Shiv2016 wrote:

nguyendinhtuong wrote:

MathRevolution wrote:

An alarm of a certain clock rings every 15 minutes. If the clock’s alarm rings first at 12:00, when will the 15th alarm ring?

A. 14:30

B. 14:00

C. 14:30

D. 15:00

E. 15:30

The 15th alarm ring will be after the first one 15 * 14 = 210 min = 3 hours 30 minutes.

Hence, the 15th alarm ring will be at: 12:00 + 3:30 = 15:30

Answer E

Hi

Can you please explain how did you get *14 ?

I solved it this way. Though not sure how great this method is.

First alarm: 12:00

Second alarm: 12:15

Third alarm: 12:30

Fourth alarm: 12:45

Fifth alarm: 13:00

After four alarms, we come to the next hour.

So I divided 15 by 4 which gives me 3.7 (so I chose E).

Shiv2016 : You were so close! These kinds of problems used to make me crazy.

First, your method was fine. (To write 15 values is doable.) You just concluded the wrong thing from not going far enough. Had you continued, you would have seen that ring #5 falls on the hour of 1, but ring #10 falls at 13:15, and ring #15 at 15:30. Neither is on the hour.

Second, the reason we use 14 can be explained in a few ways.

I can't tell exactly what isn't clear to you, so I'll try a couple and hope they help.

Idea # 1: you have to exclude the start point. The clock struck one time at 12. You have to account for that one time.

The fifteenth ring, where ring #1 has already occurred, is 1

plus [some number of rings] = 15 rings. [Some number] is 14.

Essentially, you're finding the 15th term in a sequence, using common difference plus number of terms to do it.

For the series, define common difference as .25 (of an hour):

\(A_n = A_1 + (n-1).25\)

A\(_1\) = 12

A\(_2\) = 12 + (1)(.25)

A\(_3\) = 12 = (2) (.25)

.

.

A\(_{15}\) = 12 + (14)(.25). That yields 12 + 3.5. Convert .5 to minutes, and you have 12:00 + 3:30 = 15:30.

Notice that the first term, A\(_1\) is already accounted for.

So the number of total terms is 15, but the number of terms

after the first term is

14. You then add the combined value of 14 terms to the first term.

14 terms * 15 minutes each = 210 minutes that elapse after the first strike/term.

210 minutes/60 minutes yields # of hours. 210/60 = 3.5, which is 3 hours, 30 minutes. 12:00 + 3:30 = 15:30

Idea #2: you can think about it in distance terms. Imagine fence posts and rope between them.

Imagine the distance between posts is 15 feet. There are 15 fence posts - but only 14 intervals of rope.

Try with small numbers to see the pattern. Say there are 5 fence posts with ropes of 15 ft between them.

|--15--|--15--|--15--|--15--|

1-------2------3-------4-------5

Five posts - but only

four segments between them. Fifteen fence posts? Only

fourteen segments of 15 feet each.

Think of 15 minutes as a similar distance.

Time

1_____

15_____

2_____

15_____

3_____

15_____

4_____

15_____

5_____

15_____Time

612:00___________ 12:15_________12:30________12:45_________1:00_____________1:15

Six marked times. Only five segments.

(You could finish your list the "fencepost" way, as I have started to above. Keep adding 15 minutes, write the time, and count carefully until you have 15 times written down. You'll see only 14 quarter-hours between the first and last time. Same idea.)

Hope it helps! If it doesn't, keep asking questions.

_________________

At the still point, there the dance is. -- T.S. Eliot

Formerly genxer123