Questions & Answers

Question

Answers

Answer
Verified

Hint: Assume the cost of the table is Rs x, and that of the chair is Rs y. Form two linear equations in terms of x and y using the statement of the problem. Solve the system of equations using the substitution method or elimination method or matrix method. Hence find the value of x and y satisfying both the questions. The value of x will be the cost of the table, and the value of y will be the cost of the chair.

__Complete step-by-step answer: __

Let x be the cost of the table and y be the cost of the chair.

Hence, we have the cost of two tables = 2x and the cost of three chairs = 3y.

Hence, the cost of two tables and three chairs = 2x+3y.

Also, given that the cost of two tables and three chairs = 340.

Hence, we have 2x+3y=340 (i)

Also, it is given that the cost of a table is Rs 20 more than the cost of a chair.

Hence, we have x = y+20 (ii).

Solving the system using the substitution method:

Substituting the value of x from equation (ii) in equation (i), we get

2(y+20)+3y=340

Applying distributive law, a(b+c) = ab+ac , we get,

2y+40+3y = 340

i.e. 5y+40 = 340

Subtracting 40 from both sides of the equation we get

5y+40-40 = 340-40

i.e. 5y = 300

Dividing both sides of the equation by 5, we get

$\begin{align}

& \dfrac{5y}{5}=\dfrac{300}{5} \\

& \Rightarrow y=60 \\

\end{align}$

Hence, we have y = 60.

Substituting the value of y in equation (ii), we get

x = 60+20

i.e. x = 80.

Hence the cost of a table is Rs 80, and that of a chair is Rs 60.

Note: In the above solution, we used the substitution method to find the value of x and y.

We can also solve the system by using Matrices.

Here augmented matrix A is given by

$A=\left[ \begin{matrix}

2 & 3 & 340 \\

1 & -1 & 20 \\

\end{matrix} \right]$

We use row transformations to convert the given matrix in row reduced form.

Apply ${{R}_{2}}\to 2{{R}_{2}}-{{R}_{1}}$, we get

$\begin{align}

& R=\left[ \begin{matrix}

2 & 3 & 340 \\

2-2 & -2-3 & 40-340 \\

\end{matrix} \right] \\

& \Rightarrow R=\left[ \begin{matrix}

2 & 3 & 340 \\

0 & -5 & -300 \\

\end{matrix} \right] \\

\end{align}$

where R is the row reduced form of A.

Hence, we have

-5y = -300

Dividing both sides by -5, we get

y =60

and 2x+3y=340

i.e. 2x+180 = 340

Subtracting 340 from both sides, we get

2x+180-180 = 340-180

i.e. 2x = 160

Dividing both sides by 2 we get

x = 80

Hence x = 80 and y = 60 is the solution of the given system.

Let x be the cost of the table and y be the cost of the chair.

Hence, we have the cost of two tables = 2x and the cost of three chairs = 3y.

Hence, the cost of two tables and three chairs = 2x+3y.

Also, given that the cost of two tables and three chairs = 340.

Hence, we have 2x+3y=340 (i)

Also, it is given that the cost of a table is Rs 20 more than the cost of a chair.

Hence, we have x = y+20 (ii).

Solving the system using the substitution method:

Substituting the value of x from equation (ii) in equation (i), we get

2(y+20)+3y=340

Applying distributive law, a(b+c) = ab+ac , we get,

2y+40+3y = 340

i.e. 5y+40 = 340

Subtracting 40 from both sides of the equation we get

5y+40-40 = 340-40

i.e. 5y = 300

Dividing both sides of the equation by 5, we get

$\begin{align}

& \dfrac{5y}{5}=\dfrac{300}{5} \\

& \Rightarrow y=60 \\

\end{align}$

Hence, we have y = 60.

Substituting the value of y in equation (ii), we get

x = 60+20

i.e. x = 80.

Hence the cost of a table is Rs 80, and that of a chair is Rs 60.

Note: In the above solution, we used the substitution method to find the value of x and y.

We can also solve the system by using Matrices.

Here augmented matrix A is given by

$A=\left[ \begin{matrix}

2 & 3 & 340 \\

1 & -1 & 20 \\

\end{matrix} \right]$

We use row transformations to convert the given matrix in row reduced form.

Apply ${{R}_{2}}\to 2{{R}_{2}}-{{R}_{1}}$, we get

$\begin{align}

& R=\left[ \begin{matrix}

2 & 3 & 340 \\

2-2 & -2-3 & 40-340 \\

\end{matrix} \right] \\

& \Rightarrow R=\left[ \begin{matrix}

2 & 3 & 340 \\

0 & -5 & -300 \\

\end{matrix} \right] \\

\end{align}$

where R is the row reduced form of A.

Hence, we have

-5y = -300

Dividing both sides by -5, we get

y =60

and 2x+3y=340

i.e. 2x+180 = 340

Subtracting 340 from both sides, we get

2x+180-180 = 340-180

i.e. 2x = 160

Dividing both sides by 2 we get

x = 80

Hence x = 80 and y = 60 is the solution of the given system.

×

Sorry!, This page is not available for now to bookmark.