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At 15:00 there were 20 students in the lab. At 15:03 and

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Director
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At 15:00 there were 20 students in the lab. At 15:03 and [#permalink] New post 22 Oct 2007, 22:04
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At 15:00 there were 20 students in the lab. At 15:03 and every three minutes after that 3 students entered the lab. At 15:10 and every ten minutes after that 8 students left the lab. How many students were in the lab at 15:44?

a) 7
b) 14
c) 25
d) 27
e) 30

Pls. explain.
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 [#permalink] New post 22 Oct 2007, 22:46
# of students in = [(42-3)/3 + 1]*3 = 14*3 = 42
# of students out = [(40-10)/10 + 1]*8 = 32
Net = 42-32 + 20 = 30

Ans E
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 [#permalink] New post 23 Oct 2007, 00:30
ywilfred wrote:
# of students in = [(42-3)/3 + 1]*3 = 14*3 = 42
# of students out = [(40-10)/10 + 1]*8 = 32
Net = 42-32 + 20 = 30

Ans E


ywilfred -- whats the formula u used? very fast it seems..
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 [#permalink] New post 23 Oct 2007, 00:47
yash500 wrote:
ywilfred wrote:
# of students in = [(42-3)/3 + 1]*3 = 14*3 = 42
# of students out = [(40-10)/10 + 1]*8 = 32
Net = 42-32 + 20 = 30

Ans E


ywilfred -- whats the formula u used? very fast it seems..


We know there's an inflow of students and an outflow of students.

The inflow comes at every 3 minutes after 3:03pm and each interval there are 3 students added in. So between 1544 and 1503, there are 42 minutes. So there're 14 such intervals and therefore 14*3 = 42 students.

The outflow comes at every 10 minutes after 3:10pm and each interval there are 8 students going out. From 1510 to 1540, there will be 4 such intevals and so 4*8 = 32 students are added.

The nett is 20 - 32 + 42 = 30 student.

Regarding my working... I used:

(42/3 + 1) --> the 42 comes from subtracting 1503 from 1542 because for 3 minute intervals, the last whole divisible by 3 is 1542 and the smallest whole number divisible is the start time given as 1503. I'm not sure the principal behind this (or maybe I forgot) but it worked well for me in cases such as..

How many evens between 21 and 10. I will take (20-10)/2 + 1 = 6
How many terms divisible by 3 between 15 and 1. I will take (15-3)/3 + 1 = 5
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 [#permalink] New post 23 Oct 2007, 05:30
ywilfred wrote:
yash500 wrote:
ywilfred wrote:
# of students in = [(42-3)/3 + 1]*3 = 14*3 = 42
# of students out = [(40-10)/10 + 1]*8 = 32
Net = 42-32 + 20 = 30

Ans E


ywilfred -- whats the formula u used? very fast it seems..


We know there's an inflow of students and an outflow of students.

The inflow comes at every 3 minutes after 3:03pm and each interval there are 3 students added in. So between 1544 and 1503, there are 42 minutes. So there're 14 such intervals and therefore 14*3 = 42 students.

The outflow comes at every 10 minutes after 3:10pm and each interval there are 8 students going out. From 1510 to 1540, there will be 4 such intevals and so 4*8 = 32 students are added.

The nett is 20 - 32 + 42 = 30 student.

Regarding my working... I used:

(42/3 + 1) --> the 42 comes from subtracting 1503 from 1542 because for 3 minute intervals, the last whole divisible by 3 is 1542 and the smallest whole number divisible is the start time given as 1503. I'm not sure the principal behind this (or maybe I forgot) but it worked well for me in cases such as..

How many evens between 21 and 10. I will take (20-10)/2 + 1 = 6
How many terms divisible by 3 between 15 and 1. I will take (15-3)/3 + 1 = 5



Thanks for the great explanation! Wish I could work out Math like you :)
  [#permalink] 23 Oct 2007, 05:30
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At 15:00 there were 20 students in the lab. At 15:03 and

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