A certain circular stopwatch has exactly 60-second marks and a single

Question

A certain circular stopwatch has exactly 60-second 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 seconds, what is the probability that the hand will stop less than 10 marks from the 53-second mark?

A. \(\frac{1}{6}\)

B. \(\frac{19}{60}\)

C. \(\frac{1}{3}\)

D. \(\frac{29}{60}\)

E. \(\frac{1}{2}\)

A.  \frac{1}{6}

B.  \frac{19}{60}

C.  \frac{1}{3}

D.  \frac{29}{60}

E.  \frac{1}{2}

EgmatQuantExpert · Accepted Answer

Solution

Given:  
 • A certain circular stopwatch has exactly 60-second marks and a single hand
• The hand of the watch is randomly set to one of the marks and allowed to count exactly 10 seconds

To find:  
 • The probability that the hand will stop less than 10 marks from the 53-second mark

Approach and Working:   
Considering less than 10 marks from 53, we have two limits to consider
 • More than 43
• Less than 3

Therefore, the possible values are = 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 00, 01, and 02 = total 19 favourable cases
 • The probability =  \frac{19}{60}

We can count the favourable cases in other way:
From 43 to 03, total cases excluding both = 20 – 1 = 19

Hence, the correct answer is option B.

Answer: B

chetan2u · Answer

Seems it is a Kaplan question and the wording is a bit confusing.. 
1) Easy to infer that "  the hand will stop less than 10 marks from the 53-second mark  " and then answer will be 43 to 53 = 10 as also solved by  GyMrAT

2) But here it seems to be trying to say  " WITHIN 10 marks from 53-second mark" 
so two cases
 a) TEN below 53 that is 43 to 53 and 
b) TEN above 53 that is 53 to 63 (or 03)  
ans  10+10-1=19....   MINUS 1 is because 53 is common in both the list 
so probability =  \frac{19}{60}

B

Bunuel , I know although the original question says "10 marks LESS than", the question would be better saying " WITHIN 10 marks from"

GyMrAT · Answer

Less than 10 marks from the 53-second mark, means the hand stops at {53, 52, 51, 50, 49, 48, 47, 46, 45, 44}

Since the hand is allowed to move exactly 10 marks from its starting position, hence the starting positions can be {43, 42, 41, 40, 39, 38, 37, 36, 35, 34}

Hence we have 10 available marks out of 60 marks.

Required Probability = 10/60 = 1/6

Answer A.

Thanks,
GyM

GyMrAT · Answer

Is there a stopwatch that goes in anti-clockwise direction as well?

On the GMAT, are we supposed to be strictly literal about whats specifically mentioned or are we at liberty to use normal logic, for normal everyday items?

Thanks,
GyM

EgmatQuantExpert · Answer

We are not assuming anti-clockwise movement. Assume that you are starting to count from 54 only - then also it can stop anywhere between 54 and 2, making a situation where the hand is in less than 10 marks from 53.

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A certain circular stopwatch has exactly 60-second marks and a single

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