Question

a two digit number is such that the product of the digits is 14.when 45 is added to the number ,then the digits are reversed .find tge number.​

Answer 1

Answer:

Step-by-step explanation:

Solution :-

Let the ten’s place digit be x.

And the unit’s place digits = (14/x).

Number = (10x + 14/x)

According to the Question,

⇒ 10x + 14/x + 45 = 10 × 14/x + x

⇒ x² + 5x - 14 = 0

⇒ (x+ 7)(x - 2) = 0

⇒ x = - 7, 2 (Neglecting negative sign's one)

⇒ x = 2

Number = (10x + 14/x)

= (10 × 2 + 14/2)

= 27

Hence, the required number is 27.

Answer 2

Let the ten's and unit's digits of 2digit no.be x&y

Then the required no=10x+y

No.obtained by reversing the digits=10y+x

A/q,xy=14

Also,10x+y+45=10y+x

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

45=9y-9x

9(y-x)=45

(y-x)=5________i

But,(y+x)^2=(y-x)^2+4yx

=y+x=9________ii

Adding eqns i&ii we get

y-x=5

y+x=9

2y=14

y=7

Putting y=7 in eqn is we get

x=2

Thus original no.=27