Question

A number between 10 and 100 is five times the sum of its digits. If 9 be added to it the
digits are reversed find the number.
hl 53​

Answer 1

Answer:

Let x be the first digit.

Let y be the second digit.

...

5(x+y)=10x+y ...(i)

5x+5y=10x+y

5y-y=10x-5x

4y=5x ...(ii)

...

The 10x+y is because the x digit is in the ten's column and it is worth more.

...

10x+y+9=10y+x ...(iii)

10x-x+9=10y-y

9x+9=9y

9(x+1)=9y

x+1=y ...(iv)

substitute x+1=y into (ii)

4(x+1)=5x ...(v)

4x+4=5x

4=5x-4x

x=4

substitute x=4 into (ii)

4y=5(4)

4y=20

y=20/4

y=5

...

The first digit is 4. The second digit is 5.

The number is 45.

Step-by-step explanation:

hope this will help you and you will Mark me as brainliest

Answer 2

Answer:

45

Step-by-step explanation:

This is 2-digit number as it is between 10 and 100

The number is: 10a+b

According to question:

• 10a+b=5(a+b) => 5a=4b

and

• 10a+b+9=10b+a => a+1=b

• 5a=4b

• 5a= 4(a+1)

• 5a=4a+4

• a=4

• b=4+1=5

Number is 45

Proof:

• 45=5*(4+5)
• 45+9=54