Question

the sum of a two digit number and the number formed by interchanging the digits is 110 .if 10 is subtracted from the first number, the new number is 4 more than 5 times the sums of the digits of the first number . find the first number.​

Answer 1

Assumption

Ones Digit be p

Also

Tens digit be n

Another number

= (10n + p)

Now

Interchanging the digits

Number = (10p + n)

(10n + p) + (10p + n) = 110

11n + 11p = 110

11(n + p) = 110

$\tt{\rightarrow n+p=\dfrac{110}{11}}$

n + p = 10 …….(1)

n = 10 - p …..…(2)

Hence

(10n + p) - 10 = 4 + 5(n + p)

(10n + p) - 10 = 4 + 5(10)

(10n + p) = 4 + 50 + 10

(10n + p) = 64

10(10 - p) + p = 64

100 - 10p + p = 64

100 - 9p = 64

-9p = 64  - 100

-9p = -36

$\tt{\rightarrow p=\dfrac{36}{9}}$

p = 4

p = 4

$\large{\boxed{\sf{Substitute\:the\:value\:of\:p\:in\:(2)}}}$

n = 10 - p

n = 10 - 4

n = 6

Therefore ,

Number is 6

Second number is 4

Another Number

10n + p

= 10 × 6 + 4

= 60 + 4

= 64

$\Large{\boxed{\sf{Another\:Number=64}}}$

Answer 2

Answer:

Step-by-step explanation:

Given :-

The sum of a two digit number and the number formed by interchanging the digits is 110.

If 10 is subtracted from the first number, the new number is 4 more than 5 times the sums of the digits of the first number.

To Find :-

The Numbers

Solution :-

Let the unit digit be y

And the tens digit be x.

Original number = (10x + y)

According to the Question,

(10x + y) + (10y + x) = 110

11x + 11y = 110

11(x + y) = 110

x + y = 110/11

x + y = 10…………...(1)

x = 10 - y……………(2)

(10x + y) - 10 = 4 + 5(x + y)

(10x + y) - 10 = 4 + 5(10)

(10x + y) = 4+ 50+10

(10x + y) = 64

Putting value of x, we get

10(10 - y) + y = 64

100 - 10y + y= 64

100 - 9y = 64

-9y = 64 - 100

-9y = -36

y = 36/9= 4

y = 4

Putting the value of y in eq 2

x = 10 - y

x = 10 - 4

x = 6

Original Number  

= 10x + y

= 10 × 6 + 4

= 60 + 4

= 64

Hence the number is 64.