Question

The sum of the digits of a two-digit number is 9. If the digits are reversed, the number formed
bears the ratio 3:8 to the original number. Find the original number.​

Answer 1

$\bf{\huge{\underline{\boxed{\sf{\blue{Answer:}}}}}}$

$\bf{\underline{\sf{\red{Given:}}}}$

• The sum of the digits of a two-digit number is 9.
• When the digits are reversed, the number formed bears the ratio 3:8 to the original number

$\bf{\underline{\sf{\red{To\:find :}}}}$

• The original number.

$\bf{\underline{\sf{\red{Solution:}}}}$

Let the digit in the tens place be x

Let the digit in the units place be y

Original number = 10x + y

$\bf{\underline{\sf{\green{As\:per\:first\:condition\::}}}}$

• The sum of the digits of a two-digit number is 9.

Representing it mathematically we get our first equation.

x + y = 9

$\bf{\underline{\sf{\green{As\:per\:second\:condition\::}}}}$

• When the digits are reversed, the number formed bears the ratio 3:8 to the original number

Reversed number = 10y + x

Representing the condition mathematically and solving further to get our second equation.

=> $\sf\frac{10y + x }{10x + y }$ = $\sf\frac{3}{8}$

Cross multiplying,

=> 8 ( 10y + x) = 3 ( 10x + y)

=> 80y + 8x = 30x + 3y

=> 80y - 3y = 30x - 8x

=> 30x - 8x = 80y - 3y

=> 22x = 77y

=> 22x - 77y = 0 -----> 2

Multiply equation 1 by 77,

x + y = 9 ----> 1

77 × x + 77 × y = 77 × 9

77x + 77y = 693 -----> 3

Solve equations 2 and 3 simultaneously by elimination method.

Add equation 2 to 3,

+ 22x - 77y = 0 -------> 2

+ 77x + 77y = 693 ------> 3

_________________

99x = 693

x = $\sf\frac{693}{99}$

x = 7

Substitute x = 7 in equation 1,

x + y = 9

7 + y = 9

y = 9 - 7

y = 2

Digit at tens place = x = 7

Digit at units place = y = 2

Original Number = 10x + y

Original number = 10 ( 7) + 2

Original number = 70 + 2 = 72

Answer 2

ANSWER:-

Given:

The sum of the digits of a two- digit number is 9.If the digits are reversed, the number formed bears the ratio 3:8 to the original number.

To find:

The original number.

Solution:

Let the tens digit = a

Let the units digit= b

The actual number is 10a + b

The number with the digita reversed is =10b + a.

Therefore,

a+b = 9...............(1)

$\frac{10 b+ a}{10 a+ b} = \frac{3}{8}...............(2)$

So, from equation (2), we get:

=) 8(10b+a)= 3(10a+b)

=) 80b+ 8a = 30a+ 3b

=) 30a- 8a = 3b-80b

=) 22a = -77b = 0

=) 22a = 77b

=) a= 77b/22

=) a = 7b/2

Therefore,

Substituting in equation (1), we get:

=) 7b/2 + b= 9

=) 7b +2b = 18

=) 9b= 18

=) b= 18/9

=) b= 2

Since,

a+b = 9

=) a + 2 = 9 [b= 2]

=) a= 9-2

=) a= 7

Hence,

The original number is 72.

Hope it helps ☺️