One man Buy two cows for Rs.200.He sold the cows for Rs.210.One he sold for 10% profit and another he sold for 10% loss.Then what is the cost price of each cow?

90 rs,--gives 10 perc loss
110 rs, --gives 10% profit
total=210
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soln

  • May 26th, 2006
 

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.

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yusufqtv

  • May 31st, 2006
 

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

yusufqtv

  • May 31st, 2006
 

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

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Sarvesh Kumar

  • Jun 1st, 2006
 

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.

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Star

  • Jun 1st, 2006
 

profit equation

10 = 110/100 x - 90/100y

cost price eq

200=x+y

x=95 and y =105

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grace

  • Jun 30th, 2006
 

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

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diju

  • Oct 18th, 2006
 

one he bought for 50

other for 150

first he sold for 45

second for 165

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