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Re: Submit Your Own GMAT Math Questions  Get GMAT Club Tests [#permalink]
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03 May 2013, 17:33
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Zarrolou wrote: 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. Given \(a_1=81\) , and \(a_n=a_{n1}+3\) for \(n>1\), Therefore, \(a_2=a_{1}+3\) & \(a_3=a_{2}+3\) This means the Common increment of 3, & as we know that there are 54 terms in total but if we remove \(a_1\) , then we will have 53 terms each increasing with the common increment of 3. Therefore we have \(a_54\) as 3*5381. \(a_54\) =159  81 \(\Rightarrow\) \(a_54\) =78. Now, we have first term & the last term & the common difference. so as per the properties..... > \(\frac{Last Term + First Term}{2} * Total # of Terms\) > \(\frac{7881}{2}*54\) > 27*3 \(\Rightarrow\) Hence, 81.
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04 May 2013, 22:49
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Question :Two cars move along a circular track 1.2 miles long at constant speeds. When they move in opposite directions, they meet every 15 seconds. However, when they move in same direction, once car overtakes the other car every 60 seconds. What is the speed of the faster car ? Options :A) 0.02 miles/s B) 0.03 miles/s C) 0.05 miles/s D) 0.08 miles/s E) 0.1 miles/s Answer :Explanation : Let the speed of the 2 dots be "a" (faster dot) and "b" (slower dot) miles/s respectively.  When they move in opposite directions, \(\frac{1.2}{a+b}=15\)  When they move in same direction, \(\frac{1.2}{ab}=60\)  Simplifying 2 equations, we get 15a + 15b = 1.2 and 60a  60b = 1.2  Solving 2 equations, 120a = 6 or a = 0.05 miles/s
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Last edited by dipen01 on 04 May 2013, 23:03, edited 1 time in total.



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Re: Submit Your Own GMAT Math Questions  Get GMAT Club Tests [#permalink]
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04 May 2013, 23:02
Question :How many chocolates did the two girls buy, if the sum of the cubes of the number of chocolates bought by them adds up to 189 and the result of subtracting the product of chocolates bought by them from the sum of the squares of the chocolates bought by them is 21 ? Options :A) 21 B) 5 C) 20 D) 16 E) 9 Answer :Explanation : Let the number of chocolates bought by one of the toddlers be "x" and the number of chocolates bought by second toddler be "y"  Then, \(x^3 + y^3 = 189\)......... (I)  and \(x^2 + y^2  xy = 21\)..........(II)  \(x^3 + y^3 = (x+y)(x^2  xy + y^2)\)  Thus by dividing equation (I) by equation (II) we will get the value of x + y = 9
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16 May 2013, 12:26
dipen01 wrote: Question :How many chocolates did the two girls buy, if the sum of the cubes of the number of chocolates bought by them adds up to 189 and the result of subtracting the product of chocolates bought by them from the sum of the squares of the chocolates bought by them is 21 ? Options :A) 21 B) 5 C) 20 D) 16 E) 9 Answer :Explanation : Let the number of chocolates bought by one of the toddlers be "x" and the number of chocolates bought by second toddler be "y"  Then, \(x^3 + y^3 = 189\)......... (I)  and \(x^2 + y^2  xy = 21\)..........(II)  \(x^3 + y^3 = (x+y)(x^2  xy + y^2)\)  Thus by dividing equation (I) by equation (II) we will get the value of x + y = 9 Another way to do this that involves only the first equation is to realize that x and y must be integers and that there is a max on x and y. Since \(6^3\) is 216, the max on x and y is 5. It is impossible for \(x^3 + y^3\) to equal 189 if \(x + y = 5\), so the only possible answer is 9. In fact, the only combination of positive integers that works for the first equation is 4 and 5. The question difficulty could be improved if the answers were all between 5 and 10. It could also be improved by removing the second equation from the question (and then it would require the same trick as OG13 #64).



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17 May 2013, 02:25
If x = 1! + 2! + 3! + 4! + ....+ 456!, what will be the remainder if x is divided by 7? A) 1 B) 5 C) 8 D) 5 E) 3 7! onward all the terms will be divisible by 7, hence the summation of 1! till 6! should only give a remainder when divided by 7.
1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 1 + 3 + 1 1 = 5 is the OA
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17 May 2013, 03:50
A question paper had 40 questions. Every correct answer would fetch 4 marks and every wrong answer would deduct 1 mark from the total. The unanswered questions doesn't have any impact on the total score. How many distinct scores can be obtained by a student if he/she takes the test A) 200 B) 201 C) 197 D) 195 E) 194 The highest score can be obtained by any of the student is +160 and minimum score can be obtained is 40, Hence the total number of scores can be obtained is 160+40 + 1 (zero) = 201
Now some of the scores are not possible such as 159, 158, 157, 154, 153 and 149, hence 201  6 = D) 195
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17 May 2013, 04:15
How many ways 5! can be written as summation of consecutive positive integers not necessarily starting from 1? A) 0 B) 1 C) 2 D) 3 E) 4 It can be solved using the AP of common difference of 1, that becomes lengthy. Let's observe a pattern here. 3 can be written in consecutive terms as 1+2  1 way , 5 = 2+3  1 way, 15 = 8+7, 1+2+3+4+5, 4+5+6  3 ways, we can see it is number of odd factors of the number 1 Hence 5! can only two odd factors 3 and 5, hence the number of ways should be 41 = 3 ways C)5! = 120 39+40+41, 18+19+20+21+22, 1+ 2+ 3+ 4+ 5+ 6+ 7+ 8 + 9 +10 + 11+ 12+13+14+15 Also one more strategy can be applied has 120/odd number to get number of ways to find the consecutive terms 2 ways to solve the problem
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17 May 2013, 04:36
How many ways can can 23 chocolates can be distributed into 3 students such as everyone gets at least one and none gets more than 10? A) 23C2 B) 20C2 C) 21C2  3*11C2 D) 22C23*11C2 E) 23C23*11C2 Let's say A+B+C = 23 As every body should get 1 chocolate, hence assigning 1 to each, the equation becomes A+B+C = 20, and number of ways of distributing chocolates = 22C2
Now none should get more than 10, lets assign 11 to A, then the equation becomes A+B+C = 9, hence number of ways of distribution being 11C2 > That many cases is when A gets more than 10 and similarly for all the people, the number of cases being 3*11C2
Hence number of ways being 22C23*11C2 D)
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17 May 2013, 21:05
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17 May 2013, 21:50
X is divided by a divisor leaves 19 as a remainder, and when 5X is divided by the same divisor , it leaves 29 as the remainder. What remainder would be obtained if 6X is divided by the same divisor? A) 48 B) 38 C) 19 D) 10 E) 3 As per the equation:
X = DA + 19 if D is the divisor and A is the quotient. 5X = DB + 29 if B is the quotient.
5X = 5DA + 5*19 = 5DA + 95 but the remainder should be 29, hence 9529 = 66 is the multiple of D. The factors of D are 1, 6,11, 66.
1, 6, 11 cannot be D as 19 and 29 are remainders, hence D should be 66
Hence 11X = 11DA + 19*11 = 11DA + 201. As D = 66, hence 11X should leave a remainder of (201/66) Remainder = 3
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17 May 2013, 22:52
Heres my question In how many ways can 12 girls be divided into 4 different groups, such that each group contains atleast one girl? A) 12C4 x 4! B) 12P4 C) 11C3 X 12! D) 12C4 E) 12C3 x 9C3 x 6C3 Correct Answer : C
The explanation is as follows : The question doesnt state that the groups have to be equal. A group can have any number of girls between 1 and 9. So 12C3 x 9C3 x 6C3 wont be the answer. It can be solved using the following technique             The 12 different girls can be arranged in a single line in 12! ways
To arrange them into 4 groups with atleast 1 girl in each lets consider markers which will divide the line into 4 parts. So we require 3 markers which can divide the line as follows
part1 Marker1 part2 marker2 part 3 marker3 part4
There are 11 spaces in the line of girls as shown above
IIIIIIIIIII The markers can be places in the above 11 places in 11C3 ways
Therefore the ways of dividing the girls into 4 groups is 11C3 X 12! i.e C
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17 May 2013, 23:41
a question very similar to the one above but with a twist In how many ways can 14 similar balls be dropped into 4 bins A) 14C4 x 4! B) 14P4 C) 11C3 X 12! D) 17C3 E) \(14^4\) x \(10^3\) x 7C2 Correct Answer : D
The explanation is as follows : The Balls are similar so the order or the arrangement will not matter. Only the ways the balls can be grouped is important. Also the balls have to be dropped into the bins so one scenario is that all the 14 balls are dropped into one bin and other 3 bins are empty The balls can be placed as such              
To arrange them into 4 groups such that the no. of balls in any bin varies from 014.lets consider markers which will divide the line into 4 parts. So we require 3 markers which can divide the line as follows
part1 Marker1 part2 marker2 part 3 marker3 part4
So there are 14 balls and 3 markers
III and              
In case the marker is placed as
III              we have all the 14 balls in the 4th bin and none in the other 3
When its like III              There are 13 balls in the 4th bin and 1 ball in the 3rd bin
So the ways of arranging these 17 items is \(\frac{17!}{(14!*3!)}\) as 14 balls and 3 markers are the same
In other words the balls can be dropped in 4 different bins in 17C3 ways which is the same as \(\frac{17!}{(14!*3!)}\)
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Last edited by aceacharya on 18 May 2013, 01:08, edited 2 times in total.



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18 May 2013, 00:40
aceacharya wrote: a question very similar to the one above but with a twist In how many ways can 14 similar balls be dropped into 4 bins A) 14C4 x 4! B) 14P4 C) 11C3 X 12! D) 16C3 E) 14C4 x 10C3 x 7C2 Correct Answer : D
The explanation is as follows : The Balls are similar so the order or the arrangement will not matter. Only the ways the balls can be grouped is important. Also the balls have to be dropped into the bins so one scenario is that all the 14 balls are dropped into one bin and other 3 bins are empty The balls can be placed as such              
To arrange them into 4 groups such that the no. of balls in any bin varies from 014.lets consider markers which will divide the line into 4 parts. So we require 3 markers which can divide the line as follows
part1 Marker1 part2 marker2 part 3 marker3 part4
For this we have 16 places to place the markers
IIIIIIIIIIIIIIII
In case the marker is placed as
III              we have all the 14 balls in the 4th bin and none in the other 3
When its like III              There are 13 balls in the 4th bin and 1 ball in the 3rd bin
so the 3 markers can be placed in the 16 places in 16C3 ways
In other words the balls can be dropped in 4 different bins in 16C3 ways I guess it should by 17C3, 14 balls + 3 markers.... Isn't it?
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18 May 2013, 01:00
kinjiGC wrote: aceacharya wrote: a question very similar to the one above but with a twist In how many ways can 14 similar balls be dropped into 4 bins A) 14C4 x 4! B) 14P4 C) 11C3 X 12! D) 16C3 E) 14C4 x 10C3 x 7C2 Correct Answer : D
The explanation is as follows : The Balls are similar so the order or the arrangement will not matter. Only the ways the balls can be grouped is important. Also the balls have to be dropped into the bins so one scenario is that all the 14 balls are dropped into one bin and other 3 bins are empty The balls can be placed as such              
To arrange them into 4 groups such that the no. of balls in any bin varies from 014.lets consider markers which will divide the line into 4 parts. So we require 3 markers which can divide the line as follows
part1 Marker1 part2 marker2 part 3 marker3 part4
For this we have 16 places to place the markers
IIIIIIIIIIIIIIII
In case the marker is placed as
III              we have all the 14 balls in the 4th bin and none in the other 3
When its like III              There are 13 balls in the 4th bin and 1 ball in the 3rd bin
so the 3 markers can be placed in the 16 places in 16C3 ways
In other words the balls can be dropped in 4 different bins in 16C3 ways I guess it should by 17C3, 14 balls + 3 markers.... Isn't it? Thanks for the heads up. I had got the solution wrong. Do check the corrected solution.
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18 May 2013, 08:33
Bunuel wrote: 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! Hi Bunuel, Any idea when the access to the GMAT Club tests will be granted and can you please let me know which tests I can access? I have taken some free tests and those are really good. Thanks
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18 May 2013, 22:19
Bunuel wrote: 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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18 May 2013, 23:57
One of my favorite question A and B complete a piece of the task together in D days. A takes 9 days more than the days he will take to complete the task on his own and B takes 4 more days that he would take if he completes the task on his own. What is the value of D? A) 13 B) 6 C) 4.77 D) 12 E) 15 There is one concept I want to discuss here:
Lets say A takes a days and B takes b days to complete the work.
Together they will take ab/(a+b) days to complete the work.
The difference between the days taken by A alone and days taken when A and B worked together = a  ab/(a+b) = a^2/(a+b) Similarly difference of days in case of B = b^2/(a+b)
So if the difference is given, then the combined days can be obtained as sqrt (product of diffrence) = sqrt(a^2*b^2/(a+b)^2) = ab/(a+b)
Applying the formula, the combined number of days are sqrt(9*4) = 6 days
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01 Jun 2013, 20:17
If \(x\), \(y\), and \(z\) are nonzero numbers, is \(\frac{x}{y}\) an integer? (1) \(z=\frac{154\pi}{y}\) (2) \(x=\frac{49\pi^2}{z}\) A number that can be expressed as a ratio of two numbers is rational. However, roots, \(\pi\), and endless nonrepeating decimals are irrational numbers; they cannot be written in a ratio of two numbers.
(1) expression is not sufficient, need x (2) expression is not sufficient, need y
(1) (2) expression reduces to \(\frac{7\pi}{22}\). \(\frac{22}{7}\) is commonly used to approximate \(\pi\), yet \(\frac{22}{7}\neq\pi\). Thus, \(\frac{x}{y}\) is not an integer.
The correct answer is C



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01 Jun 2013, 20:22
If \(t\) is an integer, what is \(t\)? (1) \(t^t 1\) is not positive (2) \(t! \neq 0\) (1) If \(t^t1\) is not positive, then \(t\leq0\). \(t\) can be any odd negative integer or 0. Not sufficient (2) If \(t! \neq 0\), then \(t\) can still be any positive integer. Not sufficient
(1) (2) If \(t^t 1\) is not negative and \(t! \neq 0\), then \(t=0\). \(0^0  1 = 0\) and \(0! = 1\). Sufficient
Therefore, the correct answer is C.
Last edited by mejia401 on 02 Jun 2013, 09:56, edited 1 time in total.



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01 Jun 2013, 21:56
Four Alarm Clocks bells at intervals of 3, 6, 9, and 12 seconds respectively. In 30 minutes, how many times do they bell together ? A. 50 B. 51 C. 45 D. 16 E. 60
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