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# How to do these 2 Math problems?

Registered User Posts: 595 Member
edited September 2013
http://i. imgur .com/UfsT9za.png (no spaces)
http://i. imgur. com/RXEDtpi.png
Post edited by MyRealName on

## Replies to: How to do these 2 Math problems?

• Registered User Posts: 595 Member
Answers are 80/3/26.6/26.7, and 101, respectively
• Registered User Posts: 260 Junior Member
for the 1st qs
since the car travelled 10 miles with a speed of 20m/h, so the car travelled exactly half an hour.
Likewise, for the second 10 miles, the car travelled 1/4 hour, which is 15 minutes.

So in general, the car travelled 20 miles in 1/2+1/4 hour
so we know the speed is 20/0.75 = 80/3

For the 2nd question
the fence actually is the flap places at both sides of the stretch. So you know that there will be totally 500/5 stretchs, but the fence will have one more number to be added since the fence separates the stretch into 50 pieces.

Let me know if you have any questions. :)
• Registered User Posts: 595 Member

For the 2nd question: How come you add only one more number? There are two posts placed at each end, so shouldn't you add two?
• Registered User Posts: 1,289 Senior Member
• Registered User Posts: 1,079 Senior Member
Try thinking if it this way: what if they asked you how many posts form the left-hand border of a panel? In that case, there would be 100 panels, each with one left-hand border-post. Now, over-all, does that leave out any posts? Just the one post at the right-most end of the fence. That is the only one that isn't also a left-hand border post so it is the only one you left out.
• Registered User Posts: 25,441 Senior Member
Try google searching this site for your problems. If they come from an official TCB book, chances are that they have been answered multiple times with different solutions.

For your first, there is nothing easier than applying the simple formula

2.S1.S2 / S1 + S2 when S is the speed given in the typical SAT problem. Apply it here and you get

2.20.40 / 20+40 or 800/30 and 80/3.

Takes about 5 to 10 seconds, once you know how to recognize the problem.
This discussion has been closed.