ritula wrote:

Find the value of x

\(x= \sqrt[2]{20+\sqrt[2]{20+\sqrt[2]{20}}}\)

1. 20

2. 5

3. 2

4. 8

1: \(x= \sqrt{20+\sqrt{20+\sqrt{20}}}\)

\(x= \sqrt{20+\sqrt{20+4.47}}\)

\(x= \sqrt{20+\sqrt{24.47}}\)

\(x= \sqrt{20+4.95}\)

\(x= \sqrt{24.95}\)

\(x= 4.995 = approx. 5.00\)

2: From third sqrt: sqrt 20 = 4 and a fraction of 1.

From second sqrt: sqrt (20+4.00 = 24) = 4 and a fraction of 1.

From first sqrt: sqrt (20+4.00 = 24) = 4 and a fraction of 1. which is definitely close to none other than 5.

3: Using POE:

A: it cannot be 20 cuz for 20, the value under root must be 400, which is impossible. so A is ruled out.

B: It could be 5 as done above in method 2.

3. 2 is also not possible because even if we consider first 20 under root, the value must not be smaller than 4.

4. 8 is not possible because for 8, the value under root must be 64. Even if we add up all three 20s, the sum would not be more than 60. so it is also not possible. So left with 5.

So B make sense.

_________________

Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html

Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT