Five times the sum of the digits of a two-digit number is 13 less than the original number. If you reverse the digits in the two-digit number, four times the sum of its two digits is 21 less than the reversed two-digit number.

(Hint: You can use variables to represent the digits of a number. If a two-digit number has the digit x in tens place and y in one’s place, the number will be 10x + y. Reversing the order of the digits will change their place value and the reversed number will 10y + x.)

The difference of the original two-digit number and the number with reversed digits is........

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AFOKE88 AFOKE88 · Answer 1 · 2021-05-09T09:40:42+02:00

Answer:

The difference of the original two-digit number and the number with reversed digits is 18.

Step-by-step explanation:

Since it's a two digit number, let x represent the tens digit and let y represent the units digit.

Thus, the original two digit number is;

10x + y.

The reverse two digit number is;

10y + x.

We are told that five times the sum of the digits of the two-digit number is 13 less than the original number.

Thus;

5(x + y) = (10x + y) - 13

Multiplying out the bracket gives;

5x + 5y = 10x + y - 13

Rearranging gives;

10x - 5x + y - 5y = 13

5x - 4y = 13 - - - - (3)

Also,we are told that, four times the sum of its two digits is 21 less than the reversed two-digit number. Thus;

4(x + y) = (10y + x) - 21 - - - (4)

Simplifying gives;

4x + 4y = 10y + x - 21

>> 10y - 4y - 4x + x = 21

>> 6y - 3x = 21 - - - (4)

Solving eq(3) and (4) simultaneously gives;

x = 5 and y = 3

Thus,

Original number = 53

Reversed number = 35

Difference between original and reversed number = 53 - 35 = 18