Maths, Physics and chemistry books are stored on a library shelf that can accommodate 25 books.
Maths, Physics and chemistry books are stored on a library shelf that can accommodate 25 books. Currently, 20% of the shelf spots remain empty. There are twice as many maths books as physics books and the number of physics books is 4 greater than that of chemistry books. Among all the books, 12 books are soft cover and the remaining are hard-cover. If there are a total of 7 hard-cover books among the maths and physics books. What is the probability E, that a book selected at random is either a hard cover book or a chemistry book?
Answer/Solution
9/20
Steps/Work
First phase of this problem requires you to determine how many mathematics and chemistry books are even on the shelf. To do so, you have the equations:
m + p + c = 20 (since 4/5 of the 25 spots are full of books)
m = 2p
p = 4 + c
From that, you can use Substitution to get everything down to one variable.
c = p - 4
m = 2p
p = p
Then (p - 4) + 2p + p = 20, so 4p = 24 and p = 6. That means that there are 12 math, 6 physics, and 2 chemistry books on the shelf.
With those numbers, you also know that there are 8 total hardcovers, 1 of which is chemistry. So if your goal is to get either a hardcover or a chemistry, there are 9 ways towin- either one of the 7 hardcovers that aren't chemistry or the two chemistry books. So out of the 20 total, E=9 provide the desired outcome, making the answer E.
m + p + c = 20 (since 4/5 of the 25 spots are full of books)
m = 2p
p = 4 + c
From that, you can use Substitution to get everything down to one variable.
c = p - 4
m = 2p
p = p
Then (p - 4) + 2p + p = 20, so 4p = 24 and p = 6. That means that there are 12 math, 6 physics, and 2 chemistry books on the shelf.
With those numbers, you also know that there are 8 total hardcovers, 1 of which is chemistry. So if your goal is to get either a hardcover or a chemistry, there are 9 ways towin- either one of the 7 hardcovers that aren't chemistry or the two chemistry books. So out of the 20 total, E=9 provide the desired outcome, making the answer E.