Question

The average of two 2-digit positive integers is equal to the decimal number obtained by writing one of the two-digit integers before the decimal point and the other two-digit integer after the decimal point. What is the smaller of the two integers?

Answer 1

Explanation:

Let us consider two the numbers be a and b.

Avg. of the two digits =
`(a+b)/(2)`

If it is (a.b) , then the actual value of the number is (if you see 21.45, you get it by 21 +45/100)

∴equating =
`(a+b)/(2)=a+(b)/(100)`

On solving this equation, we get 50a = 49b

We can now write this as

a = 49b - 49a

∴ a = 49 ( b - a )

This tells us a few things. First, b - a has to be a positive value, because it can't be 0 other wise a would become 0.

It cannot be negative because then a would be negative .. and we know, a is a 2 digit positive integer.

∴ b > a ..

Also, if you notice .. whatever value ( b - a ) is equal to .. say 1, 2, 3, say 15 .. whatever ...

a will always be a multiple of 49 ..

because a will be 49 × ( 1 or 2 or 3 or say 15 .. )

That means, a has to be a multiple of 49.

That said, there are only 2 such numbers within 100 which are multiples of 49.

One is 49 itself, the other is 98.

So now we have some idea , as to what a can be .. it can be either 49 or 98.

Lets start with the assumption that a = 98. What happens then ?

We know b > a .. so it means b has to be a value less than 100 but greater than 98.

The only such value is 99. ∴ We get b = 99.

But if you do the math, = 98.50 .. it doesn't satisfy the second condition, "obtained by writing one of the two-digit integers before the decimal point and the other two-digit integer after the decimal point."

So that means, a = 49 for sure.

Now that we know a, its easy to find b from :

50a = 49b

∴ b = 50

If you test it out,
`(49+50)/(2)`