2025/2026 Mathcounts, AMCs, AIMEs Competition Preparation Strategies

Hi, Thanks for visiting my blog.

E-mail me at thelinscorner@gmail.com if you want to learn with me.  :) :) :) 

Currently I'm running different levels of problem solving lessons, and it's lots of fun learning along with students from different states/countries. 

So many students are not learning smart.

Problem solving is really fun (and a lot of the times very hard, yes).

Good questions are intriguing and delicious, so come join our vibrant community and have the pleasure of finding things out on your own.

5 ways to kill your dreams from TED Talk

There is no overnight success.

Testimonials from my students/parents.   :) 

My other blogs :

thelinscorner  : Standardized test preps (ONLY the hardest problems), books, links/videos for life-time learning

Studentsspeak Magazine  (featuring students' independent projects)

Take care and have fun learning.

Don't forget other equally interesting activities/contests, which engage your creativity  and imagination. 
Some also require team work. Go for those and have fun !! 

Don't just do math.  

11 Qualities Google Looks For In Job Candidates 

In Chinese : 《專題報導》亞裔尖子生 職場難出頭？

Before going full throttle mode for competition math, please spend some time reading this
well- thought-out article from BOGTRO at AoPS "Learn How to Learn".

It will save you tons of time and numerous, unnecessary hours without a clear goal, better method in mind.

Less is more. My best students make steady, very satisfactory progress in much less time than those
counterparts who spent double, triple, or even more multiple times of prep with little to show.

It's all about "deliberate practices", "tenacity", and most of all, "the pleasure of finding things out on your own".

Take care and have fun problem solving.

I have been coaching students for many years. By now, I know to achieve stellar performance you need :
Grit (from TED talk), not only that but self-awareness (so you can fairly evaluate your own progress) and a nurturing-caring environment. (Parents need to be engaged as well.)
               
Thanks a lot !!  Mrs. Lin

"Work Smart !!" , "Deliberate practices that target your weakness ", " Relax and get fully rested.", "Pace your time well", "Every point is the same so let go of some questions first; you can always go back to them if time permits."

"It's tremendous efforts preparing for a major event on top of mounting homework and if you are the ones who want to try that, not your parents and you work diligently towards your goal, good for you !!"

"Have fun, Mathcounts changes lives, because at middle school level at least, it's one of those rare occasions that the challenges are hard, especially at the state and national level."

Now, here are the links to get you started: 

Of course use my blog.  Whenever I have time I analyze students' errors and try to find better ways (the most elegant solutions or the Harvey method I hope) to tackle a problem. Use the search button to help you target your weakness area.

Newest Mathcounts' competition problems and answer key

For state/national prep, find your weakness and work on the problems backwards, from the hardest to the easiest. 

Here are some other links/sites that are the best.

Mathcounts Mini : At the very least, finish watching and understanding most of the questions from 2010 till now and work on the follow-up sheets, since detailed solutions are provided along with some more challenging problems.

For those who are aiming for the state/national competition, you can skip the warm-up and go directly to "The Problems" used on the video as well as work on the harder problems afterward.

Art of Problem Solving 

The best place to ask for help on challenging math problems. 

Some of the best students/coaches/teachers are there to help you better your problem solving skills.

                                                             Do Not Rush !!

Awesome site!!
       
For concepts reviewing, try the following three links.
 
Mathcounts Toolbox
 
Coach Monks's Mathcounts Playbook
 
You really need to understand how each concept works for the review sheets to be useful.

To my exasperation, I have kids who mix up the formulas without gaining a true understanding and appreciation of how an elegant, seemingly simple formula can answer myriads of questions.

You don't need a lot of formulas, handbook questions, or test questions to excel.

You simply need to know how the concepts work and apply that knowledge to different problems/situations.

Hope this is helpful!!

The Grid Technique in Solving Harder Mathcounts Counting Problems : from Vinjai

The following notes are from Vinjai, a student I met online. He graciously shares and offers the tips here on how to tackle those harder Mathcounts counting problems. 

The point of the grid is to create a bijection in a problem that makes it easier to solve. Since the grid just represents a combination, it can be adapted to work with any problem whose answer is a combination.

For example, take an instance of the classic 'stars and bars' problem (also known as 'balls and urns', 'sticks and stones', etc.):

Q: How many ways are there to pick an ordered triple (a, b, c) of nonnegative integers such that a+b+c = 8? (The answer is 10C2 or 45 ways.)

Solution I: 

This problem is traditionally solved by thinking of ordering 8 stars and 2 bars. An example is:

* * * |    | * * * * *

  ^       ^       ^

  a       b       c

This corresponds to a = 3, b = 0, c = 5.

Solution II: 

But this can also be done using the grid technique:

The red path corresponds to the same arrangement: a = 3, b = 0, c = 5. The increase corresponds to the value: a goes from 0 to 3 (that is an increase of 3), b goes from 3 to 3 (that is an increase of 0), and c goes from 3 to 8 (that is an increase of 5). So a = 3, b = 0, c = 5.

Likewise, using a clever 1-1 correspondence, you can map practically any problem with an answer of nCk to fit the grid method. The major advantage of this is that it is an easier way to think about the problem (just like the example I gave may be easier to follow than the original stars and bars approach, and the example I gave in class with the dice can also be thought of in a more numerical sense).

Similar Triangles: Team question : Beginning level

9. In the figure below, quadrilateral CDEG is a square with CD = 3, and quadrilateral BEFH is a rectangle. If EB = 5, how many units is BH? Express your answer as a mixed number

Triangle BED is a 3-4-5 right triangle and is similar to triangle GEF.

BE : ED = GE : EF = 5 : 3 = 3 : FE

EF = 9/5 = BH  The answer

Harder Mathcounts State/AMC Questions: Intermediate level if you can solve in less than 2 mins.

2012 Mathcounts State Sprint #30: In rectangle ABCD, shown here, point M is the midpoint of side BC, and point N lies on CD such that DN:NC = 1:4. Segment BN intersects AM and AC at points R and S, respectively. If NS:SR:RB = x:y:z, where x, y and z are positive integers, what is the minimum possible value of x + y + z? 

Solution I :

\(\overline {AB}:\overline {NC}=5:4\) [given]

Triangle ASB is similar to triangle CSN (AAA)

\(\overline {NS}:\overline {SB}= 4 : 5\)

Let \(\overline {NS}= 4a,  \overline {SB}= 5a.\)

Draw a parallel line to \(\overline {NC}\) from M and mark the interception to \(\overline {BN}\)as T.

 \(\overline {MT}: \overline {NC}\) = 1 to 2. [\(\Delta BMT\) and \(\Delta BCN\) are similar triangles ]

\(\overline {NT} = \overline {TB}= \dfrac {4a+5a} {2}=4.5a\)

\(\overline {ST} = 0.5a\)

 \(\overline {MT} :  \overline {AB}\) = 2 to 5
[Previously we know  \(\overline {MT}: \overline {NC}\) = 1 to 2 or 2 to 4 and  \(\overline {NC}:\overline {AB}= 4 : 5\) so the ratio of the two lines  \(\overline {MT} :  \overline {AB}\) is 2 to 5.]


\(\overline {TB} = 4.5 a\)  [from previous conclusion]

Using 5 to 2 line ratio [similar triangles \(\Delta ARB\) and \(\Delta MRT\) , you get \(\overline {BR} =\dfrac {5} {7}\times 4.5a =\dfrac {22.5a} {7}\) and \(\overline {RT} =\dfrac {2} {7}\times 4.5a =\dfrac {9a} {7}\)

Thus, x : y : z = 4a : \( \dfrac {1} {2}a + \dfrac {9a} {7}\) : \(\dfrac {22.5a} {7}\) = 56 : 25 : 45

x + y + z = 126

Solution II : 
From Mathcounts Mini: Similar Triangles and Proportional Reasoning

Solution III: 
Using similar triangles ARB and CRN , you have \(\dfrac {x} {y+z}=\dfrac {5} {9}\).
9x = 5y + 5z ---- equation I

Using similar triangles ASB and CSN and you have \( \dfrac {x+y} {z}=\dfrac {5} {4}\).
4x + 4y = 5z  ---- equation II

Plug in (4x + 4y) for 5z on equation I and you have 9x = 5y + (4x + 4y) ; 5x = 9y ; x = \(\dfrac {9} {5}y\)
Plug in x = \(\dfrac {9} {5}y\) to equation II and you have z  =  \( \dfrac {56} {25}y\)

x : y : z = \(\dfrac {9} {5}y\)  : y  :  \( \dfrac {56} {25}y\) =  45 y :  25y  :  56y

45 + 25 + 56 = 126

Solution IV : Yes, there is another way that I've found even faster, saved for my private students. :D 

Solution V : from Abhinav, one of my students solving another similar question : 

Two other similar questions from 2016 AMC A, B tests : 

2016 AMC 10 A, #19 : Solution from Abhinav 

2016 AMC 10 B #19 : Solution from Abhinav 

A Skill for the 21st Century: Problem Solving by Richard Rusczyk

Does our approach to teaching math fail even the smartest kids ? 

Quotes from that article  "According to research from the University of California, Los Angeles, as many as 60 percent of all college students who intend to study a STEM (science, technology, engineering, math) subject end up transferring out. In an era when politicians and educators are beside themselves with worry over American students’ lagging math and science scores compared to the whiz kids of Shanghai and Japan, this attrition trend so troubles experts it has spawned an entire field of research on “STEM drop-out,” citing reasons from gender and race to GPAs and peer relationships."

Why Science Majors Change Their Minds (It's Just So Darn Hard) from the New York Times 

A Skill for the 21st Century: Problem Solving by Richard Rusczyk, founder of  "Art of Problem Solving".

Top 10 Skills We Wish Were Taught at School, But Usually Aren't 
from Lifehacker

Mathcounts prep

 Hi, Thanks for visiting my blog.

E-mail me at thelinscorner@gmail.com if you want to learn with me.  :) :) :) 

Currently I'm running different levels of problem solving lessons, and it's lots of fun learning along with students from different states/countries. 

Sequences and Series -- Arithmetic and Geometric Sequences

Sequences are fun to learn and not really that difficult. 
There are many similarities between arithmetic and geometric sequences, so 
learn both together. 

Enjoy !!!!! 

From Mthcounts Mini: Sequences and Series

Easier concepts:

Sequences

Arithmetic sequence/determine the nth term

Arithmetic and geometric sequences

Mathcounts strategies : review some sums 

Note : Don't just memorize, but really understand the concepts.

Harder concepts:

Sum and Average of An Evenly Space

Relationship between arithmetic sequences, mean and median

Sequences, series and patterns

Some Common Sums

Pathfinder

From Mathcounts Mini :

Counting/Paths Along a Grid

From Art of Problem Solving

Counting Paths on a Grid 

Math Principles : Paths on a Grid : Two Approaches 

Question #1: How many ways to move the dominoes on a 6 by 6 checker board if you can only move the dominoes to the right or to the bottom starting from the upper left and you can't move the dominoes diagonally? 

Solution :
You can move the dominoes 5 times to the right at most and 5 down to
the bottom at most, so the answer is \(\dfrac {\left( 5+5\right) !} {5! \times 5!}\) = 252 ways

Question # 2: How many ways can you  move from A to B if you can only move downward and to right? 

Solution : There are \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) * 2 * \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) = 9800 ways from A to B

Sum of All the Possible Arrangements of Some Numbers

Check out Mathcounts, the best middle school competition math program up to the national level.

Questions to ponder: (detailed solutions below) 

It's extremely important for you to spend some time pondering on these questions first without peeking on the solutions. 

#1: Camy made a list of every possible distinct four-digit positive integer that can be formed using each of the digits 1, 2 , 3 and 4 exactly once in each integer. What is the sum of the integers on Camy's list?

#2: Camy made a list of every possible distinct five-digit positive even integer that can be formed using each of the digits 1, 3, 4, 5 and 9 exactly once in each integer. What is the sum of the integers on Camy's list? (2004 Mathcounts Chapter Sprint #29)

#3: 2020 Mathcounts state sprint #24 













Solutions:
#1:  
Solution I: 
There are 4! = 24 ways to arrange the four digits. Since each digit appears evenly so each number will appear 24 / 4 = 6 times.
1 + 2 + 3 + 4 = 10 and 10 times 6 = 60 ; 60 (1000 +100 +10 + 1) = 60 x 1111 = 66660, which is the answer.

Solution II: 

The median of four numbers 1, 2, 3, 4 is (1 + 2 + 3 + 4) / 4 = 2.5 and there are 4! = 24 ways to arrange
the four numbers. 
2.5 (1000 + 100 + 10 + 1) x 24 = 66660 

#2:  
Solution I: 
Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely.
Again there are 4! = 24 ways to arrange the four numbers. 1 + 3 + 5 + 9 = 18 and 18 x 6 = 108 (Each number that can be moved freely appears 6 times evenly.)108 x 11110 + 4 x 24 = 1199976

Solution II: 

Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely. 
There will be 4! = 24 times the even number 4 will be used so 4 x 24 = 96
As for the remaining 4 numbers, their average (or mean) is \(\dfrac{1 + 3+ 5 + 9} {4} = 4.5\)                               
4.5 * ( 10000 + 1000 + 100 + 10) * 24 (arrangements)  + 96 = 4.5 * 11110 * 24 + 96 = 1199976

#3: The answer is 101. 

Other applicable problems: (answers below)

#1: What is the sum of all the four-digit positive integers that can be written with the digits 1, 2, 3, 4 if each digit must be used exactly once in each four-digit positive integer? (2003 Mathcounts Sprint #30)

#2: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once? (You can use a calculator for this question.) (2005 AMC-10 B)

#3: What is the sum of all the four-digit positive integers that can be written with the digits 2, 4, 6, 8 if each digit must be used exactly once in each four-digit positive integer?

#4: What is the sum of all the 5-digit positive odd integers that can be written with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly once in each five-digit positive integer?  

#5:What is the sum of all the four-digit positive integers that can be written with the digits 2, 3, 4, 5 if each digit must be used exactly once in each four-digit positive integer?

  





Answer key: 
#1: 66660
#2: \(\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8\)
4.8 * 11111 =\(\color{red}{53332.8}\) 
#3: 133320
#4: 1333272
#5: 93324

16 17 Mathcounts handbook more interesting questions that have nicer solutions

Thanks to Achuth for trying out these questions and time them as an actual Mathcounts test.  :) 

First week : warm up 1, 4, 7.  (time for 40 mins. like sprint)

Second week : warm up 2, 5, 8.

third week : workout 3 --> all right. (pair 1 to 6, 2 to 7, each time for 6 mins. as 

target) 

fourth week: workout 4 --> #95, then self correct. 

At lesson: workout 5 and other harder problems. 

These are nice questions that have various solutions, so it’s better to slow down and try them as puzzles.

Less is more and slow is fast.

If you are new to problem solving, one nice strategy is to make the question much simpler and explore ideas that come to your mind. 

Answer key down below. 

#66: A school of 100 fish swims in the ocean and comes to a very wide horizontal pipe. The fish have three choices to get to the food on the other side: swim above the pipe, through the pipe or below the pipe. If we do not consider the fish individually, in how many ways can the entire school of fish be partitioned into three groups with each group choosing a different one of the three options and with at least one fish in each group? 

 #105 When fully matured, a grape contains 80% water. After the drying process, called dehydration, the resulting raisin is only 20% water. What fraction of the original water in the grape remains after dehydration? Express your answer as a common fraction. 

 #112: Cora has five balls—two red, two blue and one yellow—which are indistinguishable except for their color. She has two containers—one red and one green. If the balls are randomly distributed between the two containers, what is the probability that the two red balls will be alone in the red container? Express your answer as a common fraction? 

 #116: A 12-foot by 12-foot square bathroom needs to be tiled with 1-foot square tiles. Two of the tiles are the wrong color. If the tiles are placed randomly, what is the probability that the two wrong-colored tiles share an edge? Express your answer as a common fraction.

#66: 4851

#105:  1/16

#112:  1/32

#116: 1/39 

Mass Points Geometry

Some of the harder/hardest questions at Mathcounts can be tackled at ease using mass point geometry
so spend some time understanding it.

Basics 

2014-15 Mathcounts handbook Mass Point Geometry Stretch
from page 39 to page 40

(Talking about motivation, yes, there are students already almost finish
this year's Mathcounts' handbook harder problems.)

From Wikipedia

From AoPS

Mass Point Geometry by Tom Rike

Another useful notes 

Videos on Mass Point :

Mass Points Geometry Part I 

Mass Points Geometry : Split Masses Part II 

Mass Points Geometry : Part III 

other videos from Youtube on Mass Points

It's much more important to fully understand how it works, the easier questions the weights align
very nicely.

The harder problems the weights are messier, not aligning nicely, so you need to find ways to may them integers (LCM) for easier solving.

Let me know if you have questions. I love to help (:D) if you've tried.

2020 Mathcounts State Prep: Simon's Favorite Factoring Trick

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.

The most common cases of Simon's Favorite Factoring Trick are:

I:  \(xy+x+y+1=\left( x+1\right) \left( y+1\right)\)

II:  \(xy-x-y+1=\left( x-1\right) \left( y-1\right)\)

It's easy to learn. Here is the best tutorial online, by none other than Richard Rusczyk.
The method Rusczyk uses at the second half is very nifty. Thanks!!

Questions to ponder:(answer key below)
#1: Both x and y are positive integers and \(x>y\). Find all positive integer(s) that \(xy+x+y=13\) 
#2: Both x and y are positive integers and \(x>y\). Find all positive integer(s) that \(2xy+2x-3y=18\)
#3: Find the length and the width of a rectangle whose area is equal to its perimeter.
#4: Twice the area of a non-square rectangle equals triple it's perimeter, what is the area of the rectangle? 













Answer key:
#1:  x = 6 and y = 1
#2: ( x, y ) = (4, 2) 
#3: Don't forget square is a kind of rectangle (but not the other way around) so there are two answers: 
4 by 4 and 3 by 6 units. 
#4: One side is 4 units and the other 12 units so the answer is 4 x 12 or 48 square units. 
There is another one, 6 by 6 that would fit if the question doesn't specify non-square rectangle. 

2012 Harder Mathcounts State Target Questions

Check out Mathcounts here -- the best competition math program for middle schoolers up to the 
state and national level. 

# 6: A semicircle and a circle are placed inside a square with sides of length 4 cm, as shown. The circle is tangent to two adjacent sides of the square and to the semicircle. The diameter of the semicircle is a side of the square. In centimeters, what is the radius of the circle? Express your answer as a decimal to the nearest hundredth. [2012 Mathcounts State Target #6]


#6:  Solution:
Using Pythagorean theory: (2 + r)² = (4-r)² + ( 2- r)²
4 + 4r + r² = 16 - 8r + r² + 4 - 4r + r²
 r² - 16 r + 16 = 0
Using the quadratic formula You have 8 ± 4√ 3
Only 8 - 4√ 3 = 1.07 works

There is a Mathcounts Mini #34 on the same question. Check that out !!

The above question looks very similar to this year's AMC-10 B #22, so try that one.
(cover the answer choices so it's more like Mathcounts)

2014 AMC-10 B problem #22 

#8: In one roll of four standard, six-sided dice, what is the probability of rolling exactly three different numbers? Express your answer as a common fraction. [2012 Mathcounts State Target #8]

Solution I : Permutation method
If order matters, there are 6 * 5 * 4 * 1 ways to choose the number, 1 being the same number as one of the previous one.
Let's say if you choose 3 1 4 1.

Now for the placement of those 4 numbers on the 4 different dice. There are 4C2 ways to place where the two "1" will
be positioned so the answer is : \(\dfrac {6\times 5\times 4\times 1\times 4C2} {6^{4}}\) = \(\dfrac{5}{9}\)

Solution II:  Combination method

There are 6C3 = 20 ways to choose the three numbers.

There are 3 ways that the number can be repeated. [For example: If you choose 1, 2, and 3, the fourth number could be 1, 2 or 3.]

There are \(\dfrac {4!} {2!}\) =12ways to arrange the chosen 4 numbers.[same method when you arrange AABC]

So the answer is\(\dfrac{20* 3 *12}{6^4}\) = \(\dfrac{5}{9}\)

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]
\(\dfrac {1} {2}\left( 2\times 10\pi + 2\times 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\pi \right)\)* \(\left( 10-10\times \dfrac {\sqrt [3] {2}} {\sqrt [3]{3}}\right)\)