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# New Set of Mixed Questions!!!

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Math Expert
Joined: 02 Sep 2009
Posts: 34497
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Kudos [?]: 79910 [28] , given: 10022

New Set of Mixed Questions!!! [#permalink]

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01 Apr 2013, 07:48
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The next set of medium/hard PS questions. I'll post OA's with detailed explanations on Friday. Please, post your solutions along with the answers.

1. The distance from the Y-axis to point K is 1/3 of the distance from the X-axis to point K. If the coordinates of K are (-3, y), what is the distance between point K and X-axis?

A. 1/2
B. 1
C. 3
D. 4.5
E. 9.

Solution: new-set-of-mixed-questions-150204-100.html#p1208436

2. What is the area of a region enclosed by |x/3|+|y/9|=10?

A. 675
B. 1350
C. 2700
D. 5400
E. 10800

Solution: new-set-of-mixed-questions-150204-100.html#p1208441

3. Three workers, A, B, and C, can complete a certain task in 10, 5 and x hours respectively. A starts working alone and 2 hours later B joins. After another 2 hours joins C. After that A, B, and C together complete the task in 15 minutes. What is the value of x?

A. 1
B. 1.25
C. 2
D. 2.5
E. 4

Solution: new-set-of-mixed-questions-150204-100.html#p1208445

4. A draining pipe can empty a pool in 4 hours. On a rainy day, when the pool is full, the draining pipe is opened and the pool is emptied in 6 hours. If rain inflow into the pool is 3 liters per hour, what is the capacity of the pool?

A. 9 liters
B. 18 liters
C. 27 liters
D. 36 liters
E. 45 iters

Solution: new-set-of-mixed-questions-150204-100.html#p1208449

5. For a certain set of numbers, if x is in the set, then both -x^2 and -x^3 are also in the set. If the number 1/2 is in the set , which of the following must also be in the set ?

I. -1/64
II. 1/64
III. 1/2^(1/3)

A. I only,
B. II only,
C. III only,
D. I and II only
E. I, II and III

Solution: new-set-of-mixed-questions-150204-100.html#p1208454

6. A team contributes total of $399 from its members. If each member contributed at least$10, and no one contributed $19, what is the greatest number of members the club could have? A. 37 B. 38 C. 39 D. 40 E. 41 Solution: new-set-of-mixed-questions-150204-100.html#p1208457 7. Mary spent 64 percent of her salary on food (including meat) and 16% of her salary on meat. What percent of the salary spent on food were not spent on meat? A. 16% B. 25% C. 32% D. 48% E. 75% Solution: new-set-of-mixed-questions-150204-100.html#p1208459 8. Usually Holly leaves home to school at 9:00, however today she left home 20 minutes later. In order to be at school on time she increased her usual speed by 20% and still was at school 15 minutes later than usual. What is her usual time from home to school? A. 15 minutes B. 20 minutes C. 25 minutes D. 30 minutes E. 210 minutes Solution: new-set-of-mixed-questions-150204-100.html#p1208462 9. If x and y are integers and x + y = -12, which of the following must be true? A. Both x and y are negative B. xy > 0 C. If y < 0, then x > 0 D. If y > 0, then x < 0 E. x - y > 0 Solution: new-set-of-mixed-questions-150204-100.html#p1208466 10. If n is a non-negative integer and the remainder when 3^n is divided by 4 is a multiple of 3, then which of the following must be true? I. n^2 divided by 4 yields the reminder of 1 II. (-2)^n is less than 0 III. n is a prime number A. I only B. II only C. III only D. I and II only E. II and III only Solution: new-set-of-mixed-questions-150204-100.html#p1208471 BONUS QUESTION: 11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue? A. 7/8 B. 3/4 C. 2/3 D. 5/8 E. 3/8 Solution: new-set-of-mixed-questions-150204-100.html#p1208473 Kudos points for each correct solution!!! _________________ Manager Joined: 05 Nov 2012 Posts: 71 Concentration: International Business, Operations Schools: Foster '15 (S) GPA: 3.65 Followers: 1 Kudos [?]: 101 [0], given: 8 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 19 Apr 2013, 21:29 Bunuel wrote: 7. Mary spent 64 percent of her salary on food (including meat) and 16% of her salary on meat. What percent of the salary spent on food were not spent on meat? A. 16% B. 25% C. 32% D. 48% E. 75% 64% of her salary on food; 16% of her salary on meat; 64%-16%=48% on food but not on meat --> 48/64=3/4=75% of the salary spent on food were not spent on meat. Answer: E. Here is another way to go after this one. Since Mary spent 16% of her salary on meat which means she spent 1/4 of the 64% of her salary on meat and the rest, 3/4 not on meat. So 75%. _________________ ___________________________________________ Consider +1 Kudos if my post helped Math Expert Joined: 02 Sep 2009 Posts: 34497 Followers: 6297 Kudos [?]: 79910 [0], given: 10022 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 20 Apr 2013, 05:15 pikachu wrote: Bunuel wrote: 2. What is the area of a region enclosed by |x/3|+|y/9|=10? A. 675 B. 1350 C. 2700 D. 5400 E. 10800 Find the x and y intercepts. When y=0, then x=30 or x=-30. When x=0, then y=90 or x=-90. So, we have 4 points: (30, 0), (-30, 0) (0, 90), (0, -90). When joining these points we get the rhombus: Attachment: 2.png The area of a rhombus is $$\frac{d_1*d_2}{2}$$ (where $$d_1$$ and $$d_2$$ are the lengths of the diagonals), thus the area of the enclosed figure is 60*180/2=5,400. Answer: D. Bunuel what about the following points that also meets the given conditions of |x/3| + | y/9| = 10 but come with a different area. (x,y) = (-3,81) (x,y) = (-3,-81) (x,y) = (3,-81) (x,y) = (3,81) In this case Area of the rectangle will be = 972 or you can have (x,y) = (6,72) (x,y) = (6,-72) (x,y) = (-6,-72) (x,y) = (-6,72) In this case the area of the rectangle will be = 1728. Can you please clarify Below is the graph of |x/3|+|y/9|=10: The points you mention are on it and not making some other figure. _________________ Director Status: My Thread Master Bschool Threads-->Krannert(Purdue),WP Carey(Arizona),Foster(Uwashngton) Joined: 28 Jun 2011 Posts: 894 Followers: 83 Kudos [?]: 210 [0], given: 57 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 28 Apr 2013, 22:13 11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue? A. 7/8 B. 3/4 C. 2/3 D. 5/8 E. 3/8 The problem can be translated to "If 5 red marbles and 3 blue marbles are arranged in a row, what is the probability that the last marble is red?" Total number of arrangements = 8!/[(5!)*(3!)] = 8*7*6/3! = 8*7 Number of arrangements such that last marble is red = Number of arrangements for 4 red marbles and 3 blue marbles = 7!/[(4!)*(3!)] = 786*5/3! = 7*5 Hence, required probability = 7*5/(887) = 5/8 _________________ Math Expert Joined: 02 Sep 2009 Posts: 34497 Followers: 6297 Kudos [?]: 79910 [0], given: 10022 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 23 May 2013, 13:26 SaraLotfy wrote: Bunuel wrote: 5. For a certain set of numbers, if x is in the set, then both -x^2 and -x^3 are also in the set. If the number 1/2 is in the set , which of the following must also be in the set ? I. -1/64 II. 1/64 III. 1/2^(1/3) A. I only, B. II only, C. III only, D. I and II only E. I, II and III Wouldn't -x^2 = +1/4 since the power is even? or should I treat this as -|X|^2? Since 1/2 is in the set, the so must be: -x^2 = -1/4; -x^3 = -1/8. Since -1/4 is in the set, the so must be: -x^3 = 1/64; Since -1/8 is in the set, the so must be: -x^2 = -1/64. The only number we cannot get is 1/2^(1/3). Answer: D. $$-x^2=-(\frac{1}{2})^2=-\frac{1}{4}$$ (its -x^2 not (-x)^2). Hope it's clear. _________________ Manager Joined: 30 May 2013 Posts: 190 Location: India Concentration: Entrepreneurship, General Management GPA: 3.82 Followers: 0 Kudos [?]: 53 [0], given: 72 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 22 Jun 2013, 23:46 g106 wrote: 3. Three workers, A, B, and C, can complete a certain task in 10, 5 and x hours respectively. A starts working alone and 2 hours later B joins. After another 2 hours joins C. After that A, B, and C together complete the task in 15 minutes. What is the value of x? A. 1 B. 1.25 C. 2 D. 2.5 E. 4 A, B, and C, can complete a certain task in 10, 5 and x hours respectively A completes 1/10 of the total work in 1 hr B Completes 1/5 of the total work in 1 hr C completes 1/x of the total work in 1 hr A starts working alone and 2 hours later B joins. After another 2 hours joins C. A completes 2/10 of the total work in 2 hrs later B joins A A&B together can complete 1/10 + 1/5 of the total work in 1 hr = 3/10 A&B work together for 2 hour and they complete 2 * 3/10 = 6/10 of the total work By the time C joins A&B 8/10 of the total work is complete To complete 2/10 of the work A, B & C together take 15 minutes (1/4 hr) 2/10 = ¼( 1/10 + 1/5+ 1/x) 8/10 = 3/10 + 1/X 5/10 = 1/X X = 2 Answer = C Hi, plz help me to find out. i dont understand this point " To complete 2/10 of the work A, B & C together take 15 minutes (1/4 hr) " How come u got 2/10? A and b joins to complete 6/10 of work then c joins. But C'r work rate is 1/X. Then how u got 2/10? Please help me Thanks in adavnce, RRSNATHAN Math Expert Joined: 02 Sep 2009 Posts: 34497 Followers: 6297 Kudos [?]: 79910 [0], given: 10022 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 23 Jun 2013, 01:39 rrsnathan wrote: g106 wrote: 3. Three workers, A, B, and C, can complete a certain task in 10, 5 and x hours respectively. A starts working alone and 2 hours later B joins. After another 2 hours joins C. After that A, B, and C together complete the task in 15 minutes. What is the value of x? A. 1 B. 1.25 C. 2 D. 2.5 E. 4 A, B, and C, can complete a certain task in 10, 5 and x hours respectively A completes 1/10 of the total work in 1 hr B Completes 1/5 of the total work in 1 hr C completes 1/x of the total work in 1 hr A starts working alone and 2 hours later B joins. After another 2 hours joins C. A completes 2/10 of the total work in 2 hrs later B joins A A&B together can complete 1/10 + 1/5 of the total work in 1 hr = 3/10 A&B work together for 2 hour and they complete 2 * 3/10 = 6/10 of the total work By the time C joins A&B 8/10 of the total work is complete To complete 2/10 of the work A, B & C together take 15 minutes (1/4 hr) 2/10 = ¼( 1/10 + 1/5+ 1/x) 8/10 = 3/10 + 1/X 5/10 = 1/X X = 2 Answer = C Hi, plz help me to find out. i dont understand this point " To complete 2/10 of the work A, B & C together take 15 minutes (1/4 hr) " How come u got 2/10? A and b joins to complete 6/10 of work then c joins. But C'r work rate is 1/X. Then how u got 2/10? Please help me Thanks in adavnce, RRSNATHAN Check here: new-set-of-mixed-questions-150204-100.html#p1208445 Hope it helps. Links to OE's are given in the original post. _________________ Math Expert Joined: 02 Sep 2009 Posts: 34497 Followers: 6297 Kudos [?]: 79910 [0], given: 10022 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 09 Jul 2013, 00:40 Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE _________________ Intern Joined: 05 May 2013 Posts: 27 GMAT 1: 730 Q50 V39 GRE 1: 1480 Q800 V680 Followers: 0 Kudos [?]: 22 [0], given: 5 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 17 Jul 2013, 04:29 Bunuel wrote: 11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue? A. 7/8 B. 3/4 C. 2/3 D. 5/8 E. 3/8 Basically we need to find the probability that the seventh marble drawn is red (so not blue). Now, the initial probability of drawing red marble is 5/8. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is 5/8. The same for blue marble: the probability of drawing blue marble is 3/8, the probability that for instance the 8th marble drawn is blue is still 3/8. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results). Answer: D. Hi Bunuel, Can you please elaborate your thinking on this qn ? I am having a hard time wrapping my head arnd this. My line of thinking was to calculate the probability that you are left with 1 or two red marbles when you are about to draw the seventh one. Not only was this approach very drawn out, i didnt get this answer. Thanks. Math Expert Joined: 02 Sep 2009 Posts: 34497 Followers: 6297 Kudos [?]: 79910 [0], given: 10022 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 17 Jul 2013, 06:12 Expert's post 2 This post was BOOKMARKED vs129 wrote: Bunuel wrote: 11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue? A. 7/8 B. 3/4 C. 2/3 D. 5/8 E. 3/8 Basically we need to find the probability that the seventh marble drawn is red (so not blue). Now, the initial probability of drawing red marble is 5/8. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is 5/8. The same for blue marble: the probability of drawing blue marble is 3/8, the probability that for instance the 8th marble drawn is blue is still 3/8. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results). Answer: D. Hi Bunuel, Can you please elaborate your thinking on this qn ? I am having a hard time wrapping my head arnd this. My line of thinking was to calculate the probability that you are left with 1 or two red marbles when you are about to draw the seventh one. Not only was this approach very drawn out, i didnt get this answer. Thanks. Below discussions might help: a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-90272.html a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html each-of-four-different-locks-has-a-matching-key-the-keys-101553.html if-40-people-get-the-chance-to-pick-a-card-from-a-canister-97015.html new-set-of-mixed-questions-150204-100.html#p1208473 a-medical-researcher-must-choose-one-of-14-patients-to-127396.html a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html _________________ Intern Joined: 26 Feb 2013 Posts: 3 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 06 Aug 2013, 17:27 the function f is defined for positive three digit integer n by f(n) = 2^x3^y5^z, where x,y,z are the hundreds, tens, and units digits of n, respectively. if m and v are three digit positive integers such that f(m) = 9 f(v), then m - v = note: 2^x denote x is power of 2 anyone please give me solution of this problem Math Expert Joined: 02 Sep 2009 Posts: 34497 Followers: 6297 Kudos [?]: 79910 [0], given: 10022 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 11 Aug 2013, 10:30 Riyad47 wrote: the function f is defined for positive three digit integer n by f(n) = 2^x3^y5^z, where x,y,z are the hundreds, tens, and units digits of n, respectively. if m and v are three digit positive integers such that f(m) = 9 f(v), then m - v = note: 2^x denote x is power of 2 anyone please give me solution of this problem Discussed here: the-function-f-is-defined-for-each-positive-three-digit-100847.html _________________ Senior Manager Joined: 10 Jul 2013 Posts: 335 Followers: 3 Kudos [?]: 277 [0], given: 102 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 11 Aug 2013, 12:14 Bunuel wrote: The next set of medium/hard PS questions. I'll post OA's with detailed explanations on Friday. Please, post your solutions along with the answers. 1. The distance from the Y-axis to point K is 1/3 of the distance from the X-axis to point K. If the coordinates of K are (-3, y), what is the distance between point K and X-axis? A. 1/2 B. 1 C. 3 D. 4.5 E. 9. Solution: new-set-of-mixed-questions-150204-100.html#p1208436 2. What is the area of a region enclosed by |x/3|+|y/9|=10? A. 675 B. 1350 C. 2700 D. 5400 E. 10800 Solution: new-set-of-mixed-questions-150204-100.html#p1208441 3. Three workers, A, B, and C, can complete a certain task in 10, 5 and x hours respectively. A starts working alone and 2 hours later B joins. After another 2 hours joins C. After that A, B, and C together complete the task in 15 minutes. What is the value of x? A. 1 B. 1.25 C. 2 D. 2.5 E. 4 Solution: new-set-of-mixed-questions-150204-100.html#p1208445 4. A draining pipe can empty a pool in 4 hours. On a rainy day, when the pool is full, the draining pipe is opened and the pool is emptied in 6 hours. If rain inflow into the pool is 3 liters per hour, what is the capacity of the pool? A. 9 liters B. 18 liters C. 27 liters D. 36 liters E. 45 iters Solution: new-set-of-mixed-questions-150204-100.html#p1208449 5. For a certain set of numbers, if x is in the set, then both -x^2 and -x^3 are also in the set. If the number 1/2 is in the set , which of the following must also be in the set ? I. -1/64 II. 1/64 III. 1/2^(1/3) A. I only, B. II only, C. III only, D. I and II only E. I, II and III Solution: new-set-of-mixed-questions-150204-100.html#p1208454 6. A team contributes total of$399 from its members. If each member contributed at least $10, and no one contributed$19, what is the greatest number of members the club could have?

A. 37
B. 38
C. 39
D. 40
E. 41

Solution: new-set-of-mixed-questions-150204-100.html#p1208457

7. Mary spent 64 percent of her salary on food (including meat) and 16% of her salary on meat. What percent of the salary spent on food were not spent on meat?

A. 16%
B. 25%
C. 32%
D. 48%
E. 75%

Solution: new-set-of-mixed-questions-150204-100.html#p1208459

8. Usually Holly leaves home to school at 9:00, however today she left home 20 minutes later. In order to be at school on time she increased her usual speed by 20% and still was at school 15 minutes later than usual. What is her usual time from home to school?

A. 15 minutes
B. 20 minutes
C. 25 minutes
D. 30 minutes
E. 210 minutes

Solution: new-set-of-mixed-questions-150204-100.html#p1208462

9. If x and y are integers and x + y = -12, which of the following must be true?

A. Both x and y are negative
B. xy > 0
C. If y < 0, then x > 0
D. If y > 0, then x < 0
E. x - y > 0

Solution: new-set-of-mixed-questions-150204-100.html#p1208466

10. If n is a non-negative integer and the remainder when 3^n is divided by 4 is a multiple of 3, then which of the following must be true?

I. n^2 divided by 4 yields the reminder of 1
II. (-2)^n is less than 0
III. n is a prime number

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

Solution: new-set-of-mixed-questions-150204-100.html#p1208471

BONUS QUESTION:
11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?

A. 7/8
B. 3/4
C. 2/3
D. 5/8
E. 3/8

Solution: new-set-of-mixed-questions-150204-100.html#p1208473

Kudos points for each correct solution!!!

really a good set of practice...thanks
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Re: New Set of Mixed Questions!!! [#permalink]

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02 Sep 2013, 12:44
Bonus question:
11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?

A. 7/8
B. 3/4
C. 2/3
D. 5/8
E. 3/8

Ans is D, probability of not drawing a blue is not dependent on the position, considering that there is no restriction or criterion defined on sequence of draws any ball can be drawn before 7th not blue ball condition. Therefore it should be Redballs/totalballs i.e 5/8.
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Re: New Set of Mixed Questions!!! [#permalink]

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25 Sep 2013, 06:02
4.
Volume of pool = V
Water added in pool BY rain in 6 hours = 3x6=18liters
Standard pump output rate (V/4) for 6 hours V6/4

Thus
V + 18 - V6/4 =0

V=36
_________________

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-----------------------
Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison
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Re: New Set of Mixed Questions!!! [#permalink]

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01 Jan 2014, 04:06
Bunuel wrote:
11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. 7/8
B. 3/4
C. 2/3
D. 5/8
E. 3/8

Basically we need to find the probability that the seventh marble drawn is red (so not blue).

Now, the initial probability of drawing red marble is 5/8. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is 5/8.

The same for blue marble: the probability of drawing blue marble is 3/8, the probability that for instance the 8th marble drawn is blue is still 3/8. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).

Hi Bunuel,
Why is the probability not changing? If we keep withdrawing marbles withoout replacing then following things are changing:-
1. total no. of marble (hence the denominator in probability)
2. No. of Red marbles left.
3. No. of blue marbles left.
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Re: New Set of Mixed Questions!!! [#permalink]

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01 Jan 2014, 05:44
Expert's post
1
This post was
BOOKMARKED
Nilabh_s wrote:
Bunuel wrote:
11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. 7/8
B. 3/4
C. 2/3
D. 5/8
E. 3/8

Basically we need to find the probability that the seventh marble drawn is red (so not blue).

Now, the initial probability of drawing red marble is 5/8. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is 5/8.

The same for blue marble: the probability of drawing blue marble is 3/8, the probability that for instance the 8th marble drawn is blue is still 3/8. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).

Hi Bunuel,
Why is the probability not changing? If we keep withdrawing marbles withoout replacing then following things are changing:-
1. total no. of marble (hence the denominator in probability)
2. No. of Red marbles left.
3. No. of blue marbles left.

Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8.

Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change?

Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade?

There is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?

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each-of-four-different-locks-has-a-matching-key-the-keys-101553.html
if-40-people-get-the-chance-to-pick-a-card-from-a-canister-97015.html
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html

Hope this helps.
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Re: New Set of Mixed Questions!!! [#permalink]

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01 Jan 2014, 11:40
2
KUDOS
11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?

A. 7/8
B. 3/4
C. 2/3
D. 5/8
E. 3/8

We fix the seventh ball to be red. Since all balls are same it doesnt matter which of the red balls.

So the rest 7 balls can be arranged in 7!/(4!*3!)= 35 ways becoz there are 4 identical red balls and 3 identical blue balls.

Without any restriction 8 balls can be arranged in 8!/(5!*3!) = 56 ways.

Therefore probability is 35/56 = 5/8.
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Joined: 08 Dec 2013
Posts: 21
Location: India
Concentration: General Management, Operations
GMAT 1: 710 Q50 V35
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Kudos [?]: 7 [0], given: 7

Re: New Set of Mixed Questions!!! [#permalink]

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01 Jan 2014, 11:52
when i say arranging i mean different possibilities or orders in which the balls can be taken out from the box.
I hope Thats understood...
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Re: New Set of Mixed Questions!!! [#permalink]

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01 Jan 2014, 19:21
Bunuel wrote:
Nilabh_s wrote:
Bunuel wrote:
.Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8.

Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change?

Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade?

There is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?

Hope this helps.

For the statement
" NOW that you pick a spade out of 5 remaining cards"

If you throw 3 cards then how are we sure than 5 spades are still remaining. Following three cards may have been thrown:-
2. All three hearts.
3. 2 spade & 1 heart & vice-versa.

Hence how can we be sure that 5 spades are till remaining?
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Re: New Set of Mixed Questions!!! [#permalink]

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02 Jan 2014, 05:51
Nilabh_s wrote:
Bunuel wrote:
Bunuel wrote:
.Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8.

Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change?

Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade?

There is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?

Hope this helps.

For the statement
" NOW that you pick a spade out of 5 remaining cards"

If you throw 3 cards then how are we sure than 5 spades are still remaining. Following three cards may have been thrown:-
2. All three hearts.
3. 2 spade & 1 heart & vice-versa.

Hence how can we be sure that 5 spades are till remaining?

Yes, but we don't know which cards are thrown and this won't affect the probability. Please follow the links in my previous post for better understanding of such problems.
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Re: New Set of Mixed Questions!!!   [#permalink] 02 Jan 2014, 05:51

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