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

It is currently 17 Oct 2018, 15:38

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

TOUGH & TRICKY SET Of PROBLEMS

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

Hide Tags

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49968
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 15 Jan 2011, 14:13
yogesh1984 wrote:
Bunuel wrote:
4. ADDITION PROBLEM:
AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1
(B) 3
(C) 7
(D) 9
(E) Cannot be determined

AB and CD are two digit integers, their sum can give us only one three digit integer of a kind of AAA it's 111.
So, A=1. 1B+CD=111
C can not be less than 9, because no to digit integer with first digit 1 (mean that it's<20) can be added to two digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111).
C=9

Answer: D.

Bunuel I have doubt here-- consider following AB as 25 CD 86 which sums to 111 and ||ly
if we consider AB as 15 & CD as 96 then also we end up with 111 so I am doubtfull with
the answer provided above....(please do correct if Am missing something :| )


SOLUTION:
AB and CD are two digit integers, their sum can give us only one 3-igit integer of a kind of AAA: 111. So, A=1 --> 1B+CD=111.

C can not be less than 9, because no 2-digit integer with first digit 1 (mean that it's <20) can be added to 2-digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111) --> C=9.

Answer: D.

So, as A=1 then AB can not be 25.

Hope it's clear.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Senior Manager
Senior Manager
avatar
Joined: 21 Mar 2010
Posts: 280
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 20 Feb 2011, 17:24
Bunuel wrote:
SOLUTION:
2. THE PRICE OF BUSHEL:
The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of \(5x\) cents per day while the price of wheat is decreasing at a constant rate of \(\sqrt{2}*x-x\) cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?

(A) $4.50
(B) $5.10
(C) $5.30
(D) $5.50
(E) $5.60

Note that we are not asked in how many days prices will cost the same.

Let \(y\) be the # of days when these two bushels will have the same price.

First let's simplify the formula given for the rate of decrease of the price of wheat: \(\sqrt{2}*x-x=1.41x-x=0.41x\), this means that the price of wheat decreases by \(0.41x\) cents per day, in \(y\) days it'll decrease by \(0.41xy\) cents;

As price of corn increases \(5x\) cents per day, in \(y\) days it'll will increase by \(5xy\) cents;

Set the equation: \(320+5xy=580-0.41xy\), solve for \(xy\) --> \(xy=48\);

The cost of a bushel of corn in \(y\) days (the # of days when these two bushels will have the same price) will be \(320+5xy=320+5*48=560\) or $5.6.

Answer: E.



Bunuel,

While solving for xy i get xy = 56

5xy - 0.41xy = 580-320
4.59xy = 260
xy = 56.6 cents

Let me know if i missed something!
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49968
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 20 Feb 2011, 17:38
mbafall2011 wrote:
Bunuel wrote:
SOLUTION:
2. THE PRICE OF BUSHEL:
The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of \(5x\) cents per day while the price of wheat is decreasing at a constant rate of \(\sqrt{2}*x-x\) cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?

(A) $4.50
(B) $5.10
(C) $5.30
(D) $5.50
(E) $5.60

Note that we are not asked in how many days prices will cost the same.

Let \(y\) be the # of days when these two bushels will have the same price.

First let's simplify the formula given for the rate of decrease of the price of wheat: \(\sqrt{2}*x-x=1.41x-x=0.41x\), this means that the price of wheat decreases by \(0.41x\) cents per day, in \(y\) days it'll decrease by \(0.41xy\) cents;

As price of corn increases \(5x\) cents per day, in \(y\) days it'll will increase by \(5xy\) cents;

Set the equation: \(320+5xy=580-0.41xy\), solve for \(xy\) --> \(xy=48\);

The cost of a bushel of corn in \(y\) days (the # of days when these two bushels will have the same price) will be \(320+5xy=320+5*48=560\) or $5.6.

Answer: E.



Bunuel,

While solving for xy i get xy = 56

5xy - 0.41xy = 580-320
4.59xy = 260
xy = 56.6 cents

Let me know if i missed something!


From 320+5xy=580-0.41xy --> 5xy+0.41xy=580-320 --> xy=48.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 06 May 2011
Posts: 10
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 05 Sep 2011, 06:26
fluke wrote:
The following answers may not be correct. If you find some discrepancy in the solution, please PM me or post it under the same topic. I took a cursory look into the OA's but am yet to match all my answers and compare the logic.
Bunuel wrote:
5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a

minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20


Let A's speed: A m/s
Let B's speed: B m/s

1/10th of a minute = 6sec
1/30th of a minute = 2sec

Total distance of the race track = 480m

First heat;
A's distance = 480m
A's speed = A m/s
A's time = t secs

B's distance = 480-48=432m
B's speed = B m/s
B's time = t+6 secs

A's time = B's time - 6 (if A took 1000 seconds; B took 1006 seconds)
\(\frac{480}{A}=\frac{432}{B}-6\) ------ 1st (\(rate*time=distance \quad or \quad time,t=D/r\))

Second heat;
A's distance = 480m
A's speed = A m/s
A's time = t secs

B's distance = 480-144=336m
B's speed = B m/s
B's time = t-2 secs

A's time = B's time + 2
\(\frac{480}{A}=\frac{336}{B}+2\) ------ 2nd

Using 1st and 2nd,
\(\frac{432}{B}-6=\frac{336}{B}+2\)

\(\frac{432}{B}-\frac{336}{B}=8\)

\(\frac{96}{B}=8\)

\(B=12\)

B's speed : 12m/s

Ans: "A"




I dont have any Question on calculations made above.
But i do have a clarification to ask.

While calculating the time of B , it appears that we have not accounted the time taken by B to cover the first 48m(first heat) or 144m (second heat).
( As if the start time taken into consideration is the point from when both A and B are in action ).
Why is that we are going by this. I mean why are we not considering the head start time provided to B.

Please correct me if i missed anything.
Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8386
Location: Pune, India
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 05 Sep 2011, 10:55
vinodmallapu wrote:
fluke wrote:
The following answers may not be correct. If you find some discrepancy in the solution, please PM me or post it under the same topic. I took a cursory look into the OA's but am yet to match all my answers and compare the logic.
Bunuel wrote:
5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a

minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20


Let A's speed: A m/s
Let B's speed: B m/s

1/10th of a minute = 6sec
1/30th of a minute = 2sec

Total distance of the race track = 480m

First heat;
A's distance = 480m
A's speed = A m/s
A's time = t secs

B's distance = 480-48=432m
B's speed = B m/s
B's time = t+6 secs

A's time = B's time - 6 (if A took 1000 seconds; B took 1006 seconds)
\(\frac{480}{A}=\frac{432}{B}-6\) ------ 1st (\(rate*time=distance \quad or \quad time,t=D/r\))

Second heat;
A's distance = 480m
A's speed = A m/s
A's time = t secs

B's distance = 480-144=336m
B's speed = B m/s
B's time = t-2 secs

A's time = B's time + 2
\(\frac{480}{A}=\frac{336}{B}+2\) ------ 2nd

Using 1st and 2nd,
\(\frac{432}{B}-6=\frac{336}{B}+2\)

\(\frac{432}{B}-\frac{336}{B}=8\)

\(\frac{96}{B}=8\)

\(B=12\)

B's speed : 12m/s

Ans: "A"




I dont have any Question on calculations made above.
But i do have a clarification to ask.

While calculating the time of B , it appears that we have not accounted the time taken by B to cover the first 48m(first heat) or 144m (second heat).
( As if the start time taken into consideration is the point from when both A and B are in action ).
Why is that we are going by this. I mean why are we not considering the head start time provided to B.

Please correct me if i missed anything.


When A gives B a head start of 48 m, it means B starts from 48 m ahead of the start line while A starts from the start line. They both start the race at the same time and run till A reaches the finish line. Thereafter, only B runs for 1/10th of a minute.
There is no head start time provided to B, only head start distance.
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8386
Location: Pune, India
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 05 Sep 2011, 11:04
2
Bunuel wrote:
5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a

minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20


Alternative Approach here:

A starts from the start line and runs till the finish line in both the races. He would take the same time in both the cases then. On the other hand, in the first race, B takes 6 secs more than A while in the second race, B takes 2 secs less than A.
So there is a time difference of 8 secs in the time taken by B in the two cases. B travels (144 - 48 =) 96 m less in the second race and taken 8 secs less in the second race. This means that in the first race, B runs a distance of 96m in 8 secs.
So speed of B = 96/8 = 12 m/sec
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49968
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 27 Feb 2012, 09:11
flokki wrote:
Bunuel wrote:
SOLUTION OF 8-10
8. THE AVERAGE TEMPERATURE:
The average of temperatures at noontime from Monday to Friday is 50; the lowest one is 45, what is the possible maximum range of the temperatures?

A. 20
B. 25
C. 40
D. 45
E. 75

Average=50, Sum of temperatures=50*5=250
As the min temperature is 45, max would be 250-4*45=70 --> The range=70(max)-45(min)=25

Answer: B.




Isnt the correct answer A. it says 45 is the lowest temperature. Doest that statement imply that all other numbers have to be greater than 45. Hence at least 45,00001?


No, it doesn't. There can be more than 1 temperatures equal to 45.

COMPLETE SOLUTION:

Given: T(min)=45 and average=50 --> sum of temperatures=50*5=250.

We want to maximize the range --> in order to maximize the range we need to maximize the highest temperature --> in order to maximize the highest temperature we should make all other temperatures as low as possible, so equal to 45 (lower limit) --> T(max)=250-4*45=70 --> Range=T(max)-T(min)=70-45=25.

Answer: B.

Hope it's clear.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Senior Manager
Senior Manager
avatar
Joined: 07 Sep 2010
Posts: 266
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 27 Mar 2012, 07:11
Hi Bunuel,
I understand the below solution. However, I was wondering if it could be solved using combination approach. Please tell me where I am going wrong.

Probability = favorable/ total outcomes.
Total outcomes = 9C2 (i.e total ways if drawing 2 balls)
Favorable = 3C2*4C1
Hence Probability = 3C2*4C1/9C2 = 1/6. which is completely wrong.
Please reply.
Thanks
H


Quote:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):



We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49968
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 27 Mar 2012, 09:08
imhimanshu wrote:
Hi Bunuel,
I understand the below solution. However, I was wondering if it could be solved using combination approach. Please tell me where I am going wrong.

Probability = favorable/ total outcomes.
Total outcomes = 9C2 (i.e total ways if drawing 2 balls)
Favorable = 3C2*4C1
Hence Probability = 3C2*4C1/9C2 = 1/6. which is completely wrong.
Please reply.
Thanks
H


Quote:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):



We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.


Since we have the case with replacement then we can not use \(C^2_9\) for it, because it gives total # of ways to select 2 different balls out of 9 without replacement.

If you wan to solve this question with combinations approach you still should consider two scenarios: \(P(RW)=2*\frac{C^1_3*C^1_2}{C^1_9*C^1_9}=\frac{4}{27}\), we are multiplying by 2 for the same reason: there are two possible wining scenarios RW and WR..

Hope it's clear.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49968
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 03 Apr 2012, 10:04
theamwan wrote:
Bunuel wrote:
SOLUTION:
3. LEAP YEAR:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

Probability of a randomly selected person NOT to be born in a leap year=3/4
Among 2 people, probability that none of them was born in a leap = 3/4*3/4=9/16. The probability at least one born in leap = 1- 9/16=7/16<1/2
So, we are looking for such n (# of people), when 1-(3/4)^n>1/2
n=3 --> 1-27/64=37/64>1/2

Thus min 3 people are needed.

Answer: C.



Hi Bunuel - Regarding this question, I believe the answer should be 2 i.e. (B).

It is because the probability of a person's birth year to be a leap yr = 1/2, this is identical to a rainy/non-rainy day situation, where a year is either a leap or normal year.


Case I - No of people = 1
Probability of atleast 1 person's birth yr falling on a leap year = 1/2 = 50%
.....INCORRECT

Case II - No of people = 2
Probability of none of guys' birth year falling on a leap year = 1/2*1/2 = 1/4
Probability of atleast one of them was born on a leap year = 1 - 1/4 = 3/4 = 75%
.....CORRECT

Hence (B).

What do you think? Thanks!


That's not correct out of 4 years 1 is leap, so the probability of a randomly selected person NOT to be born in a leap year is 3/4.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 25 Jan 2012
Posts: 1
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 03 Apr 2012, 10:17
Bunuel wrote:

That's not correct out of 4 years 1 is leap, so the probability of a randomly selected person NOT to be born in a leap year is 3/4.


But the question doesn't say that it's a set of 4 consecutive years. What if the years chosen are 2011, 2010, 2009, 2007?
IMO, a person's birth year to fall on a leap year should be a binary value - Y or N, hence a 50% probability.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49968
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 03 Apr 2012, 10:32
1
theamwan wrote:
Bunuel wrote:

That's not correct out of 4 years 1 is leap, so the probability of a randomly selected person NOT to be born in a leap year is 3/4.


But the question doesn't say that it's a set of 4 consecutive years. What if the years chosen are 2011, 2010, 2009, 2007?
IMO, a person's birth year to fall on a leap year should be a binary value - Y or N, hence a 50% probability.


Again that's not correct. Let me ask you a question: what is the probability that a person in born on Monday? Is it 1/2? No, it's 1/7.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 25 Jun 2011
Posts: 37
Location: Sydney
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 29 Jun 2012, 20:10
Bunuel wrote:
SOLUTION:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):

\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.


The question says that red and white balls are selected in two Successive draws. Doesn't this imply that white is selected AFTER red? Thus no need for x2?

Thanks,
Diana
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49968
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 30 Jun 2012, 04:17
1
dianamao wrote:
Bunuel wrote:
SOLUTION:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):

\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.


The question says that red and white balls are selected in two Successive draws. Doesn't this imply that white is selected AFTER red? Thus no need for x2?

Thanks,
Diana


No, in that case we would be asked "what is the the probability of the first ball being red and the second ball being white?"
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49968
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 10 Oct 2012, 12:05
Current Student
avatar
Joined: 12 Jan 2012
Posts: 23
GMAT 1: 720 Q49 V39
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 16 Dec 2012, 22:30
Bunuel wrote:
SOLUTION:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):

\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.


Hi Bunuel

I guess the question should be re-worded to restrict the number of draws to 2.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49968
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 16 Dec 2012, 23:08
geneticsgene wrote:
Bunuel wrote:
SOLUTION:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):

\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.


Hi Bunuel

I guess the question should be re-worded to restrict the number of draws to 2.


I think we are given that: "What is the probability of drawing a red and a white ball in two successive draws..."
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Director
Director
avatar
Status: Final Lap Up!!!
Affiliations: NYK Line
Joined: 21 Sep 2012
Posts: 957
Location: India
GMAT 1: 410 Q35 V11
GMAT 2: 530 Q44 V20
GMAT 3: 630 Q45 V31
GPA: 3.84
WE: Engineering (Transportation)
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 20 Dec 2012, 16:26
I think i made mistake since 6 can be also a probable number but is not a multiple of 4..............Bunuel can u pls help me solve this one using multiple principle
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49968
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 21 Dec 2012, 04:40
1
Archit143 wrote:
I think i made mistake since 6 can be also a probable number but is not a multiple of 4..............Bunuel can u pls help me solve this one using multiple principle


If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 25%
B. 50%
C. 62.5%
D. 72.5%
E. 75%

\(n(n + 1)(n + 2)\) is divisible by 8 in two cases:

A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. \(n+1\) is itself divisible by 8;

(Notice that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.)

Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8=0.625.

Answer: C.

Similar question: divisible-by-12-probability-121561.html

Hope it helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 29 Oct 2012
Posts: 4
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

Show Tags

New post 25 Jan 2013, 01:22
Bunuel wrote:
angel2009 wrote:
Bunuel wrote:
SOLUTION:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):

P=2*3/9*2/9=4/27

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.


I've some problem to understand this ; as the two events are mutually exclusive , independent.
So why not the answer is 3/9*2/9 = 2/27


Winning scenario consists of TWO cases RW and WR. Probability of each case is: 3/9*2/9, so 3/9*2/9+3/9*2/9=2*3/9*2/9.

Only 3/9*2/9 would be correct answer if we were asked to determine the probability of FIRST ball being red and SECOND white, OR FIRST white and SECOND red. These are the cases in which order of drawing matters.


Hey can you please help with the solution using combination formula

P(1R1W) = 3C1 x 2C1 / (9C2) = 1/6 WHY? what am i doing wrong?
GMAT Club Bot
Re: TOUGH & TRICKY SET Of PROBLEMS &nbs [#permalink] 25 Jan 2013, 01:22

Go to page   Previous    1   2   3   4    Next  [ 76 posts ] 

Display posts from previous: Sort by

TOUGH & TRICKY SET Of PROBLEMS

  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®.