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If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^

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If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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If a = 1,248 and b = 1,152, what is the value of \((a^2 - b^2) ^{\frac{1}{2}}\)?

A. 240
B. 360
C. 480
D. 600
E. 720

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Originally posted by SajjadAhmad on 14 May 2017, 10:55.
Last edited by Bunuel on 15 May 2017, 12:33, edited 1 time in total.
Corrected the typo.
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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New post 15 May 2017, 01:52
SajjadAhmad wrote:
If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^\(\frac{1}{2}\)?

A. 240
B. 360
C. 480
D. 600
E. 720

Is there any way to solve this other than doing a long multiplication?
(a^2 + b^2) can be replaced by (a+b)^2-2ab. But still one long multiplication has to be done
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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New post 15 May 2017, 02:15
(a^2 + b^2) ^1/2 -->
((a+b)(a-b))^1/2 -->
((1248+1152)(1248-1152))^1/2
((2400)(96))^1/2

Factor these two numbers to get

((100*24*24*4))^1/2 -->
10*24*2 = 480
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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New post 15 May 2017, 02:29
2
lpetroski wrote:
(a^2 + b^2) ^1/2 -->
((a+b)(a-b))^1/2 -->
((1248+1152)(1248-1152))^1/2
((2400)(96))^1/2

Factor these two numbers to get

((100*24*24*4))^1/2 -->
10*24*2 = 480


Highlights portion is Incorrect.

A^2+B^2 ≠ (A+B)(A-B)

Rather its (A+B) (A-B) = A^2-B^2

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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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New post 15 May 2017, 02:33
1
Here is what I think -->

This is an INCORRECT Question.



Is this really a Princeton Review question ?

a = 1,248 and b = 1,152

We are asked about => √a^2+b^2 =p(say)

Hence a^2+b^2 and p^2 must have the same units digit.

Units digits of a^2 => 4
Units digits of b^2 => 4
Hence Units digits of a^2 +b^2 => 8

Now, how on earth can √a^2+b^2 end with a zero(all option end in zero)?
I hope I am not missing anything :)


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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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New post 15 May 2017, 02:38
stonecold wrote:
lpetroski wrote:
(a^2 + b^2) ^1/2 -->
((a+b)(a-b))^1/2 -->
((1248+1152)(1248-1152))^1/2
((2400)(96))^1/2

Factor these two numbers to get

((100*24*24*4))^1/2 -->
10*24*2 = 480


Highlights portion is Incorrect.

A^2+B^2 ≠ (A+B)(A-B)

Rather its (A+B) (A-B) = A^2-B^2


Ohhh you're right. Was seeing how much I had retained my skills after 1 year off the books! Guess some work to do :D
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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New post 15 May 2017, 04:39
2
stonecold wrote:
Here is what I think -->

This is an INCORRECT Question.



Is this really a Princeton Review question ?

a = 1,248 and b = 1,152

We are asked about => √a^2+b^2 =p(say)

Hence a^2+b^2 and p^2 must have the same units digit.

Units digits of a^2 => 4
Units digits of b^2 => 4
Hence Units digits of a^2 +b^2 => 8

Now, how on earth can √a^2+b^2 end with a zero(all option end in zero)?
I hope I am not missing anything :)


Regards
Stone Cold


exactly my point.
also if you try to solve it via factorization, you end up with a non perfect square number.
i also verified using the calculator; the answer is in decimals; the options are no where close.

I think the question poster wanted to type in a sqr - b sqr instead of a sqr + b sqr.
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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New post 15 May 2017, 12:25
2
stonecold wrote:
Here is what I think -->

This is an INCORRECT Question.



Is this really a Princeton Review question ?

a = 1,248 and b = 1,152

We are asked about => √a^2+b^2 =p(say)

Hence a^2+b^2 and p^2 must have the same units digit.

Units digits of a^2 => 4
Units digits of b^2 => 4
Hence Units digits of a^2 +b^2 => 8

Now, how on earth can √a^2+b^2 end with a zero(all option end in zero)?
I hope I am not missing anything :)


Regards
Stone Cold


Here is the Screenshot from book
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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New post 15 May 2017, 12:32
3
SajjadAhmad wrote:
stonecold wrote:
Here is what I think -->

This is an INCORRECT Question.



Is this really a Princeton Review question ?

a = 1,248 and b = 1,152

We are asked about => √a^2+b^2 =p(say)

Hence a^2+b^2 and p^2 must have the same units digit.

Units digits of a^2 => 4
Units digits of b^2 => 4
Hence Units digits of a^2 +b^2 => 8

Now, how on earth can √a^2+b^2 end with a zero(all option end in zero)?
I hope I am not missing anything :)


Regards
Stone Cold


Here is the Screenshot from book


Yes, the question is not correct. It happens sometimes. Should be:

If a = 1,248 and b = 1,152, what is the value of (a^2 - b^2)^(1/2)?

In this case the answer will be C, as stated. Will edit the original post.
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If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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New post 15 May 2017, 17:03
Thanks Bunuel for correcting the question.

Here is how I did the modified question ==>
a^b-b^2 => (a+b)*(a-b) => (1,248 + 1,152) * (1,248 - 1,152) => 2400 * 96
Using prime factorisation =>

2400 * 96 => 2^10*3^2*5^2

Hence √a^2-b^2 => 2^5*3*5 => 160*3 => 480

Smash that C.
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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New post 18 May 2017, 00:50
If a = 1,248 and b = 1,152, what is the value of (a2−b2)12(a2−b2)12?

A. 240
B. 360
C. 480
D. 600
E. 720

my 2 cents.

It would have been ideal if I realized that this was another foil question but...I didn't.
As the numbers on AC were not close, I thought I could just use lump sum calculation.

a is about 1,200, and b is about 1,100.
If we square these, we would get something like 1,440,000 (from 12 x 12) and 1,210,000 (from 11 x 11).
Then when we subtract these, we would get 230,000.
Only C comes close this number.
Hence C.
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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New post 18 May 2017, 01:05
SajjadAhmad wrote:
If a = 1,248 and b = 1,152, what is the value of \((a^2 - b^2) ^{\frac{1}{2}}\)?

A. 240
B. 360
C. 480
D. 600
E. 720


\((a ^2 - b^2)^{\frac{1}{2}} = ((a+b)(a-b))^{\frac{1}{2}} = ((1248+1152)(1248-1152)){\frac{1}{2}} = (2400*96)^{\frac{1}{2}} = (24*24*4*100)^{\frac{1}{2}} = 24*2*10= 480\)
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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New post 19 May 2017, 06:32
SajjadAhmad wrote:
If a = 1,248 and b = 1,152, what is the value of \((a^2 - b^2) ^{\frac{1}{2}}\)?

A. 240
B. 360
C. 480
D. 600
E. 720


We can simplify the given expression:

√(a + b)(a - b) = ?

√(1248 + 1152)(1248 - 1152) = √(2400)(96) = √2400 x √96

√2400 = √100 x √24 = 10√24 = 10 x √4 x √6 = 20√6

√96 = √16 x √6 = 4√6

Thus, √2400 x √96 = (20√6)(4√6) = 80 x 6 = 480

Answer: C
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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New post 19 May 2017, 09:52
SajjadAhmad wrote:
If a = 1,248 and b = 1,152, what is the value of \((a^2 - b^2) ^{\frac{1}{2}}\)?

A. 240
B. 360
C. 480
D. 600
E. 720


\(\sqrt{(a^2 - b^2)} = \sqrt{(a + b)(a - b)}\)

Now, \(\sqrt{(a + b)(a - b)} = \sqrt{(1248 + 1152)(1248 - 1152)}\)

Or, \(\sqrt{(a + b)(a - b)} = \sqrt{2400*96}\)

Or, \(\sqrt{(a + b)(a - b)} = 480\)

Thus, the answer must be (C) 480
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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New post 31 May 2017, 05:19
1
nishantd88 wrote:
SajjadAhmad wrote:
If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^\(\frac{1}{2}\)?

A. 240
B. 360
C. 480
D. 600
E. 720

Is there any way to solve this other than doing a long multiplication?
(a^2 + b^2) can be replaced by (a+b)^2-2ab. But still one long multiplication has to be done


Hi, you don't need to do the multiplication of numbers. Do it like this,

(a^2-b^2)^1/2 = [(a-b)(a+b)]^1/2
(a-b) =2400
(a+b)=96

Now start taking perfect squares out of square root. Start with 2400, we get 24*100 where 100^1/2 = 10

So we get 10*(24*96)^1/2 = 10*2*(6*96)^1/2 = 10*2*4*(6*6)^1/2 = 10*2*4*6 = 480.

Here, 24 can be broken as 2^2 *6 and 96 can be written as 4^2*6.

You can do it by full factorisation also which will consume extra time.
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^  [#permalink]

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