Ratios.. OG12 PS-66

[Abdulla]

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?

OA is 4 to 5

[Stuart@KaplanGMAT]

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).

[Abdulla]

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).

Thanks.. but I guess this way is harder than the OG's way..which is..

F,S,T,R
S/R=8/5 ==> S=8R/5---(1)
F/S=3/4 ==> F=3S/4----(2)
T/R=3/2 ==> T=3R/2---(3)

F/T = (3S/4)/(3R/2)= 3S/4 * 2/3R = S/2R ----(4)

Substitute 1 into 4.

(8R/5)/2R = 8R/5*1/2R= 8R/10R= 4/5. here is the answer.

[Stuart@KaplanGMAT]

Thanks.. but I guess this way is harder than the OG's way..which is..

F,S,T,R
S/R=8/5 ==> S=8R/5---(1)
F/S=3/4 ==> F=3S/4----(2)
T/R=3/2 ==> T=3R/2---(3)

F/T = (3S/4)/(3R/2)= 3S/4 * 2/3R = S/2R ----(4)

Substitute 1 into 4.

(8R/5)/2R = 8R/5*1/2R= 8R/10R= 4/5. here is the answer.

I'm not sure "harder" is the right word - it's the difference of using combination and substitution and, if you take all the "explanation" out of my answer, the amount of math is actually a bit less if you use combination.

It is a good idea to be familiar with both methods of solving - on some questions combination is faster, on some substitution is the way to go.

[Abdulla]

Thanks.. but I guess this way is harder than the OG's way..which is..

F,S,T,R
S/R=8/5 ==> S=8R/5---(1)
F/S=3/4 ==> F=3S/4----(2)
T/R=3/2 ==> T=3R/2---(3)

F/T = (3S/4)/(3R/2)= 3S/4 * 2/3R = S/2R ----(4)

Substitute 1 into 4.

(8R/5)/2R = 8R/5*1/2R= 8R/10R= 4/5. here is the answer.

I'm not sure "harder" is the right word - it's the difference of using combination and substitution and, if you take all the "explanation" out of my answer, the amount of math is actually a bit less if you use combination.

It is a good idea to be familiar with both methods of solving - on some questions combination is faster, on some substitution is the way to go.

You're right, we should know both.

Thank you