digit of a two digit number is twice the tens digit when the number formed by reversing the digit is added to original number the sum is 99, find the original number?
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3
Answer
Let the two digit number is xy
y is digit at unit's place
x is digit at ten's place
⇒ Number can be written as = 10x+y
According to the conditions : y=2x ___ (1)
and
(10x+y)+(10y+x)=99
⇒11x+11y=99
⇒x+y=9 (dividing by 11)
Now x+2x=9 (from (1))
⇒3x=9
⇒x=9/3=3
y=2x=6
⇒ Number is 36
Answered by
1
Answer:
The original number is 12 times 3 = 36.
Step-by-step explanation:
Let's assume that, the tens digit is x
According to the problem, the units digit is 2x
Therefore, the original number is 10x + 2x = 12x
When the digits are reversed, the number becomes 10(2x) + x = 21x
According to the problem:
12x + 21x = 99
=> 33x = 99
∴ x = 3
∴ The original number is 12 times 3 = 36.
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