Posts tagged with: ratios

Here’s a problem from Khan Academy’s SAT practice section. Please explain this one for me.

A marine aquarium has a small tank and a large tank, each containing only red and blue fish. In each tank, the ratio of red fish to blue fish is 3 to 4. The ratio of fish in the large tank to fish in the small tank is 46 to 5. What is the ratio of blue fish in the small tank to red fish in the large tank?

A ratio of 3 red to 4 blue means the total must be a multiple of 7, so it might help you to first multiply the totals in each tank by 7. The 46:5 ratio for large tank to small tank becomes 322:35. Let’s just pretend those are the actual numbers of fish in the tank.

In the small tank, 4/7 of the 35 fish in the tank are blue, so 20 fish are blue.

In the large tank, 3/7 of the 322 fish are red, so 138 fish are red.

Therefore, the ratio of blue fish in the small tank to red fish in the big tank is 20:138, which simplifies to 10:69.

Note that while multiplying everything by 7 was helpful (in my opinion) in making the math more intuitive, you don’t need to. You can simply enter the following in your calculator and ask it to convert to a fraction:

\left(\dfrac{4}{7}\times 5\right)\div\left(\dfrac{3}{7}\times 46\right)

A bag contains mangoes that are either green or yellow. The ratio of green mangoes to yellow mangoes in the bag is 3 to 5. When two green mangoes are removed and ten yellow mangoes are added, the ratio becomes 2 to 5. How many green mangoes were originally in the bag?

Say that the original number of green mangoes is g and the original number of yellow mangoes is y. So you know that \dfrac{g}{y}=\dfrac{3}{5}.

When two green mangoes are removed, and 10 yellow mangoes are added, the ratio becomes \dfrac{g-2}{y+10}=\dfrac{2}{5}.

Now we have two equations and two variables—we can solve for g to get the answer we seek!

Manipulate the first equation to set it equal to y:


Now substitute into the other equation:


You can check your work on this easily enough. If there were originally 18 green mangoes and the ratio of green to yellow was 3 to 5, then there were originally 30 yellow mangoes. What happens if you take away 2 green mangoes and add 10 yellow mangoes? You end up with 16 green and 40 yellow, which simplifies to a ratio of 2 to 5, just like the question said.

One last note here. You didn’t provide answer choices for this question so I assume there weren’t any. However, if there were answer choices here, this would be an ideal question to backsolve—essentially to “check your work” like we just did in the last paragraph without doing the work!

Can you do test 4 section 4 number 35 please?

This is a very SAT-ish question: the math is pretty trivial, but conceptually the question is still a bit tricky.

There are a couple things you need to have nailed down to get this one right.

First, you need to recognize that they’re asking you for a ratio of two dynamic pressures, so they’re asking you for a ratio of two q‘s. Let’s call them q_1 and q_2. We’ll say q_1 is the one that corresponds to a velocity of v and q_2 is the one that corresponds to a velocity of 1.5v.

Second, you need to make sure you’re providing the ratio that’s asked for: the q of the faster fluid to the q of the slower fluid. Which fluid is faster—the one with velocity v or the one with velocity 1.5v? 1.5v is always going to be a larger number than v, so that’s the faster fluid. Therefore, we need to calculate the ratio of q_2 to q_1.


How do you do Test 2 Section 3 Number 18?

The key here is to recognize that you’re dealing with similar triangles (pro tip: similar triangle questions often take this “hourglass” form). The two angles at point B are vertical, so they must be congruent. And because segments AE and CD are parallel, you’ve got alternate interior angles for the rest. When all the angles in two triangles are congruent, those triangles are similar.

(It might be easier to see the alternate interior angles if you extend the lines…expand the image on the right.)

Anyway, now that you’ve established that these are similar triangles, you just need to use ratios to solve. If AB=10 and [latexBD=5[/latex], then each set of corresponding sides will be in the same ratio. So we can solve for BC thusly:


But wait—you’re not quite done! The question asks for CE, not BC!

CEBCBE = 4 + 8 = 12

There. Now you’re done.

Whats the best and fastest way of doing Practise test 1, section 4, question 23?

Would it be calculating the ratio of all the options and then comparing it to the Human Resources ratio?


Yeah, that’s the best and fastest way, although you can save yourself some calculator keystrokes by rounding. This isn’t just a time saver: the more keystrokes you make, the more likely you are to make an error! Rounding to the nearest million (it’s already in thousands) will work just fine. So, my calculations would look like this:

Human resources 2007 to 2010: \dfrac{4051}{5921}=0.68

Agriculture/natural resources 2007 to 2010: \dfrac{374}{488}=0.77

Education 2007 to 2010: \dfrac{2165}{3008}=0.72

Highways and transportation 2007 to 2010: \dfrac{1468}{1774}=0.83

Public safety 2007 to 2010: \dfrac{263}{464}=0.57

There you have it: the Human resources and Education budgets have the closest ratios.

Test 6 #20 from the no calculator section

The circle, you’re told, has a radius of 1, so that means its circumference is 2\pi (1) =2\pi. The arc between A and B has a length of \dfrac{\pi}{3}. How much of the circumference is that?



Please answer #23 in test 2 section 4.

For this one, let’s plug in. Let’s say Observer A observes an intensity of 16, so I = 16. Observer B would then observe an intensity of 1, so his I = 1.

Both observers are observing the same ratio signal, so the value of P will be the same for both. Because it won’t change, we can plug in for that, too. Let’s say P = 100. (You can really pick anything for this, but 100 is a good choice because you’ll end up taking its square root.)

Now solve both equations for r. I’m going to use r_A and r_B to keep track of which distance is which.

    \begin{align*}16&=\dfrac{100}{4\pi {(r_A)}^2}\\64\pi{(r_A)}^2&=100\\{(r_A)}^2&=\dfrac{100}{64\pi}\\r_A&=\sqrt{\dfrac{100}{64\pi}}\\r_A&=\dfrac{10}{8\sqrt{\pi}}\\r_A&=\dfrac{5}{4\sqrt{\pi}}\end{align*}

    \begin{align*}1&=\dfrac{100}{4\pi {(r_B)}^2}\\4\pi {(r_B)}^2&=100\\{(r_B)}^2&=\dfrac{100}{4\pi}\\{r_B}&=\sqrt{\dfrac{100}{4\pi}}\\r_B&=\dfrac{10}{2\sqrt{\pi}}\\r_B&=\dfrac{5}{\sqrt{\pi}}\end{align*}

The question asks for the ratio \dfrac{r_A}{r_B}. We can calculate that!


That frankenfraction is not as gnarly as it looks. Just remember that dividing by a fraction is the same as multiplying by its reciprocal, and simplify:



The circle graph above shows the percent of 4th graders at an elementary school who have the indicated numbers of pets in their homes. If 68 of the 4th graders have at least one pet, how many have exactly two pets?

(A) 16

“At least one” means one, two, or more than two—anything but zero pets. So that’s 30 + 20 + 35  = 85 percent of the students.

If 68 students represent 85 percent of the class, how many students represent 20%? We can do a ratio:



At a certain camp, the counselor-to-camper ratio is 2 to 9. If the camp has 18 counselors, how many campers does it have?

For more practice with ratios, click here.

In this case, all we need to do is set up our ratio and be careful to keep our units straight:

\dfrac{2\text{ counselors}}{9\text{ campers}}=\dfrac{18\text{ counselors}}{x\text{ campers}}

Cross multiply and solve:



So there are 81 campers at the camp.

A new high-tech transportation system is to be built connecting a city at sea level and a suburb 3,500 feet above sea level. The maximum allowable grade is 3 percent, which means that the track for the new system can ascend no more than 3 feet for every 100 feet of horizontal length. What is the minimum whole number of feet of track needed for the new system?

A) 113,167 ft
B) 116,615 ft
C) 116,615 ft
D) 116,667 ft
E) 120,167 ft

How many times does this track need to go up 3 feet?

3500/3 = 1166.666…

Every time it goes up 3 feet, it needs to have 100 feet of horizontal length, so multiply that by 100:

100 × 1166.666… = 116,667

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How’s everyone else doing on this quiz?

10% got 5 right
17% got 4 right
15% got 3 right
26% got 2 right
31% got 0 or 1 right


Source: Married to the Sea.

This started out as an Old MacDonald’s farm question. No wait, I thought to myself, not depressing enough.

The prize this week: You’ll get the satisfaction of knowing that you probably solved this problem in less time than I spent staring at my computer screen trying to come up with a clever prize for this week. I swear I used to be more creative.

In Wendell’s house, the ratio of unopened credit card offers to out-of-date phone books is 9 to 5. The ratio of magazines to crushed loose cigarettes is 25 to 7, and the ratio of McDonald’s Happy Meal toys to rotting, half-eaten pizzas is 3 to 2. There are 6 used-up batteries lying around for each broken VCR. The ratio of crushed loose cigarettes to McDonald’s Happy Meal toys is 5 to 8, and the ratio of used-up batteries to out-of-date phone books is 5 to 7. There are 30 magazines for each 4 broken VCRs. If there were 108 unopened credit card offers, how many rotting, half-eaten pizzas would Wendell have in his house?

Please note that, as usual for the weekend challenge, I’m taking a concept that SAT has been known to test, and extending it to an extreme to which the SAT would not go (not only in subject matter, but in scope as well). All of this is to say that these weekend challenges are meant to be fun for you, not to freak you out if you can’t get them. They’re usually a degree or two harder than what the SAT will throw your way.
Put your answers in the comments. I’ll post a solution Monday (probably late in the day). Good luck!UPDATE: Nice work, Chris. And special thanks to Catherine from kitchen table math, who pointed out that my original wording (“If there are…how many does…”) made non-integer quantities (fractional batteries??) quite unsavory. I’ve changed the wording of the last sentence a bit now for the benefit of people who find this in the archives and want to torture themselves with this question later.

Solution below the cut.

The SAT would never throw such a complex question at you, but the solution I advocate is one that might help you on the harder ratio questions the SAT will toss your way. Remember that on ratio questions, units are paramount. When you’re presented with ratios of more than two things and asked to suss out the relationship between just two of those things, the best and most elegant solution is to line up all the fractions and multiply them together, eliminating unwanted units along the way. Let me show you what I mean:

What we’re given: a bunch of ratios relating together the following things (in order of appearance):

credit card offers (CCO)
phone books (PBK)
magazines (MGZ)
cigarettes (CIG)
Happy Meal toys (HMT)
half-eaten pizzas (HEP)
batteries (BAT)
VCRs (same, duh.)

What we want: the ratio of unopened credit card offers (CCO) to rotting, half-eaten pizzas (HEP). Once we have that ratio, then we’ll deal with the 108 CCO.

How we get there: Start by listing the ratios that contain the units you want. Make sure to put CCO on top, and HEP on bottom.


We need to get rid of PBK and HMT, so let’s find some ratios we can use to do so, one at a time.


And then we just keep going. This might get monotonous (it’s a challenge question), but it’s really just the same procedure over and over again until the desired result.







Gasp…Deep breath…Simplify…

That wasn’t so bad now, was it?

Now set up a proportion to see how many HEP will exist if there are 108 CCO:
Note: Although it will get gnarly with things like fractional batteries, if you’re good with your calculator you can also do this by plugging in 108 for CCO and solving consecutive ratios until you end at 16 HEP. I’m not going to spell that way out here. For some more realistic SAT-style questions that can be solved similarly, look here.

So here’s the thing with ratios and proportions on the SAT: they’re really easy. No, seriously, where are you going? Come back! They’re easy, I swear. All you have to do is keep very close track of your units, and you’ll be good to go. That means when you set up a proportion, actually write the units next to each number. Make sure you’ve got the same units corresponding to each other before you solve, and you’re home free. Pass Go, collect your $200, and spend it all on Lik-M-Aid Fun Dip. So uh…let’s try one?

  1. A certain farm has only cows and chickens as livestock. The ratio of cows to chickens is 2 to 7. If there are 63 livestock animals on the farm, how many cows are there?
    (A) 13
    (B) 14
    (C) 16
    (D) 18
    (E) 49

The SAT writers would love for you to set up a simple proportion here and solve:

Hooray! = 18! That’s answer choice (D)! NOT SO FAST, SPANKY. You just solved for a terrifying hybrid beast, the COWNIMALKEN. Let’s look at that fraction more carefully, with the units included:

So when you casually multiplied by 63 and solved, you solved for a unit that won’t do you any good: the [(cow)(animal)]/(chicken), or COWNIMALKEN. That’s terrifying. Nature never intended it to be so. Not only are you unwisely playing God, but you’re getting an easy question wrong. Before we can solve this bad boy, we need to make sure our units line up on both sides of the equal sign. So let’s change the denominator on the left to match the one on the right. Get rid of “7 chickens” and replace it with “9 animals.” Get it? Because cows count as animals, if there are 2 cows for every 7 chickens, that means there are 2 cows for every 9 animals.

Now, we can solve: x = 14 cows. That’s choice (B). See how the units cancel out nicely when you’ve properly set up a ratio question? That should make the hairs on your neck stand on end.

Look out for this tricky crap, too:

But but but! There’s one more thing you need to watch out for. Sometimes they’ll give you units that aren’t quite as easily converted. Like so:

  1. The ratio of students to teachers at a certain school is 28 to 3. The ratio of teachers to cafeteria workers is 9 to 2. What is the ratio of cafeteria workers to students?
    (A) 1 to 42
    (B) 2 to 28
    (C) 3 to 37
    (D) 9 to 56
    (E) 3 to 14

Here, we have a few options. It’s not too hard to find a number of teachers that will work with both ratios, so I’ll leave it to you to figure out how to solve it that way if you prefer. Instead let me point out that there’s a pretty elegant solution here that comes from simply multiplying the two ratios together, essentially solving for the expression we’re looking for. Peep the skillz:

What happens to the teachers? They cancel! So multiply, and simplify:

Since the question asked for the ratio of cafeteria workers to students, just flip it and you’re done! 1 cafeteria worker to 42 students. That’s choice (A). Ah-mazing.

Mind your units, son.

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