Math, asked by jasmin9, 1 year ago

please answer this question fast. content quality required please its urgent

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Answered by Anonymous

Hi there !!

Let the digit in tens place be x

The sum of digits is 9
Digit in units place = 9-x

The original number is 10(x)+9-x = 10x+9-x

= 9x + 9 _____(i)

Given,
if the digits are interchanged , the resulting number is greater that the original number by 27
So,
by interchanging the digits,
we have,
Digit in units place = x
Digit in tens place = 9-x
The new number is 10(9-x) + x = 90 - 10x + x

= 90 - 9x _____(ii)

The following balanced equation will be formed

90 - 9x = 9x + 9 - 27

90 - 9x = 9x - 18

Transposing 9x to RHS and 18 to LHS
we have,

90 + 18 = 9x + 9x

$108 = 18x$

$x = \frac{108}{18}$

x = 6

So,

digit in tens place = x = 6

Digit in tens place = 9 - x = 9 - 6 = 3

So,

the original number is 36

jasmin9: I wanted explanation so I asked the question

jasmin9: so it should be 36 as 6 +3=9

siddhartharao77: Thanks. Actually, I wanted to answer Because i know its 36. But It is also brainliested.

siddhartharao77: Thanks for replying.

jasmin9: hmmm

Anonymous: It actually depends on the place value. If we take first 9-x in units place and x in tens place, then the answer will be 36

jasmin9: yes

siddhartharao77: Thanks.

jasmin9: it was a bit confusing in ncert so I thought to asked here and I got perfect answet

Anonymous: Now the answer is done as per your requirement :D

Answered by vinayak19

Let be two numbers =a-d and a+d
So, a-d+a+d=9
2a=9
a=9÷2
without interchanging the place =10(a-d)+(a+d)
=10a-10d+a+d
=11a-9d
Interchanging the digits=10(a+d)+(a-d)
=10a+10d+a-d
=11a+9d
Now ,according to the question
11a-9d=11a+9d+27 (11a get cancelled)
-18d=27
-d=27÷18
-d=3÷2
d= -3÷2
So the numbers are
1st number is =a-d=9÷2-(-3÷2)=12÷2=6
2nd number is =a+d=9÷2+(-3÷2)=6÷2=3
so the two digits number is =63
I HOPE THIS WILL HELP YOU MARK ME BRAINLIEST

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