a number consists of two digits whose sum is 9 if the number are reversed the new number is 3 by 8 of the first number find the number in one variable
Answers
Answer:
Let the number =ab=10a+b
Sum of digits =9
a+b=9
number −27=number with reversrd digits
10a+b−27=10b+a
9(a−b)=27
a−b=3
(a+b)+(a−b)=9+3⇒a=6,b=3
∴ Required number is 63
Answer:
With a problem like this, the best method is Trial and Error.
A number consists of two digits whose sum is nine.
What are the two-digit numbers whose digit-sum is nine?
09
18
27
36
45
54
63
72
81
90
We can cross out the first and last because 09 is not a two-digit number, and if we reverse the digits of 90, we get an illegal number.
18
27
36
45
54
63
72
81
If the digits are reversed the new number is 3/8 of the original number,
This lets us cross out the first four numbers because, when we reverse them, we end up with a bigger number.
54
63
72
81
Now, just write each of these as a fraction, with the new reversed-digit-number on top, then simplify the fraction:
45/54=5/6
36/63=?/?
27/72=?/?
18/81=?/?
Which one simplifies to three-eighths?