Mathacre JV Written 2013 Solutions - Cherokee County Schools

2013 Valentine’s Day Mathacre
Creekview High School
Junior Varsity Written Test Solutions
2013 Saint Valentine’s Day Mathacre Junior Varsity Written Test Solutions Page 1

2013 Saint Valentine’s Day Mathacre Junior Varsity Written Test Solutions Page 2

1. A parallelogram with altitude 6 and a ° angle shares its base with a 6 by 10 rectangle.
Find the area of overlap. Round to a whole number if necessary.
Solution: First, we need to find EB. Using simple trig, we can find that √ . Since
, √ . Now we have all we need to find the area of ABFD.
√ sq units. Round, and you get 29.
A) B) C) D) √ E) None of the above
2. Compute the last digit of .
Solution: Let’s notice a pattern:
We notice, that it always ends on 6, no matter what. Therefore, the answer is 6
A) 2 B) 4 C) 6 D) 8 E) 0 (zero)
3. You have 2 equations: and . Find the sum of all
possible y.
Solution: Using substitution we can find that
A) B) C) D) E)
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4. A smaller triangle DBE is inside the big isosceles triangle ABC as shown. ,
, and . Find .
B
D
E
A
C
Solution: Since triangles ABC and DBE have 3 congruent angles, they are similar. Since
they are similar and , .
.
Therefore, we can find AD.
A) B) C) ⁄ D) ⁄ E) ⁄
5. You have to pay $10.23 to get into a movie theater. How many 5c coins are in the most
the most efficient (mass-wise; assuming that all coins have the same mass) way to pay for
the ticket if you can use only 7 quarters (25c), 3 50c, and an unlimited amount of 10c, 5c,
and 1c coins.
Solution:
Therefore, the answer is:
25c x 7
50c x 3
10c x 69
5c x 1
1c x 3
The answer is 1.
A) 10 B) 1 C) 5 D) 9 E) None of the above
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6. If , find .
Solution:
A) B) C) D) E)
7. Oleg took 4 tests in Calculus and he has his grades as following: 98, 71, 98, 91. What
integer grade does he need to get on the next test so his average will be at least 90? (In
case your answer is not a whole number, round it up to a whole number)
Solution: Let’s set up and equation with a variable corresponding to the desired grade.
A) 92 B) 93 C) 94 D) 95 E) 96
8. What is the small angle (in degrees) between clock hands when they show 10:02?
Solution: The angle we need to find can be found by finding
where A is
the big angle formed by the hour hand and B is the angle formed by the minute hand.
Let’s find angle B first:
. Finding angle A is a little trickier. Contrary
to the popular mistake, it is NOT
. Since hour hand moves between hours, it
moved by a very small amount during those 2 minutes. Angle A can be found by doing
( ) . Now that we know angles A and B, we can plug it all in the
original equation:
A) 301 B) 309 C) 79 D) 71 E) None of the above
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9. Creekview holds a chess tournament. 8 finest chess players in the school are going to
play. The way they play is single elimination, which means that at after every round, if a
player loses 2 out of 3 games to a single opponent, he is eliminated. Note, that to lose, he
does not have to play all 3 games, in case an opponent wins 2 games in a row. Rounds go
as following - Round Of 8, Quarter-Finals, Semi-Finals, Finals. If every player is on the
same skill level, what is the percent chance that Oleg (who is one of the players) wins the
tournament (in real life, he would win it with 100% chance, but we omit it for the sake of
problem)?
Solution: First, let’s determine what is the chance of a player getting through one round
is. Since every player has the same skill level, the chance that he passes through a round
is ⁄ , therefore we don’t even need this information about Best of 3 because it is
irrelevant. We have a total of 4 rounds (RO8, QF, SF, F), therefore the chance that Oleg
wins is ( ) , which, converted to percentage, is 6.25%
A) 5 B) 10 C) 12.5 D) 10.5 E) 6.25
10. An amazing Olympic athlete is throwing a spear into a target. The radius of the target is
10 inches. An area with which the spear contacts the target is 0.21 sq.in. If this truly
wonderful athlete has 100% chance of hitting the target, what is the probability that he
will hit a dot with surface area 0.01 sq. in. that rests on that target?
Solution: To find the probability, we need to divide the number of plausible outcomes by
the number of all possible outcomes. First, let’s find the area of the target. .
The number of all outcomes is . The number of plausible outcomes is
. Therefore, the answer is .
A) B) C) D) E)
11. Find at if .
Solution:
A) ⁄ B) ⁄ C) ⁄ D) ⁄ E) None of the above
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12. Eric has 100ft of a barbed wire. A rectangle is made out of this wire. What is the biggest
area possible?
Solution: Since we know that the biggest area produced will always be by a square, we
need to find dimensions of a square with . . Therefore, dimensions of the
square are 25x25 ft.
A) 25 B) 125 C) 1250 D) 1225 E) None of the above
13. Find the magnitude of a vector that is the sum of vectors (5,0) and (0, 5).
Solution: Since these vectors form a right triangle, it is pretty easy to find the length of
the hypotenuse (which is the magnitude of the vector we are trying to find. (√ )
√
A) B) C) D) √ E) None of the above
14. What is the sum of all interior angles in a hexagon (polygon of 6 sides)?
Solution: since sum of interior angles in a polygon can be found using formula
where n is the number of angles in a polygon, we can find the sum of all angle by
plugging 6 into this formula. .
A) 360 B) 540 C) 720 D) 900 E) 1080
15. Find the bigger factor of the following expression: .
Solution: You can either use Vieta’s Theorem or use the quadratic formula. It factors as
, therefore the answer is
A) B) C) D) E)
16. What is the value of ?
Solution:
A) 15 B) 14 C) 10 D) 5 E) 3
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17. Creekview Grizzlies Baseball Team has a score 25:15 at the halfway mark of the season.
How many games must Grizzlies win in the second half of the season in order to have 80
Win/Loss percentage for the whole season?
Solution: Since they already played half of the season games, they have
games left to play. Let’s create an expression with a variable x responsible for the
desired number of wins. ; . is the number of games
they need to win.
A) 36 B) 37 C) 38 D) 39 E) 40
18. Find area of a circle that can be expressed by √ √ .
Solution: From this expression, we can deduct that radius of this circle equals 8. The rest
is just plugging in this number into the area formula, which is A = piR^2. A = pi*8^2 =
64pi.
A) B) C) D) E)
19. What is the value of 13 squared plus 5 factorial?
Solution:
A) 324 B) 193 C) 269 D) 289 E) None of the above
20. Mr. O went to store to buy a new (remember, the problem is set in 1976) Boston music
album. He paid $4.60 including 8% sales tax. How many full gallons of gas he could
have bought if he decided not to pay the tax and cost of a gallon of gas is $0.61 in 1976?
Solution: First, let’s find the tax he had to pay.
. At this point, it is
obvious that the amount of tax is not enough to buy even 1 gallon of gas, therefore, the
answer is .
A) 1 B) 2 C) 3 D) 4 E) 0
2013 Saint Valentine’s Day Mathacre Junior Varsity Written Test Solutions Page 8

21. Simplify the following equation and find the sum of roots of this
equation.
Solution:
⁄
A) B) C) D) ⁄ E) ⁄
22. While doing her physics lab, Julia, finds that speed of the racing car she was supposed to
measure is 0.9 m/s. However, unsurprisingly, her calculated theoretical value is 1.78.
Find percent error and round it to 1 digit after decimal.
Solution: Percent error can be expressed as
| | .
After we plug in numbers, we get | | % (already rounded)
A) B) C) D) E)
23. Anil, Eric, and Ben decided to go to a Halloween party together. Being classy guys they
are, they decided to dress up as one another, so they made imprints of their faces and
created masks based on those imprints. Before heading to the party, they said a couple of
phrases to help those trying to solve this problem. One in Eric’s costume said “I am not
Ben,” one in Ben’s said “I always say truth,” and one in Anil’s said “I am Anil.”
Considering that Ben never lies, Anil sometimes lies and sometimes does not, and Eric
always lies, find who is wearing Anil’s costume?
Solution: First, we know for sure that the one in Eric’s costume is not Ben because he
always says truth, therefore he cannot be Ben under any circumstances. He also cannot be
in Anil’s costume, therefore Ben is in the costume of himself. Since we know that the guy
in Eric’s costume cannot be Ben, he must be saying truth, therefore this is Anil, and Eric
is in Anil’s costume. Answers go as following: Ben is in Ben’s costume, Anil is in Eric’s,
and Eric is in Anil’s.
A) Anil B) Eric C) Ben D) Ben or Anil
E) Not enough information to answer the question
2013 Saint Valentine’s Day Mathacre Junior Varsity Written Test Solutions Page 9

24. Today is November 6, 2012, Presidential Election day. How many days are left from
today until the next presidential election? (Presidential elections occur on the first
Tuesday of November every 4 years, and year 2016 is a leap year)
Solution: Since next Presidential Election takes place on November 8, 2016, it is
relatively easy to calculate that there are 1462 days left.
A) 1459 B) 1460 C) 1461 D) 1462 E) 1463
25. What is the last digit of ?
Solution: To solve this problem, we need to find last digit patterns for both 6 and 2. For 2
it is 2 4 8 6. Therefore, the last digit of is 6. For 6 the last digit of its powers is
always 6. . Therefore, the last digit of is 6.
A) 2 B) 4 C) 6 D) 8 E) 0
2013 Saint Valentine’s Day Mathacre Junior Varsity Written Test Solutions Page 10