Question

A two digit no. Is such that the product of their digit is 14 . When 45 is added to the no. Then the digit are reversed find the no.

Answer 1

Answer:27

Step-by-step explanation:

Let the digit of the no. be X and Y

So according to the question:

X × Y = 14

XY = 14

Let the two digit no. be 10x + y

10x + y + 45 = 10y + X

10x - X+y - 10y= -45

9x - 9y = -45

9(x-y) = -45

x-y = -5

X= -5+y

x×y=14

-5 + y(y) = 14

-5y+y^2=14

y^2-5y-14=0

y^2-7y+2y-14=0

y(y-7)+2(y-7)=0

y=-2(neglected)

or

y=7

We cannot take a negative value so we must take y=7

XY= 14

X × 7= 14

X= 14/7

X= 2

So the required no. is 27

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Answer 2

☯ Let's consider the two digits number be x and y.

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Therefore,

• Number = 10x + y

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$\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}$

• Product of digits of a two digit number is 14.

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$:\implies\sf xy = 14\\ \\ :\implies\sf y = \dfrac{14}{x}\qquad\qquad\bigg\lgroup\bf eq\;(1)\bigg\rgroup$

• When 45 is added to the no. Then the digit are reversed.

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$:\implies\sf (10x + y) + 45 = 10y + x\\ \\ :\implies\sf 10x - x + 45 = 10y - y\\ \\ :\implies\sf 9x + 45 = 9y\\ \\ :\implies\sf 9x - 9y + 45 = 0\\ \\ :\implies\sf 9(x - y + 5) = 0\\ \\ :\implies\sf x - y + 5 = 0\qquad\qquad\bigg\lgroup\bf eq\;(2)\bigg\rgroup$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━

$\dag\;{\underline{\sf{Now,\; Putting \;eq \;(1)\; in\; eq \;(2),}}}$

$:\implies\sf x - \dfrac{14}{x} + 5 = 0\\ \\ :\implies\sf \dfrac{x^2 - 14 + 5x}{x} = 0\\ \\ :\implies\sf x^2 - 14 + 5x = 0 \times x\\ \\ :\implies\sf x^2 - 14 + 5x = 0\\ \\ :\implies\sf x^2 + 5x - 14 = 0$

$\dag\;{\underline{\sf{Splitting\;middle\;term,}}}$

$:\implies\sf x^2 + 7x - 2x - 14 = 0\\ \\ :\implies\sf x(x + 7) - 2(x + 7) = 0\\ \\ :\implies\sf (x + 7)(x - 2) = 0\\ \\ :\implies\sf x + 7 = 0\quad or \quad x - 2 = 0\\ \\ :\implies\sf x = - 7\quad or \quad x = 2$

Since, Value of x can't be negative.

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$:\implies{\underline{\boxed{\frak{\purple{x = 2}}}}}\;\bigstar$

$\dag\;{\underline{\sf{Now,\; Putting \;value \;of \;x\; in\; eq \;(1),}}}$

$:\implies\sf y = \cancel{ \dfrac{14}{2}}\\ \\ :\implies{\underline{\boxed{\frak{\pink{y = 7}}}}}\;\bigstar$

$\therefore{\underline{\sf{Hence,\;the\:two\:digit\;number\;is\; \bf{27}.}}}$