{"id":5611,"date":"2015-02-23T14:46:46","date_gmt":"2015-02-23T22:46:46","guid":{"rendered":"https:\/\/magoosh.com\/gmat\/?p=5611"},"modified":"2020-01-15T10:48:16","modified_gmt":"2020-01-15T18:48:16","slug":"gmat-math-weighted-averages","status":"publish","type":"post","link":"https:\/\/magoosh.com\/gmat\/gmat-math-weighted-averages\/","title":{"rendered":"GMAT Math: Weighted Averages"},"content":{"rendered":"

Problems range from easy to hard.<\/p>\n

1) On a ferry, there are 50 cars and 10 trucks.\u00a0 The cars have an average mass of 1200 kg and the trucks have an average mass of 3000 kg.\u00a0 What is the average mass of all 60 vehicles on the ferry?<\/p>\n

    \n\t(A) 1200 kg
    \n\t(B) 1500 kg
    \n\t(C) 1800 kg
    \n\t(D) 2100 kg
    \n\t(E) 2400 kg\n<\/ol>\n

    2) In Plutarch Enterprises, 70% of the employees are marketers, 20% are engineers, and the rest are managers.\u00a0 Marketers make an average salary of $50,000 a year, and engineers make an average of $80,000.\u00a0 What is the average salary for managers if the average for all employees is also $80,000?<\/p>\n

      \n\t(A) $80,000
      \n\t(B) $130,000
      \n\t(C) $240,000
      \n\t(D) $290,000
      \n\t(E) $320,000\n<\/ol>\n

      3) At Didymus Corporation, there are just two classes of employees: silver and gold.\u00a0 The average salary of gold employees is $56,000 higher than that of silver employees.\u00a0 If there are 120 silver employees and 160 gold employees, then the average salary for the company is how much higher than the average salary for the silver employees?<\/p>\n

        \n\t(A) $24,000
        \n\t(B) $28,000
        \n\t(C) $32,000
        \n\t(D) $36,000
        \n\t(E) $40,000\n<\/ol>\n

        4) In a company of only 20 employees, 10 employees make $80,000\/yr, 6 employees make $150,000\/yr, and the 4 highest-paid employees all make the same amount.\u00a0 If the average annual salary for the 20 employees is $175,000\/yr, then what is the annual salary of each highest-paid employee?<\/p>\n

          \n\t(A) $250,000
          \n\t(B) $300,000
          \n\t(C) $350,000
          \n\t(D) $400,000
          \n\t(E) $450,000\n<\/ol>\n

          5) In a certain apartment building, there are one-bedroom and two-bedroom apartments.\u00a0 The rental prices of the apartment depend on a number of factors, but on average, two-bedroom apartments have higher rental prices than do one-bedroom apartments.\u00a0 Let R be the average rental price for all apartments in the building.\u00a0 If R is $5,600 higher than the average rental price for all one-bedroom apartments, and if the average rental price for all two-bedroom apartments is $10,400 higher that R, then what percentage of apartments in the building are two-bedroom apartments?<\/p>\n

            \n\t(A) 26%
            \n\t(B) 35%
            \n\t(C) 39%
            \n\t(D) 42%
            \n\t(E) 52%\n<\/ol>\n

            6) At a certain upscale restaurant, there just two kinds of food items: entrees and appetizers.\u00a0 Each entr\u00e9e item costs $30, and each appetizer item costs $12.\u00a0 Last year, it had a total of 15 food items on the menu, and the average price of a food item on its menu was $18.\u00a0 This year, it added more appetizer items, and the average price of a food item on its menu dropped to $15.\u00a0 How many appetizer items did it add this year?<\/p>\n

              \n\t(A) 3
              \n\t(B) 6
              \n\t(C) 9
              \n\t(D) 12
              \n\t(E) 15\n<\/ol>\n

              Solutions will appear at the end of the article.<\/p>\n

               <\/p>\n

              Weighted average<\/h2>\n

              For the purposes of the GMAT, the weighted average situations occurs when we combine groups of different sizes and different group averages.\u00a0 For example, suppose in some parameter, suppose male employees in a company have one average score, the females have another average score, and gender non-binary\/non-conforming employees (GNB\/NC) have yet another average. If there were an equal number of men, women, and GNB\/NC folks, we could just average the three separate gender averages: that would be ridiculously easy, which is precisely why the GMAT will never present you with groups of equal size in such a question.\u00a0 Instead, the number of male, female, and GNB\/NC employees will be profoundly different, one significantly outnumbering the other, and then we will have to combine the individual gender averages to produce a total average for all employees.\u00a0 That’s a weighted average.<\/p>\n

              It is common for weighted average problems on the GMAT to have only two groups, but sometimes there are three (like the above problem), and conceivably, on a very hard problem, there could be four groups.\u00a0 Each group is a different size, each has its own average, and the job is to find the average of everyone all together.\u00a0 Or, perhaps the question will give us most of the info for the individual groups, and give us the total average for everyone, and then ask us to find the size or average of a particular group.<\/p>\n

              We have three basic approaches we can take to these question.<\/p>\n

               <\/p>\n

              Approach I: averages & sums<\/h2>\n

              Remember that, even with ordinary average<\/a> questions, thinking in terms of the sum can often be helpful.\u00a0 We can’t<\/span><\/em> add or subtract averages, but we can<\/span><\/em> add or subtract sums! Right there is the key to one approach to the weighted average situation.\u00a0 If we calculate the sums for each separate group, we can simply add these sums to get the sum of the whole group.\u00a0 Alternately, if we know the size of the total group and the total average for everyone, we can figure out the total sum for everyone, and simply subtract the sums of the individual groups in order to find what we need.<\/p>\n

              Some weighted average problems give percents, not actual counts, of individual groups.\u00a0 In that case, we could simply pick a convenient number for the size of the population, and use the sums method from there.\u00a0 For example, if group A has 20% of the population,\u00a0 group B has 40% of the population,\u00a0 and group C also has 40% of the population,\u00a0 then we could just pretend that group A has one person, groups B & C each have two people, and total population has five people.\u00a0 From this, we could calculate all our sums.<\/p>\n

              This method always works, although is not always the most convenient in more advanced problems.\u00a0 This will be demonstrated in a few of the answers below.<\/p>\n

               <\/p>\n

              Approach II: proportional & percentages<\/h2>\n

              Sometimes the information about the sizes of individual groups is given, not in absolute counts of members, but simply in percents or proportions.\u00a0 In the problems above, question #2 simply gives percents, and question #5 asks for a percent.\u00a0\u00a0 Yes, we could use Approach I, but there’s a faster way.<\/p>\n

              In Approach II, we simply multiply each group average by the percent of the population, expressed as a decimal, which that group occupies.\u00a0 When we add these products up, the sum is magically the total average for everyone.\u00a0 Suppose we have three groups, groups J and K and L which together constitute the entire population.\u00a0 Suppose this summarizes their separate information:<\/p>\n

              \"gm-wa_img1\"<\/a><\/p>\n

              The percent have to be expressed in decimal form, so that:<\/p>\n

              \"gm-wa_img2\"<\/a><\/p>\n

              Then, the total average is simply given by<\/p>\n

              \"gm-wa_img3\"<\/a><\/p>\n

              This approach will be demonstrated in #2 below.<\/p>\n

               <\/p>\n

              Approach III: proportional placement of total average<\/h2>\n

              This final method can be hard to understand at first, but if you appreciate it, it is an incredibly fast time-saver.\u00a0 This approach only works if there are exactly two groups, no more.<\/p>\n

              Suppose there are two groups in a population, group 1 and group 2, and suppose that group 1 is bigger than group 2.\u00a0 Let’s also suppose that group 1 has a lower group average, and group 2 has the higher group average.\u00a0 Of course, the total combined average of the two groups together will be between the two individual group averages.\u00a0 In fact, because group 1 is bigger, the combined average will have to be closer to group 1’s average, and further away from group 2’s average.<\/p>\n

              Now, think about a number line, with the two individual group averages and the total averages indicated on the number line.<\/p>\n

              \"gm-wa_img4\"<\/a><\/p>\n

              On that number line, I have labeled d1, the distance from the average of group #1 to the combined average, and d2, the distance from the combined average to the average of group 2.\u00a0 The ratio of these two distances is equal to a ratio of the size of the two individual groups.\u00a0 Let’s think about this.\u00a0 The bigger group, here group 1, will have more of an effect on the combined average and therefore will be closer to the combined average\u2014a smaller distance.\u00a0 Therefore, the ratio of the distances must equal the reciprocal of the same ratio of the sizes of the groups:<\/p>\n

              \"gm-wa_img5\"<\/a><\/p>\n

              Let’s say group 1 is 3 times larger than group 2.\u00a0 This means d2, the distance from group 2’s average to the combined average, would have to be 3 times bigger than d1, the distance from group 1’s average to the combined average.\u00a0 If the latter is x, then the former is 3x, and the total distance is 4x.\u00a0 If we know the averages of the two groups individually, we would simply have to divide the difference between those group averages by 4: the combined average would be one part away from group 1’s average, or three parts away from group 2’s average.<\/p>\n

              This approach is hard to explain clearly in words.\u00a0 You really have to see it demonstrated in the solutions below to understand it fully.\u00a0\u00a0 Once you understand it, though, this is an extremely fast method to solve many problems.<\/p>\n

               <\/p>\n

              Summary<\/h2>\n

              If the above article gave you any insights, you may want to give the practice problems another look.\u00a0 Remember, in your practice problems on weighted averages, practice all three of these methods.\u00a0 The more ways you have to understand any problems, the more options you will have on test day!<\/p>\n

              \"gm-wa_img6\"<\/a><\/p>\n

               <\/p>\n

              Practice problem explanations<\/h2>\n

              1) Method I: using sums<\/strong><\/p>\n

              We will divide the two masses by 1000, 1.2 and 3 respectively, to simplify calculations.\u00a0\u00a0 Note the use of the Doubling and Halving<\/a> trick in the first multiplication.<\/p>\n

              sum for cars = 50*1.2 = 100*0.6 = 60<\/p>\n

              sum for trucks = 10*3 = 30<\/p>\n

              total sum = 60 + 30 = 90<\/p>\n

              To find the total average, we need to divide this total sum by the total number of vehicles, 60.<\/p>\n

              \"gm-wa_img7\"<\/a><\/p>\n

              Since we divided masses by 1000 earlier, we need to multiply by 1000 to get the answer.\u00a0 Total average = 1500 kg.\u00a0 Answer = (B)<\/strong><\/p>\n

              Method II: proportional placement of the total average<\/strong><\/p>\n

              Cars to trucks is 5:1, so if the distance between the car’s average and truck’s average were divided into 6 parts, the car’s average is 1 part away from the total average, and the truck’s average is 5 parts away.<\/p>\n

              Well, the difference in the two group averages is 3000 \u2013 1200 = 1800 kg.\u00a0 Divide that by six: each “part” is 300 kg.\u00a0 Well, the total average must be 300 kg bigger than 1200 kg, or 5*300 kg smaller than 3000 kg.\u00a0 Either way, that’s 1500 kg.\u00a0 Answer = (B)<\/strong><\/p>\n

               <\/p>\n

              2) We will approach this use the proportion & percents approach.\u00a0 Divide all dollar amounts by 1000 for smaller numbers.\u00a0 Multiply each group average by the percent expressed as a decimal:<\/p>\n

              marketers = 0.70*50 = 35<\/p>\n

              engineers = 0.20*80 = 16<\/p>\n

              managers = 0.10*x = 0.1x<\/p>\n

              where x is the average salary for the managers.\u00a0 These three should add up to the average for all employees:<\/p>\n

              35 + 16 + 0.1x = 80<\/p>\n

              0.1x = 80 \u2013 35 \u2013 16<\/p>\n

              0.1x = 29<\/p>\n

              x = 290<\/p>\n

              Now, multiply the 1000 again, to get back to real dollar amounts.\u00a0 The average salary for managers is $290,000.\u00a0 Answer = (D) <\/strong><\/p>\n

               <\/p>\n

              3) Method I: using sums<\/strong><\/p>\n

              We can use this if we pick a value for the average salary for the silver employees.\u00a0 It actually doesn’t matter what value we pick, because averages will fall in the same relative places regardless of whether all the individual values are slid up and down the number line.\u00a0 The easiest value by far to pick is zero.\u00a0 Let’s pretend that the silver folks make $0, and the gold folks make $56.\u00a0 (I divided dollars by 1000 for simplicity).<\/p>\n

              Now, we also have to simplify the numbers of employees.\u00a0 We could reduce the number of employees as long as we preserve the relative ratio.<\/p>\n

              silver : gold = 120:160 = 12:16 = 3:4<\/p>\n

              So everything would be the same if we just had 3 silver employees and 4 gold employees.\u00a0 OK, now find the sums.<\/p>\n

              silver = 3*0 = 0<\/p>\n

              gold = 4*56<\/p>\n

              I am not even going to bother to multiply that out, because we know that the next step is to divide by 7, the total number of employees.<\/p>\n

              total average = 4*56\/7 = 4*8 = 32<\/p>\n

              The average salary is $32,000, which is $32,000 higher than the average for the silver employees.\u00a0 Answer = (C)<\/strong><\/p>\n

              Method II: proportional placement of the total average<\/strong><\/p>\n

              The ratio of silver employees to gold employees is<\/p>\n

              silver : gold = 120:160 = 12:16 = 3:4<\/p>\n

              If we divide the distance between the two averages by 7, then silver will be “four parts” away from the total average, and gold will be “three parts” away.<\/p>\n

              Well, the difference is $56,000, so that divided by 7 is $8000.\u00a0 That’s one part.\u00a0 Four parts would be $32,000, which has to be the distance from the silver average to the total average.<\/p>\n

              Answer = (C) <\/strong><\/p>\n

               <\/p>\n

              4) We will approach this using sums.\u00a0 The individual employee numbers are small.\u00a0 We will divide all dollar amounts by 1000, for easier calculations.\u00a0 Call the highest-paid employee salary x.\u00a0 Then the sums are<\/p>\n

              lowest = 10*80 = 800<\/p>\n

              middle = 6*150 = 3*300 = 900<\/p>\n

              highest = 4x<\/p>\n

              Individual sums must add up to the total sum.<\/p>\n

              800 + 900 + 4x = 3500<\/p>\n

              4x = 1800<\/p>\n

              2x = 900<\/p>\n

              x = 450<\/p>\n

              The salary of each of those four highest paid employees is $450,000.<\/p>\n

              Answer = (E)<\/strong><\/p>\n

               <\/p>\n

              5) This question is designed for an analysis involving proportional placement of the mean.\u00a0 First, observe that R is much closer to the average for one-bedroom apartments, so there must be more one-bedroom apartments and fewer two-bedroom apartments.<\/p>\n

              The ratio of the distances to R is<\/p>\n

              5600:10400 = 56:104<\/p>\n

              Cancel a factor of 8 from both 56 and 104<\/p>\n

              56:104 = 7:13<\/p>\n

              One-bedroom apartments are “13 parts” of the building, and two-bedroom apartments are “7 parts.”\u00a0 That’s a total of 7 + 13 = 20 parts in the building.\u00a0 Two-bedroom apartments constitute 7\/20 of the apartments in the building.\u00a0 Since 1\/20 = 5%, 7\/20 = 35%.<\/p>\n

              Answer = (B)<\/strong><\/p>\n

               <\/p>\n

              6) Method I: using sums<\/strong><\/p>\n

              First, last year.\u00a0 Let x be the number of entrees.\u00a0 Then (15 \u2013 x) is the number of appetizers.\u00a0 The sums are:<\/p>\n

              entrees = 30x<\/p>\n

              appetizers = (15 \u2013 x)*12 = 12*15 \u2013 12x = 6*30 \u2013 12x = 180 \u2013 12x<\/p>\n

              total = 15*18 = 30*9 = 270<\/p>\n

              Notice the use of the Doubling and Halving<\/a> trick in the second and third lines.\u00a0 The two individual sums should add up to the total sum.<\/p>\n

              30x + 180 \u2013 12x = 270<\/p>\n

              18x = 90<\/p>\n

              x = 5<\/p>\n

              They start out with 5 entrees and 10 appetizers.<\/p>\n

              Let N be the number of appetizers added, so now there are 5 entrees and (10 + N) appetizers.\u00a0 We need to solve for N.\u00a0 Again, the sums:<\/p>\n

              entrees = 5*30 = 150<\/p>\n

              appetizers = (10 + N)*12 = 120 + 12N<\/p>\n

              total = (15 + N)*15 = 225 + 15N<\/p>\n

              Again, the two individual sums should add up to the total sum.<\/p>\n

              150 + 120 + 12N = 225 + 15N<\/p>\n

              270 = 225 + 3N<\/p>\n

              45 = 3N<\/p>\n

              15 = N<\/p>\n

              They added 15 more appetizers.\u00a0 Answer = (E)<\/strong><\/p>\n

              Method I was do-able, but we had to solve for many values.<\/p>\n

              Method II: proportional placement of the total average<\/strong><\/p>\n

              Originally, the entr\u00e9e price was 30 \u2013 18 = 12 from the total average, and the appetizer price was 18 \u2013 12 = 6.\u00a0 This means there must have been twice as many appetizers as entrees.\u00a0 Therefore , with 15 items, there must have been 10 appetizers and 5 entrees.<\/p>\n

              The number of entrees doesn’t change.\u00a0 The average drops to $15, so the distance from the entr\u00e9e price is now 30 \u2013 15 = 15, and the distance from the appetizer price is now 15 \u2013 12 = 3.\u00a0 That’s a 5-to-1 ratio, which means there must be 5x as many appetizers as entrees.\u00a0 Since there still are 5 entrees, there must now be 25 appetizers, so 15 have been added.<\/p>\n

              Answer = (E)<\/strong><\/p>\n

              If you know how to employ this method, it is much more elegant.<\/p>\n

               <\/p>\n","protected":false},"excerpt":{"rendered":"

              Problems range from easy to hard. 1) On a ferry, there are 50 cars and 10 trucks.\u00a0 The cars have an average mass of 1200 kg and the trucks have an average mass of 3000 kg.\u00a0 What is the average mass of all 60 vehicles on the ferry? (A) 1200 kg (B) 1500 kg (C) […]<\/p>\n","protected":false},"author":26,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[193],"tags":[],"ppma_author":[13209],"class_list":["post-5611","post","type-post","status-publish","format-standard","hentry","category-word-problems"],"acf":[],"yoast_head":"\nGMAT Math: Weighted Averages - Magoosh Blog \u2014 GMAT\u00ae Exam<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/magoosh.com\/gmat\/gmat-math-weighted-averages\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"GMAT Math: Weighted Averages\" \/>\n<meta property=\"og:description\" content=\"Problems range from easy to hard. 1) On a ferry, there are 50 cars and 10 trucks.\u00a0 The cars have an average mass of 1200 kg and the trucks have an average mass of 3000 kg.\u00a0 What is the average mass of all 60 vehicles on the ferry? 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