💡First, I'll define the variables for the number of treats and pieces of bread brought by Jane and Wanda.Let ( J_t ) represent the number of treats Jane brings, and ( J_b ) represent the number of pieces of bread Jane brings. Similarly, let ( W_t ) represent the number of treats Wanda brings, and ( W_b ) represent the number of pieces of bread Wanda brings.From the problem statement, Jane brings 75% as many pieces of bread as treats. This can be expressed as:[J_b = 0.75 cdot J_t]Wanda brings three times as many pieces of bread as treats:[W_b = 3 cdot W_t]Given that Wanda brings 90 pieces of bread:[3 cdot W_t = 90 implies W_t = 30]The total number of pieces of bread and treats brought by both Jane and Wanda is 225:[J_b + J_t + W_b + W_t = 225]Substituting the known values and expressions:[0.75 cdot J_t + J_t + 90 + 30 = 225]Combining like terms:[1.75 cdot J_t + 120 = 225]Solving for ( J_t ):[1.75 cdot J_t = 105 implies J_t = 60]Finally, to find the ratio of the number of treats Wanda brings compared to Jane:[text{Ratio} = frac{W_t}{J_t} = frac{30}{60} = frac{1}{2}]

💡First, I need to find the average monthly salary of the 24 employees excluding the manager. Let's denote this average as ( A ).The total salary of the 24 employees is ( 24 times A ).When the manager's salary, which is Rs. 4900, is added, the total number of employees becomes 25, and the average salary increases by Rs. 100. Therefore, the new average salary is ( A + 100 ).The total salary with the manager included is ( 25 times (A + 100) ).Setting up the equation:[24A + 4900 = 25(A + 100)]Expanding and simplifying:[24A + 4900 = 25A + 2500][4900 - 2500 = 25A - 24A][2400 = A]So, the average monthly salary of the employees excluding the manager is Rs. 2400.

💡Okay, so I have this problem with three parts involving two circles and a point. Let me try to tackle each part step by step. I'll start with part (1).**Problem (1):** Draw a tangent line to ⊙M passing through point O, and find the equation of this tangent line.Alright, so first, let me recall what the circles look like. ⊙O is given by the equation x² + y² = 2. That's a circle centered at the origin (0,0) with radius √2. Then, ⊙M is given by (x + 2)² + (y + 2)² = 2. So, this circle is centered at (-2, -2) and also has a radius of √2.Point O is the origin, (0,0), and we need to find the equation of a tangent line from O to ⊙M. Since O is outside ⊙M, there should be two tangent lines. I remember that the tangent from an external point to a circle can be found using the formula involving the distance from the point to the center and the radius.Let me denote the center of ⊙M as M(-2, -2). The distance between O and M is sqrt[(-2 - 0)² + (-2 - 0)²] = sqrt[4 + 4] = sqrt[8] = 2√2. The radius of ⊙M is √2, so the length of the tangent from O to ⊙M is sqrt[(OM)² - r²] = sqrt[(8) - (2)] = sqrt[6]. Hmm, wait, but in the solution, they have sqrt[2]. Maybe I made a mistake.Wait, no, the radius is √2, so the tangent length squared is OM² - r² = 8 - 2 = 6, so the tangent length is sqrt(6). But in the solution, they have sqrt(2). Maybe I'm confusing something.Wait, actually, when finding the equation of the tangent line, we can use the condition that the distance from the center of the circle to the line is equal to the radius. So, let me try that approach.Let me assume the equation of the tangent line is y = kx, since it passes through the origin. Then, the distance from the center M(-2, -2) to this line should be equal to the radius √2.The distance from a point (x₀, y₀) to the line ax + by + c = 0 is |ax₀ + by₀ + c| / sqrt(a² + b²). So, for the line y = kx, we can rewrite it as kx - y = 0. So, a = k, b = -1, c = 0.Thus, the distance from M(-2, -2) to this line is |k*(-2) + (-1)*(-2) + 0| / sqrt(k² + 1) = |-2k + 2| / sqrt(k² + 1). This should equal the radius √2.So, |-2k + 2| / sqrt(k² + 1) = sqrt(2). Let me square both sides to eliminate the square roots:(|-2k + 2|)² = (sqrt(2))² * (sqrt(k² + 1))²Which simplifies to:(4k² - 8k + 4) = 2(k² + 1)Expanding both sides:4k² - 8k + 4 = 2k² + 2Subtract 2k² + 2 from both sides:2k² - 8k + 2 = 0Divide both sides by 2:k² - 4k + 1 = 0Now, solving this quadratic equation for k:k = [4 ± sqrt(16 - 4)] / 2 = [4 ± sqrt(12)] / 2 = [4 ± 2*sqrt(3)] / 2 = 2 ± sqrt(3)So, the slopes are k = 2 + sqrt(3) and k = 2 - sqrt(3). Therefore, the equations of the tangent lines are y = (2 + sqrt(3))x and y = (2 - sqrt(3))x.Wait, that's what the solution has. So, I think my initial confusion was about the tangent length, but using the distance formula approach worked out correctly. So, part (1) is done.**Problem (2):** If point Q is on ⊙O, and tangent lines to ⊙M passing through Q touch ⊙M at points E and F, with ∠EQF = π/3, find the coordinates of point Q.Okay, so Q is on ⊙O, which is x² + y² = 2. From Q, we draw two tangent lines to ⊙M, which touches ⊙M at E and F. The angle between these two tangents at Q is π/3, which is 60 degrees.I remember that the angle between two tangents from a point outside a circle can be related to the distance from that point to the center of the circle. The formula for the angle θ between two tangents is given by θ = 2 arcsin(r / d), where r is the radius of the circle, and d is the distance from the external point to the center.In this case, the angle is π/3, so θ = π/3. Therefore, π/3 = 2 arcsin(r / d). Let me solve for d.Divide both sides by 2: π/6 = arcsin(r / d)Take sine of both sides: sin(π/6) = r / dWe know sin(π/6) = 1/2, so 1/2 = r / d => d = 2rGiven that the radius r of ⊙M is √2, so d = 2√2.Therefore, the distance from Q to M must be 2√2.But Q is on ⊙O, which is centered at O(0,0) with radius √2. So, Q lies on both ⊙O and the circle centered at M(-2, -2) with radius 2√2.So, we can set up the equations:1. x² + y² = 2 (since Q is on ⊙O)2. (x + 2)² + (y + 2)² = (2√2)² = 8 (since Q is at distance 2√2 from M)Now, let's solve these two equations simultaneously.First, expand the second equation:(x + 2)² + (y + 2)² = 8x² + 4x + 4 + y² + 4y + 4 = 8x² + y² + 4x + 4y + 8 = 8But from the first equation, x² + y² = 2. Substitute that into the second equation:2 + 4x + 4y + 8 = 8Simplify: 4x + 4y + 10 = 8Subtract 8: 4x + 4y + 2 = 0Divide by 2: 2x + 2y + 1 = 0Simplify: x + y = -1/2So, the points Q lie on the line x + y = -1/2 and also on ⊙O: x² + y² = 2.Let me substitute y = -1/2 - x into the circle equation:x² + (-1/2 - x)² = 2Expand: x² + (1/4 + x + x²) = 2Combine like terms: 2x² + x + 1/4 = 2Subtract 2: 2x² + x - 7/4 = 0Multiply both sides by 4 to eliminate fractions: 8x² + 4x - 7 = 0Now, solve this quadratic equation for x:x = [-4 ± sqrt(16 + 224)] / 16 = [-4 ± sqrt(240)] / 16 = [-4 ± 4*sqrt(15)] / 16 = [-1 ± sqrt(15)] / 4So, x = (-1 + sqrt(15))/4 or x = (-1 - sqrt(15))/4Corresponding y values are y = -1/2 - x.So, if x = (-1 + sqrt(15))/4, then y = -1/2 - (-1 + sqrt(15))/4 = (-2/4 + 1/4 - sqrt(15)/4) = (-1/4 - sqrt(15)/4) = (-1 - sqrt(15))/4Similarly, if x = (-1 - sqrt(15))/4, then y = -1/2 - (-1 - sqrt(15))/4 = (-2/4 + 1/4 + sqrt(15)/4) = (-1/4 + sqrt(15)/4) = (-1 + sqrt(15))/4Therefore, the coordinates of Q are:Q((-1 - sqrt(15))/4, (-1 + sqrt(15))/4) and Q((-1 + sqrt(15))/4, (-1 - sqrt(15))/4)Wait, but in the solution, they have Q((-1 - sqrt(15))/4, (-1 + sqrt(15))/4) and Q((-1 + sqrt(15))/4, (-1 - sqrt(15))/4). That's the same as what I got, just written differently. So, that's correct.**Problem (3):** Draw two distinct lines through point P intersecting ⊙O at points A and B respectively, and the slope angles of lines PA and PB are complementary. Determine whether the line OP is parallel to AB, and explain your reasoning.Alright, so point P is (1,1). We draw two lines through P intersecting ⊙O at A and B. The slope angles of PA and PB are complementary, meaning if one has a slope angle θ, the other has π/2 - θ.I need to determine if OP is parallel to AB.First, let's note that OP is the line from O(0,0) to P(1,1), which has a slope of 1. So, if AB also has a slope of 1, then they are parallel.Let me try to find the coordinates of points A and B.Let me denote the lines PA and PB. Since their slope angles are complementary, their slopes are negative reciprocals. Wait, no. If the angles are complementary, say θ and π/2 - θ, then tan(θ) and tan(π/2 - θ) = cot(θ). So, the slopes would be m and 1/m, but with a sign depending on the quadrant.Wait, but in the solution, they have lines PA and PB with slopes k and -k, which are negative reciprocals if k = 1/m, but in this case, they are just negatives. Hmm, maybe I need to think differently.Wait, the slope angles are complementary, meaning that the sum of the angles is π/2. So, if one line has a slope angle θ, the other has π/2 - θ. Therefore, tan(θ) and tan(π/2 - θ) = cot(θ). So, the slopes are m and 1/m.But in the solution, they have PA as y - 1 = k(x - 1) and PB as y - 1 = -k(x - 1). So, they have slopes k and -k. That suggests that the angles are supplementary, but the problem says complementary. Hmm, maybe I need to check.Wait, complementary angles add up to π/2, so if one angle is θ, the other is π/2 - θ. So, their slopes would be tan(θ) and tan(π/2 - θ) = cot(θ). So, if one slope is m, the other is 1/m. But in the solution, they have slopes k and -k, which would mean angles θ and π - θ, which are supplementary, not complementary.Wait, maybe the problem is referring to the angles of the slopes with respect to the x-axis, so if one is θ, the other is π/2 - θ, so their slopes are tan(θ) and tan(π/2 - θ) = cot(θ). So, if one slope is m, the other is 1/m. But in the solution, they have slopes k and -k, which would correspond to angles θ and π - θ, which are supplementary, not complementary.Hmm, perhaps I need to clarify. Let me think again.If two lines have slopes m1 and m2, and their angles with the x-axis are θ1 and θ2, then if θ1 + θ2 = π/2, then tan(θ1) = m1 and tan(θ2) = m2. Since θ2 = π/2 - θ1, then tan(θ2) = tan(π/2 - θ1) = cot(θ1) = 1/tan(θ1) = 1/m1. So, m2 = 1/m1.Therefore, the slopes should be reciprocals, not negatives. So, in the solution, they have PA with slope k and PB with slope -k, which would mean that the angles are θ and π - θ, which are supplementary, not complementary. So, perhaps the solution is incorrect in this part.Wait, but let me check the solution again. They have:PA: y - 1 = k(x - 1)PB: y - 1 = -k(x - 1)So, slopes k and -k. Then, they find the points A and B by solving these lines with ⊙O: x² + y² = 2.For PA: Substitute y = k(x - 1) + 1 into x² + y² = 2:x² + [k(x - 1) + 1]^2 = 2Expanding:x² + k²(x - 1)^2 + 2k(x - 1) + 1 = 2x² + k²(x² - 2x + 1) + 2k(x - 1) + 1 = 2x² + k²x² - 2k²x + k² + 2kx - 2k + 1 = 2Combine like terms:(1 + k²)x² + (-2k² + 2k)x + (k² - 2k + 1) = 2Subtract 2:(1 + k²)x² + (-2k² + 2k)x + (k² - 2k - 1) = 0This is a quadratic in x. Since P(1,1) is on the line PA, x=1 is a root. So, we can factor it out.Let me use polynomial division or factor theorem. Let me plug x=1 into the quadratic:(1 + k²)(1) + (-2k² + 2k)(1) + (k² - 2k - 1) = 1 + k² - 2k² + 2k + k² - 2k -1 = 0Yes, it equals zero, so (x - 1) is a factor.Let me perform polynomial division or factorization.Let me write the quadratic as:(1 + k²)x² + (-2k² + 2k)x + (k² - 2k - 1) = (x - 1)(Ax + B)Expanding the right side:x(Ax + B) - 1(Ax + B) = Ax² + Bx - Ax - B = Ax² + (B - A)x - BComparing coefficients:Ax² = (1 + k²)x² => A = 1 + k²(B - A)x = (-2k² + 2k)x => B - A = -2k² + 2k-B = k² - 2k - 1 => B = -k² + 2k + 1From A = 1 + k², then B = A + (-2k² + 2k) = (1 + k²) + (-2k² + 2k) = 1 - k² + 2kBut also, B = -k² + 2k + 1, which matches.So, the quadratic factors as (x - 1)( (1 + k²)x + (-k² + 2k + 1) ) = 0Therefore, the other root is x = [k² - 2k - 1]/(1 + k²)So, x_A = [k² - 2k - 1]/(1 + k²)Similarly, for PB with slope -k, we can do the same substitution:y = -k(x - 1) + 1Substitute into x² + y² = 2:x² + [-k(x - 1) + 1]^2 = 2Expanding similarly, we'll get x_B = [k² + 2k - 1]/(1 + k²)Now, to find the slope of AB, we need the coordinates of A and B.Let me denote:A: (x_A, y_A) = ( [k² - 2k - 1]/(1 + k²), k(x_A - 1) + 1 )Similarly, B: (x_B, y_B) = ( [k² + 2k - 1]/(1 + k²), -k(x_B - 1) + 1 )Let me compute y_A and y_B.For y_A:y_A = k(x_A - 1) + 1 = k( [k² - 2k - 1]/(1 + k²) - 1 ) + 1= k( [k² - 2k - 1 - (1 + k²)] / (1 + k²) ) + 1= k( [k² - 2k - 1 - 1 - k²] / (1 + k²) ) + 1= k( [-2k - 2] / (1 + k²) ) + 1= [ -2k² - 2k ] / (1 + k²) + 1= [ -2k² - 2k + (1 + k²) ] / (1 + k² )= [ -k² - 2k + 1 ] / (1 + k² )Similarly, for y_B:y_B = -k(x_B - 1) + 1 = -k( [k² + 2k - 1]/(1 + k²) - 1 ) + 1= -k( [k² + 2k - 1 - (1 + k²)] / (1 + k²) ) + 1= -k( [k² + 2k - 1 - 1 - k²] / (1 + k²) ) + 1= -k( [2k - 2] / (1 + k²) ) + 1= [ -2k² + 2k ] / (1 + k² ) + 1= [ -2k² + 2k + (1 + k²) ] / (1 + k² )= [ -k² + 2k + 1 ] / (1 + k² )Now, we have coordinates for A and B:A: ( [k² - 2k - 1]/(1 + k²), [ -k² - 2k + 1 ] / (1 + k² ) )B: ( [k² + 2k - 1]/(1 + k²), [ -k² + 2k + 1 ] / (1 + k² ) )Now, let's compute the slope of AB.Slope k_AB = (y_B - y_A) / (x_B - x_A)Compute numerator:y_B - y_A = [ (-k² + 2k + 1) - (-k² - 2k + 1) ] / (1 + k² )= [ (-k² + 2k + 1 + k² + 2k - 1) ] / (1 + k² )= [4k] / (1 + k² )Denominator:x_B - x_A = [ (k² + 2k - 1) - (k² - 2k - 1) ] / (1 + k² )= [k² + 2k - 1 - k² + 2k + 1] / (1 + k² )= [4k] / (1 + k² )Therefore, k_AB = (4k / (1 + k² )) / (4k / (1 + k² )) = 1So, the slope of AB is 1, which is the same as the slope of OP, which is from (0,0) to (1,1), slope 1.Therefore, AB is parallel to OP.Wait, but in the solution, they have k_AB = 1, which is the same as k_OP = 1, so they are parallel. So, the answer is yes, OP is parallel to AB.But earlier, I was confused about the slopes being k and -k, which would correspond to supplementary angles, but the problem says complementary. However, in the solution, they still get AB parallel to OP. Maybe the key is that regardless of the slopes, the calculation shows that AB has slope 1, same as OP.Alternatively, perhaps the condition of complementary angles leads to the slopes being such that the product is -1, but in this case, it's leading to AB having a slope of 1. Hmm, maybe the way they set up the lines with slopes k and -k is a way to satisfy the complementary angle condition, but I'm not entirely sure. But the calculation shows that AB has slope 1, same as OP, so they are parallel.Therefore, the answer is yes, OP is parallel to AB.

💡Okay, so I have this problem about a parabola, specifically y = x². I remember that the standard form of a parabola is y = ax² + bx + c, but in this case, it's just y = x², which is a simple upward-opening parabola. The problem mentions the focus F, so I need to recall where the focus is for this parabola.I think for a parabola in the form y = ax², the focus is at (0, 1/(4a)). Since here a = 1, the focus should be at (0, 1/4). Let me confirm that. Yes, for y = x², the focus is indeed (0, 1/4). Got that down.Now, the problem says that a line passes through this focus F and intersects the parabola at points A and B. So, this line is a secant of the parabola, cutting it at two points. The distance between A and B is given as 4 units. I need to find the distance from the midpoint of chord AB to the x-axis.Hmm, okay. So, first, maybe I should find the equation of the line passing through F. Since it's a line passing through (0, 1/4), I can write it in the form y = mx + c. But since it passes through (0, 1/4), when x = 0, y = 1/4, so c = 1/4. Therefore, the equation of the line is y = mx + 1/4, where m is the slope.But I don't know the slope m. So, perhaps I can find m such that the distance between points A and B is 4. Let me think about how to do that.First, I can find the points of intersection between the line y = mx + 1/4 and the parabola y = x². Setting them equal, x² = mx + 1/4. Rearranging, x² - mx - 1/4 = 0. That's a quadratic equation in x.Let me denote this quadratic equation as x² - mx - 1/4 = 0. Let me denote the roots as x₁ and x₂, which correspond to the x-coordinates of points A and B.From quadratic equation theory, I know that the sum of roots x₁ + x₂ = m, and the product x₁x₂ = -1/4.Now, the distance between points A and B is given as 4. Let me recall that the distance between two points (x₁, y₁) and (x₂, y₂) is sqrt[(x₂ - x₁)² + (y₂ - y₁)²]. In this case, since both points lie on the line y = mx + 1/4, their y-coordinates can be expressed as y₁ = mx₁ + 1/4 and y₂ = mx₂ + 1/4.So, the distance AB is sqrt[(x₂ - x₁)² + (y₂ - y₁)²] = sqrt[(x₂ - x₁)² + (m(x₂ - x₁))²] = sqrt[(1 + m²)(x₂ - x₁)²] = |x₂ - x₁| * sqrt(1 + m²).We are given that this distance is 4. So, |x₂ - x₁| * sqrt(1 + m²) = 4.Now, I need to express |x₂ - x₁| in terms of m. From quadratic equations, I know that (x₂ - x₁)² = (x₁ + x₂)² - 4x₁x₂. So, (x₂ - x₁)² = m² - 4*(-1/4) = m² + 1. Therefore, |x₂ - x₁| = sqrt(m² + 1).Substituting back into the distance formula, we have sqrt(m² + 1) * sqrt(1 + m²) = 4. So, sqrt(m² + 1) * sqrt(m² + 1) = (sqrt(m² + 1))² = m² + 1 = 4.Therefore, m² + 1 = 4 => m² = 3 => m = sqrt(3) or m = -sqrt(3). So, the slopes of the lines are sqrt(3) and -sqrt(3). Therefore, the equations of the lines are y = sqrt(3)x + 1/4 and y = -sqrt(3)x + 1/4.Now, I need to find the midpoint of chord AB. The midpoint M will have coordinates ((x₁ + x₂)/2, (y₁ + y₂)/2). From earlier, we know that x₁ + x₂ = m, so (x₁ + x₂)/2 = m/2. Similarly, y₁ + y₂ = (mx₁ + 1/4) + (mx₂ + 1/4) = m(x₁ + x₂) + 1/2 = m*(m) + 1/2 = m² + 1/2. Therefore, (y₁ + y₂)/2 = (m² + 1/2)/2 = (m²)/2 + 1/4.So, the coordinates of the midpoint M are (m/2, (m²)/2 + 1/4). Now, since we have two possible values for m, sqrt(3) and -sqrt(3), let's compute the midpoint for each case.Case 1: m = sqrt(3)Midpoint M has coordinates (sqrt(3)/2, ( (sqrt(3))² )/2 + 1/4 ) = (sqrt(3)/2, (3)/2 + 1/4 ) = (sqrt(3)/2, 3/2 + 1/4 ) = (sqrt(3)/2, 7/4 ).Case 2: m = -sqrt(3)Midpoint M has coordinates (-sqrt(3)/2, ( (-sqrt(3))² )/2 + 1/4 ) = (-sqrt(3)/2, (3)/2 + 1/4 ) = (-sqrt(3)/2, 7/4 ).So, in both cases, the y-coordinate of the midpoint is 7/4. The x-coordinate is either sqrt(3)/2 or -sqrt(3)/2, but since we are only asked for the distance from the midpoint to the x-axis, which is the absolute value of the y-coordinate, it's 7/4.Therefore, the distance from the midpoint of chord AB to the x-axis is 7/4.Wait, let me just make sure I didn't make a mistake in calculating the midpoint's y-coordinate. So, y-coordinate is (y₁ + y₂)/2, which is (mx₁ + 1/4 + mx₂ + 1/4)/2 = (m(x₁ + x₂) + 1/2)/2. Since x₁ + x₂ = m, so m*(m) + 1/2 = m² + 1/2, and then divided by 2 is (m² + 1/2)/2 = m²/2 + 1/4. Since m² = 3, so 3/2 + 1/4 = 7/4. Yep, that seems correct.So, the distance is 7/4 units. So, the answer is 7/4.**Final Answer**The distance from the midpoint of chord ( AB ) to the ( x )-axis is (boxed{dfrac{7}{4}}).

💡First, I need to substitute the given values of ( x = 3 ) and ( y = 2 ) into the expression ( 3x^{y+1} + 4y^{x+1} ).I'll start by calculating ( x^{y+1} ). Since ( y = 2 ), ( y + 1 = 3 ). Therefore, ( x^{y+1} = 3^3 = 27 ).Next, I'll calculate ( y^{x+1} ). With ( x = 3 ), ( x + 1 = 4 ). So, ( y^{x+1} = 2^4 = 16 ).Now, I'll substitute these calculated values back into the original expression:[3x^{y+1} + 4y^{x+1} = 3 times 27 + 4 times 16][= 81 + 64][= 145]

💡Okay, so I've got this problem about two trains starting from opposite points, P and Q, and moving towards each other. They meet at some point, and then each continues to their respective destinations. I know the speeds of both trains: train A goes at 120 km/h, and train B goes at 160 km/h. The problem says that train A takes 16 hours to reach its destination after meeting train B, and I need to find out how long it takes train B to reach its destination after they meet.Alright, let me break this down. First, both trains start at the same time from points P and Q, moving towards each other. They meet somewhere in between, and then continue moving until they reach their destinations. The key here is that when they meet, they've been traveling for the same amount of time, right? So, if I can figure out how long it took them to meet, I can use that to find out how much distance each train has left to their destinations.Let's denote the time it takes for both trains to meet as 'x' hours. In that time, train A would have traveled a distance of 120x km, and train B would have traveled 160x km. So, the total distance between points P and Q would be the sum of these two distances: 120x + 160x = 280x km.Now, after meeting, train A takes 16 more hours to reach its destination. Since train A's speed is 120 km/h, the remaining distance it needs to cover after the meeting point is 120 km/h * 16 h = 1920 km. Wait, that seems like a lot. Is that right? Let me think. If train A has already traveled 120x km before meeting train B, and then it needs to cover another 1920 km to reach point Q, then the total distance between P and Q should be 120x + 1920 km.But earlier, I calculated the total distance as 280x km. So, setting these equal: 280x = 120x + 1920. Subtracting 120x from both sides gives 160x = 1920. Dividing both sides by 160, I get x = 12 hours. So, it took both trains 12 hours to meet.Now, train B has been traveling for 12 hours at 160 km/h, so it has covered 160 * 12 = 1920 km before meeting train A. The total distance between P and Q is 280x = 280 * 12 = 3360 km. Therefore, the remaining distance for train B after the meeting point is 3360 - 1920 = 1440 km.Since train B's speed is 160 km/h, the time it takes to cover 1440 km is 1440 / 160 = 9 hours. So, train B takes 9 hours to reach its destination after meeting train A.Wait, but in the initial problem, it was mentioned that train A takes 16 hours after meeting, and I got train B taking 9 hours. Does that make sense? Let me double-check my calculations.Total distance: 280x = 280 * 12 = 3360 km.Train A's remaining distance: 1920 km, which at 120 km/h takes 16 hours. That checks out.Train B's remaining distance: 3360 - 1920 = 1440 km. At 160 km/h, time = 1440 / 160 = 9 hours. Yeah, that seems correct.So, the answer should be 9 hours for train B to reach its destination after meeting train A.