SOLUTION: 1. the sum of the digits f a two-digit number is 11. when the digits are reversed, the new number is increased by 20 which is twice the original number. find the original number.

Question 395146: 1. the sum of the digits f a two-digit number is 11. when the digits are reversed, the new number is increased by 20 which is twice the original number. find the original number.
2. the units digit of a two-digit number is 3 less than its tens digit. if the number is divided by the sum of its digits, the quotient is 6 and the remainder is 8. find the number.
pls help me!
Found 2 solutions by stanbon, jsmallt9:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1. the sum of the digits f a two-digit number is 11. when the digits are reversed, the new number is increased by 20 which is twice the original number. find the original number.
---
Let the number be 10t+u.
Equations:
t + u = 11
10u+t+20 = 2(10t+u)
----
Substitute for "t" and solve for "u":
---
10u + 11-u + 20 = 20(11-u)+2u
---
9u + 31 = 220-20u+2u
-27u = -189
u = 7
---
Solve for "t":
t+u = 11
t+7 = 11
t = 4
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Original Number: 47
=============================

2. the units digit of a two-digit number is 3 less than its tens digit. if the number is divided by the sum of its digits, the quotient is 6 and the remainder is 8. find the number.
---
Equations:
u = t-3
(10t+u)/(t+u) = 6(t+u)+8
---
Substitute for "u" and solve for "t":
(10t+t-3)/(t+t-3) = 6(t+t-3)+8
----
(11t-3)/(2t-3) = 12t-18+8
----
(11t-3)/(2t-3) = 12t-10
---
11t-3 = 2(2t-3)(6t-5)
---
11t-3 = 2[12t^2-10t-18t+15)
---
11t-3 = 24t^2-56t+30
---
24t^2-67t+33 = 0
-----------------------
Comment:
This equation does not have a whole
number solution.
=====
Cheers,
Stan H.

Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
The solution to the first problem provided by another tutor is correct. The solution to the second problem has an error.

2. the units digit of a two-digit number is 3 less than its tens digit. if the number is divided by the sum of its digits, the quotient is 6 and the remainder is 8. find the number.
The proper equations which express this are:
u = t-3

We can eliminate the fractions in the second equation if we multiply both sides of the equation by (t+u):

which simplifies as follows:
10t + u = 6*(t+u) + 8
10t + u = 6t + 6u + 8
Substituting t-3 for u we get:
10t + (t-3) = 6t + 6(t-3) + 8
which simplifies to:
11t - 3 = 6t + 6t - 18 + 8
11t - 3 = 12t - 10
Subtracting 11t from each side:
-3 = t - 10
Adding 10 to each side:
7 = t
This makes u = 7-3 or 4.
The original number is 74.
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