Question

Brandy set her watch 4 seconds behind and it falls behind another 1 second every day. How many days had it been since brandy last set her watch if the watch is 22 seconds behind?

Answer 1

Its been 19 days since brandy last set her watch as the watch is 22 seconds behind.

Explanation:

We are given the following in the question:

Brandy set her watch 4 seconds behind and it falls behind another 1 second every day.

Thus, this forms an arithmetic progression.

4, 5, 6, 7, ...

First term, a = 4

Common difference, d = 1

The
`n^(th)` terms is 22. We have to find the value of n.

The
`n^(th)` terms of A.P is given by

`a_n = a + (n-1)d`

Putting values, we get,

`22 = 4 + (n-1)1\\18 = n-1\\n = 19`

Thus, its been 19 days since brandy last set her watch as the watch is 22 seconds behind.

Answer 2

18 days

Explanation:

Let x represent number of days.

We have been given that Brandy's watch falls 1 second every day. So Brandy's watch will fall
`1x` seconds behind in x days.

We are also told that Brandy set her watch 4 seconds behind, so Brady's watch will be behind by total
`x+4` seconds in x days.

Since Brady's watch is 22 seconds behind, so we will equate
`x+4` by 22 to solve for x as:

`x+4=22`

`x+4-4=22-4`

`x=18`

Therefore, Brandy set her watch 18 days ago.