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Difficult Math Problem #82 - Arithmetic, Algebra

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by 800guy » Wed Jan 03, 2007 11:59 am
5 is placed to the right of two – digit number, forming a new three – digit number. The new number is 392 more than the original two-digit number. What was the original two-digit number?

OA coming after a few people answer..
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by thankont » Wed Jan 03, 2007 11:27 pm
xy5 - 392 = xy (1) so y=3 and (1) becomes x35 - 392 = x3 so x=4
and number is 43 --> 435 = 43+392
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by gabriel » Thu Jan 04, 2007 3:23 am
me got the same answer

another way of doing it is.... let the no be AB

the eqn wuld be 10A+B+392=100A+10B+5

solving this we get A=4 and B=3 ...hence the answer
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OA

by 800guy » Fri Jan 05, 2007 10:13 am
OA

If the original number has x as the tens digit and y as the ones digit (x and y are integers less than 10) then we can set up the equation:
100x + 10y + 5 = 10x + y + 392
90x + 9y = 387
9(10x+y) = 387
10x + y = 43 ==> x = 4, y = 3
the original number is 43, the new number is 435
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by medenica » Thu Jan 11, 2007 10:06 pm
There is a lot easier way to think about these.... (for me anyhow). Ie. when you add a digit to the right, you're multiplying by ten (because you're shifting to the left) and adding the number....

So...
10x+5=x+392
9x=387
x=43. Done.
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by indu.nair » Sun Oct 14, 2012 2:21 am
i still din get it guys!
the person who answered first! how did u get y=3??
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