Problema Solution
There are two cameras that take pictures of a traffic intersection. Camera A starts taking pictures at $6$ AM and takes a picture every $11$ minutes. Camera B starts taking pictures at $7$ AM and takes pictures every $7$ minutes. Camera A and Camera B take a picture at the same time at four different times before noon. When Camera A and Camera B take their last picture together, how many minutes before noon is it?
Answer provided by our tutors
Camera A starts taking pictures at 6 AM and takes a picture every 11 minutes.
Camera A takes picture at 6:55 AM
Camera B takes picture at 7:00 AM
Camera A takes picture at 7:06 AM
lets say that after x times that Camera B takes picture, Camera A also takes picture at the same time as Camera B that is:
7x = 6 + 11y
y = (7x - 6)/11
we need to find the first integer y that satisfies the above equation:
for x = 1, y = 1/11
for x = 2, y = 8/11
for x = 3, y = 15/11
for x = 4, y = 22/11
for x = 5, y = 29/11
for x = 6, y = 36/11
for x = 7, y = 43/11
for x = 8, y = 50/11
for x = 9, y = 57/11
for x = 10, y = 64/11
for x = 11, y = 71/11
for x = 12, y = 78/11
for x = 13, y = 85/11
for x = 14, y = 92/11
for x = 15, y = 99/11 = 9
after 15*7 = 105 min the cameras will take pictures at the same time, for the first time.
the next time (second time) fill happen after:
LCM(11, 7) = 77 min
this means that after 77+105 = 182 minutes Camera A and Camera B will take a picture at the same.
third time after 2*77 + 105 = 259 min
forth time after 3*77 + 105 = 336 min
336 min = 6 hr 36 min after 7 am is 1 pm 36 min.
Camera A and Camera B take their last (forth) picture together 1 hour and 36 min after noon.