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

 It is currently 09 Dec 2018, 15:57

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 December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• Free GMAT Algebra Webinar

December 09, 2018

December 09, 2018

07:00 AM PST

09:00 AM PST

Attend this Free Algebra Webinar and learn how to master Inequalities and Absolute Value problems on GMAT.
• Free lesson on number properties

December 10, 2018

December 10, 2018

10:00 PM PST

11:00 PM PST

Practice the one most important Quant section - Integer properties, and rapidly improve your skills.

M18-10

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51035

Show Tags

16 Sep 2014, 00:03
00:00

Difficulty:

55% (hard)

Question Stats:

50% (00:55) correct 50% (00:56) wrong based on 52 sessions

HideShow timer Statistics

If a point is arbitrarily selected on a line segment, thus breaking it into two smaller segments, what is the probability that the larger segment is at least twice as long as the smaller one?

A. $$\frac{1}{4}$$
B. $$\frac{1}{3}$$
C. $$\frac{1}{2}$$
D. $$\frac{2}{3}$$
E. $$\frac{3}{4}$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 51035

Show Tags

16 Sep 2014, 00:03
Official Solution:

If a point is arbitrarily selected on a line segment, thus breaking it into two smaller segments, what is the probability that the larger segment is at least twice as long as the smaller one?

A. $$\frac{1}{4}$$
B. $$\frac{1}{3}$$
C. $$\frac{1}{2}$$
D. $$\frac{2}{3}$$
E. $$\frac{3}{4}$$

The larger segment will be at least twice as long as the smaller one if the breaking point is either in the first or in the last third of the segment. The probability of this is $$\frac{2}{3}$$.

The problem can also be solved algebraically. If the length of the original segment is 1 and the length of the first sub-segment is $$x$$ then we are looking for $$x$$ such that $$\frac{x}{1 - x} \ge 2$$ or $$\frac{1 - x}{x} \ge 2$$. The first inequality reduces to $$x \ge \frac{2}{3}$$, the second to $$x \le \frac{1}{3}$$. The probability that either $$x \ge \frac{2}{3}$$ or $$x \le \frac{1}{3}$$ is $$\frac{2}{3}$$.

_________________
Senior Manager
Joined: 31 Mar 2016
Posts: 385
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
GPA: 3.8
WE: Operations (Commercial Banking)

Show Tags

03 Aug 2016, 07:50
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Did not understand the solution is explained. Probability is total number of desired outcomes / total outcomes in terms of this usual way of solving can you explain in detail as your explanation at present is not clear to me. Thanks.
Current Student
Joined: 12 Jan 2016
Posts: 70
Location: United States
Concentration: Operations, General Management
GMAT 1: 690 Q49 V35
GPA: 3.5
WE: Supply Chain Management (Consumer Electronics)

Show Tags

03 Aug 2016, 10:24
Divide the line into 3 equal segments, lets A, B, C.

After a point is selected, for the longer segment to be 2 times as long as the smaller segment, the point has to be either in segment A or in C.

As length of A = length of B = length of C = 1/3, so, probability that the point is selected in segment A or C is 1/3 + 1/3 = 2/3.

Current Student
Joined: 08 Jan 2015
Posts: 79

Show Tags

Updated on: 10 Aug 2016, 02:01
I solved it in the following way:

Say the line is 3x length. What's the probability that the point will break the line into desired segments? It's 2x/3x, since there are two points that satisfy the condition one is at 1x length and the another one at 2x length. Frankly speaking, I didn't notice "at least", but it turns out that I got a correct answer.

In any case, if you divide a line into three segments, there will be two segment on symmetrically opposite ends that satisfy condition.

Originally posted by manlog on 09 Aug 2016, 22:09.
Last edited by manlog on 10 Aug 2016, 02:01, edited 1 time in total.
Intern
Joined: 17 Mar 2016
Posts: 15
Location: Singapore
GPA: 3.5
WE: Business Development (Energy and Utilities)

Show Tags

10 Aug 2016, 01:59
3
1
Bunuel wrote:
If a point is arbitrarily selected on a line segment, thus breaking it into two smaller segments, what is the probability that the larger segment is at least twice as long as the smaller one?

A. $$\frac{1}{4}$$
B. $$\frac{1}{3}$$
C. $$\frac{1}{2}$$
D. $$\frac{2}{3}$$
E. $$\frac{3}{4}$$

Did it in rather quick and crude method, on a scale of 100 (10 unit intervals):
10:90 - Yes (10X2 = 20)
20:80 - Yes
30:70 - Yes
40:60 - No
50:50 - No
60:40 - No
70:30 - Yes
80:20 - Yes
90:10 - Yes

So probability = 6/9 = 2/3
Intern
Joined: 07 Oct 2016
Posts: 13

Show Tags

Updated on: 28 Nov 2016, 22:15
The question has been re-edited stated below

Originally posted by gmatdemolisher1234 on 28 Nov 2016, 21:39.
Last edited by gmatdemolisher1234 on 28 Nov 2016, 22:15, edited 1 time in total.
Intern
Joined: 07 Oct 2016
Posts: 13

Show Tags

28 Nov 2016, 22:13
Bunuel wrote:
Official Solution:

If a point is arbitrarily selected on a line segment, thus breaking it into two smaller segments, what is the probability that the larger segment is at least twice as long as the smaller one?

A. $$\frac{1}{4}$$
B. $$\frac{1}{3}$$
C. $$\frac{1}{2}$$
D. $$\frac{2}{3}$$
E. $$\frac{3}{4}$$

The larger segment will be at least twice as long as the smaller one if the breaking point is either in the first or in the last third of the segment. The probability of this is $$\frac{2}{3}$$.

The problem can also be solved algebraically. If the length of the original segment is 1 and the length of the first sub-segment is $$x$$ then we are looking for $$x$$ such that $$\frac{x}{1 - x} \ge 2$$ or $$\frac{1 - x}{x} \ge 2$$. The first inequality reduces to $$x \ge \frac{2}{3}$$, the second to $$x \le \frac{1}{3}$$. The probability that either $$x \ge \frac{2}{3}$$ or $$x \le \frac{1}{3}$$ is $$\frac{2}{3}$$.

If the question said - If a point is arbitrarily selected on a line segment, thus breaking it into two smaller segments, what is the probability that the 2 segments are equal-?

Then If the length of the original segment is 1 and the length of the both sub-segment is x/(1-x) = 1, we'll get x= 1/2,

So there is only 1 point 1/2 ..

Now what do we do ?? how should we solve this???
Intern
Joined: 04 Aug 2014
Posts: 29
GMAT 1: 620 Q44 V31
GMAT 2: 620 Q47 V28
GPA: 3.2

Show Tags

27 Mar 2017, 23:40
hi brunel..

after going through the solution i got stuck at the last statement : The probability that either x≥2/3x or x≤1/3x is 2/3

how we can conclude probability is 2/3 and not 1/3.Both the options are given in the answer choices
Intern
Joined: 08 Jan 2017
Posts: 9
Location: India
Concentration: General Management, Technology
GMAT 1: 750 Q51 V40
GPA: 3.5
WE: Analyst (Retail)

Show Tags

28 Mar 2017, 00:01
sidagar wrote:
hi brunel..

after going through the solution i got stuck at the last statement : The probability that either x≥2/3x or x≤1/3x is 2/3

how we can conclude probability is 2/3 and not 1/3.Both the options are given in the answer choices

Look at the problem in this way.
Consider a line segment AB (any length)

Put two points on the line segment (say C and D), such that 2*AC=CB and 2*DB=AD
(i.e. the two points divide the line segment into 1:2 ratio)

Now, consider the question that you have to arbitrarily select a point on the segment.
Note carefully here that if you chose a point between A and C, or D and B, then it will satisfy the condition, which is that one line segment is greater than the twice of other.

Next, notice that C and D divide the line segment in a ration of 1/3 (i.e., AC=CD=DB=1/3*AB)

We also know that any point between C and D will not satisfy the condition

Therefore Probability of condition not true will be when the point is chosen between C and D, which is 1/3

Hence, probability of condition being true is 2/3 and answer is D

Hope this helps.
Current Student
Joined: 12 Feb 2015
Posts: 56
Location: India
GPA: 3.84

Show Tags

01 Sep 2017, 09:38
sidagar wrote:
hi brunel..

after going through the solution i got stuck at the last statement : The probability that either x≥2/3x or x≤1/3x is 2/3

how we can conclude probability is 2/3 and not 1/3.Both the options are given in the answer choices

Bunuel

Same question,can u please explain the last step of having probability as 2/3
Math Expert
Joined: 02 Sep 2009
Posts: 51035

Show Tags

01 Sep 2017, 09:44
himanshukamra2711 wrote:
sidagar wrote:
If a point is arbitrarily selected on a line segment, thus breaking it into two smaller segments, what is the probability that the larger segment is at least twice as long as the smaller one?

hi brunel..

after going through the solution i got stuck at the last statement : The probability that either x≥2/3x or x≤1/3x is 2/3

how we can conclude probability is 2/3 and not 1/3.Both the options are given in the answer choices

Bunuel

Same question,can u please explain the last step of having probability as 2/3

---

The larger segment will be at least twice as long as the smaller one if we cut anywhere at the two red segments. The probability of that will be 2/3.
_________________
Intern
Joined: 16 Jun 2018
Posts: 10

Show Tags

30 Aug 2018, 22:44
The problem can also be solved algebraically. If the length of the original segment is 1 and the length of the first sub-segment is $$x$$ then we are looking for $$x$$ such that $$\frac{x}{1 - x} \ge 2$$ or $$\frac{1 - x}{x} \ge 2$$. The first inequality reduces to $$x \ge \frac{2}{3}$$, the second to $$x \le \frac{1}{3}$$. The probability that either $$x \ge \frac{2}{3}$$ or $$x \le \frac{1}{3}$$ is $$\frac{2}{3}$$.

Bunuel

I do understand the role of this equation
$$\frac{x}{1 - x} \ge 2$$
but I dont understand why we have used the equation given after this where we take up the reciprocal of LHS.
And on what basis do we reject 1/3.
Re: M18-10 &nbs [#permalink] 30 Aug 2018, 22:44
Display posts from previous: Sort by

M18-10

Moderators: chetan2u, Bunuel

 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®.