Dimensional Change

There are lots of questions on dimensional change and this is a very common one.

Make sure you understand the relationship among linear, 2-D (area) and 3-D (volume) ratio.

There are many similar triangles featured in the image on the left.
Each of the two legs of the largest triangles is split into 4 equal side lengths.





                                                                                            


Question : What is the area ratio of the sum of the two white trapezoids to the largest triangle? 
\(\dfrac {\left( 3+7\right) } {16}=\dfrac {10} {16}=\dfrac{5}{8}\)  

Question: If the area of the largest triangles are 400 square units, what is the area of the blue-colored trapezoid?
\(\dfrac {5} {16}\times 400\) =125 square units 

Again, each of the two legs are split into three equal segments. 

The volume ration of the cone on the top to the middle frustum to the 
bottom frustum is 1 : 7 : 19. 
 
Make sure you understand why.










 

2019 AMC 8 problems, solutions and some thoughts

2019 AMC 8 problems and solutions, for students, by students

A student's reflection on this year's test : 

Mrs. Lin,
I did the AMC 8 yesterday, and it was actually quite easier than last year. I was reviewing my answers, and I believe I only got the last two wrong. I used stars and bars for the last one, but did 21C2 instead of 20C2. I could’ve done number 24, because geometry is really my best subject in math. I realized that I should’ve used mass points later on. It’s fine, though, because it’s still a good score. I think that many people could solve this test pretty well because in many of the last questions you could plug in the multiple choice answers and get the right answer. Also, a lot of it was just plain algebra. Question 20 was actually just an equation, which anybody who knows basic algebra can solve. I thought that I would never say this, but I honestly wish that it was harder, because I was hoping for some interesting problems. Those are the problems that get people’s gears turning; this year the problems were quite basic. I think many people will get really good scores on this test, which, along with a good thing, is also not so good because it brings down the credibility of the test.

Thanks,

some links that you can review those very basic, but extremely useful strategies on this 
year's seemingly harder, but not really last two questions. 

mass points  learn together with triangles sharing the same vertex 

dimensional change / scaling 

balls and urns, stars and bars  (lots of variations or twists on this one, so 
you need to fully understand the concept so to use it well. Be patient !!!!!) 

Hints/links or Solutions to 2014 Harder Mathcounts State Sprint and Target question

Links, notes, Hints or/and solutions to 2014 Mathcounts state harder problems.

2014, 2015 Mathcounts state are harder 

Sprint round:

#14 :
Solution I :

(7 + 8 + 9)  + (x + y + z)  is divisible by 9, so the sum of the three variables could be 3, 12, or 21.
789120 (sum of 3 for the last three digits) works for 8 but not for 7.
21 is too big to distribute among x, y and z (all numbers are district),
thus only x + y + z = 12 works and z is an even number
__ __ 0 does't work (can't have 6 6 0 and the other pairs all have 7, 8 or 9)
264 works (789264 is the number)

Solution II : 
789000 divided by the LCM of 7, 8 and 9, which is 504 = 1565.47...
Try 504 * 1566 = 789264 (it works)
The answer is 264.

#18:
Watch this video from Mathcounts mini and use the same method for the first question,
you'll be able to get the answer. It's still tricky, though.

#23 : Drop the heights of the two isosceles triangles and use similar triangles to get the length of FC.
Then solve.

#24: 
The key is to see 2¹⁰ is 1024 or about 10³

2³⁰ = ( 2¹⁰ )³  or about (10³  )³about 10⁹ so the answer is 10 digit.

#25:
As you can see, there are two Pythagorean Triples : 9-12-15 and 9-40-41.
Base (40-12) = 28 gives you the smallest area.
The answer is 28 * 18 = 504

#26 : Let there be A, B, C three winners. There are 4 cases to distribute the prizes.
A     B    C
1      1     5    There are 7C1 * 6C1 * \( \dfrac {3!} {2!}\) = 126 ways -- [you can skip the last part for C
because it's 5C5 = 1]

1       2    4    There are 7C1* 6C2 * 3! = 630

1      3     3    There are 7C1 * 6C3 * \( \dfrac {3!} {2!}\) = 420

2      2    3     There are 7C2 * 5C2 * 3 (same as above)

Add them up and the answer is 1806.

If you can't see why it's \( \dfrac {3!} {2!}\) when there is one repeat, try using easier case to help you understand.

What about A, B two winners and 4 prizes ?
There are 2 cases, 1 3 or 2 2, and you'll see how it's done.

#27 : Read this and you'll be able to solve this question at ease, just be careful with the sign change.
Vieta's Formula and the Identity Theory

#28: There are various methods to solve this question.
I use binomial expansion :
\(11^{12}=\left( 13-2\right)^{12}=12C0*13^{12}\)+ \(12C1*13^{11}*2^{1}\)+... \(12C11*13^{1}*2^{11}\)+ \(12C12*2^{12}\) Most of the terms will be evenly divided by 13 except the last term, which is \(2^{12}\) or 4096, which, when divided by 13, leaves a remainder of 1.

Solution II :
\(11\equiv -2\left ( mod13\right)\) ; \((-2)^{12}\equiv 4096\equiv 1\left ( mod13\right)\)

Solution III :  
Or use Fermat's Little Theorem (Thanks, Spencer !!)
\(11^{13-1}\equiv 11^{12}\equiv1 (mod 13)\)

Target Round : 

#3: Lune of Hippocrates : in seconds solved question.
^__^

#6: This question is very similar to this Mathcounts Mini.
My students should get a virtual bump if they got this question wrong.

#8: Solution I : by TMM (Thanks a bunch !!)
Using similar triangles and Pythagorean Theorem.

The height of the cone, which can be found using the Pythagorean  is . 
Usingthediagram below, let  be the radius of the top cone and let  be the height of the topcone. 
Let  be the slant height of the top cone.


Drawing the radius as shown in the diagram, we have two right triangles. Since the bases of the top cone and the original cone are parallel, the two right triangles are similar. So we have the proportionCross multiplying yields This is what we need.

Next, the volume of the original cone is simply . 
The volume of the top cone is .
From the given information, we know that We simply substitute the value of  from above to yield We will leave it as is for now so the decimals don't get messy.

We get  and .

The lateral surface area of the frustum is equal to the lateral surface area of the original cone minus the lateral surface area of the top cone. The surface area of the original cone is simply 
. 
The surface area of the top cone is . 
So our lateral surface area is 

All we have left is to add the two bases. The total area of thebases is . So our final answer is 

Solution II : 
Using dimensional change and ratio, proportion.

Cut the cone and observe the shape.

The circumference of the larger circle is 20pi (10 is the radius) and the base of
the cone circle circumference is 10pi (5 is the radius), which means that the cut-off cone shape is a half circle because it's \(\dfrac {10\pi } {20\pi }\) or \(\dfrac {1 } {2 }\) of the larger circle. (180 degrees)

To find the part that is the area of the frustum not including the top and bottom circles,
you use the area of the half circle minus the area of the smaller half circle.

Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the
two radius is \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).

Using this ratio, we can get the radius of the smaller circle as 10 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\) and the radius of the top circle of the frustum as 5 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).

Now we can solve this :
 \(\dfrac {1 } {2 }\)\(\left[ 10^{2}\pi -\left( 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\right) ^{2}\pi \right] \) + \(5^{2}\pi +\left( 5\times \dfrac {\sqrt [3] {2}} {\sqrt {3}}\right) ^{2}\pi \) = about 176 (after you round up)ional change and ratio, proportion.

Cut the cone and observe the shape.

The circumference of the larger circle is 20pi (10 is the radius) and the base of
the cone circle circumference is 10pi (5 is the radius), which means that the cut-off cone shape is a half circle because it's \(\dfrac {10\pi } {20\pi }\) or \(\dfrac {1 } {2 }\) of the larger circle. (180 degrees)

To find the part that is the area of the frustum not including the top and bottom circles,
you use the area of the half circle minus the area of the smaller half circle.

Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the
two radius is \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).

Using this ratio, we can get the radius of the smaller circle as 10 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\) and the radius of the top circle of the frustum as 5 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).

Now we can solve this :
 \(\dfrac {1 } {2 }\)\(\left[ 10^{2}\pi -\left( 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\right) ^{2}\pi \right] \) + \(5^{2}\pi +\left( 5\times \dfrac {\sqrt [3] {2}} {\sqrt {3}}\right) ^{2}\pi \) = about 176 (after you round up)

Solution III : Another way to find the surface area of the Frustum is : 
median of the two half circle [same as median of the two bases] * the height [difference of the two radius]
\(\dfrac {1} {2}\left( 2\times 10\pi + 2\times 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\pi \right)\)* \(\left( 10-10\times \dfrac {\sqrt [3] {2}} {\sqrt [3]{3}}\right)\)

Dimensional Change questions I:

Questions written by Willie, a volunteer.  Answer key and detailed solutions below.

1a. There is a regular cylinder, which has a height equal to its radius. If the radius and height are both increased by 50%, by what % does the total volume of the cylinder increase?

1b. If the radius and height are both decreased by 10%, by what % does the total volume of the cylinder decrease?

1c. If the radius is increased by 20% and the height is decreased by 40%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

1d. If the radius is increased by 40% and the height is decreased by 20%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

1e. If the height is increased by 125%, what % does the radius need to be decreased by for the volume to remain the same?

2. If the side of a cube is increased by 50%, by what % does the total surface area of the cube increase?

3a. If the volume of a cube increases by 72.8%, by what % does the total surface area of the cube increase?

3b. By what % did the side length of the cube increase?

4. You have a collection of cylinders, all having a radius of 5. The first cylinder has a height of 2, the second has a height of 4, the third a height of 6, etc. The last cylinder has a height of 50. What is the sum of the volumes of all the cylinders (express your answer in terms of pi)?

Answer key: (Each question should not take you more than 30 seconds to solve if you really understand the concepts involved.)

 

 

1a.  The volume of a cylinder is πr²x h (height). The radius itself will be squared and the height stays at constant ratio. The volume will increased thus (1.5)³ - 1³ -- the original 100% of the volume = 2.375

=237.5%

1b.  Like the previous question: 1³ - 0.9³ [when it's discount/percentage decrease, you use the 100% or 1 - the discount/decrease percentage] = 0.271 =  27.1% decrease

1c.  1.2² [100% + 20% increase = 1.2] x 0.6 [100% -40% = 0.6] = 0.864  or  

86.4% of the original volume

1d.  1.4² [100% + 40% increase = 1.4] x 0.8 [100% -20% = 0.8] = 1.568 = 156.8% of the original volume

1e.  When the height of a cylinder is increased 125%, the total volume is is 225% of the original cylinder, or 9/4.

Since the radius is used two times (or squared), it has to decrease 4/9^1/2 = 2/3 for the new cylinder to have the same volume as the old one. [9/4 times 4/9 = 1 or the original volume.]

1 - (2/3) = 1/3 = 0.3 = 33.3%

2. Surface area is 2-D so 1.5² - 1 = 1.25 = 125% increase

3a. If a volume of a cube is increased by 72.8 percent, it's 172.8% or 1.728 of the original volume. Now you are going from 3-D (volume) to 2-D (surface area). 1.728^2/3 = 1.44 or 44% increase. [Don't forget to minus 1 (the original volume) since it is asking you the percentage increase.]

3b. From surface area, you can get the side increase by using 1.44^1/2 = 1.2, so 20% increase.

Or you can also use 1.728^1/3 = 1.2;  1.2 - 1 = 20%

4. The volume of a cylinder is πr²x h . (2 + 4 + 6 + ...50) x 5²π = (25 x 26) x 25π =16250π

2013 Mathcounts State Harder Problems

 You can download this year's Mathcounts state competition questions here.

Trickier 2013 Mathcounts State Sprint Round questions :
Sprint #14:  
From Varun: 
Assume the term "everything" refers to all terms in the given set.
1 is a divisor of everything, so it must be first.
Everything is a divisor of 12, so it must be last.
The remaining numbers left are 2, 3, 4, and 6.
2 and 3 must come before 6, and 2 must come before 4.
Therefore, we can list out the possibilities for the middle four digits:
2,3,4,6
2,3,6,4
3,2,4,6
3,2,6,4
2,4,3,6
There are 5 ways--therefore 5 is the answer.

From Vinjai:
Here's how I did #14:
First, notice that 1 must be the first element of the set and 12 must be the last one.
So that leaves only 2,3,4,6 to arrange.
We can quickly list them out.
The restrictions are that 2 must be before 4, 3 must be before 6, and 2 must be before 6:
2,3,4,6          3,2,4,6          4,2,3,6            6,2,3,4
2,3,6,4          3,2,6,4          4,2,6,3            6,2,4,3
2,4,3,6          3,4,2,6          4,3,2,6            6,3,2,4
2,4,6,3          3,4,6,2          4,3,6,2            6,3,4,2
2,6,3,4          3,6,2,4          4,6,2,3            6,4,2,3
2,6,4,3          3,6,4,2          4,6,3,2            6,4,3,2

Only the bold ones work. So, the answer is 5.

#17: Common dimensional change problem
\(\overline {ZY}:\overline {WV}=5:8\) -- line ratio
The volume ratio of the smaller cone to the larger cone is thus \(5^{3}: 8^{3}\).
The volume of the frustum is the volume of the larger cone minus the volume of the smaller cone
= \(\dfrac {8^{3}-5^{3}} {8^{3}}\times \dfrac {1} {3}\times 8^{2}\times 32\times \pi\) = 516\(\pi\)

More problems to practice from Mathcounts Mini 

#24:  The answer is \(\dfrac {1} {21}\).

#28: Hats off to students who can get this in time !! Wow!!
From Vinjai:

For #28, there might be a nicer way but here's how I did it when I took the sprint round:

# 4's     # 3's     # 2's       # 1's     # ways

   1         2          0            0          3

   1         1          1            1          24

   1         1          0            3          20

   1         0          3            0          4

   1         0          2            2          30

   1         0          1            4          30

   0         2          2            0          6

   0         2          1            2          30

   0         2          0            4          15

   0         1          3            1          20

   0         1          2            3          60

   0         0          3            4          35

TOTAL: 277

#29: 

\(\Delta ADE\) is similar to \(\Delta ABC\)

Let the two sides of the rectangle be x and y (see image on the left)

\(\dfrac {x} {21}=\dfrac {8-y} {8}\)
x =\(\dfrac {21\left( 8-y\right) } {8}\)

xy =  \(\dfrac {21\left( 8-y\right) } {8}\)  * y = \(\dfrac {-21y\left( y-8\right) } {8}\) =
\(\dfrac {-2l\left( y-4\right) ^{2}+21\times 16} {8}\)

From the previous equation you know when y = 4, the area \(\dfrac {21\times 16} {8}\)is the largest. The answer is 42. 





Here is a proof to demonstrate that the largest area of a rectangle inscribed in a triangle is
half of the area of that triangle.

#30:
Solution I :
If (x, y) are the coordinates of the center of rotational points, it will be equal distance from A and A' as well as from B and B'.
Use distance formula, consolidate/simplify and solve the two equations, you'll get the answer (4, 1).

Solution II:

How to find the center of rotation from Youtube.

From AoPS using the same question

To sum up:
First, connect the corresponding points, in this case A to A' and B to B'.
Second, find the equation of the perpendicular bisector of line \(\overline {AA'}\), which is
y =  - x + 5
and \(\overline {BB'}\), which is y = 5x - 19
The interception of the two lines is the center of rotation.
The answer is (4, 1).

2013 Mathcounts Target :
#3:
RT = D, unit conversions and different rates are tested here:

Make Joy's rate (speed) uphill be x m/s, his downhill speed be 2x m/s.

It takes Greg 3000 seconds (time) to reach the starting point and that is also what it takes Joy to
ride up to the hill and down to the same point.

\(\dfrac {7000} {x}+\dfrac {10000} {2x}=3000\) \(\rightarrow\) x = 4 m/s 

#8:

Using "finding the height to the hypotenuse".( click to review)

 you get \(\overline {CD}=\dfrac {7\times 24} {25}\).

Using similar triangles ACB and ADC, you get  \(\overline {AD}=\dfrac {576} {25}\).
[\(\dfrac {24} {x}=\dfrac {25} {24}\)]

Using angle bisector (click to review),

you have \(\overline {AC}:\overline {AD}=\overline {CE}:\overline {ED}= 24: \dfrac {576} {25}\) = 600 : 576 = 25 : 24

\(\rightarrow\)\(\overline {ED}= \overline {CD}\times \dfrac {24} {24+25}\) = \(\dfrac {7\times 24} {25}\times \dfrac {24} {24+25}\) = \(\dfrac {576} {175}\)

2013 Mathcounts School and Chapter Harder Problems

You can now download and discuss with your friends this year's school and chapter problems.
Here is the link to the official Mathcounts website.

Some more challenging problems from this year's Mathcounts school/or chapter problems.

2013 school team #10 : Three concepts are testing here :
Hint: 
a. If you get rid of the remainder, the numbers will be evenly divided into 192, so you are looking at
those factors of 192 - 12 = 180

b. To leave a remainder of 12, those factors of 180 that are included in the Set must be smaller than 12, otherwise, you can further divide it.

c. To find the median, make sure to line up the numbers from the smallest to the largest and find the middle numbers. If there are even numbers of factors larger than 12, average the middle two. Otherwise, the middle number is the answer.

\(180=2^{2}\times 3^{2}\times 5 \) so there are (2 + 1) (2 + 1) (1 + 1) = 18 factors

The list on the left side gives you the first 9 and if you times those numbers with "5", you get 9 other factors,which are 5, 10, 20, 15, 18, 60, 45, 90 and 180.

Discard the factors that are smaller or equal to 12 and list all the other factors in order and find the median.

The answer is "36".



2013 Chapter Sprint:
#21: Dimensional change problem : The height of the top pyramid is \(\dfrac {2} {3}\) of the larger
pyramid so its volume is \(\left( \dfrac {2} {3}\right) ^{3}\) of the larger pyramid.

\(\left( \dfrac {2} {3}\right) ^{3}\times \dfrac {1} {3}\times \left( \dfrac {36} {4}\right) ^{2}\times 12
= \) \(96  cm^{3}\)

# 24:  According to the given:   \(xyz=720\)   and   \(2( xy+yz+zx)= 484 \) so
\(( xy+yz+zx )= 242\)

Since x, y and z are all integers, you factor 720 and see if it will come up with the same x, y and z values
for the second condition.

Problem writer(s) are very smart using this number because the numbers "6", "10", "12" would give you
a surface area of 252. (not right)

The three corrrect numbers are "8", "9", and "10" so the answer is \(\sqrt {8^{2}+9^{2}+10^{2}}=\) \(7\sqrt {5}\)

#25: Geometric probability: Explanations to similar questions and more practices below. 

Probability with geometry representations form Aops.

Geometric probability from "Cut the Knots".

#26: This one is similar to 2002 AMC-10B #21, so try that question to get more practices. 
2002 AMC-10B #21 link 

#27:
\(\dfrac {1} {A}+\dfrac {1} {B}=\dfrac {1} {2}\)
\(\dfrac {1} {B}+\dfrac {1} {C}=\dfrac {1} {3}\)
\(\dfrac {1} {C}+\dfrac {1} {A}=\dfrac {1} {4}\)
Add them up and you have  \(2 * (\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {12}\)

\((\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {24}\)

\(\dfrac {1} {\dfrac {1} {A}+\dfrac {1} {B}+\dfrac {1} {C}} = \)\(\dfrac {{24}} {13}\) hours

#28: Hint : the nth triangular number is the sum of the first "n" natural numbers and \(\dfrac {n\left( n+1\right) } {2}\) is how you use to find the sum.
From there, you should be able to find how many numbers will be evenly divided by "7".

#29 : Circle questions are very tricky so make sure to find more problems to practice accuracy.

#30 :  
Solution I: 
Read the solution that is provided by Mathcounts.org here.
Solution II:
Case 1 : \(x-1 > 0\rightarrow x > 1\) Times ( x - 1) on both sides and you have
\(x^{2}-1>8\) so x > 3 or x < -3 (discard)

Case 2: \(x-1 < 0\) so \(x < 1\) \(\rightarrow x^{2}-1 < 8\) [You need to change the sign since it's negative.]-3 < x < 3. Combined with x < 1 you have the range as -3 < x < 1
The answer is 60%.

                                               2013 Mathcounts Target #7 and 8: 

Target question #8 is very similar to 2011 chapter team #10. 
It just asks differently.   
Read the explanations provided on the Mathcounts official website.
They are explained very well.
Let me know if there are other easier ways to tackle those problems.

Hope this is helpful !! Thanks a lot !! Good luck on Mathcounts state.

Dimentional Change Questions III: Similar Shapes

There are numerous similar triangle questions on Mathcounts.

Here are the basics:

If two triangles are similar, their corresponding angles are congruent and their corresponding sides will have the same ratio or proportion.

Δ ABC and ΔDEF are similar. \(\frac{AB}{DE}\) = \(\frac{AC}{DF}\) = \(\frac{BC}{EF}\)= their height ratio = their perimeter ratio.







Once you know the linear ratio, you can just square the linear ratio to get the area ratio and cube the linear ratio to get the volume ratio. 

Practice Similarity of Triangles here.  Read the notes as well as work on the practice problems.  There is instant feedback online. 

Other practice sheets on Similar Triangles                                                        

Many students have trouble solving this problem when the two similar triangles are superimposed. 

Just make sure you are comparing smaller triangular base with larger triangular base and smaller triangular side with corresponding larger triangular side, etc... In this case:

\(\frac{BC}{DE}\)= \(\frac{AB}{AD}\) = \(\frac{AC}{AE}\)




Questions to ponder (Solutions below)

#1: Find the area ratio of Δ ABC to trapezoid BCDE to DEGF to FGIH. You can easily get those ratios using similar triangle properties. All the points are equally spaced and line \(\overline{BC}\)// \(\overline{DE}\) // \(\overline{FG}\) // \(\overline{HI}\). 

#2: Find the volume of the cone ABC to Frustum BCDE to DEGF to FGIH. Again, you can use the similar cone, dimensional change property to easily get those ratios.Same conditions as the previous question.

 

Answer key: 

#1:

 #2:

Rate, Time, and Distance Question

Question: Harder SAT question: Esther drove to work in the morning at an average speed of 45 miles per hour. She returned home in the evening along the same route and averaged 30 miles per hour. If Esther spent a total of one hour commuting to and from work, how many miles did Esther drive to work in the morning?

Solutions I:
Let t be the time it took Esther to drive to work and (1-t ) be the time it took him to return home.

Since rate times time = distance, we can set the equation as 45 t = 30 (1- t), 75t = 30 so
t = \(\frac{\Large{2}}{\Large{5}}\) Plug in t to the previous equation : 45 * \(\frac{\Large{2}}{\Large{5}}\) = 18 miles ,which is the answer.

Solution II: 
Let D be the distance from Esther's home to work. 

\(\frac{\Large{D}}{\Large{45}}\) +  \(\frac{\Large{D}}{\Large{30}}\) = 1 (hour)
Times 90 both sides to get rid of the denominator and you have 2D + 3D = 90 
D = 18 miles

Solution III:

The rate ratio between driving to work and returning home is 45 : 30 or 3 :  2.

Since rate and time are inversely related (rt = d), the time ratio between the two is 2 : 3. 
45*\(\frac{\Large{2}}{\Large{5}}\) * 1 (hour) = 18 miles

Solution IV: (from Varun in FL)
The average speed can be found by harmonic mean: \(\dfrac {2\times 45\times 30}{45+35}\) = 36 mph. She drives for a total of 1 hour, so she drives 36 miles. Since she drives the same distance both ways, the distance to work is \(\dfrac {36} {2}\)=18 miles.

Question 1991 Mathcounts National #28 : A man is running through a train tunnel. When he is \(\frac{\Large{2}}{\Large{5}}\) of the way through, he nears a train that is approaching the tunnel from behind him at a speed of 60 mph. Whether he runs ahead or runs back, he will reach an end of the tunnel at the same time the train reaches that end. At what rate, in miles per hour, is he running? (Assume he runs at a constant rate.)

Solution : When the man went back \(\frac{\Large{2}}{\Large{5}}\) of the length of the tunnel, the train is at the beginning of the tunnel. 

Same thing happened if the men went ahead \(\frac{\Large{2}}{\Large{5}}\) of the length of the tunnel. The time it took the train to run through the length of the tunnel, the man could only run \(\frac{\Large{1}}{\Large{5}}\) of the same distance. 

So the speed of the man is \(\frac{\Large{1}}{\Large{5}}\) of the train's speed, which is \(\frac{\Large{1}}{\Large{5}}\) * 60 or 12 mph.  










Question #3: Sally drives to her aunt's house, which is 84 miles away, at 40 miles per hour and comes back home at 60 miles per hour. What is her average speed? 
Solution I: 
To find average speed, you use total distance over total time it takes Sally to drive. 
It takes Sally \(\frac{\Large{84}}{\Large{40}}\) + \(\frac{\Large{84}}{\Large{60}}\), or total 3.5 hours 
to drive 84 * 2 = 168 miles. 
\(\frac{\Large{168}}{\Large{3.5}}\) = 48 mph

Solution II: 
Similar to dimensional change question, as long as the segments are constant, the distance Sally drives is extraneous. Thus, it's much easier to use \(\frac{\Large{2}}{\frac{\Large{1}}{\Large{40}} - \frac{\Large{1}}{\Large{60}}}\)or \(\frac{\Large{2*40*60}}{\Large{40+60}}\) = 48 mph


Question #4: If Sally drives to her work at 40 miles per hour, she will be 15 minutes late. If she drives to her work at 60 miles per hour, she will be 15 minutes early, what speed will she arrive at her work on time?  How far away is her office ? 
Solution I: 
Let t be the time (hpm) Sally drives to her work on time. According to the given, we can set up 
the following equation: 
40 (t + \(\frac{\Large{1}}{\Large{4}}\)) = 60 ( t- \(\frac{\Large{1}}{\Large{4}}\) )
40 t + 10 = 60 t - 15 ;      25 = 20t ;       t = 1.25
Plug in to get the distance as 40 (1.25 + 0.25) = 60 miles

To get the average, use total distance over total time she spent \(\rightarrow\) \(\frac{D}{T}\) = \(\frac{\Large{120}}{\Large{1.5 + 1}}\) = 48 mph

Solution II :
Again, have you noticed that if  both time (late or early) to arrive to the destination is the same, the easiest
method is to still use the harmonic mean. \(\rightarrow\) \(\frac{\Large{2*a * b}}{\Large{a + b}}\)
a and b are the rates. \(\frac{\Large{2*40*60}}{\Large{40+60}}\) = 48 mph