Question

solve question 2and3 with full explanation. ​

Answer 1

Step-by-step explanation:

→ 1) .

Let the tenth place digit be ' x '

and ones place digit be ' y '

So the two digit number be ' 10 x+y '

A/c to the first condition , 'Sum of two digits is 7',

$=>x+y=7...(1)$

A/c to the second condition 'If the digits are reversed the new number is equal to 3 less than 4 times the original number'

$=>10y+x=4(10x+y)-3\\\\=>10y+x=40x+4y-3\\\\=>6y-39x+3=0\\\\=>2y-13x+1=0\\\\=>13x-2y=1...(2)$

Now,solve 2*(1)+(2) ,

  $=>2(x+y)+(13x-2y)=2(7)+1\\\\=>2x+2y+13x-2y=14+1\\\\=>15x=15\\\\=>x=1$

Sub. x value in (1) , we get ,

       $=>(1)+y=7\\\\=>y=6$

So the  number is ,

          $=>10(1)+6\\\\=>16$

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→ 2 ) .

When the digits are interchanged , the number  would be :-  10 y + x

A/c to the condition , 'The sum of a two digit number and the number obtained by interchanging the digits is 110'

$=>10x+y+10y+x=110\\\\=>11x+11y=110\\\\=>x+y=10...(1)$

A/c to the 2nd condition , 'Difference of digits is 6'

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

Now solve (1) + (2) , Then we get ,

$=>(x-y)+(x+y)=10+6\\\\=>2x=16\\\\=>x=8$

Now sub. x value in (2) , we get ,

              $=>y=2$

So the number is ,

                 $=>10x+y\\\\=>10(8)+2\\\\=>80+2\\\\=>82$

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