Question

a number consists of two digits, whose sum is 9. if the digits are reversed , the new number is 3/8 of the first​

Answer 1

Answer:

The sum of the two digits of a two digit number is 9. Let the two digits be x and y

x+y=9 ———Eq-(1)

Let the tens digit be x and units digit be y .

The number is 10x+y

If we reverse the digitas the number we get =10y+x.

Condition given in the problem is that the New number is 3/8 of the original number. If we put this condition into equation it will look like 10y+x=(3/8)(10x+y)

By solving the above equation we get

80y+8x=30x+3y. By rearranging the equation we get

30x+3y=80y+8x.

22x−77y=0 by dividing the left hand and right hand side by 11 we get

2x−7y=0 ————Eq-(2)

Eq - (1)×2 =2x+2y=18

Eq - (2)×1 =2x−7y=0

By subtracting the equation 2 from equation 1 we get

9y=18.

y=18/9=2.

By substituting the value of y in equation 1 we get value of x=7.

The original number is 72, if we interchange the digits we get 27. 27 is 3/8th of 72

Step-by-step explanation:

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