# Help with pattern to numbers and average speed.

• Dec 12th 2009, 02:06 AM
Mukilab
Help with pattern to numbers and average speed.
if nyone cn plese do this s fst s possible.

What should be the value of the fourth column?

oh and
Rudolph travels at 993 mph on Christmas Eve in order to deliver presents to all the children.

He returns home, over exactly the same distance, at 331 mph.

What is his average speed over the entire journey?

i got p to 662= $\frac{D}{T}$
• Dec 12th 2009, 02:15 AM
11rdc11
Quote:

Originally Posted by Mukilab
if nyone cn plese do this s fst s possible.

What should be the value of the fourth column?

• Dec 12th 2009, 02:23 AM
Mukilab
Quote:

Originally Posted by 11rdc11

the nswer for the second one is 496.5
• Dec 12th 2009, 02:27 AM
11rdc11
Quote:

Originally Posted by Mukilab

the nswer for the second one is 496.5

I assigned a variable to each color

a= red
b=yellow
c=blue

Now make three equations with the given

a + 2b +c = 78
2a + b + c = 77
3a +b = 81

Now solve these equations by elimnation.

Then plug the values you found for a,b,c into the equation for the final line

a + b + 2c = ?
• Dec 12th 2009, 02:33 AM
Mukilab
yes, sorry, i calmed down a bit after posting and was able to work it out with ease. any answers to my second query?

I used substitution though
• Dec 12th 2009, 02:35 AM
mr fantastic
Quote:

Originally Posted by Mukilab
[snip]
Rudolph travels at 993 mph on Christmas Eve in order to deliver presents to all the children.

He returns home, over exactly the same distance, at 331 mph.

What is his average speed over the entire journey?

i got p to 662= $\frac{D}{T}$

Let distance each way be x.

Then t = (x/993) + (x/331) = (4x)/993.

Then average speed = (2x)/[(4x)/993] = 993/2 mph.
• Dec 12th 2009, 02:49 AM
Mukilab
Quote:

Originally Posted by mr fantastic
Let distance each way be x.

Then t = (x/993) + (x/331) = (4x)/993.

Then average speed = (2x)/[(4x)/993] = 993/2 mph.

how did you keep the 993 there?

I'm assuming you accounted for the 2 distances and two times to get 4x...
• Dec 12th 2009, 02:52 AM
mr fantastic
Quote:

Originally Posted by Mukilab
how did you keep the 993 there?

I'm assuming you accounted for the 2 distances and two times to get 4x...

993 is the common denominator for the total time of the trip. You should know that average speed = distance/time. Distance is 2x. Time is as calculated.
• Dec 12th 2009, 02:53 AM
Mukilab
what would be the common denominator for 33.3 recrring and 50?
• Dec 12th 2009, 03:02 AM
mr fantastic
Quote:

Originally Posted by Mukilab
what would be the common denominator for 33.3 recrring and 50?

What has this got to do with the original question? Post new questions in a new thread. (This question belongs in Pre-algebra).
• Dec 12th 2009, 03:07 AM
Mukilab
Quote:

Originally Posted by mr fantastic
What has this got to do with the original question? Post new questions in a new thread. (This question belongs in Pre-algebra).

the same question basically
A delivery lorry covers its outward journey at 50mph.

It returns, over exactly the same distance at 33.3333mph.

What is the lorry’s average speed over the entire journey?
• Dec 12th 2009, 03:14 AM
mr fantastic
Quote:

Originally Posted by Mukilab
the same question basically
A delivery lorry covers its outward journey at 50mph.

It returns, over exactly the same distance at 33.3333mph.

What is the lorry’s average speed over the entire journey?

You have been shown in detail how to do this sort of question. The only thing different are the numbers. You're expected to learn something from the help you get. Use a calculator if you cannot do the basic arithmetic.