Question

a two digit number is such that product is 14 if45 is added to the number the digits interchange their places find the number

Answer 1

Let the ten’s digit be x and ones digit be y
Given (xy) = 14 … (1)
The number is (10x + y)
Given that when 45 is added to the number the digits gets interchanged.
Hence 10x + y + 45 = 10y + x
9x – 9y + 45 = 0
x– y + 5 = 0 … (2)
From (1) and (2), we get
x – (14/x) + 5 = 0
x2 + 5x – 14 = 0
x2 + 7x – 2x – 14 = 0
x(x + 7) – 2(x + 7) = 0
(x + 7)(x – 2) = 0
x = - 7 or x = 2
Since the digits cannot be negative, x = 2
xy = 14
Therefore, y = 7
The number is (10x + y) = 27
May this helful

Answer 2

Answer: 27

Step-by-step explanation:

Suppose the tens and unit digit be m and p

Therefore  

mp = 14

Required number = (10m + p)

On reversing  = (10p + m)

Make an equation we get  

(10m + p) + 45 = (10p + m)    

9(p - m) = 45

p - m = $\bf\huge\frac{45}{9}$

p - m = 5 ……. (Eqn 1)  

Hence

(p + m)^2 - (p - m)^2 = 4mp

(p + m) = $\bf\huge\sqrt{(p - m)^2 + 4 mp}$  

(p + m) = $\bf\huge\sqrt{25 + 4\times 14}$

(p + m) = $\bf\huge\sqrt{81}$

p + m = 9 …… (2)

Adding (1) and (2) we get

2p = 14

p = $\bf\huge{14}{2}$

p = 7

Put value of p in (2) we get

7 + m = 9

m = 9 - 7

m = 2

Therefore

m = 2 and p = 7 and required number is 27