Solve the following question with correct explanation with full solution : Sum of the digit of a two digit number is 4 when we interchange the digits it is found that the resulting number is greater than the original number by 18. What is the two digit number ?
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Answers
Given data : Sum of the digit of a two digit number is 4 when we interchange the digits it is found that the resulting number is greater than the original number by 18.
To find : What is the two digit number ?
Solution : Sum of the two digit number is 4.
Let, unit/one's place digit be x
Hence, unit/one's place digit is 4 - x
Units/one's place digit, x = x ----{1}
Ten's place digit, y = 4 - x ----{2}
So, here,
Number is represented by 10y + x
Hence,
➜ Original number = 10y + x
➜ Original number = 10 * (4 - x) + x ----{3}
➜ Original number = 40 - 10x + x
➜ Original number = 40 - 9x
Now, after interchanging the digits;
➜ New number = 10x + y
➜ New number = 10x + (4 - x) ----{4}
➜ New number = 10x + 4 - x
➜ New number = 9x + 4
The resulting number is greater than the original number by 18.
➜ New number = Original number + 18
➜ New number - Original number = 18
➜ (9x + 4) - (40 - 9x) = 18
➜ 9x + 4 - 40 + 9x = 18
➜ 9x + 9x + 4 - 40 = 18
➜ 18x - 36 = 18
➜ 18x = 18 + 36
➜ 18x = 54
➜ x = 54/18
➜ x = 3
{Hence, unit/one's place digit is 3}
Put value of x in eq. {3}
➜ Original number = 10 * (4 - x) + x
➜ Original number = 10 * (4 - 3 ) + 3
➜ Original number = 10 * 1 + 3
➜ Original number = 10 + 3
➜ Original number = 13
Answer : Hence, the two digit number is 13.
Answer:
Let the digits at tens place and ones place: x and 9−x respectively.
∴ original number =10x+(9−x)
=9x+9
Now Interchange the digits: Digit at ones place and tens place: x and 9−x respectively.
∴ New number: 10(9−x)+x
=90−10x+x
=90−9x
AS per the question
New number = Original number +27
90−9x=9x+9+27
90−9x=9x+36
18x=54
x=
18
54
x= 3
Digit at tens place ⇒3 and one's place : 6
∴ Two digit number is 36
Like this this sum can be done