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?
+128474 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 !!!!!!
+1667 The answer is 93.
9 is 3 times more than 3, and when you subtract 54 from 93, you get 39.
+128474 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 !!!!!!
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
