A Trussian's weight, in keils, can be calculated by taking the square

\(\sqrt{age} + 3 = \sqrt{2*age + 17}\)

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

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

Check other Age Problems HERE.
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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

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

Dear Bunuel, Let us call the Trussian's current age a. Why my equation incorrect? \(17-3\sqrt{a}=2a\)
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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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Algebraic way is given 3 posts above: https://gmatclub.com/forum/a-trussian-s ... l#p1624365
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Thank you Sir.
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Current weight = x
Current age = y

x = \(\sqrt{y}\)

Given

\(\sqrt{2y + 17}\) = \(\sqrt{y}\) + 3

Squaring on both sides

2y + 17 = y + 9 + 6\(\sqrt{y}\)

Simplifying further,

y - 6\(\sqrt{y}\) + 8 = 0

\(\sqrt{y}\) = 4 or \(\sqrt{y}\) = 2

Therefore y = 16 or 4

Option C
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