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What is (1331)^(1/2)/(396^(1/2) + 275^(1/2))

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What is (1331)^(1/2)/(396^(1/2) + 275^(1/2))  [#permalink]

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New post 13 Feb 2016, 04:00
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A
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Question Stats:

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What is \(\frac{\sqrt{1331}}{\sqrt{396} + \sqrt{275}}\) ?

A \(\frac{1}{2}\)

B 1

C \(\sqrt{3}\)

D 2

E \(\frac{2\sqrt3}{2}\)
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Re: What is (1331)^(1/2)/(396^(1/2) + 275^(1/2))  [#permalink]

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New post 13 Feb 2016, 08:48
11sqrt(11)/(6sqrt(11) + 5sqrt(11)) = 11/(6+5) = 1

Answer: B
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Re: What is (1331)^(1/2)/(396^(1/2) + 275^(1/2))  [#permalink]

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New post 13 Feb 2016, 12:39
Vyshak wrote:
11sqrt(11)/(6sqrt(11) + 5sqrt(11)) = 11/(6+5) = 1

Answer: B


hi vyshak
u just wrote down the answer.
you didn't explain how u found out that 396 is 36 * 11
and 1331 is 11 * 121.
Did it just pop out from your head?
then you must be a savant.
but I'm not.

I'll appreciate further explanation.
thanks.
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Re: What is (1331)^(1/2)/(396^(1/2) + 275^(1/2))  [#permalink]

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New post 13 Feb 2016, 13:26
1
The gap between the potential answers is pretty large, so I used approximations to find the answer.

sqrt(1331):
20^2 = 400
30^2 = 900
40^2 = 1600
1600 - 900 = 700
700/2 = 350
900 + 350 = 1250
The top part is between 35-40, so let's do a litmus test with 36^2 -> 1296. This value is 35 off from 1331, so let's go with 36+ (a value just above 36) as our top value.

sqrt(396):
20^2 = 400, clearly our value will be just below 20^2, so let's tag this one as 20-.

sqrt(275):
15^2 = 225 (look up the rules for squaring values that end in 5 to do this calculation in < 5 seconds)
We still have a lot of gap between 225 and 275, so let's try 16^2.
16^2 = 2^8 = 256 (memorized powers of 2)
Let's go with 16+ for this value

Plugging in all the approximations:
36+ / (20- + 16+) -> approximately 36+ /~36 -> 1.
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Re: What is (1331)^(1/2)/(396^(1/2) + 275^(1/2))  [#permalink]

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New post 13 Feb 2016, 21:29
1
Nez wrote:
What is \(\frac{\sqrt{1331}}{\sqrt{396} + \sqrt{275}}\) ?

A \(\frac{1}{2}\)

B 1

C \(\sqrt{3}\)

D 2

E \(\frac{2\sqrt3}{2}\)


Hi Nez,
1331 is 11^3. I knew the value and its always better to know squares upto 30 and cubes upto 10.
It will be very useful for you too if you can learn it.
Now. since 1331 = 11^3 = 11\(\sqrt11\), we have a \(\sqrt11\) in the numerator. I next look forward to see if the denominator also can be reduced to some value in terms of \(\sqrt11\). 396 and 275 are small integers and can be easily determined to be 36*11 (6\(\sqrt11\)) and 25*11 (5\(\sqrt11\)). That's how I did it.
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Re: What is (1331)^(1/2)/(396^(1/2) + 275^(1/2))  [#permalink]

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New post 14 Feb 2016, 23:29
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1
Nez wrote:
What is \(\frac{\sqrt{1331}}{\sqrt{396} + \sqrt{275}}\) ?

A \(\frac{1}{2}\)

B 1

C \(\sqrt{3}\)

D 2

E \(\frac{2\sqrt3}{2}\)



You can also just approximate.

\(\sqrt{1331}\) will be between \(30^2 = 900\) and \(40^2 = 1600\). We know that \(35^2 = 1225\). So \(\sqrt{1331}\) is about 36.

\(\sqrt{396}\) is about 20 because \(20^2 = 400\).

\(\sqrt{275}\) is about 16 because \(16^2 = 256\).

So in the denominator you get about 20 + 16 = about 36

So the fraction is approximately 1 (both numerator and denominator are about 36).

Answer (B)

All other options are very different from 1.
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Re: What is (1331)^(1/2)/(396^(1/2) + 275^(1/2))  [#permalink]

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New post 15 Feb 2016, 02:22
Nez wrote:
What is \(\frac{\sqrt{1331}}{\sqrt{396} + \sqrt{275}}\) ?

A \(\frac{1}{2}\)

B 1

C \(\sqrt{3}\)

D 2

E \(\frac{2\sqrt3}{2}\)


well I had to look into dictionary to know the meaning of savant. :wink:
coming to the problem, almost all options are in simplest form(Integers, Simple fractions of numbers or multiples of root 3).
By these options we can understand that given large numbers in roots will get simplified by cancelling one or the other factor of them.
The smallest term is 275. since this is a multiple of 5 we can divide 275 by 5 to get 55.
Thus we can express 275 as \(5^2*11\)
Thus \(\sqrt{275}=5\sqrt{11}\)

Since there is no option in answer choices which contains \(\sqrt{11}\)
we can guess that this is term which is common among all other terms and is going to get cancelled.
Thus divide 396 by 11 and we get \(36=6^2\) and \(\sqrt{396}=6\sqrt{11}\)

similarly do the same for 1331 and get its square root as \(11\sqrt{11}\)

thus whole fraction becomes

\((11\sqrt{11})\)
-------------------------
\((6\sqrt{11} + 5\sqrt{11})\)

=\(\frac{11}{(6+5)}\) = 1

I hope this helps!
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Re: What is (1331)^(1/2)/(396^(1/2) + 275^(1/2))  [#permalink]

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New post 18 Feb 2017, 08:02
Nezdem wrote:
What is \(\frac{\sqrt{1331}}{\sqrt{396} + \sqrt{275}}\) ?

A \(\frac{1}{2}\)

B 1

C \(\sqrt{3}\)

D 2

E \(\frac{2\sqrt3}{2}\)


\(\sqrt{1331} = 11\sqrt{11}\)

\(\sqrt{396} = 6\sqrt{11}\)

\(\sqrt{275} = 5\sqrt{11}\)


\(\frac{\sqrt{1331}}{\sqrt{396} + \sqrt{275}}\)

= \(11\sqrt{11}/(6\sqrt{11}+5\sqrt{11})\)

= \(11\sqrt{11}/11\sqrt{11}\)

= \(1\)

Hence, answer will be (B) 1
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Re: What is (1331)^(1/2)/(396^(1/2) + 275^(1/2))  [#permalink]

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