Problem 1
Andy and Betsy both live in Mathville. Andy leaves Mathville on his bicycle at $1: 30$, traveling due north at a steady 8 miles per hour. Betsy leaves on her bicycle from the same point at 2:30, traveling due east at a steady 12 miles per hour. At what time will they be exactly the same distance from their common starting point?
(A) 3:30
(B) $3: 45$
(C) $4: 00$
(D) $4: 15$
(E) 4:30
Solution :
Let $h$ be the number of hours after Andy starts. Andy travels $8 h$ miles, and Betsy has traveled $12(h-1)$ miles since she started one hour later. Setting them equal:
$$
8 h=12(h-1) \Rightarrow 8 h=12 h-12 \Rightarrow 4 h=12 \Rightarrow h=3
$$
Since Andy started at 1:30, the catch-up time is 4:30. Answer: (E).
Alternatively, from Betsy's perspective: $8(h+1)=12 h \Rightarrow 8 h+8=12 h \Rightarrow h=2$ Same result: (E) $4: 30$.
Problem 2
A box contains 10 pounds of a nut mix that is 50 percent peanuts, 20 percent cashews, and 30 percent almonds. A second nut mix containing 20 percent peanuts, 40 percent cashews, and 40 percent almonds is added to the box resulting in a new nut mix that is 40 percent peanuts. How many pounds of cashews are now in the box?
(A) 3.5
(B) 4
(C) 4.5
(D) 5
(E) 6
Solution :
We are given $0.2(10)=2$ pounds of cashews in the first box.
Denote the pounds of nuts in the second nut mix as $x$.
$$
\begin{gathered}
5+0.2 x=0.4(10+x) \
0.2 x=1 \
x=5
\end{gathered}
$$
Thus, we have 5 pounds of the second mix.
$$
0.4(5)+2=2+2=\text { (B) } 4
$$
Problem 3
A team of students is going to compete against a team of teachers in a trivia contest. The total number of students and teachers is 15 . Ash, a cousin of one of the students, wants to join the contest. If Ash plays with the students, the average age on that team will increase from 12 to 14 . If Ash plays with the teachers, the average age on that team will decrease from 55 to 52 . How old is Ash?
(A) 28
(B) 29
(C) 30
(D) 32
(E) 33
Solution :
When Ash joins a team, the team's average is pulled towards his age. Let $A$ be Ash's age and $N$ be the number of people on the student team. This means that there are $15-N$ people in the teacher team. Let us write an expression for the change in the average for each team.
The students originally had an average of 12 , which became 14 when Ash joined, so there was an increase of 2 . The term $A-12$ represents how much older Ash is compared to the average of the students'. If we divide this by $N+1$, which is the number of people on the student team when Ash joins, we get the average change per team member once Ash is added. Therefore,
$$
\frac{A-12}{N+1}=2 .
$$
Similarly, for teachers, the average was originally 55 , which decreased by 3 to become 52 when Ash joined. Intuitively, $55-A$ represents how much younger Ash is than the average age of the teachers. Dividing this by the expression $(15-N)+1$, which is the new total number of people on the teacher team, represents the average change per team member once Ash joins. We can write the equation
$$
\frac{55-A}{16-N}=3 .
$$
To solve the system, multiply equation (1) by $N+1$, and similarly multiply equation (2) by $16-N$. Then add the equations together, canceling $A$, leaving equation $43=50-N$. From this we get $N=7$ and $A=28$.
Problem 4:
Agnes writes the following four statements on a blank piece of paper.
Each statement is either true or false. How many false statements did Agnes write on the paper?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Solution
We first number all the statements:
1) At least one of these statements is true. 2) At least two of these statements are true. 3) At least two of these statements are false. 4) At least one of these statements is false.
We can immediately see that statement 4 must be true, as it would contradict itself if it were false. Similarly, statement 1 must be true, as all the other statements must be false if it were false, which is contradictory because statement 4 is true. Since both 1 and 4 are true, statement 2 has to be true. Therefore, statement 3 is the only false statement, making the answer (B) 1.
Problem 5
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Problem 6:
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