A change making machine contains $1$ rupee, $2$ rupee , and $5$ rupee coins. The total number of coins is $300$. The amount is $Rs.960$ . If the number of $1$ rupee and $2$ rupee coins are interchanged, the value comes down by $Rs.40$. The total number of $5$ rupee coins is:
A. $100$
B. $140$
C. $60$
D. $150$
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.
Last updated date: 17th Sep 2023
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Total views: 359.1k
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