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Question from nasr, a teacher:

Four apples and two oranges cost Rs. 30, and one apple and 3 oranges costs Rs.15. How much does each apple and each oranges cost?


I am going to solve this problem using the technique called the method of elimination but using two presentations, once as a story and then using algebra.


You go to a fruit dealer's stand and there is a sign that says

4 apples and 2 oranges: Rs. 30
1 apples and 3 oranges: Rs. 15

You look at the second line on the sign and go to the bins and get 1 apple and 3 oranges and put Rs.15 beside them on the counter. You repeat this 4 times and now you have 4 apples and 12 oranges with Rs. 60 there for payment.

Then you realize that you don't want any apples so, looking at the first line of the sign you put 4 apples and 2 oranges back in the bins and remove Rs. 30 from the payment.

What remains on the counter is 10 oranges and Rs. 30. Now it is easy to see that each orange costs Rs. 3. Looking again at the second line on the sign, 3 oranges cost $3 \times 3 = \mbox{ Rs. }9$ and since $15 - 9 = 6,$ each apple costs Rs. 6.


Suppose each apple costs Rs. $x$ and each orange costs Rs. $y.$ The fact that four apples and two oranges cost Rs. 30, and one apple and 3 oranges costs Rs.15 can then be written as

$4 x + 2 y = 30$
$x + 3y = 15.$

Multiplying the second equation by $4$ gives the equations

$4 x + 2 y = 30$
$4x + 12y = 60.$

Now subtract the first equation from the second equation to get

$10 y = 30$

and dividing both sides by $10$ yields

$y = 3$

Substitute $y = 3$ into the equation $x + 3y = 15$ to obtain $x + 3 \times 3 = 15$ and hence $x = 6.$

I hope this helps,

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