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| Filter | Status | Condition | Details |
|---|---|---|---|
| HTTP status | PASS | download_http_code = 200 | HTTP 200 |
| Age cutoff | PASS | download_stamp > now() - 6 MONTH | 2.6 months ago |
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| Property | Value |
|---|---|
| URL | http://mathcentral.uregina.ca/QQ/database/QQ.09.07/h/marina3.html |
| Last Crawled | 2026-01-25 15:48:07 (2 months ago) |
| First Indexed | 2018-09-09 18:26:04 (7 years ago) |
| HTTP Status Code | 200 |
| Meta Title | Cows, chickens, heads and feet - Math Central |
| Meta Description | null |
| Meta Canonical | null |
| Boilerpipe Text | Marina,
For years I taught a mathematics course to students who were studying to be teachers. This problem is one I used very often as an example of a problem which is much easier solved without algebra than it is with algebra. It even bothers me to write down an algebraic approach as it clashes with my sense of the aesthetics in mathematics, but here goes.
Let
c
be the number of cows and
h
the number of chickens (hens). There are 65 heads so
c
+
h
= 65.
Each cow has 4 legs, each hen has 2 legs and there are 226 legs in total so
4
c
+ 2
h
= 226.
Multiply both sides of the first equation by 2 and subtract it from the second equation to get
2
c
= 96 and thus
c
= 48.
From the first equation it then follows that
h
= 65 - 48 = 17.
Harley |
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|  Question from marina, a parent: I am helping my 6th grader son solve this problem. I found similar problem in the Q\&A. I understood the simple answer without using an equation. I want to solve this using the equation.This is the question about chicken and cows. There are cows and chickens in the farm. There are 65 heads and 226 legs. How many chicken are there? Thank you. marina | | |
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| Marina, For years I taught a mathematics course to students who were studying to be teachers. This problem is one I used very often as an example of a problem which is much easier solved without algebra than it is with algebra. It even bothers me to write down an algebraic approach as it clashes with my sense of the aesthetics in mathematics, but here goes. Let **c** be the number of cows and **h** the number of chickens (hens). There are 65 heads so **c** + **h** = 65. Each cow has 4 legs, each hen has 2 legs and there are 226 legs in total so 4**c** + 2**h** = 226. Multiply both sides of the first equation by 2 and subtract it from the second equation to get 2**c** = 96 and thus **c** = 48. From the first equation it then follows that **h** = 65 - 48 = 17. Harley | | |
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| Unparsed URL | ca,uregina!mathcentral,/QQ/database/QQ.09.07/h/marina3.html h80 |