4. Word problems for linear equations

The perimeter of the rectangular cardboard is $68 \ cm$. The length of the rectangular cardboard is $20 \ cm$ more than the breadth. Find the length and breadth of the rectangular cardboard.

 

 

 

Let $x$ be the breadth of the rectangular cardboard.

 

Let $20 + x$ be the length of the rectangular cardboard.

 

Now, let us frame the equation.

 

Perimeter of the rectangular cardboard $= 2(l + b)$

 

$68 = 2(20 + x + x)$

 

$68 = 2(2x + 20)$

 

$\frac{68}{2} = 2x + 20$

 

$34 = 2x + 20$

 

$34 - 20 = 2x$

 

$14 = 2x$

 

$\frac{14}{2} = x$

 

$7 = x$

 

Breadth $= x = 7 \ cm$.

 

Length $= x + 20 = 27 \ cm$.

 

Therefore, the length of the rectangular cardboard is $27 \ cm$, and the breadth of the rectangular cardboard is $7 \ cm$.

Mary bought $2$ pens and $3$ erasers for the cost of $₹35$. Frame the equation and also find the cost of one pen when the cost of one eraser is $₹5$.

 

Solution:

 

Let $x$ denote the cost of $1$ pen.

 

Let $y$ denote the cost of $1$ eraser.

 

Let us frame the equation.

 

$2x + 3y = 35$

 

We shall find the cost of $1$ pen when the cost of $1$ eraser is $₹5$.

 

That is, substituting $y = 5$ in the above equation, we have:

 

$2x + 3(5) = 35$

 

$2x + 15 = 35$

 

$2x = 35 - 15$

 

$2x = 20$

 

$x = \frac{20}{2}$

 

$x = 10$

 

Therefore, the cost of $1$ pen is $₹10$.

 

2. The cost of $5$ chocolates is $₹15$ times biscuits. The sum of cost of chocolates and biscuits is $₹40$. Find the cost of a chocolate and a biscuit.

 

Solution:

 

Let $x$ denote the cost of a chocolate.

 

Let $y$ denote the cost of the biscuit.

 

Let us frame the equation.

 

$5x = 15y$

 

$x = \frac{15}{5}y$

 

$x = 3y$ ---- ($1$)

 

$x + y = 40$ ---- ($2$)

 

Substituting equation ($1$) in ($2$), we have:

 

$3y + y = 40$

 

$4y = 40$

 

$y = \frac{40}{4}$

 

$y = 10$

 

Put $y = 10$ in equation ($1$), we have:

 

$x = 3 \times 10$

 

$x = 30$

 

Thus, the cost of $1$ chocolate is $₹30$, and the cost of $1$ biscuit is $₹10$.