Answer

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Hint: The question is related to the linear equation in three variables. Try to make three equations using the information given in the problem statement and solve them simultaneously.

Complete step-by-step answer:

In the question, the amount given that is changed is \[Rs.960\] in the form of $1$ rupee, $2$ rupee , and $5$ rupee coins. It is also given that the total number of coins is $300$ . It is also given that if the number of $1$ rupee and $2$ rupee coins are interchanged, the value comes down by $Rs.40$ . So, we will consider $x$ as the number of $1$ rupee coins , $y$ as the number of $2$ rupee coins, and $z$ as the number of $5$ rupee coins.

Now, in the first case, it is given that a sum of $Rs.960$ is changed in the form of $1$ rupee, $2$ rupee , and $5$ rupee coins. So, the amount in the form of $1$ rupee coins is equal to $1\times x=Rs.x$ . Also, the amount in the form of $2$ rupee coins $2\times y=Rs.2y$ . And, the amount in the form of $5$ rupee coins $5\times z=Rs.5z$ . So, the total amount will be $Rs.\left( x+2y+5z \right)$ . But it is given that the total amount changed is \[Rs.960\]. So. $x+2y+5z=960..........(i)$

Now, we have considered $x$ as the number of $1$ rupee coins, $y$ as the number of $2$ rupee coins, and $z$ as the number of $5$ rupee coins. So, the total number of coins will be equal to $x+y+z$ . But it is given that the total number of coins is equal to $300$. So, $x+y+z=300.....(ii)$

Now, it is given that if the number of $1$ rupee and $2$ rupee coins are interchanged, the value comes down by $Rs.40$. So, $y+2x+5z=920.....(iii)$

Now, we will solve the linear equations to find the values of $x$ , $y$ and $z$ .

From equation$(ii)$ , we have $x+y+z=300$

$\Rightarrow z=300-x-y$

On substituting $z=300-x-y$ in equation$(i)$ , we get

$x+2y+5\left( 300-x-y \right)=960$

$\Rightarrow x+2y+1500-5x-5y=960$

$\Rightarrow 1500-960=4x+3y$

$\Rightarrow 540=4x+3y$

$\Rightarrow x=\dfrac{540-3y}{4}$

Now, substituting $x=\dfrac{540-3y}{4}$ in $z=300-x-y$, we get:

$z=300-\dfrac{540-3y}{4}-y$

\[\Rightarrow z=\dfrac{1200-540+3y-4y}{4}\]

\[\Rightarrow z=\dfrac{660-y}{4}\]

Now, substituting $x=\dfrac{540-3y}{4}$ and \[z=\dfrac{660-y}{4}\] in equation $(iii)$ , we get:

\[2\left( \dfrac{540-3y}{4} \right)+y+5\left( \dfrac{660-y}{4} \right)=920\]

\[\Rightarrow \dfrac{1080-6y}{4}+y+\dfrac{3300-5y}{4}=920\]

\[\Rightarrow \dfrac{1080-6y+4y+3300-5y}{4}=920\]

\[\Rightarrow 4380-7y=3680\]

\[\Rightarrow 7y=4380-3680=700\]

\[\Rightarrow y=100\]

Now, we need to find the number of $5$ rupee coins, i.e. we need to find the value of $z$ .

Substituting $y=100$ in \[z=\dfrac{660-y}{4}\] , we get:

$z=\dfrac{660-100}{4}=\dfrac{560}{4}=140$

Hence, the number of $5$ rupee coins is $140$ . So, option B, is the correct option.

Note: While solving the equations, make sure that the substitutions are done correctly and no sign mistakes are present. Sign mistakes can cause the final answer to be wrong.

Complete step-by-step answer:

In the question, the amount given that is changed is \[Rs.960\] in the form of $1$ rupee, $2$ rupee , and $5$ rupee coins. It is also given that the total number of coins is $300$ . It is also given that if the number of $1$ rupee and $2$ rupee coins are interchanged, the value comes down by $Rs.40$ . So, we will consider $x$ as the number of $1$ rupee coins , $y$ as the number of $2$ rupee coins, and $z$ as the number of $5$ rupee coins.

Now, in the first case, it is given that a sum of $Rs.960$ is changed in the form of $1$ rupee, $2$ rupee , and $5$ rupee coins. So, the amount in the form of $1$ rupee coins is equal to $1\times x=Rs.x$ . Also, the amount in the form of $2$ rupee coins $2\times y=Rs.2y$ . And, the amount in the form of $5$ rupee coins $5\times z=Rs.5z$ . So, the total amount will be $Rs.\left( x+2y+5z \right)$ . But it is given that the total amount changed is \[Rs.960\]. So. $x+2y+5z=960..........(i)$

Now, we have considered $x$ as the number of $1$ rupee coins, $y$ as the number of $2$ rupee coins, and $z$ as the number of $5$ rupee coins. So, the total number of coins will be equal to $x+y+z$ . But it is given that the total number of coins is equal to $300$. So, $x+y+z=300.....(ii)$

Now, it is given that if the number of $1$ rupee and $2$ rupee coins are interchanged, the value comes down by $Rs.40$. So, $y+2x+5z=920.....(iii)$

Now, we will solve the linear equations to find the values of $x$ , $y$ and $z$ .

From equation$(ii)$ , we have $x+y+z=300$

$\Rightarrow z=300-x-y$

On substituting $z=300-x-y$ in equation$(i)$ , we get

$x+2y+5\left( 300-x-y \right)=960$

$\Rightarrow x+2y+1500-5x-5y=960$

$\Rightarrow 1500-960=4x+3y$

$\Rightarrow 540=4x+3y$

$\Rightarrow x=\dfrac{540-3y}{4}$

Now, substituting $x=\dfrac{540-3y}{4}$ in $z=300-x-y$, we get:

$z=300-\dfrac{540-3y}{4}-y$

\[\Rightarrow z=\dfrac{1200-540+3y-4y}{4}\]

\[\Rightarrow z=\dfrac{660-y}{4}\]

Now, substituting $x=\dfrac{540-3y}{4}$ and \[z=\dfrac{660-y}{4}\] in equation $(iii)$ , we get:

\[2\left( \dfrac{540-3y}{4} \right)+y+5\left( \dfrac{660-y}{4} \right)=920\]

\[\Rightarrow \dfrac{1080-6y}{4}+y+\dfrac{3300-5y}{4}=920\]

\[\Rightarrow \dfrac{1080-6y+4y+3300-5y}{4}=920\]

\[\Rightarrow 4380-7y=3680\]

\[\Rightarrow 7y=4380-3680=700\]

\[\Rightarrow y=100\]

Now, we need to find the number of $5$ rupee coins, i.e. we need to find the value of $z$ .

Substituting $y=100$ in \[z=\dfrac{660-y}{4}\] , we get:

$z=\dfrac{660-100}{4}=\dfrac{560}{4}=140$

Hence, the number of $5$ rupee coins is $140$ . So, option B, is the correct option.

Note: While solving the equations, make sure that the substitutions are done correctly and no sign mistakes are present. Sign mistakes can cause the final answer to be wrong.

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