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√2450 =

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Joined: 02 Sep 2009
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√2450 = [#permalink]

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New post 11 Mar 2018, 21:01
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

90% (01:20) correct 10% (01:16) wrong based on 88 sessions

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Senior Manager
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Joined: 07 Jul 2012
Posts: 335
Location: India
Concentration: Finance, Accounting
GPA: 3.5
Re: √2450 = [#permalink]

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New post 11 Mar 2018, 21:06
35\(\sqrt{2}\)= \(\sqrt{35*35*2}\)= \(\sqrt{2450}\)

Answer: E.

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Re: √2450 = [#permalink]

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New post 12 Mar 2018, 01:52
We try to simplify the number 2450 by finding its prime factors.

\(\sqrt{2450} = \sqrt{2*5^2*7^2} = 5*7\sqrt{2} = 35\sqrt{2}\)

Hence, answer is E.
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Re: √2450 = [#permalink]

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New post 12 Mar 2018, 09:11
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Top Contributor
Bunuel wrote:
√2450 =

A. 25√2
B. 35
C. 28√2
D. 49
E. 35√2


Here's another approach...

2450 is a little bit less than 2500

So, √2450 will be a little bit less than √2500

√2500 = 50, so the correct answer will be a little bit less than 50

IMPORTANT: Before test day, you should memorize the following approximations:
√2 ≈ 1.4
√3 ≈ 1.7
√5 ≈ 2.2

Now let's use this info to check the answer choices
A. 25√2 ≈ (25)(1.4) ≈ 35 ELIMINATE

B. 35 ELIMINATE

C. 28√2 ≈ (28)(1.4) ≈ 40 ELIMINATE

D. 49 This answer choice is close to 50. HOWEVER, we can see that 49² will have units digit 1, while 2450 has units digit 0. ELIMINATE

E. 35√2
By the process of elimination, the correct answer must be E
If we want to confirm, we can recognize the following:
35√2 = (√35²)(√2)
= (√1225)(√2)
= √2450
BINGO!!
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Re: √2450 = [#permalink]

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New post 12 Mar 2018, 09:25
Bunuel wrote:
√2450 =

A. 25√2
B. 35
C. 28√2
D. 49
E. 35√2


\(\sqrt{2450} = \sqrt{2^1 * 5^2 * 7^2}\)

Or, \(\sqrt{2450} = 35\sqrt{2}\), hence answer will be (E)

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Re: √2450 = [#permalink]

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New post 13 Mar 2018, 17:02
Bunuel wrote:
√2450 =

A. 25√2
B. 35
C. 28√2
D. 49
E. 35√2


We see that 2450 ends with 50 which means it’s divisible by 5^2 = 25. Since 2450/25 = 98 and 98 = 49 x 2, we have 2450 = 25 x 49 x 2. Thus

√2450 = √(25 x 49 x 2) = √25 x √49 x √2 = 5 x 7 x √2 = 35√2

Answer: E
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Re: √2450 =   [#permalink] 13 Mar 2018, 17:02
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