<p>Post any SAT math problem you struggle with, or any questions related to the math concepts in SAT math problems.
I'm a grad student willing to help. I'll answer detailed answers in this thread.</p>
<p>Someone sent me his problems via private message. Please don’t do, since other people may have the same questions.
Here are the two problems I answered:</p>
<p>1) 5, 6, 5, 4, 7, x, 6. What is the least value of x for which the average (arthmetic mean) of the list of numbers above is equal to the median of the list?</p>
<p>2) There are 99 adults and 1 child in a room. How many of the adults must leave the room so that 2 percent of all the people remaining are children?</p>
<p>And my answer:
For the first problem:
The problem asks for the average and the median. So the first thing to do is to write them down.
Average is (5 + 6 + 5 + 4 + 7 + x + 6)/7 = (33 + x)/7.
Since there are 7 terms, median is the 4th term of the list, in ascending order (4, 5, 5, 6, 6, 7 without x). Realize that the median has to be between 5 and 6, because if you add a small x (< 5), then 5 will be in 4th position, and if you add a large x (> 6), then 6 will be in 4th position.</p>
<p>Let’s re-read the question: We want the LEAST VALUE of x and the equality between average and median. Since we know that the least value for the median is 5, let’s ask if it is possible to find a small x such that average is equal to 5. We can, with x = 2 (since (33+2)/7 = 35/7 = 5). Is it possible to find a smaller x? No, because the median will still be the same while the average will decrease.
Hence the least value of x is x = 2.</p>
<p>For the second problem:
The difficulty is to translate the problem in mathematical terms. Ask yourself what are the numbers that matter here and give them a mathematical name: The number of adults is X and the number of child is Y.
So, we have X = 99 and Y = 1, and we can only change the X (the adults leave the room).
We want to find a new X’ such that the proportion of children is 2%: We want that Y / (X’ + Y) = 2%. We know that Y = 1, so: 1 / (X’ + 1) = 2%.
With a few manipulations, you end up with X’ = 49. Now, this is now what is asked, always re-read the question when you think you’re done. We are asked how many of the adults must leave the room. Since there are 99 at first and 49 at the end, the answer is that 50 adults must leave the room.</p>
<p>Another way to see it is to notice that we can’t change the number of children, there will always be only 1 child. Since we want it to represent 2% of the population, we must have a total of people of 50, which means 49 adults, thus 50 adults leave the room.
It’s quicker that way, but the former method will be useful for other problems.</p>