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# New Set of Mixed Questions!!!

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Math Expert
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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!!! _________________ Senior Manager Joined: 16 Dec 2011 Posts: 433 Followers: 11 Kudos [?]: 214 [3] , given: 70 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 04 Apr 2013, 09:44 3 This post received KUDOS 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 When 3^n is divided by 4, remainder is a multiple of 3 ==> n is odd. I. For all odd n’s, remainder will be 1 if n^2 is divided by 4. Correct. II. As n is odd, (-2)^n is –ve. Correct. III. n can be 9 or 15 or some other composite odd number. Incorrect. Correct answer is D. Senior Manager Joined: 16 Dec 2011 Posts: 433 Followers: 11 Kudos [?]: 214 [3] , given: 70 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 04 Apr 2013, 09:45 3 This post received KUDOS 1 This post was BOOKMARKED 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 Total number of marbles is 8. For any given position, probability for red marble is 5/8 and blue marble is 3/8. So, the probability that the seventh marble drawn is NOT blue is 5/8 Correct answer is D. Math Expert Joined: 02 Sep 2009 Posts: 38864 Followers: 7729 Kudos [?]: 106089 [0], given: 11607 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 05 Apr 2013, 04:39 Expert's post 9 This post was BOOKMARKED SOLUTIONS: 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 Point K has the coordinates (-3, y) means that it's somewhere on the line x=-3. Hence the distance from this point to the Y-axis is 3 units. Since the distance from the Y-axis to point K is 1/3 of the distance from the X-axis to point K, then the distance from K to the X-axis is 9 units. Answer: E. _________________ Math Expert Joined: 02 Sep 2009 Posts: 38864 Followers: 7729 Kudos [?]: 106089 [3] , given: 11607 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 05 Apr 2013, 04:51 3 This post received KUDOS Expert's post 15 This post was BOOKMARKED 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 [ 7.85 KiB | Viewed 24274 times ] 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. _________________ Math Expert Joined: 02 Sep 2009 Posts: 38864 Followers: 7729 Kudos [?]: 106089 [2] , given: 11607 New Set of Mixed Questions!!! [#permalink] ### Show Tags 05 Apr 2013, 05:06 2 This post received KUDOS Expert's post 19 This post was BOOKMARKED 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 After 2 hours $$2*\frac{1}{10}=\frac{1}{5}$$ of the taks will be done (as only A works); After 4 hours $$\frac{1}{5}+2*(\frac{1}{10}+\frac{1}{5})=\frac{4}{5}$$ of the task will be done and $$\frac{1}{5}$$ will be left to be done; We are told that $$\frac{1}{5}$$th of the task is done in 15 minutes (1/4th of an hour) by all three workers: $$\frac{1}{4}*(\frac{1}{10}+\frac{1}{5}+\frac{1}{x})=\frac{1}{5}$$. From which we can find that $$x=2$$ hours. Answer: C _________________ Math Expert Joined: 02 Sep 2009 Posts: 38864 Followers: 7729 Kudos [?]: 106089 [2] , given: 11607 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 05 Apr 2013, 05:22 2 This post received KUDOS Expert's post 15 This post was BOOKMARKED 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 liters Let the rate of the draining pipe be $$x$$ liters per hour. Then the capacity of the tank will be $$C=time*rate=4x$$; Now, when raining, the net outflow is x-3 liters per hour, and we are told that at this new rate the pool is emptied in 6 hours. So, the capacity of the pool also equals to $$C=time*rate=6(x-3)$$; Thus we have: $$4x=6(x-3)$$ --> $$x=9$$ --> $$C=4x=36$$. Answer: D. _________________ Math Expert Joined: 02 Sep 2009 Posts: 38864 Followers: 7729 Kudos [?]: 106089 [0], given: 11607 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 05 Apr 2013, 05:33 Expert's post 4 This post was BOOKMARKED 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 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. _________________ Math Expert Joined: 02 Sep 2009 Posts: 38864 Followers: 7729 Kudos [?]: 106089 [0], given: 11607 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 05 Apr 2013, 05:44 Expert's post 3 This post was BOOKMARKED 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

Obviously the team could not have 40 or more members, since $10*40=$400>$399. What about 39? If 37 members contributes$10 each ($10*37=$370) and the remaining two members contributed for example, $11 and$18, respectively then the team would have is 37+1+1=39.

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

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05 Apr 2013, 05:49
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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.

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

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05 Apr 2013, 05:53
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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

Let the usual speed be $$s$$ and usual time $$t$$ minutes, then as the distance covered is the same we will have: $$st=1.2s*(t-20+15)$$ --> $$t=30$$ minutes.

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

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05 Apr 2013, 05:57
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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

Look at option D: if y is positive, then x must be negative in order the sum of x and y to be negative.

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

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05 Apr 2013, 06:06
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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

3^0=1 --> the remainder when 1 is divided by 4 is 1;
3^1=3 --> the remainder when 3 is divided by 4 is 3;
3^2=9 --> the remainder when 9 is divided by 4 is 1;
3^3=27 --> the remainder when 27 is divided by 4 is 3;
...

We can see that in order the condition to hold true n must be odd.

I. n^2 divided by 4 yields the reminder of 1 --> odd^2 divided by 4 always yields the reminder of 1. So, this statement must be true.

II. (-2)^n is less than 0 --> (-2)^odd<0. So, this statement must be true.

III. n is a prime number. Not necessarily true.

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

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05 Apr 2013, 06:07
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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).

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

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05 Apr 2013, 06:11
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Links to the solutions are in the original post here: new-set-of-mixed-questions-150204.html#p1206328
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Re: New Set of Mixed Questions!!! [#permalink]

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05 Apr 2013, 06:23
Kudos points given for each correct solution.

Note that I cannot award more than 5 Kudos to the same person per day, so those of you who have more than 5 correct solutions please PM me tomorrow the links for which I owe you kudos points.

Thank you.

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

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06 Apr 2013, 02:51

bunuel-signature-collection-all-in-one-with-solutions-146628.html#p1208799

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

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07 Apr 2013, 05:51
1
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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 ?

Hi Bunuel,

For the above question, can you please let me know where I am going wrong.

I'm able to solve till here-
Since 1/2 is in the set, the so must be:
-x^2 = -1/4;
-x^3 = -1/8.
[b]Doubt 1-

However, what I failed to understand is why set should continue beyond -1/8 as we don't know the number of elements in the set, and since this is a must be true question, the below solution seems redundant to me.

Bunuel wrote:
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).

Doubt 2-
I am hopeful that I lack some understanding the above solution will most probably be correct. In that case, I want to understand what will be maximum no of elements in this set. Will this be an Infinite set.

Doubt 3-
Will the answer behaved differently had this question been "Could be true"

Kindly elaborate.

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

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07 Apr 2013, 13:37
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imhimanshu wrote:
Doubt 1-

However, what I failed to understand is why set should continue beyond -1/8 as we don't know the number of elements in the set, and since this is a must be true question, the below solution seems redundant to me.
`

The logic is, if 1/2 is in the set, then -1/8 MUST be in the set. Knowing that -1/8 MUST be in the set, by the rule of formation of the set, (-1/8)ˆ3 and (-1/8)ˆ2 MUST also be in the set.. and so on.

imhimanshu wrote:
Doubt 2-
I am hopeful that I lack some understanding the above solution will most probably be correct. In that case, I want to understand what will be maximum no of elements in this set. Will this be an Infinite set.

Given that 1/2 (different from 0) is part of the set, then Yes, this is an infinite set!

imhimanshu wrote:
Doubt 3-
Will the answer behaved differently had this question been "Could be true"
[/b]

"could be true" problems try to play on POSSIBILITIES when a lack of some sort of restriction allows the items to be true, somehow. If the question was posed as "could be true", then ANY real number could be a part of the set. Think of it this way...

Could a real number X be a part of the set? Yes, if cbrt(-x) is part of the set (hypothesis), then x COULD be part of the set.

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

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19 Apr 2013, 12:50
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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.

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.

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

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19 Apr 2013, 13:55
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Bunuel wrote:
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 liters

Let the rate of the draining pipe be $$x$$ liters per hour. Then the capacity of the tank will be $$C=time*rate=4x$$;

Now, when raining, the net outflow is x-3 liters per hour, and we are told that at this new rate the pool is emptied in 6 hours. So, the capacity of the pool also equals to $$C=time*rate=6(x-3)$$;

Thus we have: $$4x=6(x-3)$$ --> $$x=9$$ --> $$C=4x=36$$.

Bunuel,

Here is how I solved the problem. I must admit though I got it incorrect the first time I tried and when the answer choices didn't showed my answer I gave it another shot.

Let x be the capacity of the pool. The rate at which the pipe drains the pool is x/4.

Now on the rainy day the pull is already full and its raining at the rate of 3 lit/hr. The total inflow of rain in 6 hrs is 6*3 = 18 lit. So the pipe must drain x+18 lit in 6 hrs. The new rate is (x+18)/6 and we are given that these two rates are equal. Hence

x/4 = (x+18)/6 --> solving for x gives 36lits

Is this approach correct. I must say this didn't strike the first time I tried.
_________________

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Re: New Set of Mixed Questions!!!   [#permalink] 19 Apr 2013, 13:55

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