Question

The sum of digits of a two digite number is 16new number by reversing the digite is greater than the original number by 18 .Find the original number

Answer 1

ⓗⓔⓨ ⓕⓡⓘⓔⓝⓓ

ⓗⓔⓡⓔ ⓘⓢ ⓨⓞⓤⓡ

ⓐⓝⓢⓦⓔⓡ



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⊙ Let the ones digit of the first number be x

⊙ Then ten's digit is ( 16-x )


So, the number is

10 ( 16-x) +x

160-10x+x

160-9x



Now , the number obtained by reversing the digits


10x + 16-x

9x+16


Now ,


New number - Original number = 18

( 9x+16 ) - (160-9x ) = 18

9x+16-160+9x = 18

18x -144 = 18

18x = 144+18

18x = 162

x = 162/18 = 9


so, the original number is 160-9x


=
$160 - (9 \times 9) \\ \\ 160 - 81 = 79$



Hope it helps you


$<b > prabhudutt$

Answer 2

✺QUESTION.


The sum of digits of a two digite number is 16new number by reversing the digite is greater than the original number by 18 .Find the original number




✺ANSWER.


Let the number be= 10X+Y

where X is tens place and y is ones place


so according to given values.



✦SUM OF TWO DIGITS NUMBER


First number. is X


Second number is Y



then,



X+Y= 16


X=16-Y..........eq(1)


ON REVERSE THE DIGITS


Reverse number (10 Y+X)


✦Reverse number - original number = 18✦



SO, (10Y+X) -(10X +Y)=18



10Y +X - 10X -Y=18



9Y-9X=18


Y - X=2...................eq(2)



✦From eq 1 And 2



Y - (16-Y)=2


Y - 16+Y=2


2Y=2+16


2Y=18


Y=9

and X= 16-Y
=16-9
=7




✦HENCE

X= 7and Y= 9

and Original number will be

10X+Y= 10(7)+9=70+9=79



and new number = 97




✦VERIFICATION .(MORE CLEARNESS)


Reverse number - original number = 18



=>97-79

=18