AMC 10 Weekly Practice Round 2

Let the original cost price be dollars, and the original profit rate be .

According to the problem:

Solving this equation gives:

Answer: .

Total cost of the purchase: dollars

Total quantity available for sale:

Let the retail price per kilogram be dollars.

Then:

Answer: The retail price per kilogram should be dollars.

After exchanging, the salt concentration becomes:

Volume of exchanged solution:

Each of containers A and B had of solution exchanged.

Let the cost price be dollars. Then the equation is:

Solving this gives:

Answer: The cost price of the goods was dollars.

Let the mass of solution be , then the mass of solution is , and the mass of solution is .

The total salt content equation is:

Solving gives:

Then:

Answer: The masses of the saline solutions with concentrations of , , and are , , and , respectively.

Let the number of boys in School A be , then the number of boys in School B is .

According to the given ratio, we can set up the proportion:

Using the property of proportions (cross multiplication), we get:

Solving the equation gives:

Therefore, School A has boys,

and School B has boys.

The total number of students in both schools is: .

Let there be motorcycles and cars.

According to the problem:

Multiply both sides by :

Solving gives:

Therefore, the ratio of motorcycles to cars is

Let the number of boys in Grade 7 be and the number of girls be .

Then, in Grade 8, there are boys and girls.

The total number of students in Grade 9 is .

So, the total number of students across all three grades is:

Let the number of boys in Grade 9 be .

According to the problem:

Solving this gives:

So, the number of girls in Grade is:

Thus, the total number of girls across all grades is:

The fraction of girls among all students is:

Answer: .

Let the number of students participating from School A and School B be and respectively.

According to the problem, the number of students who did **not** win awards in each school is:

Solving the equation gives:

Therefore, the total number of participants from both schools is: students.

Let the original amount of fruit in Bag A and Bag B be grams and grams, respectively.

From Bag A, was taken out, so the remaining amount is

From Bag B, grams were taken out, so the remaining amount is

Thus,

Next, half of the remaining fruit in Bag B was taken out, so the remaining amount is

It is also given that this remaining amount equals :

Solving this equation:

Substitute back:

So the total original weight of the fruit is:

Answer: The original total weight of the fruit was grams.

There are numbers in the first group and numbers in the second group.

According to the problem,

Simplifying gives:

which leads to:

Therefore, the number of elements in the second group is of the number in the first group.

Let the number of s in the original data set be . Numbers in the original data set have sum , and the data set with the s removed has sum . Thus

Solving this equation for yields . For example, the data set could consist of copies of and copies of .

First, estimate the value of :

If the missing number is (the largest possible value), then the average would be

;

If the missing number is (the smallest possible value), then the average would be

.

This shows that the actual average, , must lie between and .

That means can only be or .

Moreover, since the fraction must be derived from a total divided by ,

and the denominator must be divisible by , we conclude that .

Now calculate:

.

This should equal the sum from to minus the missing number.

The sum from to is ,

so the missing number is:

.

The average of these numbers is:

So the average is around 20, which suggests that is around 40.

Since 3 numbers were erased, the number of remaining numbers should be a multiple of 9.

Therefore, .

The sum of numbers from 1 to 39 is:

After erasing 3 numbers, 36 remain. The total sum of the remaining numbers is:

So the sum of the 3 erased numbers is:

Among the prime numbers less than 39, the maximum possible sum of two primes that add up to no more than 64 is: .

Let the adjusted average scores of the first, second, and third prize winners be , , and , respectively.

According to the problem:

Simplifying:

Therefore:

Answer: After the adjustment, the average score of the first prize winners is points higher than that of the second prize winners.

Let travel at km per hour, and travel at km per hour.

Solving gives: .

travels at km per hour.

Let be the time taken by from departure until meeting .

Let ’s speed be , and ’s speed be .

When and meet, the ratio of the distances they have traveled is , which simplifies to .

Here, is the time from ’s departure to the meeting.

Let the distance between and be .

Then .

Let be the time Andy took to run the first lap.

From the problem, Andy’s speed for the first lap is:

Increasing his speed by for the second lap, the speed becomes:

Thus, the time taken for the second lap is:

Since the second lap is seconds faster than the first lap:

which gives seconds.

Therefore, Andy took seconds to run the first lap.

Casework:

(1) If the number of days is even, then for it is equivalent to riding $$

50 \text{km per day}.$$

Let be the number of days such that at the end of day , B’s distance from is twice ’s distance from .

We have the equation: which gives not a valid solution.

(2) If the number of days is odd, then except for the last day (when rides km), rides km per day on the previous days.

Let be the number of days such that at the end of day , ’s distance from is twice ’s distance from .

We have the equation which gives a valid solution.

Thus, at the end of the 15th day, ’s distance from is twice ’s distance from .

Let the distance between two adjacent flags be , so the circumference of the track is .

Since , let and .

For to overtake , must run more laps than .

Then:

Thus, the time taken for to overtake is:

In this time, covers a distance of:

For to overtake exactly at a flag position, the distance covered must be an integer, which requires to be even.

Thus: (the maximum is extra laps).

However, when , both runners return to the starting point.

Therefore, overtakes exactly at a flag position times.

Let the train’s speed before the increase be km/h, and the distance between the two cities be .

From the problem, the following equations can be set up:

It is easy to solve and obtain , .

Let A walk meters per minute, then B’s speed is meters per minute.

Let minutes be the time B takes to walk from to .

From the problem, we have:

Solving the system gives:

Verification shows , satisfy the original equations.