Question

A two-digit number is such that the product of the digits is 12. When 36 is added to the number the digits interchange their places. Find the two-digit number.

Answer 1

Answer:

दस के अंक को x और इकाई के अंक को y होने दें। नंबर = 10x + y xy = 12 = y = … (i) इसके अलावा, 10x + y + 36 = 10y + x 9x-9y + 36 = 0 x-y + 4 = 0… (ii) से (i) और (ii), x + 4 =  x2 + 4x-12 = 0 (x + 6) (x-2) = 0 x = 2 या x = -6 X = -6 को अस्वीकार करते हुए, हमारे पास x = 2 है। y =  =  = 6

ANSWER

Let the ten's digit of the number be x

It is given that the product of digits is 12

Unit's digit=

x

12

Number=10x+

x

12

If 36 is added to the number the digitis interchange their places

∴10x+

x

12

+36=10×

x

12

+x

⇒10x+

x

12

+36=

x

120

+x⇒9x−

x

108

+36=0⇒x

2

+4x−12=0 (divided throughout by 9)

hence, required quadratic equation is x

2

+4x−12=0

Step-by-step explanation:

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Answer 2

$\boxed{ \boxed{ \bf \dag \: Note : - }}$

Here, we have to find out a two-digit number, so we assume ten's digit of the number as x and then apply all conditions to get required quadratic equation.

$\boxed{\boxed{ \bf SOLUTION}}$

Let, the ten's digit of the number be x.

According to the question,

Product of the digits = 12

$\rm \: i.e. \: Ten's \: digit \times Unit's \: digit = 12$

$\implies \: \rm \: Unit's \: digit = \frac{12}{x} \: \: \: \: \: \: \: [ \because \: ten's \: digit = x]$

$\therefore \: \rm \: Two - digit \: number = 10x + \frac{12}{x}$

Also, it is given that if 36 is added to the number, the digits get interchange.

$\therefore \: \rm \: 10x + \frac{12}{x} + 36 = 10 \times \frac{12}{x} + x$

$\implies \: \rm \: 10x {}^{2} + 12 + 36x = 120 + x {}^{2}$

$\implies \: \rm \: 9x {}^{2} - 108 + 36x = 0$

$\implies \: \rm \: {x}^{2} + 4x - 12 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: [diving \: both \: sides \: by \: 9]$

which is the required quadratic equation.

By factorisation, we get

$\rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{2} + 6x - 2x - 12 = 0$

$\implies \: \rm \: x(x + 6) - 2(x + 6) = 0$

$\implies \: \rm \: (x + 6)(x - 2) = 0$

$\implies \: \rm \: x + 6 = 0 \: or \: x - 2 = 0$

$\implies \: \rm \: x = - 6 \: or \: x = 2$

But a digit can never be negative. So, x = 2.

Hence, the required two-digit number

$\: \: \: \: \: \: \: \: \: \: \rm \: = 10 \times 2 + \frac{12}{2} = 20 + 6 = \boxed{\underline{ \underline{ \red{26.}}}}$