Question

cos
length and breadth.
1. The sum of the digits of a two-digit number is 14. The number is nineteen times the digit at its
one's place. Find the number.
both the numerator​

Answer 1

Answer :-

Let the unit's digit be x

Ten's digit be y

→ Number = 10y + x

Sum of digits = 14

→ x + y = 14 -i

According to the question :-

→ Number = 19 × unit's digit

→ 10y + x = 19x

→ 10y = 19x - x

→ 10y = 18x

→ 5y = 9x

→ 5y - 9x = 0 -ii

Multiplying equation i with 9 :-

→ 9 ( x + y ) = 14 × 9

→ 9x + 9y = 126 -iii

Adding equation iii and ii :-

→ 5y - 9x + 9x + 9y = 126 + 0

→ 5y + 9y = 126

→ 14y = 126

→ y = 9

Substituting the value in equation i :-

→ x + y = 14

→ x + 9 = 14

→ x = 5

Number = 10y + x = 90 + 5 = 95

Required number = 95

Answer 2

A N S W E R :

• The number both the numerator is 95.

Given :

• The sum of the digits of a two-digit number is 14

• The number is nineteen times the digit at its one's place

To find :

• Find the number ?

Solution :

• Let the number's one's place be x and tens place be y

So,

The number will be 10y + x (equation 1)

Sum of the digits of that number is 14,

• x + y = 14 (equation 2)

Also known that,

• The number is nineteen times the digit at its one's place,

=> 10y = 18x

=> 5y = 9x

=> y = 9x/5 (equation 3)

Substituting the values,

• equation (3) in equation (2),

Now, We have,

=> x + 9x/5 = 14

=> On solving this, we have,

=> 14x/5 = 14

=> x = 14 × 5 ÷ 14

=> x = 5

Substituting the value, of

• x in equation (2),

Now, We have,

=> 5 + y = 14

=> y = 14 - 5

=> y = 9

We got, x = 5, y = 9

Now, Put this values in equation (1)

=> 10(9) + 5

=> 90 + 5

=> 95

Hence,

• The number both the numerator is 95.