Question

the one digit of 2 digit number is twice the tens digit when the number formed by reversing the digit is added to the original number the sum is 99 find the original number

Answer 1

$\orange{hey\:mate}$⤵
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⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬
ʟᴇᴛ ᴛʜᴇ ᴅɪɢɪᴛ ᴀᴛ ᴛᴇɴꜱ ᴩʟᴀᴄᴇ ʙᴇ x
ᴀɴᴅ ᴅɪɢɪᴛ ᴀᴛ ᴏɴᴇꜱ ᴩʟᴀᴄᴇ = 2x
ᴏʀɪɢɪɴᴀʟ ɴᴜᴍʙᴇʀ=10x+2x
=12x
ᴀꜰᴛᴇʀ ʀᴇᴠᴇʀꜱᴇᴅ
ɴᴜᴍʙᴇʀ=10(2x)+x
=20x+x
=21x.
ᴀᴛq
12x+21x=99
33x=99
x=3
ʜᴇɴꜱᴇ ᴛʜᴇ ᴅɪɢɪᴛ ᴀᴛ ᴏɴᴄᴇ ᴩʟᴀᴄᴇ=2x=2×3=6
ᴛʜᴇ ᴅɪɢɪᴛ ᴀᴛ ᴛᴇɴꜱ ᴩʟᴀᴄᴇ=x=3
ᴛʜᴇ ʀᴇqᴜɪʀᴇᴅ ᴏʀɪɢɪɴᴀʟ ɴᴜᴍʙᴇʀ ɪꜱ 63
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$\orange{hope\:it\:helps}$

Answer 2

Answer:

Let the tens digit be y and the ones digit be x.

The original number = 10y + x

The reverse number = 10x + y

It is given that ones digit is twice the tens digit :]

➳ x = 2y ............[Equation (i)]

According to question now,

➳ 10x + y + 10y + x = 99

➳ 11x + 11y = 99

➳ 11 (x + y) = 99

➳ x + y = 99/11

➳ x + y = 9

➳ y = 9 - x.........[Equation (ii)]

Now, Substituting equation (ii) in equation (i) we get :

➳ x = 2 (9 - x)

➳ x = 18 - 2x

➳ 3x = 18

➳ x = 18/3

➳ x = 6

Putting x = 6 in equation (ii) we get :

➳ y = 9 - x

➳ y = 9 - 6

➳ y = 3

Therefore,

The original number = 10y + x = 10(3) + 6 = 30 + 6 = 36