sum of the digits of a two-digit number is 11 when we interchange the digits it is found that the resulting new number is greater than the original number by 63 find the two digit number.
Answers
Given
Sum of the digits of a two-digit numbers = 11
Let unit’s digit of a 2-digit number be x
Then ten’s digit will be 11 – x
So, number = x + 10 (11 – x)
= x + 110 – 10x
= 110 – 9x
Now, by interchanging the digits, we get,
One’s digit of a new number = 11 – x
And ten’s digit will be = x
Hence, number = 11 – x + 10x
= 11 + 9x
According to the condition,
11 + 9x – (110 – 9x) = 63
11 + 9x – 110 + 9x = 63
18x = 63 – 11 + 110
18x = 162
We get,
x = 162 / 18
x = 9
So, original number = 110 – 9x
= 110 – 9 × 9
= 110 – 81
= 29
Therefore, the original number is 29
Answer:
Let the number be 10x+y
on interchanging the digits the number will be 10y+x
a/q = x+y = 11
= y = 11-x - call this as equation 1
again a/q = 10y+x+63 = 10x+y
= 10y-y+x-10x+63 = 0
= 9y-9x+63 = 0
= 9(y-x+7) = 0
= y-x+7 = 0/9
= y-x+7 = 0 - call it as equation 2
put value of equation 1 in equation 2:-
= 11-x-x+7 = 0
= 11-2x+7 = 0
= 11+7 = 2x
= 18 = 2x
= 18/2 = x
= 9 = x
put x = 9 in equation 1 ,we get :-
= y = 11-x
= 11-9
= 2
the two digit number will be :-
= 10x+y (put values of x and y)
= 10*9+2
= 90+2
= 92
Verification :- the sum of the two digits is 11
= 9+2=11
on interchanging the digit it is found that the resulting new number is greater than the original number by 63
= 92-29
= 63
Credit goes to @Khushigautam07
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