Math, asked by unknown9326, 1 year ago

The sum of the digits of a two–digit number is 7.The number obtained by iterchanging the digits exceeds the original number by 27.Find the number

Answers

Answered by akshitht03
0

Answer:


Step-by-step explanation:

Let the number be 10x+y.

So,

x+y=7..............(i)

Also,

10y+x=10x+y+27

9x-9y=-27

x-y=-3...............(ii)

Solving (i) and (ii) together,

x+y=7

x-y=-3

2x=4

x=2

y=5

The number=10(2)+(5)

                   =25


akshitht03: plz mark it the brainliest answer
Answered by Yuvrajpaul
0
Hey mate! Heres your answer⬇️

Let the two digits of the number be x,y and x be on in ten's position.

The number become (10x + y). eq(1)


Given
x+y=7

So y=7-x eq(2)
put these values in eq 1

We get the number as
(10x+7-x)
= (9x+7)

Now,Given that the digits are interchanged
so new eq becomes (10y+x)
but y=7-x

So the new eq becomes (10(7-x) +x)

that is

70 -9x

Given that
70-9x = 9x +7 +27

18x = 70-7-27
18x= 36
So
x=2...

and we know y=7-x
y= 5

The number is
10x+y

That is

10×2 + 5

= 25

Hope it helped
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