Crack DAT Math

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

KFChicken

New Member
10+ Year Member
15+ Year Member
Joined
Sep 9, 2008
Messages
3
Reaction score
0
I am so pissed at all the geometry problems on there. Is there a lot of GEomery problems on the DAT? I am lost and I hate this so much. their #2 is so confusing as wel2) How long wil 2 trains 520m and 280m long and travelling 38 and 30km/hr, respectively, tae to aps one another completely whenthey are travelling int he same direction? It pisses my brain off... please help... i have my DAT on 9/30

Members don't see this ad.
 
I am so pissed at all the geometry problems on there. Is there a lot of GEomery problems on the DAT? I am lost and I hate this so much. their #2 is so confusing as wel2) How long wil 2 trains 520m and 280m long and travelling 38 and 30km/hr, respectively, tae to aps one another completely whenthey are travelling int he same direction? It pisses my brain off... please help... i have my DAT on 9/30

You think you've got it bad I have my DAT in four days! :barf:


I'm wondering the same thing so bump.
 
I am so pissed at all the geometry problems on there. Is there a lot of GEomery problems on the DAT? I am lost and I hate this so much. their #2 is so confusing as wel2) How long wil 2 trains 520m and 280m long and travelling 38 and 30km/hr, respectively, tae to aps one another completely whenthey are travelling int he same direction? It pisses my brain off... please help... i have my DAT on 9/30


Those train and related problems are easy. Let me break it down for you since Im bored.

Start with:
Distance = Rate * Time
They ask for how long so solve for time:
Time = Distance / Rate

Total distance is 520+280 = .800 km (remember units)
Total rate = 38-30 = 8 km/hr

(You subtract the rates since they are traveling in the same direction, if they were going in opposite directions you add the rates. You can easily remember this if you recall traveling in a car and seeing the relative speeds of cars traveling in the same direction as your car vs. cars traveling in opposite directions. That is, cars going in the same direction relative to you go slow, and those going opposite to you appear really fast.)

Time = Distance / Rate = .800/8 = .1 hr = .1*60 min = 6 min

Hope this helps.
 
Pretty sure the above is wrong because you do not add the lengths of the trains, you need to calculate how long it will take for the back of the 520 train to clear the front of the 280 train. That means assuming that both trains start in the same position, you need to calcultae how long it will take the 520 train to travle 280m farther than the 80 train. This way it will have cleared the 280 train completely. To solve this you set up the equation,

38*x - 30*x = .280

Where x is the number of hours and .280 is the number of additional KM needed to travel. Solving for X gives .035 which is in hoursand converting to minutes it would be 2.1 minutes.

(P.S. on a side note the question doesnt say whether the front of the trains are in the same place or the bac of the trains are at the same place. Above was for the back of the trains start at the same place. If the fronts are, then the 520 train has to go 520m farther than the 280 so the equation would be 38x-30x=.52 which would equal .065 and converted to minutes is 3.9)
 
Members don't see this ad :)
I'm with FuzzyChicken on this. I think the trains start out such that they don't overlap.

Please explain what fuzzychicked did, why is she adding to calculte the total distance of both trains. I calculated for no overlap on bottom. What was wrong with the way I did it???
 
more CHEM q. 107
which specie is paramagnetic?
a. Ti +4 has[Ar] configuration
b. Al 3+ has [Ne] config
c. Ne is noble gas
d. F- has [Ne] config
e. Fe +2 3d6 (d subshell holds 10 e-, and the electrons are unpaired therefore is paramagnetic) ANS.

i dont k now why Fe2+ doesnt have Cr configuration instead???on the solution they said Fe2+ has Fe2+

Cr will still be unpaired.
 
Please explain what fuzzychicked did, why is she adding to calculte the total distance of both trains. I calculated for no overlap on bottom. What was wrong with the way I did it???

[==TRAIN A==]
....................[====TRAIN B====]


That's how they start out. Train B travels faster by 8mph. Not only does the front of train B have to clear the front of train A (distance = 280m) but then the back of train B has to clear the front of train A (distance = 520m) for a total of 800m. Just using that picture, the back of train B has to end up at the front of train A. You can see that the distance from the back of train B to the front of train A equals the distance of the two trains added together.
 
sorry.i' was trying to post a new thread.

it's my thread!!! lol... jkThanks everyone... I hate CRack DAT Math!... anyone have any opinion on how to study for Geometry? I don't want to miss all 4 questions on Geometry. Thanks...KFC
 
[==TRAIN A==]
....................[====TRAIN B====]


That's how they start out. Train B travels faster by 8mph. Not only does the front of train B have to clear the front of train A (distance = 280m) but then the back of train B has to clear the front of train A (distance = 520m) for a total of 800m. Just using that picture, the back of train B has to end up at the front of train A. You can see that the distance from the back of train B to the front of train A equals the distance of the two trains added together.

Oh, i didnt even think of the trains being aligned like this. Hmph stupid question. Thanks for explaining.
 
I think their intent was for a more simpler anwer.

Both trains start out head to head as in any race.... Longer train .520 is faster
Shorter train is .280 and is slower .....

The tail of the faster/longer train has to pass the head of the slower train, it needs to make up the difference of .520 the differential in speed is 8km/hr, thus a simple answer of .520/8
 
Last edited:
To answer your other question, YES - there are a lot of geometry problems! Study angle relationships and inscribed shapes, if you can.
 
how bout those pain in the butt trig functions and rules lol the crap i forgot about since i last took algebra in high school.
 
yep, all of those... nothing TOO complicated, but you definitely need to know how to find a sin, cos, secant, etc..
If I were going to do it again I'd memorize the most common ones - sin/cos of 30/45/60 degrees, that sort of thing.
 
Top