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How to Go From a 48 to 51 in GMAT Quant [#permalink]
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How to Go From a 48 to 51 in GMAT QuantPeople often ask – how do we go from 48 to 51 in Quant? This question is very hard to answer since we don’t have a step by step plan – do theory from here – do questions from there – take a test from here – read posts from there etc. Today and in the next few weeks, we will discuss how to go from 48 to 51 in Quant. Above Q48, the waters are pretty choppy! Questions are hard less because of the content and more because they look so unique – even though they’re testing the same concepts. Training yourself to see familiarity in the obscure is difficult, and that happens from seeing a lot of problems. There is barely any scope for making silly mistakes – you must run through all simple questions quickly and neatly, leaving you plenty of time to think through the tougher ones. It’s important to have enough time and keep your cool, which is easier to do if you have more time. The question for today is: how do you handle simple questions quickly? We have mentioned many times that most GMAT Quant questions do not need Algebra. We can easily solve them by just analyzing while reading the question stem! Here is how we can do that: Question: School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer?(A) 20 (B) 40 (C) 60 (D) 80 (E) 100 Solution: This is a pretty simple nontricky PS question. To solve it, most people use an algebraic method which looks something like this. Girls in school A : Girls in school B = 4 : 3 So number of girls in school \(A = 4n\) and number of girls in school \(B = 3n\) Since in school A, 40% students are girls and 60% are boys, number of boys is \(6n\). Since in school B, 60% students are girls and 40% students are boys, number of boys is \(2n\). If we transfer 20 boys from school A, we are left with \(6n  20\) and when 20 boys are added to school B, we get \(2n + 20\) boys in school B. \(\frac{(6n  20)}{(2n + 20)} = \frac{5}{3}\) You get \(n = 20\). Boys at school A after transfer = \(6*20  20 = 100\) Boys at school B after transfer = \(2*20 + 20 = 60\) Difference = 40 Answer (B). This question is discussed HERE. This method gives you the correct answer, obviously, but it does take quite a bit of time. On the other hand, this is what should go through your mind while reading the question if you are focused on using logic: “School A is 40% girls and school B is 60% girls.” School A – 40% girls School B – 60% girls “The ratio of the number of girls at school A to the number of girls at school B is 4:3” When we read this line, we should take a step back to the previous line with the % figures. We see that school A has more girls than school B (4:3) but its % of girls is lower (only 40% compared to 60% in B). This means that school A has more students than school B. Say something like school A has 200 students while school B has 100 (use easy numbers). So school A has 80 girls while school B has 60 girls. This gives us a ratio of 4:3. (If you do not get 4:3 on your first try, you should tweak the assumed numbers a bit but you should stick to simple numbers.) Then verify the rest of the data against these numbers and get your answer. School A has 120 boys and school B has 40 boys. Transfer 20 boys from school A to school B to get 100 boys in school A and 60 boys in school B giving us a difference of 40 boys. This takes lesser time but requires some ingenuity. That could be the difference between Q48 and Q51. Hope this gave you some ideas. Try the reasoning approach on other simple questions. With practice, you can save a ton of time! ______________________________________________________________________________________________________ Part 2 Part 3 Part 4 Part 5 Part 6 Part 7
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Re: How to Go From a 48 to 51 in GMAT Quant [#permalink]
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01 Jul 2015, 02:57
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Another method of saving time on simple questions – use data given in one statement to examine the other!Now you might think we have lost it! After all, you know very well that in Data Sufficiency questions of GMAT, you must examine each statement independently. You CANNOT use data from one to analyze the other – absolutely correct. So you should ignore the other statement completely while examining one – hmm, not necessarily! Sometimes, one statement could give us ideas about the next one such that we could save time while examining it. Needless to say, we need to be very careful but it certainly is a useful strategy. Also, it could help us verify that our calculations are correct. Here is why… When we say DS question, think of a puzzle. The question stem gives you the statement of a puzzle ending with something like “What is the value of x?” or “Is x 7?” etc. You have to answer the question asked in the puzzle. Think of the two statements that come with the question as clues to the puzzle. So the puzzlemaster gives you the first clue (statement 1) and asks you: can you answer the question now? If you are able to, your answer is either (A) or (D). Then he tells you to ignore the first clue and gives you another clue (statement 2). Again he asks you: can you answer the question now? Again, you may or may not able to. If you are able to, your answer will be (B) or (D) depending on how you fared in statement 1. If you are unable to answer the question, he tells you to consider both statements together and then try to answer. If you are able to, your answer is (C). The point to note here is that both clues lead you to answer the same puzzle. Say if the puzzle is: What is x? If clue 1 tells you that x is 6, clue 2 cannot tell you that x is 9. They both must lead you to the same value of x. Clue 1 could tell you that x is either 6 or 8 and clue 2 could tell you that x is either 8 or 9. In this case, when we use both clues together, we find that x must be 8 to satisfy both. Hence the statements never contradict each other. This means, if we get possible values of x from statement 1, we know that statement 2 will also give us at least one of those values. This is how one statement could give us a starting point for the next one. Now that you understand the “why”, let’s go on to “how”, using a question. Question: If K is a positive integer less than 10 and N = 4,321 + K, what is the value of K?Statement 1: N is divisible by 3 Statement 2: N is divisible by 7 Solution:Given: \(N = 4321 + K\) \(1 \leq K \leq 10\) So N could range from 4322 (when \(K = 1\)) to 4331 (when \(K = 10\)). To find the value of K, we need to find the unique value of N. Statement 1 tells us that N is divisible by 3. 4321 is not divisible by 3 since the sum of its digits is 4+3+2+1 = 10. It is 1 more than a multiple of 3. So the next multiple of 3 will be 4323. Hence N could be 4323. But there are some other multiples of 3 which could be the value of N. After 4323, 4326 and 4329 could also be the values of N since they are multiples of 3 too. We know this because if A is a multiple of 3, A+3, A+6, A+9, A3, A6 etc are also multiples of 3. So since 4323 is a multiple of 3, 4326 and 4329 will also be multiples of 3. We did not get a unique value for N so statement 1 alone is not sufficient. Now let’s go on to statement 2. This tells us that N must be a multiple of 7. In 10 consecutive numbers, there will be either one multiple of 7 or two multiples of 7. If there is only one multiple of 7 in the range 4322 to 4331, statement 2 alone will be sufficient to give us the value of N. If there are two multiples of 7 in this range, then statement 2 alone will not be sufficient. Recall that from statement 1, we already know that N will take one of three values: 4323, 4326 or 4329. Let’s check for 4326 because it is in the middle. If 4326 is divisible by 7, there will be no other multiple of 7 in the range 4322 to 4331 because the closest multiples of 7 to 4326 will be 4326 – 7 and 4326 + 7. When we divide 4326 by 7, we find that it is divisible. This means that statement 2 gives us a single value of N. Hence statement 2 alone is sufficient. Hypothetically, what if we had found that 4326 is not divisible by 7? Then we would have known that either 4323 or 4329 must be a multiple of 7. In both cases, statement 2 would have given us 2 multiples of 7 because both 4330 (7 more than 4323) and 4322 (7 less than 4329) are in the possible range. Then we would have known that the answer will be (C) i.e. we will need both statements to answer the question since the possible values from the two statements will have only one overlap in either case. Note that what we gleaned from statement 1 helped us quickly examine statement 2 and get to the answer right away. But this is an advanced technique and you should use it only if you understand it very well. Else, it is best to stick to completely ignoring one statement while working on the other. This question is discussed HERE.
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Re: How to Go From a 48 to 51 in GMAT Quant [#permalink]
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Let’s get back to strategies that will help us reach the coveted 51 in Quant. First, take a look at Part I and Part II of this blog series. Since the Quant section is not a Math test, you need conceptual understanding and then some ingenuity for the hard questions (since they look unique). Today we look at a Quant problem which is very easy if the method “strikes”. Else, it can be a little daunting. What we will do is look at a “brute force” method for times when the textbook method is not easily identifiable. Question: What is \(\frac{0.99999999}{1.0001}  \frac{0.99999991}{1.0003}\)?(A) \(10{(8)}\) (B) \(3*10^{(8)}\) (C) \(3*10^{(4)}\) (D) \(2*10^{(4)}\) (E) \(10^{(4)}\) Solution: Usually, when we have decimals such as 0.99999999 or 1.0001, we round them off to 1 without thinking twice. The issue here is that all numbers are very close to 1 so if we round them off to 1, we will get \(\frac{1}{1}  \frac{1}{1} = 0\). This, obviously, doesn’t work and we need to work with the complicated numbers only. Now here is the official method, something a Math professor will give us: Method 1: For simplification, you will need to use \(a^2  b^2 = (a  b)(a + b)\) Note that 0.99999999 is 0.00000001 less than 1 and 1.0001 is 0.0001 more than 1. 0.00000001 is the square of 0.0001. \(\frac{0.99999999}{1.0001}  \frac{0.99999991}{1.0003}\) \(\frac{1  0.00000001}{1 + .0001}  \frac{1  0.00000009}{1 + 0.0003}\) \(\frac{1^2  0.0001^2}{1 + 0.0001}  \frac{1^2  0.0003^2}{1 + 0.0003}\) \(\frac{(1  0.0001)(1 + 0.0001)}{1 + 0.0001}  \frac{(1  0.0003)(1 + 0.0003)}{1 + 0.0003}\) \((1  0.0001)  (1  0.0003)\) \(0.0002 = 2*10^{4}\) All in all, the question only required us to recall something we learnt in 7th standard: \(a^2  b^2 = (a  b)(a + b)\) Does it mean it is a very simple question? Not really. The problem is that it is hard to identify that all you need is this formula and that you need to bring the terms in this format. So here is a “brute force” method that people came up with and that we can use when Math fails us: Method 2: The fractions are quite complicated but the options are not fractions. This means that we are able to get rid of the denominator somehow. This brings an idea to mind: 0.99999999 might be a multiple of 1.0001. But how do we find ‘which multiple’? For that, we need to use some pattern recognition. \(9*1.0001 = 9.0009\) \(99*1.0001 = 99.0099\) and so on till \(9999*1.0001 = 9999.9999\) (to get eight 9's) Now since the decimal is 4 digits to the left i.e. the number is actually 0.99999999, \(0.9999 * 1.0001 = 0.99999999\) This means \(\frac{0.99999999}{1.0001} = 0.9999\) On the same lines, we might guess that 0.99999991 is a multiple of 1.0003. To find ‘which multiple’, we might need to think even harder now. Note that something needs to multiply 1.0003 to give something ending in 1. Perhaps this multiple ends with a 7 because 3*7 ends in 1. And sure, \(9997*1.0003 = 9999.9991\) which gives us \(0.9997 * 1.0003 = 0.99999991\) This gives us \(\frac{0.99999991}{1.0003} = 0.9997\) Thus, the problem boils down to \(0.9999  0.9997 = 0.0002\) We encourage you to look for some more brute force methods. This question is discussed HERE.
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Re: How to Go From a 48 to 51 in GMAT Quant [#permalink]
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01 Jul 2015, 03:35
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Another point to keep in mind while targeting Q50+ in GMAT: don’t buy complex official solutions. Most GMAT questions can be solved in a few steps. The point is that sometimes it is hard to identify those “few steps” and we keep going round and round in circles for a while till we arrive at the answer. The way to hit 51 is to look for simple solutions for difficult questions. The best example of this would be question number 148 of Official Guide 12. The question is tough, no doubt about it but just because it is tough, don’t think that the solution needs to be tough too – you don’t have to live with the solution provided. If, even after reading the solution a couple of times, you know that if you try the question again in a week, you won’t be able to solve it on your own, this means you need to review either the concept or the solution. If the given solution is too complex and you almost have to learn it up step by step, it means you need a better solution. The next step of the solution should be apparent to you – you should be able to solve it on your own within two minutes. Also, even if one method looks good, try to find other ways of solving the question. Often, there are multiple good ways of solving a particular question. Here is a question similar to question number 148 of OG12. Let me give a few good methods of solving it: Question: If x, y, and k are positive numbers such that \(\frac{x}{(x+y)}*20 + \frac{y}{(x+y)}*40 = k\), and if \(x < y\), which of the following could be the value of k?(A) 15 (B) 20 (C) 25 (D) 35 (E) 40 Solution: One solution you have in the OG. Three more are provided here: Method 2: AlgebraNote the “could be” in the question. This means that k can take multiple values and one of them is provided here. \(20*\frac{x}{(x+y)} + 40*\frac{y}{(x+y)} = k\) \(20*\frac{(x+2y)}{(x+y)} = k\) \(20*(\frac{(x + y)}{(x + y)} + \frac{y}{(x + y)}) = k\) \(20*(1 + \frac{y}{(x + y)}) = k\) Now since y is greater than x, \(\frac{y}{x+y}\) will be more than 1/2 but definitely less than 1 (x and y are positive numbers). So the value of k will lie in the range \(20*(1 + \frac{1}{2}) < k < 20*(1 + 1)\) i.e. \(30 < k < 40\) Only option (D) falls in this range. Answer (D)Method 3: Weighted AverageDoes this equation remind you of something: \(20*\frac{x}{(x+y)} + 40*\frac{y}{(x+y)} = k\)? If you are a weighted average fan like me, you will notice that this is just the weighted average formula applied: \(C_{avg} = \frac{(C_1*w_1 + C_2*w_2)}{(w_1 + w_2)}\) Where \(C_{avg} = k\) \(C_1 = 20\) \(C_2 = 40\) \(w_1 = x\) \(w_2 = y\) \(k = \frac{(20*x + 40*y)}{(x + y)}\) It might be hard to see this on your own but the point is that if you do see it, the return is very high. We know that the average of two quantities will lie in between them. So k must lie between 20 and 40. Also, we are given that x is less than y i.e. weight given to 20 is less than the weight given to 40. So the weighted average will lie toward 40. Between 30 and 40, there is only option (D) Hence, answer (D).Method 4: Plugging NumbersNow, what if neither of the above given methods work for you during the test and your mind goes blank? Then you can pick some numbers to get an idea of the kind of values you will get. This is absolute brute force and may not always work out but it will give you a fighting chance of getting the correct answer. \(20*\frac{x}{(x+y)}+ 40*\frac{y}{(x+y)} = k\)  Say, \(x = 1\), \(y = 3\) (x and y are positive numbers and x < y)Then \(20*\frac{1}{(1+3)} + 40*\frac{3}{(1+3)} = k = 35\)  Say, \(x = 2\), \(y = 3\) (when you assume numbers, assume those which make the denominator a factor of 20 and 40 for ease of calculations. So assume numbers such that \(x+y\) is 4 or 5 or 10 etc)Then \(20*\frac{2}{(2+3)} + 40*\frac{3}{(2+3)} = k = 32\)  Say, \(x = 1\), \(y = 4\)Then \(20*\frac{1}{(1+4)}+ 40*\frac{4}{(1+4)} = k = 36\) Even if you do not get 35, note that the other values of k lie in the 30s. So your best bet would be to mark answer as (D). Hope you see that there are many different ways of solving a given question, so you don’t usually require complex solutions. Practice on! Similar OG question is discussed HERE.
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Re: How to Go From a 48 to 51 in GMAT Quant [#permalink]
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Bunuel wrote: How to Go From a 48 to 51 in GMAT QuantPeople often ask – how do we go from 48 to 51 in Quant? This question is very hard to answer since we don’t have a step by step plan – do theory from here – do questions from there – take a test from here – read posts from there etc. Today and in the next few weeks, we will discuss how to go from 48 to 51 in Quant. Above Q48, the waters are pretty choppy! Questions are hard less because of the content and more because they look so unique – even though they’re testing the same concepts. Training yourself to see familiarity in the obscure is difficult, and that happens from seeing a lot of problems. There is barely any scope for making silly mistakes – you must run through all simple questions quickly and neatly, leaving you plenty of time to think through the tougher ones. It’s important to have enough time and keep your cool, which is easier to do if you have more time. The question for today is: how do you handle simple questions quickly? We have mentioned many times that most GMAT Quant questions do not need Algebra. We can easily solve them by just analyzing while reading the question stem! Here is how we can do that: Question: School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer?(A) 20 (B) 40 (C) 60 (D) 80 (E) 100 Solution: This is a pretty simple nontricky PS question. To solve it, most people use an algebraic method which looks something like this. Girls in school A : Girls in school B = 4 : 3 So number of girls in school \(A = 4n\) and number of girls in school \(B = 3n\) Since in school A, 40% students are girls and 60% are boys, number of boys is \(6n\). Since in school B, 60% students are girls and 40% students are boys, number of boys is \(2n\). If we transfer 20 boys from school A, we are left with \(6n  20\) and when 20 boys are added to school B, we get \(2n + 20\) boys in school B. \(\frac{(6n  20)}{(2n + 20)} = \frac{5}{3}\) You get \(n = 20\). Boys at school A after transfer = \(6*20  20 = 100\) Boys at school B after transfer = \(2*20 + 20 = 60\) Difference = 40 Answer (B). This question is discussed HERE. This method gives you the correct answer, obviously, but it does take quite a bit of time. On the other hand, this is what should go through your mind while reading the question if you are focused on using logic: “School A is 40% girls and school B is 60% girls.” School A – 40% girls School B – 60% girls “The ratio of the number of girls at school A to the number of girls at school B is 4:3” When we read this line, we should take a step back to the previous line with the % figures. We see that school A has more girls than school B (4:3) but its % of girls is lower (only 40% compared to 60% in B). This means that school A has more students than school B. Say something like school A has 200 students while school B has 100 (use easy numbers). So school A has 80 girls while school B has 60 girls. This gives us a ratio of 4:3. (If you do not get 4:3 on your first try, you should tweak the assumed numbers a bit but you should stick to simple numbers.) Then verify the rest of the data against these numbers and get your answer. School A has 120 boys and school B has 40 boys. Transfer 20 boys from school A to school B to get 100 boys in school A and 60 boys in school B giving us a difference of 40 boys. This takes lesser time but requires some ingenuity. That could be the difference between Q48 and Q51. Hope this gave you some ideas. Try the reasoning approach on other simple questions. With practice, you can save a ton of time! ______________________________________________________________________________________________________ Part 2 Part 3 Part 4 How about taking number of students at A as 400 and at B as 200 no. of boys will be 240 and 80 respectively hence the difference after transfer will be 220100=110 How about taking number of students at A as 100 and at B as 50 no. of boys will be 60 and 20 respectively hence the difference after transfer will be 4040=0 it's actually tough to get ratio as 5:3 after transfer if it doesn't set well there are chances of panic during the exam.



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Re: How to Go From a 48 to 51 in GMAT Quant [#permalink]
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I have seen with my own experience and from expert's views on various forums that one can get a score of Q50 while still missing 1015 questions. However, with regards to Q51 I have not seen such more or less conclusive opinion. I do remember reading a post on BTG (I think by a chicago based tutor) that one can't hope to score Q51 while answering more than 5 questions wrong. Do the experts agree yet on the following? "How many question one can't afford to get wrong while still hoping to score Q51?" Thanks!
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01 Jun 2016, 03:17
Today, we bring another tip for you to help get that dream score of 51 – if you must write down the data given, write down all of it! Let us explain. If you think that you will need to jot down the data given in the question and then solve it on your scratch pad (instead of in your mind), you must jot down every single detail. It is easy to overlook small things which are difficult to express algebraically such as ‘x is an integer’. These details are often critical and could make all the difference between an ‘unsolvable’ question and a ‘solvable within 2 minutes’ one. Once you start solving the question on your scratch pad, you will not refer back to the original question again and again and hence might forget these details. Have them along with the rest of the data. Read every word of the question carefully, and ensure that it is consolidated on your scratch pad. For example, look at this question: A set of five positive integers has an arithmetic mean of 150. A particular number among the five exceeds another by 100. The rest of the three numbers lie between these two numbers and are equal. How many different values can the largest number among the five take? It is a difficult question because it incorporates statistics as well as maxmin – both tricky topics. On top of it, people often overlook the ‘are equal’ part of the question here. The reason for that is that they are actively looking for implications of the sentences and the moment they read “The rest three numbers lie between these two numbers”, they go back to the previous sentence which tells us “A particular number among the five exceeds another by 100”. They then make a note of the fact that 100 is the range of the five positive integers. In all this excitement, they miss the three critical words “and are equal”. Ensure that when you go to the sentence above, you pick the next sentence from the point where you left it. Another thing to note here is that all numbers are positive integers. This information will be critical to us. Let’s demonstrate how you will solve this question after incorporating all the information given. Question: A set of five positive integers has an arithmetic mean of 150. A particular number among the five exceeds another by 100. The rest of the three numbers lie between these two numbers and are equal. How many different values can the largest number among the five take?(A) 18 (B) 19 (C) 21 (D) 42 (E) 59 Solution:Let’s assume that the 5 natural numbers in increasing order are: a, b, b, b, a+100 We are given that \(a < b < a+100\). Also, we are given that a and b are positive integers. This information is critical – we will see later why. The average of the 5 numbers is \(\frac{(a+b+b+b+a+100)}{5} = 150\) \((a+b+b+b+a+100) = 5*150\) \(2a+3b = 650\) We need to find the number of distinct values that a can take because a+100 will also take the same number of distinct values. Now there are two methods to proceed. Let’s discuss both of them. Method 1: Pure Algebra – Write b in terms of a and plug it in the inequality \(b = \frac{(650 – 2a)}{3}\) \(a < \frac{(650 – 2a)}{3} < a+100\) \(3a < 650 – 2a < 3a + 300\) Now split it into two inequalities: \(3a < 650 – 2a\) and \(650 – 2a < 3a + 300\) Inequality 1: \(3a < 650 – 2a\) \(5a < 650\) \(a < 130\) Inequality 2: \(650 – 2a < 3a + 300\) \(5a > 350\) \(a > 70\) So we get that \(70 < a < 130\). Since a is an integer, can we say that a can take all values from 71 to 129? No. What we are forgetting is that b is also an integer. We know that \(b = \frac{(650 – 2a)}{3}\) For which values will be get b as an integer? Note that 650 is not divisible by 3. You need to add 1 to it or subtract 2 out of it to make it divisible by 3. So a should be of the form 3x+1. \(b = \frac{(650 – 2*(3x+1))}{3} = \frac{(648 – 6x)}{3} = 216 – 2x\) Here, for any positive integer x, b will be an integer. From 71 to 129, we have the following numbers which are of the form 3x+1: 73, 76, 79, 82, 85, … 127 This is an Arithmetic Progression. How many terms are there here? Last term = First term + (n – 1)*Common Difference \(127 = 73 + (n – 1)*3\) \(n = 19\) a will take 19 distinct values so the last term i.e. (a+100) will also take 19 distinct values. Method 2: Using Transition PointsNote that \(a < b < a+100\) Since a < b, let’s find the point where a = b, i.e. the transition point \(2a + 3a = 650\) \(a = 130 = b\) But b must be greater than a. If we increase b by 1, we need to decrease a by 3 to keep the average same. But decreasing a by 3 decreases the largest number i.e. a+100 by 3 too; so we need to increase b by another 1. We get \(a = 127\) and \(b = 132\). This give us the numbers as 127, 132, 132, 132, 227. Here the average is 150 Since \(b < a+100\), let’s find the point where \(b = a+100\) \(2a + 3(a+100) = 650\) \(a = 70\), \(b = 170\) But b must be less than a+100. If we decrease b by 1, we need to increase a by 3 to keep the average same. But increasing a by 3 increases the largest number, i.e. a+100 by 3 too, so we need to decrease b by another 1. We get \(a = 73\) and \(b = 168\). This gives us the numbers as 73, 168, 168, 168, 173. Here the average is 150 Values of a will be: 73, 76, 79, ….127 (Difference of 3 to make b an integer) This is an Arithmetic Progression. Last term = First term + (n – 1)*Common difference \(127 = 73 + (n – 1)*3\) \(n = 19\) a will take 19 distinct values so the last term i.e. (a+100) will also take 19 distinct values. Answer (B) This question is discussed HERE.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: How to Go From a 48 to 51 in GMAT Quant [#permalink]
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01 Jun 2016, 03:25
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Today’s post the next part in our “How to Go From 48 to 51 in Quant” series. Again, we will learn a technique that can be employed by the testtaker at an advanced stage of preparation – requiring one to understand the situations in which one can use this simplifying technique. We all love to use the plugin method on GMAT Quant questions. We have an equation given, and if the answer choices are the possible values of x, we just plug in these values to find the one that satisfies the equation. But what if the answer choices are all complicated values? What if it seems that five times the calculation (in the worst case) will be far more time consuming than actually solving the given equation? Then one is torn between using the favorite plugin method and using algebra. Let’s take an example to review the methods we can use to solve the question and learn how to simplify the plugin process by approximating the five available options: Question: If 4x−4=2x+30, which of the following could be a value of x?(A) –35/3 (B) −21/2 (C) −13/3 (D) 11/5 (E) 47/5 Solution: This question is an ideal candidate for the “plugin” method. Here, you have the absolute value equation with the potential values of x given in the answer choices. The problem is that the values of x given are fractional. Of course, if we do plan to solve the equation rather than “plugin”, we can still solve it using our holistic approach rather than pure algebra. Let’s take a look at that now, and later we will discuss the trick to making the answer choices easier for us to plug in. Method 1:\(4x – 4 = 2x + 30\) \(4 * x – 1 = 2 * x + 15\) \(2 * x – 1 = x + 15\) This is how we rephrase the equation in our words: twice the distance of x from 1 should be equal to the distance of x from 15. ——————(15) —————————————————(0)——(1)——————There are two ways to find the value of x: Case 1: x could be between 15 and 1 such that the distance between them is split in the ratio 2:1. or Case 2: x could be to the right of 1 such that the distance between x and 15 is twice the distance between x and 1. Let’s examine both of these cases in further detail: Case 1: The distance from 15 to 1 is of 16 units – this can be split into 3 sections of 16/3 units each. So, the distance of x from 1 should be 16/3, which would make the distance of x from 15 two times 16/3, i.e. 32/3. So, x should be at a point 16/3 away from 1 toward the left. \(x = 1 – \frac{16}{3} = \frac{13}{3}\) This is one of our answer choices and, hence, the correct answer. Normally, we would just move on to the next question at this point, but had we not found 13/3 in the answer options, we would have moved on to Case 2: Case 2: The distance between 15 and 1 is 16 units. x should be an additional 16 units to the right of 1, so the distance between x and 1 is 16 and the distance between x and 15 is two times 16, i.e. 32. This means that x should be 16 units to the right of 1, i.e. x = 17. If you would not have found 13/3 in the answer choices, then you would have found 17. Now let’s move on to see how we can make the plugin method work for us in this case by examining each answer choice we are given: Method 2:\(4x – 4 = 2x + 30\) \(2 * x – 1 = x + 15\) (A) 35/3 It is difficult to solve for \(x = \frac{35}{3}\) to see if both sides match. Instead, let’s solve for the closest integer, 12. \(2 * 12 – 1 = 12 + 15\) On the lefthand side, you will get 26, but on the righthand side, you will get 3. These values are far away from each other, so x cannot be 35/3. As the value of x approaches the point where the equation holds – i.e. where the two sides are equal to each other – the gap between the value of the two sides keeps reducing. With such a huge gap between the value of the two sides in this case, it is unlikely that a small adjustment of 35/3 from 12 will bring the two sides to be equal. (B) 21/2 For this answer choice, let’s solve for the nearest integer, x = 10. \(2 * 10 – 1 = 10 + 15\) On the lefthand side, you will get 22; on the righthand side, you will get 5. Once again, these values are far away from each other and, hence, x will not be 21/2. (C) 13/3 For this answer choice, let’s solve for x = 4. \(2 * 4 1 = 4 + 15\) On the lefthand side, you will get 10; on the righthand side, you will get 11. Here, there is a possibility that x can equal 13/3, as the two sides are so close to one another – plug in the actual value of 13/3 and you will see that the lefthand side of the equation does, in fact, equal the righthand side. Therefore, C is the correct answer. This question is discussed HERE. Basically, we approximated the answer choices we were given and shortlisted the one that gave us very close values. We checked for that and found that it is the answer. We can also solve this question using pure algebra (taking positive and negative signs of absolute values) but in my opinion, the holistic Method 1 is almost always better than that. Out of the two methods discussed above, you can pick the one you like better, but note that Method 2 does have limited applications – only if you are given the actual values of x, can you use it. Method 1 is far more generic for absolute value questions.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: How to Go From a 48 to 51 in GMAT Quant [#permalink]
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Both a testtaker at the 48 level and one at the 51 level in the GMAT Quant section, are conceptually strong – given an unlimited time frame, both will be able to solve most GMAT questions correctly. The difference lies in the two things a testtaker at the 51 level does skillfully: 1. Uses holistic, bigpicture methods to solve Quant questions. 2. Handles questions he or she finds difficult in a timely manner.We have been discussing holistic methods on this blog for a long time now and will continue discussing them. (Before you continue reading, be sure to check out parts I, II, III, IV, V and VI of this series.) Today we will focus on “handling the hard questions in a timely manner.” Note that we do not say “solving the hard questions in a timely manner.” Occasionally, one might be required to make a quick call and choose to guess and move on – but again, that is not the focus of this post. We are actually going to talk about the “lightbulb” moment that helps us save on time. There are many such moments for the 51 level testtaker – in fact, the 51 scorers often have time left over after attempting all these questions. Test takers at the 48 level will also eventually reach the same conclusions but might need much more time. That will put pressure on them the next time they look at the ticking clock, and once their cool is lost, “silly errors” will start creeping in. So it isn’t about just that one question – one can end up botching many other questions too. There are many steps that can be easily avoided by a lightbulb moment early on. This is especially true for Data Sufficiency questions. Let’s take an official example: Question: Pam owns an inventory of unopened packages of corn and rice, which she has purchased for $17 and $13 per package, respectively. How many packages of corn does she have ?Statement 1: She has $282 worth of packages. Statement 2: She has twice as many packages of corn as of rice. Solution:A high scorer will easily recognize that this question is based on the concept of “integral solutions to an equation in two variables.” Since, in such real world examples, x and y cannot be negative or fractional, these equations usually have a finite number of solutions. After we find one solution, we will quickly know how many solutions the equation has, but getting the first set of values that satisfy the equation requires a little bit of brute force. The good thing here is that this is a Data Sufficiency question – you don’t need to find the actual solution. The only thing we need is to establish that there is a single solution only. (Obviously, there has to be a solution since Pam does own $282 worth of packages.) So, the testtaker will start working on finding the first solution. We are told: Price of a packet of corn = $17 Price of a packet of rice = $13 Say Pam has “x” packets of corn and “y” packets of rice. Statement 1: She has $282 worth of packagesUsing Statement 1, we know that 17x + 13y = 282. We are looking for the integer values of x and y. If x = 0, y will be 21.something (not an integer) If x = 1, y = 20.something If x = 2, y = 19.something If x = 3, y = 17.something This is where the 51 level scorer stops because they never lose sight of the big picture. The “lightbulb” switches on, and now he or she knows that there will be only one set of values that can satisfy this equation. Why? Because y will be less than 17 in the first set of values that satisfies this equation. So if we want to get the next set that satisfies, we will need to subtract y by 17 (and add 13 to x), which will make y negative. So in any case, there will be a unique solution to this equation. We don’t actually need to find the solution and hence, nothing will be gained by continuing these calculations. Statement 1 is sufficient. Statement 2: She has twice as many packages of corn as of rice.Statement 2 gives us no information on the total number of packages or the total amount spent. Hence, we cannot find the total number of packages of corn using this information alone. Therefore, our answer is A. This question is discussed HERE. I hope you see how you can be alert to what you want to handle these Quant questions in a timely manner.
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New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: How to Go From a 48 to 51 in GMAT Quant [#permalink]
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10 Jun 2017, 04:10
Hi Bunuel, Thanks for this! Very very helpful. In the 7th part  could you please explain this  'This is where the 51 level scorer stops because they never lose sight of the big picture. The “lightbulb” switches on, and now he or she knows that there will be only one set of values that can satisfy this equation. Why? Because y will be less than 17 in the first set of values that satisfies this equation. So if we want to get the next set that satisfies, we will need to subtract y by 17 (and add 13 to x), which will make y negative.' ?? Additionally, could you please help me with my approach? Whether I could use this across questions? Statement 1: 17x + 13y = 282 > Here, without actually calculating the values for x and y, we can establish that this equation must have only one solution since: Statement 2 provides us with x=2y => x2y = 0 [ Without using this statement, we know that this statement must be true ] and since, (a/b) =/= (c/d) => eq 1 will only have one such solution. And so, the answer is A. Please let me know ! Bunuel wrote: Both a testtaker at the 48 level and one at the 51 level in the GMAT Quant section, are conceptually strong – given an unlimited time frame, both will be able to solve most GMAT questions correctly. The difference lies in the two things a testtaker at the 51 level does skillfully: 1. Uses holistic, bigpicture methods to solve Quant questions. 2. Handles questions he or she finds difficult in a timely manner.We have been discussing holistic methods on this blog for a long time now and will continue discussing them. (Before you continue reading, be sure to check out parts I, II, III, IV, V and VI of this series.) Today we will focus on “handling the hard questions in a timely manner.” Note that we do not say “solving the hard questions in a timely manner.” Occasionally, one might be required to make a quick call and choose to guess and move on – but again, that is not the focus of this post. We are actually going to talk about the “lightbulb” moment that helps us save on time. There are many such moments for the 51 level testtaker – in fact, the 51 scorers often have time left over after attempting all these questions. Test takers at the 48 level will also eventually reach the same conclusions but might need much more time. That will put pressure on them the next time they look at the ticking clock, and once their cool is lost, “silly errors” will start creeping in. So it isn’t about just that one question – one can end up botching many other questions too. There are many steps that can be easily avoided by a lightbulb moment early on. This is especially true for Data Sufficiency questions. Let’s take an official example: Question: Pam owns an inventory of unopened packages of corn and rice, which she has purchased for $17 and $13 per package, respectively. How many packages of corn does she have ?Statement 1: She has $282 worth of packages. Statement 2: She has twice as many packages of corn as of rice. Solution:A high scorer will easily recognize that this question is based on the concept of “integral solutions to an equation in two variables.” Since, in such real world examples, x and y cannot be negative or fractional, these equations usually have a finite number of solutions. After we find one solution, we will quickly know how many solutions the equation has, but getting the first set of values that satisfy the equation requires a little bit of brute force. The good thing here is that this is a Data Sufficiency question – you don’t need to find the actual solution. The only thing we need is to establish that there is a single solution only. (Obviously, there has to be a solution since Pam does own $282 worth of packages.) So, the testtaker will start working on finding the first solution. We are told: Price of a packet of corn = $17 Price of a packet of rice = $13 Say Pam has “x” packets of corn and “y” packets of rice. Statement 1: She has $282 worth of packagesUsing Statement 1, we know that 17x + 13y = 282. We are looking for the integer values of x and y. If x = 0, y will be 21.something (not an integer) If x = 1, y = 20.something If x = 2, y = 19.something If x = 3, y = 17.something This is where the 51 level scorer stops because they never lose sight of the big picture. The “lightbulb” switches on, and now he or she knows that there will be only one set of values that can satisfy this equation. Why? Because y will be less than 17 in the first set of values that satisfies this equation. So if we want to get the next set that satisfies, we will need to subtract y by 17 (and add 13 to x), which will make y negative. So in any case, there will be a unique solution to this equation. We don’t actually need to find the solution and hence, nothing will be gained by continuing these calculations. Statement 1 is sufficient. Statement 2: She has twice as many packages of corn as of rice.Statement 2 gives us no information on the total number of packages or the total amount spent. Hence, we cannot find the total number of packages of corn using this information alone. Therefore, our answer is A. This question is discussed HERE. I hope you see how you can be alert to what you want to handle these Quant questions in a timely manner.



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Re: How to Go From a 48 to 51 in GMAT Quant [#permalink]
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10 Aug 2017, 19:43
PMZ21 wrote: [ Without using this statement, we know that this statement must be true ] and since, (a/b) =/= (c/d) => eq 1 will only have one such solution.
What does this mean? I don't follow.




Re: How to Go From a 48 to 51 in GMAT Quant
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