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Submit Your Own GMAT Math Questions  Get GMAT Club Tests
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27 Mar 2013, 05:59
Submit Your Own GMAT Math Questions  Get GMAT Club TestsGet quick access to the GMAT Club tests!Have you done thousands of questions and dream about them at night? Take a shot of creating your own!We are going to release a new version/next gen of the tests this summer and would like to get some questions into the pool. The opportunity: Post your DS or PS questions in this thread and we will give you 5 kudos. Once you get 25 kudos (5 questions or 4 questions and profile update) are enough to get access to GMAT Club tests for 3 months! Terms & Conditions:  Your question has to be unique and not copyrighted by anyone else. If you choose to use a text or passage of text you must paraphrase it or modify it to the point that we will not be able to trace it or otherwise your question will be disqualified.
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01 Apr 2013, 09:22
Aww..... Guess I should have waited for this thread to arrive...
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01 Apr 2013, 20:50
Great idea. This is probably a great way to ingrain things for question writers. My Quant strength is exponents, so I'll get things started with one of those:
If \(\frac{4^{3x}}{16^{3x/2}} = (2)(2^{3x})\), what is x?
A) x = 1/3 B) x = 2/3 C) x = 1 D) x = 3/2 E) x = 3



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01 Apr 2013, 21:07
vandygrad11 wrote: Great idea. This is probably a great way to ingrain things for question writers. My Quant strength is exponents, so I'll get things started with one of those:
If \(\frac{4^{3x}}{16^{3x/2}} = (2)(2^{3x})\), what is x?
A) x = 1/3 B) x = 2/3 C) x = 1 D) x = 3/2 E) x = 3 Here's the explanation to this question: You should quickly realize that the bases can and should all be 2's. So, this is one methodology to solve:
\(\frac{4^{3x}}{16^{3x/2}} = (2)(2^{3x})\)
\(\frac{(2^2)^{3x}}{(2^4)^{3x/2}} = \frac{2^1}{2^{3x}}\)
\(\frac{2^{6x}}{2^{6x}} = \frac{2^1}{2^{3x}}\)
\(2^1=2^{3x}\)
\(1=3x\)
\(\frac{1}{3}=x\)



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02 Apr 2013, 03:02
For the 2008 season, a certain sports league had eight teams. For the 2010 season, the number of teams was increased to ten. In a season, each team plays exactly two matches with each of the other teams. If the revenue per match for the 2008 season was $1 million and the revenue per match for the 2010 season was $1.25 million, what was the percentage increase in total revenues from all matches from the 2008 season to the 2010 season? A)25% B)50% C)75% D)100% E)125% No. of teams in 2008 = 8 No. of matches in 2008 = \(8P_2\) = 56
Total revenue in 2008 = 56*1 million = $56 million
No. of teams in 2010 = 10 No. of matches in 2010 = \(10P_2\) = 90
Total revenue in 2008 = 90*1.25 million = $112.5 million
Percentage increase = \(\frac{112.556}{56}*100\) \(\approx\) 100%
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Updated on: 02 Apr 2013, 15:13
There is a crack at the bottom of a tank. Before the crack appeared, pipe A could fill the tank in 2 hours. Now it takes 30 min longer. How long will the crack take to empty a full tank when pipe A is closed ? A)15 hours B)8 hours C)10 hours D)12 hours E)15 hours Shortcut : If the pipe can fill a tank in 'a' hours but takes 'x' hours longer due to a leak, then the time taken by the leak to empty the tank fully is \(a(1+a/x)\)
Or Before the crack appeared , Pipe A could fill \(1/2\) the tank in an hour. Now it takes 2 hours 30 mins to fill the tank. That is \(5/2\) hours. Hence, it fills \(2/5 th\) of the tank every hour. The crack at the bottom accounts for this reduction in the amount . The crack therefore drains \(1/22/5 =54/10 =1/10 th\)of the tank every hour Hence , total time taken to drain the tank is C)10 hours
Originally posted by thinktank on 02 Apr 2013, 13:28.
Last edited by thinktank on 02 Apr 2013, 15:13, edited 1 time in total.



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02 Apr 2013, 14:56
thinktank wrote: There is a crack at the bottom of a tank. Before the crack appeared, pipe A could fill the tank in 2 hours. Now it takes 30 min longer. How long will the crack take to empty a full tank when pipe A is closed ? A)15 hours B)8 hours C)10 hours D)12 hours E)15 hours I'm not aware as to how to post the Answer under REVEAL here.. can someone edit this post plz ? Solution : Shortcut : If the pipe can fill a tank in 'a' hours but takes 'x' hours longer due to a leak, then the time taken by the leak to empty the tank fully is \(a(1+a/x)\)
Or Before the crack appeared , Pipe A could fill \(1/2\) the tank in an hour. Now it takes 2 hours 30 mins to fill the tank. That is \(5/2\) hours. Hence, it fills \(2/5 th\) of the tank every hour. The crack at the bottom accounts for this reduction in the amount . The crack therefore drains \(1/22/5 =54/10 =1/10 th\)of the tank every hour Hence , total time taken to drain the tank is C)10 hours Done. Mark the text you want to hide and press "Spoiler" button.
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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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02 Apr 2013, 15:13
Here is the one from my side: DS If x and y are positive, is \(2x^2/17+y^2/4 < 1\)? (1) \(2y>x^2\) (2) \(x^2 < 9y^2\) 1.Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. 2.Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. 3.BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. 4.EACH statement ALONE is sufficient. 5.Statements (1) and (2) TOGETHER are NOT sufficient. Ans. 2 This can be solved graphically. The equation in the question stem is the equation of an ellipse with xintercept ~ 3 and yintercept 2 and represents the area inside the ellipse. The statement 1 is the equation of the area enclosed inside a parabola and only some of the enclosed area will intersect with the enclosed area of ellipse. NOT SUFFICIENT. The statement 2 is the equation of the enclosed area of a circle with radius 3 and will therefore completely enclose ellipse as xintercept of ellipse < 3 and yintercept of ellipse is also less than 3. SUFFICIENT.



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03 Apr 2013, 21:07
vandygrad11 wrote: vandygrad11 wrote: Great idea. This is probably a great way to ingrain things for question writers. My Quant strength is exponents, so I'll get things started with one of those:
If \(\frac{4^{3x}}{16^{3x/2}} = (2)(2^{3x})\), what is x?
A) x = 1/3 B) x = 2/3 C) x = 1 D) x = 3/2 E) x = 3 Here's the explanation to this question: You should quickly realize that the bases can and should all be 2's. So, this is one methodology to solve:
\(\frac{4^{3x}}{16^{3x/2}} = (2)(2^{3x})\)
\(\frac{(2^2)^{3x}}{(2^4)^{3x/2}} = \frac{2^1}{2^{3x}}\)
\(\frac{2^{6x}}{2^{6x}} = \frac{2^1}{2^{3x}}\)
\(2^1=2^{3x}\)
\(1=3x\)
\(\frac{1}{3}=x\) Kudos for the question but just looking at it first time I thought of a easier way. Correct me if i am wrong \(\frac{4^{3x}}{16^{3x/2}} = (2)(2^{3x})\) \(\frac{(4)^{3x}}{(4)^{2*3x/2}} = 2^{13x}\) Numerator and denominator are equal after cancelling out the 2 \(1 = 2^{13x}\) \(2^0 = 2^{13x}\) \(13x = 0\) \(\frac{1}{3}=x\)



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Updated on: 04 Apr 2013, 05:56
Highest prime factor of (\(97^3\) \(57^3\)  \(40^3\)) is 1) 97 2) 19 3) 17 4) 5 5) 3 Answer : Option 1.
Explanation : We know, \(a^3\) +\(b^3\) + \(c^3\) = 3abc if a+b+c = 0. Here a+b+c = 97  57  40 = 0. So, The given expression can be simplified to = 3 * 97 * 57 * 40. So,the highest prime factor would be 97.
Originally posted by subhendu009 on 04 Apr 2013, 05:51.
Last edited by subhendu009 on 04 Apr 2013, 05:56, edited 1 time in total.



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04 Apr 2013, 06:45
Here is a question from my side. A Train left Mumbai for Pune at noon sharp. Two hours later, another train started from Mumbai in the same direction. The second train passed the first one at 8 P.M. Find the average speed of the two trains over the journey if the sum of their average speeds is 70 kmph a) 28.50 kmph b) 30 kmph c) 34.26 kmph d) 35 kmph e) 55 kmph Regards, Naren. Answer : C
Solution : Let the first train be T1 and second train be T2
Speed T1 = A Time T1 = 8 Speed T2 = B Time T2 = 6
Since Distance is constant and time is in ratio 4/3, their speed must be in ratio 3/4 So Speed T1 = 3x and Speed T2 = 4x
We know sum of their speeds is 70 > A + B = 70 > 3x + 4x = 70 > 7x = 70 > x = 10 Speed T1 = 30 and Speed T2 = 40
Average speed = (2 X speed1 X speed2)/(Speed1 + Speed2) > (2 X 30 X 40) / (30 + 40) > 2400 / 70 > 34.26
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Updated on: 07 Apr 2013, 09:53
A and B together can do a work in 20 days. B and C together can finish it in 25 days. If A does double the work in a day than C, then the number of days in which C alone can finish it is? A) 70 days B) 80 days C) 85 days D) 100 days E) 120 days Regards, Naren Answer = D
Solution : Let the work be 100% A and B together can do a work in 20 days. that mean in a day they are doing 100/20 = 5 % of work B and C together can finish it in 25 days. that means in a day they are doing 100/25 = 4% of work B + A = 5 B + C = 4 A does double the work in a day than C. So A = 2C B + 2C = 5 B + C = 4 Upon Solving we get C = 1 that mean C does 1% of the work in a day, Hence to complete the whole work it would take him 100 days.
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Originally posted by Narenn on 04 Apr 2013, 07:05.
Last edited by Narenn on 07 Apr 2013, 09:53, edited 1 time in total.



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05 Apr 2013, 13:50
By what percentage must a trader mark his goods up if he has to make a profit of \(20%\) after providing a discount of \(25%\) on his marked price ? A)25% B)30% C)50% D)60% E)70% D A handy shortcut : When ever a trader marks his goods up by \(x%\) and offers a discount of \(y%\) then the profit/loss percentage he makes is given by \(xy(xy/100)\) +1 if u think this approach would help you save precious time



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06 Apr 2013, 06:48
Here is one more : this time on Permutations and Combinations Five persons entered the lift cabin on the ground floor of an 8 floor house. Suppose each of them can leave the cabin independently at any floor beginning with the first. Find the total number of ways in which each of the five persons can leave the cabin i) At any one of the 7 floors ii) At different floors. A) i) 7^5 ii) 7! B) i) 35 ii) 7! C) i) 5^7 ii) 7! D) i) 5^7 ii) 2520 E) i) 7^5 ii) 2520 Correct Choice is E Let the five persons be b,c,d,e,f I) b can leave the cabin at any one of the seven floors. So he has 7 options Similarly each of c,d,e,f also has 7 options. Thus the total number of ways in which each of the five persons can leave the cabin at any of the seven floors is 7 X 7 X 7 X 7 X7 = 7^5 II) b can leave the cabin in 7 ways. c can leave the cabin in 6 ways, since he can not leave at where b left. In the same way d has 5, e has 4, and f has 3 way. Hence total number of ways = 7 X 6 X 5 X 4 X 3 = 2520
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Updated on: 13 Apr 2013, 04:08
(1/2)^3 * 64^(1/2) a) (1/64) b) 1 c) (1/64) d) (1) e) (64) = (2)^3 * (1/64)^1/2
= 8 * (1/8) = 1
Originally posted by shivanigs on 13 Apr 2013, 03:31.
Last edited by shivanigs on 13 Apr 2013, 04:08, edited 3 times in total.



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13 Apr 2013, 03:46
/x3/ + /x2/ = 0. What is the value of x? Modulus hence 2 scenarios,positive and negative:
1. x  3 + x  2 = 0 2. x + 3  x +2 = 0
2x  5 = 0 2x +5 = 0
2x = 5 2x = 5
x = 5/2 x = 5/2



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13 Apr 2013, 04:34
A and B can complete building a house together in 30 days. If A can build the house alone in 90 days, how long will it take for B to build a house by himself? a) 45 days b) 60 days c) 15 days d) 30 days e) 60 days Let x be the number of days it takes B to build a house.
So,(1/90) + (1/x) = (1/30)
(1/x) = (1/30)  (1/90)
(1/x) = (3  1)/90
(1/x) = 2/90
(1/x) = (1/45)  B does (1/45) th of the work in 1 hour.
So amount of time of time it takes him to build 1 whole house,
Rate * Time = Work
1/45 * Time = 1
Time = 45 days



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13 Apr 2013, 05:20
Lucky owned a car worth $40,000.Happy bought the car from Lucky for $ 55,000.Happy then sold it to Smiley for $50,000. Lucky started to miss his car and bought it back from Smiley for $45,000. What is Lucky's net profit/loss %? Note : The answer choices are in the mixed fraction format a) 18(2/11)% profit b) 37(1/2)% profit c) 12(2/4)% loss d) 9(1/11)% loss e) 25% profit Lucky's original cost for the car = $ 40,000
Happy paid him $ 55,000,so Lucky's profit amount = $ 15,000
Lucky then paid Smiley $ 45,000, hence Lucky's net profit now = $ 10,000
Therefore Lucky's profit % = (10,000/40,000) * 100 = 25%



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03 May 2013, 12:26
Here is my question about Patterns/Series Given \(a_1=81\) and \(a_n=a_{n1}+3\) for \(n>1\), what is the value of the sum of the first 54 elements: \(a_1+a_2+...+a_{53}+a_{54}\) ? A)0 B)78 C)81 D)81 E)78 OA: C)81
the sequence is as follow: \(a_1=81\) \(a_2=78\) ... and this will reach 0 is 27 passages \(a_{28}=0\) and then will become positive and every term will balance its negative correspondent.
\(a_{27}=3\) will be balanced by \(a_{29}=3\) => sum=0 and so on... This process continues for all terms, if the question were "What is the sum of the first 55 terms the balance will be perfect and the sum would be 0.
But here we are asked the sum of the 54 terms: the very first term will not be balanced! \(a_{1}=81\), \(a_{2}=78\), ..., \(a_{54}=78\)
The sum is 81.
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