Question

8. A man buys 11 9. A fruit vendor buys oranges at 6 for 10 and an equal number at 10 for 20. He mixes them and sells at 21.60 per dozen. Find his gain or loss per cent. man sold a camera for 3 1710 losing 5%. At what price should he sell the camera in order to​

Answer 1

Answer:

Consider the two types of oranges separately and find the cost price of six type 2 orange using the unitary method. Now, add the cost price of six oranges of type 1 and six oranges of type 2 to get the total cost price of a dozen mixtures of two types of oranges. If selling price is greater than cost price then apply the formula gain % = S.P−C.PC.P×100%S.P−C.PC.P×100% and if cost price is more then apply the formula loss % = 

C.P−S.PC.P×100%C.P−S.PC.P×100%

. Here C.P = Cost Price and S.P = Selling Price.

Complete step-by-step solution:

Now, here we have been given that the cost price of six type 1 oranges is Rs. 10 and that of 10 type two oranges is Rs. 20. The selling price of the mixture of the two types of oranges is Rs. 21.60 per dozen. So we consider that 6 oranges of both types are present in the mixture. Let us find the total C.P of 6 oranges of both types.

(1) For type 1 orange it is already given that the cost price of 6 oranges is Rs. 10.

(2) For type 2 oranges it is given that the cost price of 10 oranges is Rs. 20. So applying the unitary method we have,

⇒⇒ Cost price of 6 oranges of type 2 = Rs. 2010×62010×6

⇒⇒ Cost price of 6 oranges of type 2 = Rs. 12

Therefore, if we mix 6 oranges of both types to find the total cost price of a mixture of dozen oranges we get,

⇒⇒ Total cost price of mixture of oranges = Rs. (10 + 12) = Rs. 22

Clearly we can see that the Cost price of the mixture is more that the Selling price of the mixture so the fruit vendor will suffer loss given as loss % = 

C.P−S.PC.P×100%C.P−S.PC.P×100%

, where C.P = Cost Price and S.P = Selling Price. Therefore we get,

⇒⇒ Loss % = 22−21.6022×100%22−21.6022×100%

∴∴ Loss % = 1.818 %

Hence, the fruit vendor suffers a loss of 1.818 %.