Question

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Answer 1

Let the number of bananas in lot A is x and the number of bananas in lot B is y.

(1) Given that the cost of 3 bananas in Lot A = 2

           Then the cost of 1 banana in Lot A = 3/2.

    Given that the cost of 1 banana in Lot B = 1.

The total cost of lot A and lot B = 400.

(2x/3) + y = 400

2x + 3y = 1200    ----- (1)



(2) Given that Cost of 1 banana in lot A = 1. 

     Cost of 5 bananas in lot B = 4

     Then the cost of 1 banana in Lot B = 4/5.

Total cost of Lot A and Lot B = 460

        x + 4/5 y = 960

        5x + 4y = 2300     ---- (2)



On solving (1)*5 and (2)*2, we get

10x + 8y = 4600

10x + 15y = 6000
------------------------

          -7y = -1400

            y = 200      ------------ (3).


Substitute (3) in (1), we get

2x + 3(200) = 1200

2x + 600 = 1200

2x = 600

x = 300.


Therefore, Number of bananas in Lot A = 300

                   The number of bananas in Lot B = 200.

Total Number of bananas = 300 + 200 = 500.


The total number of bananas he had = 500.


Hope this helps!

Answer 2

this is the way to solve