The sum of the digits of a two-digit number is 9. If the digits are reversed, the number formed
bears the ratio 3:8 to the original number. Find the original number.
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The sum of the digits of a two-digit number is 9. If the digits are reversed, the number formed
bears the ratio 3:8 to the original number. Find the original number.
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Verified answer
Let the digit in the tens place be x
Let the digit in the units place be y
Original number = 10x + y
Representing it mathematically we get our first equation.
x + y = 9
Reversed number = 10y + x
Representing the condition mathematically and solving further to get our second equation.
=> =
Cross multiplying,
=> 8 ( 10y + x) = 3 ( 10x + y)
=> 80y + 8x = 30x + 3y
=> 80y - 3y = 30x - 8x
=> 30x - 8x = 80y - 3y
=> 22x = 77y
=> 22x - 77y = 0 -----> 2
Multiply equation 1 by 77,
x + y = 9 ----> 1
77 × x + 77 × y = 77 × 9
77x + 77y = 693 -----> 3
Solve equations 2 and 3 simultaneously by elimination method.
Add equation 2 to 3,
+ 22x - 77y = 0 -------> 2
+ 77x + 77y = 693 ------> 3
_________________
99x = 693
x =
x = 7
Substitute x = 7 in equation 1,
x + y = 9
7 + y = 9
y = 9 - 7
y = 2
Digit at tens place = x = 7
Digit at units place = y = 2
Original Number = 10x + y
Original number = 10 ( 7) + 2
Original number = 70 + 2 = 72