A farmer sold a calf and a cow for Rs. 760 Thereby making a profit of 25% on the calf and 10% on the cow. By selling them for Rs. 767.5 he would have raised a profit of 10% on the calf and 25% on the cow. Find the cost of each.
[Dont convert the numbers into decimals.]
Answers
Answer :
Let us assume that the C.P of a calf is Rs. x
Also, let the C.P of cow be Rs. y.
Now,
(100 + 25x / 100) + (100 + 10y/ 100) = 760
.°. 125x/100 + 110y/100 = 767.5 .. (1)
Now, As per the next condition,
(100+10x/100) = (100+25y/100)
.°. 110x/100 = 125y/100.. (2)
Now, Let's add the eq. (1) and (2)
125x/100 + 110y/100 = 767.5
+ 110x/100 + 125y/100 = 760
__________________________
=> 235x/100 + 235y/100 = 1527.5
Dividing both the sides by 235/100, we get :
x + y = 650... (3)
Similarly, Subtract this 2 eq.
110x/100 + 125y/100 = 760
- 125x/100 + 110y/100 = 767.5
___________________________
=> 15x/100 - 15y/100 = - 7.5
Dividing both the sides by 15/100, we get :
x - y = - 5 .. (4)
Now,
We got eq. (3) and (4)
Adding eq. (3) and (4),
x + y = 650
+ x - y = -5
______________
=> 2x = 645
x = 322.5
Substituting value of x in eq. (3)
327.5 + y = 650
y = 650 - 322.5
y = 327.5
Thus,
Cost of the calf is Rs. 322.5
Cost of the cow is Rs. 327.5
___________________________
Step-by-step explanation:
l S.P. = (1.25y + 1.1x)
Given that the selling price is 760.
So, (1.25y + 1.1x) = 760
⇒1.1x = 760 - 1.25y
⇒x = 760/1.1 - 1.25y/1.1 ...(1)
S.P. of cow after 25% profit = (100+profit)% of CP
= (100+25)% of x
= 1.25x
S.P. of calf after 10% profit
= 110% of y
= 1.1y
Total S.P. = (1.25x + 1.1y)
Given that the selling price is 767.50.
So, (1.25x + 1.1y) = 767.50 ...(2)
Putting the value of 'x' from (1) in (2), we get:
1.25(760/1.1 - 1.25y/1.1) + 1.1y = 767.50
⇒ 863.64 - 1.42y + 1.1y = 767.50
⇒ 1.42y - 1.1y = 863.64 - 767.50
⇒ 0.32y = 96.14
⇒ y = 96.14/0.32 ≈ 300
Putting the value of y in eq.(2), we get:
1.25x + 1.1(300) = 767.50
⇒ 1.25x = 767.50 - 330 = 437.50
⇒ x = 437.50/1.25 = 350
x = Rs 350
y = Rs 300