Question

In a two digit number, digit at the ten's place is twice the digit at units's place. if the number obtained by interchange the digit is added to the original number, the sum is 66. Find the number

Answer 1

Originally Answered: One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two digit number and add the resulting number to the original number, you get 88. What is the original number?

Let's call the first digit a and the second digit b. Also, let's say b is the larger digit.

We know from the question:

10a + b + 10b + a = 88

=> 11a + 11b = 88

and

b = 3a => 3a - b = 0

Let's make a system of equations and solve it:

11a + 11b = 88

3a - b = 0

Multiplying the second equation by 11 gives us the following:

11a + 11b = 88

33a - 11b = 0

As we now get rid of b, let's solve a:

44a = 88

a = 2

Let's put the value of a in place of a in the second equation:

3*2 = b

=> b = 6

So the two numbers are 26 and 62.

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Answer 2

Answer:

Let the units digit be x.

tens digit=2x

original no.=2x*10 + x=21x

no. formed after interchanging digits=10*x+2x=12x

According to the question,

21x+12x=66

33x=66

x=66/33=2

therefore no.=10*2*2 + 2=42