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# The coach of a cricket team buys one bat and 2 balls for Rs.300. later he buys another 2 bats and 3 balls of the same kind for Rs.525. Represent this situation algebraically and solve it by graphical method. Also find out how much money the coach will pay for the purchase of one bat and one ball.

Last updated date: 20th Jun 2024
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Hint: Assume the cost of one bat be x rupees and the cost of one ball be y rupees. Then we will formulate the given conditions into equations involving x and y and then solve the equations by elimination method and determining the value of x and y.

Let the cost of one bat be x rupees and the cost of one ball be y rupees.
In Algebraic form:
Then the equation for one bat and 2 balls costing Rs.300 will be represented as:
$x + 2y = 300{\text{ }}...{\text{(1)}}$
$\Rightarrow x = 300 - 2y{\text{ }}...{\text{(2)}}$
And the other equation for 2 bats and 3 balls of the same kind costing Rs.525 can be represented as:
$2x + 3y = 525{\text{ }}...{\text{(3)}}$
In graphical method: In this, we will substitute 2 to 3 values of y and get the table following for:
$x+2y = 300$
 x 0 300 y 150 0

$2x + 3y = 525$
 x 0 262.5 y 175 0

Now, putting the value of equation (2) in equation (3), we will get:
$2(300 - 2y) + 3y = 525 \\ \Rightarrow y = 75 \\$
And substituting the value of $y = 75$ in (2), we will get $x = 150$.

So, the coach will pay Rs.150 for a bat and Rs.75 for a ball.

Note: Whenever there are 2 equations in 2 variables, the easiest method to solve them is by the elimination method.
Put the value of one variable from one equation into the second equation to get the value of the second variable.
In the graph, the point at which both the lines are intersecting will give the values for x and y.
It can also be verified by solving the equations algebraically.