If ∛(12167)*√(16384) = x*11.5, then what is the value of x ?

Question

If  \sqrt {12167}*\sqrt{16384} = x*11.5 , then what is the value of x ?

A. 204
B. 234
C. 256
D. 272
E. 286

yashikaaggarwal · Answer

³√12167*√16384 = x*11.5
Taking Cube and Square root of respective numbers.
23*128 = x*11.5
Equating both sides
X = 256

Answer is C

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Superman249 · Answer

IMO:- C

12167 = 23^3 (Prime factor only 23)
16384= 2^14

so 23 * 2^7 = x * 11.5
=> 2944 = x * 11.5
=> x = 256

rishovnits91 · Answer

Hi  Bunuel , is there a quick way to find out the cube roots and square roots?

SiddharthR · Answer

I don't think there is a "quick" way to find the square and cube roots here. I think that just comes with practice by factorizing numbers.

The difficulty I faced in this problem was trying to find factors for cube root(12167). This took way too much time and i believe wasn't the greatest of approaches.

I looked at the answer choices and realized I needed an integer. When we solve the right hand side of the equation we get an improper fraction "23/2", where 23 is a prime.

On the left hand side, we also have sqrt(16384) which is definitely divisible by 2. So, we can find the prime factors for that quickly. Since the end result is an integer, the only number not accounted for is "23" and that's how i found out that cube root(12167) is actually 23.

This problem was nice !

rishovnits91 · Answer

Ok thank you for your explanation.

dave13 · Answer

To solve this question you need a radical approach

approach question to yourself and not to approach yourself to a question

\sqrt {12} is between \sqrt {8} and \sqrt {27} so i pick smaller number which is \sqrt {8} = 2 So first digit is 2

And last digit has to end with digit 3 because 3^3 =27 So \sqrt {12167} = 23 \sqrt{16384} is between 120^2 12^2 =144 *100 =14400 and 130^2 ( 13^2*100 = 169*100 ) since 130^2 ( \sqrt{16900} is greater than \sqrt{16384} so i go for a smaller number , i.e 12 So first two digits are 12 And the last digit has to be either 8 because 8^2 = 64 or 2^2 =4 , So Since \sqrt{16384} is closer to \sqrt{16900} than to \sqrt{14400} i pick 8 as the last digit. \sqrt{16384} = 128 23*128= x*11.5 C

hr1212 · Answer

How I approached this was by doing minimal calculation and trying to get some hints from these numbers to approximate cube roots and square roots, 120^2 < 16384 < 130^2 => So it could be either 122 or 128 due to the last digit being 4. As the number is weighted towards 130^2, I assumed the square to be of number 128, which brings us to => \sqrt {12167}*\sqrt{16384} = x*11.5 \sqrt {12167}*2^8 = x*23/2 \sqrt {12167}*2^9/23 = x Next observation was that 12167 should be divisible by 23 for us to get an integer answer and 3^3 results into a number with last digit 7, so 23^3 would likely be 12167. Additional hints which supported my approximation here was that 2^9 is one of the option choice, 20^3 < 12167 < 30^3, so it's unlikely that there could be any other number in this range with it's cube ending in 7, so the obvious answer choice becomes 256. IMO: C

pappal · Answer

from choices its clear that x is an integer
square root term can be derived by simply factorization of the term giving 2^14.
for x to be an integer the cube root term must be a multiple of 11.5
so to start with 23 we can reach to the requisite.
hope it helps

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If ∛(12167)√(16384) = x11.5, then what is the value of x ?

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