Arithmetic problem

[philomina]

Please help me in finding answer for the following as early as possible as my test on the same is in two days time. I am totally confused.


A farmer has free range chickens. They are all over his farm and he has no easy way of counting them. However, he has devised a clever system. One day he rounded up 50 chickens, put metal rings around their legs and let them go. The next day he rounded up 50 chickens and found that 6 of them had rings on their legs. Assuming that there was no change in the number of chickens between the two days, what was his estimate of the number of chickens on the farm. Give your answer to the nearest whole chicken.
The options in the answer are-
300
367
417
1500
2200

 

[Loren]

Please help me in finding answer for the following as early as possible as my test on the same is in two days time. I am totally confused.


A farmer has free range chickens. They are all over his farm and he has no easy way of counting them. However, he has devised a clever system. One day he rounded up 50 chickens, put metal rings around their legs and let them go. The next day he rounded up 50 chickens and found that 6 of them had rings on their legs. Assuming that there was no change in the number of chickens between the two days, what was his estimate of the number of chickens on the farm. Give your answer to the nearest whole chicken.
The options in the answer are-
300
367
417
1500
2200

On the second round-up 6 out of 50 had leg bands. Six out of 50 is 12%. The assumption is that this second round-up represents the same ratio of banded to non-banded as the first round-up. On the first round up, the farmer put bands on the chickens, therefore, 12% of the total population is 50.

Some might set the problem up as a proportion...
\(\displaystyle \frac{total.popultion}{no.in.first.capture}=\frac{no.in.2nd.capture}{no.of.banded.in.2nd.capture}\)