Math, asked by rasheydamsel, 10 months ago

what is a two digit number that is increased by 20% when the order of its digit is reversed

Answers

Answered by VIGYAS
1

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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Answered by jeevansyriaccom
0

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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