write a two digit number with the sum of its the number got by interchanging the digits is 27 more than the original number what is the number
Answers
Answer:
Let’s represent the tens digit with ‘a,” and the ones digit with ‘b.’
The value of this number is 10a + b; this number is 27 more than the number formed by the same digits in the other order, so 10a + b = 10b + a + 27.
The digit sum is 11; a + b=11, or b = 11 - a.
10a + b = 10b + a + 27
10a + (11 - a) = 10(11 - a) + a + 27
10a + 11 - a = 110 - 10a + a + 27
9a + 11 = 110 - 9a + 27
9a + 11 = 137 - 9a
18a + 11 = 137
18a = 126
a = 126/18
a = 7
b = 11 - 7 = 4
So the original number is 74.
The sum of the digits is 7+4=11.
And if you reverse the digits you get 47; 47 + 27 = 74.
Answer:
Let the units place be x
then the tens digit be 7 - X
number formed by these digits = 10 x t's digit+u's digit
=> 10(7-x)+x
70 x -10 x +x
70 x -9x
when the digits are interchanged than,
t's digit= x
u's digit = 7-x
the new no. formed = 10x + 7-X
= 9 x +7
given that the no. exceeds by 27
new no. - given no.
9x +7 - (70-9x)=27
9 x +7- 70 + 9x =27
18x -63= 27
18 x = 63+27
18 x= 90
x= 90/18
=5
the no. = 70-9x
=> 70-45
=> 25 ans
verify : 2+5 = 7