clock hands in same position; hours for C to move sand

[eouirai]

1) At what time after 4:00 will the hour and minute hand first be in the same position?

~I'm not sure how to start this problem.

2) Trucks A, B, and C, working together, can move a load of sand in t hours. When working alone, it takes truck A one extra hour to move the sand; B, six extra hours; and C, t extra hours. Find t.

~I set the equations as:
1/A + 1/B + 1/C = 1/t
1/A= 1/(t=1)
1/B= 1/(t+6)
1/C= 1/(t+8) and i substituted so
1/(t+1) +1/(t+6) + 1/(t+8) =1/t.
But I keep getting the wrong answer so can you tell what im doing wrong?

 

[skeeter]

Re: 2 really annoying problems

1) At what time after 4:00 will the hour and minute hand first be in the same position?

basic equation is ...

$\displaystyle \theta = \theta_i + \omega t$

let the 12:00 position be $\displaystyle \theta = 0$

for the minute hand ...
$\displaystyle \theta_i = 0$
$\displaystyle \omega = 2\pi$ rad/hr

for the hour hand ...
$\displaystyle \theta_i = \frac{2\pi}{3}$
$\displaystyle \omega = \frac{\pi}{6}$ rad/hr

$\displaystyle \theta$ 's will be equal when they are at the same position


2) Trucks A, B, and C, working together, can move a load of sand in t hours. When working alone, it takes truck A one extra hour to move the sand; B, six extra hours; and C, t extra hours. Find t.

~I set the equations as:
1/A + 1/B + 1/C = 1/t
1/A= 1/(t+1)
1/B= 1/(t+6)
1/C= 1/(2t) and i substituted so
1/(t+1) +1/(t+6) + 1/(2t) =1/t.
But I keep getting the wrong answer so can you tell what im doing wrong?

fixed ... try again.

 

[eouirai]

Re: 2 really annoying problems

Thanks, but I'm still a bit confused on the first problem...

 

[skeeter]

$\displaystyle 2\pi t = \frac{2\pi}{3} + \frac{\pi}{6} t$

solve for t.

 

[Denis]

1/(t+1) + 1/(t+6) + 1/(2t) =1/t.

Before using LCM, do this:
1/(t+1) + 1/(t+6) = 1/t - 1/(2t)
1/(t+1) + 1/(t+6) = 1/(2t)

Now use LCM on the left side; then crisscross multiply. Cuts the work by at least half.

 

[eouirai]

How did you set up the equation for the clock problem? What if it was after 5 o clock instead of after 4.

 

[TchrWill]

At what time after 4:00 will the hour and minute hand first be in the same position?

~I'm not sure how to start this problem.

1--The hour hand moves 1 minute every .50º or 1M/.5º.
2--The minute hand moves 1 minute every 6º or 1M/6º.
3--When the two hands are coincident, the hour hand has moved Xº and the minute hand has moved (120 + X)º.
4--So, in T minutes, the two movements are (120 + X)1/6 = X/.5.
5--This leads to 60 + .5X = 6X or X = 10.909º.
6--Therefore, the hour hand has moved 10.909º, the minute hand 130.909º, and the minute hand has moved 20 + 10.909090(1/6) = 21.818 minutes past 4 o'clock.

Alternatively:
1--Let the speed (rate) of the hour hand be "r" and the speed (rate) of the minute hand be "12r" in minutes per minute.
2--Let the distance the hands move be measured by the minutes of the hour.
3--We are looking for the distance (in minutes) that the minute hand must move to overtake, and be coincident with, the hour hand.
4--Let this distance (or time in minutes) be "x."
5--Let the distance the hour hand must move in the same time period be (x - 20), the hour hand already being located at the "20 minutes after the hour" point.
6--As time is equal to distance divided by speed, the time for the minute hand to reach the point of coincidence is given by x/12r.
7--Similarly, the time for the hour hand to reach the same point is given by (x - 20)/r.
8--SInce the two hands each move the same amount of time, we can write x/12r = (x -20)/r or x = (12/11)20 = 21 9/11 = 21.8181 minutes.
9--Therefore, the two hands coincide at 04:21:49.09 or 4h-21m-49.09s after 12.
10--It is worth noting that in the expression x = (12/11)20, the quantity 20 represents the number of minutes that the hour hand is ahead of the minute hand to start out with.
11--Therefore, the value of 12/11 represents the number of minutes required for the minute hand to overtake the hour hand, per minute of head start.
12--For example, if we were seeking the time after 9 o'clock that the two hands overlapped, we need only multiply (12/11)45 = 49.0909 to derive the time of 09:49.0909 or 09h-49m-5.45s after 12.

Trucks A, B, and C, working together, can move a load of sand in t hours. When working alone, it takes truck A one extra hour to move the sand; B, six extra hours; and C, t extra hours. Find t.

The following variation of the 3 people version should give you a clue to how to solve the sand moving problem.

It takes Al 5 hours to paint a shed, Ben 10 hours and Charlie 15 hours. How long would it take all three to paint the shed working together?

1--A can paint the shed in 5 hours.
2--B can paint the shed in 10 hours.
3--C can paint the shed in 15 hours.
4--A's rate of painting is 1 shed per A hours (5 hours) or 1/A (1/5) shed/hour.
5--B's rate of painting is 1 shed per B hours (10 hours) or 1/B (1/10) shed/hour.
6--C's rate of painting is 1 shed per C hours (15 hours) or 1/C (1/15 shed/hour.
7--Their combined rate of painting is therefore 1/A + 1/B + 1/C = (AC + BC + AB)/ABC = (1/5 + 1/10 + 1/15) = (11/30 sheds /hour.
8--Therefore, the time required for all of them to paint the 1 shed working together is 1 shed/(AC+BC+AB)/ABC sheds/hour = ABC/(AC+BC+AB) = 5(10)15/[5(15)+10(15)+5(10) = 30/11 hours = 2.7272 hours = 2hr-43min-38.18sec.

Note - The time required to complete a single "specific task" by three individuals working together, who can complete the task individually in A, B, and C units of time is ABC/(AC + BC + AB).