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algebraic method needed

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Source: — Problem Solving |
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by sgarnepudi » Sat Nov 08, 2008 10:15 pm
let fathers age is 10x+y and sons age is 10y+x

Ages an year ago are F = 10x+y-1 and S = 10y+x-1
F-S = 9x-9y which is equal to the age of the son i.e.
9x-9y = 10y+x-1 => 8x-19y+1=0 .. substitute values in x and y to make the total 0 and it happens at y=3 and x=7

so F=73 and S=37
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by c_numb » Sat Nov 08, 2008 10:44 pm
thanks
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by raunekk » Sun Nov 09, 2008 12:03 am
substitute values in x and y to make the total 0 and it happens at y=3 and x=7


is there any other way other ten guessing numbers...


can we form one more equation????
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by vittalgmat » Sun Nov 09, 2008 12:30 am
raunekk wrote:
substitute values in x and y to make the total 0 and it happens at y=3 and x=7


is there any other way other ten guessing numbers...


can we form one more equation????

cant think of any.. at this time. However, you could use the answer choices to figure out the answer, starting from D.

HT Helps
-V
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by aj5105 » Sun Nov 09, 2008 1:35 am
nice qn !

Solved till 8x-19y+1=0 & started wondering ! :D
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by cartera » Sun Nov 09, 2008 6:31 am
sgarnepudi wrote:let fathers age is 10x+y and sons age is 10y+x

Ages an year ago are F = 10x+y-1 and S = 10y+x-1
F-S = 9x-9y which is equal to the age of the son i.e.
9x-9y = 10y+x-1 => 8x-19y+1=0 .. substitute values in x and y to make the total 0 and it happens at y=3 and x=7

so F=73 and S=37
But it says a year ago, the father was twice as ols as his son, so it should be:

10x+y-1=20y+2x-1; 8x=21y, how do you solve this?
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Re: algebraic method needed

by vivek.kapoor83 » Sun Nov 09, 2008 7:06 am
c_numb wrote:If you reverse the digits of my age, you have the age of my son. A year ago, I was twice his age. How old are we both now?

To be honest ...the fastest way to solve is..try to put the choices and u ll be able to slove it fastest. Rather than assuming present age etc.
Pick each choice and see if that choice completes the above condition.
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by c_numb » Sun Nov 09, 2008 8:49 am
v.kapoor - u are right.
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