Abdulla wrote:At a certain school, the ratio of the number of second grades to the number of the fourth grades is 8 to 5, and the ratio of the number of the first grades to the number of the second grades is 3 to 4. If the ratio of the number of the third grades to the number of the fourth grades is 3 to 2, what is the ratio of the number of the first grades to the number of the third grades?
Let's let a=1st, b=2nd, c=3rd and d=4th, just so we don't get numbers and letters confused.
So, algebraically:
b/d = 8/5
a/b = 3/4
c/d = 3/2
Q: what's a/c?
Let's start with our two equations involving a and c. If we divide the 2nd equation by the third, we get:
(a/b)/(c/d) = (3/4)/(3/2)
Using our fraction division rules, we invert and multiply on both sides:
(a/b)*(d/c) = (3/4)*(2/3)
rearranging the left side to isolate a/c (which is what we're solving for) and simplifying the right side:
(a/c)(d/b) = 6/12
isolating a/c on the left:
(a/c) = 1/2(b/d)
From the first equation, we know that b/d = 8/5, so we now sub in:
(a/c) = (1/2)(8/5)
(a/c) = 8/10 = 4/5
* * *
We could also solve by picking numbers (a great alternative to doing the algebra), but unless you find a great number right away you'll end up working with fractions instead of integers as you proceed. I haven't worked it out fully, but I think that picking b=240 is the smallest number that gives you integers all the way through (d=120 might work as well).
