Solve If You Are A Genius (From 1869). Harvard Admissions Question Algebra Problem

The Harvard admissions exam in 1869 included this algebra problem: “A man bought a watch, a chain, and a locket for 6. The watch and locket together cost three times as much as the chain, and the chain and locket together cost half as much as the watch. What was the price of each?” Can you solve it? The video presents a solution.

My blog post for this video
http://wp.me/p6aMk-4Vi

The Harvard 1869 Test
http://graphics8.nytimes.com/packages/pdf/education/harvardexam.pdf

Via New York Times
http://thechoice.blogs.nytimes.com/2011/03/31/remembering-when-college-was-a-buyers-bazaar/?_r=0

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19 thoughts on “Solve If You Are A Genius (From 1869). Harvard Admissions Question Algebra Problem

  1. Let w= watch , c =chain and L= locket,
    (1)w + L + c = 216 , total cost of all three items
    (2) w+ L = 3 c , the watch and locket together cost three times as much as the chain
    (3) w +c = w/2 , the watch and chain cost half as much as the watch
    substituting equation (2) into equation (1) gives : 3 c + c= 216 or 4 c= 216 or c=54
    substituting equation (3) into equation (1) gives : w + w/2= 216 or 3/2 w = 216 or w = 144
    substituting c and w into equation (1) to find L gives: L = 216-144 – 54 or 216-198 or L= 18
    cost of all three items : watch $ 144
    chain $ 54
    locket $ 18

  2. in my opinion, the question was way clearer when it was normally asked in writing. after i read the question i solved the problem and watched the explanation. i got the amswers right but couldent follow the explanation

  3. I'm drunk and stoned. And I needed 2 minutes to get my head right. isn't it much easier this way. Part one. 4*A=216 A=54. Chain. Part 2 (2+1)*X=216 X=72. So 2X =144 Watch. And X-A=18 Locket. :-)

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