Explanation:
Okay, so I need to find the coordinates of point P on the line y = -x + 6 such that P is equidistant from points A(10, -10) and O(0, 0). Hmm, let me think about how to approach this.

First, I remember that if a point is equidistant from two other points, it lies on the perpendicular bisector of the segment connecting those two points. Maybe I can use that concept here. But wait, the point P isn't just any point equidistant from A and O—it has to lie on the line y = -x + 6. So perhaps I need to find the intersection of the perpendicular bisector of AO with the given line?

Let me verify. The perpendicular bisector of AO should consist of all points equidistant from A and O. Therefore, the point P we are looking for must lie on both the perpendicular bisector of AO and the line y = -x + 6. That makes sense. So I need to find the equation of the perpendicular bisector of AO first.

Alright, step one: find the midpoint of AO. The midpoint formula is ((x₁ + x₂)/2, (y₁ + y₂)/2). So for points A(10, -10) and O(0, 0), the midpoint M would be ((10 + 0)/2, (-10 + 0)/2) = (5, -5). Okay, so the midpoint is (5, -5).

Next, I need the slope of AO to find the slope of the perpendicular bisector. The slope of AO is (y₂ - y₁)/(x₂ - x₁) = (0 - (-10))/(0 - 10) = (10)/(-10) = -1. Therefore, the slope of AO is -1. The perpendicular bisector will have a slope that is the negative reciprocal of that. The negative reciprocal of -1 is 1. So the slope of the perpendicular bisector is 1.

Now, with the midpoint (5, -5) and the slope 1, I can write the equation of the perpendicular bisector. Using point-slope form: y - y₁ = m(x - x₁). Substituting in the midpoint coordinates and the slope: y - (-5) = 1*(x - 5), which simplifies to y + 5 = x - 5. Then, subtracting 5 from both sides: y = x - 10. So the equation of the perpendicular bisector is y = x - 10.

Now, the point P lies on both this perpendicular bisector and the line y = -x + 6. Therefore, to find P, I need to solve these two equations simultaneously.

So set y = x - 10 equal to y = -x + 6:

x - 10 = -x + 6

Adding x to both sides: 2x - 10 = 6

Adding 10 to both sides: 2x = 16

Dividing by 2: x = 8

Then, substitute x = 8 into one of the equations to find y. Let's use y = -x + 6:

y = -8 + 6 = -2

Therefore, the coordinates of P would be (8, -2). Wait, let me check if this is correct.

First, verify that (8, -2) lies on the line y = -x + 6. Plugging x = 8: y = -8 + 6 = -2, which matches. Good.

Now check if it's equidistant from A(10, -10) and O(0, 0). Let's compute PA and PO.

PA: distance between (8, -2) and (10, -10). Using the distance formula:

√[(10 - 8)^2 + (-10 - (-2))^2] = √[(2)^2 + (-8)^2] = √[4 + 64] = √68.

PO: distance between (8, -2) and (0, 0):

√[(0 - 8)^2 + (0 - (-2))^2] = √[(-8)^2 + (2)^2] = √[64 + 4] = √68.

Yes, both distances are √68. So PA = PO. Therefore, (8, -2) is indeed equidistant from A and O and lies on the given line. Therefore, the coordinates of P are (8, -2).

Wait, but let me think again. I used the perpendicular bisector approach. Is there another way to approach this? Maybe by setting up equations directly?

Alternatively, suppose P is (x, y) on the line y = -x + 6. Then PA = PO.

So PA² = (x - 10)^2 + (y + 10)^2

PO² = (x - 0)^2 + (y - 0)^2 = x² + y²

Set them equal:

(x - 10)^2 + (y + 10)^2 = x² + y²

Expanding the left side:

(x² - 20x + 100) + (y² + 20y + 100) = x² + y²

Combine like terms:

x² - 20x + 100 + y² + 20y + 100 = x² + y²

Subtract x² + y² from both sides:

-20x + 100 + 20y + 100 = 0

Simplify:

-20x + 20y + 200 = 0

Divide both sides by 20:

- x + y + 10 = 0

Which simplifies to y = x - 10. Wait, that's the same equation as the perpendicular bisector. So this confirms that the set of points equidistant from A and O is y = x - 10. Therefore, solving y = x -10 and y = -x + 6 gives x = 8, y = -2. So same result. Therefore, seems correct.

But maybe I should check if there's another intersection point? Wait, two lines usually intersect at one point unless they are parallel. Since the slopes are 1 and -1, which are not equal, so they must intersect at exactly one point. So there is only one solution, which is (8, -2). Therefore, answer is (8, -2).

Alternatively, maybe I can parametrize the line y = -x + 6. Let me try that approach.

Let me let x = t, then y = -t + 6. Therefore, P is (t, -t + 6). Then compute PA² and PO² and set them equal.

PA² = (t - 10)^2 + (-t + 6 + 10)^2 = (t -10)^2 + (-t +16)^2

PO² = (t - 0)^2 + (-t + 6 - 0)^2 = t² + (-t + 6)^2

Set PA² = PO²:

(t -10)^2 + (-t +16)^2 = t² + (-t +6)^2

Compute each term:

(t -10)^2 = t² -20t +100

(-t +16)^2 = t² -32t +256

Left side: t² -20t +100 + t² -32t +256 = 2t² -52t + 356

Right side: t² + (-t +6)^2 = t² + t² -12t +36 = 2t² -12t +36

Set equal:

2t² -52t + 356 = 2t² -12t +36

Subtract 2t² from both sides:

-52t +356 = -12t +36

Bring all terms to left side:

-52t +356 +12t -36 = 0 => -40t + 320 = 0 => -40t = -320 => t = 8

So t = 8, therefore x = 8, y = -8 +6 = -2. So same answer again. So P is (8, -2).

Therefore, regardless of method, it's (8, -2). So that's the answer.

Wait, but just to make sure I didn't make any calculation mistakes. Let me check again.

First, midpoint of AO: ( (10 + 0)/2, (-10 + 0)/2 ) = (5, -5). Correct.

Slope of AO: ( -10 - 0 ) / (10 - 0 ) = -10/10 = -1. Correct. Perpendicular slope is 1. Correct. Equation of the perpendicular bisector: y - (-5) = 1*(x - 5) => y + 5 = x -5 => y = x -10. Correct. Then intersection with y = -x +6. Setting x -10 = -x +6 => 2x = 16 => x =8. y = -8 +6 = -2. Correct.

Another approach with coordinates gives the same result. So answer seems correct.

Alternatively, maybe visualize the problem. The line y = -x +6 is a straight line with slope -1, crossing the y-axis at 6 and x-axis at 6. The point O is at origin, point A is at (10, -10). The point P is somewhere on that line such that it's equidistant to O and A. Since the perpendicular bisector of AO is y = x -10, which is a line with slope 1. So intersecting with y = -x +6. The intersection point is (8, -2).

Alternatively, if I plot mentally, the line y = -x +6 is going from upper left to lower right. The perpendicular bisector y = x -10 is going from lower left to upper right. They intersect at (8, -2). Let me check if that point is on the line y = -x +6. At x=8, y = -8 +6 = -2. Correct. And on the bisector y = x -10: 8 -10 = -2. Correct.

Another way to check: if I move from the midpoint (5, -5) along the perpendicular bisector direction (slope 1), so moving in direction (1,1). How far? The point P is (8, -2). The vector from midpoint to P is (8 -5, -2 - (-5)) = (3, 3), which is in direction (1,1), so that's along the bisector. Similarly, moving from midpoint in direction (1,1) by 3 units. So that seems consistent.

Alternatively, distance from midpoint to P: sqrt(3² +3²) = sqrt(18). But maybe that's not necessary.

Alternatively, maybe there's a formula for distance? But no, since I already checked the distances PA and PO are both sqrt(68). Which is approx 8.246. Let me check with coordinates:

PA: from (8, -2) to (10, -10). The difference in x is 2, difference in y is -8. The distance squared is 4 + 64 = 68. Correct. PO: from (8, -2) to (0,0): difference x 8, y 2. Squared distance 64 + 4 = 68. Correct.

So all checks out. Therefore, the coordinates of P are (8, -2).

I think that's solid. I can't see any mistakes here. So the answer should be (8, -2).

**Final Answer**
The coordinates of point $ P $ are \boxed{(8, -2)}.

To find the coordinates of point $ P $ on the line $ y = -x + 6 $ such that $ P $ is equidistant from points $ A(10, -10) $ and $ O(0, 0) $, we follow these key geometric and algebraic steps:

---

### **Step 1: Understand the Problem**
We are looking for a point $ P(a, b) $ that satisfies two conditions:
1. It lies on the line $ y = -x + 6 $, so $ b = -a + 6 $
2. It is equidistant from the points $ A(10, -10) $ and $ O(0, 0) $, meaning $ PA = PO $

---

### **Step 2: Use the Perpendicular Bisector of Segment $ AO $**
Since $ P $ must lie equidistant from $ A $ and $ O $, it must lie on the **perpendicular bisector** of segment $ AO $.

- **Midpoint of $ AO $**:

$$
M = \left( \frac{10 + 0}{2}, \frac{-10 + 0}{2} \right) = (5, -5)
$$

- **Slope of $ AO $**:

$$
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-10 - 0}{10 - 0} = \frac{-10}{10} = -1
$$

- So, the slope of the **perpendicular bisector** is the negative reciprocal of $ -1 $, which is $ 1 $

- Using the point-slope form with the midpoint $ M(5, -5) $:

$$
y - (-5) = 1(x - 5) \Rightarrow y + 5 = x - 5 \Rightarrow y = x - 10
$$

This is the **equation of the perpendicular bisector** of $ AO $.

---

### **Step 3: Find the Intersection of the Two Lines**
Point $ P $ lies on both:
- The line $ y = -x + 6 $, and
- The perpendicular bisector $ y = x - 10 $

Set the equations equal and solve for $ x $:

$$
x - 10 = -x + 6 \Rightarrow 2x = 16 \Rightarrow x = 8
$$

Substitute back to find $ y $:

$$
y = -x + 6 = -8 + 6 = -2
$$

---

### **Step 4: Verify the Result**
We now verify that $ P = (8, -2) $ is:
- On the line: $ y = -x + 6 $ → $ -2 = -8 + 6 $ ✅
- Equidistant from $ A(10, -10) $ and $ O(0, 0) $

Compute the distances:
- $ PA^2 = (8 - 10)^2 + (-2 - (-10))^2 = 4 + 64 = 68 $
- $ PO^2 = (8 - 0)^2 + (-2 - 0)^2 = 64 + 4 = 68 $

Since $ PA = PO $, the point is equidistant from both points. ✅

---

### **Final Answer**

$$
\boxed{(8, -2)}
$$

[Assistant]
Explanation:
Okay, so I need to find the coordinates of point P on the line y = -x + 6 such that P is equidistant from points A(10, -10) and O(0, 0). Hmm, let me think about how to approach this.

First, I remember that if a point is equidistant from two other points, it lies on the perpendicular bisector of the segment connecting those two points. Maybe I can use that concept here. But wait, the point P isn't just any point equidistant from A and O—it has to lie on the line y = -x + 6. So perhaps I need to find the intersection of the perpendicular bisector of AO with the given line?

Let me verify. The perpendicular bisector of AO should consist of all points equidistant from A and O. Therefore, the point P we are looking for must lie on both the perpendicular bisector of AO and the line y = -x + 6. That makes sense. So I need to find the equation of the perpendicular bisector of AO first.

Alright, step one: find the midpoint of AO. The midpoint formula is ((x₁ + x₂)/2, (y₁ + y₂)/2). So for points A(10, -10) and O(0, 0), the midpoint M would be ((10 + 0)/2, (-10 + 0)/2) = (5, -5). Okay, so the midpoint is (5, -5).

Next, I need the slope of AO to find the slope of the perpendicular bisector. The slope of AO is (y₂ - y₁)/(x₂ - x₁) = (0 - (-10))/(0 - 10) = (10)/(-10) = -1. Therefore, the slope of AO is -1. The perpendicular bisector will have a slope that is the negative reciprocal of that. The negative reciprocal of -1 is 1. So the slope of the perpendicular bisector is 1.

Now, with the midpoint (5, -5) and the slope 1, I can write the equation of the perpendicular bisector. Using point-slope form: y - y₁ = m(x - x₁). Substituting in the midpoint coordinates and the slope: y - (-5) = 1*(x - 5), which simplifies to y + 5 = x - 5. Then, subtracting 5 from both sides: y = x - 10. So the equation of the perpendicular bisector is y = x - 10.

Now, the point P lies on both this perpendicular bisector and the line y = -x + 6. Therefore, to find P, I need to solve these two equations simultaneously.

So set y = x - 10 equal to y = -x + 6:

x - 10 = -x + 6

Adding x to both sides: 2x - 10 = 6

Adding 10 to both sides: 2x = 16

Dividing by 2: x = 8

Then, substitute x = 8 into one of the equations to find y. Let's use y = -x + 6:

y = -8 + 6 = -2

Therefore, the coordinates of P would be (8, -2). Wait, let me check if this is correct.

First, verify that (8, -2) lies on the line y = -x + 6. Plugging x = 8: y = -8 + 6 = -2, which matches. Good.

Now check if it's equidistant from A(10, -10) and O(0, 0). Let's compute PA and PO.

PA: distance between (8, -2) and (10, -10). Using the distance formula:

√[(10 - 8)^2 + (-10 - (-2))^2] = √[(2)^2 + (-8)^2] = √[4 + 64] = √68.

PO: distance between (8, -2) and (0, 0):

√[(0 - 8)^2 + (0 - (-2))^2] = √[(-8)^2 + (2)^2] = √[64 + 4] = √68.

Yes, both distances are √68. So PA = PO. Therefore, (8, -2) is indeed equidistant from A and O and lies on the given line. Therefore, the coordinates of P are (8, -2).

Wait, but let me think again. I used the perpendicular bisector approach. Is there another way to approach this? Maybe by setting up equations directly?

Alternatively, suppose P is (x, y) on the line y = -x + 6. Then PA = PO.

So PA² = (x - 10)^2 + (y + 10)^2

PO² = (x - 0)^2 + (y - 0)^2 = x² + y²

Set them equal:

(x - 10)^2 + (y + 10)^2 = x² + y²

Expanding the left side:

(x² - 20x + 100) + (y² + 20y + 100) = x² + y²

Combine like terms:

x² - 20x + 100 + y² + 20y + 100 = x² + y²

Subtract x² + y² from both sides:

-20x + 100 + 20y + 100 = 0

Simplify:

-20x + 20y + 200 = 0

Divide both sides by 20:

- x + y + 10 = 0

Which simplifies to y = x - 10. Wait, that's the same equation as the perpendicular bisector. So this confirms that the set of points equidistant from A and O is y = x - 10. Therefore, solving y = x -10 and y = -x + 6 gives x = 8, y = -2. So same result. Therefore, seems correct.

But maybe I should check if there's another intersection point? Wait, two lines usually intersect at one point unless they are parallel. Since the slopes are 1 and -1, which are not equal, so they must intersect at exactly one point. So there is only one solution, which is (8, -2). Therefore, answer is (8, -2).

Alternatively, maybe I can parametrize the line y = -x + 6. Let me try that approach.

Let me let x = t, then y = -t + 6. Therefore, P is (t, -t + 6). Then compute PA² and PO² and set them equal.

PA² = (t - 10)^2 + (-t + 6 + 10)^2 = (t -10)^2 + (-t +16)^2

PO² = (t - 0)^2 + (-t + 6 - 0)^2 = t² + (-t + 6)^2

Set PA² = PO²:

(t -10)^2 + (-t +16)^2 = t² + (-t +6)^2

Compute each term:

(t -10)^2 = t² -20t +100

(-t +16)^2 = t² -32t +256

Left side: t² -20t +100 + t² -32t +256 = 2t² -52t + 356

Right side: t² + (-t +6)^2 = t² + t² -12t +36 = 2t² -12t +36

Set equal:

2t² -52t + 356 = 2t² -12t +36

Subtract 2t² from both sides:

-52t +356 = -12t +36

Bring all terms to left side:

-52t +356 +12t -36 = 0 => -40t + 320 = 0 => -40t = -320 => t = 8

So t = 8, therefore x = 8, y = -8 +6 = -2. So same answer again. So P is (8, -2).

Therefore, regardless of method, it's (8, -2). So that's the answer.

Wait, but just to make sure I didn't make any calculation mistakes. Let me check again.

First, midpoint of AO: ( (10 + 0)/2, (-10 + 0)/2 ) = (5, -5). Correct.

Slope of AO: ( -10 - 0 ) / (10 - 0 ) = -10/10 = -1. Correct. Perpendicular slope is 1. Correct. Equation of the perpendicular bisector: y - (-5) = 1*(x - 5) => y + 5 = x -5 => y = x -10. Correct. Then intersection with y = -x +6. Setting x -10 = -x +6 => 2x = 16 => x =8. y = -8 +6 = -2. Correct.

Another approach with coordinates gives the same result. So answer seems correct.

Alternatively, maybe visualize the problem. The line y = -x +6 is a straight line with slope -1, crossing the y-axis at 6 and x-axis at 6. The point O is at origin, point A is at (10, -10). The point P is somewhere on that line such that it's equidistant to O and A. Since the perpendicular bisector of AO is y = x -10, which is a line with slope 1. So intersecting with y = -x +6. The intersection point is (8, -2).

Alternatively, if I plot mentally, the line y = -x +6 is going from upper left to lower right. The perpendicular bisector y = x -10 is going from lower left to upper right. They intersect at (8, -2). Let me check if that point is on the line y = -x +6. At x=8, y = -8 +6 = -2. Correct. And on the bisector y = x -10: 8 -10 = -2. Correct.

Another way to check: if I move from the midpoint (5, -5) along the perpendicular bisector direction (slope 1), so moving in direction (1,1). How far? The point P is (8, -2). The vector from midpoint to P is (8 -5, -2 - (-5)) = (3, 3), which is in direction (1,1), so that's along the bisector. Similarly, moving from midpoint in direction (1,1) by 3 units. So that seems consistent.

Alternatively, distance from midpoint to P: sqrt(3² +3²) = sqrt(18). But maybe that's not necessary.

Alternatively, maybe there's a formula for distance? But no, since I already checked the distances PA and PO are both sqrt(68). Which is approx 8.246. Let me check with coordinates:

PA: from (8, -2) to (10, -10). The difference in x is 2, difference in y is -8. The distance squared is 4 + 64 = 68. Correct. PO: from (8, -2) to (0,0): difference x 8, y 2. Squared distance 64 + 4 = 68. Correct.

So all checks out. Therefore, the coordinates of P are (8, -2).

I think that's solid. I can't see any mistakes here. So the answer should be (8, -2).

**Final Answer**
The coordinates of point $ P $ are \boxed{(8, -2)}.

To find the coordinates of point $ P $ on the line $ y = -x + 6 $ such that $ P $ is equidistant from points $ A(10, -10) $ and $ O(0, 0) $, we follow these key geometric and algebraic steps:

---

### **Step 1: Understand the Problem**
We are looking for a point $ P(a, b) $ that satisfies two conditions:
1. It lies on the line $ y = -x + 6 $, so $ b = -a + 6 $
2. It is equidistant from the points $ A(10, -10) $ and $ O(0, 0) $, meaning $ PA = PO $

---

### **Step 2: Use the Perpendicular Bisector of Segment $ AO $**
Since $ P $ must lie equidistant from $ A $ and $ O $, it must lie on the **perpendicular bisector** of segment $ AO $.

- **Midpoint of $ AO $**:

$$
M = \left( \frac{10 + 0}{2}, \frac{-10 + 0}{2} \right) = (5, -5)
$$

- **Slope of $ AO $**:

$$
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-10 - 0}{10 - 0} = \frac{-10}{10} = -1
$$

- So, the slope of the **perpendicular bisector** is the negative reciprocal of $ -1 $, which is $ 1 $

- Using the point-slope form with the midpoint $ M(5, -5) $:

$$
y - (-5) = 1(x - 5) \Rightarrow y + 5 = x - 5 \Rightarrow y = x - 10
$$

This is the **equation of the perpendicular bisector** of $ AO $.

---

### **Step 3: Find the Intersection of the Two Lines**
Point $ P $ lies on both:
- The line $ y = -x + 6 $, and
- The perpendicular bisector $ y = x - 10 $

Set the equations equal and solve for $ x $:

$$
x - 10 = -x + 6 \Rightarrow 2x = 16 \Rightarrow x = 8
$$

Substitute back to find $ y $:

$$
y = -x + 6 = -8 + 6 = -2
$$

---

### **Step 4: Verify the Result**
We now verify that $ P = (8, -2) $ is:
- On the line: $ y = -x + 6 $ → $ -2 = -8 + 6 $ ✅
- Equidistant from $ A(10, -10) $ and $ O(0, 0) $

Compute the distances:
- $ PA^2 = (8 - 10)^2 + (-2 - (-10))^2 = 4 + 64 = 68 $
- $ PO^2 = (8 - 0)^2 + (-2 - 0)^2 = 64 + 4 = 68 $

Since $ PA = PO $, the point is equidistant from both points. ✅

---

### **Final Answer**

$$
\boxed{(8, -2)}
$$

Solution:
If $P$ is equidistant from $A$ and $O$, it must lie on the perpendicular bisector of $AO$. Since $A$ has coordinates $(10,-10)$ and $O$ has coordinates $(0,0)$, $AO$ has slope $\frac{-10-0}{10-0}=-1$.  The perpendicular bisector of $AO$ must have slope $-\frac{1}{-1}=1$, and must also pass through the midpoint of $AO$, which is $(5,-5)$. Therefore, the perpendicular bisector has equation $y-(-5)=x-5$ or $y=x-10$.

$P$ is the point of intersection of the lines $y=x-10$ and the line $y=-x+6$.  Setting these equations equal and solving for $x$ yields $-x+6=x-10 \Rightarrow x=8$.  It follows that $y=-8+6=-2$ and $P=(x,y)=\boxed{(8,-2)}$.


Explanation:
Okay, let me try to figure out this problem. So, the question is asking for which grade the number of male bus riders is closest to 135% of the number of female bus riders. There are four grades: 9th, 10th, 11th, and 12th. For each grade, we have percentages of males and females who ride the bus. The survey includes 300 males and 300 females in each grade.

First, I need to understand what exactly the question is asking. It wants the grade where the number of male bus riders is closest to 135% of the female bus riders. So, for each grade, I need to calculate the actual number of male riders and the actual number of female riders, then check how close the male number is to 135% of the female number.

Let me break it down step by step.

Starting with the 9th grade. The percentage of male bus riders is 41.1%, and for females, it's 39.4%. Since there are 300 males and 300 females surveyed in each grade, I should calculate the actual number of male and female riders.

For males in 9th grade: 41.1% of 300. To calculate that, I can convert the percentage to decimal by dividing by 100, so 41.1% is 0.411. Then multiply by 300. Let me compute that: 0.411 * 300. Hmm, 0.4 * 300 is 120, and 0.011 * 300 is 3.3, so total is 120 + 3.3 = 123.3. Since we can't have a fraction of a person, but the question probably expects exact numbers, maybe we can keep the decimal for calculation purposes and then check the difference.

Similarly, for females in 9th grade: 39.4% of 300. Converting to decimal: 0.394. Multiply by 300: 0.394 * 300. Let me calculate that. 0.4 * 300 is 120, but since it's 0.394, which is 0.4 - 0.006. So 120 - (0.006 * 300) = 120 - 1.8 = 118.2.

So, male riders: 123.3, female riders: 118.2. Now, the question is, how does 123.3 compare to 135% of 118.2? Let's compute 135% of female riders.

135% is 1.35 times the female number. So, 1.35 * 118.2. Let me compute that. First, 1 * 118.2 is 118.2, 0.35 * 118.2. Let me compute 0.3 * 118.2 = 35.46, and 0.05 * 118.2 = 5.91. Adding those: 35.46 + 5.91 = 41.37. So total is 118.2 + 41.37 = 159.57.

So, 135% of female riders is 159.57. But the actual number of male riders is 123.3. The difference between 123.3 and 159.57 is 159.57 - 123.3 = 36.27. So, the male riders are 36.27 less than 135% of female riders.

But maybe I need to check if I did that correctly. Wait, the question says "closest to 135% of the number of its female bus riders." So, is there a way that the male riders are closest to 1.35 times the female riders? So, perhaps for each grade, compute 1.35 times female riders and see how close male riders are to that. Then find which grade has the smallest difference.

So, let me do that for each grade. Maybe I should make a table or list for each grade.

Starting with 9th grade:

Male riders: 123.3

Female riders: 118.2

1.35 * female = 159.57

Difference between male and 1.35*female: |123.3 - 159.57| = 36.27

Now moving to 10th grade:

Males: 34.4% of 300.

So 0.344 * 300. Let me calculate that. 0.3 * 300 = 90, 0.04 * 300 = 12, 0.004 * 300 = 1.2. Adding those up: 90 + 12 = 102, +1.2 = 103.2. So male riders: 103.2

Females: 33.1% of 300. Which is 0.331 * 300. Let me compute that. 0.3 * 300 = 90, 0.03 * 300 = 9, 0.001 * 300 = 0.3. So 90 + 9 + 0.3 = 99.3. So female riders: 99.3

Compute 1.35 * female: 1.35 * 99.3. Let me calculate that.

First, 1 * 99.3 = 99.3, 0.35 * 99.3. Let's compute 0.3 * 99.3 = 29.79, 0.05 * 99.3 = 4.965. Adding those gives 29.79 + 4.965 = 34.755. Total 1.35 * female is 99.3 + 34.755 = 134.055.

So, male riders is 103.2, and 1.35*female is approx 134.055. The difference is |103.2 - 134.055| = 30.855.

So, difference here is about 30.86, which is less than the 9th grade's difference of 36.27. So 10th grade has a smaller difference.

Now moving to 11th grade:

Males: 20.6% of 300. Let me compute that.

0.206 * 300. 0.2 * 300 is 60, 0.006 * 300 = 1.8. So 60 + 1.8 = 61.8. So male riders: 61.8

Females: 13.8% of 300. 0.138 * 300. Let me compute that. 0.1 * 300 = 30, 0.03 * 300 = 9, 0.008*300=2.4. So total is 30 + 9 + 2.4 = 41.4. So female riders: 41.4

Compute 1.35 * female: 1.35 * 41.4. Let me calculate:

First, 1 * 41.4 = 41.4

0.35 * 41.4: 0.3*41.4 = 12.42, 0.05*41.4 = 2.07. Adding those gives 12.42 + 2.07 = 14.49. So total is 41.4 + 14.49 = 55.89

Now, male riders is 61.8. The difference between 61.8 and 55.89 is |61.8 - 55.89| = 5.91.

That's a much smaller difference compared to previous grades. So 11th grade has a difference of about 5.91.

Now, checking 12th grade:

Males: 11.6% of 300. 0.116 * 300. Let's compute. 0.1*300=30, 0.01*300=3, 0.006*300=1.8. So total is 30+3+1.8=34.8. So male riders: 34.8

Females: 8.6% of 300. 0.086 * 300. Let me calculate that. 0.08*300=24, 0.006*300=1.8. So total is 24 + 1.8 = 25.8. Female riders: 25.8

Compute 1.35 * female: 1.35 * 25.8. Let me compute that.

First, 1 * 25.8 = 25.8.

0.35 * 25.8: 0.3*25.8=7.74, 0.05*25.8=1.29. Sum is 7.74 + 1.29 = 9.03. So total is 25.8 + 9.03 = 34.83.

So, male riders is 34.8, and 1.35*female is 34.83. The difference is |34.8 - 34.83| = 0.03. That's a very small difference.

So, comparing all the differences:

9th grade: 36.27

10th grade: 30.86

11th grade: 5.91

12th grade: 0.03

So, the smallest difference is for 12th grade. Therefore, the answer should be 12th grade? Wait, but the question says "closest to 135% of the number of its female bus riders." So, the difference between male riders and 1.35*female riders is smallest for 12th grade. So, is that the answer?

Wait, but let me confirm my calculations again for 12th grade. Because the difference is only 0.03, which is really small. Let me check that again.

For 12th grade:

Male riders: 11.6% of 300. Let me check 11.6% is 0.116. 0.116 * 300 = 34.8. That's correct.

Female riders: 8.6% of 300. 0.086 * 300 = 25.8. Correct.

Then 1.35 * female riders: 1.35 * 25.8. Let me compute that again.

Alternatively, 25.8 * 1.35. Let me do 25.8 * 1.35. Maybe in another way:

First, 25.8 * 1 = 25.8

25.8 * 0.3 = 7.74

25.8 * 0.05 = 1.29

So adding up: 25.8 + 7.74 + 1.29 = 25.8 + 9.03 = 34.83. So that's correct. So male riders is 34.8, which is 34.8 vs 34.83. The difference is 0.03.

So that's extremely close. So according to that, 12th grade is the closest. But wait, the problem says "the number of its male bus riders closest to 135% of the number of its female bus riders." So, that would mean that 135% of female riders is 34.83, and male riders is 34.8, which is just 0.03 less. So that's the closest.

Wait, but maybe I made an error here. Because if you think about the percentages, the male percentage is 11.6% and female is 8.6%. So, 11.6% is higher than 8.6%, but 135% of female is 1.35 * 8.6% = 11.51%? Wait, wait, hold on. Wait, maybe there is another way to think about it. Maybe I can calculate what percentage the male riders are of the female riders, and see how close it is to 135%.

Wait, but the question is specifically about the number of male riders being closest to 135% of female riders. So, for example, if I take the actual numbers, which are absolute numbers, not percentages. So, even though the male percentage is 11.6% and female is 8.6%, the actual numbers are 34.8 vs 25.8.

So, 135% of female is 1.35 * 25.8 = 34.83, which is very close to the actual male riders of 34.8. So that's correct.

But let me check if I did the calculation for 11th grade again, because I thought 11th grade difference was 5.91. Let me check again:

For 11th grade:

Male riders: 20.6% of 300 = 61.8

Female riders: 13.8% of 300 = 41.4

1.35 * female riders: 1.35 * 41.4. Let me compute that again.

Alternatively, 41.4 * 1.35. Let me think of 41.4 * 1.35.

Well, 41.4 * 1.35. Let me compute 41.4 * 1.35.

First, note that 1.35 is 27/20. So 41.4 * 27 / 20. Let me compute 41.4 * 27 first.

But maybe decimal multiplication is easier.

41.4 * 1 = 41.4

41.4 * 0.3 = 12.42

41.4 * 0.05 = 2.07

Adding those: 41.4 + 12.42 + 2.07 = 55.89. Which matches what I had before. So 1.35 * female is 55.89. Male is 61.8. The difference is 61.8 - 55.89 = 5.91. Correct.

So, 11th grade difference is 5.91, 12th grade is 0.03, so 12th grade is much closer.

But then why does the problem say "closest to 135% of the number of its female bus riders"? Because 12th grade's male riders are almost exactly 135% of female riders.

But let me check if there is a mistake here. Because 135% of female riders would be 1.35 * 25.8 = 34.83, and male riders is 34.8. So the difference is 0.03. So that's 34.8 vs 34.83, so almost the same.

But maybe the problem is expecting integer numbers? Because you can't have a fraction of a person. Wait, but the question says "the number of its male bus riders closest to 135% of the number of its female bus riders." So, the actual numbers are 34.8 and 25.8. But in reality, the numbers should be integers, since you can't have a fraction of a person.

Wait, but the problem says that there were 300 males and 300 females surveyed. So, when they say 300 males and 300 females in each grade, that must mean that the percentages are exact? Wait, but the percentages are given with one decimal place. So, for example, 11.6% of 300 males. Let me check if that is an exact number.

11.6% of 300. Well, 10% is 30, 1% is 3, so 11.6% is 11*3 + 0.6*3 = 33 + 1.8 = 34.8. So, that is exactly 34.8. But you can't have 0.8 of a person, but in the context of the problem, maybe they are using percentages that result in decimal numbers for calculation purposes, and we can treat them as exact for the purpose of calculating the difference.

Alternatively, maybe the survey had exact numbers? For example, if they had 34.8 male riders, but in reality, they had 35 or 34. However, since the problem gives us percentages with decimal points, perhaps we are intended to use them as exact, even if they result in decimal numbers for the actual counts. So, in that case, proceeding with decimals is okay.

Therefore, according to the calculations, 12th grade has the closest difference, with only 0.03 difference. But let me check again the problem statement to make sure I didn't misread anything.

Wait, the problem says "the number of its male bus riders closest to 135% of the number of its female bus riders." So, for the 12th grade, male riders is 34.8, which is 34.8 compared to 135% of female riders (which is 34.83). So, they are extremely close.

But let me check the other grades again to make sure I didn't miscalculate.

For 9th grade:

Male: 41.1% of 300 = 123.3

Female: 39.4% of 300 = 118.2

135% of female = 1.35 * 118.2 = 159.57. Difference: 159.57 - 123.3 = 36.27. Correct.

For 10th grade:

Male: 34.4% of 300 = 103.2

Female: 33.1% of 300 = 99.3

135% of female = 1.35 * 99.3 = 134.055. Difference: 134.055 - 103.2 = 30.855. Correct.

For 11th grade:

Male: 20.6% of 300 = 61.8

Female: 13.8% of 300 = 41.4

1.35 * 41.4 = 55.89. Difference: 61.8 - 55.89 = 5.91. Correct.

For 12th grade:

Male: 11.6% of 300 = 34.8

Female: 8.6% of 300 = 25.8

1.35 * 25.8 = 34.83. Difference: 34.8 - 34.83 = -0.03, absolute value 0.03. Correct.

So, the differences are:

9th: 36.27

10th: 30.86

11th: 5.91

12th: 0.03

So, the smallest difference is 12th grade. So, the answer should be 12th grade.

But wait, the answer seems counterintuitive. Because 12th grade has very low percentages for both males and females. But maybe it's correct. Let me think again.

If the number of male riders is 34.8 and 135% of female is 34.83, which is really close. So, the answer is 12th grade.

But let me check once again for calculation errors. For example, when calculating 1.35 * 25.8.

Alternatively, 25.8 * 1.35. Let me do this multiplication step by step:

First, 25.8 * 1 = 25.8

25.8 * 0.3 = 7.74

25.8 * 0.05 = 1.29

Adding up: 25.8 + 7.74 = 33.54; 33.54 + 1.29 = 34.83. Correct.

So, 34.83 vs 34.8. The difference is 0.03. That's correct.

So, according to this, the answer is 12th grade.

But maybe the problem expects us to round the numbers? Wait, for instance, if we consider that the actual number of riders must be whole numbers, so for example, the number of male riders is 34.8, which would be approximately 35, and female riders is 25.8, approx 26. Then, 135% of 26 is 1.35 * 26 = 35.1. Which is very close to 35. So, the difference is 0.1.

But in that case, with rounding, 12th grade still has the closest.

Alternatively, maybe I need to check if the male number is closest to 135% of female number, not the absolute difference, but maybe some other measure. But no, the question says "closest to," so the absolute difference is the correct measure.

Alternatively, maybe I misread the question? Let me check again.

The question says: "For which grade is the number of its male bus riders closest to 135% of the number of its female bus riders?"

So, that means, compute for each grade, male = 135% of female? Or how close is male to 135% of female?

Yes, so for each grade, we calculate 135% of female riders and see how close the male riders are.

Yes, so the difference is as computed.

So, with the 0.03 difference, 12th grade is the answer.

But I want to check if maybe the question is in percentages. Wait, the percentages given are for the whole survey, but the question is about absolute numbers.

But the percentages are given per grade. For example, in 12th grade, male percentage is 11.6%, so that's 11.6% of 300 males, which is 34.8. Similarly for females.

So, maybe the answer is 12th grade.

But this seems very close. Maybe the answer is 11th grade? Because I recall that in some cases, people might misinterpret the question. Let me check.

Wait, let me think again. Maybe I need to check for which grade the ratio of male to female riders is closest to 1.35.

Wait, that's a different approach. Because 135% of female is male. So, male / female = 1.35. So, maybe instead of calculating 1.35 * female and comparing to male, we can compute the ratio male / female and see how close it is to 1.35.

Let me try that approach for each grade.

For 9th grade:

male / female = 123.3 / 118.2 ≈ 1.043. Which is about 104.3%, which is way less than 135%.

For 10th grade:

male / female = 103.2 / 99.3 ≈ 1.039. Which is about 103.9%, still less than 135%.

For 11th grade:

male / female = 61.8 / 41.4 ≈ 1.493. Which is about 149.3%, which is higher than 135%.

For 12th grade:

male / female = 34.8 / 25.8 ≈ 1.349. Which is about 134.9%, which is very close to 135%.

So, here, the ratio is 1.349, which is 134.9%, which is extremely close to 135%. Therefore, 12th grade is the answer.

So, this confirms the previous result. Therefore, the answer is 12th grade.

Therefore, even if I approach it through ratios, it's the same result.

But maybe I made a mistake with the decimal points? Let me check the 12th grade ratio again:

Male: 34.8, female: 25.8. 34.8 divided by 25.8. Let me compute that.

Dividing 34.8 by 25.8.

First, note that 25.8 * 1.35 = 34.83, as before.

So, 34.8 / 25.8 ≈ 1.349.

Which is 1.349, which is 134.9%.

So, that's just 0.1% less than 135%, which is very close.

So, in terms of percentage difference, this is the closest.

Therefore, the answer is 12th grade.

But let me check once again with exact fractions. Maybe there's some rounding that I didn't consider.

Let me compute male / female for 12th grade:

Male riders: 11.6% of 300.

But 11.6% is equal to 11.6 / 100. So, 11.6% of 300 = 11.6/100 * 300 = 11.6 * 3 = 34.8.

Female riders: 8.6% of 300 = 8.6/100 * 300 = 8.6 * 3 = 25.8.

So, male / female = 34.8 / 25.8.

Multiply numerator and denominator by 10 to eliminate decimals: 348 / 258.

Divide numerator and denominator by 6: 348 ÷ 6 = 58, 258 ÷ 6 = 43. So, 58 / 43 ≈ 1.3488... which is approximately 1.3488, which is about 134.88%, so 134.88% is the ratio, which is 0.12% less than 135%.

Therefore, this is the closest to 135%.

Therefore, the answer is 12th grade.

But just to be thorough, let me check the other grades' ratios to see how far they are from 135%.

For 11th grade, male / female was approximately 1.493, which is 149.3%, which is 14.3% higher than 135%.

For 10th grade, 1.039, which is 3.9% less than 135%.

For 9th grade, 1.043, which is 4.3% less than 135%.

So, the 12th grade is the closest, with only 0.12% difference.

So, this seems correct.

Therefore, the answer is the 12th grade.

But wait, looking back at the problem, the answer choices are 9th, 10th, 11th, 12th.

But let me check again if there is any possible mistake. I wonder if the question is asking for the number of male riders to be closest to 135% of the female riders, meaning that if you take the female riders and multiply by 1.35, then find which grade's male riders are closest to that number.

Which is exactly what I did. And for 12th grade, male riders = 34.8, and 1.35 * female riders = 34.83, so difference is 0.03, which is the smallest.

So, all signs point to 12th grade being the correct answer.

But maybe the problem is in the way the question is phrased? For example, maybe "closest to 135% of the number of its female bus riders"—so, maybe they want the male riders to be closest to 135% of the female riders, meaning that the value of male riders minus 1.35*female is closest to zero.

Which is exactly what I computed.

Alternatively, maybe the problem is in the interpretation of "percent of bus riders"—if the percentages are of the total students, but the problem says "percent of bus riders in a survey of Central H.S. students; 300 males and 300 females in each grade were surveyed."

So, the percentages are of the males and females surveyed. So, for example, for males, the 41.1% is the percentage of males who ride the bus, so that's the percentage of the 300 males.

Therefore, my calculations are correct.

Therefore, I think the correct answer is the 12th grade.

But just to make sure, let me check once more with exact numbers:

For 12th grade:

Male riders: 34.8

Female riders: 25.8

Compute 1.35 * female:

1.35 * 25.8 = ?

Let me calculate 25.8 * 1.35:

Breakdown:

First, multiply 25.8 by 1.35:

Compute 25.8 * 1 + 25.8 * 0.3 + 25.8 * 0.05

= 25.8 + 7.74 + 1.29 = 25.8 + 7.74 = 33.54 + 1.29 = 34.83.

Therefore, 1.35 * female = 34.83.

Male is 34.8.

Difference is 34.83 - 34.8 = 0.03.

So, that's a very small difference.

Therefore, 12th grade is the answer.

But I wonder why this seems so close. Maybe because the percentages are set up that way.

Alternatively, maybe the question has a typo, but assuming the numbers are correct, 12th grade is the answer.

Therefore, the answer is 12th grade.

**Final Answer**
\boxed{12}

To solve this problem, we are asked to determine which grade has the **number of its male bus riders that is closest to 135% of the number of its female bus riders**. The survey includes 300 males and 300 females in each grade. For each grade, we are provided with percentage estimates for the male and female riders.

---

### Step 1: Convert Percentages to Actual Counts

For each grade, the actual number of male and female bus riders is calculated by multiplying the given percentage by 300 (the number of students surveyed in each group).

Let’s compute the values for each grade:

#### **9th Grade**
- **Males:** $ 41.1\% \times 300 = 0.411 \times 300 = 123.3 $
- **Females:** $ 39.4\% \times 300 = 0.394 \times 300 = 118.2 $
- $ 135\% \text{ of females} = 1.35 \times 118.2 = 159.57 $
- **Difference:** $ |123.3 - 159.57| = 36.27 $

#### **10th Grade**
- **Males:** $ 34.4\% \times 300 = 0.344 \times 300 = 103.2 $
- **Females:** $ 33.1\% \times 300 = 0.331 \times 300 = 99.3 $
- $ 135\% \text{ of females} = 1.35 \times 99.3 = 134.055 $
- **Difference:** $ |103.2 - 134.055| = 30.855 $

#### **11th Grade**
- **Males:** $ 20.6\% \times 300 = 0.206 \times 300 = 61.8 $
- **Females:** $ 13.8\% \times 300 = 0.138 \times 300 = 41.4 $
- $ 135\% \text{ of females} = 1.35 \times 41.4 = 55.89 $
- **Difference:** $ |61.8 - 55.89| = 5.91 $

#### **12th Grade**
- **Males:** $ 11.6\% \times 300 = 0.116 \times 300 = 34.8 $
- **Females:** $ 8.6\% \times 300 = 0.086 \times 300 = 25.8 $
- $ 135\% \text{ of females} = 1.35 \times 25.8 = 34.83 $
- **Difference:** $ |34.8 - 34.83| = 0.03 $

---

### Step 2: Compare the Differences

From the above calculations, we can see the **absolute differences** between the number of male riders and 135% of female riders for each grade:

- **9th Grade:** 36.27
- **10th Grade:** 30.86
- **11th Grade:** 5.91
- **12th Grade:** 0.03

---

### Step 3: Identify the Closest Match

The smallest absolute difference is for the **12th grade**, with a difference of just **0.03**, which is extremely close. This indicates that the number of **male bus riders in 12th grade is nearly identical to 135% of the number of female riders**, making it the **closest match**.

---

### Final Answer

$$
\boxed{12}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. So, the question is asking for which grade the number of male bus riders is closest to 135% of the number of female bus riders. There are four grades: 9th, 10th, 11th, and 12th. For each grade, we have percentages of males and females who ride the bus. The survey includes 300 males and 300 females in each grade.

First, I need to understand what exactly the question is asking. It wants the grade where the number of male bus riders is closest to 135% of the female bus riders. So, for each grade, I need to calculate the actual number of male riders and the actual number of female riders, then check how close the male number is to 135% of the female number.

Let me break it down step by step.

Starting with the 9th grade. The percentage of male bus riders is 41.1%, and for females, it's 39.4%. Since there are 300 males and 300 females surveyed in each grade, I should calculate the actual number of male and female riders.

For males in 9th grade: 41.1% of 300. To calculate that, I can convert the percentage to decimal by dividing by 100, so 41.1% is 0.411. Then multiply by 300. Let me compute that: 0.411 * 300. Hmm, 0.4 * 300 is 120, and 0.011 * 300 is 3.3, so total is 120 + 3.3 = 123.3. Since we can't have a fraction of a person, but the question probably expects exact numbers, maybe we can keep the decimal for calculation purposes and then check the difference.

Similarly, for females in 9th grade: 39.4% of 300. Converting to decimal: 0.394. Multiply by 300: 0.394 * 300. Let me calculate that. 0.4 * 300 is 120, but since it's 0.394, which is 0.4 - 0.006. So 120 - (0.006 * 300) = 120 - 1.8 = 118.2.

So, male riders: 123.3, female riders: 118.2. Now, the question is, how does 123.3 compare to 135% of 118.2? Let's compute 135% of female riders.

135% is 1.35 times the female number. So, 1.35 * 118.2. Let me compute that. First, 1 * 118.2 is 118.2, 0.35 * 118.2. Let me compute 0.3 * 118.2 = 35.46, and 0.05 * 118.2 = 5.91. Adding those: 35.46 + 5.91 = 41.37. So total is 118.2 + 41.37 = 159.57.

So, 135% of female riders is 159.57. But the actual number of male riders is 123.3. The difference between 123.3 and 159.57 is 159.57 - 123.3 = 36.27. So, the male riders are 36.27 less than 135% of female riders.

But maybe I need to check if I did that correctly. Wait, the question says "closest to 135% of the number of its female bus riders." So, is there a way that the male riders are closest to 1.35 times the female riders? So, perhaps for each grade, compute 1.35 times female riders and see how close male riders are to that. Then find which grade has the smallest difference.

So, let me do that for each grade. Maybe I should make a table or list for each grade.

Starting with 9th grade:

Male riders: 123.3

Female riders: 118.2

1.35 * female = 159.57

Difference between male and 1.35*female: |123.3 - 159.57| = 36.27

Now moving to 10th grade:

Males: 34.4% of 300.

So 0.344 * 300. Let me calculate that. 0.3 * 300 = 90, 0.04 * 300 = 12, 0.004 * 300 = 1.2. Adding those up: 90 + 12 = 102, +1.2 = 103.2. So male riders: 103.2

Females: 33.1% of 300. Which is 0.331 * 300. Let me compute that. 0.3 * 300 = 90, 0.03 * 300 = 9, 0.001 * 300 = 0.3. So 90 + 9 + 0.3 = 99.3. So female riders: 99.3

Compute 1.35 * female: 1.35 * 99.3. Let me calculate that.

First, 1 * 99.3 = 99.3, 0.35 * 99.3. Let's compute 0.3 * 99.3 = 29.79, 0.05 * 99.3 = 4.965. Adding those gives 29.79 + 4.965 = 34.755. Total 1.35 * female is 99.3 + 34.755 = 134.055.

So, male riders is 103.2, and 1.35*female is approx 134.055. The difference is |103.2 - 134.055| = 30.855.

So, difference here is about 30.86, which is less than the 9th grade's difference of 36.27. So 10th grade has a smaller difference.

Now moving to 11th grade:

Males: 20.6% of 300. Let me compute that.

0.206 * 300. 0.2 * 300 is 60, 0.006 * 300 = 1.8. So 60 + 1.8 = 61.8. So male riders: 61.8

Females: 13.8% of 300. 0.138 * 300. Let me compute that. 0.1 * 300 = 30, 0.03 * 300 = 9, 0.008*300=2.4. So total is 30 + 9 + 2.4 = 41.4. So female riders: 41.4

Compute 1.35 * female: 1.35 * 41.4. Let me calculate:

First, 1 * 41.4 = 41.4

0.35 * 41.4: 0.3*41.4 = 12.42, 0.05*41.4 = 2.07. Adding those gives 12.42 + 2.07 = 14.49. So total is 41.4 + 14.49 = 55.89

Now, male riders is 61.8. The difference between 61.8 and 55.89 is |61.8 - 55.89| = 5.91.

That's a much smaller difference compared to previous grades. So 11th grade has a difference of about 5.91.

Now, checking 12th grade:

Males: 11.6% of 300. 0.116 * 300. Let's compute. 0.1*300=30, 0.01*300=3, 0.006*300=1.8. So total is 30+3+1.8=34.8. So male riders: 34.8

Females: 8.6% of 300. 0.086 * 300. Let me calculate that. 0.08*300=24, 0.006*300=1.8. So total is 24 + 1.8 = 25.8. Female riders: 25.8

Compute 1.35 * female: 1.35 * 25.8. Let me compute that.

First, 1 * 25.8 = 25.8.

0.35 * 25.8: 0.3*25.8=7.74, 0.05*25.8=1.29. Sum is 7.74 + 1.29 = 9.03. So total is 25.8 + 9.03 = 34.83.

So, male riders is 34.8, and 1.35*female is 34.83. The difference is |34.8 - 34.83| = 0.03. That's a very small difference.

So, comparing all the differences:

9th grade: 36.27

10th grade: 30.86

11th grade: 5.91

12th grade: 0.03

So, the smallest difference is for 12th grade. Therefore, the answer should be 12th grade? Wait, but the question says "closest to 135% of the number of its female bus riders." So, the difference between male riders and 1.35*female riders is smallest for 12th grade. So, is that the answer?

Wait, but let me confirm my calculations again for 12th grade. Because the difference is only 0.03, which is really small. Let me check that again.

For 12th grade:

Male riders: 11.6% of 300. Let me check 11.6% is 0.116. 0.116 * 300 = 34.8. That's correct.

Female riders: 8.6% of 300. 0.086 * 300 = 25.8. Correct.

Then 1.35 * female riders: 1.35 * 25.8. Let me compute that again.

Alternatively, 25.8 * 1.35. Let me do 25.8 * 1.35. Maybe in another way:

First, 25.8 * 1 = 25.8

25.8 * 0.3 = 7.74

25.8 * 0.05 = 1.29

So adding up: 25.8 + 7.74 + 1.29 = 25.8 + 9.03 = 34.83. So that's correct. So male riders is 34.8, which is 34.8 vs 34.83. The difference is 0.03.

So that's extremely close. So according to that, 12th grade is the closest. But wait, the problem says "the number of its male bus riders closest to 135% of the number of its female bus riders." So, that would mean that 135% of female riders is 34.83, and male riders is 34.8, which is just 0.03 less. So that's the closest.

Wait, but maybe I made an error here. Because if you think about the percentages, the male percentage is 11.6% and female is 8.6%. So, 11.6% is higher than 8.6%, but 135% of female is 1.35 * 8.6% = 11.51%? Wait, wait, hold on. Wait, maybe there is another way to think about it. Maybe I can calculate what percentage the male riders are of the female riders, and see how close it is to 135%.

Wait, but the question is specifically about the number of male riders being closest to 135% of female riders. So, for example, if I take the actual numbers, which are absolute numbers, not percentages. So, even though the male percentage is 11.6% and female is 8.6%, the actual numbers are 34.8 vs 25.8.

So, 135% of female is 1.35 * 25.8 = 34.83, which is very close to the actual male riders of 34.8. So that's correct.

But let me check if I did the calculation for 11th grade again, because I thought 11th grade difference was 5.91. Let me check again:

For 11th grade:

Male riders: 20.6% of 300 = 61.8

Female riders: 13.8% of 300 = 41.4

1.35 * female riders: 1.35 * 41.4. Let me compute that again.

Alternatively, 41.4 * 1.35. Let me think of 41.4 * 1.35.

Well, 41.4 * 1.35. Let me compute 41.4 * 1.35.

First, note that 1.35 is 27/20. So 41.4 * 27 / 20. Let me compute 41.4 * 27 first.

But maybe decimal multiplication is easier.

41.4 * 1 = 41.4

41.4 * 0.3 = 12.42

41.4 * 0.05 = 2.07

Adding those: 41.4 + 12.42 + 2.07 = 55.89. Which matches what I had before. So 1.35 * female is 55.89. Male is 61.8. The difference is 61.8 - 55.89 = 5.91. Correct.

So, 11th grade difference is 5.91, 12th grade is 0.03, so 12th grade is much closer.

But then why does the problem say "closest to 135% of the number of its female bus riders"? Because 12th grade's male riders are almost exactly 135% of female riders.

But let me check if there is a mistake here. Because 135% of female riders would be 1.35 * 25.8 = 34.83, and male riders is 34.8. So the difference is 0.03. So that's 34.8 vs 34.83, so almost the same.

But maybe the problem is expecting integer numbers? Because you can't have a fraction of a person. Wait, but the question says "the number of its male bus riders closest to 135% of the number of its female bus riders." So, the actual numbers are 34.8 and 25.8. But in reality, the numbers should be integers, since you can't have a fraction of a person.

Wait, but the problem says that there were 300 males and 300 females surveyed. So, when they say 300 males and 300 females in each grade, that must mean that the percentages are exact? Wait, but the percentages are given with one decimal place. So, for example, 11.6% of 300 males. Let me check if that is an exact number.

11.6% of 300. Well, 10% is 30, 1% is 3, so 11.6% is 11*3 + 0.6*3 = 33 + 1.8 = 34.8. So, that is exactly 34.8. But you can't have 0.8 of a person, but in the context of the problem, maybe they are using percentages that result in decimal numbers for calculation purposes, and we can treat them as exact for the purpose of calculating the difference.

Alternatively, maybe the survey had exact numbers? For example, if they had 34.8 male riders, but in reality, they had 35 or 34. However, since the problem gives us percentages with decimal points, perhaps we are intended to use them as exact, even if they result in decimal numbers for the actual counts. So, in that case, proceeding with decimals is okay.

Therefore, according to the calculations, 12th grade has the closest difference, with only 0.03 difference. But let me check again the problem statement to make sure I didn't misread anything.

Wait, the problem says "the number of its male bus riders closest to 135% of the number of its female bus riders." So, for the 12th grade, male riders is 34.8, which is 34.8 compared to 135% of female riders (which is 34.83). So, they are extremely close.

But let me check the other grades again to make sure I didn't miscalculate.

For 9th grade:

Male: 41.1% of 300 = 123.3

Female: 39.4% of 300 = 118.2

135% of female = 1.35 * 118.2 = 159.57. Difference: 159.57 - 123.3 = 36.27. Correct.

For 10th grade:

Male: 34.4% of 300 = 103.2

Female: 33.1% of 300 = 99.3

135% of female = 1.35 * 99.3 = 134.055. Difference: 134.055 - 103.2 = 30.855. Correct.

For 11th grade:

Male: 20.6% of 300 = 61.8

Female: 13.8% of 300 = 41.4

1.35 * 41.4 = 55.89. Difference: 61.8 - 55.89 = 5.91. Correct.

For 12th grade:

Male: 11.6% of 300 = 34.8

Female: 8.6% of 300 = 25.8

1.35 * 25.8 = 34.83. Difference: 34.8 - 34.83 = -0.03, absolute value 0.03. Correct.

So, the differences are:

9th: 36.27

10th: 30.86

11th: 5.91

12th: 0.03

So, the smallest difference is 12th grade. So, the answer should be 12th grade.

But wait, the answer seems counterintuitive. Because 12th grade has very low percentages for both males and females. But maybe it's correct. Let me think again.

If the number of male riders is 34.8 and 135% of female is 34.83, which is really close. So, the answer is 12th grade.

But let me check once again for calculation errors. For example, when calculating 1.35 * 25.8.

Alternatively, 25.8 * 1.35. Let me do this multiplication step by step:

First, 25.8 * 1 = 25.8

25.8 * 0.3 = 7.74

25.8 * 0.05 = 1.29

Adding up: 25.8 + 7.74 = 33.54; 33.54 + 1.29 = 34.83. Correct.

So, 34.83 vs 34.8. The difference is 0.03. That's correct.

So, according to this, the answer is 12th grade.

But maybe the problem expects us to round the numbers? Wait, for instance, if we consider that the actual number of riders must be whole numbers, so for example, the number of male riders is 34.8, which would be approximately 35, and female riders is 25.8, approx 26. Then, 135% of 26 is 1.35 * 26 = 35.1. Which is very close to 35. So, the difference is 0.1.

But in that case, with rounding, 12th grade still has the closest.

Alternatively, maybe I need to check if the male number is closest to 135% of female number, not the absolute difference, but maybe some other measure. But no, the question says "closest to," so the absolute difference is the correct measure.

Alternatively, maybe I misread the question? Let me check again.

The question says: "For which grade is the number of its male bus riders closest to 135% of the number of its female bus riders?"

So, that means, compute for each grade, male = 135% of female? Or how close is male to 135% of female?

Yes, so for each grade, we calculate 135% of female riders and see how close the male riders are.

Yes, so the difference is as computed.

So, with the 0.03 difference, 12th grade is the answer.

But I want to check if maybe the question is in percentages. Wait, the percentages given are for the whole survey, but the question is about absolute numbers.

But the percentages are given per grade. For example, in 12th grade, male percentage is 11.6%, so that's 11.6% of 300 males, which is 34.8. Similarly for females.

So, maybe the answer is 12th grade.

But this seems very close. Maybe the answer is 11th grade? Because I recall that in some cases, people might misinterpret the question. Let me check.

Wait, let me think again. Maybe I need to check for which grade the ratio of male to female riders is closest to 1.35.

Wait, that's a different approach. Because 135% of female is male. So, male / female = 1.35. So, maybe instead of calculating 1.35 * female and comparing to male, we can compute the ratio male / female and see how close it is to 1.35.

Let me try that approach for each grade.

For 9th grade:

male / female = 123.3 / 118.2 ≈ 1.043. Which is about 104.3%, which is way less than 135%.

For 10th grade:

male / female = 103.2 / 99.3 ≈ 1.039. Which is about 103.9%, still less than 135%.

For 11th grade:

male / female = 61.8 / 41.4 ≈ 1.493. Which is about 149.3%, which is higher than 135%.

For 12th grade:

male / female = 34.8 / 25.8 ≈ 1.349. Which is about 134.9%, which is very close to 135%.

So, here, the ratio is 1.349, which is 134.9%, which is extremely close to 135%. Therefore, 12th grade is the answer.

So, this confirms the previous result. Therefore, the answer is 12th grade.

Therefore, even if I approach it through ratios, it's the same result.

But maybe I made a mistake with the decimal points? Let me check the 12th grade ratio again:

Male: 34.8, female: 25.8. 34.8 divided by 25.8. Let me compute that.

Dividing 34.8 by 25.8.

First, note that 25.8 * 1.35 = 34.83, as before.

So, 34.8 / 25.8 ≈ 1.349.

Which is 1.349, which is 134.9%.

So, that's just 0.1% less than 135%, which is very close.

So, in terms of percentage difference, this is the closest.

Therefore, the answer is 12th grade.

But let me check once again with exact fractions. Maybe there's some rounding that I didn't consider.

Let me compute male / female for 12th grade:

Male riders: 11.6% of 300.

But 11.6% is equal to 11.6 / 100. So, 11.6% of 300 = 11.6/100 * 300 = 11.6 * 3 = 34.8.

Female riders: 8.6% of 300 = 8.6/100 * 300 = 8.6 * 3 = 25.8.

So, male / female = 34.8 / 25.8.

Multiply numerator and denominator by 10 to eliminate decimals: 348 / 258.

Divide numerator and denominator by 6: 348 ÷ 6 = 58, 258 ÷ 6 = 43. So, 58 / 43 ≈ 1.3488... which is approximately 1.3488, which is about 134.88%, so 134.88% is the ratio, which is 0.12% less than 135%.

Therefore, this is the closest to 135%.

Therefore, the answer is 12th grade.

But just to be thorough, let me check the other grades' ratios to see how far they are from 135%.

For 11th grade, male / female was approximately 1.493, which is 149.3%, which is 14.3% higher than 135%.

For 10th grade, 1.039, which is 3.9% less than 135%.

For 9th grade, 1.043, which is 4.3% less than 135%.

So, the 12th grade is the closest, with only 0.12% difference.

So, this seems correct.

Therefore, the answer is the 12th grade.

But wait, looking back at the problem, the answer choices are 9th, 10th, 11th, 12th.

But let me check again if there is any possible mistake. I wonder if the question is asking for the number of male riders to be closest to 135% of the female riders, meaning that if you take the female riders and multiply by 1.35, then find which grade's male riders are closest to that number.

Which is exactly what I did. And for 12th grade, male riders = 34.8, and 1.35 * female riders = 34.83, so difference is 0.03, which is the smallest.

So, all signs point to 12th grade being the correct answer.

But maybe the problem is in the way the question is phrased? For example, maybe "closest to 135% of the number of its female bus riders"—so, maybe they want the male riders to be closest to 135% of the female riders, meaning that the value of male riders minus 1.35*female is closest to zero.

Which is exactly what I computed.

Alternatively, maybe the problem is in the interpretation of "percent of bus riders"—if the percentages are of the total students, but the problem says "percent of bus riders in a survey of Central H.S. students; 300 males and 300 females in each grade were surveyed."

So, the percentages are of the males and females surveyed. So, for example, for males, the 41.1% is the percentage of males who ride the bus, so that's the percentage of the 300 males.

Therefore, my calculations are correct.

Therefore, I think the correct answer is the 12th grade.

But just to make sure, let me check once more with exact numbers:

For 12th grade:

Male riders: 34.8

Female riders: 25.8

Compute 1.35 * female:

1.35 * 25.8 = ?

Let me calculate 25.8 * 1.35:

Breakdown:

First, multiply 25.8 by 1.35:

Compute 25.8 * 1 + 25.8 * 0.3 + 25.8 * 0.05

= 25.8 + 7.74 + 1.29 = 25.8 + 7.74 = 33.54 + 1.29 = 34.83.

Therefore, 1.35 * female = 34.83.

Male is 34.8.

Difference is 34.83 - 34.8 = 0.03.

So, that's a very small difference.

Therefore, 12th grade is the answer.

But I wonder why this seems so close. Maybe because the percentages are set up that way.

Alternatively, maybe the question has a typo, but assuming the numbers are correct, 12th grade is the answer.

Therefore, the answer is 12th grade.

**Final Answer**
\boxed{12}

To solve this problem, we are asked to determine which grade has the **number of its male bus riders that is closest to 135% of the number of its female bus riders**. The survey includes 300 males and 300 females in each grade. For each grade, we are provided with percentage estimates for the male and female riders.

---

### Step 1: Convert Percentages to Actual Counts

For each grade, the actual number of male and female bus riders is calculated by multiplying the given percentage by 300 (the number of students surveyed in each group).

Let’s compute the values for each grade:

#### **9th Grade**
- **Males:** $ 41.1\% \times 300 = 0.411 \times 300 = 123.3 $
- **Females:** $ 39.4\% \times 300 = 0.394 \times 300 = 118.2 $
- $ 135\% \text{ of females} = 1.35 \times 118.2 = 159.57 $
- **Difference:** $ |123.3 - 159.57| = 36.27 $

#### **10th Grade**
- **Males:** $ 34.4\% \times 300 = 0.344 \times 300 = 103.2 $
- **Females:** $ 33.1\% \times 300 = 0.331 \times 300 = 99.3 $
- $ 135\% \text{ of females} = 1.35 \times 99.3 = 134.055 $
- **Difference:** $ |103.2 - 134.055| = 30.855 $

#### **11th Grade**
- **Males:** $ 20.6\% \times 300 = 0.206 \times 300 = 61.8 $
- **Females:** $ 13.8\% \times 300 = 0.138 \times 300 = 41.4 $
- $ 135\% \text{ of females} = 1.35 \times 41.4 = 55.89 $
- **Difference:** $ |61.8 - 55.89| = 5.91 $

#### **12th Grade**
- **Males:** $ 11.6\% \times 300 = 0.116 \times 300 = 34.8 $
- **Females:** $ 8.6\% \times 300 = 0.086 \times 300 = 25.8 $
- $ 135\% \text{ of females} = 1.35 \times 25.8 = 34.83 $
- **Difference:** $ |34.8 - 34.83| = 0.03 $

---

### Step 2: Compare the Differences

From the above calculations, we can see the **absolute differences** between the number of male riders and 135% of female riders for each grade:

- **9th Grade:** 36.27
- **10th Grade:** 30.86
- **11th Grade:** 5.91
- **12th Grade:** 0.03

---

### Step 3: Identify the Closest Match

The smallest absolute difference is for the **12th grade**, with a difference of just **0.03**, which is extremely close. This indicates that the number of **male bus riders in 12th grade is nearly identical to 135% of the number of female riders**, making it the **closest match**.

---

### Final Answer

$$
\boxed{12}
$$

Solution: The grade level in which the number of male bus riders is closest to $135\%$ of the number of female bus riders is the same grade level in which the percentage of male bus riders is closest to $135\%$ of the percentage of female bus riders. To find what percentage is $135\%$ of the percentage of female bus riders, we multiply each percentage by $1.35.$ In ninth grade, the percentage is $39.4\cdot1.35=53.19$ percent. In tenth grade, the percentage is $33.1\cdot1.35=44.685$ percent. In eleventh grade, the percentage is $13.8\cdot1.35=18.63$ percent. In twelfth grade, the percentage is $8.6\cdot1.35=11.61$ percent. From inspection, we see that $135\%$ of the percentage of female bus riders in twelfth grade is closest to the percentage of male bus riders in twelfth grade. So the answer is $\boxed{12}.$