Question

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 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

34

Step-by-step explanation:

Let’s consider the digit at tens place as x

And let the digit at unit place be y

The number is 10 × x + y × 1 = 10x + y

So, reversing the number = 10 × y + x × 1 = 10y + y

Then according to the first condition, we have

10x + y = 4 (x + y) + 6

10x – 4x + y – 4y = 6

6x – 3y = 6

2x – y = 2 … (i)

And according to the second condition, we have

10x + y + 9 = 10y + x

10x – x + y – 10y = -9

9x – 9y = -9

x – y = -1 … (ii)

Now, subtracting (ii) from (i) we have

2x – y = 2

x – y = -1

(-)–(+)–(+)—

x = 3

Substituting the value of x in (i), we get

2(3) – y = 2

6 – y = 2

y = 6 – 2

y = 4

Therefore, the number is 10 x 3 + 4 = 30 + 4 = 34.