Question

The sum of the digits of two digit number is 9. The fraction formed by taking 3 less than number as denominator and 3 more tan number formed by reversing digits of number as numerator is 25/8. Find the number

Answer 1

Answer:

$\huge\mathfrak\red{Hello\: Dude}$

Let ,

The digit in units place be x and digit in ten's place be y

x+y = 9 \ x = 9-y

∴ Original number =  x+10x

∴ New number = 10x+y+9

∴10x +y+9=3(10y+x)

∴10x+y+9= 30y+3x

∴ 10x-3x+9= 30y-y

∴7x+9=29y

∴7x+9-29y = 0

∴7(9-y) +9-29y = 0

∴63-7y+9-29y = 0

∴72-36y = 0

∴72=36y

∴y= 72/36

∴y=2

Now ,

∴x+y = 9

∴x+2 = 9

∴x=9-2=7

∴x=7

∴ Now , put x =7 ,y= 2 in eqaution .

∴Hence theOriginal number  is = 10y+x =10(2) +7= 27

Answer 2

Answer:

Required number is 27.

Step-by-step explanation:

Let,

The digit in units place be x and digit in ten's place be y

Therefore the number is yx

Given,

The sum of the digits of two digit number is 9

$x + y = 9 \\ x = 9 - y$

Therefore original number = (x + 10)x

New number = 10x+y+9

According to question,

$10x+y + 9=3(10y+x)$

$10x + y + 9 = 30y + 3x$

$10x - 3x + 9 = 30y - y$

$7x - 9 = 29y$

$7x + 9 - 29y = 0$

$7(9 - y) + 9 - 29y = 0 \\ 63 - 7y + 9 - 29y = 0 \\ - 29y - 7y = - 63 - 9 \\ - 36y = - 72 \\ y = \frac{ - 72}{ - 36} \\ y = 2$

Now,

x + y = 9

So,

$x + 2 = 9 \\ x = 9 - 2 \\ x = 7$

Therefore

$y = 2$ and $x = 7$

Required original number is 27.

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

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