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?

41 minutes before noon

Step-by-step explanation:

The given parameters are;

The time camera A starts taking pictures = 6 AM

The frequency of picture taking by camera A = Once every 11 minutes

The time camera B starts taking pictures = 7 AM

The frequency of picture taking by camera B = Once every 7 minutes

The number of times both cameras take a picture at the same time before noon = 4 times

Let the time the two cameras first take a picture the same time be x, we have;

11·y - 60 = x

7·z  = x

Taking the number of times after 7 camera A snaps and noting that the first snap is 6 minutes after 7, we have

11·b + 6  = x

7·z = x

x is a factor of 7 and 11·b + 6 and x is some minutes after 7

By using Excel, to create a series of values for Camera A based,  on 11·b + 6, and dividing the results by 7 we have the factors of 7 at;

28, 105, 182, and 259 minutes after 7

Given that there are 60 minutes in one hour, we have;

259/60 = 4 hours 19 minutes, which is 11:19 a.m. or 41 minutes before noon.

the answer is 200

step-by-step explanation:

the way the question is worded, it seems to imply the $264 interest is simple interest, not compound interest.

the simple interest for 2 years is $264; the simple interest for 1 year is $132.

$132 interest on a $600 investment means an interest rate of (132/600)*100 = 22%.

the value of $200 at 22% interest compounded annually for three years is

200%281.22%29%5e3

the amount of interest is that value, minus the original $200.

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