The sum of a two digit number and the number formed by interchanging the digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sums of the digits of the first number. Find the first number.
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13
case 1
10x+y+10y+x=110
11x+11y=110
x+y=10-----------------I
case 2
10x+y-10=5(x+y)+4
5x-4y=14--------------II
From I and II
5x-4y=14
x+y=10 x 4
9x =54
x=6
y=4
The number is 64
Answered by
4
Let the digit at ones place = y
Let the digit at tens place = x
original number = 10x + y and
number formed by reversing the digits = 10y + x
therefore, the equation becomes,
Refer to the attachment for equation........1
on subtracting 10 from the original number , we get = 10x + y -10
and
4 more than 5 times the sum of digits = 5(x + y) + 4
now, the equation becomes:
Refer to the attachment for the equation..2
on solving (1) and (2), by the elimination method, we get x = 6 and y = 4.
Thus, the original number = 64.
Let the digit at tens place = x
original number = 10x + y and
number formed by reversing the digits = 10y + x
therefore, the equation becomes,
Refer to the attachment for equation........1
on subtracting 10 from the original number , we get = 10x + y -10
and
4 more than 5 times the sum of digits = 5(x + y) + 4
now, the equation becomes:
Refer to the attachment for the equation..2
on solving (1) and (2), by the elimination method, we get x = 6 and y = 4.
Thus, the original number = 64.
Attachments:
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