Question

five times the sum of the digits of a two digit number is 9 less than the number formed by reversing its digits.if 4 times the value of the digit is ones place is equal to half of the place value of the digit at tens place,find the number​

Answer 1

Answer:-

Let the number be (10x + y).

Given:

Case - 1 :

5 times the sum of the digits of the number is 9 less than the number formed by reversing digits.

→ 5(x + y) = (10y + x) - 9

→ 5x + 5y = 10y + x - 9

→ 5x - x = 10y - 5y - 9

→ 4x = 5y - 9

→ x = (5y - 9)/4 -- equation (1).

Case - 2 :

And, 4 times the value of the digits at ones place is equal to half the place value of digit at tens place (10x/2).

→ 4(y) = 1/2 * (10x)

→ 4y = 5x

Putting the value of x from equation - (1) we get,

→ 4y = 5 [ (5y - 9) / 4 ]

On cross multiplication we get,

→ 16y = 25y - 45

→ 16y - 25y = - 45

→ - 9y = - 45

→ y = ( - 45)/( - 9)

→ y = 5

Substitute the value of y in equation (1)

→ x = (5y - 9)/4

→ x = [ 5(5) - 9 ] / 4

→ x = 25 - 9/4

→ x = 16/4

→ x = 4

→ Number = 10(4) + 5 = 40 + 5 = 45.

Hence the required number is 45.

Answer 2

Step-by-step explanation:

Here , the tens place be x and once place be y.

the required number be 10x + y .

five times the sum of the digits of a two digit number is 9 less than the number formed by reversing its digits

According to the question

i.e 5(x + y) = (10y + x - 9)

or, 5x + 5y = 10y + x - 9

or, 4x = 5y - 9

or, x = (5y - 9)/4.............1

and

4 times the value of the digit is ones place is equal to half of the place value of the digit at tens place

According to the question.

4y = 10x/2

or, 4y = 5x............2

Now, put the value of x in equation (2) , we get

4y = 5(5y - 9)/4

or, 16y =25y -45

or, 9y = 45

or, y = 5

put the value of y in equation (1) , we get

x =(5×5 - 9)/4

or, x = 16/4

:.x = 4

now (x, y) = (4, 5)

thus,

the required number be 10x + y = 10×4 +5

so, the required number be =45.