Question

the sum of the digits of a two digit number is 15 if the number formed by reversing the digit is less than the original number by 27 find the original number​

Answer 1

Answer:

Let the unit's place=x

Then the ten's place=15−x

∴ original number=10(15−x)+x=150−10x+x=150−9x

By reversing the digits, we get

New number=10x+(15−x)=10x+15−x=9x−15

According to the problem,

original number−New number=27

⇒150−9x−9x+15=27

⇒−18x+165=27

⇒−18x=27−165=−108

⇒x=  −108 /-18 =6

Hence original number=150−9x=150−9×6=150−54=96

Answer 2

Given:-

• The sum of two digit number is 15
• The number formed by reversing the digit is less than the original number by 27.

To Find:-

• The number

Assumption:-

• Let the number in unit place be x
• Number in tens place = 15 - x

Solution:-

As we have number in unit place as x and in tens place as (15 - x),

The original number = 10(15 - x) + x

Original number = 150 - 10x + x

Original number = 150 - 9x

Now,

On reversing the number, x comes in the tens place and (15 - x) comes in ones place,

New number = 10x + (15 - x)

New number = 10x + 15 - x

New number = 15 + 9x

Now,

ATQ,

It is said that the new number formed by reversing the place of digits is less than the original number by 27.

Hence,

(15 - 9x) - (15 + 9x) = 27

=> 150 - 9x - 15 - 9x = 27

=> 135 - 18x = 27

=> -18x = 27 - 135

=> -18x = -108

=> x = -108/-18

=> x = 6

Now,

The numbers is as follows,

Number in units place = x = 6

Number in tens place = 15 - x = 15 - 6 = 9

Therefore the number becomes 96.

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