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New Set of Mixed Questions!!! [#permalink]
01 Apr 2013, 06:48

20

This post received KUDOS

Expert's post

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?

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?

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

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

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?

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?

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

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?

Re: New Set of Mixed Questions!!! [#permalink]
19 Apr 2013, 20:29

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

Re: New Set of Mixed Questions!!! [#permalink]
20 Apr 2013, 04:15

Expert's post

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.

Re: New Set of Mixed Questions!!! [#permalink]
28 Apr 2013, 21:13

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
_________________

Re: New Set of Mixed Questions!!! [#permalink]
23 May 2013, 12:26

Expert's post

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).

Re: New Set of Mixed Questions!!! [#permalink]
22 Jun 2013, 22:46

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?

Re: New Set of Mixed Questions!!! [#permalink]
23 Jun 2013, 00:39

Expert's post

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?

Re: New Set of Mixed Questions!!! [#permalink]
17 Jul 2013, 03:29

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.

Re: New Set of Mixed Questions!!! [#permalink]
17 Jul 2013, 05:12

Expert's post

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.

Re: New Set of Mixed Questions!!! [#permalink]
06 Aug 2013, 16:27

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

Re: New Set of Mixed Questions!!! [#permalink]
11 Aug 2013, 09:30

Expert's post

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

Re: New Set of Mixed Questions!!! [#permalink]
11 Aug 2013, 11:14

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?

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?

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

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

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?

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?

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

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?

Re: New Set of Mixed Questions!!! [#permalink]
02 Sep 2013, 11:44

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.
_________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos

Re: New Set of Mixed Questions!!! [#permalink]
25 Sep 2013, 05:02

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
_________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos

Re: New Set of Mixed Questions!!! [#permalink]
01 Jan 2014, 03:06

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).

Answer: D.

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.

Re: New Set of Mixed Questions!!! [#permalink]
01 Jan 2014, 04:44

Expert's post

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).

Answer: D.

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?

Re: New Set of Mixed Questions!!! [#permalink]
01 Jan 2014, 10:40

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

answer:

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.

Re: New Set of Mixed Questions!!! [#permalink]
01 Jan 2014, 18:21

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:- 1. All 3 spades. 2. All three hearts. 3. 2 spade & 1 heart & vice-versa.

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

Re: New Set of Mixed Questions!!! [#permalink]
02 Jan 2014, 04:51

Expert's post

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:- 1. All 3 spades. 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.
_________________