Mathematics
A number of two digits exceeds four times the sum of its digits by 6 and it is increased by 9 on reversing the digits. Find the number.
Answer
Let's consider the digits at ten's place as x and let the digit at unit's place be y.
Number = 10 × x + y = 10x + y,
On reversing digits the number is = 10 × y + x = 10y + x.
According to first condition, we have
⇒ 10x + y - [4(x + y)] = 6
⇒ 10x + y - 4x - 4y = 6
⇒ 10x - 4x + y - 4y = 6
⇒ 6x - 3y = 6
⇒ 2x - y = 2 ……(i)
According to second condition, we have
⇒ 10x + y + 9 = 10y + x
⇒ 10x - x + y - 10y = -9
⇒ 9x - 9y = -9
⇒ x - y = -1
⇒ y - x = 1 ……(ii)
Adding eq. (i) and (ii) we get,
⇒ 2x - y + (y - x) = 2 + 1
⇒ x = 3.
Substituting value of x in (ii) we get,
⇒ y - 3 = 1
⇒ y = 4.
Number = 10 × x + y = 10 × 3 + 4 = 30 + 4 = 34.
Hence, number = 34.
Related Questions
In a two digit number the sum of the digits is 7. If the number with the order of the digits reversed is 28 greater than twice the unit's digits of the original number, find the number.
When a two digit number is divided by the sum of its digits the quotient is 8. If the ten's digit is diminished by three times the unit's digit, the remainder is 1. What is the number?