In a two-digit number, the digit in the tens place value is 3 times the digit in the units place value. When the number is decreased by 54, the digits are reversed. What is the number?

Guest Dec 1, 2015

#2**+5 **

Let the unit digit = x

Then, the tens digit = 3x

And we have the following equation

[3x(10) + x] - 54 = [ x(10) + 3x ] simplify

30x + x - 54 = 10x + 3x

31x - 54 = 13 x

18x - 54 = 0

18x = 54 divide both sides by 18

x = 3 and 3x = 3(3) = 9

So

The larger number is 3(3)(10) + 3 = 93

And the smaller number is 10(3) + 3(3) = 39

And their difference = 54

Just as Anonymous4338 found !!!!!!

CPhill
Dec 1, 2015

#1**+5 **

The answer is 93.

9 is 3 times more than 3, and when you subtract 54 from 93, you get 39.

Anonymous4338
Dec 1, 2015

#2**+5 **

Best Answer

Let the unit digit = x

Then, the tens digit = 3x

And we have the following equation

[3x(10) + x] - 54 = [ x(10) + 3x ] simplify

30x + x - 54 = 10x + 3x

31x - 54 = 13 x

18x - 54 = 0

18x = 54 divide both sides by 18

x = 3 and 3x = 3(3) = 9

So

The larger number is 3(3)(10) + 3 = 93

And the smaller number is 10(3) + 3(3) = 39

And their difference = 54

Just as Anonymous4338 found !!!!!!

CPhill
Dec 1, 2015

#3**+5 **

In a two-digit number, the digit in the tens place value is 3 times the digit in the units place value. When the number is decreased by 54, the digits are reversed. What is the number?

Let the first digit =n, then we have the 2nd digit=3n. But we have:

(3n*10) + n - 54=10n + 3n

30n + n - 54=13n

31n - 13n=54

18n=54

n=54/18

n=3 the first number

3 X 3=9 the second number, therefore the number is:

=93, because 93 - 54=39

Guest Dec 1, 2015