Question

A farmer sold a cow and a calf for Rs. 12750 thereby making a profit of 25% on
the cow and 10% on the calf. By selling them for Rs. 11925, he would have realised
a profit of 10% on the cow and 25% on the calf. Find the cost price of the cow and
the calf.​

Answer 1

Step-by-step explanation:

Let C.P. of a calf be Rs 'x' and C.P. of a cow be Rs 'y'.

Now according to the question;

125x/100 + 110y/100 = 760 ..............(1)

110x/100 + 125y/100 = 767.50 ..............(2)

Now by solving (1) and (2), we get,

x = 300 and y = 350

therefore,

the cost of calf is Rs 300.

the cost of cow is Rs 350. Like? Yes (19) | No (3) N

Answer 2

Answer:

Let ' x' and y' be the cost price of the cow and the calf respectively.

Step-by-step explanation:

By questions , we have

(x+25% of x) + (y +10% of y ) = 12750.

( x+25| 100× x ) + ( y+10|100× y = 12750

x+ 25x| 100 + y + 10y |100 = 12750

100x + 25x + 100y + 10 y | 100 = 12750

125x + 110y = 12750× 100

125x + 110y = 1275000 ( equation 1)

And ,

( x + 10% of x ) + ( y + 25% of y ) = 11925

(x + 10|100 × x ) + ( y + 25|100 × y ) = 11925

x + 10x | 100 + y + 25y|100 = 11925

100x + 10x + 100y + 25y|100 = 11925

110x + 125y = 11925× 100

110x + 125y = 1192500 ( equation 2 )

From eqn. 2 ,

110x + 125y = 1192500

110x = 1192500 - 125y

x = 1192500 - 125y | 110

Sustituting the value of x = 1192500 - 125y | 110 in eqn . 1 , we get

125 × 1192500 - 125y | 110 + 110y = 1275000

29812500 - 3125y | 22 +110y = 1275000

29812500 - 3152y + 2420y | 22 = 1275000

29812500 - 705y = 1275000 × 22

29812500 - 705y = 28050000

- 705y = 28050000 - 29812500

- 705y = - 1762500

y = - 1762500 | - 705

y = 2500

Substituting the value of y = 2500 in eqn. 1

125x + 110 × 2500 = 1275000

125x + 275000 = 1275000

125x = 1275000 - 275000

125x = 1000000

x = 1000000 | 125

x = 8000

Hence the cost price of the cow and the calf are rupees 8000 and rupees 2500.