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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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Updated on: 15 May 2017, 12:33
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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) ^
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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)^22ab. 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) ^
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15 May 2017, 02:15
(a^2 + b^2) ^1/2 > ((a+b)(ab))^1/2 > ((1248+1152)(12481152))^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) ^
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15 May 2017, 02:29
lpetroski wrote: (a^2 + b^2) ^1/2 > ((a+b)(ab))^1/2 > ((1248+1152)(12481152))^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)(AB)
Rather its (A+B) (AB) = A^2B^2
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^
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15 May 2017, 02:33
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
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^
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15 May 2017, 02:38
stonecold wrote: lpetroski wrote: (a^2 + b^2) ^1/2 > ((a+b)(ab))^1/2 > ((1248+1152)(12481152))^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)(AB)
Rather its (A+B) (AB) = A^2B^2Ohhh 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) ^
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15 May 2017, 04:39
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) ^
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15 May 2017, 12:25
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) ^
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15 May 2017, 12:32
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) ^
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15 May 2017, 17:03
Thanks Bunuel for correcting the question. Here is how I did the modified question ==> a^bb^2 => (a+b)*(ab) => (1,248 + 1,152) * (1,248  1,152) => 2400 * 96 Using prime factorisation => 2400 * 96 => 2^10*3^2*5^2 Hence √a^2b^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) ^
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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) ^
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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 \((a ^2  b^2)^{\frac{1}{2}} = ((a+b)(ab))^{\frac{1}{2}} = ((1248+1152)(12481152)){\frac{1}{2}} = (2400*96)^{\frac{1}{2}} = (24*24*4*100)^{\frac{1}{2}} = 24*2*10= 480\)
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Re: If a = 1,248 and b = 1,152, what is the value of (a^2 + b^2) ^
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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) ^
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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) ^
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31 May 2017, 05:19
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)^22ab. 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^2b^2)^1/2 = [(ab)(a+b)]^1/2 (ab) =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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