If the tens digit of a two-digit number is subtracted from the units digit, the difference is 8. The number with the digits reversed is 10 more than nine times the units digit of the original number. Find the number.
Let the two digit number be 10A + B.
Then, from "If the tens digit of a two-digit number is subtracted from the units digit, the difference is 8.", or
B - A = 8.
From "The number with the digits reversed is 10 more than nine times the units digit of the original number.", 10B + A = 9B + 10.
Substituting A = B - 8 into 10B + A = 9B + 10 yields 10B + B - 8 = 9B + 10 or 11B - 8 = 9B + 10 or 2B = 18 making B = 9.
From B - A = 8, 9 - A = 8 making A = 1 or the number = 19.
Let unit digit = x
Then tens did it = 8+x
Original no.is 10(8+x)+x
80+10x+x=80+11x
On reversing : 10x+8+x =11x+8
11x+8=10+9x
11x-9x=10-8
2x=2
X=1
No.is 80+11x=80+11=91
To solve this problem, let's assume the original number is represented as "10a + b", where "a" represents the tens digit and "b" represents the units digit.
According to the problem, if the tens digit of the number is subtracted from the units digit, the difference is 8. Mathematically, this can be represented as:
b - a = 8 ----(1)
The problem also states that the number with the digits reversed (i.e., "10b + a") is 10 more than nine times the units digit of the original number. Mathematically, this can be represented as:
10b + a = 9b + 10 ----(2)
Now, we have a system of two equations with two variables. Let's solve it to find the values of "a" and "b".
From equation (1), we can express "a" in terms of "b" by rearranging the equation:
a = b - 8 ----(3)
Substituting equation (3) into equation (2), we get:
10b + (b - 8) = 9b + 10
Combining like terms, we have:
11b - 8 = 9b + 10
Simplifying further, we get:
2b = 18
Dividing both sides by 2, we find:
b = 9
Now, substitute the value of "b" back into equation (3) to find the value of "a":
a = 9 - 8 = 1
Therefore, the original number is 10a + b = 10(1) + 9 = 19.