The tens digit of a two-digit number is two-thirds that of the units. Reversing the digits gives rise to a new

Question

The tens digit of a two-digit number is two-thirds that of the units. Reversing the digits gives rise to a new number that is 18 more than the original. Find the original number.

Answer 1

46 is the number you are looking for

Answer 2

Let the number be xy
x= 2/3 y
10y + x = 18 + 10x + y
substitute the first equation into the second
10 y + 2/3 y = 18 + 10 x 2/3 y + y
32/3 y = 18 + 23/3 y
18 = 9/3 y
18 = 3y
y = 6
x = 4

The original number is 46

Answer 3

The original number can be written

10a + b where a and b are respectively the tens and the units digits.

We have a = 2/3b
And on reversing digits

10b +a = (10a + b) + 18


Solving these equations gives you a = 4 and b = 6
Therefore the number was 46

Answer 4

let the original number be 10x+y,x and y being the ten's and umot's digits respectively
by the problem.
x=2y/3.......(1)
10y+x=10x+y+18....(2)
From eqn 2,we get
10y+x-10x-y=18
=> -9x+9y=18
=>-x+y=2
putting the value of x =2y/3 in this we get
-2y/3+y=2
=>y/3=2
y=6
hence,x=2*6/3=4
Therefore the original number is 46.

Answer 5

n = 10x + y

x = (2/3)y

10y + x = n + 18

Three equations, three unknowns.

10y + x = 10x + y + 18
9y - 9x = 18
y - x = 2

y - (2/3)y = 2
3y - 2y = 6
y = 6

x = 4

Therefore,

n = 10x + y = 10(4) + 6 = 46

The original number is 46.

Answer 6

a=first digit b=2nd digit ----------------------------- a=(2/3)b a-(2/3)b=0 ---------------------------- 10b+a-18=10a+b -9a+9b-18=0 ------------------------------ -9a+6b=0 -9a+9b-18=0 ------------ 0-3b-18=0 b=6 -------------------------------------- a=4