what is a two digit number that is increased by 20% when the order of its digit is reversed
Answers
Answer:
The sum of digits of required number is 9
Consider the provided information.
A two digit number is increased by 20% when its digits are reversed.
Let 10x + y is the original two digit number
Now reverse the order.
10y + x is the number with digits reversed
It is given that the number increase by 20% when its digits are reversed.
Therefore,
1.2(10x + y) = 10y + x
12x +1.2y= 10y + x
11x=8.8y
x=0.8y
By hit and trial we can find that at x=4 we get y=5 (Both are integer)
Hence, the original number is 45.
The sum of its digits is 4+5=9
Hence, the sum of digits of required number is 9
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Answer:The sum of digits of required number is 9.
Step-by-step explanation:
Consider the provided information.
A two digit number is increased by 20% when its digits are reversed.
Let 10x + y is the original two digit number
Now reverse the order.
10y + x is the number with digits reversed
It is given that the number increase by 20% when its digits are reversed.
Therefore,
1.2(10x + y) = 10y + x
12x +1.2y= 10y + x
11x=8.8y
x=0.8y
By hit and trial we can find that at x=4 we get y=5 (Both are integer)
Hence, the original number is 45.
The sum of its digits is 4+5=9
Hence, the sum of digits of required number is 9
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