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# Two numbers are such that their sum is 54 and product is

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Senior Manager
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Two numbers are such that their sum is 54 and product is [#permalink]  29 Jan 2004, 14:05
Two numbers are such that their sum is 54 and product is 629. Find the difference of the numbers.

a. 20
b. 25
c. 40
d. 46
e. None of these
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****GMAT Loco****
Este examen me conduce jodiendo loco

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I got A ; 20 by plugging numbers
took alomost 10 mins to solve

Look simple but takes long time to solve.
Any easy methods to solve this question
Senior Manager
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Right off the bat we can set up an equation. If we let the two numbers = x1, and x2.
x1 + x2 = 54.........(is a given)
x1 * x2 = 629 ....(is a given)
Hence, x1^2 - 54x1 + 629 = 0
Solve quadratic for x1....I got stuck here though...
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a) 20, the factors of 629 are 17 and 37. So 17+37=54 and 17*37=629
Senior Manager
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ofAns was (e)
But I guess you are right
nice
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Pls include reasoning along with all answer posts.
****GMAT Loco****
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sunniboy007 wrote:
ofAns was (e)
But I guess you are right
nice

What is the source of the question?
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one more method.. let the 1st number is x and 2nd number is 54-x
then their product is x*(54-x)=629..factorise..
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shubhangi

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Another method.

(a-b) ^2 = (a+b)^2-4ab
= 54^2-4*629 = 400
a-b = 20
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hey Anand.. would u please tell me how did u get 4ab.. is'nt 2ab..
(a-b)^2= (a+b)^2-2ab..
or i've forgotten the formula..pl..tell me ..if am wrong..
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shubhangi

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Numbers that make nine (via multiplication):
1 9
3 3
7 7

Of these, only 7 and 7 will sum to make 4.

We need to multiply to the 630 ballpark. sqrt 630 is about 25.

Take 27 and 27. No good.

17 and 37 should be the next thing to try.

To get a 51 in quant (and I did not get a 51 in quant), you probably need to be able to anser the question with a similar fast thought process.

Wasting 3 or 4 minutes on questions like this is a recipe for a low score.
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Hi stoolfi,

You are scaring me man! If you didnt get 51 in Qauntitative test then I know what to expect.

Anand.

Hi Shubangi,

we know a+b and ab but we want to know a-b.

1. (a-b)^2 = a^2+b^2-2ab
2. (a+b)^2 = a^2+b^2+2ab

so from eq 1 a^2+b^2 = (a-b)^2+2ab
Substitute this in eq 2

(a+b)^2 = (a-b)^2+2ab+2ab
so (a-b)^2 = (a+b)^2-4ab
I hope it is clear.
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