Ex.3.1 Q2 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

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Question

The coach of a cricket team buys \(3\) bats and 6 balls for \(₹\,3900.\) Later, she buys another bat and \(3\) more balls of the same kind for ₹ \(1300.\) Represent this situation algebraically and geometrically.

 Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.1 | Question 2

Text Solution

What is Known?

(i) Three bats and six balls for ₹ \(3900\)

(ii) One bat and three balls for ₹ \(1300\)

What is Unknown?

Represent the situation geometrically and algebraically

Reasoning:

Assuming the cost of one bat as ₹ \(x\) and the cost of one ball as ₹ \(y,\) two linear equations can be formed for the above situation.

Steps:

The cost of \(3\) bats and 6 balls is ₹ \(3900.\)

Mathematically:

\[\begin{align}3x + 6y&= 3900\\3(x + 2y)& = 3900\\x + 2y& = 1300\end{align}\]

Also, the cost of \(1\) bat and \(3\) balls is ₹ \(1300.\)

Mathematically:

\[x + 3y = 1300\]

Algebraic representation where \(x\) and \(y\) are cost of bat and ball respectively.

\[\begin{align}x + 2y &= 1300 \qquad(1)\\x + 3y &= 1300\qquad(2)\end{align}\]

Therefore, the algebraic representation for equation \(1\) is:

\[\begin{align}x + 2y &= 1300\\2y &= 1300-x\\y &= \frac{{1300 - x}}{2}\end{align}\]

And, the algebraic representation for equation \(2\) is:

\[\begin{align}x + 3y &= 1300\\3y &= 1300-x\\y &= \frac{{1300 - x}}{3}\end{align}\]

Let us represent these equations graphically. For this, we need at least two solutions for each equation. We give these solutions in table shown below.

\(x\) \(700\) \(500\)
\(y = \frac{{1300 - x}}{2}\) \(300\) \(400\)

 

\(x\) \(400\) \(700\)
\(y = \frac{{1300 - x}}{3}\) \(300\) \(200\)

The graphical representation is as follows.

Unit: \(1\,\rm{cm} =\)\(100.\)

The answer is \((1300, \,0)\)

  
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