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22 Jun 2014, 11:12
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D
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Difficulty:
5%
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Question Stats:
93% (01:20) correct
7%
(02:16)
wrong
based on 81
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An old man distributed all the gold coins he had to his two sons into two different numbers such that the difference between the squares of the two numbers is 36 times the difference between the two numbers. How many coins did the old man have?
(A) 24
(B) 26
(C) 30
(D) 36
(E) 40
Source: Math Bible Nova
(A) 24
(B) 26
(C) 30
(D) 36
(E) 40
Source: Math Bible Nova
22 Jun 2014, 11:27
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An old man distributed all the gold coins he had to his two sons into two different numbers such that the difference between the squares of the two numbers is 36 times the difference between the two numbers. How many coins did the old man have?
(A) 24
(B) 26
(C) 30
(D) 36
(E) 40
Source: Math Bible Nova
(A) 24
(B) 26
(C) 30
(D) 36
(E) 40
Source: Math Bible Nova
Let the number of coins one son got be x and the number of coins another got be y. Total = x + y.
x^2 - y^2 = 36(x - y) --> x + y = 36.
Answer: D.
15 Jul 2014, 22:28
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Let the coins distributed be x & y
We require to find x+y
\(x^2 - y^2 = 36 (x-y)\)
x+y = 36
Answer = D
We require to find x+y
\(x^2 - y^2 = 36 (x-y)\)
x+y = 36
Answer = D
