Old McDonald divide his hens evenly between his two...

[AAPL]

Old McDonald divided his hens between his two daughters, Pamela and Paris. After one week, during which the flock of Pamela had grown by 60 hens and the flock of Paris had grown by 10 hens, both daughters sold all of their hens. Pamela sold each hen for 80 cents and Paris sold each hen for 30 cents. How many hens did old McDonald originally have, if Pamela's revenue from selling her flock was four times as much as Paris revenue?

A. 160
B. 180
C. 200
D. 220
E. 240

The OA is B.

Pamela = x hens.
Paris = x hens.

after the increase,

Pamela = x+60 and Paris = x+10, right?

Then, Pamela revenue is four times Paris revenue, hence
$$(x+60)80=4(x+10)30$$
Solving for x we have, x=90, then old McDonald had 180 hens.

Is there a strategic approach to this question? Can any experts help?

[Jeff@TargetTestPrep]

Old McDonald divided his hens between his two daughters, Pamela and Paris. After one week, during which the flock of Pamela had grown by 60 hens and the flock of Paris had grown by 10 hens, both daughters sold all of their hens. Pamela sold each hen for 80 cents and Paris sold each hen for 30 cents. How many hens did old McDonald originally have, if Pamela's revenue from selling her flock was four times as much as Paris revenue?

A. 160
B. 180
C. 200
D. 220
E. 240

We can let the total number of hens = n and create the equation:

80(n/2 + 60) = 4[30(n/2 + 10)]

40n + 4800 = 60n + 1200

3600 = 20n

180 = n

Answer: B