2009 DAT QRC

Started by cobbby
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cobbby

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Hello
Anybody has the explanation for 2009 DAT QR SECTION specially # 4,6,14?
Thanks
 
I don't have the exam on me, but that exam is riddled with errors - just a heads up (Not saying those questions have errors, but if you think something is inccorectly marked - it may be).
 
Hello
Anybody has the explanation for 2009 DAT QR SECTION specially # 4,6,14?
Thanks


4.
tan(2θ) = AB/DB
AB = DB*tan(2θ)
AB = DB*2tanθ/(1-tan^2(θ))
AB = DB*2(DE/FE)/(1-(DE/FE)^2)

Right now we have to find DE, FE and DB
x^2 + y^2 = 25
x^2 = 25 - y^2 = 25 - 3^2 = 16
x = ±4
So point D is (4, 3). Therefore FE is 4, DE is 3 and DB is 12 - 4 = 8

AB = 8*2(3/4)/(1-(3/4)^2) = 192/7
AC = AB + BC = 192/7 + 3 = 30 3/7

AC starts from y = 0 and goes up 30 3/7 so the y is 30 3/7

6.
The goal is to find the line equation for DE.

Slope of AB = -3/2
Thus slope of BC, CD and DE are 3/2, -3/2 and 3/2 respectively. This is due to the nature of incident and refraction angles.

BC: y = 3/2x + b
b= 0 - 3/2(2) = -3
y = 3/2x -3

Point C coordinate is (x, 10)
Finding x: 10 = 3/2x -3 , x = 26/3
C (26/3, 10)

CD: y = -3/2x + b
b = 10 + 3/2(26/3) = 23
y = -3/2x + 23

Point D coordinate is (10, y)
Finding y: y = -3/2(10) + 23, y = 8
D (10, 8)

Note: it's not possible for CD to hit the bottom of the box.

DE: y = 3/2x + b
b = 8 - 3/2(10) = -7
y = 3/2x - 7
Point E coordinate is (x, 0)
Finding x: 0 = 3/2x - 7, x = 14/3

14.
You use reference angle here
OP = SQRT(8^2+15^2) = 17
Cos(α) = - cos(γ) = -(8/17)

Cosine of any angle in second quadrant is negative.
 

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Dayum, dude went hero mode. Super generous post

4.
tan(2θ) = AB/DB
AB = DB*tan(2θ)
AB = DB*2tanθ/(1-tan^2(θ))
AB = DB*2(DE/FE)/(1-(DE/FE)^2)

Right now we have to find DE, FE and DB
x^2 + y^2 = 25
x^2 = 25 - y^2 = 25 - 3^2 = 16
x = ±4
So point D is (4, 3). Therefore FE is 4, DE is 3 and DB is 12 - 4 = 8

AB = 8*2(3/4)/(1-(3/4)^2) = 192/7
AC = AB + BC = 192/7 + 3 = 30 3/7

AC starts from y = 0 and goes up 30 3/7 so the y is 30 3/7

6.
The goal is to find the line equation for DE.

Slope of AB = -3/2
Thus slope of BC, CD and DE are 3/2, -3/2 and 3/2 respectively. This is due to the nature of incident and refraction angles.

BC: y = 3/2x + b
b= 0 - 3/2(2) = -3
y = 3/2x -3

Point C coordinate is (x, 10)
Finding x: 10 = 3/2x -3 , x = 26/3
C (26/3, 10)

CD: y = -3/2x + b
b = 10 + 3/2(26/3) = 23
y = -3/2x + 23

Point D coordinate is (10, y)
Finding y: y = -3/2(10) + 23, y = 8
D (10, 8)

Note: it's not possible for CD to hit the bottom of the box.

DE: y = 3/2x + b
b = 8 - 3/2(10) = -7
y = 3/2x - 7
Point E coordinate is (x, 0)
Finding x: 0 = 3/2x - 7, x = 14/3

14.
You use reference angle here
OP = SQRT(8^2+15^2) = 17
Cos(α) = - cos(γ) = -(8/17)

Cosine of any angle in second quadrant is negative.
 
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I wouldn't bother with this problem. There is no way you can finish them in 1 minute. Furthermore, every problem is weigh the same. Focus on the fundamentals and concepts that are taught in Math Destroyer. Don't focus on something that is almost impossible to do in 1 minute.
 
I wouldn't bother with this problem. There is no way you can finish them in 1 minute. Furthermore, every problem is weigh the same. Focus on the fundamentals and concepts that are taught in Math Destroyer. Don't focus on something that is almost impossible to do in 1 minute.
Ya, its pretty in depth. I would still understand it, but its unlikely to see something of this magnitude on the real DAT