A few days ago
I don’t know how to solve this math problem?
Please explain to me how to solve this word problem. I don’t just want the answer… I want to know how the answer was found.
The tens digit of a two-digit number is three greater than the ones digit. The different between twice the original number and half the number obtained after reversing the digits is 108. What is the original number?
Thank you in advance!
Top 1 Answers
A few days ago
Favorite Answer
If the ones digit is X, then the two-digit number is:
(3+X)*10+X
If you reverse the digits, you get:
X*10+3+X
So
2*((3+X)*10+X)-(X*10+3+X)/2
=108
Solve that for X:
2*((3+X)*10+X)-(X*10+3+X)/2
=108
therefore
2*(30+11*X)-(X*10+3+X)/2=108
60+22*X-(X*10+3+X)/2=108
60+22*X-(X*11+3)/2=108
60+22*X-5.5*X-1.5=108
58.5+16.5*X=108
X=(108-58.5)/16.5
X=3
So the original number is
(3+X)*10+X=
(3+3)*10+3=
63
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