Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 15 Sep 2014, 01:52

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

TOUGH & TRICKY SET Of PROBLEMS

Author Message
TAGS:
Intern
Joined: 31 Aug 2009
Posts: 17
Schools: University of Warwick
Followers: 0

Kudos [?]: 18 [0], given: 3

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]  30 Dec 2009, 03:25
there is an errata on solution of 4. the answer requested is the value of C not CD.

Thus the answer is 9, option D

Megan

Bunuel wrote:
As asked I'm combining all the problems from Tough & tricky problems in single thread. Here are the first ten questions. Next set and solutions to these ten will follow in couple of hours.

1. THE SUM OF EVEN INTEGERS:
The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159

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 2^1/2*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 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 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 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 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 7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE: What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3? A) 1.4 B) sqrt (2) C) 1.7 D) sqrt (3) E) 2.0 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 9. PROBABILITY OF INTEGER BEING DIVISIBLE BY 8: If n is an integer from 1 to 96 (inclusive), what is the probability for n*(n+1)*(n+2) being divisible by 8? A. 25% B 50% C 62.5% D. 72.5% E. 75% 10. SUM OF INTEGERS: If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is: A. A+1 inquiry B. A+5 C A+25 D 2A E. 5A THE OA WITH SOLUTIONS WILL BE PROVIDED. Intern Joined: 31 Aug 2009 Posts: 17 Schools: University of Warwick Followers: 0 Kudos [?]: 18 [0], given: 3 Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] 30 Dec 2009, 03:42 The problem number 9 was very tough. I would appreciate if someone can provide an alternative explaination to it. I don't quite understand the one that is already given. Senior Manager Joined: 25 Jul 2009 Posts: 331 Followers: 1 Kudos [?]: 26 [0], given: 0 Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] 09 Jan 2010, 04:10 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*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 Math Expert Joined: 02 Sep 2009 Posts: 23538 Followers: 3482 Kudos [?]: 26116 [0], given: 2705 Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] 09 Jan 2010, 04:49 Expert's post 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. _________________ Manager Joined: 25 Dec 2009 Posts: 102 Followers: 1 Kudos [?]: 25 [0], given: 3 Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] 19 Jan 2010, 05:43 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 2^1/2*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 # of days when these two bushels will have the same price.
First 2^1/2*x-x=0.41x
3.2+5xy=5.8-0.41xy solving for xy=0.48
The cost of a bushel of corn=3.2+5*0.48=5.6

How do you make this Bunuel , I dont get it

Quote:
First 2^1/2*x-x=0.41x
Math Expert
Joined: 02 Sep 2009
Posts: 23538
Followers: 3482

Kudos [?]: 26116 [0], given: 2705

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]  19 Jan 2010, 06:19
Expert's post
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 2^1/2*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 # of days when these two bushels will have the same price. First 2^1/2*x-x=0.41x 3.2+5xy=5.8-0.41xy solving for xy=0.48 The cost of a bushel of corn=3.2+5*0.48=5.6 Answer: E. How do you make this Bunuel , I dont get it Quote: First 2^1/2*x-x=0.41x Yes it's kind of confusing, partly because I didn't use the formula formatting. Edited the original solution. So here it is: 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: 3.2+5xy=5.8-0.41xy, solve for xy --> xy=0.48; The cost of a bushel of corn in y days (the # of days when these two bushels will have the same price) will be 3.2+5xy=3.2+5*0.48=5.6. Answer: E. Hope it's clear. Tell me if it needs more clarification. _________________ Manager Joined: 25 Dec 2009 Posts: 102 Followers: 1 Kudos [?]: 25 [0], given: 3 Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] 19 Jan 2010, 06:43 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): Quote: 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. The probability of drawing a red and a white ball has to be( 2/9)*(3/9) and then the solution follows. How come it is (2*3/9) / (2/9), i dont get it again . Will continue again with rest of the questions. P.S. I have never liked questions who have just been tough but no brainerbut the one in the list here are some of the best questions to train on many tricks and fundamentals I believe. More I see the stuff on quant , more I admire the work of the people engaged with the site. Bravo.. and cheers from cold Paris but grâce à GMAT for keeping me hot Last edited by GMATMadeeasy on 19 Jan 2010, 07:44, edited 1 time in total. Manager Joined: 25 Dec 2009 Posts: 102 Followers: 1 Kudos [?]: 25 [0], given: 3 Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] 19 Jan 2010, 07:42 Bunuel wrote: GMATMadeeasy 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 2^1/2*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 # of days when these two bushels will have the same price.
First 2^1/2*x-x=0.41x
3.2+5xy=5.8-0.41xy solving for xy=0.48
The cost of a bushel of corn=3.2+5*0.48=5.6

How do you make this Bunuel , I dont get it

Quote:
First 2^1/2*x-x=0.41x

Yes it's kind of confusing, partly because I didn't use the formula formatting. Edited the original solution. So here it is:

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: 3.2+5xy=5.8-0.41xy, solve for xy --> xy=0.48;

The cost of a bushel of corn in y days (the # of days when these two bushels will have the same price) will be 3.2+5xy=3.2+5*0.48=5.6.

Hope it's clear. Tell me if it needs more clarification.

My problem was in understanding how to get .41x . It is crystal clear now.
Thank you so much.
Math Expert
Joined: 02 Sep 2009
Posts: 23538
Followers: 3482

Kudos [?]: 26116 [0], given: 2705

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]  19 Jan 2010, 10:52
Expert's post
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):

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

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

The probability of drawing a red and a white ball has to be( 2/9)*(3/9) and then the solution follows. How come it is (2*3/9) / (2/9), i dont get it again .

Will continue again with rest of the questions.

P.S. I have never liked questions who have just been tough but no brainerbut the one in the list here are some of the best questions to train on many tricks and fundamentals I believe. More I see the stuff on quant , more I admire the work of the people engaged with the site. Bravo.. and cheers from cold Paris but grâce à GMAT for keeping me hot

We want out of two balls one to be red (R) and another white (W).

There are two ways this could happen: we can draw FIRST ball red and SECOND white OR FIRST ball white and SECOND red.

Probability of each case is: \frac{3}{9}*\frac{2}{9}, so P=\frac{3}{9}*\frac{2}{9}+\frac{2}{9}*\frac{3}{9}=2*\frac{3}{9}*\frac{2}{9}.

One more thing worth of mentioning: \frac{3}{9}*\frac{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.

Hope it's clear.
_________________
Manager
Joined: 25 Dec 2009
Posts: 102
Followers: 1

Kudos [?]: 25 [0], given: 3

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]  19 Jan 2010, 12:08
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):

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

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

The probability of drawing a red and a white ball has to be( 2/9)*(3/9) and then the solution follows. How come it is (2*3/9) / (2/9), i dont get it again .

Will continue again with rest of the questions.

P.S. I have never liked questions who have just been tough but no brainerbut the one in the list here are some of the best questions to train on many tricks and fundamentals I believe. More I see the stuff on quant , more I admire the work of the people engaged with the site. Bravo.. and cheers from cold Paris but grâce à GMAT for keeping me hot [/quote]

We want out of two balls one to be red (R) and another white (W).

There are two ways this could happen: we can draw FIRST ball red and SECOND white OR FIRST ball white and SECOND red.

Probability of each case is: \frac{3}{9}*\frac{2}{9}, so P=\frac{3}{9}*\frac{2}{9}+\frac{2}{9}*\frac{3}{9}=2*\frac{3}{9}*\frac{2}{9}.

One more thing worth of mentioning: \frac{3}{9}*\frac{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.

Hope it's clear
Perfect. That is how I solved it.

Next attack : Inequality : Will be going through your links to flex "inequality muscles" . Can't thank enough for the efforts you guys have put.
Manager
Joined: 14 Feb 2010
Posts: 62
Location: Tokyo
Followers: 1

Kudos [?]: 10 [0], given: 15

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]  15 Feb 2010, 17:42
This is fantastic, Bunuel! Thank you so much!
Intern
Joined: 03 Feb 2010
Posts: 4
Location: İstanbul Turkey
Followers: 0

Kudos [?]: 0 [0], given: 2

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]  07 Mar 2010, 06:18
Gents,

An easier way to solve this prob is thru basic geometry if anyone interested. An illustration will explain better but if you draw a (perpendicular) line between the center of the circle to the line y = 3/4*x - 3, you get a 3,4,5 right triangle where the longest edge is 4cm long. 4/5*3 minus r of the circle (1) is equal to 1.4, the least possible distance.

Edit: oppps already done w/ illustration!
Intern
Affiliations: AIESEC Member
Joined: 02 Mar 2010
Posts: 19
Location: Gevgelija, Macedonia
Schools: Rotterdam School of Management
Followers: 1

Kudos [?]: 1 [0], given: 36

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]  01 Apr 2010, 07:45
thanx man,

this is very usefull,

Kudos.
Intern
Joined: 13 Jul 2010
Posts: 4
Location: India
Followers: 0

Kudos [?]: 0 [0], given: 2

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]  29 Jul 2010, 03:10
thanks brunel.. good collection!
Intern
Joined: 22 May 2010
Posts: 12
WE 1: Private Equity
WE 2: Management Consulting
Followers: 1

Kudos [?]: 15 [0], given: 9

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]  30 Jul 2010, 13:41
Nice one, thanks for this challenge.
Manager
Joined: 27 May 2010
Posts: 102
Followers: 2

Kudos [?]: 6 [0], given: 13

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]  16 Aug 2010, 07:14
thanks for the post.
Intern
Joined: 26 Mar 2010
Posts: 48
Followers: 0

Kudos [?]: 7 [0], given: 5

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]  08 Sep 2010, 12:20
Thanks for the tough problems!
Senior Manager
Joined: 20 Jul 2010
Posts: 271
Followers: 2

Kudos [?]: 39 [0], given: 9

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]  09 Sep 2010, 10:52
thanks for posting this
_________________

If you like my post, consider giving me some KUDOS !!!!! Like you I need them

Senior Manager
Joined: 18 Jun 2010
Posts: 302
Schools: Chicago Booth Class of 2013
Followers: 20

Kudos [?]: 127 [0], given: 194

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]  23 Sep 2010, 13:03
Bunuel wrote:
Vyacheslav wrote:
7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0

I find useful to roughly draw such questions. You can instantly recognize that line y=3/4*x-3 and axises form a right triangle with sides 3 and 4, on Y-axis and X-axis respectively. The hypotenuse lies on the line y=3/4*x-3 and equals 5 (Pythagorean Triple). The closest distance from any point to line is perpendicular from that point to line. Perpendicular to the hypotenuse (from origin in this case) will always divide the triangle into two triangles with the same properties as the original triangle. So, lets perpendicular will x, then

\frac{x}{3}=\frac{4}{5} and x=3*\frac{4}{5}=\frac{12}{5}=2,4

To find distance from point on circle to line we should substract lenth of radius from lenth of perpendicular: 2,4-1=1,4

Unfortunately, I cannot attach illustration. But if you draw this problem (even roughly), you will find it really easy to solve.

Excellent! +1.

For those who don't know the distance formula (which is in fact very rarely tested) this is the easiest and most elegant solution.

Bunuel,
Could you please provide the formula that is rarely tested? I want to be prepared in case I face such a question and the trick with triangles does not work due to some mistery
Math Expert
Joined: 02 Sep 2009
Posts: 23538
Followers: 3482

Kudos [?]: 26116 [0], given: 2705

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]  23 Sep 2010, 13:14
Expert's post
Financier wrote:
Bunuel wrote:
Vyacheslav wrote:
7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0

I find useful to roughly draw such questions. You can instantly recognize that line y=3/4*x-3 and axises form a right triangle with sides 3 and 4, on Y-axis and X-axis respectively. The hypotenuse lies on the line y=3/4*x-3 and equals 5 (Pythagorean Triple). The closest distance from any point to line is perpendicular from that point to line. Perpendicular to the hypotenuse (from origin in this case) will always divide the triangle into two triangles with the same properties as the original triangle. So, lets perpendicular will x, then

\frac{x}{3}=\frac{4}{5} and x=3*\frac{4}{5}=\frac{12}{5}=2,4

To find distance from point on circle to line we should substract lenth of radius from lenth of perpendicular: 2,4-1=1,4

Unfortunately, I cannot attach illustration. But if you draw this problem (even roughly), you will find it really easy to solve.

Excellent! +1.

For those who don't know the distance formula (which is in fact very rarely tested) this is the easiest and most elegant solution.

Bunuel,
Could you please provide the formula that is rarely tested? I want to be prepared in case I face such a question and the trick with triangles does not work due to some mistery

tough-tricky-set-of-problms-85211.html#p638361

Two solutions: one with the formula you mentioned another with triangle.
_________________
Re: TOUGH & TRICKY SET Of PROBLMS   [#permalink] 23 Sep 2010, 13:14
Similar topics Replies Last post
Similar
Topics:
31 Tricky problem sets 48 04 Oct 2013, 21:47
Sets tough Problem Try it...... 5 03 Jan 2012, 04:48
Tough and tricky 5: Race 5 11 Oct 2009, 17:14
3 Tough and tricky 4: addition problem 9 11 Oct 2009, 16:42
Tough and tricky 2: 4 11 Oct 2009, 16:17
Display posts from previous: Sort by