# I need help on a few math problems. 1. The sum of the two digits of a positive integer is 12. When the digits were reversed, the new number was 54 greater than the original. What is the product of...

I need help on a few math problems.

1. The sum of the two digits of a positive integer is 12. When the digits were reversed, the new number was 54 greater than the original. What is the product of the digits of the original number minus the original number?

2. How many possible answer combinations are there to a nine question multiple choice test with five options per question.

3. 1000 ml of a saline solution contained 52% salt. How much water should be added to make the solution 51% water?

### 3 Answers | Add Yours

Oh, to make sure you understand which one the answer is for the third part, we would end up with y, 1019.61 ml. But, the problem is asking for how much should be added. That's x. So, the answer for the third problem is x, 19.61 ml.

For the first one, one needs to consider that for two digit numbers, like 32, we can write that as:

32 = 10*3 + 1*2

As in 10 the the "tens" place value and 1 for the "ones" place value

Therefore, for an unknown 2 digit number, that would be written as:

10*x + 1*y or 10x + y

Now, the problem says that the sum of the digits is 12. That wouldn't include the 10 and 1, only the digits. So:

x+y = 12

Now, the problem says, "when the digits were reversed". So, that means the second digit would be:

10y + x

The problem says that this number is 54 greater than the first number. So, then, we would have:

10x + y + 54 = 10y + x

9x - 9y = -54

So, we have a system of equations:

x+y = 12

9x - 9y = -54

We can solve this various ways. Using linear combination, we can multiply the first equation by 9 then add the equations:

9x+9y = 108

9x-9y = -54

18x = 54

x = 3

So, then, if x = 3, we can substitute this into the first equation:

3+y = 12

y = 9

So, the numbers would be 39 and 93. And, when one checks, 93 is in fact 54 greater than 39.

For the second one, assuming one is talking specifically of "combinations" for the test, that means order of the questions doesn't make a difference. So, all 9 questions would be included on the test. So, that would only be one combination. When one considers the answers as well, we still get the same thing. Since order doesn't make a difference, we could order the answers different on every question. Since they would still be all the same answers, only ordered differently, it still still be only one combination.

For the last one, we can make another system of equations. We don't know how much water is added; that's x. And, we don't know how much water we would end up with; that's y. So, one equation would be:

1000 + x = y

For the other equation, trying to provide a formula, it would be:

Amount#1 * %#1 + Amount#2 * %#2 + Amount#3 * %#3 + . . . = Amount(total) * %total

So, for this problem, we would have:

1000*0.52 + x*0 = y*0.51

We use "0" for the percent of the water, since pure water has no salt in it.

So, simplifying each equation:

1000 + x = y

520 = 0.51y

Solving the second equation for y:

y = 1019.61 ml

Plugging this back into the first equation, we get:

x = 19.61 ml

1. The sum of the two digits of a positive integer is 12. When the digits were reversed, the new number was 54 greater than the original. What is the product of the digits of the original number minus the original number?

the number is 93 because 9+3=12

93-39=54

27-93= -66

You just have to break it apart and test different choices.