GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Sep 2019, 11:30 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # New Set of Mixed Questions!!!

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 58142
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

8
25
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 36603 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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58142
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

5
25
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: $$4x=6(x-3)$$ --> $$x=9$$ --> $$C=4x=36$$.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58142
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

1
9
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).

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58142
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

2
3
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58142
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

3
9
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58142
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

4
23
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58142
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

3
1
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58142
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58142
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

10
26
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).

_________________
Senior Manager  Joined: 07 Sep 2010
Posts: 251
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

1
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
Manager  Joined: 02 Jan 2013
Posts: 55
GMAT 1: 750 Q51 V40 GPA: 3.2
WE: Consulting (Consulting)
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

1
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.

KUDOS +1 IF THIS HELPED YOU
Manager  Joined: 05 Nov 2012
Posts: 64
Concentration: International Business, Operations
Schools: Foster '15 (S)
GPA: 3.65
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

1
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.

Can you please clarify
_________________
___________________________________________
Consider +1 Kudos if my post helped
Manager  Joined: 05 Nov 2012
Posts: 64
Concentration: International Business, Operations
Schools: Foster '15 (S)
GPA: 3.65
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

1
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.
_________________
___________________________________________
Consider +1 Kudos if my post helped
Math Expert V
Joined: 02 Sep 2009
Posts: 58142
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

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.

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.
_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58142
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

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).

$$-x^2=-(\frac{1}{2})^2=-\frac{1}{4}$$ (its -x^2 not (-x)^2).

Hope it's clear.
_________________
Intern  Joined: 23 Jun 2013
Posts: 36
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

1
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.
Math Expert V
Joined: 02 Sep 2009
Posts: 58142
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

1
2
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?

Similar questions to practice:
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
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html

Hope this helps.
_________________
Intern  Joined: 08 Dec 2013
Posts: 19
Location: India
Concentration: General Management, Operations
GMAT 1: 710 Q50 V35 GMAT 2: 730 Q50 V38 GPA: 3.46
WE: Engineering (Manufacturing)
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

4
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.
Intern  Joined: 23 Jun 2013
Posts: 36
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58142
Re: New Set of Mixed Questions!!!  [#permalink]

### Show Tags

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.
_________________ Re: New Set of Mixed Questions!!!   [#permalink] 02 Jan 2014, 05:51

Go to page   Previous    1   2   3   4   5   6   7    Next  [ 138 posts ]

Display posts from previous: Sort by

# New Set of Mixed Questions!!!

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  