Statement (2)<\/span>: if the zoo acquired 16 more mammals, the ratio of mammals to birds would be 3:1<\/p>\n 2) In a certain large company, the ratio of college graduates with a graduate degree to non-college graduates is 1:8, and ratio of college graduates without a graduate degree to non-college graduates is 2:3.\u00a0 If one picks a random college graduate at this large company, what is the probability this college graduate has a graduate degree?<\/p>\n 3) Dan’s car gets 32 miles per gallon.\u00a0 If gas costs $4\/gallon, then how many miles can Dan’s car go on $50 of gas?<\/p>\n 4) For a certain concert, the price of balcony tickets was exactly half the price of orchestra tickets.\u00a0 The ratio of balcony\u00a0to orchestra\u00a0tickets sold was 3:2.\u00a0 What was the price of one orchestra ticket?<\/p>\n Statement (1)<\/span>: the total revenue taken in from tickets of both kinds was $4200<\/p>\n Statement (2)<\/span>: the difference between the number of balcony\u00a0tickets sold and the number of orchestra\u00a0tickets sold was 40<\/p>\n 5) At a certain high school, there are three sports: baseball, basketball, and football.\u00a0 Some athletes at this school play two of these three, but no athlete plays in all three.\u00a0 At this school, the ratio of (all baseball players) to (all basketball players) to (all football players) is 15:12:8.\u00a0 How many athletes at this school play baseball?<\/p>\n Statement (1)<\/span>: 40 athletes play both baseball and football, and 75 play football only and no other sport<\/p>\n Statement (2)<\/span>: 60 athletes play only baseball and no other sport<\/p>\n 6)\u00a0 In a certain class, the ratio of girls to boys is 5:4.\u00a0 How many girls are there?<\/p>\n Statement (1)<\/span>: If four new boys joined the class, the number of boys would increase by 20%.<\/p>\n Statement (2)<\/span>: If the number of girls increases by 50%, then after such an increase, the probability that a randomly chosen student would be a boy would be 8\/23<\/p>\n 7) If 28 passes to a show cost $420, then at the same rate, how much will 42 passes cost?<\/p>\n 8) Metropolitan Concert Hall was half full on Tuesday night.\u00a0 How many seats are in the Hall?<\/p>\n Statement (1)<\/span>: If the number of people in the Hall increased by 20% from Tuesday night to Wednesday night, then the Hall would be 60% full on Wednesday night.<\/p>\n Statement (2)<\/span>: If 20 more people showed up on Tuesday night, that would have increased the number of people in the Hall by 4%.<\/p>\n Answers will come at the end of the article.<\/p>\n Probably ratios and proportions have been on your radar since some time in grade school or middle school, when they are introduced.\u00a0 You remember there are a lot of “mathy” facts related to these things, but it’s a bit blurry.\u00a0 Here are a few quick facts as reminders.<\/p>\n 1. What is a ratio?\u00a0 Fundamentally, a ratio is a fraction<\/a>, and is subject to all the laws of fractions.\u00a0\u00a0 “Ratios” and “fractions” are mathematically identical.<\/p>\n 2. What is a proportion?\u00a0 Is it the same as a ratio?\u00a0 No, a proportion is NOT the same as a ratio.\u00a0 Whereas a ratio is single fraction by itself (e.g. 1\/3), a proportion is an equation that sets two ratios equal to each other (e.g. 1\/3 = 4\/12).\u00a0\u00a0 See the fraction post<\/a> for a refresher on what you can and can’t cancel in a proportion.<\/p>\n 3.\u00a0 When geometric figures are similar<\/a>, the sides are proportional.\u00a0 Geometric similarity is a topic rife with ratios and proportions.\u00a0 One helpful idea discussed in that post is the idea of a scale factor<\/b>, which, it turns out, is helpful in many many proportional situations well beyond anything having to do with geometry.<\/p>\n 4.\u00a0 Percents<\/a> and probabilities<\/a> are specialized cases of ratios, and either gives you very much the same kind of information.<\/p>\n Those are four simple ideas, and there’s one more, but it takes a little setting up to express.\u00a0\u00a0 Let’s divide the mathematical information that can appear in a problem into two categories.\u00a0 The first category is “ratio\u00a0 information”, and this includes any statements about percents or probability information.\u00a0\u00a0 The second I’ll call, for lack of a better term, “count information” — not a percent or ratio, but an actual count — i.e. how many people or animals or whatever; it could be how many in any particular group, the sum or difference of multiple groups, or how many are in the whole population.\u00a0\u00a0 This leads us the final important simple idea:<\/p>\n 5.\u00a0 To get count information as an output, you need some count information as an input.\u00a0 If you all you have is ratio information as an input, it is impossible to get count information as an output.<\/p>\n <\/p>\n Suppose, in some population, there are three kinds of things, A & B & C, and they are in a proportion of 3:8:4.\u00a0 That’s ratio information.\u00a0 Suppose we want to know either the percent that A makes of the whole, or the ratio of A to the whole.\u00a0 That’s also ratio information, so we should be able to calculate it from that given ratio.<\/p>\n It can be very helpful to understand ratios in terms of “parts”.\u00a0 For every 3 parts of A, there are 8 parts of B and 4 parts of C.\u00a0 To get the whole in the same ratio, we simply add up the parts — 3 + 8 + 4 = 15 parts.\u00a0 Keep in mind, we have zero information about the actual size of the population — we have no count information.\u00a0 This simply indicates the size of the population in the same ratios, so we could say A to the whole is 3\/15 = 1\/5, and B to the whole is 8\/15, and C to the whole is 4\/15.\u00a0 This means that A is 1\/5 of the whole, or 20%.<\/p>\n Similarly, suppose a class is made up of people with brown eyes and people with blue eyes. 4\/7 of the class have brown eyes.\u00a0Those with brown eyes are four parts, and the whole is seven parts, so those with blue eyes must be the other three parts.\u00a0 Proceeding, we see that those with blue eyes must be 3\/7 of the class, the ratio of blue eyes to brown eyes must be 3:4.\u00a0 Thinking about “parts” and portioning can be a powerful way to expand the information you get from any given ratio.<\/p>\n <\/p>\n If you had any realizations while reading this blog, you may want to go back and give the practice problems another glance, before proceeding to the solutions.\u00a0 Here’s another practice question, involving two less-than-lovable baseball teams:<\/p>\n 9) http:\/\/gmat.magoosh.com\/questions\/60<\/a><\/p>\n If you have any questions about this article, please let us know in the comment section at the bottom!<\/p>\n <\/p>\n 1) A short way to do this problem.\u00a0 The prompt gives us ratio information.\u00a0 Each statement gives use some kind of count information, so each must be sufficient on its own.\u00a0\u00a0 From that alone, we can conclude: answer = D<\/b>.\u00a0 This is all we have to do for Data Sufficiency.<\/p>\n Here are the details, if you would like to see them.<\/p>\n Statement (1)<\/span>: there are twelve more mammals in the zoo than there are reptiles<\/p>\n From the ratio in the prompt, we know mammals are 11 “parts” and reptiles are 8 “parts”, so mammals have three more “parts” than do reptiles.\u00a0 If this difference of three “parts” consists of 12 mammals, that must mean there are four animals in each “part.”\u00a0 We have five bird “parts”, and if each counts as four animals, that’s 5*4 = 20 birds.\u00a0 This statement, alone and by itself, is sufficient<\/b>.<\/p>\n Statement (2)<\/span>: if the zoo acquired 16 more mammals, the ratio of mammals to birds would be 3:1<\/p>\n Let’s say there are x animals in a “part” — this means there are currently 11x mammals and 5x birds.\u00a0 Suppose we add 16 mammals.\u00a0 Then the ratio of (11x + 16) mammals to 5x birds is 3:1. —-<\/p>\n (11x + 16)\/(5x) = 3\/1 = 3<\/p>\n 11x + 16 = 3*(5x) = 15x<\/p>\n 16 = 15x\u00a0 \u2013 11x<\/p>\n 16 = 4x<\/p>\n 4 = x<\/p>\n So there are four animals in a “part”.\u00a0 The birds have five parts, 5x, so that’s 20 birds.\u00a0\u00a0 This statement, alone and by itself, is sufficient<\/b>.<\/p>\n Both statements sufficient.\u00a0 Answer = D<\/b>.<\/p>\n 2) We are given ratio information, and we are asked for ratio information: a probability.\u00a0 That’s fine.\u00a0 For simplicity, let’s use the abbreviations:<\/p>\n P = college graduates with a graduate degree<\/p>\n Q = college graduates without a graduate degree<\/p>\n R = non-college graduates<\/p>\n The two ratios we are given is<\/p>\n P:R = 1:8<\/p>\n Q:R = 2:3<\/p>\n We have combine the ratios, by making the common term the same.\u00a0\u00a0 The non-college graduates, R, are the common member, accounting for 8 parts in the first ratio, and 3 parts in the second, so we have to multiply the first ratio by 3\/3, and the second by 8\/8.<\/p>\n P:R = 1:8 = 3:24<\/p>\n Q:R = 2:3 = 16:24<\/p>\n P:Q:R = 3:16:24<\/p>\n Now, the probability about which the question asks\u00a0 is about only college graduates, so ignore the non-college graduates, and just focus on the ratio among college graduates:<\/p>\n P:Q = 3:16<\/p>\n There are 3 + 16\u00a0 = 19 parts in total, and of those, 3 are the folks in P, so that’s a probability of 3\/19<\/b>.\u00a0 Answer = (D)<\/b>.<\/p>\n 3) To get from dollars to gallons, we have to start with the $50, and divide<\/i> by ($4\/gallon) — that cancels the unit of dollars, and leaves us with gallons —– 50\/4 gallons.\u00a0 (For the moment, I’ll leave it as that un-simplified fraction).<\/p>\n Now, we need to get from gallons to distance.\u00a0 We multiply<\/i> by (32 miles\/gallon), to cancel the units of gallons, and leave only miles —-<\/p>\n (50\/4) gallons*(32 miles\/gallon) = (32*50)\/4 miles = (8*50) miles = 400 miles<\/p>\n Notice, we didn’t do anything with the four in the denominator until we got a 32 in the numerator with which we could cancel.<\/p>\n Answer = (C)<\/b><\/p>\n 4) We know the price of a balcony ticket, B = (1\/2)*C, or 2B = C, where C = the price of orchestra ticket.\u00a0 The 3:2 ratio tells us that, for some n, the concert sold 3n balcony tickets and 2n orchestra\u00a0tickets.<\/p>\n Statement (1)<\/span>: the total revenue taken in from tickets of both kinds was $4200<\/p>\n We know total revenue would be (3n) balcony ticket plus (2n) orchestra tickets:<\/p>\n (3n)*B + (2n)*C = (3n)*B + (2n)* 2B = 3nB + 4nB = 7nB = $4200,\u00a0 or nB = 600.\u00a0 The problem with this — we have two variables, n and B.\u00a0 In one extreme case, we could say n = 1 (sold 3 balcony ticket and 2 orchestra tickets), and a balcony ticket cost $600.\u00a0 At another extreme, we could say n = 600 (sold 1800 balcony ticket and 1200 orchestra tickets) and a balcony ticket cost $1.\u00a0 This information alone does allow us to determine a definitive answer to the prompt question.\u00a0\u00a0 This statement, alone and by itself, is insufficient<\/b>.<\/p>\n Statement (2)<\/span>: the difference between the number of orchestra tickets sold and the number of balcony tickets sold was 40<\/p>\n Well, balcony tickets are 3 parts, and orchestra tickets are two parts, so there’s one part of difference between them.\u00a0 The statement lets us know: one part = 40, so that allows us to figure out: we sold 3*40 = 120 balcony tickets, and 2*40 = 80 orchestra tickets.\u00a0 That’s great, but unfortunately, with this statement, we get absolutely no financial information, so we can’t solve for a price.\u00a0\u00a0 This statement, alone and by itself, is insufficient<\/b>.<\/p>\n Combined statements<\/span>: now, put this altogether.\u00a0 From the first statement, we got down to the equation nB = 600, and the second statement, in essence, tells us n = 40.\u00a0 Therefore, B = 600\/40 = 60\/4 = $15. \u00a0That’s the price of a balcony ticket.\u00a0 The price of an orchestra ticket is twice that, $30. \u00a0With both pieces of information, we were able to solve for this.\u00a0\u00a0 Together, the statements are sufficient<\/b>.<\/p>\n Statement are sufficient together but not individually.\u00a0 Answer = C<\/b><\/p>\n 5) This is a tricky one.\u00a0 The ratio 15:12:8 double-counts some students.\u00a0\u00a0 In terms of a triple Venn diagram:<\/p>\n So, in that diagram<\/p>\n a = folks who play baseball only<\/p>\n b = folks who play football only<\/p>\n c = folks who play basketball only<\/p>\n d = folks who play baseball and football only<\/p>\n e = folks who play baseball and basketball only<\/p>\n f = folks who play football and basketball only<\/p>\n g = folks who play all three sports<\/p>\n We know from the prompt that g = 0, but at the outset, that’s still six unknowns!!<\/p>\n Now, notice:<\/p>\n everyone who plays baseball = a + d + e<\/p>\n everyone who plays basketball = c + e + f<\/p>\n everyone who plays football = b + d + f<\/p>\n So, the ratio given in the problem is:<\/p>\n (a + d + e):(c + e + f):(b + d + f) = 15:12:8<\/p>\n Of course, these are all fractions, so we can’t simplify: there is no way to simplify.<\/p>\n We would like to find the total number of baseball players, a + d + e.<\/p>\n Statement #1<\/span> tells us that d = 40 and b = 75.\u00a0 Thus<\/p>\n total number of baseball players = a + 40 + e<\/p>\n We don’t have enough information to calculate this, and we don’t have enough ratio information to solve.\u00a0\u00a0 This statement, alone and by itself, is insufficient<\/b>.<\/p>\n Statement #2<\/span> tells us a = 60.\u00a0 Thus<\/p>\n total number of baseball players = 60 + d + e<\/p>\n We don’t have enough information to calculate this.\u00a0 This statement, alone and by itself, is insufficient<\/b>.<\/p>\n Combined statements<\/span>: Now we know a = 60, b = 75, and d = 40.<\/p>\n total number of baseball players = 60 + 40 + e= 100 + e<\/p>\n total number of football players = 75 + 40 + f = 115 + f<\/p>\n We know the ratio of these two quantities, baseball to football, is 15:8.\u00a0 Unfortunately, that would give us only one equation for two unknowns, e & f.\u00a0 If we can’t solve for these, then we can solve for the total number of baseball players.\u00a0\u00a0 Even with both statements combined, we cannot determine the answer.<\/p>\n Both statements combined are insufficient.\u00a0\u00a0 Answer = E<\/b><\/p>\n 6) Statement (1)<\/span>: If four new boys joined the class, the number of boys would increase by 20%<\/i>.\u00a0 This means, those four new boys count as 20% of the original boys, which means 2 new boys would be 10% of the original boys, which means there must have been 20 original boys.\u00a0\u00a0 If we know how many boys, we can use the ratio to calculate how many girls.\u00a0\u00a0 This statement, alone and by itself, is sufficient<\/b>.<\/p>\n Statement (2)<\/span>: If the number of girls increases by 50%, then after such an increase, the probability that a randomly chosen student would be a boy would be 8\/23<\/i>.\u00a0 The prompt gives us “ratio information”, and this statement also gives us more “ratio information”.\u00a0 We have absolutely no “count information”, so we can’t figure out the count or number of anything.\u00a0\u00a0 This statement, alone and by itself, is insufficient<\/b>.<\/p>\n First statement sufficient, second insufficient.\u00a0 Answer = A<\/b><\/p>\n 7) We can solve this question by setting up a proportion:<\/p>\n Divide the left fraction, numerator and denominator, by 7:<\/p>\n Now, cancel the factor of 4:<\/p>\n For this multiplication, use the doubling & halving trick<\/a> — double 15 is 30, and half of 42 is 21:<\/p>\n x = (15)*(42)\u00a0 = 30*21 = $630<\/b><\/p>\n Answer = C<\/b><\/p>\n 8) The prompt gives us “ratio information”.<\/p>\n![\"grap_img1\"](\"https:\/\/magoosh.com\/gmat\/files\/2013\/09\/grap_img1.png\")
<\/a><\/p>\n\n\t(A) 61.25 miles
\n\t(B) 256 miles
\n\t(C) 400 miles
\n\t(D) 1600 miles
\n\t(E) 6400 miles\n<\/ul>\n\n\t(A) $500
\n\t(B) $560
\n\t(C) $630
\n\t(D) $700
\n\t(E) $840\n<\/ul>\nRatios and proportions: a review<\/h2>\n
Ratios and portioning<\/h2>\n
Summary<\/h2>\n
Practice problem explanations<\/h2>\n
<\/a><\/p>\n<\/a><\/p>\n<\/a><\/p>\n<\/a><\/p>\n