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# Veritas Prep Blog

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Veritas Prep Representative
Joined: 21 Jan 2010
Posts: 99
Own Kudos [?]: 655 [2]
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Veritas Prep Representative
Joined: 21 Jan 2010
Posts: 99
Own Kudos [?]: 655 [0]
Given Kudos: 2
Veritas Prep Representative
Joined: 21 Jan 2010
Posts: 99
Own Kudos [?]: 655 [0]
Given Kudos: 2
Veritas Prep Representative
Joined: 21 Jan 2010
Posts: 99
Own Kudos [?]: 655 [0]
Given Kudos: 2
This Blog post was imported into the forum automatically. We hope you found it helpful. Please use the Kudos button if you did, or please PM/DM me if you found it disruptive and I will take care of it. -BB
Veritas Prep Representative
Joined: 21 Jan 2010
Posts: 99
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GMAT Tip of the Week: Today's Date in Geometry History [#permalink]
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This Blog post was imported into the forum automatically. We hope you found it helpful. Please use the Kudos button if you did, or please PM/DM me if you found it disruptive and I will take care of it. -BB
Veritas Prep Representative
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Figuring Out the Topic of Discussion on the GMAT [#permalink]
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 FROM Veritas Prep Blog: Figuring Out the Topic of Discussion on the GMAT You must have come across questions which you thought tested one concept but later found out could be easily dealt with using another concept.  Often, crafty little mixture problems belong to this category. For example:Mark is playing poker at a casino. Mark starts playing with 140 chips, 20% of which are \$100 chips and 80% of which are \$20 chips. For his first bet, Mark places chips, 10% of which are \$100 chips, in the center of the table. If 70% of Mark’s remaining chips are \$20 chips, how much money did Mark bet?You can view this as a word problem where you assume the number of chips and then go splitting them up or you can view this as a mixtures problem even though it doesn’t use words such as ‘mixture’, ‘solution’, ‘combined’ etc. As we have seen enough number of times, our mixture problems are solved in seconds using the weighted average concept.The question discussed here also belongs to the same category – looks super tricky but can be easily solved with weighted averages formula. But we have seen plenty and more of such questions in our blog posts. Today we will take a look at a different type of sinister question and I suggest you to think about the concept being tested in that before trying to solve it.Question: Mark owns four low quality watches. Watch1 loses 15 minutes every hour. Watch2 gains 15 minutes every hour relative to watch1 (that is, as watch1 moves from 12:00 to 1:00, watch2 moves from 12:00 to 1:15). Watch3 loses 20 minutes every hour relative to watch2. Finally, watch4 gains 20 minutes every hour relative to watch3. If Mark resets all four watches to the correct time at 12 noon, what time will watch4 show at 12 midnight that day?(A)10:00(B)10:34(C)11:02(D)11:48(E)12:20Before we look at the solution, think about the concept being tested here – clocks? Circular motion?Neither!Solution: Note that when giving data about watch1, you are told how it varies with the actual time. Data about all other watches tells us about the time they show relative to the incorrect watches. The concept being tested here is Relative Speed!What do we mean by “gains 15 mins” or “loses 20 mins” etc? When a watch gains 15 mins every hour, it means that even though it should show that one hour has passed, it shows that 1 hr 15 mins have passed. So the watch runs faster than it should. Hence the speed of the watch is more than the speed of a correct watch. Now the question is how much more? The minute hand of the correct watch travels one full circle in one hour. The minute hand of the incorrect watch travels one full circle and then a quarter circle in one hour (to show that 1 hour 15 mins have passed even when only an hour has passed). So it is 5/4 times the speed of a correct watch. On the same lines, let’s analyze each watch.Say the speed of a correct watch is s.- “Watch1 loses 15 minutes every hour. “Watch1 covers only three quarters of the circle in an hour.Speed of watch1 = (3/4)*s- “Watch2 gains 15 minutes every hour relative to watch1 (that is, as watch1 moves from 12:00 to 1:00, watch2 moves from 12:00 to 1:15).”Now we have the speed of watch2 relative to speed of watch1. Speed of watch2 is (5/4) times the speed of watch1.Speed of watch2 = (5/4)*(3/4)s = (15/16)*s- “Watch3 loses 20 minutes every hour relative to watch2.”Watch3 loses 20 mins every hour means its speed is (2/3)rd the speed of watch2Speed of watch3 = (2/3)*(15/16)*s = (5/8)*s- “Finally, watch4 gains 20 minutes every hour relative to watch3.”Speed of watch4 = (4/3)*Speed of watch3 = (4/3)*(5/8)*s = (5/6)*sSo the speed of watch4 is (5/6)th the speed of a correct watch. So if a correct watch shows that 6 hours have passed, watch4 will show that 5 hours have passed. If a correct watch shows that 12 hours have passed, watch4 will show that 10 hours have passed. From 12 noon to 12 midnight, a correct watch would have covered 12 hours. Watch4 will cover 10 hours and will show the time as 10:00.Answer (A)Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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Veritas Prep Representative
Joined: 21 Jan 2010
Posts: 99
Own Kudos [?]: 655 [1]
Given Kudos: 2
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This Blog post was imported into the forum automatically. We hope you found it helpful. Please use the Kudos button if you did, or please PM/DM me if you found it disruptive and I will take care of it. -BB
Veritas Prep Representative
Joined: 21 Jan 2010
Posts: 99
Own Kudos [?]: 655 [1]
Given Kudos: 2
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This Blog post was imported into the forum automatically. We hope you found it helpful. Please use the Kudos button if you did, or please PM/DM me if you found it disruptive and I will take care of it. -BB
Veritas Prep Representative
Joined: 21 Jan 2010
Posts: 99
Own Kudos [?]: 655 [1]
Given Kudos: 2
Making the Best of Application Season: A Guide for High School Juniors [#permalink]
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This Blog post was imported into the forum automatically. We hope you found it helpful. Please use the Kudos button if you did, or please PM/DM me if you found it disruptive and I will take care of it. -BB
Veritas Prep Representative
Joined: 21 Jan 2010
Posts: 99
Own Kudos [?]: 655 [0]
Given Kudos: 2
SAT Tip of the Week: 6 Ways to Master Academic Tone on the Essay [#permalink]
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Why Does the GMAT Test Geometry? [#permalink]
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This Blog post was imported into the forum automatically. We hope you found it helpful. Please use the Kudos button if you did, or please PM/DM me if you found it disruptive and I will take care of it. -BB
Veritas Prep Representative
Joined: 21 Jan 2010
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This Blog post was imported into the forum automatically. We hope you found it helpful. Please use the Kudos button if you did, or please PM/DM me if you found it disruptive and I will take care of it. -BB
Veritas Prep Representative
Joined: 21 Jan 2010
Posts: 99
Own Kudos [?]: 655 [2]
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This Blog post was imported into the forum automatically. We hope you found it helpful. Please use the Kudos button if you did, or please PM/DM me if you found it disruptive and I will take care of it. -BB
Veritas Prep Representative
Joined: 21 Jan 2010
Posts: 99
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Finding the Last Two Digits on GMAT Quant Questions - Part I [#permalink]
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 FROM Veritas Prep Blog: Finding the Last Two Digits on GMAT Quant Questions - Part I We all know how to find the last digit using cyclicity when we are given a number raised to a power. Last digit of a number depends only on the last digit of the base.  You must be quite familiar with something like this -Last Digit of Base:0 – Last digit of expression with any power will be 0.1 – Last digit of expression with any power will be 1.2 – 2, 4, 8, 6, 2, 4, 8, 6… Cyclicity is 4.3 – 3, 9, 7, 1, 3, 9, 7, 1… Cyclicity is 4.4 – 4, 6, 4, 6, 4, 6, 4, 6… Cyclicity is 2.5 -  Last digit of expression with any power will be 5.6 – Last digit of expression with any power will be 6.7 – 7, 9, 3, 1, 7, 9, 3, 1… Cyclicity is 4.8 – 8, 4, 2, 6, 8, 4, 2, 6… Cyclicity is 4.9 – 9, 1, 9, 1, 9, 1, 9, 1… Cyclicity is 2.Cyclicity is nothing but pattern recognition. You see that when you multiply 2 by  itself, there is a pattern of last digit which goes 2, 4, 8, 6, 2, 4, 8, 6 and so on. We can use the same principle for when a question asks us for the last two digits of the expression. Let me remind you first that here at QWQW, we sometimes flirt with the lines that define GMAT scope. Obviously, we do point out whenever we are indulging and that’s exactly what we are going to do in this post. We are carrying on for the love of Math and the Q51 score.The last two digits of the base decide the last two digits of the expression. For example,Example 1: Let’s look at powers of 11.11^1 = 1111^2 = 12111^3 = 133111^4 = …41 (we should just multiply the last two digits together and ignore the rest)11^5 = …5111^6 = …6111^7 = …71Note that the last two digits are displaying a pattern depending on the power. So we expect the cyclicity here to be 10.11^8 = …8111^9 = …9111^(10) = …0111^(11) = …1111^(12) = …21and so on. So the last two digits should go from 11, 21 to 91, 01 and then go to 11 again. The cycle of 10 starts from power of 1, 11, 21 etc. This means that 11^(46) should have last two digits as 61, 11^(92) should have last two digits as 21 and 11^(168) should have last two digits as 81.Let’s look at some other numbers now:Example 2: Say, we need the last two digits of 6^{58}6^1 = 6 (No second last digit)6^2 = 366^3 = 2166^4 = …96 (Just multiply the last two digits)6^5 = …766^6 = …566^7 = …36and hence starts the cycle again:3, 1, 9, 7, 5, 3, 1, 9, 7, 5 and so on.The new cycle with tens digit of 3 begins at the powers of 2, 7, 12, 17, 22, 27 etc. So the new cycle will also begin at power of 57 and 6^58 will have 1 as the tens digit.Example 3: How about the last two digits of 7^102?7^1 = 7 (No second last digit)7^2 = 497^3 = 3437^4 = …017^5 = …077^6 = …497^7 = …43We see a cyclicity of 4 here: 49, 43, 01, 07, 49, 43, 01, 07 … and so on. The new cycle begins at 2, 6, 10, 14 i.e. even powers which are not multiples of 4. So a new cycle will begin at 102 too. So the last two digits of 7^(102) will be 49.Now there can be many variations in the questions asking us to find the last two digits. We will use different concepts for different question types. Today we saw how to use pattern recognition. We will look at some other methods next week.Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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Veritas Prep Representative
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Posts: 99
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How to Write Breakthrough Application Essays for Kellogg [#permalink]
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Veritas Prep Representative
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Posts: 99
Own Kudos [?]: 655 [0]
Given Kudos: 2
SAT Tip of the Week: 5 Sources to Prepare Essay Examples [#permalink]
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Veritas Prep Representative
Joined: 21 Jan 2010
Posts: 99
Own Kudos [?]: 655 [0]
Given Kudos: 2
This Blog post was imported into the forum automatically. We hope you found it helpful. Please use the Kudos button if you did, or please PM/DM me if you found it disruptive and I will take care of it. -BB
Veritas Prep Representative
Joined: 21 Jan 2010
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Mind The Gap Year: 6 Considerations for a High School Senior [#permalink]
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Veritas Prep Representative
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Posts: 99
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Given Kudos: 2
GMAT Tip of the Week: Serial and Sufficiency [#permalink]
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