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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Tricky problem sets [#permalink]
08 Oct 2013, 07:03

1

This post received KUDOS

Qoofi wrote:

fozzzy wrote:

New problem added.. OA later.. Kudos for right solution

A drawer in a darkened room contains 100 red socks, 80 green socks, 60 blue socks and 40 black socks. A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn. What is the smallest number of socks that must be selected to guarantee that the selection contains at least 10 pairs? ( A pair of socks is two socks of the same color. No sock may be counted in more than one pair).

a) 21 b) 23 c) 24 d) 30 e) 50

Hi

Kind of simple one. We need 10 pairs of socks.

To find the smallest number of socks that must be selected :

We can get 9 pairs of socks from selecting 18 socks ( 9*2) To get the 10th pair, let choose 4 more socks. In this 4 socks, we would get one of the 4 colors of socks. But one sock of a color doesn't make it a pair

So choose 1 more socks to make it to a pair of sock, which is the 10th Pair of sock.

So total number of socks select = 18+4+1 = 23

Option B

Hope it is clear

Cheers Qoofi

@qoofi , Although we have arrived at the same answer, I can spot a flaw in your explanation. Picking 18 socks do not guarantee 9 pairs.. Let me demonstrate.

Picking 16 socks will surely result in me getting 8 pairs. But the 17th and 18th sock i pick could be of different colors. So it will not form the 9th pair ( Say 8 pairs + a red sock and a black sock).

However after picking 16 pairs of socks, picking 7 more would surely give me 2 pairs. (4 of each, then 2 more greens and a red or 3 greens etc.) _________________

Re: Tricky problem sets [#permalink]
08 Oct 2013, 09:24

goforsriram wrote:

Qoofi wrote:

fozzzy wrote:

New problem added.. OA later.. Kudos for right solution

A drawer in a darkened room contains 100 red socks, 80 green socks, 60 blue socks and 40 black socks. A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn. What is the smallest number of socks that must be selected to guarantee that the selection contains at least 10 pairs? ( A pair of socks is two socks of the same color. No sock may be counted in more than one pair).

a) 21 b) 23 c) 24 d) 30 e) 50

Hi

Kind of simple one. We need 10 pairs of socks.

To find the smallest number of socks that must be selected :

We can get 9 pairs of socks from selecting 18 socks ( 9*2) To get the 10th pair, let choose 4 more socks. In this 4 socks, we would get one of the 4 colors of socks. But one sock of a color doesn't make it a pair

So choose 1 more socks to make it to a pair of sock, which is the 10th Pair of sock.

So total number of socks select = 18+4+1 = 23

Option B

Hope it is clear

Cheers Qoofi

@qoofi , Although we have arrived at the same answer, I can spot a flaw in your explanation. Picking 18 socks do not guarantee 9 pairs.. Let me demonstrate.

Picking 16 socks will surely result in me getting 8 pairs. But the 17th and 18th sock i pick could be of different colors. So it will not form the 9th pair ( Say 8 pairs + a red sock and a black sock).

However after picking 16 pairs of socks, picking 7 more would surely give me 2 pairs. (4 of each, then 2 more greens and a red or 3 greens etc.)

I just took a case. Yup you are right. It worked for this question. thanks for pointing it out.

Cheers Qoofi _________________

I'm telling this because you don't get it. You think you get it which is not the same as actually getting it. Get it?

Re: Tricky problem sets [#permalink]
08 Oct 2013, 09:35

Hey bud , sorry about all the confusion, I am once again making the same mistake you did, when I assume that 16 socks give 8 pairs.. the correct number is actually from the formula got by induction , i.e 16+3 = 19 socks. 16 socks would have given me 7 pairs for sure, and then 3 more socks make sure i get the 8th pair _________________

Re: Tricky problem sets [#permalink]
09 Oct 2013, 04:07

regarding the second question...

A drawer in a darkened room contains 100 red socks, 80 green socks, 60 blue socks and 40 black socks. A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn. What is the smallest number of socks that must be selected to guarantee that the selection contains at least 10 pairs? (A pair of socks is two socks of the same color. No sock may be counted in more than one pair).

If we require two pairs, then it suffices to select 7 socks. Any set of 7 socks must contain a pair; if we remove this pair( individual sock pair), then the remaining 5 socks will contain a second pair as shown above. On the other hand, 6 socks might contain 3 greens, 1 black,1 red and 1 blue - hence only one pair. Thus 7 socks is the smallest number to guarantee two pairs.

Similar reasoning shows that we must draw 9 socks to guarantee 3 pairs, and in general, 2p+3 socks to guarantee p pairs. Thus 23 socks are needed to guarantee 10 pairs. _________________

Re: Tricky problem sets [#permalink]
09 Oct 2013, 07:31

OA later. Kudos for right solutions.

The mean,median,unique mode , and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is

Re: Tricky problem sets [#permalink]
09 Oct 2013, 08:41

1

This post received KUDOS

fozzzy wrote:

OA later. Kudos for right solutions.

The mean,median,unique mode , and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is

A) 11 B) 12 C) 13 D) 14 E) 15

PS: unique mode means only one mode, not more than one mode.

If 15 is the largest then the smallest number must be 7. When 8 is the mean and median and mode , it is not possible to write such a combination Note: In every set you form, there must be at least two 8s as it is the median value. e.g. 7 7 8 8 8 8 15 , is the set with the lowest sum with median and mode 8. But the mean is not 8

If 14 is the largest number, smallest number is 6.

So 6 6 6 8 8 8 8 14 is a valid set. Hence is the answer

The method I used was to pair numbers in 2s and see if they can bring the mean back to 8. For eg. When i picked 6 and 14 , its sum exceeds 16 by 4. So 4 less than 16 is 12 which can be formed by 2 6s. Now that there are 3 6s, there must be 4 8s because the mode is 8 _________________

Re: Tricky problem sets [#permalink]
21 Oct 2013, 06:34

1

This post received KUDOS

New problem added... Kudos for right solution.. OA later

Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and than ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of yan's distance from his home to his distance from the stadium?

Re: Tricky problem sets [#permalink]
21 Oct 2013, 10:24

1

This post received KUDOS

fozzzy wrote:

New problem added... Kudos for right solution.. OA later

Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and than ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of yan's distance from his home to his distance from the stadium?

Lets say S1 is the speed of walking & S2 is the speed of cycling. Given S2 = 7S1

Let x be the distance between Yan & stadium and y be the distance between Yan & home. To fine y / x ? Method 1 : Yan to Stadium by walking Time t = y / S1

Method 2 : Yan to Home (Walk) + Home to stadium (Cycle)

Time t = x / S1 + (x+y) / S2, which equals to x / S1 + (x+y) / 7S1

Since time is same, we can equate! y / S1 = x / S1 + (x+y) / 7S1

Solving y/x = 3/4 Option B

Hope it is clear Cheers Qoofi _________________

I'm telling this because you don't get it. You think you get it which is not the same as actually getting it. Get it?

Re: Tricky problem sets [#permalink]
22 Oct 2013, 08:10

fozzzy wrote:

New problem added... Kudos for right solution.. OA later

Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and than ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of yan's distance from his home to his distance from the stadium?

Low GPA MBA Acceptance Rate Analysis Many applicants worry about applying to business school if they have a low GPA. I analyzed the low GPA MBA acceptance rate at...

In out-of-the-way places of the heart, Where your thoughts never think to wander, This beginning has been quietly forming, Waiting until you were ready to emerge. For a long...