Question

A man purchased two items A & B and invested Rs 50 & Rs 75 on their repairing respectively. If he earns profit of 10% on A and 12% on B, overall profit earned by him is is Rs 84. But if he earns 20% on A & 10% on B, overall profit earned by him is 14% of total price of items. Find initial total purchasing price of both items.

Answer 1

Step-by-step explanation:

ask me if any doubt..........

Answer 2

The initial purchasing price for both the items are Rs. 250 and Rs. 375 respectively.

Given:

Amount invested on item A= Rs. 50

Amount invested on item B= Rs. 75

Profit earned on A and B= 10% and 12% respectively

Total profit in this case= Rs. 84

Profit earned on A and B in another case= 20% and 10% respectively

Total profit in this case= 14% of total price

To Find:

Initial total purchasing price of both items.

Solution:

Let the price of item A be Rs. x and price of item B be Rs. y.

Case-1

Price of A including repairs= 50 + x

Price of B including repairs= 75 + y

Total profit earned= $\frac{10}{100} (50+x)+\frac{12}{100} (75+y)$

It is given that total profit in this case was Rs. 84. So,

$\frac{10}{100} (50+x)+\frac{12}{100} (75+y)=84\\or, 5+\frac{10x}{100} +9+\frac{12y}{100} =84\\or, \frac{10x+ 12y}{100} =70\\or, 10x+12y= 7000...............................(eq.i)$

Case-2

Total profit earned= $\frac{20}{100} (50+x)+\frac{10}{100} (75+y)$

It is given that total profit in this case was 14% of total price.

Total price = (50+x)+(75+y)= 125+x+y

So,

$\frac{20}{100} (50+x)+\frac{10}{100} (75+y)= \frac{14}{100} (125+x+y)\\or, 10+ \frac{20x}{100} +7.5+\frac{10y}{100} = 17.5 + \frac{14x}{100} + \frac{14y}{100} \\or, \frac{6x}{100} = \frac{4y}{100} \\or,3x=2y\\or, x= \frac{2}{3} y...........................(eq.ii)$

Putting value of x from eq. ii in eq.i, we get,

$10x+18x=7000\\or, x= \frac{7000}{28} = 250\\\\\\So, y=\frac{3}{2}X250= 375$

Hence, the initial cost price of A and B were Rs. 250 and Rs. 375 respectively.

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