GMAT Ratios 700 Lvl

" "On a certain sight-seeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the sight-seeing tour? (1) On the sight-seeing tour, the ratio of the number of children to the number of men was 5 to 11. (2) The number of women on the sight-seeing tour was less than 30.

" "Given woman: children=5:2 1). children: man=5:11, you agree it is insufficient 2). W<30, you also agree it alone is insufficient Together, w:c:m = 25:10:22 (all have to be integers!) thus w=25 and m=22. Answer is C

" "Malik's recipe for 4 servings of a certain dish requires 3/2 cups of pasta. According to this recipe, what is the number of cups of pasta that Malik will use the next time he prepares this dish? (1) The next time he prepares this dish, Malik will make half as many servings as he did the last time he prepared the dish. (2) Malik used 6 cups of pasta the last time he prepared this dish.

" " Premise: one serving includes a certain number of dishes.(we don't know the exact number),and a dish requires 3/2 cups of pasta.( it means 4Y=mX, and X=3/2pasta. ) Question: nY require how many cups of pasta? 1). if Malik make X servings next time. He did prepare 2X dishes last time. 2). Malik used 6 cups of pasta the last time he prepared this dish.(it means 2X=6). In this case, either condition one or condition two cannot deduce the final answer in that the decisive factors m, n are unknown. As a result, the correct answer is C.

" "A certain galaxy is known to comprise approximately 4 x 1011 stars. Of every 50 million of these stars, one is larger in mass than our sun. Approximately how many stars in this galaxy are larger than the sun? 800 1,250 8,000 12,000 80,000

" "50 million can be represented in scientific notation as 5 x 107. Restating this figure in scientific notation will enable us to simplify the division required to solve the problem. If one out of every 5 x 107 stars is larger than the sun, we must divide the total number of stars by this figure to find the solution: 4 x 1011 5 x 107 = 4/5 x 10(11-7) = 0.8 x 104 The final step is to move the decimal point of 0.8 four places to the right, with a result of 8,000. The correct answer is C.

" "A foreign language club at Washington Middle School consists of n students, 2/5 of whom are boys. All of the students in the club study exactly one foreign language. 1/3 of the girls in the club study Spanish and 5/6 of the remaining girls study French. If the rest of the girls in the club study German, how many girls in the club, in terms of n, study German? 2n/5 n/3 n/5 2n/15 n/15

" "Boys = 2n/5, girls = 3n/5 3n Girls studying Spanish = × 5 1 3 n = 5 3n Girls not studying Spanish = - 5 n 5 2n = 5 2n Girls studying French = × 5 5 6 n = 3 Girls studying German = (all girls) - (girls studying Spanish) - (girls studying French) 3n Girls studying German = - 5 n 5 n - 3 2n Girls studying German = - 5 n 3 n = 15 The correct answer is E.

" "3/5 of a certain class left on a field trip. 1/3 of the students who stayed behind did not want to go on the field trip (all the others did want to go). When another vehicle was located, 1/2 of the students who did want to go on the field trip but had been left behind were able to join. What fraction of the class ended up going on the field trip? ½ 2/3 11/15 23/30 4/5

" "For a fraction question that makes no reference to specific values, it is best to assign a smart number as the ""whole value"" in the problem. In this case we'll use 30 since that is the least common denominator of all the fractions mentioned in the problem. If there are 30 students in the class, 3/5 or 18, left for the field trip. This means that 12 students were left behind. 1/3 of the 12 students who stayed behind, or 4 students, didn't want to go on the field trip. This means that 8 of the 12 who stayed behind did want to go on the field trip. When the second vehicle was located, half of these 8 students or 4, were able to join the other 18 who had left already. That means that 22 of the 30 students ended up going on the trip. 22/30 reduces to 11/15 so the correct answer is C.

" "A lemonade stand sold only small and large cups of lemonade on Tuesday. 3/5 of the cups sold were small and the rest were large. If the large cups were sold for 7/6 as much as the small cups, what fraction of Tuesday's total revenue was from the sale of large cups? 7/16 7/15 10/21 17/35 ½

" "For a fraction word problem with no actual values for the total, it is best to plug numbers to solve. Since 3/5 of the total cups sold were small and 2/5 were large, we can arbitrarily assign 5 as the number of cups sold. Total cups sold = 5 Small cups sold = 3 Large cups sold = 2 Since the large cups were sold at 7/6 as much per cup as the small cups, we know: Pricelarge = (7/6)Pricesmall Let's assign a price of 6 cents per cup to the small cup. Price of small cup = 6 cents Price of large cup = 7 cents Now we can calculate revenue per cup type: Large cup sales = quantity × cost = 2 × 7 = 14 cents Small cup sales = quantity × cost = 3 × 6 = 18 cents Total sales = 32 cents The fraction of total revenue from large cup sales = 14/32 = 7/16. The correct answer is A.

" "The ratio by weight, measured in pounds, of books to clothes to electronics in Jorge's suitcase initially stands at 8 to 5 to 3. Jorge then removes 4 pounds of clothing from his suitcase, thereby doubling the ratio of books to clothes. Approximately how much do the electronics in the suitcase weigh, to the nearest pound? 3 4 5 6 7

" "Initially the ratio of B: C: E can be written as 8x: 5x: 3x. (Recall that ratios always employ a common multiplier to calculate the actual numbers.) After removing 4 pounds of clothing, the ratio of books to clothes is doubled. To double a ratio, we double just the first number; in this case, doubling 8 to 5 yields a new ratio of 16 to 5. This can be expressed as follows: books clothing = 8x 5x - 4 = 16 5 Cross multiply to solve for x: 40x = 80x - 64 40x = 64 x = 8/5 The question asks for the approximate weight of the electronics in the suitcase. Since there are 3x pounds of electronics there are 3 × (8/5) = 24/5 or approximately 5 pounds of electronics in the suitcase. The correct answer is C.

" "Joe, Bob and Dan worked in the ratio of 1:2:4 hours, respectively. How many hours did Bob work? (1) Together, Joe, Bob and Dan worked a total of 49 hours. (2) Dan worked 21 hours more than Joe.

" "It is useful to think of the ratio as 1x : 2x : 4x, where x is the ""missing multiplier"" that you use to find the actual numbers involved. For example, if x = 1, then the numbers of hours worked by the three men are 1, 2, and 4. If x = 2, then the numbers are 2, 4, and 8. If x = 11, then the numbers are 11, 22, and 44. Notice that these numbers all retain the original ratio. If we knew the multiplier, we could figure out the number of hours any of the men worked. So we can rephrase the question as, ""What is the missing multiplier?"" SUFFICIENT: Since the three men worked a total of 49 hours and since 1x + 2x + 4x = 7x, we know that 7x = 49. Therefore, x = 7. Since Bob worked 2x hours, we know he worked 2(7) = 14 hours. SUFFICIENT: This statement tells us that 4x = 1x + 21. Therefore, 3x = 21 and x = 7. Since Bob worked 2x hours, we know he worked 2(7) = 14 hours. The correct answer is D.

" "Each employee of Company Z is an employee of either Division X or Division Y, but not both. If each division has some part-time employees, is the ratio of the number of full-time employees to the number of part-time employees greater for Division X than for Company Z? (1) The ratio of the number of full-time employees to the number of part-time employees is less for Division Y than for Company Z. (2) More than half the full-time employees of Company Z are employees of Division X, and more than half of the part-time employees of Company Z are employees of Division Y.

" "Let Xf/Yf is the full time in Division X/Y, and Xp/Yp is part time in X/Y, X,Y, and Z are number of employees in X, Y, and Z. X=Xf+Xp Y=Yf+Yp Z=X+Y For 1, Yf/Yp<Zf/Zp, as a compensation, Xf/Xp should be greater than Zf/Zp For2, More than ?Zf /less than ?of the Zp should be greater than Zf/Zp Answer is D

" "If Pool Y currently contains more water than Pool X, and if Pool X is currently filled to 2/7 of its capacity, what percent of the water currently in Pool Y needs to be transferred to Pool X if Pool X and Pool Y are to have equal volumes of water? (1) If all the water currently in Pool Y were transferred to Pool X, Pool X would be filled to 6/7 of its capacity. (2) Pool X has a capacity of 14,000 gallons.

" "Let x represent the amount of water in Pool X, and y represent the amount of water in Pool Y. If we let z represent the proportion of Pool Y's current volume that needs to be transferred to Pool X, we can set up the following equation and solve for z: (water currently in Pool X) + (water transferred) = (water currently in Pool Y) - (water transferred) x + zy = y - zy x + 2zy = y 2zy = y - x z = y 2y - x 2y z = 1 2 ( 1 - x y ) So, the value of z depends only on the ratio of the water currently in Pool X to the water currently in Pool Y, or: x y . The rephrased question is: ""What is x y ?"" Remember that x and y do NOT represent the capacities of either pool, but rather the ACTUAL AMOUNTS of water in each pool. (1) SUFFICIENT: if we let X represent the capacity of Pool X, then the amount of water in Pool X is (2/7)X. So, x = (2/7)X. We can calculate the total amount of water in Pool Y, or y, as follows: y = (6/7)X - (2/7)X = (4/7)X. We can see that Pool Y has twice as much water as Pool X, or 2x = y, or x y = 1/2. (2) INSUFFICIENT: This gives no information about the amount of water in Pool Y. The correct answer is A.

" "At the beginning of the year, the ratio of juniors to seniors in high school X was 3 to 4. During the year, 10 juniors and twice as many seniors transferred to another high school, while no new students joined high school X. If, at the end of the year, the ratio of juniors to seniors was 4 to 5, how many seniors were there in high school X at the beginning of the year? 80 90 100 110 120

" "Let's denote the number of juniors and seniors at the beginning of the year as j and s, respectively. At the beginning of the year, the ratio of juniors to seniors was 3 to 4: j/s = 3/4. Therefore, j = 0.75s At the end of the year, there were (j - 10) juniors and (s - 20) seniors. Additionally, we know that the ratio of juniors to seniors at the end of the year was 4 to 5. Therefore, we can create the following equation: j - 10 s - 20 = 4 5 Let's solve this equation by substituting j = 0.75s: j - 10 s - 20 = 0.8 (j - 10) = 0.8(s - 20) (0.75s - 10) = 0.8s - 16 0.8s - 0.75s = 16 - 10 0.05s = 6 s = 120 Thus, there were 120 seniors at the beginning of the year. The correct answer is E.

" "A certain ball team has an equal number of right- and left-handed players. On a certain day, two-thirds of the players were absent from practice. Of the players at practice that day, one-third were left handed. What is the ratio of the number of right-handed players who were not at practice that day to the number of lefthanded players who were not at practice? 1/3 2/3 5/7 7/5 3/2

" "Since the problem deals with fractions, it would be best to pick a smart number to represent the number of ball players. The question involves thirds, so the number we pick should be divisible by 3. Let's say that we have 9 right-handed players and 9 left-handed players (remember, the question states that there are equal numbers of righties and lefties). Two-thirds of the players are absent from practice, so that is (2/3)(18) = 12. This leaves 6 players at practice. Of these 6 players, one-third were left-handed. This yields (1/3)(6) = 2 left-handed players at practice and 9 - 2 = 7 left-handed players NOT at practice. Since 2 of the 6 players at practice are lefties, 6 - 2 = 4 players at practice must be righties, leaving 9 - 4 = 5 righties NOT at practice. The question asks us for the ratio of the number of righties not at practice to the number of lefties not at practice. This must be 5 : 7 or 5/7. The correct answer is C.

1000.

" "The question asks us to find the ratio of gross revenue of computers to printers, given that the price of a computer is five times the price of a printer. We will prove that the statements are insufficient either singly or together by finding two examples that satisfy all the criteria but give two different ratios for the gross revenue of computers to printers. (1) INSUFFICIENT: Statement (1) says that the ratio of computers to printers sold in the first half of 2003 was in the ratio of 3 to 2, so let's assume they sold 3 computers and 2 printers. Using an example price of

7 in the piggy bank.

" "The question asks us to solve for the ratio of pennies (p) to dimes (d). INSUFFICIENT: This tells us that the ratio of nickels (n) to dimes (d) is 3:2. This gives us no information about the ratio of pennies to dimes. INSUFFICIENT: This tells us that there is

" "The ratio of boys to girls in Class A is 3 to 4. The ratio of boys to girls in Class B is 4 to 5. If the two classes were combined, the ratio of boys to girls in the combined class would be 17 to 22. If the number of boys in Page 16 Class B is one less than the number of boys in Class A, and if the number of girls in Class B is two less than the number of girls in Class A, how many girls are in Class A? 8 9 10 11 12

" "The ratio of boys to girls in Class A is 3 to 4. We can represent this as an equation: b/g = 3/4. We can isolate the boys: 4b = 3g b = (3/4)g Let's call the number of boys in Class B x, and the number of girls in Class B y. We know that the number of boys in Class B is one less than the number of boys in Class A. Therefore, x = b - 1. We also know that the number of girls in Class B is two less than the number of girls in Class A. Therefore, y = g - 2. We can substitute these in the combined class equation: The combined class has a boy/girl ratio of 17 to 22: (b + x)/(g + y) = 17/22. (b + b - 1)/(g + g - 2) = 17/22 (2b - 1)/(2g - 2) = 17/22 Cross-multiplying yields: 44b - 22 = 34g - 34 Since we know that b = (3/4)g, we can replace the b: 44(3/4)g - 22 = 34g - 34 33g - 22 = 34g - 34 12 = g Alternatively, because the numbers in the ratios and the answer choices are so low, we can try some real numbers. The ratio of boys to girls in Class A is 3:4, so here are some possible numbers of boys and girls in Class A: B:G 3:4 6:8 9:12 The ratio of boys to girls in Class B is 4:5, so here are some possible numbers of boys and girls in Class A: B:G 4:5 8:10 12:15 We were told that there is one more boy in Class A than Class B, and two more girls in Class A than Class B. If we look at our possibilities above, we see that this information matches the case when we have 9 boys and 12 girls in Class A and 8 boys and 10 girls in Class B. Further, we see we would have 9 + 8 = 17 boys and 12 + 10 = 22 girls in a combined class, so we have the correct 17:22 ratio for a combined class. We know now there are 12 girls in Class A. The correct answer is E.

" "In a certain pet shop, the ratio of dogs to cats to bunnies in stock is 3 : 5 : 7. If the shop carries 48 cats and bunnies total in stock, how many dogs are there? 12 13 14 15 16

" "The ratio of dogs to cats to bunnies (Dogs: Cats: Bunnies) can be expressed as 3x: 5x : 7x. Here, x represents an ""unknown multiplier."" In order to solve the problem, we must determine the value of the unknown multiplier. Cats + Bunnies = 48 5x + 7x = 48 12x = 48 x = 4 Now that we know that the value of x (the unknown multiplier) is 4, we can determine the number of dogs. Dogs = 3x = 3(4) = 12 The correct answer is A.

" "Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How many of the animals are cows? (1) the farm has more than twice as many cows at it has pigs. (2) the farm has more than 12 pigs

" "The total number of pigs and cows is 40. For1, C>2P For 2, P>12 Combine 1 and 2, if P=13, C is 14;if P=14, C is 12, it is impossible. Answer is C

" "Miguel is mixing up a salad dressing. Regardless of the number of servings, the recipe requires that 5/8 of the finished dressing mix be olive oil, 1/4 vinegar, and the remainder an even mixture of salt, pepper and sugar. If Miguel accidentally doubles the vinegar and forgets the sugar altogether, what proportion of the botched dressing will be olive oil? 15/29 5/8 5/16 ½ 13/27

" "This problem can be solved most easily by picking smart numbers and assigning values to the portion of each ingredient in the dressing. A smart number in this case would be one that enables you to add and subtract ingredients without having to deal with fractions or decimals. In a fraction problem, the 'smart number' is typically based on the least common denominator among the given fractions. The two fractions given, 5/8 and 1/4, have a least common denominator of 8. However, we must also consider the equal parts salt, pepper and sugar. Because 1/4 = 2/8, the total proportion of oil and vinegar combined is 5/8 + 2/8 = 7/8. The remaining 1/8 of the recipe is split three ways: 1/24 each of salt, pepper, and sugar. 24 is therefore our least common denominator, suggesting that we should regard the salad dressing as consisting of 24 units. Let's call them cups for simplicity, but any unit of measure would do. If properly mixed, the dressing would consist of 5/8 × 24 = 15 cups of olive oil 1/4 × 24 = 6 cups of vinegar 1/24 × 24 = 1 cup of salt 1/24 × 24 = 1 cup of sugar 1/24 × 24 = 1 cup of pepper Miguel accidentally doubled the vinegar and omitted the sugar. The composition of his bad salad dressing would therefore be 15 cups of olive oil 12 cups of vinegar 1 cup of salt 1 cup of pepper The total number of cups in the bad dressing equals 29. Olive oil comprises 15/29 of the final mix. The correct answer is A.

" "Harold and Millicent are getting married and need to combine their already-full libraries. If Harold, who has 1/2 as many books as Millicent, brings 1/3 of his books to their new home, then Millicent will have enough room to bring 1/2 of her books to their new home. What fraction of Millicent's old library capacity is the new home's library capacity? ½ 2/3 ¾ 4/5 5/6

" "This problem never tells us how many books there are in any of the libraries. We can, therefore, pick numbers to represent the quantities in this problem. It is a good idea to pick Smart Numbers, i.e. numbers that are multiples of the common denominator of the fractions given in the problem. In this problem, Harold brings 1/3 of his books while Millicent brings 1/2. The denominators, 2 and 3, multiply to 6, so let's set Harold's library capacity to 6 books. The problem tells us Millicent has twice as many books, so her library capacity is 12 books. We use these numbers to calculate the size of the new home's library capacity. 1/3 of Harold's 6-book library equals 2 books. 1/2 of Millicent's 12-book library equals 6 books. Together, they bring a combined 8 books to fill their new library. The fraction we are asked for, (new home's library capacity) / (Millicent's old library capacity), therefore, is 8/12, which simplifies to 2/3. The correct answer is B.

" "In a certain solution consisting of only two chemicals, for every 3 milliliters of Chemical A, there are 7 milliliters of Chemical B. After 10 milliliters of Chemical C are added to this solution, what is the ratio of the quantities of Chemical A to Chemical C? (1) Before Chemical C was added, there were 50 milliliters of solution. (2) After Chemical C was added, there were 60 milliliters of solution.

" "To determine the ratio of Chemical A to Chemical C, we need to find the amount of each in the solution. The question stem already tells us that there are 10 milliliters of Chemical C in the final solution. We also know that the original solution consists of only Chemicals A and B in the ratio of 3 to 7. Thus, we simply need the original volume of the solution to determine the amount of Chemical A contained in it. SUFFICIENT: This tells us that original solution was 50 milliliters. Thus, there must have been 15 milliliters of Chemical A (to 35 milliliters of Chemical B). The ratio of A to C is 15 to 10 (or 3 to 2). SUFFICIENT: This tells us that the final solution was 60 milliliters. We know that this includes 10 milliliters of Chemical C. This means the original solution contained 50 milliliters. Thus, there must have been 15 milliliters of Chemical A (to 35 milliliter of Chemical B). The ratio of A to C is 15 to 10 (or 3 to 2). The correct answer is D.

" "Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 30 white marbles. How many red marbles could be in bag A? 1 3 4 6 8

" "We are told that bag B contains red and white marbles in the ration 1:4. This implies that WB, the number of white marbles in bag B, must be a multiple of 4. What can we say about WA, the number of white marbles in bag A? We are given two ratios involving the white marbles in bag A. The fact that the ratio of red to white marbles in bag A is 1:3 implies that WA must be a multiple of 3. The fact that the ratio of white to blue marbles in bag A is 2:3 implies that WA must be a multiple of 2. Since WA is both a multiple of 2 and a multiple of 3, it must be a multiple of 6. We are told that WA + WB = 30. We have already figured out that WA must be a multiple of 6 and that WB must be a multiple of 4. So all we need to do now is to test each candidate value of WA (i.e. 6, 12, 18, and 24) to see whether, when plugged into WA + WB = 30, it yields a value for WB that is a multiple of 4. It turns out that WA = 6 and WA = 18 are the only values that meet this criterion. Recall that the ratio of red to white marbles in bag A is 1:3. If there are 6 white marbles in bag A, there are 2 red marbles. If there are 18 white marbles in bag A, there are 6 red marbles. Thus, the number of red marbles in bag A is either 2 or 6. Only one answer choice matches either of these numbers. The correct answer is D.

" "Three business partners shared all the proceeds from the sale of their privately held company. If the partner with the largest share received exactly 5/8 of the total proceeds, how much money did the partner with the smallest share receive from the sale? (1) The partner with the smallest share received from the sale exactly 1/5 the amount received by the partner with the second largest share. (2) The partner with the second largest share received from the sale exactly half of the two million dollars received by the partner with the largest share.

" "We can rewrite the information in the question as an equation representing the T, the total dollar value of the sale: L + M + S = T L = the dollar amount received by the partner with the largest share M = the dollar amount received by the partner with the middle (second largest) share S = the dollar amount received by the partner with the smallest share We are also told in the question that L = (5/8)T. Thus we can rewrite the equation as follows: (5/8)T + M + S = T. Since the question asks us the value of S, we can simplify the equation again as follows: S = M + (3/8)T Thus, in order to solve for S, we will need to determine the value of both M and T. The question can be rephrased as, what is the value of M + (3/8)T? NOT SUFFICIENT: The first statement tells us that S = (1/5)M. This gives us no information about T so statement one alone is not sufficient. SUFFICIENT: The second statement tells us that M = (1/2)L =

" "John's front lawn is 1/3 the size of his back lawn. If John mows 1/2 of his front lawn and 2/3 of his back lawn, what fraction of his lawn is left unmowed? 1/6 1/3 3/8 ½ 5/8

" "We can solve this problem by choosing a smart number to represent the size of the back lawn. In this case, we want to choose a number that is a multiple of 2 and 3 (the denominators of the fractions given in the problem). This way, it will be easy to split the lawn into halves and thirds. Let's assume the size of the back lawn is 6. (size back lawn) = 6 units (size front lawn) = (1/3)(size back lawn) = 2 units (size total lawn) = (size back lawn) + (size front lawn) = 8 units Now we can use these numbers to calculate how much of each lawn has been mowed: Front lawn: (1/2)(2) = 1 unit Back lawn: (2/3)(6) = 4 unit So, in total, 5 units of lawn have been mowed. This represents 5/8 of the total, meaning 3/8 of the lawn is left unmowed. Alternatively, this problem can be solved using an algebraic approach. Let's assume the size of the front lawn is x and size of the back lawn is y. So, John has mowed (1/2)x and (2/3)y, for a total of (1/2)x + (2/3)y. We also know that x = (1/3)y. Substituting for x gives: (1/2)x + (2/3)y (1/2)(1/3)y + (2/3)y (1/6)y + (2/3)y (5/6)y = lawn mowed The total lawn is the sum of the front and back, x + y. Again, substituting for x gives (1/3)y + y = (4/3)y. So, the fraction of the total lawn mowed is: lawn mowed total lawn (5/6)y (4/3)y = (5/6) (4/3) = (5/6) × (3/4) = 15/24 = 5/8. This leaves 3/8 unmowed. The correct answer is C.

" "At Jefferson Elementary School, the number of teachers and students (kindergarten through sixth grade) totals 510. The ratio of students to teachers is 16 to 1. Kindergarten students make up 1/5 of the student population and fifth and sixth graders account for 1/3 of the remainder. Students in first and second grades account for 1/4 of all the students. If there are an equal number of students in the third and fourth grades, then the number of students in third grade is how many greater or fewer than the number of students in kindergarten? 12 greater 17 fewer 28 fewer 36 fewer 44 fewer

" "We know that the student to teacher ratio at the school is 16 to 1, and the total number of people is 510. Therefore: Number of students = (16/17)(510) = 480 Number of teachers = (1/17)(510) = 30 Kindergarten students make up 1/5 of the student population, so: Number of kindergarten students = (1/5)(480) = 96 Fifth and sixth graders account for 1/3 of the remainder (after kindergarten students are subtracted from the total), therefore: Number of 5th and 6th grade students = (1/3)(480 - 96) = (1/3)(384) = 128 Students in first and second grades account for 1/4 of all the students, so: Number of 1st and 2nd grade students = (1/4)(480) = 120 So far, we have accounted for every grade but the 3rd and 4th grades, so they must consist of the students left over: Number of 3rd and 4th grade students = Total students - students in other grades Number of 3rd and 4th grade students = 480 - 96 - 128 - 120 = 136 If there are an equal number of students in the third and fourth grades, then: Number of 3rd grade students = 136/2 = 68 The number of students in third grade is 68, which is fewer than 96, the number of students in kindergarten. The number of students in 3rd grade is thus 96 - 68 = 28 fewer than the number of kindergarten students. The correct answer is C.

Which of the following fractions is at least twice as great as 11/50? 2/5 11/34 43/99 8/21 9/20

"First, let us rephrase the question. Since we need to find the fraction that is at least twice greater than 11/50, we are looking for a fraction that is equal to or greater than 22/50. Further, to facilitate our analysis, note that we can come up with an easy benchmark value for this fraction by doubling both the numerator and the denominator and thus expressing it as a percent: 22/50 = 44/100 = 44%. Thus, we can rephrase the question: "Which of the following is greater than or equal to 44%?" Now, let's analyze each of the fractions in the answer choices using benchmark values: 2/5: This fraction can be represented as 40%, which is less than 44%. 11/34: This value is slightly less than 11/33 or 1/3. Therefore, it is smaller than 44%. 43/99: Note that the fraction 43/99 is smaller than 44/100, since fractions get smaller if the same number (in this case integer 1) is subtracted from both the numerator and the denominator. 8/21: We know that 8/21 is a little less than 8/20 or 2/5. Thus, 8/21 is less than 44%. 9/20: Finally, note that by multiplying the numerator and the denominator by 5, we can represent this fraction as 45/100, thus concluding that this fraction is greater than 44%. The correct answer is E.