1 cost 50

other cost 150

let cost of cow A x;

cost of cow B y;

given x+y 210;.........................................1

10 profit on A means he got x/10;

10 loss on B means he got -y/10;

given total profit 5 ;

therefore x/10-y/10 5 of(210);

>x-y 105;...............................................2

solving 1 & 2

we get x 157.2 y 52.8;

> cost of cow A 157.2 & cost of B 52.8

actual price-> A B

Selling price->Sa Sb

Sa+Sb 210

combined profit-> 5

So A+B 200

Substitute for Sa+Sb 210

[(200-B)+(200-B)10/100 ] +

[(200-A)-(200-A)10/100 ] 210

gets A 150

so B 50

Original Price of the cows: (a + b) 210 - (5 of 210)

a + b 210 - 10.5

a + b 199.5 -------> (Eq.1)

a 199.5 - b & b 199.5 - a

profit on a 10 ie (199.5 - b)/10

loss on b 10 ie (199.5 - a)/10

altogether former made a profit of 5

so [(199.5 - b)/10] - [(199.5 - a)/10] 10.5

199.5 - b - 199.5 + a 10 * 10.5

a - b 105 -----> (Eq.2)

solve Eqs (1) & (2)

a + b 199.5

a - b 105 Adding (1) & (2)

2a 304.5 > a 152.25

subs in any one eq u'll get b 47.25 so (152.25 + 15.225 (10 profit) + 47.25 - 4.725(10 loss) 210