Question

in a two digit number sum of its digit is7 of the number is reversed the new numbers 28 greater than twice the the unit digit of the original number​

Answer 1

Gɪᴠᴇɴ :-

In a two digit number sum of its digit is7 of the number is reversed the new numbers 28 greater than twice the the unit digit of the original 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 = 7

➮ x + y = 7

➮ x = 7 - y. --(1)

According to 2nd condition :-

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

• Reversed number = 28 + 2(Unit digit)

➮ (10y + x) = 28 + 2(y)

➮ 10y + x = 28 + 2y

➮ 10y - 2y + x = 28

➮ 8y + x = 28

➮ x = 28 - 8y. --(2)

From (1) and (2) , we get,

➮ (7 - y) = (28 - 8y)

➮ 8y - y = 28 - 7

➮ 7y = 21

➮ y = 21/7

➮ y = 3

Put y = 3 in (1) , we get,

➮ x = 7 - y

➮ x = 7 - 3

➮ x = 4

Hence,

• Tens place digit = x = 4
• Ones place digit = y = 3

Therefore,

• Original number (10x + y) = 43
• Reversed number (10y + x) = 34

Answer 2

Given :-

in a two digit number sum of its digit is7 of the number is reversed the new numbers 28 greater than twice the the unit digit of the original number

To Find :-

• Orignal number
• Reserved number

Solution :-

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

Then,

According to 1st condition :-

• tens place digit + once place digit = 7

↪️ x + y = 7

↪️ x = 7 - y --> (1)

According to 2nd condition :-

• orignal number = (10x + y)

• Reserved number = (10y + x)

• Reserved number = 28 + 2 ( Unit digit )

↪️ (10y + x) = 28 + 2 (y)

↪️ 10y + x = 28 + 2y

↪️ 10y - 2y + x = 28

↪️ 8y + x = 28

↪️ x = 28 - 8y -->(2)

From (1) and (2) we get :-

↪️ x = 7 - y

↪️ x = 7 - 3

↪️ x = 4

Hence,

• Tens place digit = x = 4
• once place digit = y = 3

Therefore,

• orignal number = (10x + y) = 43
• Reserved number = (10y + x) = 34

Thanks