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two digit word problems

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by dikku07 » Wed May 20, 2009 9:42 am
Hi All,
here are 2 similar problems. I guess i'm doing something wrong somewhere that my answers are not matching OA. Can someone pls solve these? thanks in advance.

Problem 1- The sum of the digits of a two-digit number is 8. The number obtained by interchanging the two digits exceeds the given number by 36. Find the number.

problem 2- The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits of the number. Find the number.
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by scoobydooby » Wed May 20, 2009 9:56 am
1) let the 10s digit be x and units be y. original number 10x+y
x+y=8.........a
also, 10y+x-(10x+y)=36
=>9y-9x=36
=>y-x=4......b

adding a and b, 2y=12=>y=6, x=2 or the number is 26


2) let the 10s digit be x and units be y. original number 10x+y
x+y=9........a
also, 9(10x+y)=2(10y+x)
=>88x=11y
=>y=8x
putting in a we get x+8x=9x=9
=>x=1, y=8
the number is 18
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by dikku07 » Wed May 20, 2009 10:05 am
these are correct OAs. 26 and 18.

But I don't get one thing. for equation a, you are taking x+y=8. when it says sum of 2 digits why not 10x+y=8?

I was taking 10x+Y and thats why not getting right OA.
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by Katrusya » Wed May 20, 2009 10:16 am
10x+y - you'll get the number itself
but x+y - is the sum of digits.

For instance, you have the number 123
Sum of digits would be 1+2+3=6
But number itself you can present as 1*100+2*10+3=123
Hope it helps!
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by dikku07 » Wed May 20, 2009 10:20 am
ok. got it. made word translation mistake. thanks alot
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