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A Trussian's weight, in keils, can be calculated by taking the square
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03 Nov 2014, 09:09
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Re: A Trussian's weight, in keils, can be calculated by taking the square
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03 Nov 2014, 09:12



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Re: A Trussian's weight, in keils, can be calculated by taking the square
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03 Nov 2014, 14:18
Bunuel wrote: Tough and Tricky questions: Word Problems. 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 Kudos for a correct solution.\(\sqrt{age} + 3 = \sqrt{2*age + 17}\) 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
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Re: A Trussian's weight, in keils, can be calculated by taking the square
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03 Nov 2014, 21:44
Bunuel wrote: Tough and Tricky questions: Word Problems. 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 Kudos for a correct solution.Age ..................... Weight x ............................ \(\sqrt{x}\) ..................... Current age/weight 2x+17 ......................\(\sqrt{2x+17}\) ................ (Seventeen years after he is twice as old as he is now age/weight) \(\sqrt{2x+17}  3 = \sqrt{x}\) Only 16 is a perfect square which fits in Answer = C
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Re: A Trussian's weight, in keils, can be calculated by taking the square
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21 Dec 2015, 03:54
I solved this one back solving, Found it was easier by this way.
Let x be the age in years
X = 2x + 17
For back solving I like to start from the middle one C) (16 x 2) + 17 = 49 (which brings an integer of 7, can be correct, lets check the others) D) 51 (it will increase by 2 always) E) 53 B) 47 A) 45
C is the only answer which brings an integer, therefore its the right answer



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Re: A Trussian's weight, in keils, can be calculated by taking the square
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30 Dec 2015, 08:57
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 \(a6\sqrt{a}+8=0\) To make it look more familiar we can substitute \(\sqrt{a}=x\), now the equation looks like: \(x^26x+8=0\) Factor it \((x2)(x4)=0\) \(x=2\) 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
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A Trussian's weight, in keils, can be calculated by taking the square
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30 Mar 2017, 19:13
Bunuel wrote: Tough and Tricky questions: Word Problems. 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 Kudos for a correct solution.Dear Bunuel, Let us call the Trussian's current age a. Why my equation incorrect? \(173\sqrt{a}=2a\)
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Re: A Trussian's weight, in keils, can be calculated by taking the square
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31 Mar 2017, 04:58
ziyuen wrote: Bunuel wrote: Tough and Tricky questions: Word Problems. 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 Kudos for a correct solution.Dear Bunuel, Let us call the Trussian's current age a. Why my equation incorrect? \(173\sqrt{a}=2a\) This is how you form your equation: \(Weight = \sqrt{Age}\) We need to find his current age so let's say it is y years. Then, his current weight is \(\sqrt{y}\) "seventeen years after he is twice as old as now" is \((2y + 17)\). At that time his weight will be \(\sqrt{(2y + 17)}\) \(\sqrt{y} + 3 = \sqrt{(2y + 17)}\) Try out the options now to see which value satisfies the equation. y = 16 does. Answer (C)
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A Trussian's weight, in keils, can be calculated by taking the square
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16 Dec 2017, 07:23
VeritasPrepKarishma wrote: ziyuen wrote: Bunuel wrote: Tough and Tricky questions: Word Problems. 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 Kudos for a correct solution.Dear Bunuel, Let us call the Trussian's current age a. Why my equation incorrect? \(173\sqrt{a}=2a\) This is how you form your equation: \(Weight = \sqrt{Age}\) We need to find his current age so let's say it is y years. Then, his current weight is \(\sqrt{y}\) "seventeen years after he is twice as old as now" is \((2y + 17)\). At that time his weight will be \(\sqrt{(2y + 17)}\) \(\sqrt{y} + 3 = \sqrt{(2y + 17)}\) Try out the options now to see which value satisfies the equation. y = 16 does. Answer (C) Dear Karishma, Is there any way to solve this equation algebraically ( A is the present age ) \(\sqrt{2A +17} \sqrt{A}=3\) I was thinking along the lines of \(\sqrt{2A +17}\) =\(\sqrt{2A}\) +\(\sqrt{17}\)obviously this is NOT true. Square root is not distributive over addition.Is there any other way for simplification ? Thank you.
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Re: A Trussian's weight, in keils, can be calculated by taking the square
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16 Dec 2017, 07:52



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Re: A Trussian's weight, in keils, can be calculated by taking the square
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16 Dec 2017, 07:58
Bunuel wrote: stne wrote: Dear Karishma,
Is there any way to solve this equation algebraically ( A is the present age ) \(\sqrt{2A +17} \sqrt{A}=3\)
I was thinking along the lines of \(\sqrt{2A +17}\) =\(\sqrt{2A}\) +\(\sqrt{17}\)obviously this is NOT true. Square root is not distributive over addition.
Is there any other way for simplification ? Thank you. Algebraic way is given 3 posts above: https://gmatclub.com/forum/atrussians ... l#p1624365Thank you Sir.
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