For all the genius people, answer this???

Question

Two clocks commence striking at a certain hour simultaneously,but the frequency of chiming of the first is different frm dat of the second.
The third stroke of the first is coincident wid the fourth stroke of the second.
If at a certain hour,the first strikes three times after the second has stopped striking,then what is the hour?

options:
a.9 b.10 c.11 d.12

all OF YOU , please EXPLAIN YOUR ANSWER IN DETAIL ???

DATA IS ABSOLUTELY SUFFICIENT

PLEASE EXPLAIN UR ANSWER
lil girl give more explanation

Answer 1

So in a certain amount of time, call it x, the first clock strikes 3 times while the second strikes four times, thus the second clock strikes faster. Now we know that the difference between the total strike times must be enough for the first clock to strike 3 times, so write an equation:
Let y be the hour, then the time for the first clock to strike its chimes is y*x/3 and for the second clock, y*x/4, we need the difference to be 3*x/3 = x so,

y*x/3 - y*x/4 = x

solve this for y gives 12, so this happens at 12 o'clock.

Answer 2

Answer is d. It could have been a,b,c, or e (all of the above).. but:

If we assume the first strike is on the hour itself, then it makes the MCQ question strange because it provides partial answer: 9,10,11 among 8,9,10,11 (see bandf illustration with 2 mor rows for 10 and 11 seconds). Unless, the rules is that you may choose more than one answers. If so, answers are a,b,c.

However, considering the mechanics behind, it is more realistic to assume that at the hour, the striking mechanisms is activated and repeatedly activated for the required number of times. So, 'commence striking ... simultaneously' doesn't mean the first chimings sounded simultaneously, but the mechanisms started simultaneously while the faster one produces the chime earlier than the second ones. So, we get answers as a,b,c,d, or e (all of the above).

However, upon more careful reading, '... has stopped striking...' implies that there is an amount of striking time to produce the chime. This clears things up. So, it is after a full completion of the last chime of the faster clock, that there are 3 more chimes, thus answers are 12, 13, 14, 15 chimes, but most clock will strike once for 1300hr. So d is the answer.

Still, this MCQ question is not flawless because it involves too much thinking time for a MCQ question.

Answer 3

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On second thought, I think it could happen at ANY OF THE ABOVE.
9, 10, 11, or 12.

The reason is,
At 9 o'clock, the two start chiming. After the faster clock has chimed 9 times, the slower clock has only chimed 3/4 * 9 times = 6 and 3/4 times. But there's no such thing as 3/4ths of a chime, so it's really only chimed 6 times! It's CLOSE to chiming the 7th time, but it's not there yet. Therefore there are three still to go.

Likewise, at 10 o'clock, when the fast clock has chimed 10 times, the slow clock has only chimed 3/4 * 10 times = 7.5 times. Once again, the slow clock has three more to go!

After the fast clock chimes 11, the slow clock has only chimed 3/4 * 11 times, = 8 1/4 times, so there are once again 3 more to be heard.

And finally with 12, the slow clock will have only chimed 3/4*12 = 9 times. So yet again, there are three remaining.

So my answer is "All of the above" or rather, "Any of the above".

Very tricky question. It shows the dangers of treating a discrete operation--like the chiming of a clock--as if it were a continuous function.

Answer 4

My thought is this:

take it in beats. If they start striking simultaneously and then are coincident after 3 strokes of the first and 4 of the second then at times (arbitrary but we can call it seconds if we like):

0,2,4,6,8,10,12,14,16,18,20... the second will strike
0,3,6,9,12,15,18,21, ... the first will strike

(assuming they strike at even frequendy intervals)
This means that at time:
14 the second will have struck 8, the first will have only struck 5, leaving 3 strikes to go. The answer is 8.

However, at time 16 the second will have struck 9, the first will have only struck 6, leaving 3 strikes to go. The answer is also 9.

However, at time 18, the second will have struck 10, the first will only have struck (coincidentally again) 7, leaving 3 to go. The answer is also 10.

I am probably missing a trick here as it is multiple choice with 3 answers that I think would work...

If it is when they both strike coincidentally and have 3 strikes, I would say B. The first hour it could be is 8...

Answer 5

The ratio is 3:4 then right? So that means every 4 the second does, the first has to do 1 more. So that means that the first has to chime 4 X 3 times in order to be 3 ahead of the first. So that makes it d) 12.

3:4
6:8
9:12- here the difference is 3 chimes.

Answer 6

If clock 1 equals 3 and clock 2 equals 4, the answer has to be divisible by both 3 and 4. If clock 1 has to strike 3 more time after clock 2 has finished then: 4A=3A+3 "A" being the number of times they stroked. This equals 3. Therefore, the time is 12.

Answer 7

OK, Here comes the mathematical calculation that leads to d.12

t1 is the time interval between 2 striks of clock #1, t2 is same for clock #2. n is the hour to be solved.

"The third stroke of the first is coincident wid the fourth stroke of the second". Thus 3*t1=4*t2 then t1=4/3*t2

"If at a certain hour,the first strikes three times after the second has stopped striking". Thus (n-3)*t1 = n*t2

thus (n-3)*4/3*t2 = n*t2 -> 4n - 12 = 3n -> n = 12

Answer 8

it's 12 O'clock
For every 4 strokes of the 2nd clock, the first only strikes 3.
So if it were twelve o'clock, the 2nd clock would be done striking when the first was only on 9, so it would chime 3 more times to reach 12.

Answer 9

The answer is d. 12 o clock

reason:

from the given data we get,
the first one stikes 3 times while the second strikes 4 times.
So on a scale, we can see that when the second one stikes 12 times, the first one have completed striking 9 times.
And if the first one strikes three more times after the first stopped striking, then it is 12 o clock.. and that is the answer :)

cool :)

Answer 10

The answer is d)12

For every four strikes of the second (2) clock, the first (1) clock strikes three times. When clock (2) has stroke 4 times, clock (1) has stroke 3 times. When clock (2) has stroke 8 times, clock (1) has stroke 6 times. When clock (2) has stroke 12 times, clock (1) has stroke 9 times, and it will strike 3 more times.