The sum of the digit of a number lying between 10 and 100 is 9. If the number is multiplied by 7 it becomes. Four times the number obtained by writing the digit in reverse order compare the digit of the number
Answers
Let the digits be 'x' and 'y'
Sum of the digits = x + y = 9. ------- (equation 1)
The digit in tens place is 'x' therefore it is expressed as '10x' and the digit in ones place is 'y' therefore it is expressed as '1y' ( this is for the original number)
When the digits are reversed then 'y' becomes '10y' and 'x' becomes '1x'
(10x+y)(7) = 4(10y+x)
70x+7y = 40y + 4x
66x - 33y = 0. -------------- ( equation 2)
Multiply equation 1 with 33
33x + 33y = 297
Now add equation 2 and equation 1
66x + 33y + 33x - 33y = 297
99x = 297
X = 297/99
x = 3
Substitute the value of x = 3 in equation 1
3 + y = 9
y = 6
The original number = 36
The reversed number = 63
Answer:
Let the digits be 'x' and 'y'
Sum of the digits = x + y = 9. ------- (equation 1)
The digit in tens place is 'x' therefore it is expressed as '10x' and the digit in ones place is 'y' therefore it is expressed as '1y' ( this is for the original number)
When the digits are reversed then 'y' becomes '10y' and 'x' becomes '1x'
(10x+y)(7) = 4(10y+x)
70x+7y = 40y + 4x
66x - 33y = 0. -------------- ( equation 2)
Multiply equation 1 with 33
33x + 33y = 297
Now add equation 2 and equation 1
66x + 33y + 33x - 33y = 297
99x = 297
X = 297/99
x = 3
Substitute the value of x = 3 in equation 1
3 + y = 9
y = 6
The original number = 36
The reversed number = 63