BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Stopwatch probability

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 7
Joined: Fri May 21, 2010 12:31 am
Thanked: 1 times

Stopwatch probability

by Tanjello » Fri May 21, 2010 12:44 am
A certain circular stopwatch has exactly 60 sec marks and a single hand. If the hand of the watch is randomly set to one of the marks and allowed to count exactly 10 secs, what is the probability that the hand will stop less than 10 marks from the 53-sec mark?

A) 1/6
B) 19/60
C) 1/3
D) 29/60
E) 1/2

I think the answer is B (53+/- 9 means 18 ways, plus stopping at 53), thus 19/60. But it took me a while to realize the answer. What is the best way to approach this problem during a timed test? Thanks!
Top
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 am
Location: India
Thanked: 267 times
Followed by:80 members
GMAT Score:760

by sanju09 » Fri May 21, 2010 1:10 am
Tanjello wrote:A certain circular stopwatch has exactly 60 sec marks and a single hand. If the hand of the watch is randomly set to one of the marks and allowed to count exactly 10 secs, what is the probability that the hand will stop less than 10 marks from the 53-sec mark?

A) 1/6
B) 19/60
C) 1/3
D) 29/60
E) 1/2

I think the answer is B (53+/- 9 means 18 ways, plus stopping at 53), thus 19/60. But it took me a while to realize the answer. What is the best way to approach this problem during a timed test? Thanks!
Less than 10 marks from the 53-sec mark are the marks that are more than 43 and less than 3 in the continued cycle. Hence, following are the possible second-marks where the hand could stop to meet the inevitability:

44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 1, and 2; nineteen in total, and so are the various possibilities for the hand to start from.

Required probability = [spoiler]19/60.

B[/spoiler]
Last edited by sanju09 on Fri May 21, 2010 1:14 am, edited 1 time in total.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Top
Legendary Member
Posts: 576
Joined: Sat Mar 13, 2010 8:31 pm
Thanked: 97 times
Followed by:1 members

by liferocks » Fri May 21, 2010 1:14 am
sanju09 wrote:
Tanjello wrote:A certain circular stopwatch has exactly 60 sec marks and a single hand. If the hand of the watch is randomly set to one of the marks and allowed to count exactly 10 secs, what is the probability that the hand will stop less than 10 marks from the 53-sec mark?

A) 1/6
B) 19/60
C) 1/3
D) 29/60
E) 1/2

I think the answer is B (53+/- 9 means 18 ways, plus stopping at 53), thus 19/60. But it took me a while to realize the answer. What is the best way to approach this problem during a timed test? Thanks!
Less than 10 marks from the 53-sec mark are the marks that are more than 43 and less than 3 in the continued cycle. Hence, following are the possible second-marks where the hand could stop to meet the inevitability:

44, 45, 46, 47, 48, 49, 50, 1, 2, and 3 ten in total, and so are the various possibilities for the hand to start from.

Required probability = [spoiler]10/60 = 1/6.

A[/spoiler]
Why did u excluded the marks 51,52,54,55,56,57,58,59,0 ?
"If you don't know where you are going, any road will get you there."
Lewis Carroll
Top
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 am
Location: India
Thanked: 267 times
Followed by:80 members
GMAT Score:760

by sanju09 » Fri May 21, 2010 1:16 am
liferocks wrote:
sanju09 wrote:
Tanjello wrote:A certain circular stopwatch has exactly 60 sec marks and a single hand. If the hand of the watch is randomly set to one of the marks and allowed to count exactly 10 secs, what is the probability that the hand will stop less than 10 marks from the 53-sec mark?

A) 1/6
B) 19/60
C) 1/3
D) 29/60
E) 1/2

I think the answer is B (53+/- 9 means 18 ways, plus stopping at 53), thus 19/60. But it took me a while to realize the answer. What is the best way to approach this problem during a timed test? Thanks!
Less than 10 marks from the 53-sec mark are the marks that are more than 43 and less than 3 in the continued cycle. Hence, following are the possible second-marks where the hand could stop to meet the inevitability:

44, 45, 46, 47, 48, 49, 50, 1, 2, and 3 ten in total, and so are the various possibilities for the hand to start from.

Required probability = [spoiler]10/60 = 1/6.

A[/spoiler]
Why did u excluded the marks 51,52,54,55,56,57,58,59,0 ?
edited my friend, edited, you didn't give me enough time dear, earlier I stopped my watch at 50 in place of 60, mistakenly, normally happens after the fifth one, you see
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Top
Senior | Next Rank: 100 Posts
Posts: 85
Joined: Sat Aug 15, 2009 4:36 am
Thanked: 19 times

by gmatjedi » Fri May 21, 2010 3:48 pm
my approach:

10 up and 10 down from 53
=20
but note that you include 53 twice going up and going down
so you need to eliminate a duplicate 53
=19
which is out of 60 options
Top
Post new topic Post Reply
• Page 1 of 1