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

It is currently 12 Dec 2018, 04:21

Close

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
Your Progress

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.

Close

Request Expert Reply

Confirm Cancel
Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
  • The winning strategy for 700+ on the GMAT

     December 13, 2018

     December 13, 2018

     08:00 AM PST

     09:00 AM PST

    What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
  • GMATbuster's Weekly GMAT Quant Quiz, Tomorrow, Saturday at 9 AM PST

     December 14, 2018

     December 14, 2018

     09:00 AM PST

     10:00 AM PST

    10 Questions will be posted on the forum and we will post a reply in this Topic with a link to each question. There are prizes for the winners.

Each of the cards in a deck of 50 cards has a number from 1 to 20

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

Hide Tags

Manager
Manager
User avatar
G
Joined: 22 Jun 2016
Posts: 245
Reviews Badge
Each of the cards in a deck of 50 cards has a number from 1 to 20  [#permalink]

Show Tags

New post 25 Jun 2016, 04:23
3
2
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

51% (02:18) correct 49% (02:34) wrong based on 142 sessions

HideShow timer Statistics

Each of the cards in a deck of 50 cards has a number from 1 to 20 written on it in either black, red or blue ink. If one card is to be selected at random from the deck, what is the probability that the card selected will either have an odd number or be written in red ink?

(1) The probability that the card will both have an off number and be written in red ink is 0.

(2) The probability that the card will have an odd number minus the probability that the card will be written in red ink is 0.4



P.S. Give a kudo or bookmark the question if you are unable to solve it and I will send you the solution in your inbox :)

_________________

P.S. Don't forget to give Kudos :)

Manager
Manager
User avatar
G
Joined: 22 Jun 2016
Posts: 245
Reviews Badge
Re: Each of the cards in a deck of 50 cards has a number from 1 to 20  [#permalink]

Show Tags

New post 26 Jun 2016, 04:56
3
For those of you who were not able to answer this, here is the answer.
We know there are cards written with numbers in different probability colours. We are asked for the probability of randomly selecting a card that has either an odd or is written in red. Let us assume this prob to be x.

So, x= (prob of selecting a card with an odd number) + (prob of selecting a card written in red) - (prob of selecting a card with an odd number written in red)
or we can write the above as: x = P(odd) + P(red) - P(odd and red).

Statement 1: P(odd and red) = 0

Insufficient as we don't know P(odd) + P(red).

Statement 2: P(odd) - P(red) = 0.4

This can be true for various combination of P(odd) and P(red), for ex. P(odd)=0.5 and P(red)=0.1 or P(odd)=0.6 and P(red)=0.2 etc.

Hence, statement 2 is also insufficient!

Combining 1 and 2 also we do not have exact value of P(odd) and P(red). We just have values for P(odd)-P(red) and P(odd and red).

Therefore, the answer will be E.

--------------------------------------

P.S. Don't forget to give kudos :)
_________________

P.S. Don't forget to give Kudos :)

Manager
Manager
User avatar
Joined: 14 May 2014
Posts: 60
Location: India
GMAT 1: 680 Q49 V31
GPA: 3.44
Re: Each of the cards in a deck of 50 cards has a number from 1 to 20  [#permalink]

Show Tags

New post 26 Jun 2016, 07:07
14101992 wrote:
For those of you who were not able to answer this, here is the answer.
We know there are cards written with numbers in different probability colours. We are asked for the probability of randomly selecting a card that has either an odd or is written in red. Let us assume this prob to be x.

So, x= (prob of selecting a card with an odd number) + (prob of selecting a card written in red) - (prob of selecting a card with an odd number written in red)
or we can write the above as: x = P(odd) + P(red) - P(odd and red).

Statement 1: P(odd and red) = 0

Insufficient as we don't know P(odd) + P(red).

Statement 2: P(odd) - P(red) = 0.4

This can be true for various combination of P(odd) and P(red), for ex. P(odd)=0.5 and P(red)=0.1 or P(odd)=0.6 and P(red)=0.2 etc.

Hence, statement 2 is also insufficient!

Combining 1 and 2 also we do not have exact value of P(odd) and P(red). We just have values for P(odd)-P(red) and P(odd and red).

Therefore, the answer will be E.

--------------------------------------

P.S. Don't forget to give kudos :)



Quite not understood the logic
x = P(odd) + P(red) - P(odd and red).
1) says P(odd and red)=0
so x = P(odd) + P(red) - 0= x = P(odd) + P(red)

2) says
P(odd)- P(Red)=0.4
P(odd)= 0.5 so P(red)=0.1

combine 1 and 2

x= 0.5+0.1=0.6

are we not getting a value here?
Current Student
User avatar
Joined: 18 Oct 2014
Posts: 846
Location: United States
GMAT 1: 660 Q49 V31
GPA: 3.98
GMAT ToolKit User
Re: Each of the cards in a deck of 50 cards has a number from 1 to 20  [#permalink]

Show Tags

New post 26 Jun 2016, 09:03
RatneshS wrote:
14101992 wrote:
For those of you who were not able to answer this, here is the answer.
We know there are cards written with numbers in different probability colours. We are asked for the probability of randomly selecting a card that has either an odd or is written in red. Let us assume this prob to be x.

So, x= (prob of selecting a card with an odd number) + (prob of selecting a card written in red) - (prob of selecting a card with an odd number written in red)
or we can write the above as: x = P(odd) + P(red) - P(odd and red).

Statement 1: P(odd and red) = 0

Insufficient as we don't know P(odd) + P(red).

Statement 2: P(odd) - P(red) = 0.4

This can be true for various combination of P(odd) and P(red), for ex. P(odd)=0.5 and P(red)=0.1 or P(odd)=0.6 and P(red)=0.2 etc.

Hence, statement 2 is also insufficient!

Combining 1 and 2 also we do not have exact value of P(odd) and P(red). We just have values for P(odd)-P(red) and P(odd and red).

Therefore, the answer will be E.

--------------------------------------

P.S. Don't forget to give kudos :)



Quite not understood the logic
x = P(odd) + P(red) - P(odd and red).
1) says P(odd and red)=0
so x = P(odd) + P(red) - 0= x = P(odd) + P(red)

2) says
P(odd)- P(Red)=0.4
P(odd)= 0.5 so P(red)=0.1

combine 1 and 2

x= 0.5+0.1=0.6

are we not getting a value here?


Hi! RatneshS,

Statement 2 can have many combinations:-
P(odd)- P(Red)=0.4
P(odd)= 0.5 so P(red)=0.1
or P(odd)= 0.2 so P(red)=0.2 and so on.

We are not getting a unique value by combining both statement.
_________________

I welcome critical analysis of my post!! That will help me reach 700+

Intern
Intern
avatar
B
Joined: 05 May 2017
Posts: 1
Re: Each of the cards in a deck of 50 cards has a number from 1 to 20  [#permalink]

Show Tags

New post 22 May 2017, 10:14
Divyadisha wrote:
RatneshS wrote:
14101992 wrote:
For those of you who were not able to answer this, here is the answer.
We know there are cards written with numbers in different probability colours. We are asked for the probability of randomly selecting a card that has either an odd or is written in red. Let us assume this prob to be x.

So, x= (prob of selecting a card with an odd number) + (prob of selecting a card written in red) - (prob of selecting a card with an odd number written in red)
or we can write the above as: x = P(odd) + P(red) - P(odd and red).

Statement 1: P(odd and red) = 0

Insufficient as we don't know P(odd) + P(red).

Statement 2: P(odd) - P(red) = 0.4

This can be true for various combination of P(odd) and P(red), for ex. P(odd)=0.5 and P(red)=0.1 or P(odd)=0.6 and P(red)=0.2 etc.

Hence, statement 2 is also insufficient!

Combining 1 and 2 also we do not have exact value of P(odd) and P(red). We just have values for P(odd)-P(red) and P(odd and red).

Therefore, the answer will be E.

--------------------------------------

P.S. Don't forget to give kudos :)



Quite not understood the logic
x = P(odd) + P(red) - P(odd and red).
1) says P(odd and red)=0
so x = P(odd) + P(red) - 0= x = P(odd) + P(red)

2) says
P(odd)- P(Red)=0.4
P(odd)= 0.5 so P(red)=0.1

combine 1 and 2

x= 0.5+0.1=0.6

are we not getting a value here?


Hi! RatneshS,

Statement 2 can have many combinations:-
P(odd)- P(Red)=0.4
P(odd)= 0.5 so P(red)=0.1
or P(odd)= 0.2 so P(red)=0.2 and so on.

We are not getting a unique value by combining both statement.


Odd numbers between 1 and 20 is 10. So P(odd) = 10/50= 0.2
Am I doing something wrong?
Intern
Intern
User avatar
B
Joined: 01 Jan 2018
Posts: 44
Re: Each of the cards in a deck of 50 cards has a number from 1 to 20  [#permalink]

Show Tags

New post 24 Apr 2018, 04:21
Hi,

Can someone explain this question in a simpler way? I'm sorry, I'm not able to understand the logic or correct me if I'm making some wrong assumptions.

Statement 1 says:

The probability that the card will both have an odd number and be written in red ink is 0.

Which means that the red cards are all even. Had there been one red card, which is odd then the probability wouldn't be 0.

So my understanding is that, all the red cards are even. So 10 cards.
then in the remaining 40 cards it'd be 20 blue cards of even and odd. 20 black cards of even and odd.

The probability of odd cards is 30/50 and probability of red card is 10/50.

So is the final answer 4/5? and is statement 1 not sufficient to answer?

:dream:
Manager
Manager
User avatar
S
Joined: 23 May 2017
Posts: 241
Concentration: Finance, Accounting
WE: Programming (Energy and Utilities)
Re: Each of the cards in a deck of 50 cards has a number from 1 to 20  [#permalink]

Show Tags

New post 24 Apr 2018, 04:50
Each of the cards in a deck of 50 cards has a number from 1 to 20 written on it in either black, red or blue ink. If one card is to be selected at random from the deck, what is the probability that the card selected will either have an odd number or be written in red ink?

(1) The probability that the card will both have an off number and be written in red ink is 0.
P(O) . P(R) = 0

(2) The probability that the card will have an odd number minus the probability that the card will be written in red ink is 0.4
P(O) - P(R) = 0.4

Clearly 1 and 2 individually not sufficient.
combining both

P(O) - P(R) = 0.4 : Squaring both sides we get
\([P(O)]^2\) + \([P(R)]^2\) - 2\(P(O)\).\(P(R)\) = \(0.16\) from statement 2 we know P(O).P(R) = 0

hence
\([P(O)]^2\) + \([P(R)]^2\) = \(0.16\) = re-writing the LHS

\([ P(O) + P(R) ]^2\) -2\(P(O)\).\(P(R)\) = \(0.16\)

\(P(O).P(R) = 0\) from statement 2

\([ P(O) + P(R) ]^2\) = \(0.16\)

\([ P(O) + P(R) ]\) = \(0.4\) - which is a definite answer


What is wrong with my approach?
_________________

If you like the post, please award me Kudos!! It motivates me

Intern
Intern
avatar
B
Joined: 23 Jan 2017
Posts: 5
Location: United States
Schools: HBS '21 (S)
GMAT 1: 710 Q48 V39
GPA: 2.9
Re: Each of the cards in a deck of 50 cards has a number from 1 to 20  [#permalink]

Show Tags

New post 07 May 2018, 22:30
14101992 wrote:
For those of you who were not able to answer this, here is the answer.
We know there are cards written with numbers in different probability colours. We are asked for the probability of randomly selecting a card that has either an odd or is written in red. Let us assume this prob to be x.

So, x= (prob of selecting a card with an odd number) + (prob of selecting a card written in red) - (prob of selecting a card with an odd number written in red)
or we can write the above as: x = P(odd) + P(red) - P(odd and red).

Statement 1: P(odd and red) = 0

Insufficient as we don't know P(odd) + P(red).

Statement 2: P(odd) - P(red) = 0.4

This can be true for various combination of P(odd) and P(red), for ex. P(odd)=0.5 and P(red)=0.1 or P(odd)=0.6 and P(red)=0.2 etc.

Hence, statement 2 is also insufficient!

Combining 1 and 2 also we do not have exact value of P(odd) and P(red). We just have values for P(odd)-P(red) and P(odd and red).

Therefore, the answer will be E.

--------------------------------------

P.S. Don't forget to give kudos :)



Correct me if I'm wrong:

P(odd) - P(red) = 0.4
squaring both sides:
[P(odd) - P(red)]^2 = 0.4^2
P(odd)^2 + P(red)^2 -2P(odd)xP(red) = 0.4^2
Since P(odd and red) = 0 (from statement 1): 2P(odd)xP(red) = 0
P(odd)^2 + P(red)^2 = 0.4^2

Now we have to calculate P(odd) + P(red)
squaring this expression we get:
P(odd)^2 + P(red)^2 + 2P(odd)xP(red)
Since P(odd and red) = 0 (from statement 1): 2P(odd)xP(red) = 0; we are left with
P(odd)^2 + P(red)^2
which we have proved above to be 0.4^2
Intern
Intern
avatar
B
Joined: 04 Feb 2018
Posts: 5
GMAT 1: 690 Q42 V41
Re: Each of the cards in a deck of 50 cards has a number from 1 to 20  [#permalink]

Show Tags

New post 25 Jun 2018, 17:42
14101992 wrote:
For those of you who were not able to answer this, here is the answer.
We know there are cards written with numbers in different probability colours. We are asked for the probability of randomly selecting a card that has either an odd or is written in red. Let us assume this prob to be x.

So, x= (prob of selecting a card with an odd number) + (prob of selecting a card written in red) - (prob of selecting a card with an odd number written in red)
or we can write the above as: x = P(odd) + P(red) - P(odd and red).

Statement 1: P(odd and red) = 0

Insufficient as we don't know P(odd) + P(red).

Statement 2: P(odd) - P(red) = 0.4

This can be true for various combination of P(odd) and P(red), for ex. P(odd)=0.5 and P(red)=0.1 or P(odd)=0.6 and P(red)=0.2 etc.

Hence, statement 2 is also insufficient!

Combining 1 and 2 also we do not have exact value of P(odd) and P(red). We just have values for P(odd)-P(red) and P(odd and red).

Therefore, the answer will be E.

--------------------------------------

P.S. Don't forget to give kudos :)





I agree with the rest of the comments.
What you want is P(odd n red) = P(odd) + P(red) - P(odd u red) [note: n stands for intersect, u stands for union]
Statement 1 clearly tells us that P(odd u red) is 0. So we need to know what P(odd) + P(red) is.

In statement 2, I think you meant to to exploit that some people wouldn't know the actual formula. There are several other questions like this, so it's not a bad question.
But if you randomly select a card that must be either even or odd, and we are given a finite number of evens and odds, I can tell you that P(odd)=(.5)
From there:
P(odd)- P(red) = .4
.5 - P(red) = .4
-P(red) = -.1
P(red) = .1

Thus P(red n odd) = .5 + .1 - 0 = .6
There is only one answer, so C is correct.
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 7102
Re: Each of the cards in a deck of 50 cards has a number from 1 to 20  [#permalink]

Show Tags

New post 25 Jun 2018, 18:26
1
14101992 wrote:
Each of the cards in a deck of 50 cards has a number from 1 to 20 written on it in either black, red or blue ink. If one card is to be selected at random from the deck, what is the probability that the card selected will either have an odd number or be written in red ink?

(1) The probability that the card will both have an off number and be written in red ink is 0.

(2) The probability that the card will have an odd number minus the probability that the card will be written in red ink is 0.4



P.S. Give a kudo or bookmark the question if you are unable to solve it and I will send you the solution in your inbox :)



Hi...

Lot of people going wrong ....
Reason :- Assuming something that is not given.
It is not given that even and odd are equal in number. It may just be possible that all are odd or all are even.
Don't get into P(O), P(R) etc, the meaning should get you to the correct answer.

(1) The probability that the card will both have an odd number and be written in red ink is 0.
It just tells us that there is no overlap and thus no odd number is written in red ink
Insufficient

(2) The probability that the card will have an odd number minus the probability that the card will be written in red ink is 0.4
So (ODD cards)/Total - (Red cards)/total =0.4
So O/50 - R/50 = 0.4
O-R=0.4*20=8
O can be 10 and R-2 or O-50 & R-42 etc
Insufficient

Combined
Nothing new..
Now the values can be O-10 , R-2 etc
Only that O +R cannot be >50
Insufficient
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


GMAT online Tutor

Intern
Intern
avatar
B
Joined: 04 Feb 2018
Posts: 5
GMAT 1: 690 Q42 V41
Re: Each of the cards in a deck of 50 cards has a number from 1 to 20  [#permalink]

Show Tags

New post 25 Jun 2018, 18:48
chetan2u wrote:
14101992 wrote:
Each of the cards in a deck of 50 cards has a number from 1 to 20 written on it in either black, red or blue ink. If one card is to be selected at random from the deck, what is the probability that the card selected will either have an odd number or be written in red ink?

(1) The probability that the card will both have an off number and be written in red ink is 0.

(2) The probability that the card will have an odd number minus the probability that the card will be written in red ink is 0.4



P.S. Give a kudo or bookmark the question if you are unable to solve it and I will send you the solution in your inbox :)



Hi...

Lot of people going wrong ....
Reason :- Assuming something that is not given.
It is not given that even and odd are equal in number. It may just be possible that all are odd or all are even.
Don't get into P(O), P(R) etc, the meaning should get you to the correct answer.

(1) The probability that the card will both have an odd number and be written in red ink is 0.
It just tells us that there is no overlap and thus no odd number is written in red ink
Insufficient

(2) The probability that the card will have an odd number minus the probability that the card will be written in red ink is 0.4
So (ODD cards)/Total - (Red cards)/total =0.4
So O/50 - R/50 = 0.4
O-R=0.4*20=8
O can be 10 and R-2 or O-50 & R-42 etc
Insufficient

Combined
Nothing new..
Now the values can be O-10 , R-2 etc
Only that O +R cannot be >50
Insufficient


Oh wow, my apologies to the author.
Intern
Intern
avatar
B
Joined: 21 May 2017
Posts: 42
Re: Each of the cards in a deck of 50 cards has a number from 1 to 20  [#permalink]

Show Tags

New post 25 Jun 2018, 23:50
1
Leo8 wrote:
Each of the cards in a deck of 50 cards has a number from 1 to 20 written on it in either black, red or blue ink. If one card is to be selected at random from the deck, what is the probability that the card selected will either have an odd number or be written in red ink?

(1) The probability that the card will both have an off number and be written in red ink is 0.
P(O) . P(R) = 0

(2) The probability that the card will have an odd number minus the probability that the card will be written in red ink is 0.4
P(O) - P(R) = 0.4

Clearly 1 and 2 individually not sufficient.
combining both

P(O) - P(R) = 0.4 : Squaring both sides we get
\([P(O)]^2\) + \([P(R)]^2\) - 2\(P(O)\).\(P(R)\) = \(0.16\) from statement 2 we know P(O).P(R) = 0

hence
\([P(O)]^2\) + \([P(R)]^2\) = \(0.16\) = re-writing the LHS

\([ P(O) + P(R) ]^2\) -2\(P(O)\).\(P(R)\) = \(0.16\)

\(P(O).P(R) = 0\) from statement 2

\([ P(O) + P(R) ]^2\) = \(0.16\)

\([ P(O) + P(R) ]\) = \(0.4\) - which is a definite answer


What is wrong with my approach?
The probability that the card will both have an odd number and be written in red ink is 0.

This statement does not mean P(O).P(R) = 0.
This statement means that P(O|R) = 0.
It means that there is no odd numbered card with red ink.
P(O|R) = P(O intersection R)/P(R)
O interesection R = 0

Let's take an example.
Say, there are 10 items - 5 blue marbles and 5 red chairs. Let's pick a random item now.
Probability of picking a blue object = P(B) = 1/2
Probability of picking a chair = P(C) = 1/2
Now, probability of picking a blue chair is not equal to P(B).P(C). That would give us 1/4, whereas, the probability of picking a blue chair is ZERO.

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app
GMAT Club Bot
Re: Each of the cards in a deck of 50 cards has a number from 1 to 20 &nbs [#permalink] 25 Jun 2018, 23:50
Display posts from previous: Sort by

Each of the cards in a deck of 50 cards has a number from 1 to 20

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


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.