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1) Two digit number is formed when unit place digit is multiplied by 1 and add to ten's digit which is multiplied by 10.

Algebra digit problems

Number formed(original number) = 10 x ten’s digit (x) + unit digit (y)Reversed number = 10 x unit digit ( y) + ten’s digit (x ) |

1) The digit at the ten’s place of a two digit number is twice the digit at the unit’s place. If the sum of this number and the number formed by reversing the digits is 66. Find the number .

Let the unit place digit = x

Ten’s place digit = 2x

Unit digit = 2x

Ten’s digit = x

As the sum of the number is 66.

21x + 12 x = 66

33x = 66

X = 2

So unit digit = 2

Number = 21x = 21(2)

2) The sum of digits of two digit number is 12. If the new number formed by reversing the digits is less than the original number by 54 .Find the original number.

Let the unit digit = x

Ten’s digit = 12 –x

Original number = 10( 12 –x ) + x

= 120 – 10x + x

Unit digit = 12 – x

Ten’s digit = x

New number = 10x + 12 – x

As new number is less than the original number by 54

9x +12 = 120 – 9x -54

9x +12 = 66 – 9x

9x + 9x = 66 -12

18 x = 54

X = 3

Original number = 120 – 9x

= 120 – 9(3)

= 120 – 27

3) The sum of a two digit number and the number obtained by reversing the order of the digit is 165. If the digits are differ by 3, find the original number.

Let unit digit = x

Ten’s digit = x + 3

Original number = 10(x + 3) +x

= 10x + 30 + x

Unit digit = x +3

Ten’s digit = x

New number = 10x + x + 3

As the sum of original number and the new number = 165

11x + 30 + 11x + 3 = 165

22x + 33 = 165

22x = 132

X = 6

Original number = 11 x + 30 = 11(6) +30

= 66 + 30

= 96

Here you may get answer as 69 also.

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