Question

The sum of a two digit number and the numerator formed by interchaning the digit 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

Answer:

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=

11

110

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=

9

36

p = 4

p = 4

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

Number=64

AnotherNumber=64

Answer 2

AnswEr :

Let the Ones Digit be x & Tens Digit be y.

⋆ No. Formed will be : (10y + x)

⋆ No. Formed by Interchanging : (10x + y)

• First Part of the Question :

⇒ Original No. + Interchanging No. = 110

⇒ (10y + x) + (10x + y) = 110

⇒ 11y + 11x = 110

⇒ 11(y + x) = 110

• Dividing Both term by 11

⇒ y + x = 10 ⠀⠀⠀⠀⠀⠀⠀⠀⠀—eq. ( I )

⇒ x = 10 - y ⠀⠀⠀⠀⠀⠀⠀⠀⠀—eq. ( II )

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• Second Part of the Question :

⇒ New No. - 10 = 4 + 5 × (Sum of Digits)

⇒ (10y + x) - 10 = 4 + 5 × (y + x)

⇒ 10y + x - 10 = 4 + (5 × 10)

• From eq.( I )

⇒ 10y + x - 10 = 4 + 50

⇒ 10y + x = 54 + 10

• putting the value of x from eq.( I )

⇒ 10y + 10 - y = 54 + 10

⇒ 10y - y = 54 + 10 - 10

⇒ 9y = 54

• Dividing Both term by 9

⇒ y = 6

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• Putting the Value of y in eq. ( II ) :

⇒ x = 10 - y

⇒ x = 10 - 6

⇒ x = 4

━━━━━━━━━━━━━━━━━━━━━━━━

• Original Number :

↠ (10y + x)

↠ 10(6) + 4

↠ 60 + 4

↠ 64

∴ Original Number formed will be 64.