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Re: Submit Your Own GMAT Math Questions - Get GMAT Club Tests [#permalink]

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02 Jun 2013, 11:37

kinjiGC wrote:

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

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

Assuming the student has an answer for every question on the test, the possible scores are even less than 195 because the change in score is five points for each wrong (or correct) answer. If \(+4\) for correct and \(-1\) for incorrect, then it's a total change of \(5\) points. If the score is adjusted by five points for answer, then we should only count the multiple of 5s\(\frac{106--40}{5}+1 = 41\) Besides the order of problems one gets right or wrong can vary, but there is still only a total of 40 questions.

There are 41 scores possible, unless I didn't account for the unanswered questions correctly. Please explain logic behind answer too please, Thanks!

Re: Submit Your Own GMAT Math Questions - Get GMAT Club Tests [#permalink]

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02 Jun 2013, 22:03

mejia401 wrote:

kinjiGC wrote:

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

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

Assuming the student has an answer for every question on the test, the possible scores are even less than 195 because the change in score is five points for each wrong (or correct) answer. If \(+4\) for correct and \(-1\) for incorrect, then it's a total change of \(5\) points. If the score is adjusted by five points for answer, then we should only count the multiple of 5s\(\frac{106--40}{5}+1 = 41\) Besides the order of problems one gets right or wrong can vary, but there is still only a total of 40 questions.

There are 41 scores possible, unless I didn't account for the unanswered questions correctly. Please explain logic behind answer too please, Thanks!

I didn't understand why you are targeting the question from change of marks. The question here asks how many different scores are possible.

If the student gets all of them incorrect, the score can be -40 and max being +160 where the student answers everything correctly, but some of the scores between -40 and + 160 is not possible, for example +159 as in that case he has to get all 40 questions correct and 1 incorrect which is not possible. hence you need to deduct that many scores which are not possible.

Re: Submit Your Own GMAT Math Questions - Get GMAT Club Tests [#permalink]

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03 Jun 2013, 00:21

It seems I am late to the party. Anyways better Late than Never. Here is my 2 cents If x, y and z are three positive integers and x + y +z = 6 & z>1, then what is the probability that x equals 1? (A) \(\frac{1}{2}\) (B) \(\frac{1}{3}\) (C) \(\frac{1}{4}\) (D) \(\frac{3}{7}\) (E) \(\frac{4}{7}\)

X, Y & Z all Positive Integers and Z >1 Possible scenario's:- X Y Z 1 3 2 2 2 2 3 1 2 1 2 3 2 1 3 1 1 4 Favorable Scenarios = 3 Total Scenarios = 6 Thus probability = 3/6 = \(\frac{1}{2}\)

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Max value of X can be 60.9 Max value of Y can be 65.9 Max value of X+Y can be 126.8 If we had the leeway to round off, the answer would be 127. But in this case we don't have such leeway, so the answer will be 126

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Re: Submit Your Own GMAT Math Questions - Get GMAT Club Tests [#permalink]

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03 Jun 2013, 00:29

fameatop wrote:

It seems I am late to the party. Anyways better Late than Never. Here is my 2 cents If x, y and z are three positive integers and x + y +z = 6 & z>1, then what is the probability that x equals 1? (A) \(\frac{1}{2}\) (B) \(\frac{1}{3}\) (C) \(\frac{1}{4}\) (D) \(\frac{3}{7}\) (E) \(\frac{4}{7}\)

X, Y & Z all Positive Integers and Z >1 Possible scenario's:- X Y Z 1 3 2 2 2 2 3 1 2 1 2 3 2 1 3 1 1 4 Favorable Scenarios = 3 Total Scenarios = 6 Thus probability = 3/6 = \(\frac{1}{2}\)

A shorter method can be as below:

X+Y+Z = 6, where X, Y => 1 and Z > 1, so minimum value Z can have is 2 Distribute 1 to X and Y each and Z = 2

Then X+Y+Z = 2 now to find out how many cases can have X=1, distribute 2 to Y and Z, that is Y+Z=2 -> 3C1 -> 3 ways All the cases, X+Y+Z = 2 -> 4C2 - 6

Max value of X can be 60.9 Max value of Y can be 65.9 Max value of X+Y can be 126.8 If we had the leeway to round off, the answer would be 127. But in this case we don't have such leeway, so the answer will be 126

The expression to be maximized is 2X+5 and should be an integer value, so 2X should be an integer. Hence X can be 60.5 , the value would be 126 D)
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03 Jun 2013, 00:36

Tom wants to write GMAT, but because of unavailability of seats in his hometown he decides to schedule it in a nearby town that he is not fully aware of. On Test day he travels 10 mile due south followed by 20 miles west & finally 10 miles south again to reach the incorrect test center rather than travelling 10 mile due north followed by 20 miles east & finally 10 miles north again to reach the correct test center. What is the minimum distance Tom needs to travel so that he can appear for the test? (A) 80 (B) 55 (C) 57 (D) 67 (E) 68

If we draw the traveled distance using straight lines, we will get Isosceles Right angle triangle with sides 40 & 40. We are asked to find the Hypotenuse of this triangle. Hypotenuse will be 40\(\sqrt{2}\) = 56.56= 57 miles

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Last edited by fameatop on 03 Jun 2013, 00:55, edited 1 time in total.

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03 Jun 2013, 00:37

A cunning trader bought unknown quantities of two varieties of salt costing 5 & 8$ per ounce and mixes them with 18 ounces of sand, which is available free of cost, to produce adulterated salt, which the trader will sell at below market price. At what price should the trader sell the mixture so that he can make 12.5% profit?

(A) If sand is not mixed with the salt, then the cost price of the mixture is 50% more than the final cost price of adulterated mixture or 33% more than the final sales price of adulterated mixture.

In this question we have to find the cost price of adulterated mixture (2 types of salt & sand) Let's assume that the cost price of adulterated mixture is 8c, so selling price will be 9c

Statement1:-If only two types of salts are mixed the cost price will be = 1.5 x cost price of adulterated mixture = 1.5 x 8c = 12c. From this we can find that the 5 & 8$ salts are mixed in a ratio of (8-12c) & (12c-5). Because the value of C is not known we can't find the value of this ratio. Thus insufficient

Statement2:-Two salts are mixed in a ratio of 2:1. From this we can find that if 5 & 8$ salts are mixed in a ratio of 2:1 , the cost price will be 6$/ounce

But we want the cost price of the mixture of salts & sand. We know that 18 ounces of sand is used. In order to calculate the final cost price we should know either the amount of the salts, which is mixed with sand, or the ratio of salts & sand. But we don't have the Exact amount of the salts or the the ratio of salts & sand. Thus insufficient

Statement 1&2:-Combing both we will get the answer Cost price of adulterated mixture will be 4$ or selling price will be 4.5$

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03 Jun 2013, 23:59

Simple concept but a very Challenging question An inlet pipe fills tank A at a rate of 25 liter per minutes. 1st outlet pipe drawing water at a rate of 5 liter per minute from tank A is connected to bathroom & 2nd outlet pipe is connected to Tank B. The 2nd outlet pipe will draw water, at a rate of 15 liter per minute, only if the water level in Tank A is > 50%. One outlet pipe drawing water at a rate of 5 liter per minute from tank B is connected to Kitchen. Capacity of Tank A & B is 800 & 600 liter respectively. Both the tanks are empty at the beginning & all the taps are turned on since the beginning. How much less time (in minutes) will Tank B take to fill completely than the time taken by Tank A to fill completely from the moment Tank A starts transferring water to Tank B? (A) 20 (B) 40 (C) 60 (D) 80 (E) 100

Net Filling rate for Tank A before it reaches 50% of capacity = (25-5) = 20 lit/min Time taken by Tank A to fill initial 50% of capacity = 400/(25-5) = 20 min Net Filling rate for Tank A after it reaches 50% of capacity = (25-5-15) = 5 lit/min Time taken by Tank A to fill the remaining 50% of capacity = 400/(5) = 80 min

Net Filling rate for Tank B = (15-5) = 10 lit/min Time taken by Tank B to fill completely = 600/10 = 60 min

Thus Tank B take will take (80-60) 20 min less to fill completely than the time taken by Tank A to fill completely from the moment Tank A starts transferring water to Tank B

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Last edited by fameatop on 07 Jun 2013, 18:58, edited 1 time in total.

Re: Submit Your Own GMAT Math Questions - Get GMAT Club Tests [#permalink]

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04 Jun 2013, 00:20

Simple concept but very Tricky An inlet pipe starts filling an empty tank A at a rate of 25 liter per minute at 11:00 am, while an outlet pipe draws water from tank A & fills empty tank B at a rate of 20 liter per minute only if the water level in Tank A is > 20%. Likewise, an outlet pipe draws water from tank B & fills empty tank C at a rate of 15 liter per minute only if the water level in Tank B is > 40%. Lastly, an outlet pipe draws water from tank C & fills empty tank D at a rate of 10 liter per minute only if the water level in Tank C is > 50%. Capacity of Tank A, B, C & D is 1000, 800, 600 & 400 liter respectively. At what time all the tanks will be completely filled? (A) 12:24 noon (B) 1:40 pm (C) 1:48 pm (D) 4:56 pm (E) 5:40 pm

Time taken by Tank A to fill initial 20% of capacity = 200/25 = 8 min Net Filling rate for Tank A after it reaches 20% of capacity = (25-20) = 5 lit/min Time taken by Tank A to fill the remaining 80% of capacity = 800/5 = 160 min

Tank A will take the longest to fill. By the time Tank A is full all the other tanks will be full. Thus the time at which all the tanks will be full= 8+160 = 168 min = 2 hrs 48 minutes.

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04 Jun 2013, 01:25

A water heater boils the water in 15 minutes on Low heat or in 10 minutes on Medium heat or in 5 minutes on High heat. On Sunday, water heater heats the water in 8 minutes. For how long heater was on Medium Heat mode? (A) Capacity of Water heater is 25 liters. (B) Ratio of efficiency on Low, Medium & High heat is 2:3:6

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07 Jun 2013, 19:43

A Train starts traveling from station A towards station B. The moment train starts traveling a Radar, situated at station B, emits a wave that travels towards the moving train at a rate of 50km/hr. The wave is reflected back to station B as soon as it hits the front of the train and this process goes on till train reaches station B. What is the total distance covered by the wave? (A) Speed of the train is 25 Km/hr . (B) Train takes 1 Hr to reach station B

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07 Jun 2013, 19:50

A guy standing at the platform sees a Train coming toward him . In order to touch the rear end of the train of the train, he starts running away from the train. At what speed should he run so that he could touch the rear end of the train? (A) Speed of the train is 40 km/hr. (B) The moment guy sees the front of the train, the distance between the guy & the train is 10 Km

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13 Jun 2013, 19:26

prateekbhatt wrote:

There are c number of cats and d number of dogs in a certain apartment complex. If there is at least one dog in the complex, is the number of dogs greater than the number of cats? 1- c/d<2 2- c^2/d^2>1

Hi Prateek,

This is a Veritas question and thus a copyrighted material.
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05 Jul 2013, 02:18

Parallelogram ABCD is to be constructed in the xy-plane such that the sides are parallel to x and y axis. The x and y coordinates of A, B, C and D are to be integers that lies between the lines |x-1| = 5 and |y+5| = 6. How many different parallelograms with these properties could be constructed? (A) 10 (B) 110 (C) 1980 (D) 9900 (E) 4980

How many zeros exist in the end of the product of alphabets(a-z) where a = 4^4, b = 5^5, c = 6^6, ....z = 29^29?

a) 99 b) 100 c) 101 d)102 e)103

I think, It is 100.

A zero is created when 2 is multiplied by 5. So we should find out how many multiplications of 2 and 5 are possible in the expression. Obviously 2 will occur more frequently than does 5. So number of 5 in the above expression will determine the number of zeros.

5 can be found in following terms

5^5 = 5 times

10^10 = 10 times

15^15 = 15 times

20^20 = 20 times

25^25 = 50 times

Total = 100 times. So 100 zeros should be there.
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