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Re: A certain circular stopwatch has exactly 60-second marks and a single [#permalink]
EgmatQuantExpert wrote:

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



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
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Re: A certain circular stopwatch has exactly 60-second marks and a single [#permalink]
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GyMrAT wrote:
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


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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Re: A certain circular stopwatch has exactly 60-second marks and a single [#permalink]
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Bunuel wrote:
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. \(\frac{1}{2}\)



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"
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Re: A certain circular stopwatch has exactly 60-second marks and a single [#permalink]
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EgmatQuantExpert wrote:

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




IF IT IIS MENTIONED 10 LESS THAN 53, SO WHY MORE THAN 53 IS CONSIDERED?

Hence, the correct answer is option B.

Answer: B
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Re: A certain circular stopwatch has exactly 60-second marks and a single [#permalink]
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Re: A certain circular stopwatch has exactly 60-second marks and a single [#permalink]
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