SOLUTION: it takes Mars 1.88 years to travel around the sun.It takes Mercury 0.24 years to orbit the sun. If the planets are in a straight line with the sun now, how long will it be before t
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Question 87072: it takes Mars 1.88 years to travel around the sun.It takes Mercury 0.24 years to orbit the sun. If the planets are in a straight line with the sun now, how long will it be before this is the case again? The answer must be accurate to the nearest hundredth of a year. Found 2 solutions by stanbon, Edwin McCravy:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! it takes Mars 1.88 years to travel around the sun.It takes Mercury 0.24 years to orbit the sun. If the planets are in a straight line with the sun now, how long will it be before this is the case again? The answer must be accurate to the nearest hundredth of a year.
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You need the least common multiple of 1.88 and 0.24
In 282 years mars will have orbited 150 times
In 282 years mercury will have orbited 1175 times.
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Cheers,
Stan H.
it takes Mars 1.88 years to travel around the sun.It takes Mercury 0.24 years
to orbit the sun. If the planets are in a straight line with the sun now, how
long will it be before this is the case again? The answer must be accurate to
the nearest hundredth of a year.
We must find the smallest positive integer N
such that when we multiply both 1.88 and .24 by N
we get a positive integer, that is
1.88N = P , whener P is a positive integer
and
0.24N = Q , whener Q is also a positive integer
Multiply both equations by 100 to remove decimals:
188N = 100P
24N = 100Q
Solve each for N
N =
N =
N =
N =
So we have
=
Divide both sides by 25
=
Cross multiply:
=
Divide both sides by 6Q
= =
The smallest positive integer values of P and Q
which make that equation true are P = 47 and
Q = 6.
Substitute 47 for P in
1.88N = P
1.88N = 47
N =
N = 25
As a check:
Substitute 6 for Q in
0.24N = Q
0.24N = 6
N =
N = 25
So the answer is 25 years. That's the next time when
both planets will have each completed a whole number
of revolutions.
Edwin