when the digits of a two-digit number are reversed, the new number is 9 more than the original number, and the sum of the digits of the original number is 11. what is the original number?
can you please help me set up the equation?
2007-02-07 04:34:14 · 9 answers · asked by smoovstella319 2 in Science & Mathematics ➔ Mathematics
A(10) + B = B(10) + A - 9
A+B = 11
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10A - A = 10B - B - 9
9A = 9B - 9
A = B -1
A+B = 11
(B-1)+B = 11
2B-1 = 11
2B = 12
B = 6
A+6 = 11
A=5
2007-02-07 04:39:45 · answer #1 · answered by Grant d 4 · 0⤊ 0⤋
Let x be the first digit and y be the second digit.
Then the number is
10x + y
When the digits are reversed, y will be the first digit and x the second digit, so the number after reversing the digits will be:
10y + x.
Since the number will be 9 more than the original number,
10y + x = 10x + y + 9
9y - 9x = 9
Dividing both sides by 9,
y - x = 1
y = 1 + x ..........equation (1)
Since the sum of digits in the original number is 11,
x + y = 11
y = 11 - x ...........equation (2)
From eqn. 1 and 2
1 + x = 11 - x
2x = 10
x = 5
Hence, y = 11 - 5 = 6
Thus the original number is
10x + y
= 10*5 + 6
= 50 + 6
= 56.
2007-02-07 12:53:16 · answer #2 · answered by Gopal Paudel 1 · 0⤊ 0⤋
Here's a two-digit number with digits a & b: 10a + b
Reverse the digits: 10b + a
Write the two equations:
the new number is 9 more than the original number:
10b + a = 10a + b + 9
9b = 9a + 9
b = a + 1
the sum of the digits of the original number is 11:
a + b = 11
Substitute b = a + 1 into a + b = 11:
a + a + 1 = 11
a = 5
b = 6
original number is 56
2007-02-07 12:42:13 · answer #3 · answered by ? 4 · 0⤊ 0⤋
56
Let XY be the original number. Just like 32 means 3 tens and 2 ones, we have 10X + Y for the smaller number. If we switch it, we get YX as a two digit number, or 10Y plus X. Now, it is nine more. So, our first equation is
(1) (10X + Y) + 9 = (10Y + X).
The sum of the digits is eleven, so that
(2) X + Y = 11.
That is your second equation.
2007-02-07 12:36:46 · answer #4 · answered by Asking&Receiving 3 · 0⤊ 0⤋
original number is 56. reverse the digits to make 65. 65 is 9 more than 56.
5+6=11
65-56=9
2007-02-07 12:38:28 · answer #5 · answered by Anonymous · 0⤊ 0⤋
56.
56 reversed is 65, 65-56=9, and 5+6=11
You know that the digits had to be next to each other (because it was nine) and that they had to add up to 11.
2007-02-07 12:39:13 · answer #6 · answered by Julia 3 · 0⤊ 0⤋
x is digit 1
y is digit 2
(y*10) + x = 9 + (x*10) + y
==> 9*y - 9*x = 9
x + y = 11
solves x=5 y=6
2007-02-07 12:43:27 · answer #7 · answered by ANC 2 · 0⤊ 0⤋
Let T and U be the tens and units digits of the original. Then 10T+U is its value, and the value of the reverse number is 10U+T. That makes your equations
10U+T = 9 + 10T+U and
T + U = 11.
2007-02-07 12:40:34 · answer #8 · answered by Philo 7 · 0⤊ 0⤋
y=17x+33-92
y=7
2007-02-07 12:36:21 · answer #9 · answered by Jessi-is-me 2 · 0⤊ 4⤋