Harvard Admission Exam circa 1869

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Algebra

Problem 1. Reduce the following expression to its simplest form:

(9⁢a2⁢b2-4⁢b4)⁢(a2-b2)-(3⁢a⁢b-2⁢b2)⋅(3⁢a⁢[a2+b2]-2⁢b⁢[b2+3⁢a⁢b-a2])⁢b open paren 9 a squared b squared minus 4 b to the fourth power close paren times open paren a squared minus b squared close paren minus open paren 3 a b minus 2 b squared close paren times open paren 3 a times open bracket a squared plus b squared close bracket minus 2 b times open bracket b squared plus 3 a b minus a squared close bracket close paren times b

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Problem 2. Divide 36⁢a2+1-64⁢a4-12⁢a 36 a squared plus 1 minus 64 a to the fourth power minus 12 a by 6⁢a-1-8⁢a2. 6 a minus 1 minus 8 a raised to the 2 period power

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Problem 3. Explain why when different powers of the same quantity are multiplied together their exponents are added.

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Problem 4. Reduce to one fraction with the lowest possible denominator:

3⁢a+2⁢ba+b-25⁢a2-b2a2-b2=a2⁢b the fraction with numerator 3 a plus 2 b and denominator a plus b minus the fraction with numerator 25 a squared minus b squared and denominator a squared minus b squared equals a over 2 b

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Problem 5. Divide x+y x plus y by x2+x⁢yx-y the fraction with numerator x squared plus x y and denominator x minus y and reduce the answer to its lowest terms.

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Problem 6. Solve for x x in terms of a a, b b, and c c from

a-2⁢xb=c⁢x-b⁢ca the fraction with numerator a minus 2 x and denominator b equals the fraction with numerator c x minus b c and denominator a ,

and then find x x when a=2 a equals 2, b=-1 b equals negative 1, c=3 c equals 3

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Problem 7. A man bought a watch, a chain, and a locket for $216. 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. Find the price of each item.

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Problem 8. Solve the equation 5⁢xx+12-8-3⁢x3⁢x-1=1. the fraction with numerator 5 x and denominator x plus 12 minus the fraction with numerator 8 minus 3 x and denominator 3 x minus 1 equals 1 period .

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Problem 9. Find (a-b)6 open paren a minus b close paren to the sixth power and (x⁢y-x22⁢y)6 open paren x y minus the fraction with numerator x squared and denominator 2 y close paren to the sixth power by the Binomial Theorem.

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