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At exactly what time past 7:00 will the minute and hour [#permalink]
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11 Jul 2003, 18:41
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At exactly what time past 7:00 will the minute and hour hands of an accurate working clock be precisely perpendicular to each other for the first time? (A) 20 13/21 minutes past 7:00 (B) 20 13/17 minutes past 7:00 (C) 21 3/23 minutes past 7:00 (D) 21 9/11 minutes past 7:00 (E) 22 4/9 minutes past 7:00
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



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Re: At exactly what time past 7:00 will the minute and hour [#permalink]
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11 Jul 2003, 22:36
Dear AkamaiBrah,
Thank you for not only posting such a nice questions, but also soving
the tough ones. Wish I had a teahcer like you when I was in school.
Here is my ans: D
A1 = hour angle
A2 = minute angle
A1A2 = 90
Assume the minutes = M
A1 = (7*60+M)/12*60 )*360
A2 = M*360/60
Solving all three equation , results in the answer D



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Re: At exactly what time past 7:00 will the minute and hour [#permalink]
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11 Jul 2003, 22:38
BTW:
Which one is correct? "Such a nice questions" or "Such nice questions" or "Such a nice question"



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Re: At exactly what time past 7:00 will the minute and hour [#permalink]
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11 Jul 2003, 23:04
210 + M/12  M = 90
Let M be the degrees covered by minute hand from zero. (7:00 pm is the reference point)
initially difference is 210 deg. Hour hand will move at the rate of M/12, so after M minutes hour hand will be at 210+M/12 and minute hand will be at M. The difference should be 90. Find M and divide by 6, which will give you the minutes.
Answer comes out to D



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Re: At exactly what time past 7:00 will the minute and hour [#permalink]
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12 Jul 2003, 03:56
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OKies my idea of solution,
I found it the fastest way,
Speed of hour hand/min = 360/12*60 = 1/2.
Speed of minuted hand/min = 360/60 = 6.
Relative speed = 11/2 degrees/min
Now at 7 the two are 210 degrees apart.
From 210 degress to 90 degrees. Minute has to gain 120 over the hour hand.
So time required will be = 120/Relative speed = 120*2/11 = 21 9/11 = D



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Re: At exactly what time past 7:00 will the minute and hour [#permalink]
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12 Jul 2003, 07:49
The answer is D. Evensflow did it the fastest although all of the solutions were fine. The key is computing the rate of each hand and solving for when they are 90 deg apart.
Good job all.
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



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At exactly what time past 7:00 will the minute and hour [#permalink]
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03 Sep 2014, 12:33
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At 7:00, hour hand must be at 210 degree from 12 (clock wise). Minute hand of the clock moves with 6 degree per minute speed and hour hand moves at 1/2 degree per minute. Therefore in x min hour hand will travel (1/2)*x degree further 210 degree from initial position; meanwhile minute hand must cover 6x degrees. Difference between minute and hour hand is required 90 degree. Formed following equation and solved for x. \((\frac{1}{2}x + 210)  6x = 90\) Solve: x=240/11 Ans D.
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At exactly what time past 7:00 will the minute and hour [#permalink]
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12 May 2015, 13:02
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Anotehr way to solve this problem is to use the formula that we have for the angle of the clock needels.
\(angle of clock needels =\frac{(60H + 11M)}{2}\) where H is the hour and the M is the minutes. We can imply that the H= 7 because the strating point is 7:00 any way and just need to find the M in the equation with making teh equation equal to 90 since that is what is being asked. Do the math and you will see that \(M=21\frac{9}{11}\)
So not bad to learn the formula at beggiging I was struguling with teh question and trying to go figure out some other stuff, but than I remmeberd on this formula and applied it.
hope it helps



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Re: At exactly what time past 7:00 will the minute and hour [#permalink]
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20 May 2015, 19:10
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PiyushK wrote: At 7:00, minute hand must be at 210 degree from 12 (clock wise). Minute hand of the clock moves with 6 degree per minute speed and hour hand moves at 1/2 degree per minute. Therefore in x min hour hand will travel (1/2)*x degree further 210 degree from initial position; meanwhile minute hand must cover 6x degrees. Difference between minute and hour hand is required 90 degree. Formed following equation and solved for x.
\((\frac{1}{2}x + 210)  6x = 90\) Solve: x=240/11
Ans D. Thanks for the solution. This cleared up the confusions I had in my head, bigtime!



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Re: At exactly what time past 7:00 will the minute and hour [#permalink]
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19 Sep 2015, 17:34
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I solved this problem in this way:
We know that: hour indicator moves 5 minutes every 60 minutes >1/12 to minute indicator moves 1 minute every minute>1/1 And that to be parallel they need to be 15 minutes awa from each other....then because the hour time is at 7hrs=35min and at 0 for minutes.
(35+t/12) reflects the position after t minutes of the hour indicator. t reflect the position of the minutes indcator after t minutes.
then we have to solve for t:
(35+t/12)t=15 t=240/11 >t=21+9/11



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Re: At exactly what time past 7:00 will the minute and hour [#permalink]
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27 Dec 2015, 10:10
The clock face is 360 deg/60 min (1 min = 6 deg)
The minute hand travels 360 deg / hr (M_sp) The hour hand travels 30 deg / hrs (H_sp)
Original location of minute hand = 0 deg Original location of hour hand = (360/12) ×7= 210 deg
After time T, location of hour hand would be 210 + T.H_sp = 210 + 30T (H_loc), and location of minute hand = 0 + T.M_sp = 0 + 360T (M_loc)
After T hours: H_loc  M_local = 90
210 + 30T  360T = 90
120 = 330T
T = 4/11 hrs
(Note that T converted to minutes is also the final location of the minute hand and therefore the answer to the question)
Convert to minutes: 240/11 = 21 9/11 Answer



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Re: At exactly what time past 7:00 will the minute and hour [#permalink]
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30 Jan 2016, 07:14
AkamaiBrah wrote: At exactly what time past 7:00 will the minute and hour hands of an accurate working clock be precisely perpendicular to each other for the first time?
(A) 20 13/21 minutes past 7:00 (B) 20 13/17 minutes past 7:00 (C) 21 3/23 minutes past 7:00 (D) 21 9/11 minutes past 7:00 (E) 22 4/9 minutes past 7:00 5.5 is the angle between minute n hour, this is what I was taught...so shouldn't it be solve by dividing 90 with 5.5?



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At exactly what time past 7:00 will the minute and hour [#permalink]
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30 Jan 2016, 07:27
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baharemustafa wrote: AkamaiBrah wrote: At exactly what time past 7:00 will the minute and hour hands of an accurate working clock be precisely perpendicular to each other for the first time?
(A) 20 13/21 minutes past 7:00 (B) 20 13/17 minutes past 7:00 (C) 21 3/23 minutes past 7:00 (D) 21 9/11 minutes past 7:00 (E) 22 4/9 minutes past 7:00 5.5 is the angle between minute n hour, this is what I was taught...so shouldn't it be solve by dividing 90 with 5.5? That would have been the case if your initial difference between the hour and the minute hand was = 0 degrees or in other words, both minute and hour hands were at the same location. But as per the question, you are asked for time AFTER 7:00. At 7:00, the angle between the hour and the minute hand is 210 degrees. you need to take this into account as well. So in order for the difference to decrease to 90 degrees, the minute hand must eat away this difference of 21090 = 120 degree at the rate of 5.5 degrees per minute > 120/5.5 = 21 9/11 minutes. Thus, D is the correct answer. Hope this helps. P.S.: If what you are saying was correct, then you should have calculated 16 4/11 minutes and this is not present in the given options. This should have told you that you are making a mistake somewhere.



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At exactly what time past 7:00 will the minute and hour [#permalink]
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05 Jun 2016, 17:10
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AkamaiBrah wrote: At exactly what time past 7:00 will the minute and hour hands of an accurate working clock be precisely perpendicular to each other for the first time?
(A) 20 13/21 minutes past 7:00 (B) 20 13/17 minutes past 7:00 (C) 21 3/23 minutes past 7:00 (D) 21 9/11 minutes past 7:00 (E) 22 4/9 minutes past 7:00 Assume, 1 deg = 1 Km time taken by hour hand to cover 60km (deg) = 60 km /12 hrs = 5km/hr time taken by min hand to cover 60km (deg) = 60 km/ 1 hr = 60 km/hr Relative velocity between hour hand and minute hand = 60  5 = 55 km/hr At 7:00 the distance between both hands is 35 km (35 deg). Distance between both hands when perpendicular is 15 km (15 deg) the time taken for both hands to be perpendicular = \(\frac{(3515)}{55}\) km/hr = \(\frac{(20*60)}{55}\) km/min Answer = 21 9/11



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At exactly what time past 7:00 will the minute and hour [#permalink]
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06 Jun 2016, 09:48
At exactly what time past 7:00 will the minute and hour hands of an accurate working clock be precisely perpendicular to each other for the first time?
(A) 20 13/21 minutes past 7:00 (B) 20 13/17 minutes past 7:00 (C) 21 3/23 minutes past 7:00 (D) 21 9/11 minutes past 7:00 (E) 22 4/9 minutes past 7:00
let m=location of minute hand 35+m/12=location of hour hand 35+m/12m=15 m=21 9/11 minutes past 7 D
Last edited by gracie on 06 Dec 2017, 16:39, edited 2 times in total.



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Re: At exactly what time past 7:00 will the minute and hour [#permalink]
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at 7:00 the hour hand is 7/12 of the way around. Every minute it goes 1/720 of the way around. the minute hand goes 1/60 of the way around every minute. If you think about it perpendicular times will always be 1/4 of the way around apart. So we can setup this equation. 7/12+(1x/720)=(1/4)+(x/60)



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Re: At exactly what time past 7:00 will the minute and hour [#permalink]
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21 Aug 2016, 10:23
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The initial separation between the two hands is 150 degrees and we want to make the separation 270 degrees i.e. increase the separation by 120 degrees. Every one minute accounts for the minute hand moving by 6degrees and the hour hand moving by 1/2 degrees so effectively the distance increased between the two is 11/2 (as they both move in the same direction, we would subtract the two).
Assuming it takes x minutes, we have
x*11/2=120, which gives the answer D



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Re: At exactly what time past 7:00 will the minute and hour [#permalink]
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21 Aug 2016, 20:54



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Re: At exactly what time past 7:00 will the minute and hour [#permalink]
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13 Dec 2016, 10:16
kpadma wrote: BTW:
Which one is correct? "Such a nice questions" or "Such nice questions" or "Such a nice question" Such nice questions would be correct as the subject the next part of the sentence is plural ("tough ones")



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Re: At exactly what time past 7:00 will the minute and hour [#permalink]
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19 Sep 2017, 03:59
Engr2012 wrote: baharemustafa wrote: AkamaiBrah wrote: At exactly what time past 7:00 will the minute and hour hands of an accurate working clock be precisely perpendicular to each other for the first time?
(A) 20 13/21 minutes past 7:00 (B) 20 13/17 minutes past 7:00 (C) 21 3/23 minutes past 7:00 (D) 21 9/11 minutes past 7:00 (E) 22 4/9 minutes past 7:00 5.5 is the angle between minute n hour, this is what I was taught...so shouldn't it be solve by dividing 90 with 5.5? That would have been the case if your initial difference between the hour and the minute hand was = 0 degrees or in other words, both minute and hour hands were at the same location. But as per the question, you are asked for time AFTER 7:00. At 7:00, the angle between the hour and the minute hand is 210 degrees. you need to take this into account as well. So in order for the difference to decrease to 90 degrees, the minute hand must eat away this difference of 21090 = 120 degree at the rate of 5.5 degrees per minute > 120/5.5 = 21 9/11 minutes. Thus, D is the correct answer. Hope this helps. P.S.: If what you are saying was correct, then you should have calculated 16 4/11 minutes and this is not present in the given options. This should have told you that you are making a mistake somewhere. Hi Why shouldnt I consider the initial distance between the hour and minute hand be 150 degrees?




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