he buys 2 cows for 200 Rs. That is one cow costs 100Rs. He sells the 2 cows for 210Rs. But he sells one cow for 10 profit ie 110Rs. Hence he sells the other cow for 100Rs.

That means he get niether loss/profit on the other cow.

Question: One man Buy two cows for Rs.200.He sold the cows for Rs.210.....

Answer neither profit nor loss is wrong approach

Here is the answer

Let the Cost price of Cow1 : x

So Cost price of Cow2 is :(200-x)

Selling price of Cow1: 110 of x ------>eq1

Selling price of Cow2: 90 of (200-x) ------>eq2

eq1 +eq2 210

(110/100)*x + (90/100)*(200-x) 210

Solve for x U will find x 150

SO cost price of Cow1 is: Rs 150

There fore Cost price of COw2 is :Rs 50

Question: One man Buy two cows for Rs.200.He sold the cows for Rs.210.....

Answer neither profit nor loss is wrong approach

Here is the answer

Let the Cost price of Cow1 : x

So Cost price of Cow2 is :(200-x)

Selling price of Cow1: 110 of x ------>eq1

Selling price of Cow2: 90 of (200-x) ------>eq2

eq1 +eq2 210

(110/100)*x + (90/100)*(200-x) 210

Solve for x U will find x 150

SO cost price of Cow1 is: 150

There fore Cost price of COw2 is :50

suppose cost of the cows are x and y respectively.then

x+y 200-----eq1

man sold one cow with 10 profit. then

x+x/10 11x/10

another cow with 10 loss. then

y-y/10 9y/10

He sold the cows for Rs.210.So

11x/10+9y/10 210------eq2

solving eq1 and eq2 we get

x 150 and y 50

So cost of the cows while buying are 150 and 50. While he sold the two cows on a cost of 165 and 45 respectively.

profit equation

10 110/100 x - 90/100y

cost price eq

200 x+y

x 95 and y 105

cost price of two cows x+y 200

if 10 profit for 100 then for x x/10

similarly for y y/10

(x/10) - (y/10) 10

solving x 150

y 50

one he bought for 50

other for 150

first he sold for 45

second for 165