the middle digit of a three digit number is 0 and the sum of the other two digits is 11. If the number obtained by reversing the digits exceeds the original number by 495?
Answers
Answer:308
Explanation:
Let the number in 100s digit be x,
and the number in 1s digit be y,
Then,the number would be x0y : 100x+y
Condition 1: Sum of the digits : x+y=11.....(i)
If it is reversed ,it would be y0x : 100y+x,
Given The difference between them: 100x+y -(100y+x) = -495
100x+y-100y-x =-495
99x-99y = -495 (divided by 99)
x-y=-5.....(ii)
Adding (i) and (ii), we get,
(x+y)+(x-y)=11+(-5)
x+y+x-y=11-5
2x=6
∴ x=3
Substituting value of x in (i),
x+y=11
3+y=11
y=11-3
∴ y=8
∴The number is 308
Answer:
Let the original number be a0b
Given sum of digits = a + 0 + b = 11
=> a + b = 11
=> a = 11 - b (eq. I)
Given that the original number when reversed gives a new number with exceeds it by 495, we can write the equation:
100b + a = 100a + b + 495
=> 99b - 99a = 495
=> 11b - 11a = 55 (eq. II)
Substituting for a from eq. I,
11b - 11(11-b) = 55
=> 11b - 121 +11b = 55
=> 22b = 176
= > b = 176/22 = 8
Therefore,
a = 11 - 8 = 3.
Therefore,
the original number is 308.
Check: 803 - 308 = 495. And 3 + 0 + 8 = 11.