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Joined: Sun Dec 09, 2012 1:25 pm
Posts: 24
A clock is 18 minutes slow, but is gaining 7½ seconds every hour.
(a) How long will it take for the clock to show the correct time?
Another clock is 30 minutes fast, but is losing 11¼ seconds per hour.
(b) How many minutes fast will the second clock be when the first clock shows the right
time?

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Joined: Sat Dec 03, 2011 3:14 pm
Posts: 688
a) the first clock

needs to gain 18 X 60 secs to show correct time
18 X 60 = 1080 seconds
7.5 secs gained in = 1 hour
so 1080 / 7.5 = 144 hours or 6 days needed to show correct time

b) clock 2

11.25 secs lost = 1 hour
30 X 60 = 1800 secs to lose
1800 / 11.25 = 160 hours or 6 days 16 hours
so 16 hours extra than required time to coincide with clock 1
11.25 X 16 = 180 seconds

So second clock will be 180 seconds or 3 minutes fast when the first clock shows the right time

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Joined: Tue Dec 18, 2012 10:59 am
Posts: 3579
Please tell me this question is not pitched at 10 year old kids!

The rate things are going they will be expecting them to do calibration formula for gcse!

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Joined: Sun Dec 09, 2012 1:25 pm
Posts: 24
leanmeamum wrote:
a) the first clock

needs to gain 18 X 60 secs to show correct time
18 X 60 = 1080 seconds
7.5 secs gained in = 1 hour
so 1080 / 7.5 = 144 hours or 6 days needed to show correct time

b) clock 2

11.25 secs lost = 1 hour
30 X 60 = 1800 secs to lose
1800 / 11.25 = 160 hours or 6 days 16 hours
so 16 hours extra than required time to coincide with clock 1
11.25 X 16 = 180 seconds

So second clock will be 180 seconds or 3 minutes fast when the first clock shows the right time

Thanks a lot. This is from St Paul's Girls School past paper. Can you please advise where I can find similar questions online or any book to practice?

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Joined: Mon Aug 22, 2011 8:20 pm
Posts: 1706
Location: Warwickshire
leanmeamum wrote:
a) the first clock

needs to gain 18 X 60 secs to show correct time
18 X 60 = 1080 seconds
7.5 secs gained in = 1 hour
so 1080 / 7.5 = 144 hours or 6 days needed to show correct time

write this as
18 x 60
---------
7.5

and it can cancel to
18 x 8
which is probably simpler than trying to divide by 1080 by 7.5

leanmeamum wrote:
b) clock 2

11.25 secs lost = 1 hour
30 X 60 = 1800 secs to lose
1800 / 11.25 = 160 hours or 6 days 16 hours
so 16 hours extra than required time to coincide with clock 1
11.25 X 16 = 180 seconds

So second clock will be 180 seconds or 3 minutes fast when the first clock shows the right time

Or you could write it as the number of seconds this one will lose in the time the first one is gaining - i.e. seconds per hour x number of hours / seconds in a minute
(converting 11.25 back to a fraction)
45 x 144 hours / 60
---
4

or

45 x 144
----------
4 x 60
which simply cancels down to
3 x 9
which you can subtract from 30 to give 3.

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Joined: Mon Feb 12, 2007 1:21 pm
Posts: 13993
These questions are pretty tedious and I would have thought the school could have come up with something more interesting for the children to solve.

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Joined: Sat Dec 03, 2011 3:14 pm
Posts: 688
Try Manchester Grammar School Paper 2 which are quite similar

Thanks Okanagan for showing that method as well. I would use that for my calculations but sometimes children get confused with cross multiplication and division so I try to show them the sums as ratios which they seem to find easier to follow.

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