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 \(₹10\).