At exactly what time past 7:00 will the minute and hour

Question

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

evensflow · Accepted Answer

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

PiyushK · Answer

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.

brstorewala · Answer

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

kzivrev · Answer

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

fcojcruz · Answer

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

baharemustafa · Answer

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?

ENGRTOMBA2018 · Answer

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 210-90 = 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.

vgk92 · Answer

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{(35-15)}{55}  km/hr =  \frac{(20*60)}{55}  km/min 
Answer = 21 9/11

sinhap07 · Answer

Hi
Why shouldnt I consider the initial distance between the hour and minute hand be 150 degrees?

Bunuel · Answer

The hands of the clock move so that 150 degree angle increases and 210 degree angle decreases, so that will be the angle which will be 90 degree.

siddyj94 · Answer

Bunuel   karishma  - Can we look at this question from an angle of relative change as at 7.00 PM the change in the angle would be 210 deg and to make it perpendicular post 7 PM as per the question stem the total relative change would be 300 (210+90).

Which can be derived as 300/360*11/2 which is the total change in the hourly and the minute hand of the clock.

If yes, i have solved the question as 300/360*11/2 which is coming at 55/12 and i am stuck here!

Thanks in advance.

KarishmaB · Answer

Yes, you can. Here is the link to the relevant video: https://youtu.be/dGqC_6DIKkY Here is what it says: the minute hand covers 11/2 degrees per minute relative to the hour hand. Note that position of the hour hand and the minute hand at 7. The larger angle between them is 210 degrees which needs to be brought down to 90 degrees (Think why - the hour hand will barely move but the minute hand will need to cover most of the difference between them and when it is closer to the hour hand, at some time, angle between them will be 90 degrees) Relative to the hour hand, the minute hand has to cover 210 - 90 = 120 degrees. Time taken for that will be \frac{120}{(11/2)} = \frac{240}{11} = 21(\frac{9}{11}) mins Answer (D)

EMPOWERgmatRichC · Answer

Hi All,

When dealing with a clock face, there are several pieces of information to consider: 
1) A clock face is 360 degrees, so each "hour mark" represents 30 degrees and each "minute mark" represents 6 degrees. 
2) As the minute hand moves, the hour hand ALSO moves (a little). Over the course of 1 full hour of time, the minute hand will move 360 degrees and the hour hand will move 30 degrees. 
3) For each MINUTE that the minute hand moves, the hour hand moves 0.5 DEGREES.

To find the first time AFTER 7:00 that the hour hand and minute hand are PERPENDICULAR (meaning that their positions are 90 degrees apart), you might find it useful to work in small increments. From the answer choices, we see that the correct answer has to be fairly close to 7:20, so we'll 'work up' to there:

At 7:00... 
the minute hand is at 0 degrees 
the hour hand is at 210 degrees

At 7:20... 
the minute hand is at 120 degrees 
the hour hand has moved to 210 + 10 = 220 degrees 
So the arms are getting closer to where we want them to be, but they're not there yet (they're 100 degrees apart)...

At 7:21... 
the minute hand is at 126 degrees 
the hour hand has moved to 220 + 0.5 = 220.5 degrees 
So the arms are getting closer to where we want them to be, but they're not there yet (now they're 94.5 degrees apart)...

At 7:22... 
the minute hand is at 132 degrees 
the hour hand has moved to 220.5 + 0.5 = 221 degrees 
Here, the minute hand has 'overshot' just a little bit (now they're 89 degrees apart)... Thus, the correct answer has to be just a little 'earlier' than 7:22

Final Answer:  D

GMAT assassins aren't born, they're made, 
Rich

GMATinsight · Answer

CONCEPT OF CLOCK

Please find teh concept and video solution here.

https://www.youtube.com/watch?v=81zVnu6h6SQ

Gmathub · Answer

This questions can be simply solved by the concept of relative speed. The distance will be in terms of degrees and speed will be in terms of degrees/minutes.

See the video solution:

https://www.youtube.com/watch?v=k7Zey14 ... GL&index=7

Visit www.gmathub.com for personalized gmat training.

donu · Answer

Quite frankly, if I saw this on the real test, I'd probably skip it lol. But here's a solution.

The initial gap between the minute hand and hour hand is 210 degrees at 7:00. We want to get the gap to be 90 degrees (perpendicular). So we need to shave off 120 degrees.

Every 5 min, the minute hand moves 30 degrees while the hour hand moves 2.5 degrees.

So 5 min after 7, the gap has closed be 30 - 2.5 = 27.5 degrees.

So we need to find how many minutes it will take to shave off 120 degrees.

We know:

5 min ---> 27.5 degrees
x min ---> 120 degrees

x = 120 * 5 / 27.5 
x= 600/27.5
x = 120/5.5
x = 21 9/11 (D)

nivivacious · Answer

Please don't waste time like I did trying to figure formulas etc. Think of Min hand and Hour hand as two diff people who are traveling in the same direction. 
You need to figure relative speed

Given: A travels 360 kms in 1 hour (I'm taking the liberty of assuming degree here as distance. Clock has 360 degrees)
since the question is of minutes, and not hours, divide this by 60
so relative speed is 6m/min
Now 360 degrees of a clock is divided in 12 hours. So each 'slot' is 30 degrees
Therefore hour, the hour hand travels 30 kms in 1 hour 
Convert this to minutes, so relative speed is 30/60 ie 1/2
Now at 7 we have to cover 30x7 = 210
We need them perpendicular so 210-90=120

This is a relative speed question where they're moving in the same direction so we subtract the speeds
6-1/2
Therefore 120/(6-1/2) ie 210/11 ie D

BinodBhai · Answer

Hello  KarishmaB ,

why I am not able to solve this via a method I learnt from DI review
Minute hand speed is 360
Hour hand speed is 30
relative speed is 330

Distance to cover is 90 (perpendicular)

So 90/330 will give me the time.. but my answer is not matching

At exactly what time past 7:00 will the minute and hour

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At exactly what time past 7:00 will the minute and hour

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