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Let the numbers be 2x,3x,5x
Sum of their squares is
38x^2= 1368
X =6
2x=12

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Bunuel
If three numbers are in the ratio of 2 : 3 : 5, and the sum of their squares is 1368, what is the value of the first number?

A. 6
B. 12
C. 18
D. 24
E. 30
Solution:

We can let the three numbers be 2x, 3x, and 5x and create the equation:

(2x)^2 + (3x)^2 + (5x)^2 = 1368

4x^2 + 9x^2 + 25x^2 = 1368

38x^2 = 1368

x^2 = 36

x = 6 or x = -6

Because the answer choices are all positive, we can eliminate x = -6.

Since the first number is 2x, the first number is 2(6) = 12.

Answer: B
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Bunuel
If three numbers are in the ratio of 2 : 3 : 5, and the sum of their squares is 1368, what is the value of the first number?

A. 6
B. 12
C. 18
D. 24
E. 30
\(2x^2 + 3x^2 + 5x^2 = 1368 \)

So, \(x^2 = 36\)

Or, \(x = 6\)

So, The first number is \(2*6 = 12\), Answer must be (B)
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