slingfox
If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned, but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. If no chickens are sold or bought, the farmer will be exactly on schedule. How many chickens does the farmer have?
A. 60
B. 120
C. 240
D. 275
E. 300
This question is similar to work-rate questions. The
key is always
calculate how much work each person/machine does in 1 unit of time.
For this question, we have:
number of chickens = X
stock feed = T (days)
It means X chickens can be fed in T days -->
1 chicken eats in 1 day = 1/(XT)1st scenario, we sell 75 chickens, we havenumber of chickens = X - 75
stock feed = T + 20 (days)
--> 1 chicken eats 1 day = 1/[(X-75)(T+20)]
Because the amount of food each chicken eats in 1 day is the same:--> 1/XT = 1/[(X-75)(T+20)]
--> XT = XT+ 20X - 75T - 1500
-->
20X - 75T - 1500 = 02nd scenario: we buy 100 chickensnumber of chickens = X + 100
stock feed = T - 15 (days)
--> 1 chicken eats 1 day = 1/[(X+100)(T-15)]
Because the amount of food each chicken eats in 1 day is the same:--> 1/XT = 1/[(X+100)(T-15)]
--> XT = XT -15X +100T - 1500
-->
15X +100T - 1500 = 0Solve 2 equations
20X - 75T - 1500 = 0
15X +100T - 1500 = 0Clearly, X = 300
Hence, E is correct.
We have