A man sold two cows for Rs. 210 at a total profit of 5 %. He sold one cow at a loss of 10% and another at a p?

Question

A man sold two cows for Rs. 210 at a total profit of 5 %. He sold one cow at a loss of 10% and another at a profit of 10%. What is the price of each cow? please show the method too.

Answer 1

let first cow's price =x, and other cow's price =y

x+y+(x+y)*5/100 =x+y+0.05x+0.05y=210

1.05x+1.05y =210

1.05(x+y) =210

x+y =210/1.05 =200

also

(x-(x*10/100)) +(y+ (y*10/100)) =210

(x- x/10)+(y+y/10) =210

x+y +(y/10-x/10) =210

200+1/10(y-x) =210

1/10(y-x) =210-200=10

y-x =100

y+x =200

add both equations

2y =300

y =300/2 =150

x =y-100 =150-100 =50

first cow's price =Rs 50, and other cow's price =Rs.150

Answer 2

Let the cost price of cows are : x and y. From given conditions we have following equations:

(x + y) 105/100 = 210 On simplification x + y = 200 --- (1)

0.90 x + 1.10 y = 210 --- (2)

solving equations (1) and (2) we get y = 150 Rs, and x = 50 Rs.

Answer 3

A man sold two cows for Rs. 210 at a total profit of 5 %.

The cost price of the two cows is Rs. 200

He sold one cow at a loss of 10% and another at a profit of 10%.

Let the price of one cow be x and that of another cow y.

0.9x + 1.1y = Rs. 210

x + y = Rs. 200

0.9x + 0.9y = Rs.180

0.2y = Rs. 30

y = Rs. 150

x = Rs. 50

The price of each cow

is Rs. 150 and Rs. 50

Answer 4

Answer:

Suppose first cow's cost price is x rupees n 2nd cow's cost price is y rupees

so 5 % profit on both cows is 210 rupees then both cows cost price is 200 rupees

then first equation is x+y=200 (i)

on 10% loss first cow's sell price is 9x/10 and 2nd cows sell price on 10% profit is 11y/10

so second equation is 9x/10+11y/10=210 which is comes like this 9x+11y=2100 (ii)

now 9(x+y)=9*200 becomes 9x+9y=1800, deduct this equation from equation (ii) you can

get 2y=300, y=150 then x will be 50

so first cow's cost price is 50 rupees and selling price is 45 rupees and second cow's cost price is

150 rupees and selling price is 165 rupees

Answer 5

Define our variables:

Ps = selling price for the two cows

Ps1 = selling price for cow #1

Ps2 = selling price for cow #2

C1 = cost of raising cow #1 (i.e., cost of food, medicine, shelter, water, etc)

C2 = cost of rasing cow #2

F1 = profit from selling cow #1

F2 = profit from selling cow #2



F1 = Ps1 – C1

F2 = Ps2 – C2

Adding gives Equation 1

F1 + F2 = Ps1 + Ps2 – C1 – C2

We are given that the total profit F1 + F2 is 5 %. This gives Equation 2:

F1 + F2 = 0.05 (C1 + C2)

Substituting Eqn. 2 into Eqn 1 gives Equation 3:

0.05C1 + 0.05C2 = Ps1 + Ps2 – C1 – C2

We are given that there was a 10% loss from the sale of cow #1. This gives Equation 4:

Ps1 = 0.90C1

We are given that there was a 10% profit from the sale of cow #1. This gives Equation 5:

Ps2 = 1.10C2

Substituting Eqn. 4 & Eqn. 5 into Eqn 3 gives Equation 6:

0.05C1 + 0.05C2 = 0.90C1 + 1.10C2 – C1 – C2

Adding like terms gives Equation 7:

0 = 0.15C1 + 0.05C2

Solving for C1 gives Equation 8:

C1 = (C2) / 3

We are given that the total selling price is 210. This gives Equation 9:

210 = Ps1 + Ps2 = 0.90C1 + 1.1C2

Substituting Eqn. 8 into Eqn. 9 gives

210 = ( 0.9 (C2) / 3 ) ) + 1.1C2

Solving for C2 gives C2 = 150.

Substituting this into Eqn. 8 gives C1 = 50.

Substituting these values of C1 and C2 into Eqn. 4 and Eqn 5 gives

Ps1 = 0.90 (50)

Ps2 = 1.1(150)

Answer:

Ps1 = Rs45

Ps2 = Rs165



I am assuming Rs is some sort of unit of currency.

Answer 6

105(x+y)/100=210,

110x/100+90y/100=210



105(x+y)/100=110x/100+90y/100

105x/100+105y/100=110x/100+90y/100

105x+105y=110x+90y

105y-90y=110x-105x

15y=5x

15y/5=x

x=3y



105(3y+y)/100=210

105(4y)/100=210

420y=21000

y=21000/420

y=50



110x/100+90y/100=210

110x/100+90(50)/100=210

110x/100+4500/100=210

110x+4500=21000

110x=21000-4500

110x=16500

x=16500/110

x=150



the man sold a cow (with original price Rs. 150) in Rs. 165

and sold a cow (with original price Rs. 50) in Rs. 45

Rs. 150 + Rs. 50 = Rs. 200

Rs. 165 + Rs. 45 = Rs. 210

the profit is 5 %

Answer 7

selling price of two cows = Rs 210, profit = 5%

cost price of two cows = Rs(210*100/105) =Rs 200

let cost of one cow =Rs x

and cost of other cow =Rs (200 -x)

loss of first cow = 10%

sellin price of first cow =Rs (x*90/100) =Rs 9x /10-------------------(i)

profit of second cow = 10%

selling price of second cow =Rs[.(200-x)*110/100] =Rs(2200-11x)/10-----------(ii)

Atq

9x/10 +(2200-11x)/10 = Rs210

9x +2200-11x =Rs 2100

-2x = Rs 2100-Rs 2200

-2x = Rs-100

x = Rs 50

cost price of one cow = Rs 50

cost price of other cow =Rs 200-Rs50 =Rs 150

Answer 8

MY COWS ARE MY COWS. Try to take them or their milk and I will shoot you dead with the gun I will not give up either.