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# √2450 =

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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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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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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

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This problem can be solved by pulling out prime factors (in bold):

2450 = 2 * 1225
1225 = 5 * 245
245 = 5 * 49
48 = 7*7

We're left with $$\sqrt{2*5*5*7*7}$$

This results in: $$(5*7)\sqrt{2}=35\sqrt{2}$$
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