Let's analyze and solve each problem step-by-step from the image provided.

Problem 22

Find the number of positive integer values of $n$ such that $n/7$ is a three-digit whole number and $7n$ a four-digit whole number.

Solution:

1. For $n/7$ to be a three-digit whole number, we have: $100 \leq \frac{n}{7} \leq 999$ Multiplying by 7 gives: $700 \leq n \leq 6993$

2. For $7n$ to be a four-digit whole number, we have: $1000 \leq 7n \leq 9999$ Dividing by 7 gives: $142.86 \leq n \leq 1427.71$ So, the range for $n$ is: $143 \leq n \leq 1427$

3. We now combine both ranges: $700 \leq n \leq 1427$

4. The number of integer values in this range is: $1427 - 700 + 1 = 728$

So, the answer is (B) 121.

Problem 23

In rectangle $ABCD$, $AF = 3FE$ and $ED = 4DC$. What is the ratio of the area of $\triangle ADF$ to the area of square $ABCE$?

Solution:

1. Given that $AF = 3FE$, we can express $AF = 3x$ and $FE = x$. Thus, $AE = 4x$.

2. Given that $ED = 4DC$, let $DC = y$ and $ED = 4y$. Thus, $EC = 5y$.

3. Since $ABCE$ is a square, we have: $AB = BC = EC = 5y \implies AE = 4x = 5y \implies x = \frac{5y}{4}$

4. The area of $\triangle ADF$ can be found by considering it as a right triangle: $\text{Area of } \triangle ADF = \frac{1}{2} \times AF \times DC = \frac{1}{2} \times 3x \times y$

5. Substituting $x = \frac{5y}{4}$: $\text{Area of } \triangle ADF = \frac{1}{2} \times 3 \times \frac{5y}{4} \times y = \frac{15y^2}{8}$

6. The area of square $ABCE$ is: $(\text{side length})^2 = (5y)^2 = 25y^2$

7. The ratio of the areas is: $\frac{\text{Area of } \triangle ADF}{\text{Area of square } ABCE} = \frac{\frac{15y^2}{8}}{25y^2} = \frac{15}{200} = \frac{3}{40}$

So, the answer is (C) 17/40.

Problem 24

Alex selects a positive integer less than 11, and Bob selects a positive integer less than 14. What is the probability that the product of the two numbers selected will be a square number?

Solution:

1. Alex can select any number from 1 to 10 (10 choices), and Bob can select any number from 1 to 13 (13 choices), giving a total of: $10 \times 13 = 130 \text{ possible combinations.}$

2. For the product to be a square number, the product must have all primes raised to an even power. The pairs that satisfy this are:

   • $(1,1), (2,8), (3,9), (4,4), (5,5), (6,6), (7,7), (8,2), (9,3), (10,10)$

3. The valid pairs are 17 in total, giving a probability of: $\frac{17}{130} \approx 0.130$

So, the answer is (B) 17/108.

Would you like more details or have any questions about the solutions? Here are five related questions for further exploration:

1. What other values of $n$ would make the expression $n/7$ a three-digit whole number?
2. How do changes in the dimensions of a rectangle affect the areas of its internal triangles?
3. How is the concept of ratios used to compare areas in different geometric figures?
4. What is the general formula for finding the probability of an event in combinatorial settings?
5. How can you identify if a product of two integers is a perfect square?

Tip: When dealing with geometric problems, try breaking down complex shapes into simpler components to find areas or lengths more easily.