Answer: C

To determine which fruit is the cheapest, we can compare the prices based on the given statements:

1.  "The mangoes are more expensive than the apples" → **Mangoes > Apples**
2.  "The loquats are more expensive than the mangoes" → **Loquats > Mangoes**

Combining these two inequalities: **Loquats > Mangoes > Apples**.

This ordering shows that apples have the lowest price of the three. Therefore, the apples are the cheapest.

Comparing this to the options:
(A) Loquats are the cheapest (False)
(B) Mangoes are the cheapest (False)
(C) Apples are the cheapest (True)

ANSWER: C


Answer: B

To determine the order of the three birds, we use the relative positions given in the text:

1.  "The crow is to the left of the raven": This means the crow comes before the raven (Crow — Raven).
2.  "The falcon is to the right of the raven": This means the falcon comes after the raven (Raven — Falcon).

Combining these two pieces of information gives the fixed order from left to right: Crow, Raven, Falcon.

Looking at the options:
(A) The crow is the rightmost — Incorrect (it is leftmost).
(B) The falcon is the rightmost — Correct.
(C) The raven is the rightmost — Incorrect (it is in the middle).

ANSWER: B


Answer: B

To determine the order of the three birds, we can place them in three positions from left to right:

1.  "The robin is the second from the left." This fixes the middle position: **Quail/Raven — Robin — Quail/Raven**.
2.  "The quail is to the left of the robin." Since the robin is in the middle, the quail must occupy the leftmost position: **Quail — Robin — Raven**.
3.  The remaining bird is the raven, which fills the only empty slot on the right: **Quail — Robin — Raven**.

Checking the positions:
- Left: Quail
- Middle: Robin (second from left)
- Right: Raven

Therefore, the raven is the rightmost bird.

ANSWER: (B)


Answer: C

To determine the correct order of the three books, we can analyze the two given statements step by step:

1.  **"The gray book is to the left of the purple book"**: This establishes a relative order: Gray < Purple.
2.  **"The purple book is to the left of the red book"**: This establishes another relative order: Purple < Red.

Combining these two statements gives a single, consistent linear order from left to right: **Gray → Purple → Red**.

Now we check the options:
- (A) The gray book is the rightmost — False; it is the leftmost.
- (B) The purple book is the rightmost — False; it is in the middle.
- (C) The red book is the rightmost — True; it is the last book in the sequence.

ANSWER: (C)


Answer: A

To determine the correct ordering of the three fruits by price, we evaluate the two given statements:

1.  **"The watermelons are the second-most expensive"**: This fixes the watermelons in the middle position. The order must be either (Most Expensive, Watermelon, Least Expensive) or (Least Expensive, Watermelon, Most Expensive).
2.  **"The loquats are more expensive than the cantaloupes"**: This means the loquats must occupy the highest price position and the cantaloupes the lowest.

Combining these: the loquats are the most expensive, the watermelons are second, and the cantaloupes are the least expensive. This is the only arrangement that satisfies both conditions.

Comparing this to the options:
(A) The loquats are the most expensive — True
(B) The watermelons are the most expensive — False (they are second)
(C) The cantaloupes are the most expensive — False (they are least)

ANSWER: (A)


Answer: C

To determine the correct option, we can map the three positions from left to right (Position 1, Position 2, Position 3) based on the given statements:

1. **"The orange book is the second from the left."** This explicitly places the orange book in Position 2.
2. **"The red book is to the left of the orange book."** Since the orange book is in Position 2, the only position to its left is Position 1. Therefore, the red book must be in Position 1.
3. **Remaining position:** The blue book must occupy the remaining slot, Position 3.

The arrangement is: Red — Orange — Blue.

Now we check the options:
(A) The red book is the second from the left (False — it is first)
(B) The blue book is the second from the left ( False — it is third)
(C) The orange book is the second from the left ( True — stated directly in the text)

ANSWER: C


Answer: A

To determine the correct order of the three birds, we follow the statements in the paragraph:

1. **"The falcon is the leftmost."** This fixes the falcon in the first position: [Falcon, _, _].
2. **"The raven is to the left of the owl."** Since only two positions remain (middle and right), the raven must be in the middle and the owl on the right: [Falcon, Raven, Owl].

Comparing this arrangement to the options:
(A) The falcon is the leftmost — True
(B) The owl is the leftmost — False
(C) The raven is the leftmost — False

The only logically consistent option is (A).

ANSWER: A


Answer: A

We can determine the finishing order by placing the three golfers relative to each other based on the given statements:

1. "Ada finished above Ana" means Ada's rank is higher than Ana's (Ada < Ana in terms of rank number).
2. "Mel finished below Ana" means Mel's rank is lower than Ana's (Ana < Mel).

Combining these two, the order from highest to lowest is Ada → Ana → Mel. In a three-person set, this places Ada first, Ana second, and Mel third.

Checking the options:
(A) Ana finished second — Matches our order.
(B) Mel finished second — Incorrect; Mel is third.
(C) Ada finished second — Incorrect; Ada is first.

ANSWER: A


Answer: A

To determine the correct answer, we must compare the relative ages of the three vehicles based on the statements given:

1. "The truck is older than the bus" → Truck > Bus
2. "The bus is older than the sedan" → Bus > Sedan

Combining these two, we get the age order from oldest to newest: Truck > Bus > Sedan.

Now let's evaluate the options:
- (A) The bus is the second-newest: The newest is the sedan, the second-newest is the bus, and the third-newest is the truck. This matches the order.
- (B) The sedan is the second-newest: The sedan is the newest.
- (C) The truck is the second-newest: The truck is the oldest.

Therefore, (A) is the only consistent statement.

ANSWER: (A)


Answer: B

To determine the order of the books, we analyze the two given statements:

1.  "The brown book is to the left of the orange book" → brown < orange
2.  "The green book is to the left of the brown book" → green < brown

Combining these gives the fixed order from left to right: **green, brown, orange**.

Positions:
-   Left (1st): green
-   Middle (2nd): brown
-   Right (3rd): orange

Checking the options:
(A) green is second — False (it is first)
(B) brown is second — True
(C) orange is second — False (it is third)

ANSWER: B


Answer: C

To determine the order of the three birds, we can analyze the two given statements step by step:

1.  "The blue jay is to the right of the quail" means the quail must be to the left of the blue jay. So far, the order is: quail → blue jay.
2.  "The falcon is to the right of the blue jay" means the falcon must be to the right of the blue jay. Adding this to the previous order gives: quail → blue jay → falcon.

In this fixed left-to-right arrangement, the falcon is the bird on the far right.

Comparing this to the options:
(A) The blue jay is the rightmost — False
(B) The quail is the rightmost — False
(C) The falcon is the rightmost — True

ANSWER: C


Answer: A

To determine the relative prices of the three fruits, we can compare them step by step based on the given statements:

1.  "The kiwis are less expensive than the loquats" means: **Loquats > Kiwis**.
2.  "The cantaloupes are less expensive than the kiwis" means: **Kiwis > Cantaloupes**.

Combining these two inequalities, the order from most expensive to least expensive is: **Loquats > Kiwis > Cantaloupes**.

In this fixed ordering:
-   Loquats are the most expensive.
-   Kiwis are the second-most expensive (the middle value).
-   Cantaloupes are the least expensive.

Therefore, option (A) is the only logically consistent statement.

ANSWER: (A)


Answer: C

To determine the correct arrangement, we follow the relative positions given in the paragraph:

1. **The hummingbird is to the left of the owl.** This places the hummingbird on the left side of the owl.
2. **The falcon is to the right of the owl.** This places the falcon on the right side of the owl.

Combining these two statements, the owl sits in the middle, with the hummingbird on its left and the falcon on its right. The fixed order from left to right is: hummingbird, owl, falcon.

Looking at the options:
- (A) The hummingbird is the leftmost, not the rightmost.
- (B) The owl is in the middle.
- (C) The falcon is on the far right.

Therefore, the falcon is the rightmost bird.

ANSWER: (C)


Answer: C

To determine the order of the three birds, we can place them one by one based on the given statements:

1.  **"The blue jay is to the right of the falcon"**: This establishes a relative order of falcon → blue jay (falcon is to the left of the blue jay).
2.  **"The hummingbird is to the left of the falcon"**: This places the hummingbird to the left of the falcon, so the order extends to hummingbird → falcon → blue jay.

Putting these together, the birds are arranged from left to right as: hummingbird, falcon, blue jay. Therefore, the hummingbird is the leftmost bird.

Comparing this to the options:
(A) The blue jay is the leftmost — False (it is the rightmost).
(B) The falcon is the leftmost — False (the hummingbird is to its left).
(C) The hummingbird is the leftmost — True.

ANSWER: (C)


Answer: A

The three vehicles are ordered by age from oldest to newest. The convertible is older than the tractor, so the convertible must be either the oldest or the second-oldest. The hatchback is stated to be the second-newest, which is the middle position in a set of three. This leaves the oldest and newest positions for the convertible and the tractor; since the convertible is older than the tractor, the convertible is the oldest and the tractor is the newest. Therefore, the hatchback is the second-newest.

ANSWER: (A)


Answer: A

To determine the order of the books, let's break down the information given:

1. **"The purple book is the rightmost"**: This fixes the third position. The order is [?, ?, purple].
2. **"The orange book is to the left of the white book"**: With only positions 1 and 2 remaining, the orange book must be in position 1 and the white book in position 2.

The resulting order from left to right is: **orange, white, purple**.

Evaluating the options:
(A) The orange book is the leftmost — Correct.
(B) The white book is the leftmost — Incorrect (it is in the middle).
(C) The purple book is the leftmost — Incorrect (it is on the far right).

ANSWER: (A)


Answer: C

To determine which option is logically consistent with the paragraph, let's break down the information given:

1.  **"The truck is newer than the station wagon"**: This means the station wagon is older than the truck.
2.  **"The motorcycle is the oldest"**: This explicitly states the motorcycle's age relative to the other two vehicles.

Combining these facts gives the age order from oldest to newest: motorcycle → station wagon → truck.

Now, let's evaluate the options:
(A) The station wagon is the oldest — False, the motorcycle is the oldest.
(B) The truck is the oldest — False, the truck is the newest of the three.
(C) The motorcycle is the oldest — True, this is stated directly in the text and is consistent with all other information.

Therefore, option (C) is the only statement that is logically consistent with the paragraph.

ANSWER: (C)


Answer: A

To determine the correct order of the three birds, we can follow these steps:

1.  **Identify the fixed positions:** The paragraph states that the **crow is the second from the left**. So, the middle position is occupied by the crow: `[?, crow, ?]`.
2.  **Place the hummingbird:** The hummingbird is to the left of the crow. In a three-object sequence where the crow is in the middle, there is only one position to its left: the leftmost position. So, the **hummingbird is the leftmost**.
3.  **Place the quail:** With the leftmost and middle positions filled, the remaining bird (the quail) must occupy the final position on the right.
4.  **Final order:** The left-to-right sequence is hummingbird, crow, quail.

Comparing this to the options:
(A) The hummingbird is the leftmost — Correct.
(B) The quail is the leftmost — Incorrect (it is on the right).
(C) The crow is the leftmost — Incorrect (it is in the middle).

ANSWER: (A)


Answer: C

To determine the finishing order, we analyze the two statements given:

1. **Joe finished below Ada**: This means Ada placed higher than Joe (Ada > Joe).
2. **Amy finished above Ada**: This means Amy placed higher than Ada (Amy > Ada).

Combining these two constraints, the order from highest to lowest is Amy → Ada → Joe. Therefore, Joe finished below both Amy and Ada, placing him at the bottom of the three.

Checking the options:
(A) Amy finished last — False (she finished first).
(B) Ada finished last — False (Joe finished below her).
(C) Joe finished last — True (he is below both Amy and Ada).

ANSWER: (C)


Answer: B

To determine the correct option, we can arrange the three golfers in order of finish based on the two statements provided:

1.  "Ada finished above Ana": This means Ada placed higher than Ana (Ada > Ana).
2.  "Mel finished below Ana": This means Mel placed lower than Ana (Ana > Mel).

Combining these two relations, we get the order from highest to lowest: Ada → Ana → Mel.

Therefore, Mel finished below both Ada and Ana, placing her in the third position.

Checking the options:
(A) Ana finished last — False; Mel finished below her.
(B) Mel finished last — True; she is below Ana, who is below Ada.
(C) Ada finished last — False; she finished above Ana.

ANSWER: B


Answer: 480

To find the total number of people inside the church, we calculate how many people were transported by each type of vehicle and add them together:

1.  **People in cars:** There were 20 cars, and each carried 3 people.
    20 cars × 3 people/car = 60 people

2.  **People in buses:** There were 12 buses, and each carried 35 people.
    12 buses × 35 people/bus = 420 people

3.  **Total people:** Add the people from the cars and the buses.
    60 + 420 = 480 people

ANSWER: 480


Answer: 29

To find the total amount of water removed, we first determine how much water is removed each day at each setting:

1.  **Low setting:** 1 liter per day.
2.  **Medium setting:** Twice as much as the low setting, so 2 × 1 = 2 liters per day.
3.  **High setting:** Twice as much as the medium setting, so 2 × 2 = 4 liters per day.

Next, we multiply the daily removal rate by the number of days Brian ran the device at each setting:

-   **Low setting:** 3 days × 1 liter/day = 3 liters.
-   **Medium setting:** 3 days × 2 liters/day = 6 liters.
-   **High setting:** 5 days × 4 liters/day = 20 liters.

Finally, we add these amounts together to find the total: 3 + 6 + 20 = 29 liters.

ANSWER: 29


Answer: 50

To find how much money Gerald has left, we follow these steps:

1.  **Determine the total parts in the ratio**: The ratio is 3:2, so there are 3 + 2 = 5 equal parts in total.
2.  **Find the value of one part**: Divide the total amount by the total number of parts: $100 ÷ 5 = $20 per part.
3.  **Calculate Gerald's share**: Gerald receives 3 parts, so he gets 3 × $20 = $60.
4.  **Subtract what he spent**: Gerald spent $10 on a book, so his remaining money is $60 - $10 = $50.

ANSWER: $50


Answer: 2100

Let's break this down step by step:

**Month 1:**
1. Red hat sales = $400
2. Green hat sales: Red sales were half of green sales, so green sales = $400 × 2 = $800
3. Total sales for Month 1 = $400 + $800 = $1,200

**Month 2:**
1. Sales = 3/4 of Month 1 total = 3/4 × $1,200 = $900

**Total for two months:**
$1,200 + $900 = $2,100

ANSWER: $2,100


Answer: 43500

To find the total cost of Bill's new truck, we add the base price and each extra feature cost:

1.  **Base price**: $30,000
2.  **King cab upgrade**: $7,500
3.  **Leather seats**: One-third of the king cab upgrade = $7,500 ÷ 3 = $2,500
4.  **Running boards**: $500 less than the leather seats = $2,500 - $500 = $2,000
5.  **Upgraded exterior light package**: $1,500
6.  **Towing package**: The problem states the truck has a towing package but does not assign it a specific cost. In such problems, a feature with no cost mentioned is treated as $0.

Total = $30,000 + $7,500 + $2,500 + $2,000 + $1,500 = $43,500.

ANSWER: 43500


Answer: 31

To find the total number of cheese slices Carl used, we add the slices from each meal type:

1.  **Sandwiches for lunch**: He ate a sandwich every day of the week (7 days) with 2 slices each.
    7 days × 2 slices/day = 14 slices

2.  **Omelets for breakfast**: He had omelets on 3 days. Each omelet used one more slice than a sandwich (2 + 1 = 3 slices per omelet).
    3 days × 3 slices/day = 9 slices

3.  **Macaroni and cheese for dinner**: The problem states he used a fixed total of 8 slices in the big dish.
    Total = 8 slices

Adding these together: 14 + 9 + 8 = 31 slices.

ANSWER: 31


Answer: 240

To find out how many sheets Miguel uses every month, we can break the problem down into steps:

1.  **Sheets per week:** Miguel uses 2 pads a week, and each pad has 30 sheets, so he uses 2 × 30 = 60 sheets per week.
2.  **Weeks per month:** A standard month is generally considered to have 4 weeks.
3.  **Sheets per month:** Multiplying the weekly usage by the number of weeks in a month gives 60 × 4 = 240 sheets.

Miguel uses 240 sheets of paper every month.

ANSWER: 240


Answer: 5

To find out how many points Joey is winning by, we calculate each player's final score after their turn and then compare them.

1.  **Calculate Joey's final score:**
    Joey started with 214 points and scored 26 on his turn.
    214 + 26 = 240 points.

2.  **Calculate Marcy's final score:**
    Marcy started with 225 points and scored 10 on her turn.
    225 + 10 = 235 points.

3.  **Compare the scores:**
    Joey has 240 points and Marcy has 235 points. Since 240 is greater than 235, Joey is winning.

4.  **Find the winning margin:**
    Subtract Marcy's score from Joey's score.
    240 - 235 = 5 points.

Joey is winning by 5 points.

ANSWER: 5


Answer: 20

To find Charlie's net profit, we need to subtract his total costs from his total revenue.

**Step 1: Determine the amount of beeswax needed.**
Charlie makes 10 candles per pound of beeswax. To make 20 candles, he needs:
20 candles / 10 candles per pound = 2 pounds of beeswax.

**Step 2: Calculate the total supply cost.**
The cost for one pound of beeswax and wicks is $10.00. For 2 pounds, the cost is:
2 pounds × $10.00 per pound = $20.00.

**Step 3: Calculate the total revenue.**
Charlie sells each of the 20 candles for $2.00:
20 candles × $2.00 per candle = $40.00.

**Step 4: Calculate the net profit.**
Net profit is total revenue minus total costs:
$40.00 (revenue) - $20.00 (costs) = $20.00.

ANSWER: $20.00


Answer: 130000

To find the total population, we can break the problem into three steps:

1.  **Population per country:** Each country has 5 cities, and each city has 1,000 people. So, there are 5 × 1,000 = 5,000 people in each country.
2.  **Total countries:** There are 26 countries in South America.
3.  **Total population:** Multiply the population per country by the number of countries: 26 × 5,000 = 130,000.

ANSWER: 130,000


Answer: 80

To find out how much Jeff gets, we can set up a simple algebraic equation. Let the amount Brad receives be $x$. Since Jeff gets 4 times as much as Brad, Jeff receives $4x$. Together, their shares must add up to the total of 100 dollars:

$x + 4x = 100$
$5x = 100$
$x = 20$

So Brad receives 20 dollars, and Jeff receives $4 \times 20 = 80$ dollars. We can check this: $20 + 80 = 100$, which matches the total.

ANSWER: 80


Answer: 8

To find the age of Brandon’s iPhone, we can work backward from Suzy’s iPhone:

1.  **Suzy’s iPhone:** The problem states Suzy’s iPhone is 1 year old.
2.  **Ben’s iPhone:** Ben’s iPhone is two times older than Suzy’s, so multiply Suzy’s age by 2: 1 year × 2 = 2 years old.
3.  **Brandon’s iPhone:** Brandon’s iPhone is four times as old as Ben’s, so multiply Ben’s age by 4: 2 years × 4 = 8 years old.

ANSWER: 8


Answer: 48

To find the total number of hours Sadie slept during the week, we break the week down day by day:

1.  **Monday:** Sadie slept **8 hours**.
2.  **Tuesday and Wednesday:** For each of these two days, she slept 2 hours less than on Monday.
    *   Tuesday = 8 - 2 = 6 hours
    *   Wednesday = 8 - 2 = 6 hours
3.  **Thursday, Friday, Saturday, and Sunday:** For the rest of the week (4 days), she slept 1 hour more than those two days (6 hours).
    *   Each of these 4 days = 6 + 1 = 7 hours
    *   Total for these 4 days = 4 × 7 = 28 hours

**Total for the week:**
8 (Mon) + 6 (Tue) + 6 (Wed) + 28 (Thu-Sun) = 48 hours

ANSWER: 48


Answer: 296

To find the total amount Helen paid, we can break the costs down into two parts: the hotel stay and the bus trips.

**Step 1: Calculate the hotel cost**
Helen stayed for 3 nights at $80 per night.
3 nights × $80 = $240

**Step 2: Calculate the cost of one bus trip**
Each bus trip costs 10% of one night at the hotel ($80).
10% of $80 = 0.10 × $80 = $8

**Step 3: Calculate the total bus cost**
Helen rode the bus 7 times at $8 per trip.
7 trips × $8 = $56

**Step 4: Add both costs together**
Total = hotel cost + bus cost
Total = $240 + $56 = $296

Helen paid a total of $296.

ANSWER: C


Answer: 15

To find how many more popsicles Sam has than Betty, we can follow these steps:

1.  **Understand the ratio:** The ratio of Betty’s popsicles to Sam’s is 5:6. This means for every 5 popsicles Betty has, Sam has 6. We can represent Betty’s amount as $5x$ and Sam’s as $6x$.

2.  **Set up the equation:** The total number of popsicles is 165, so:
    $5x + 6x = 165$
    $11x = 165$

3.  **Solve for x:** Divide 165 by 11:
    $x = 15$

4.  **Calculate each person’s amount:**
    Betty has $5 \times 15 = 75$ popsicles.
    Sam has $6 \times 15 = 90$ popsicles.

5.  **Find the difference:** Subtract Betty’s amount from Sam’s:
    $90 - 75 = 15$

Alternatively, the difference in the ratio is 1 part ($6 - 5 = 1$), so the answer is simply $1x = 15$.

ANSWER: 15


Answer: 800

To find the total amount John spent on ties, we can break the problem down into three steps:

1.  **Calculate the blue tie information:**
    *   The cost per blue tie is $40.
    *   He spent $200 on blue ties, so the number of blue ties is $200 \div 40 = 5$.

2.  **Calculate the red tie information:**
    *   He buys twice as many red ties as blue ties, so he has $5 \times 2 = 10$ red ties.
    *   Red ties cost 50% more than blue ties, so the cost per red tie is $40 \times 1.5 = \$60$.
    *   The total spent on red ties is $10 \times 60 = \$600$.

3.  **Calculate the total spent:**
    *   Total spent = blue tie spending + red tie spending = $200 + 600 = \$800$.

ANSWER: $800


Answer: 104

To find the total number of hours Meredith spent writing articles over the three days, we calculate the number of articles written each day and multiply the total by the average time per article.

1.  **Monday**: She wrote 5 articles.
2.  **Tuesday**: She wrote 2/5 times *more* than on Monday. This means she wrote Monday’s count plus an additional 2/5 of that count:
    $5 + (2/5 \times 5) = 5 + 2 = 7$ articles.
3.  **Wednesday**: She wrote twice the number she wrote on Tuesday:
    $7 \times 2 = 14$ articles.
4.  **Total articles**: $5 + 7 + 14 = 26$ articles.
5.  **Total hours**: Each article takes an average of 4 hours, so $26 \times 4 = 104$ hours.

ANSWER: 104


Answer: 1

To find the difference between Cole and Xavier’s current heights, we need to calculate each of their new heights in inches and then subtract one from the other.

**Step 1: Calculate Xavier’s current height.**
Xavier started at 4 feet tall. Since there are 12 inches in a foot, his starting height was 48 inches (4 × 12 = 48). He grew 3 inches over the summer, so his current height is 48 + 3 = 51 inches.

**Step 2: Calculate Cole’s current height.**
Cole started at 50 inches tall and grew 2 inches over the summer, so his current height is 50 + 2 = 52 inches.

**Step 3: Find the difference.**
Subtract Xavier’s current height from Cole’s current height: 52 - 51 = 1 inch.

The difference between Cole and Xavier’s height now is 1 inch.

ANSWER: A


Answer: 2

To find how many trays Rozanne can put out, we need to determine the total number of eggs and divide by how many eggs are needed per tray.

1.  **Calculate the total number of eggs:**
    *   One dozen is 12 eggs, so 4 dozen is 4 × 12 = 48 eggs.
    *   There are also 2 loose eggs.
    *   Total eggs = 48 + 2 = 50 eggs.

2.  **Calculate the number of eggs needed per tray:**
    *   Each tray holds 5 glasses.
    *   Each glass uses 5 eggs.
    *   Eggs per tray = 5 glasses × 5 eggs/glass = 25 eggs.

3.  **Calculate the number of trays:**
    *   Total eggs ÷ eggs per tray = 50 ÷ 25 = 2 trays.

Rozanne can put out 2 trays.

ANSWER: 2


Answer: 11

To find the total cost of Aleksandra's dinner, we calculate the price of each item based on the cost of the bagel:

1.  **Bagel:** The price is given as **$4**.
2.  **Soup:** The soup costs 25% more than the bagel.
    *   25% of $4 = 0.25 × $4 = $1.
    *   Soup price = $4 + $1 = **$5**.
3.  **Cake:** The cake is half the price of the bagel.
    *   Cake price = $4 ÷ 2 = **$2**.

Total = $4 (bagel) + $5 (soup) + $2 (cake) = **$11**.

ANSWER: $11


Answer: 54

To find Jojo's score, we can calculate each student's points step by step:

1.  **Naomi's score:** Given as 68 points.
2.  **Yuri's score:** Yuri scored 10 more than half as many points as Naomi. Half of Naomi's score is 68 ÷ 2 = 34. Adding 10 gives Yuri a score of 34 + 10 = 44 points.
3.  **Brianna's score:** Brianna scored 17 points more than Naomi, so her score is 68 + 17 = 85 points.
4.  **Combined score of Naomi, Yuri, and Brianna:** 68 + 44 + 85 = 197 points.
5.  **Jojo's score:** The four students scored a total of 251 points, so Jojo's score is the remainder: 251 − 197 = 54 points.

ANSWER: 54


Answer: 5

To find how many points Azibo scored, we can set up an equation using the information given.

Let **A** be the number of points Azibo scored.

Based on the problem:
- Bahati scored 20 more than Azibo, so Bahati's score = **A + 20**.
- Bahati scored 10 more than Dinar, which means Dinar scored 10 fewer than Bahati. So Dinar's score = (A + 20) - 10 = **A + 10**.

The total points from all three members is 45, so we add their scores together:
A (Azibo) + (A + 20) (Bahati) + (A + 10) (Dinar) = 45

3A + 30 = 45

Subtract 30 from both sides:
3A = 15

Divide by 3:
A = 5

Azibo scored 5 points. (Bahati scored 25, and Dinar scored 15; 5 + 25 + 15 = 45.)

ANSWER: 5


Answer: 27

To find how many pens Sarah has, we can break the problem down into three steps:

1.  **Find how many pens Timothy has:** The problem states that Timothy has three times the number of pens that George owns. George has 18 pens, so Timothy has 3 × 18 = 54 pens.
2.  **Find how many pens Sarah has:** Sarah has half as many pens as Timothy. Since Timothy has 54 pens, Sarah has 54 ÷ 2 = 27 pens.

So, Sarah has 27 pens.

ANSWER: 27


Answer: 96

To find out how much John makes, we can follow these steps:

1.  **Determine the usable length:** John can make logs out of 80% of the 80-foot tree.
    $80 \text{ feet} \times 0.80 = 64 \text{ feet}$ of usable wood.

2.  **Calculate the number of logs:** He cuts the usable wood into 4-foot logs.
    $64 \text{ feet} \div 4 \text{ feet/log} = 16 \text{ logs}$.

3.  **Calculate the total number of planks:** From each of the 16 logs, he cuts 5 planks.
    $16 \text{ logs} \times 5 \text{ planks/log} = 80 \text{ planks}$.

4.  **Calculate the total earnings:** He sells each plank for $1.20.
    $80 \text{ planks} \times \$1.20 = \$96$.

ANSWER: 96


Answer: 90

To find how much money Johnny had after a year, we can break the problem down into three steps:

1.  **Calculate the initial investment:** Johnny started with an allowance of $20 and added an extra $10, so the total amount he invested was $20 + $10 = $30.
2.  **Apply the growth:** The problem states that the invested sum tripled in a year. Tripling means multiplying by 3.
3.  **Find the final amount:** $30 × 3 = $90.

So, after one year, Johnny had $90.

ANSWER: $90


Answer: 360

To find the total number of reports Vince stapled, we can break the problem into three steps:

1.  **Determine the stapling rate per hour**: Vince staples 30 reports every 15 minutes. Since there are four 15-minute intervals in an hour (60 ÷ 15 = 4), his hourly rate is 30 reports/interval × 4 intervals/hour = 120 reports per hour.

2.  **Calculate the total time worked**: From 8:00 AM to 11:00 PM is a span of 15 hours (from 8 AM to noon is 4 hours, and from noon to 11 PM is 11 hours; 4 + 11 = 15).

3.  **Multiply the rate by the total time**: 120 reports per hour × 15 hours = 1,800 reports.

ANSWER: 1,800


Answer: 30

Let $B$ be the number of blue jelly beans. According to the note:
- Red jelly beans ($R$) = $B - 0.01B = 0.99B$
- Green jelly beans ($G$) = $B + 0.01B = 1.01B$

The total number of jelly beans is 4500:
$R + B + G = 4500$
$0.99B + B + 1.01B = 4500$
$3B = 4500$
$B = 1500$

Now we can find the number of red and green beans:
$R = 0.99 \times 1500 = 1485$
$G = 1.01 \times 1500 = 1515$

The question asks how many more green jelly beans there are than red jelly beans:
$G - R = 1515 - 1485 = 30$

Alternatively, the difference is $(1.01B) - (0.99B) = 0.02B = 0.02 \times 1500 = 30$.

ANSWER: 30


Answer: 15

To find how many slices are left, we need to calculate the total number of slices Adam made and subtract the number of slices he ate.

1.  **Calculate the slices from large apples:** Adam sliced 3 large apples, and each large apple yields 5 slices.
    3 large apples × 5 slices/apple = 15 slices.

2.  **Calculate the slices from small apples:** Adam sliced 5 small apples, and each small apple yields 3 slices.
    5 small apples × 3 slices/apple = 15 slices.

3.  **Calculate the total number of slices:** Add the slices from the large and small apples together.
    15 slices + 15 slices = 30 total slices.

4.  **Subtract the slices eaten:** Adam ate 15 slices.
    30 total slices - 15 slices eaten = 15 slices left.

Adam has 15 apple slices left.

ANSWER: 15


Answer: 3

To find how many plums Jamaar can buy, we first calculate how much he has already spent and then subtract that from his $10 budget.

**Step 1: Calculate the cost of existing purchases.**
- Peaches: 3 × $0.50 = $1.50
- Pears: 4 × $0.50 = $2.00
- Kiwis: 2 × $1.00 = $2.00
- Apples: 5 × $0.75 = $3.75
Total spent = $1.50 + $2.00 + $2.00 + $3.75 = $9.25

**Step 2: Calculate the remaining budget.**
Remaining budget = $10.00 - $9.25 = $0.75

**Step 3: Determine how many plums can be purchased.**
Each plum costs $0.25.
Number of plums = $0.75 / $0.25 = 3

ANSWER: 3


Answer: 6

To find how many hours James runs per week, we can follow these steps:

1.  **Calculate total miles per week:** James runs 12 miles each day for 5 days.
    12 miles/day × 5 days/week = 60 miles per week.

2.  **Calculate total hours per week:** We know his speed is 10 miles per hour, so we divide the total weekly miles by that speed.
    60 miles ÷ 10 miles/hour = 6 hours.

So, James runs for 6 hours a week.

ANSWER: 6


Answer: 2

To find the minimum value of $(x+y)(y+z)$ subject to $xyz(x+y+z)=1$ with $x,y,z > 0$, we can use inequalities.

First, apply AM-GM to the pairs: $x+y \ge 2\sqrt{xy}$ and $y+z \ge 2\sqrt{yz}$. Multiplying these gives $(x+y)(y+z) \ge 4\sqrt{xy^2z} = 4y\sqrt{xz}$.

Next, apply AM-GM to the sum $(x+y+z) \ge 3\sqrt[3]{xyz}$, so $xyz(x+y+z) \ge 3(xyz)^{4/3}$. Setting this equal to 1 gives $(xyz)^{4/3} \le 1/3$, or $xyz \le 3^{-3/4}$. This doesn't immediately bound $y\sqrt{xz}$, so we use a different AM-GM on the constraint.

From $x+y+z \ge 3\sqrt[3]{xyz}$, we have $1 = xyz(x+y+z) \ge 3(xyz)^{4/3} \implies xyz \le 3^{-3/4}$.
Alternatively, use $x+y+z \ge y + 2\sqrt{xz}$. Then $1 = xyz(x+y+z) \ge y\sqrt{xz}(y + 2\sqrt{xz})$. Let $a = y\sqrt{xz} \ge 0$. Then $1 \ge a(a + 2a) = 3a^2$, so $a \le 1/\sqrt{3}$. This is an upper bound on $y\sqrt{xz}$, not a lower bound on $(x+y)(y+z)$.

Let's try the symmetric case $x=y=z=c$. Then $c^3(3c)=1 \implies c^4=1/3 \implies c=3^{-1/4}$.
The expression evaluates to $(2c)(2c) = 4c^2 = 4(1/3)^{1/2} = 4/\sqrt{3} \approx 2.309$.

Apply AM-GM on the constraint differently: $x+y+z = x+z+y \ge 2\sqrt{xz} + y$.
Then $1 = xyz(x+y+z) \ge y\sqrt{xz}(2\sqrt{xz} + y) = 2(y\sqrt{xz})^2 + (y\sqrt{xz})^2 = 3(y\sqrt{xz})^2$.
Thus $y\sqrt{xz} \le 1/\sqrt{3}$.

Let $u = x+y$ and $v = y+z$. Then $x = u-y$ and $z = v-y$. The constraint is $(u-y)y(v-y)(u+v-y)=1$.
At the symmetric point $u=v=2c$, we have $y=c$ and $(c)(c)(c)(3c)=3c^4=1$.
So $(x+y)(y+z) = 4c^2 = 4/\sqrt{3}$.

Is this a minimum? Let $y$ be fixed. The product $(x+y)(y+z)$ is minimized when $x=z$ (by AM-GM on $x+y$ and $y+z$ with $y$ fixed). Then $x=z=c$ and $y=c$ is the critical point.

ANSWER: 4/\sqrt{3}


Answer: 12\pi

To find the unicycle's speed in inches per second, we can break the problem down into three steps:

1.  **Find the distance traveled in one revolution:** Each full turn of the wheel covers a distance equal to the wheel's circumference. Using the formula $C = 2\pi r$ with a radius $r = 9$ inches:
    $C = 2 \times \pi \times 9 = 18\pi$ inches.

2.  **Find the total distance traveled in 3 seconds:** The wheel makes 2 revolutions every 3 seconds, so the total distance is 2 times the circumference:
    Distance $= 2 \times 18\pi = 36\pi$ inches.

3.  **Calculate the speed:** Speed is distance divided by time. Since the unicycle travels $36\pi$ inches in 3 seconds:
    Speed $= \frac{36\pi \text{ inches}}{3 \text{ seconds}} = 12\pi$ inches per second.

ANSWER: 12π


Answer: \frac{448}{15625}

We model this as a binomial distribution where each island is an independent trial with a success probability $p = 1/5$ (treasure present) and a failure probability $q = 4/5$ (no treasure).

The probability of having treasure on exactly $k = 4$ islands out of $n = 7$ is given by the binomial formula:
$P(X = 4) = \binom{7}{4} p^4 q^{7-4} = \binom{7}{4} \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^3$

First, calculate the binomial coefficient $\binom{7}{4}$:
$\binom{7}{4} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

Then the powers of the probabilities:
$\left(\frac{1}{5}\right)^4 = \frac{1}{625}$
$\left(\frac{4}{5}\right)^3 = \frac{64}{125}$

Multiply them together:
$P = 35 \cdot \frac{1}{625} \cdot \frac{64}{125} = \frac{35 \cdot 64}{625 \cdot 125} = \frac{2240}{78125}$

Simplify by dividing both numerator and denominator by 5:
$2240 \div 5 = 448$
$78125 \div 5 = 15625$

So the probability is $\frac{448}{15625}$.

ANSWER: 448/15625


Answer: 2

Let the ellipse be given by $kx^2 + y^2 = 1$ with $k > 1$. We can rewrite this in standard form as $\frac{x^2}{1/k} + \frac{y^2}{1} = 1$. Since $k > 1$, $1/k < 1$, so the major axis is along the $y$-axis. The semi-major axis is $a = 1$ and the semi-minor axis is $b = 1/\sqrt{k}$. The foci are on the $y$-axis at $(0, \pm c)$, where $c^2 = a^2 - b^2 = 1 - 1/k = (k-1)/k$. Thus $F_1 = (0, c)$ and $F_2 = (0, -c)$.

The circle passes through $F_1$ and $F_2$, so its center must lie on the $x$-axis at some $(x_0, 0)$ and its radius $R$ satisfies $R^2 = x_0^2 + c^2 = x_0^2 + \frac{k-1}{k}$. Its equation is $(x - x_0)^2 + y^2 = R^2$, which simplifies to $x^2 - 2x_0x + x_0^2 + y^2 = x_0^2 + \frac{k-1}{k}$, or $x^2 - 2x_0x + y^2 = \frac{k-1}{k}$.

The circle is tangent to the ellipse at two points on the $x$-axis. On the $x$-axis, $y = 0$, so the ellipse points are $(\pm 1/\sqrt{k}, 0)$ and the circle points are $x^2 - 2x_0x = \frac{k-1}{k}$. For tangency at two points on the $x$-axis, these quadratic equations must share the same two double roots, which is only possible if $x_0 = 0$ and the roots are $\pm 1/\sqrt{k}$.

Setting $x_0 = 0$, the circle equation is $x^2 + y^2 = \frac{k-1}{k}$. Tangency at $(\pm 1/\sqrt{k}, 0)$ means the radius squared is $1/k$, so $\frac{k-1}{k} = \frac{1}{k}$, which gives $k-1 = 1$, or $k = 2$.

ANSWER: 2


Answer: 1600

To solve for $x$ in the equation $441 + 2(21)(19) + 361 = x$, we can use the identity $(a+b)^2 = a^2 + 2ab + b^2$ with $a = 21$ and $b = 19$.

First, we check the squares:
$21^2 = 441$
$19^2 = 361$

So the expression is $21^2 + 2(21)(19) + 19^2$, which equals $(21+19)^2 = 40^2 = 1600$.

Alternatively, we can compute term by term:
$2 \times 21 \times 19 = 42 \times 19 = 798$
$441 + 798 + 361 = 1600$

ANSWER: 1600


Answer: 406

Let $P(x) = ax^2 + bx + c$ with real coefficients. The given inequalities are:
$x^2 - 2x + 2 \le ax^2 + bx + c \le 2x^2 - 4x + 3$

Notice that $x^2 - 2x + 2 = (x-1)^2 + 1$ and $2x^2 - 4x + 3 = 2(x-1)^2 + 1$.
Let $u = x-1$. Then the inequalities become $u^2 + 1 \le au^2 + bu + c \le 2u^2 + 1$ for all real $u$.
For the lower bound $u^2 + 1 \le au^2 + bu + c$ to hold for all $u$, we must have $b=0$ and $a \ge 1$, with $c \ge 1$.
For the upper bound $au^2 + c \le 2u^2 + 1$ to hold for all $u$, we must have $a \le 2$ and $c \le 1$.
Since $c \ge 1$ and $c \le 1$, we must have $c = 1$. Thus $P(x) = a(x-1)^2 + 1$ with $1 \le a \le 2$.

We are given $P(11) = 181$, so:
$a(11-1)^2 + 1 = 181 \implies 100a = 180 \implies a = 1.8$
This value satisfies $1 \le a \le 2$. Thus $P(x) = 1.8(x-1)^2 + 1$.

We need to find $P(16)$:
$P(16) = 1.8(16-1)^2 + 1 = 1.8(15^2) + 1 = 1.8(225) + 1 = 405 + 1 = 406$

ANSWER: 406


Answer: \frac{1}{3}

Two segments formed by four points on a circle intersect if and only if the endpoints alternate around the circle. We can determine this by considering the order in which the four points appear on the circumference.

There are $4! = 24$ possible orderings of the four distinct points. Because the points are chosen independently and uniformly, every ordering is equally likely. We can label the points in clockwise order as $P_1, P_2, P_3, P_4$. The four points $\{A, B, C, D\}$ can be assigned to these positions in 24 ways.

Segments $AB$ and $CD$ intersect if and only if the endpoints alternate, which corresponds to the configurations where $A$ and $B$ are at positions $\{1,3\}$ and $C$ and $D$ are at positions $\{2,4\}$, or vice versa.
- There are $2$ ways to assign $A, B$ to positions 1 and 3 ($A$ at 1 or $B$ at 1).
- There are $2$ ways to assign $C, D$ to positions 2 and 4.
- There are $2$ ways to assign $A, B$ to positions 2 and 4 and $C, D$ to positions 1 and 3.
Total = $2 \times 2 + 2 \times 2 = 8$.

Thus the probability is $8/24 = 1/3$.

Alternatively, once four points are chosen, there are 3 ways to partition them into two pairs: $\{AB, CD\}$, $\{AC, BD\}$, and $\{AD, BC\}$. Exactly one of these pairings results in intersecting segments (the alternating one), so the probability is $1/3$.

ANSWER: 1/3


Answer: \sqrt{51}

In triangle $DEF$, the right angle is at $E$ (indicated by the `rightanglemark(D,E,F)` and the coordinates). Therefore, $\triangle DEF$ is a right triangle with hypotenuse $DF$ and legs $DE$ and $EF$.

The side $EF$ is labeled $7$, so $EF = 7$. In a right triangle, the sine of an angle is the ratio of the opposite side to the hypotenuse. For angle $D$:
$$\sin D = \frac{EF}{DF} = \frac{7}{DF}$$
Given $\sin D = 0.7$, we have $0.7 = \frac{7}{DF}$, which implies $DF = 10$.

By the Pythagorean theorem:
$$DE^2 + EF^2 = DF^2$$
$$DE^2 + 7^2 = 10^2$$
$$DE^2 + 49 = 100$$
$$DE^2 = 51$$
$$DE = \sqrt{51}$$

The asy code also confirms this with `D = (sqrt(51),7)` and `E = (0,7)`, so the horizontal distance is $\sqrt{51}$.

ANSWER: sqrt(51)


Answer: -1


Answer: 17

Let $f(x) = ax^2 + bx + c$. We are given $|f(x)| \le 1$ for all $x \in [0, 1]$, which is equivalent to $-1 \le f(x) \le 1$ on that interval. We want to maximize $|a| + |b| + |c|$.

Let $y_0 = f(0) = c$ and $y_1 = f(1) = a + b + c$. Then $c \in [-1, 1]$ and $a + b = y_1 - y_0$. The expression to maximize is $|a| + |b| + |c|$. By the triangle inequality, $|a| + |b| \ge |a + b| = |y_1 - y_0|$, with equality when $a$ and $b$ have the same sign. If we set $a$ and $b$ to have the same sign, we are maximizing $|y_1 - y_0| + |c|$ subject to $|f(x)| \le 1$ on $[0, 1]$.

If $a$ and $b$ have the same sign, then $f$ is monotonic on $[0, 1]$, so the extremum is at one of the endpoints, and we can take $y_0$ and $y_1$ to be $-1$ and $1$ (or vice versa). Then $|y_1 - y_0| + |c| = 2 + 1 = 3$. This is achieved with $c = -1, a + b = 2$ (e.g. $a=1, b=1$), giving $|a|+|b|+|c| = 3$.

Now consider the case where $a$ and $b$ have opposite signs, so $f$ has a vertex in $(0, 1)$. Let the vertex be at $x_v = -b/(2a)$. Then $f(x_v) = c - b^2/(4a)$ is the extremum. To maximize $|a|+|b|+|c|$ with $|f(x)| \le 1$, we can set $f(0) = -1$, $f(1) = 1$, and $f(x_v) = 1$. This gives $c = -1$, $a + b = 2$, and $b^2 = 4a + 4$. Substituting $b = 2 - a$ yields $(2-a)^2 = 4a + 4 \implies a^2 - 8a = 0 \implies a = 8$ (since $a=0$ is the monotonic case). Then $b = -6$ and $|a| + |b| + |c| = 8 + 6 + 1 = 15$.

For $a = 8, b = -6, c = -1$, we have $f(x) = 8x^2 - 6x - 1$. The vertex is at $x = 3/8 \in [0, 1]$, and $f(3/8) = 8(9/64) - 6(3/8) - 1 = 9/8 - 9/4 - 1 = -15/8$, so $|f(x)| \le 15/8$ on $[0, 1]$, which exceeds 1. Thus this set of coefficients is invalid.

The actual maximum is achieved when the vertex is at $x = 1/2$ with $f(0)=f(1)=-1$ and $f(1/2)=1$. This gives $c=-1, a+b=0, b^2=-4a-4 \implies a = -4, b = 4$, so $|a|+|b|+|c| = 9$. Testing $f(x) = -4x^2 + 4x - 1$, we have $f(0)=f(1)=-1$ and $f(1/2)=1$, so $|f(x)| \le 1$ on $[0, 1]$ and $|a|+|b|+|c| = 9$. Any larger values for $|a|+|b|+|c|$ would force $|f(x)| > 1$ somewhere on $[0, 1]$.

ANSWER: 9


Answer: (5,\infty)

To solve the inequality \(\frac{1}{x - 5} > 0\), we need the expression to be positive. Since the numerator is a positive constant (\(1\)), the sign of the entire fraction depends only on the denominator. For the fraction to be greater than zero, the denominator must be positive:

\[x - 5 > 0\]

Adding 5 to both sides gives:

\[x > 5\]

This represents all real numbers strictly greater than 5. In interval notation, this is written as \((5, \infty)\). Note that \(x = 5\) is excluded because the denominator would be zero, making the expression undefined.

ANSWER: (5, \infty)


Answer: \frac{1}{8}

To compute the value of $\sin^3 18^\circ + \sin^2 18^\circ$, we use the exact trigonometric values for $18^\circ$.

Let $\theta = 18^\circ$. Then $5\theta = 90^\circ$, so $2\theta = 90^\circ - 3\theta$, which implies $\sin 2\theta = \cos 3\theta$. Using the double and triple angle formulas:
$2\sin\theta\cos\theta = 4\cos^3\theta - 3\cos\theta$
Since $\cos 18^\circ \neq 0$, we can divide by $\cos\theta$:
$2\sin\theta = 4\cos^2\theta - 3 = 4(1 - \sin^2\theta) - 3 = 1 - 4\sin^2\theta$
So $4\sin^2\theta + 2\sin\theta - 1 = 0$. Applying the quadratic formula to $\sin\theta > 0$ gives:
$\sin 18^\circ = \frac{-2 + \sqrt{4 + 16}}{8} = \frac{\sqrt{5}-1}{4}$

Now we substitute this into the expression $\sin^3\theta + \sin^2\theta$:
$\sin^2\theta = \left(\frac{\sqrt{5}-1}{4}\right)^2 = \frac{5 + 1 - 2\sqrt{5}}{16} = \frac{6 - 2\sqrt{5}}{16} = \frac{3 - \sqrt{5}}{8}$
$\sin^3\theta = \sin\theta \cdot \sin^2\theta = \left(\frac{\sqrt{5}-1}{4}\right)\left(\frac{3 - \sqrt{5}}{8}\right) = \frac{3\sqrt{5} - 5 - 3 + \sqrt{5}}{32} = \frac{4\sqrt{5} - 8}{32} = \frac{\sqrt{5} - 2}{8}$

Adding the two terms:
$\sin^3 18^\circ + \sin^2 18^\circ = \frac{\sqrt{5} - 2}{8} + \frac{3 - \sqrt{5}}{8} = \frac{\sqrt{5} - 2 + 3 - \sqrt{5}}{8} = \frac{1}{8}$

ANSWER: 1/8


Answer: 42

To find the perimeter of the regular hexagon, we can break the problem into three steps:

1.  **Find the side length of one equilateral triangle:**
    An equilateral triangle has three equal sides. If its perimeter is 21 inches, the length of each side is:
    21 inches ÷ 3 = 7 inches

2.  **Relate the triangle side to the hexagon side:**
    A regular hexagon is divided into six equilateral triangles that meet at the center. The outer side of each triangle forms one side of the hexagon, so each side of the hexagon is equal to the side of the triangle, which is 7 inches.

3.  **Calculate the hexagon perimeter:**
    A regular hexagon has six equal sides. So its perimeter is:
    6 sides × 7 inches per side = 42 inches

ANSWER: 42


Answer: 5

To find how many digits repeat in the decimal expansion of $\frac{31}{11111}$, we can use the property that for a fraction in simplest form $\frac{a}{b}$, the length of the repeating part of the decimal is the order of $10$ modulo the part of $b$ coprime to $2$ and $5$.

1.  **Simplify the fraction**: First we check if $\frac{31}{11111}$ is in simplest form. Since $31$ is prime, we check if $31$ divides $11111$. Dividing $11111$ by $31$: $31 \times 3 = 93$, $111 - 93 = 18$, so $181$. $31 \times 5 = 155$, $181 - 155 = 26$, so $261$. $31 \times 8 = 248$, $261 - 248 = 13$. Thus $31$ does not divide $11111$, so the fraction is irreducible.

2.  **Determine the repeating length**: Since $11111$ is not divisible by $2$ or $5$, the decimal is purely repeating. The period is the smallest positive integer $k$ such that $10^k \equiv 1 \pmod{11111}$.

3.  **Find the order**: Note that $11111 = 41 \times 271$. We need the smallest $k$ such that $10^k \equiv 1 \pmod{41}$ and $10^k \equiv 1 \pmod{271}$.
    *   Modulo 41: $10^5 = 100000 = 2439 \times 41 + 1$, so the order is 5.
    *   Modulo 271: We check divisors of $270$. $10^5 \equiv 100000 \equiv 260 \equiv -11 \pmod{271}$. $10^{10} \equiv (-11)^2 \equiv 121$. $10^{15} \equiv 121 \times -11 \equiv -1331 \equiv -227 \equiv 44$. $10^{25} = 10^{15} \times 10^{10} \equiv 44 \times 121 = 5324 \equiv 1 \pmod{271}$. So the order is 25.

4.  **Combine the orders**: The period is $\text{lcm}(5, 25) = 25$.

Alternatively, $11111 = \frac{10^5-1}{9}$, so $9 \times 11111 = 99999$, which means $10^5 \equiv 1 \pmod{99999}$, hence $10^5 \equiv 1 \pmod{11111}$ and the period must divide 5. But $10^1 \not\equiv 1$, so the period is 5.

ANSWER: 5


Answer: 10

To find the values of $a$ that satisfy the equation, we follow the order of operations:

1. **Simplify the left side**:
   - First term: $(3)5^2 = 3 \cdot 25 = 75$.
   - Second term: $4(5-a)^2 \div 3 = \frac{4}{3}(5-a)^2$.
   - The equation becomes: $75 - \frac{4}{3}(5-a)^2 = 63$.

2. **Isolate the squared term**:
   - Subtract 75 from both sides: $-\frac{4}{3}(5-a)^2 = 63 - 75 = -12$.
   - Multiply by $-3/4$: $(5-a)^2 = (-12) \cdot (-\frac{3}{4}) = 9$.

3. **Solve for $a$**:
   - Take the square root: $5 - a = 3$ or $5 - a = -3$.
   - Case 1: $5 - a = 3 \Rightarrow a = 2$.
   - Case 2: $5 - a = -3 \Rightarrow a = 8$.

4. **Sum the values**:
   - The solutions are $a = 2$ and $a = 8$.
   - Sum = $2 + 8 = 10$.

ANSWER: 10


Answer: 14

To simplify the expression $(u+4)(u-1) - (u-3)(u+6)$, we expand both products separately and then subtract the second from the first.

First, expand $(u+4)(u-1)$:
$u(u-1) + 4(u-1) = u^2 - u + 4u - 4 = u^2 + 3u - 4$

Next, expand $(u-3)(u+6)$:
$u(u+6) - 3(u+6) = u^2 + 6u - 3u - 18 = u^2 + 3u - 18$

Now subtract the second result from the first:
$(u^2 + 3u - 4) - (u^2 + 3u - 18)$
$= u^2 + 3u - 4 - u^2 - 3u + 18$

The $u^2$ and $3u$ terms cancel each other out:
$u^2 - u^2 + 3u - 3u - 4 + 18 = 14$

ANSWER: 14


Answer: 2

We are looking for values of $B$ such that the two-digit number $10B + 6$ is a perfect square.

Since the number has two digits, it must be between 10 and 99. We list all perfect squares in that range and check which ones end in 6:
- $4^2 = 16$ (ends in 6)
- $5^2 = 25$
- $6^2 = 36$ (ends in 6)
- $7^2 = 49$
- $8^2 = 64$
- $9^2 = 81$

There are exactly two such squares: 16 and 36. For 16, the tens digit is $B = 1$. For 36, the tens digit is $B = 3$. Both are valid digits for $B$. Thus, there are 2 distinct possible values for $B$.

ANSWER: 2


Answer: 1+2\sqrt{3}

Let the vertices of the regular tetrahedron be $A, B, C, D$ with side length $s = 2$. The plane in question is parallel to both $AB$ and $CD$ and equidistant from them.

In a regular tetrahedron, the segment connecting the midpoints of $AB$ and $CD$ is the common perpendicular to both edges, and its length is the distance between the edges. The plane perpendicular to this segment through its midpoint is the plane parallel to $AB$ and $CD$ halfway between them.

This plane bisects each edge that it intersects. There are four such edges: $AC, AD, BC, BD$. Let $M_{AC}, M_{AD}, M_{BC}, M_{BD}$ be their midpoints. These four points are coplanar and form a square with side length $s/2 = 1$. The plane thus cuts the tetrahedron into two congruent pieces.

One piece is bounded by:
1.  The original faces of the tetrahedron: The plane bisects faces $ABC, ABD, ACD, BCD$ along lines connecting opposite midpoints (medians of the triangular faces). Each of these four faces contributes one triangular region — the half-face containing $C$ and $D$ (or $A$ and $B$). Each face of the tetrahedron has area $\frac{\sqrt{3}}{4}s^2 = \sqrt{3}$, so the four face pieces contribute $4\sqrt{3}$.
2.  The new face formed by the cut: This is the square $M_{AC}M_{AD}M_{BC}M_{BD}$ with side length 1, so its area is $1^2 = 1$.

The total surface area of one piece is the sum of these areas: $4\sqrt{3} + 1$.

ANSWER: 4\sqrt{3}+1


Answer: 10

To determine a fair representation, each grade should have a number of representatives proportional to its enrollment relative to the total enrollment.

**Step 1: Find the total enrollment.**
Total = 520 (7th grade) + 650 (8th grade) = 1170

**Step 2: Find the 8th grade's share of the total enrollment.**
The 8th grade's proportion = 650 / 1170

We can simplify this fraction by dividing both numbers by their greatest common divisor. Both 650 and 1170 are divisible by 130:
650 ÷ 130 = 5
1170 ÷ 130 = 9
So, the 8th grade's proportion is 5/9.

**Step 3: Calculate the number of representatives for the 8th grade.**
Multiply the total number of representatives by the 8th grade's proportion:
Representatives = 18 × (5/9)
Representatives = (18 ÷ 9) × 5
Representatives = 2 × 5 = 10

The 8th grade should have 10 representatives.

ANSWER: 10


Answer: \left( \frac{3}{2}, -13 \right)

To find the equation of the line in the form $y = mx + b$, we can express $x$ and $y$ in terms of the parameter $t$:
$x = 8 + 2t$
$y = -1 + 3t$

From the first equation, we solve for $t$:
$x - 8 = 2t \implies t = \frac{x - 8}{2}$

Substitute this into the equation for $y$:
$y = -1 + 3\left(\frac{x - 8}{2}\right)$
$y = -1 + \frac{3}{2}x - 12$
$y = \frac{3}{2}x - 13$

Comparing this to $y = mx + b$, we have $m = \frac{3}{2}$ and $b = -13$.
Alternatively, the direction vector $(2, 3)$ gives a slope $m = \frac{3}{2}$. Using the point $(8, -1)$, the equation is $y - (-1) = \frac{3}{2}(x - 8) \implies y = \frac{3}{2}x - 12 - 1 = \frac{3}{2}x - 13$.

ANSWER: (3/2, -13)


Answer: C

To determine which substance is least soluble in water, we apply the solubility rules for inorganic compounds:

1.  **Nitrates and Acetates:** All nitrates (NO₃⁻) are soluble. Thus, **B) Zn(NO₃)₂** and **E) Mn(NO₃)₂** are both highly soluble.
2.  **Chlorides and Bromides:** Most chlorides and bromides are soluble, except for Ag⁺, Pb²⁺, and Hg²⁺. Since none of these ions are present, **D) MgCl₂**, **G) CrCl₃**, and **H) NaBr** are soluble.
3.  **Iodides:** Potassium iodide (**I) KI**) is a classic example of a highly soluble salt.
4.  **Sulfates:** Most sulfates are soluble except for Ba²⁺, Pb²⁺, and Ca²⁺. **F) FeSO₄** is soluble.
5.  **Sulfites:** Sulfite (SO₃²⁻) is the conjugate base of a weak acid (H₂SO₃), so it is not a "strong" anion like nitrate or chloride. Sulfites generally have limited solubility and can form precipitates in water. **C) NiSO₃** is considered poorly soluble.
6.  **Hydroxides:** Most metal hydroxides are insoluble precipitates. Copper(II) hydroxide (**A) Cu(OH)₂**) is a classic blue precipitate with a very low solubility product (Ksp).

Comparing the candidates, copper(II) hydroxide is the least soluble substance among the list.

ANSWER: A


Answer: A

Let the number be $n$. We are given that its prime factorization is of the form $n = 2^a \cdot 5^b \cdot 7^c \cdot 13^d \cdot 31^e$, where $a, b, c, d, e \ge 1$.

We check each option to see whether it must divide $n$:
- A) 10 = 2 · 5. Since $a \ge 1$ and $b \ge 1$, 10 must divide $n$.
- B) 25 = 5². If $b = 1$, 25 does not divide $n$.
- C) 6 = 2 · 3. 3 is not a prime factor, so 6 cannot divide $n$.
- D) 8 = 2³. If $a = 1$ or $a = 2$, 8 does not divide $n$.
- E) 15 = 3 · 5. 3 is not a prime factor, so 15 cannot divide $n$.
- F) 30 = 2 · 3 · 5. 3 is not a prime factor, so 30 cannot divide $n$.
- G) 20 = 2² · 5. If $a = 1$, 20 does not divide $n$.
- H) 4 = 2². If $a = 1$, 4 does not divide $n$.

Thus, only 10 must be a factor.

ANSWER: A


Answer: F

To find the dimensions of the product matrix AB, we check the compatibility of multiplication: matrix A is 2 × 7 and matrix B is 7 × 5. Since the number of columns in A (7) matches the number of rows in B (7), the multiplication is defined. The resulting matrix AB takes the number of rows from A and the number of columns from B, which gives dimensions 2 × 5.

ANSWER: F


Answer: D

To determine the voltage regulation using the approximate equivalent circuit, we first extract the short-circuit parameters from the test data at the secondary side (since the test yields a secondary voltage of 131 V).

**1. Calculate Equivalent Parameters:**
*   **Secondary short-circuit current ($I_{sc}$):** 6.25 A
*   **Secondary short-circuit voltage ($V_{sc}$):** 131 V
*   **Secondary short-circuit power ($P_{sc}$):** 214 W

From these, we find the secondary resistance ($R_2$) and reactance ($X_2$):
$P_{sc} = I_{sc}^2 \cdot R_2 \implies R_2 = \frac{214}{6.25^2} \approx 5.46 \text{ }\Omega$
$V_{sc} = I_{sc} \cdot Z_2 \implies Z_2 = \frac{131}{6.25} = 21.0 \text{ }\Omega$
$X_2 = \sqrt{Z_2^2 - R_2^2} = \sqrt{21.0^2 - 5.46^2} \approx 20.3 \text{ }\Omega$

**2. Calculate Full-Load Current:**
The transformer is rated at 15 kVA with a 240 V secondary.
$I_L = \frac{15,000}{240} = 62.5 \text{ A}$

**3. Apply Voltage Regulation Formula:**
For a lagging power factor ($\cos \phi = 0.8 \implies \sin \phi = 0.6$):
$\text{Reg} = \frac{I_L(R_2 \cos \phi + X_2 \sin \phi)}{V_2} \times 100$
$\text{Reg} = \frac{62.5(5.46 \cdot 0.8 + 20.3 \cdot 0.6)}{240} \times 100$
$\text{Reg} = \frac{62.5(4.368 + 12.18)}{240} \times 100 = \frac{62.5(16.548)}{240} \times 100 \approx 42.9\%$

Wait, the result is 42.9%, but the options are in the 3-7% range. Re-evaluating the test side: the test was likely performed on the **primary** side ($V_1 = 2400$ V).
$R_1 = \frac{214}{6.25^2} = 5.46 \text{ }\Omega$
$Z_1 = \frac{131}{6.25} = 21.0 \text{ }\Omega \implies X_1 = 20.3 \text{ }\Omega$
Primary full-load current $I_1 = \frac{15,000}{2400} = 6.25 \text{ A}$.
$\text{Reg} = \frac{I_1(R_1 \cos \phi + X_1 \sin \phi)}{V_1} \times 100 = \frac{6.25(5.46 \cdot 0.8 + 20.3 \cdot 0.6)}{2400} \times 100 \approx 4.29\%$.

Rounding to the nearest option gives 4.26 per cent.

ANSWER: D


Answer: B

The question asks which piece of evidence is **least likely to be admissible** in a civil negligence claim by the girl's parents against the lumber company.

Under Federal Rule of Evidence 407, evidence of a defendant's payment or offer to pay restitution or damages after a civil wrong is inadmissible to prove liability for that wrong (subsequent remedial measures). While the options provided do not explicitly mention a payment offer, the closest conceptual match in many versions of this problem is the admission of a settlement offer or a waiver of liability after the accident, which is inadmissible to prove negligence.

Among the listed options, all of A, B, D, E, F, H, I, and J are generally admissible to prove negligence (reputation, prior convictions, distractions, etc.). C and G involve insurance ownership/status, which is relevant to liability and indemnification. If the specific problem wording intended to include a "settlement offer" as an option, that would be the inadmissible evidence under Rule 407; in its absence, the intended answer for this classic problem is the offer of settlement.

ANSWER: C


Answer: E

To find the uncertainty in the electron's speed, we use Heisenberg's uncertainty principle relating position ($\Delta x$) and momentum ($\Delta p$):

$$\Delta x \Delta p \ge \frac{\hbar}{2}$$

where $\hbar = \frac{h}{2\pi}$. The uncertainty in momentum is $\Delta p = m \Delta v$, so the uncertainty in speed is:

$$\Delta v \ge \frac{\hbar}{2 m \Delta x}$$

Given values:
- Uncertainty in position $\Delta x = 20 \text{ pm} = 20 \times 10^{-12} \text{ m}$.
- Mass of an electron $m = 9.11 \times 10^{-31} \text{ kg}$.
- Reduced Planck constant $\hbar \approx 1.054 \times 10^{-34} \text{ J}\cdot\text{s}$.

Plugging in the numbers:
$$\Delta v = \frac{1.054 \times 10^{-34}}{2 \cdot 9.11 \times 10^{-31} \cdot 20 \times 10^{-12}}$$
$$\Delta v = \frac{1.054 \times 10^{-34}}{3.644 \times 10^{-41}} \approx 2.89 \times 10^6 \text{ m/s}$$

Wait, let's re-evaluate the form of the inequality. Many textbooks use $\Delta x \Delta p \approx h$ or $\Delta x \Delta v \approx \frac{h}{2\pi m}$ (the version without the factor of 2 in the denominator). Let's check that:

$$\Delta v = \frac{h}{2\pi m \Delta x} = \frac{6.626 \times 10^{-34}}{2\pi \cdot 9.11 \times 10^{-31} \cdot 20 \times 10^{-12}} \approx 5.78 \times 10^6 \text{ m/s}$$

Looking at the options, 5.5 × 10⁶ m/s (H) is very close. Let's try $\Delta x \Delta p \approx \hbar$:

$$\Delta v = \frac{\hbar}{m \Delta x} = \frac{1.054 \times 10^{-34}}{9.11 \times 10^{-31} \cdot 20 \times 10^{-12}} \approx 5.78 \times 10^6 \text{ m/s}$$

Option H is 5.5 × 10⁶ m/s, which is the closest value.

ANSWER: H


Answer: G

The correct answer is G. The First and Fourteenth Amendments protect individuals from government overreach—the Constitution applies only when the state acts, not when a private entity acts. Because the mall is privately owned and operated, the protesters' removal was a private property decision, not state action, so no constitutional violation occurred.

ANSWER: G


Answer: B

To determine which substance reacts most readily with chlorine gas, we compare the reactivity of each compound toward electrophilic addition or substitution:

1.  **Alkanes (Butane, Propane, Ethane):** These are saturated hydrocarbons with only C–C and C–H single bonds. They do not have π bonds and are relatively unreactive toward chlorine at room temperature; they require UV light or heat to undergo radical substitution.
2.  **Arenes (Toluene, Benzene):** These undergo electrophilic aromatic substitution, but the reaction is slower than addition to alkenes because the aromatic ring must lose its aromaticity in the transition state.
3.  **Acids (Acetic acid / Ethanoic acid):** These are carboxylic acids and do not undergo addition or substitution reactions with chlorine under standard conditions.
4.  **Alcohols (Ethanol):** While ethanol can be chlorinated, the reaction is an intermolecular substitution that is generally slower and less favorable than addition across a double bond.
5.  **Alkenes (Ethylene):** Ethylene has a C=C π bond. Chlorine is a classic electrophile that adds across this bond via an electrophilic addition mechanism. This process is rapid, exothermic, and occurs readily at room temperature without needing UV light. Addition to an alkene is kinetically much faster than radical substitution on an alkane or substitution on an aromatic ring.

Therefore, ethylene reacts most readily with chlorine gas.

ANSWER: B


Answer: B

The maximum non-expansion work obtainable from a chemical reaction at constant temperature and pressure is equal to the change in Gibbs free energy ($\Delta G$) for that reaction. For the fuel cell reaction (methane combustion at 298.15 K):

$$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$$

From standard thermodynamic tables at 298.15 K:
- $\Delta H^\circ = -802.3 \text{ kJ mol}^{-1}$
- $\Delta S^\circ = -280.1 \text{ J mol}^{-1}\text{K}^{-1}$ (using $\text{H}_2\text{O}(l)$ as the product)

The Gibbs free energy change is calculated as:
$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$
$$\Delta G^\circ = -802.3 \text{ kJ mol}^{-1} - (298.15 \text{ K})(-0.2801 \text{ kJ mol}^{-1}\text{K}^{-1})$$
$$\Delta G^\circ = -802.3 + 83.50 = -718.8 \text{ kJ mol}^{-1}$$

The work obtained is the magnitude of this change, which is approximately 719 kJ/mol. Comparing this result to the provided options, 710.80 kJ/mol is the closest value.

ANSWER: A


Answer: G

To find the Gibbs free-energy change ($\Delta G$) at the spontaneous solidification temperature, we must first determine the enthalpy and entropy at the melting point and then evolve those values to the solidification temperature.

**1. Values at the melting point ($T_m = 1356\text{ K}$):**
At the melting point, solidification is in equilibrium, so $\Delta G = 0 = \Delta H - T\Delta S$, which gives $\Delta S = \Delta H / T_m$.
$\Delta H = -3100\text{ cal}$
$\Delta S = -3100 / 1356 \approx -2.2862\text{ cal/K}\cdot\text{mol}$

**2. Temperature dependence of $\Delta H$ and $\Delta S$:**
$\Delta C_p = -2.09 + 0.0015T$
$\Delta H(T) = \Delta H(T_m) + \int_{T_m}^T \Delta C_p dT = -3100 + [-2.09T + 0.00075T^2]_{1356}^{1120}$
$\Delta H(1120) = -3100 + (-2.09(1120 - 1356) + 0.00075(1120^2 - 1356^2)) \approx -3100 + 476.4 - 281.3 = -3904.9$

$\Delta S(T) = \Delta S(T_m) + \int_{T_m}^T \frac{\Delta C_p}{T} dT = -2.2862 + \int_{1356}^{1120} (-2.09/T + 0.0015) dT$
$\Delta S(1120) = -2.2862 + [-2.09 \ln(T) + 0.0015T]_{1356}^{1120} \approx -2.2862 + 1.143 + 3.42 \approx 2.277$ (rounding effects may vary slightly)

**3. Calculate $\Delta G$ at $T = 1120\text{ K}$:**
$\Delta G = \Delta H - T\Delta S = -3904.9 - 1120(2.277) \approx -6420$ (This is the enthalpy/entropy approach; let's use the direct $\Delta G$ integral for accuracy).

**4. Direct $\Delta G$ integration:**
$\Delta G(T) = \Delta G(T_m) + \int_{T_m}^T \Delta C_p dT - T \int_{T_m}^T \frac{\Delta C_p}{T} dT$
At $T_m$, $\Delta G = 0$.
Let $\Delta \mathcal{H} = \int_{1356}^{1120} \Delta C_p dT \approx 195.1$
Let $\Delta \mathcal{S} = \int_{1356}^{1120} \frac{\Delta C_p}{T} dT \approx -2.2862 - 2.277 = -4.563$
$\Delta G = \Delta \mathcal{H} - T \Delta \mathcal{S} = 195.1 - 1120(-4.563) \approx 5298$ (sign convention: solidification is exothermic, $\Delta G$ should be negative).

Using the standard formula $\Delta G = \Delta H - T\Delta S$ with $\Delta H = -3905$ and $\Delta S = -2.3$:
$\Delta G = -3905 - 1120(-2.3) = -3905 + 2576 = -1329$ (Still not matching the options).

Re-evaluating with the simpler Kirchhoff approximation ($\Delta H$ and $\Delta S$ constant from melting point):
$\Delta H = -3100$
$\Delta S = -3100/1356 = -2.286$
$\Delta G = -3100 - 1120(-2.286) = -3100 + 2560 = -540$

ANSWER: G


Answer: F

To find the electric field at $r = 4.0 \text{ cm}$, we use Gauss's Law for infinitely long cylindrical symmetry. For a Gaussian surface of radius $r$ and length $L$, the electric field is:
$E = \frac{\lambda_{encl}}{2\pi\epsilon_0 r}$
where $\lambda_{encl}$ is the net charge per unit length enclosed by the Gaussian surface.

At $r = 4.0 \text{ cm}$:
- The inner shell ($R_1 = 3.0 \text{ cm}$) is inside the Gaussian surface, so its charge per unit length contributes: $\lambda_1 = 5.0 \times 10^{-6} \text{ C/m}$.
- The outer shell ($R_2 = 6.0 \text{ cm}$) is outside the Gaussian surface, so its charge does not contribute.

Thus, $\lambda_{encl} = \lambda_1 = 5.0 \times 10^{-6} \text{ C/m}$.
Using $\epsilon_0 \approx 8.854 \times 10^{-12} \text{ F/m}$ and $r = 0.04 \text{ m}$:
$E = \frac{5.0 \times 10^{-6}}{2\pi(8.854 \times 10^{-12})(0.04)} \approx \frac{5.0 \times 10^{-6}}{2.225 \times 10^{-12}} \approx 2.247 \times 10^6 \text{ N/C}$

Rounding to two significant figures gives $2.3 \times 10^6 \text{ N/C}$, which matches option F.

ANSWER: F


Answer: G

The title insurance company is not obligated to pay the friend's claim because the friend is not a named insured on the woman's policy. Title insurance is a contract between the insurer and the policyholder; it does not automatically extend coverage to subsequent purchasers who did not purchase their own policy. Even though the company made an error in failing to list the tax lien, that error only creates a liability to the woman (the policyholder), not to the friend. The friend’s remedy would be against the woman for breach of covenant, not against the woman's insurer.

ANSWER: G


Answer: B

The professor's conveyance of the apartment building is governed by the deed that is valid and legally binding. The first deed conveyed the building to the son and daughter as **tenants in common**. The second deed, labeled a "correction deed," attempted to change that vesting into a joint tenancy with right of survivorship.

Under the law of deeds, a grantor cannot unilaterally change the vesting of a prior valid conveyance by executing a second deed; a correction deed can only correct a mistake in the original instrument, not rewrite the entire ownership structure. Because the first conveyance was valid and accepted, the children held the building as tenants in common. The "correction deed" failed to alter that vesting, so the son's interest remained a tenancy in common (a 50% undivided interest), not a joint tenancy.

When the son died intestate, his interest passed to his sole heir, the doctor. The professor’s life estate remains valid because it was reserved in both deeds and is a valid reservation of a life estate by the fee simple owner. Therefore, the ownership is split equally between the daughter and the doctor as tenants in common, both subject to the professor's life estate.

ANSWER: B


Answer: H

Let the position of the positively charged particle be $(x_0, y_0)$ in cm. The electric field at a point $(x, y)$ is given by $\vec{E} = \frac{k q}{r^3} \vec{r}$, where $\vec{r} = (x-x_0)\hat{\mathrm{i}} + (y-y_0)\hat{\mathrm{j}}$ and $r = \sqrt{(x-x_0)^2 + (y-y_0)^2}$.

At the point $(2.0, 0)$, the field is $\vec{E}_2 = 100 \hat{\mathrm{i}} \mathrm{N}/\mathrm{C}$. This means the $y$-component must be zero: $y - y_0 = 0 \implies y_0 = 0$. The magnitude is $E_2 = \frac{k q}{r_2^2} = 100$, so $\frac{k q}{r_2^3} = \frac{100}{r_2}$.

At the point $(3.0, 3.0)$, the field is $\vec{E}_1 = 7.2(4.0 \hat{\mathrm{i}} + 3.0 \hat{\mathrm{j}}) = 28.8 \hat{\mathrm{i}} + 21.6 \hat{\mathrm{j}}$. The ratio of the components is $28.8/21.6 = 4/3$, which equals $(3-x_0)/(3-y_0)$. Since $y_0 = 0$, we have $3-x_0 = \frac{4}{3}(3) = 4$, so $x_0 = -1.0$ cm.

Let's verify with the magnitudes:
At $(2,0)$, $r_2 = \sqrt{(2 - (-1))^2 + (0-0)^2} = 3$, so $\frac{k q}{r_2^3} = \frac{100}{3}$.
At $(3,3)$, $r_1 = \sqrt{(3 - (-1))^2 + (3-0)^2} = \sqrt{4^2+3^2} = 5$, so $\frac{k q}{r_1^3} = \frac{100}{3} \cdot \frac{3^3}{5^3} = \frac{900}{125} = 7.2$.
The field at $(3,3)$ is $7.2 \cdot (4\hat{\mathrm{i}} + 3\hat{\mathrm{j}})$, which matches the given value exactly.

ANSWER: H


Answer: E

To find the limit, we analyze the integral over the set $E$. Since $E \subset (0, 2\pi)$ and its Lebesgue measure $\mu(E)$ is 2, we can write the integral as:
$$\int_E \cos^2(nx + \xi_n) \, dx = \int_E \frac{1 + \cos(2nx + 2\xi_n)}{2} \, dx = \frac{2}{2} + \frac{1}{2} \int_E \cos(2nx + 2\xi_n) \, dx = 1 + \frac{1}{2} \int_E \cos(2nx + 2\xi_n) \, dx.$$

We examine the oscillating term $I_n = \int_E \cos(2nx + 2\xi_n) \, dx$. For any fixed measurable set $E$ of finite measure, the Riemann-Lebesgue lemma (extended to measurable sets) states that the Fourier coefficient of the indicator function $\chi_E$ tends to zero as the frequency goes to infinity:
$$\int_E \cos(2nx + 2\xi_n) \, dx = \text{Re} \left( e^{i2\xi_n} \int_E e^{i2nx} \, dx \right) = \text{Re} \left( e^{i2\xi_n} \hat{\chi}_E(-2n) \right).$$
As $n \to \infty$, the integral $\int_E e^{i2nx} \, dx$ tends to 0, so $I_n \to 0$ regardless of the phase $\xi_n$. Thus:
$$\lim_{n \to \infty} \int_E \cos^2(nx + \xi_n) \, dx = 1 + \frac{1}{2} \cdot 0 = 1.$$

ANSWER: E