Question

ana
(ii) The sum of the digits of a two-digit number is 9. Also, nine times this number is
twice the number obtained by reversing the order of the digits. Find the number.​

Answer 1

ɢɪᴠᴇɴ :-

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

ᴛᴏ ғɪɴᴅ :-

• Original number
• Reversed number

Sᴏʟᴜᴛɪᴏɴ :-

Let the digit at tens place be x and ones place be y

then,

According to 1st condition :-

• Tens place digit + Ones place digit = 9

➭ x + y = 9

➭ x = 9 - y. --(1)

According to 2nd condition :-

• Original number = (10x + y)
• Reversed number = (10y + x)

• 9(Original number) = 2(Reversed number)

➭ 9(10x + y) = 2(10y + x)

➭ 90x + 9y = 20y + 2x

➭ 90x - 2x + 9y - 20y = 0

➭ 88x - 11y = 0

➭ 11(8x - y) = 0

➭ 8x - y = 0

➭ y = 8x. --(2)

Put value of (2) in (1) , we get,

➭ x = 9 - y

➭ x = 9 - 8x

➭ x + 8x = 9

➭ 9x = 9

➭ x = 9/9

➭ x = 1

Put x = 1 in (1) , we get,

➭ x = 9 - y

➭ 1 = 9 - y

➭ y = 9 - 1

➭ y = 8

Hence,

• Tens place digit = x = 1
• Ones place digit = y = 8

Therefore,

• Original no (10x + y) = 18
• Reversed no (10y + x) = 81

Answer 2

Given:

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

To find:

The number.

Solution:

Let the one digit and tens digit of the number be x and y.

Number= 10y + x

No. after reversing the digits = 10x + y

According to the question:

➻ x + y = 9 $\dashrightarrow\:$(1)

➻ 9(10y + x) = 2(10x + y)

➻ 90x + 9y = 20y + 2x

➻ 88y - 11x = 0

➻ -x + 8y = 0$\dashrightarrow\:$(2)

Adding eq 1 and 2,

We get,

➻ 9y = 9

➻ y = 1$\dashrightarrow\:$(3)

Putting all values in eq 1

➻ x + y = 9

➻ x + 1 = 9

➻ x = 9 - 1

➻ x = 8

Hence, the number is

➻ 10y + x

➻ 10 × 1 + 8

$\dashrightarrow\: \underline{\boxed{\bf{\orange{8}}}}$