# 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. $150$

Last updated date: 31st Mar 2023

•

Total views: 306.9k

•

Views today: 5.83k

Answer

Verified

306.9k+ views

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.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India