. Note: While solving this question we can assume the number of RS.50 notes as x and RS.100 notes as (25-x). By this substitution we get a linear equation in one variable. We can find the value of x by solving the linear equation in one variable.
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