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03 Nov 2014, 09:09
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C
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Difficulty:
55%
(hard)
Question Stats:
71% (02:53) correct
29%
(02:51)
wrong
based on 410
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A Trussian's weight, in keils, can be calculated by taking the square root of his age in years. A Trussian teenager now weighs three keils less than he will seventeen years after he is twice as old as he is now. How old is he now?
(A) 14
(B) 15
(C) 16
(D) 17
(E) 18
(A) 14
(B) 15
(C) 16
(D) 17
(E) 18
30 Dec 2015, 08:57
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As shown above, backsolving can work for this problem. For similar problems where the numbers might not be as friendly to work with, we can also follow an algebraic approach.
Given:
Current weight is 3 less than what his weight will be when his age is doubled plus 17
Let \(a\) = current age
then current weight is \(\sqrt{a}\)
\(\sqrt{a}+3=\sqrt{2a+17}\)
Square both sides
\(a+6\sqrt{a}+9=2a+17\)
Gather terms on one side
\(a-6\sqrt{a}+8=0\)
To make it look more familiar we can substitute \(\sqrt{a}=x\), now the equation looks like:
\(x^2-6x+8=0\)
Factor it
\((x-2)(x-4)=0\)
\(x=2\) or \(x=4\)
Substituting \(a\) back in \((x=\sqrt{a})\)
\(\sqrt{a}=2\) or \(\sqrt{a}=4\)
\(a=4\) or \(a=16\)
Looking at the answer choices, only 16 is there.
Answer C
Given:
Current weight is 3 less than what his weight will be when his age is doubled plus 17
Let \(a\) = current age
then current weight is \(\sqrt{a}\)
\(\sqrt{a}+3=\sqrt{2a+17}\)
Square both sides
\(a+6\sqrt{a}+9=2a+17\)
Gather terms on one side
\(a-6\sqrt{a}+8=0\)
To make it look more familiar we can substitute \(\sqrt{a}=x\), now the equation looks like:
\(x^2-6x+8=0\)
Factor it
\((x-2)(x-4)=0\)
\(x=2\) or \(x=4\)
Substituting \(a\) back in \((x=\sqrt{a})\)
\(\sqrt{a}=2\) or \(\sqrt{a}=4\)
\(a=4\) or \(a=16\)
Looking at the answer choices, only 16 is there.
Answer C
03 Nov 2014, 14:18
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Bunuel
\(\sqrt{age} + 3 = \sqrt{2*age + 17}\)
Start plugging in numbers, answer choice C would give an integer age. Lets try C
\(\sqrt{16} + 3 = \sqrt{2*16 + 17}\)
\(4 + 3 = \sqrt{49}\)
\(7 = 7\)
Correct answer is C
