Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
May 1810:00 AM PDT
-11:00 AM PDT
Join us in a live GMAT practice session and solve 25 challenging GMAT questions with other test takers in timed conditions, covering GMAT Quant, Data Sufficiency, Data Insights, Reading Comprehension, and Critical Reasoning questions.
May 1212:00 PM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products. Ends 6/1
May 1501:00 PM IST
-11:00 AM IST
Start your journey with a fully customized action plan and work with a dedicated mentor to achieve a 735+ score.- May 19
12:00 PM PDT
-01:00 PM PDT
Scoring 329 on the GRE is not always about using more books, more courses, or a longer study plan. In this episode of GRE Success Talks, Ashutosh shares his GRE preparation strategy, study plan, and test-day experience, explaining how he kept his prep.... - Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
10 May 2019, 00:45
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
43% (02:34) correct
57%
(02:40)
wrong
based on 94
sessions
History
Date
Time
Result
Not Attempted Yet
There are 10 horses, named Horse 1, Horse 2, ..., Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse k runs one lap in exactly k minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time S > 0, in minutes, at which all 10 horses will again simultaneously be at the starting point is S = 2520. Let T > 0 be the least time, in minutes, such that at least 5 of the horses are again at the starting point. What is the sum of the digits of T?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
10 May 2019, 01:35
Kudos
Bookmarks
Bunuel
Here is my approach. ?
Time taken by all the 10 horses for 1 lap are the following:
- • Horse 1: 1 min
• Horse 2: 2 min = 2^1
• Horse 3: 3 min = 3^1
• Horse 4: 4 min = 2^2
• Horse 5: 5 min = 5^1
• Horse 6: 6 min = 2^1 * 3^1
• Horse 7: 7 min = 7^1
• Horse 8: 8 min = 2^3
• Horse 9: 9 min = 3^2
• Horse 10: 10 min = 2^1 * 5^1
Each of these horses will reach the starting point after exactly these many minutes written above.
- • Now a common time at which all the 10 would meet is nothing but the a common multiple of all the number.
- o That would give us the a number at which all the horses will be present at the starting point.
- The lowest multiple is the LCM of (1,2,3,4,5,6,7,8,9,10) = 2^3 * 3^2 * 5*7 = 2520
And that is what is given in the question too.
Now, if we want to find the least time T, for at least 5 horses meeting at the starting point, then we need to consider 5 horses out of those ten, for which we will get the least value of the LCM.
So, we can choose:
- • Horse 1: 1 min
• Horse 2: 2 min = 2^1
• Horse 3: 3 min = 3^1
• Horse 4: 4 min = 2^2
Now, if take horse 5, then in that case:
- • Horse 5: 5 min = 5^1
However, a better alternate would be to take:
- • Horse 6: 6 min = 2^1 * 3^1
Since we are asked sum of digits that will be 3. (Option B)
03 Mar 2025, 16:21
Kudos
Bookmarks
You need to know the time at which atleast 5 horses are at the starting line, now Horse 1 takes 1 min for a lap, so it will be at the starting position every minute.
Along with H1,
at 2 mins, there will be only H2
at 3 mins => H3
at 4 => H2 and H4
..and so on
Realize that the factors of the minutes you choose should at least be 5 or more, because the no. of factors of minutes are nothing but no. of horses
Now, for 5 factors least no. is \(2^4 = 16\) minutes and for 6 factors least no. is \(2^2 * 3 = 12\) minutes
You need to find the least time T, therefore T = 12; H1, H2, H3, H4, and H6 will be at starting line.
Sum of digits of T = 1 + 2 = 3
Answer B.
Along with H1,
at 2 mins, there will be only H2
at 3 mins => H3
at 4 => H2 and H4
..and so on
Realize that the factors of the minutes you choose should at least be 5 or more, because the no. of factors of minutes are nothing but no. of horses
Now, for 5 factors least no. is \(2^4 = 16\) minutes and for 6 factors least no. is \(2^2 * 3 = 12\) minutes
You need to find the least time T, therefore T = 12; H1, H2, H3, H4, and H6 will be at starting line.
Sum of digits of T = 1 + 2 = 3
Answer B.
