MOCK AMC I RESULTS

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Lucky707

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Lucky707

#1 h Jan 29, 2005, 11:15 PM • 2 Y

Y by Adventure10, Mango247

Here are the results with a question-by-question analysis.

NOTE: If you have not taken the test and would like to, please see the other thread for the questions as this thread has the questions ALONG with the answers.

TOTAL NUMBER OF CONTESTANTS: 15

MOCK AMC I-
CREATED BY: Lucky707 and Singthesorrow
DATE ADMINISTERED: January 30, 2005

Rules:
(Standard AMC Rules)
1. You have 75 minutes to complete the test.
2. Only non-programmable calculators are allowed.
3. 6 points for a correct answer, 2.5 points for a question left unanswered and 0 points for an incorrect answer so it is in your best interest NOT to guess.
4. All other standard AMC Rules apply... We've had plenty of Mock AMCs so you should know the rules by now.
5. At 5:15, you must submit the test. If you do not, it will not count. I will allow a few extra minutes for slow modems, etc. but please submit the test at 5:15.
6. My AOL is hezgotzbeenz if anyone has any questions during the test.

1. The I.Q. of a Martian varies directly with the square of the number of eyes it has. If a Martian with 5 eyes has an I.Q. of 75, find the IQ of a Martian with 8 eyes.
(A) 120
(B) 128
(C) 144
(D) 192
(E) 200

Generally well done, those who missed this question did so because they didn't read it and multiplied 8 by 15 to get 120.
ANSWER: D
AVERAGE SCORE: 5.2 (13 correct/15 attempted)

2. The number of cubic feet in the volume of a cube is the same as the number of square inches of its surface area. What is the length of an edge of it expressed in feet? (NOTE: 12 inches = 1 foot)
(A) 6
(B) 36
(C) 72
(D) 108
(E) 864

Also fairly well done. Those who missed this one did not convert at some point down the line.
ANSWER: E
AVERAGE SCORE: 4.8 (12/15)

3. The first two terms of a geometric progression are $\sqrt{3}$ and the fourth root of three. What is the fifth term?
(A) $\frac{1}{\sqrt{3}}$
(B) $\sqrt{3}$
(C) fourth root of 3
(D) fourth root of $\frac{1}{3}$
(E) tenth root of 3

This was somewhat of a trick question. Those who were tricked by it chose E but if we look at the common ratio and work it out as usual, we get the correct answer of A.
CORRECT ANSWER: A
AVERAGE SCORE: 4.8 (12/15)

4. A leaky faucet drips at a steady rate. If we measure from the instant the first drop hits the bottom, to the instant the last drop falls, it takes 36 seconds for 9 drops to fall. How many seconds would it take for 12 drops to fall?
(A) 42 seconds
(B) 44 seconds
(C) 48 seconds
(D) 49.5 seconds
(E) 54 seconds

This question was nasty. So many places where one could make an error. Here is the solution:
We measure from the instant the first drop falls to the instant the last drop falls. 36 seconds for 9 drops to fall means 36/8 seconds between drops hitting the ground. That's the first place one can make a mistake. The other place is multiplying by 12 instead of 11. The answer is 49.5
ANSWER: D
AVERAGE SCORE: 3.0 (7/14)

5. A triangle has sides of length 15, 20 and 25 units. It is inscribed in a circle. What is the circle's radius?
(A) $21/2$
(B) $43/3$
(C) $64/7$
(D) $25/2$
(E) $25$

This question was very well done. After realizing that the triangle is a right triangle all one needs to do is divide the hypotenuse by 2 to get the answer.
ANSWER: D
AVERAGE SCORE: 6.0 (15/15)

6. How many zeroes are at the end of the product of the first fifty positive integers?
(A) 5
(B) 12
(C) 17
(D) 35
(E) 47

I don't think the average would be this high in real life for this type of question but I was sure that AOPSers could steamroll through this question... and steamroll they did!
ANSWER: B
AVERAGE SCORE: 6.0 (15/15)

7. If $100! = (100)(99)(98)...(2)(1)$ , then what is the largest value of $x$ such that $2^x$ divides $100!$ evenly?
(A) 50
(B) 97
(C) 98
(D) 99
(E) 100

See comment above. 6 and 7 were quite similar actually. Nobody made an addition error so everyone got this right.
ANSWER: B
AVERAGE SCORE: 6.0 (15/15)

8. I have a red shirt, a blue shirt and a green shirt. I also have blue jeans, black jeans and khakis. How many possible combinations can I make if I absolutely hate wearing the blue shirt with the blue jeans?
(A) 5
(B) 8
(C) 14
(D) 16
(E) 17

This one was a gimme.
ANSWER: B
AVERAGE SCORE: 6.0 (15/15)

9. Let $x$ be the smallest perfect square that has 140 as a divisor. Then
(a) $0 < x < 1000$
(b) $1000 \leq x < 2000$
(c) $2000 \leq x < 3000$
(d) $3000 \leq x < 4000$
(e) $4000 \leq x < 5000$

Wow... there were a lot of prime factorization questions early on. Realizing that 140 = 7*5*2^2.... we find that 7^2*5^2*2^2 will be a perfect square. And that value is 4900, making the answer E.
ANSWER: E
AVERAGE SCORE: 6.0 (15/15)

10. What is the sum of the first twenty powers of 2 (2^1 + 2^2 + 2^3 ... + 2^20)?
(A) $2^{19} - 2$
(B) $2^{20} - 2$
(C) $2^{21} + 2$
(D) $2^{21} - 2$
(E) None of the above

This question can easily be done by induction... or if you want to stay honest (which is not a good idea during the AMC)... geometric series.
ANSWER: D
AVERAGE SCORE: 5.2 (13/15)

11. $9ax+7y=12$ and $4x+2by=9$ represent the same line. $a$ and $b$ can both be expressed as fractions in lowest terms. What is the numerator of $a$ plus the denominator of $b$ equal to?
(A) 212
(B) 144
(C) 72
(D) 48
(E) 24

Generally those who attempted this question got it right. One person got it wrong and I'm assuming it's because he forgot to reduce.
ANSWER: E
AVERAGE SCORE: 5.1 (12/13)

12. A square of side length 8 has its corners cut off to make a regular octagon. Find the area of the octagon to the nearest whole number.
(A) 32
(B) 48
(C) 53
(D) 55
(E) 59

Anyone who attempted this question got it right. The trick was not to be confused and make the cut along the midpoints or else it would be a hexagon. Luckily, nobody fell for that trick.
ANSWER: C
AVERAGE SCORE: 5.5 (13/13)

13. A piece of string is cut in two at a random point. What is the probability that the larger piece is at least $x$ times as large as the shorter piece? (where $x \geq 1$ )
(A) $\frac{2}{1+\sqrt{5}}$
(B) $\frac{x^2+x-1}{x}$
(C) $\frac{2}{x+1}$
(D) $\frac{3}{x+2}$
(E) None of the Above

This is the type of question thta is far easier when it is a multiple choice question than when it is a written question. After a little bit of trial and error, one eventually arrives at the correct answer.
CORRECT ANSWER: C
AVERAGE SCORE: 4.5 (10/12)

14. Let $S$ denote the sum of all distinct four-digit numbers that contain only 1,2,3,4 or 5, each at most once. Then
(A) $S \leq 200,000$
(B) $200,000 < S \leq 300,000$
(C) $300,000 < S \leq 400,000$
(D) $400,000 < S \leq 500,000$
(E) $500,000 < S$

A disguised combinatorics problem here. The answer is C but the number itself is so very close to D that it is easy to make a mistake if using the wrong method. Luckily, nobody made THAT mistake (Even though few got it wrong).
ANSWER: C
AVERAGE SCORE: 3.9 (8/11)

15. How many digits are in $a^b$ if $b=\sqrt{9530}$ and $\log a$ ~ $2.8356$
(A) 277
(B) 278
(C) 282
(D) 283
(E) 284

Once again, a question AOPS knows how to do but I can't say the same about the rest of the mathematical world. If you know how to do this, which almost everyone did, then you will get the correct answer (unless you forget to take the floor, which nobody did!)
ANSWER: A
AVERAGE SCORE: 5.5 (13/13)

16. If $\frac{1}{(a)(b+1)} + \frac{1}{(b)(a+1)} = \frac{1}{(a+1)(b+1)}$ , what is the value of $\frac{1}{a} + \frac{1}{b}$ ?
(A) $-1$
(B) $1$
(C) $2 \sqrt{5}$
(D) $-2 \sqrt{5}$
(E) Undefined

This was a nice question. Anyone who attempted it got it right but it wasn't exactly easy.
ANSWER: A
AVERAGE SCORE: 4.8 (10/10)

17. Let the parabola $x^2-11x+28$ meet the $x$ -axis at two distinct points, $A$ and $B$ . Let the vertex of said parabola be $C$ . The area of triangle $ABC$ can be expressed as $\frac{f}{g}$ , where $f$ and $g$ are relatively prime. $f$ and $g$ are both...
(A) Prime (but not in the form $2^n-1$ )
(B) Perfect Squares
(C) Perfect Cubes
(D) In the form $2^n-1$
(E) None of the Above

A question dealing with co-ordinates. After finding the vertex of the parabola, it becomes pretty straightforward. The answer is 27/8, which corresponds to C.
ANSWER: C
AVERAGE SCORE: 5.0 (12/14)

18. Let $x$ be the number of positive integral factors of $90^9$ . The sum of the digits of $x$ equals
(A) 7
(B) 10
(C) 13
(D) 16
(E) None of the Above

ANSWER: B (I believe it was 1900)
AVERAGE SCORE: 5.3 (12/12)

19. Bette visits her friend Keith and the nreturns home by the same route. She always walks 2 km/h when going uphill, 6 km/h when going downhill, and 3 km/h when on level ground. If her total walking time is 6 hours, then the total distance she walks, in km, is
(A) 9
(B) 12
(C) 18
(D) 22
(E) 36

I like this question. One must realize that she will walk the same distance uphill and downhill. If one finds her average speed uphill and downhill, it will be 3 km/h... the same as on level ground! Thus, in 6 hours, she walks 18 km.
ANSWER: C
AVERAGE SCORE: 3.7 (8/12)

20. If $JKLMN$ is a regular pentagon with side length $a$ and diagonal length $b$ , what is the value of $\frac{b}{a} - \frac{a}{b}$ ?
(A) $\frac{3}{4}$
(B) 1
(C) $\frac{4}{3}$
(D) 2
(E) Depends on $a$

One could approach this using Ptolemy's Theorem, trigonometry or just having seen it before. E was intended to throw some people off but wasn't that effective.
ANSWER: B
AVERAGE SCORE: 4.7 (10/11)

21. Let $P$ be a point outside circle $O$ . Draw two tangents from $P$ such that they intersect $O$ at $A$ and $B$ . If $AP=BP=7$ and the larger of the two arcs $AB$ is twice as large as the smaller of the two arcs, find the length of $AB$ .
(A) $6$
(B) $7 \sqrt{3}$
(C) $\frac{11 \sqrt{3}}{2}$
(D) $7$
(E) None of the Above

I came up with this geometry question myself after looking at a random theorem.
ANSWER: D
AVERAGE SCORE: 4.3 (9/11)

22. The system of equations
$4y+gx=34$
$\frac{g^4x^2-16y^2}{g^2x+4y} = -10$
has how many solutions where $g$ , $x$ and $y$ are all integers?
(A) $0$
(B) $1$
(C) $2$
(D) $4$
(E) None of the Above

I presented this question for my math class and wow was it hard. This is actually a spinoff of an old Cayley question but I like my version better. The only solution is (2,3,7)
ANSWER: B
AVERAGE SCORE: 3.0 (5/9)

23. How many arrangements of the letters of the word "INDEPENDENT" have no adjacent E's?
(A) $\frac{(2)(7!)}{5!}$
(B) $\frac{11!}{9!-2!}$
(C) $\frac{11!}{3!}$
(D) $\frac{11!}{3!2!}$
(E) None of the Above

I swear I made up options B, C and D off the top of my head. Question was well done after people figured out that I was randomly putting factorials wherever I wanted.
ANSWER: E
AVERAGE SCORE: 4.4 (9/10)

24. If $f(x-y)=f(x)f(y)$ for all $x$ and $y$ , and $f(x)$ never equals $0$ , then $f(3)$ equals...
(A) $-3$
(B) $3$
(C) $9$
(D) :pm: 1
(E) None of the above

This one was a killer. The only person to get this correct was towersfreak2006. Most people determined f(x)^2=1 correctly but then mistakenly thought f(x) = :pm: 1

But f(x) is a function. How can f(x) = :pm: 1

The correct method involves setting y = x/2 and working from there. Try it yourself. The correct answer is simply 1.

ANSWER: E
AVERAGE SCORE: 1.2 (1/10)

25. Pieces of paper numbered from $1$ to $100$ are each placed in one of three hats such that there is at least one piece of paper in each hat. In how many ways can this be done such that if two hats are selected and a card is taken from each, then the knowledge of their sum alone is always sufficient to identify the third hat?
(A) $6$
(B) $12$
(C) $18$
(D) $24$
(E) None of the Above

Nobody attempted this question. I put it on here to simulate an extremely difficult question 25 and to prevent any perfects. I felt that if you can get perfect on AMC10, you don't really need practice for it.

SOURCE OF THIS QUESTION: IMO 2000 B1 ... yeah, don't feel upset if you didn't solve it... it was only supposed to be a simulation. Nobody got anywhere near perfect anyway.

And for whitehorseking88, here's what the question is asking: Let's say you had 3 hats and you knew the sum of the numbers in all 3. Let's say the sums were A, B and C respectively. Now if you randomly pick two pieces of paper out of any two hats and are told their sum, you are able to identify which hat has sum A, which hat has sum B and which hat has sum C. Now how many possible sums A, B and C exist for this to work?

(HINT: Try sums like 1 and 2...)

ANSWER: I won't tell ya... IMO 2000 if you really need to know. kalva.demon.co.uk
AVERAGE SCORE: 2.5 (0 correct/ 0 attempted)

Results up soon.

This post has been edited 1 time. Last edited by Lucky707, Jan 29, 2005, 11:25 PM

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Lucky707

113 posts

Lucky707

#2 h Jan 29, 2005, 11:20 PM • 2 Y

Y by Adventure10, Mango247

Results -

TOTAL NUMBER OF CONTESTANTS: 15
AVERAGE SCORE: 116.43

AIME CUTOFF: 120 (Hey, why not? Sure... two people have scores of 119 but 8/15 make it... it works out so nicely)

FIRST: alan - 134.5
SECOND: towersfreak2006 - 133.5
*frankthetank - 133.5
FOURTH: Danbert - 128.5
FIFTH: *d343seven - 125
white_horse_king88 - 125
SEVENTH: keta - 124
EIGHTH: interesting_move - 122.5

Barely missed the cut - *rep123max (119) and *Yrael (119)

* - took it early

If you need to see where you rank, the other scores were (without names): 103.5, 100.5, 100.5, 92, 85.5

pkothari13 - unofficial score 122.5

This post has been edited 2 times. Last edited by Lucky707, Jan 30, 2005, 12:56 AM

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1234567890

1553 posts

1234567890

#3 h Jan 29, 2005, 11:22 PM • 2 Y

Y by Adventure10, Mango247

i'd like to respectfully disagree with #4. It deals with gravity, and does factor that in. This would definitely not be an AMC problem, for test writers usually avoid all physics.
If you use kinematic equations, you will see the answer is none.

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h_s_potter2002

1306 posts

h_s_potter2002

#4 h Jan 29, 2005, 11:34 PM • 2 Y

Y by Adventure10, Mango247

I hate stupid mistakes soooooooo much. I could have gotten 18 extra points. And one of the stupid mistakes was me writing down the wrong answer! I just hope I remember to check my work, and look over my answer on the real AMC this Tuesday.

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eryaman

1130 posts

eryaman

#5 h Jan 30, 2005, 12:04 AM • 2 Y

Y by Adventure10, Mango247

shoot. I couldn't make it in time to take it

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white_horse_king88

1779 posts

white_horse_king88

#6 h Jan 30, 2005, 12:29 AM • 2 Y

Y by Adventure10, Mango247

Actually, it wouldn't matter if gravity was there or not because he specifies the 9 drops from the point the first one hits to the last one falls. That's a period, so hypothetically, he already included gravity as a factor. Now, all you have to do is figure out the time from one drop to the next and multiply that by 11.

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pkothari13

709 posts

pkothari13

#7 h Jan 30, 2005, 12:38 AM • 2 Y

Y by Adventure10, Mango247

shoooooot! i did only the distance going one way on number 19.... yes.. stupid mistakes will soon kill us all...! GAH!

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Lucky707

113 posts

Lucky707

#8 h Jan 30, 2005, 12:51 AM • 2 Y

Y by Adventure10, Mango247

Erm yes... I gave you 115.5 before factoring in the INDEPENDENT... but when I added 6 for the question I forgot to take away the original 2.5 that I gave you. So I gave you 2.5 points extra. I caught it after making a spreadsheet of all the scores however.

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alan

228 posts

alan

#9 h Jan 30, 2005, 1:52 AM • 2 Y

Y by Adventure10, Mango247

Rep123max wrote:

I'm extremely stupid.

well, at least you didnt read #10 as "What is the sum of 2 (2^1 + 2^2 + 2^3 ... + 2^20)"

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keta

244 posts

keta

#10 h Jan 30, 2005, 2:32 AM • 2 Y

Y by Adventure10, Mango247

Argh,.... Number 19 about the walking one really got me. I only calculated the distance one way like pkothari13 and foolishly choose A. $9$ . Oh well....

Also, to everyone who took this, would you rank this mock AMC as harder or easier than the typical AMC 10? It seemed a little more difficult to me since there was alot of number theory and functional analysis not found on the AMC 10. (Let's not mention #25-Lucky707, how could you do this to us!!???

JK).

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pkothari13

709 posts

pkothari13

#11 h Jan 30, 2005, 2:41 AM • 2 Y

Y by Adventure10, Mango247

alan wrote:

Rep123max wrote:

I'm extremely stupid.

well, at least you didnt read #10 as "What is the sum of 2 (2^1 + 2^2 + 2^3 ... + 2^20)"

omg! dude, thats what i thought all throughout the test... and then after spending like 5-10 minutes on it, i realized what it meant!

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pkothari13

709 posts

pkothari13

#12 h Jan 30, 2005, 2:43 AM • 3 Y

Y by Adventure10, Adventure10, Mango247

hey Lucky707, do you know if i got number 25 right?

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pkothari13

709 posts

pkothari13

#13 h Jan 30, 2005, 2:45 AM • 2 Y

Y by Adventure10, Mango247

lol oh yeah, how did all of you do that INDEPENDENT one.. I just guessed on that one hoping i'd get it right....

guess that didnt work out...

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Lucky707

113 posts

Lucky707

#14 h Jan 30, 2005, 2:48 AM • 2 Y

Y by Adventure10, Mango247

Yes you got #25 right.

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d343seven

237 posts

d343seven

#15 h Jan 30, 2005, 6:18 AM • 2 Y

Y by Adventure10, Mango247

when i took it i had an answer for the INDEPENDENT one. what's the actual answer?
i don't wanna do it now, but my answer was something like 11! - something

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The Original Pi Guy

172 posts

The Original Pi Guy

#16 h Jan 30, 2005, 3:26 PM • 2 Y

Y by Adventure10, Mango247

Could someone explain to me how you do number 15?

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plokoon51

135 posts

plokoon51

#17 h Jan 30, 2005, 4:03 PM • 2 Y

Y by Adventure10, Mango247

Could someone explain to me how to come up with the answer to #22?

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chenwb

67 posts

chenwb

#18 h Jan 30, 2005, 4:14 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

For the INDEPENDENT one, I had:

Lay down the 8 "non-E" letters first. There are $\frac{8}{3!2!3!}$ ways to do this (analogous to the number of possible words from rearranging MISSISSIPPI, if you've seen this before). Insert a wedge between each letter and at the ends. There are 9 wedges. Choosing 3 wedges to place the Es can occur in $\binom{9}{3}$ ways. The answer is then $\frac{8}{3!2!3!} \cdot \binom{9}{3}$

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plokoon51

135 posts

plokoon51

#19 h Jan 30, 2005, 4:17 PM • 2 Y

Y by Adventure10, Mango247

never mind on #22. STUPID MISTAKES!!! they will be the cause of the end of all civilization

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chenwb

67 posts

chenwb

#20 h Jan 30, 2005, 4:21 PM • 2 Y

Y by Adventure10, Mango247

For 15:

We take $[\log{a^b}]$ , which will give us the greatest power of 10 less than $a^b$ . To evaluate this, we simply rewrite as $b\log{a}$ , and multiply out, leaving 276. Adding 1 for the extra digit (this is like how 1002, whose greatest power of 10 less than it is $10^3$ , has 3+1=4 digits)., we have the final answer: 277 digits, or choice A.