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Question 48327: The sum of the digits of a two-digit no. is 11. If the digits are reversed the new no. is 7 more than twice the original no. Find the original no.
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! First, lets figure it out by "looking":
2 digits add to 11, so they must be one of the following 4 possibilities:
29: 2+9 --> reversed is 92
38: 3+8 --> reversed is 83
47: 4+7 --> reversed is 74
56: 5+6 --> reversed is 65
and twice 38 is 76. Then add 7 is 83. So the answer is 38.
Right then...mathematically:
Let the first number be xy and we are told that (x+y) = 11.
Also, x is the number of tens and y is the number of units. By this we mean that xy really means (10x+y).
If we reverse it, to yx, when then have the number (10y+x).
We are told that "reversed number is twice original + 7", so we need to convert this English to maths:
(reversed number) = twice(original number) + 7
(10y+x) = 2(10x+y) + 7
10y+x = 20x+2y + 7
8y-19x = 7
So we have 2 equations now:
x+y=11
8y-19x = 7
scale up the first one by 8 --> 8x+8y = 88, giving us
8x+8y = 88
8y-19x = 7
Re-write the order:
8y+8x = 88
8y-19x = 7
Now lets subtract the equations, to get rid of the y terms.
27x = 81
x = 81/27
x = 3
so from x+y=11 we have
3+y=11
--> y=8
So the original number, xy is 38.
jon.
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