Question

a number has two digits whose sum is 10 f 54 is added to the number its digits get interchange find the number
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Answer 1

Answer:

Step-by-step explanation:

qhlet the no be xy

A/Q

x+y=10-----(1)

xy+54=yx

=>10x +y +54=10y+x

=>10x+y-10y-x=-54

=>9x-9y= -54

=>x-y= -54/9

=>x-y=-6-------(2)

Now, (1)+(2)=>x+y+x-y=10-6

=>2x=4

=>x=2

(1)=>2+y=10

=>y=8

therefore the reqd. two digit no is 28

Answer 2

Answer:

28

Step-by-step explanation:

Let the number in the one's place be 'b' and the number in the ten's place be 'a'.

Then, the number be 10a + b

and the sum of the numbers is a + b.

As per the question,

a + b = 10 ----> (1)

If 54 is added to the number,

10a + b + 54 = 10b + a

10a - a + b - 10b = -54

{The + and - signs change due to transposition i.e., when we shift numbers or variables to either side of the = }

- 9b + 9a = -54

b - a = 6 ---->(2)

Now, we add equations (1) and (2).

a + b + b - a = 10+6

2b = 16

b = 16/2

b = 8

Next, we substitute b = 8 in equation (1).

a + b = 10

a + 8 = 10

a = 10 - 8

a = 2

Hence, a = 2 and b = 8.

Therefore, the number is,

10a + b = 10*2 + 8

= 20 + 8

= 28

The number is 28.