The digits of a two digit number differ by 7. If the digits are intechanged and the resulting number is added to the original number, we get 27. Find the original number.
Answers
Complete step-by-step solution:
As we know that in a two-digit number there are two places. One’s place and tens place.
Like if ab is a two-digit number that a will be in ten’s place and b will be in one’s place. So, the number ab will be equal to 10a + b.
So, now the digits of the required number will be x and y.
Let x will be at ten’s place and y will be at one’s place.
So, the original number will be 10x + y.
Now as we know that the difference of the two digits of the number is 7.
So, x – y = 7 (1)
Now as given in the question that if the digits are interchanged and then added to the original number then the sum will be equal to 121.
So, if the digits are interchanged then y will be at ten’s place and x will be at one’s place.
So, the interchanged number will be 10y + x.
Now according to the question.
10x + y +10y + x = 121
11x + 11y = 121
On dividing both sides of the above equation by 11. We get,
x + y = 11 (2)
Now we had to solve equation 1 and equation 2 to get the value of x and y.
So, adding equation 1 and 2 . We get,
x – y + x + y = 7 + 11
2x = 18
Dividing both sides of the above equation by 2. We get,
x = 9
Now putting the value of x in equation 2. We get,
9 + y = 11
y = 11 – 9 = 2
So, the original number will be 10x + y = 10*9 + 2 = 92.
Hence, the original number is 92.
⭐Note
In these types of questions the major mistake that can happen is, if xy is a two-digit number where x will be at ten’s place and y will be in one’s place. Then we can say that the reverse of xy is yx but we cannot write xy + yx = 121 because xy and yx are the numbers that are formed when the two digits are placed together but while forming an equation we should understand that the number xy is formed when we add y with ten times of x (10x +y) because digit at tens place should be multiplied by the 10.