Alicia earns dollars per hour, of which
is deducted to pay local taxes. How
many cents per hour of Alicia's wages are used to pay local taxes?
On the AMC 12, each correct answer is worth points, each incorrect answer is
worth points, and each problem left unanswered is worth
of the points. If Charlyn problems unanswered, how many of the remaining problems
must she answer correctly in order to score at least
For how many ordered pairs of positive integers
? Bertha has daughters and no sons. Some of her daughters have daughters, and the rest have none. Bertha has a total of daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no children?
The graph of the line is shown. Which of the following is true?
, , ,
. Which of the following is the largest?
A game is played with tokens according to the following rules. In each round, the player with the most tokens gives one token to each of the other players and also places one token into a discard pile. The game ends when some player runs out of tokens. Players , and start with
many rounds will there be in the game?
Solution , and tokens, respectively. How
In the overlapping triangles
and of ,
and and intersect at
are right angles,
sharing common , , ,
. What is the difference between the areas
A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars would increase sales. If the diameter of the jars is increased by without altering the volume, by what percent must the height be decreased?
The sum of consecutive integers is . What is their median?
The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is
cents. If she had one more quarter, the average value would be cents. How
many dimes does she have in her purse? Solution
and are on the line , ? intersect at . What is the length of
be the set of points
be in the coordinate plane, where each of
, , or
. How many distinct lines pass through at least two members of ?
A sequence of three real numbers forms an arithmetic progression with a first term of . If is added to the second term and is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term in the geometric progression?
Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run
They next meet after Sally has run meters. meters past their first meeting point. Each
girl runs at a constant speed. What is the length of the track in meters?
The set of all real numbers for which
is defined is . What is the value of
Let be a function with the following properties:
, for any positive integer .
What is the value of ?
Square has side length
. A semicircle with diameter is constructed inside the square, and the tangent to the semicircle from
intersects side at
. What is the length of ?
are externally tangent to each other, and internally tangent to . Circles and
are congruent. Circle has radius and passes through . What is the radius of circle
? the center of
Select numbers and between and independently and at random, and
let be their sum. Let and be the results when and , respectively, are
? rounded to the nearest integer. What is the probability that
If , what is the value of
Three mutually tangent spheres of radius rest on a horizontal plane. A sphere of
radius rests on them. What is the distance from the plane to the top of the larger
has real coefficients with
zeroes , and with distinct complex and real, , and
Which of the following quantities can be a nonzero number?
A plane contains points and
radius in the plane that cover . Let be the union of all disks of . What is the area of ?
For each integer
product , let denote the base- number , where . The can be expressed as and are positive integers
and is as small as possible. What is the value of
20 dollars is the same as 2000 cents, and is
of 2000 Problem 2
She gets points for the problems she didn't answer. She must
get problems right to score at least 100
Every integer value of leads to an integer solution for Since
must be positive,
must be positive,
equation solutions to the values for y, which mean there are
Since Bertha has 6 daughters, Bertha has
none have daughters. Of Bertha's daughters,
so do not have daughters. granddaughters, of which have daughters,
Therefore, of Bertha's daughters and granddaughters,
do not have
Draw a tree diagram and see that the answer can be found in the sum of 6 + 6 granddaughters, 5 + 5 daughters, and 4 more daughters Problem 5
It looks like it has a slope of
and is shifted
is the largest. Problem 7 Look at a set of 3 rounds, where the players have
, , and tokens. Each of the players will gain two tokens from the others and give away 3 tokens, so overall, each player will lose 1 token.
Therefore, after 12 sets of 3 rounds, or 36 rounds, the players will have 3, 2, and 1 tokens, repectively. After 1 more round, player will give away his last 3 tokens and the game will stop .
Solution 1 Since
we find that and , . By
alternate interior angles and AA~,
, with side length ratio . Their heights also have
the same ratio, and since the two heights add up to , we have
that areas, and . . Subtracting the Solution 2
that represent the area of figure . Note and . .
When the diameter is increased by , it is increased by , so the area of the base is increased by .
To keep the volume the same, the height must be
which is a reduction of the original height,
The median of a sequence is the middle number of the sequence when the sequence is arranged in order. Since the integers are consecutive, the median is also the mean, so the median is
Let the total value (in cents) of the coins Paula has originally be , and the number of coins she has be .
Then and .Substituting yields
. It is easy to see now that Paula has 3
quarters, 1 nickel, so she has dimes.
If the new coin was worth 20 cents, adding it would not change the mean at all. The additional 5 cents raise the mean by 1, thus the new number of coins must be 5. Therefore initially there were 4 coins worth a total of cents. As in the previous solution, we conclude that the only way to get 80 cents using 4 coins is 25+25+25+5.
Solution 1 Let's count them by cases:
? Case 1
: The line is horizontal or vertical, clearly .
? Case 2: The line has slope , with through and additional ones one unit
above or below those. These total .
? Case 3: The only remaining lines pass through two points, a vertex and a non-vertex
point on the opposite side. Thus we have each vertex pairing up with two points on the two opposites sides, giving lines.
These add up to .
Solution 2 There are ways to pick two points, but we've clearly overcounted all of the lines which pass through three points. In fact, each line which passes through
three points will have been counted times, so we have to subtract for
each of these lines. Quick counting yields horizontal, vertical, and
lines, so the answer is distinct lines.
Let be the common difference. Then are the terms of the
geometric progression. Since the middle term is the geometric mean of the other two terms,
, and the third term . The smallest possible value occurs when
Solution 1 Call the length of the race track . When they meet at the first meeting point, Brenda has run meters, while Sally has run meters. By the second
meters. meeting point, Sally has run meters, while Brenda has run
Since they run at a constant speed, we can set up
that . . Cross-multiplying, we get
Solution 2 The total distance the girls run between the start and the first meeting is one half of
the track length.
The total distance they run between the two meetings is the track length.
As the girls run at constant speeds, the interval between the meetings is twice as long as the interval between the start and the first meeting.
Thus between the meetings Brenda will run length of the track is meters meters. Therefore the
We know that the domain of
and , where is a constant, is .
. By the definition of logarithms
, we then .
Let the point of tangency be . By the
The on and yields . Thus .
Solution 2 Clearly,
. Thus, the sides of
and are in arithmetic progression. Thus it is
to the triangle
since , .
since is the center of the larger circle of radius . Using
, the Pythagorean Theorem on
Now using the on ,
We can apply .
The four circles have curvatures , and .
Simplifying, we get
. The probability that
and is . Notice that the sum ranges from to
with a symmetric distribution across ,
and we want . Thus the chance is . 2.
4. . The probability that and is
, but automatically. Hence the chance
, which makes
. This is the same as the previous case. . We recognize that this is equivalent to the first case.
Our answer is .
Use areas to deal with this continuous probability problem. Set up a unit square with values of on x-axis and on y-axis.
if then this will work because . Similarly then this will work because in order for this to happen, and are
making , and . Each of these triangles in each greater than
the unit square has area of 1/8.
The only case left is when . Then each of and must be 1 and 0, in any order. These cut off squares of area 1/2 from the upper left and lower right corners of the unit square.
Then the area producing the desired result is 3/4. Since the area of the unit square is 1, the probability is
This is an infinite geometric series, which sums
. Using the formula .
, we have
the heights up, we get . Adding Problem 23
We have to evaluate the answer choices and use process of elimination:
? : We are given that
then . , so . If one of the roots is zero,
? : By , we know that is the sum of all of the roots of .
Since that is real, , and , so . ?
of : All of the coefficients are real. For sake of contradiction suppose none are zero. Then for each complex root , its
product is equal to zero. is also a root. So the roots should pair up, but we have an odd number of imaginary roots! This gives us the contradiction, and therefore the
? : We are given that . Since the coefficients are real, it follows
that if a root is complex, its conjugate is also a root; and the sum of the imaginary parts of complex conjugates is zero. Hence the RHS is zero.
There is, however, no reason to believe that quantity is
should be zero (in fact, that ).
, and there is no evidence that is a root of
As the red circles move about segment , they cover the area we are looking for. On the left side, the circle must move around pivoted on . On the right side, the circle must move pivoted on However, at the top and bottom, the circle must lie on both A and B, giving us our upper and lower bounds.
This egg-like shape is .
The area of the region can be found by dividing it into several sectors, namely
This is an infinite geometric series
with common ratio
term , and initial
Alternatively, we could have used the algebraic manipulation for repeating decimals,
Some factors cancel, (after all,
isn't one of the answer choices)
Since the only factor in the numerator that goes into Therefore the answer is . is , is minimized.