answer：First, I need to determine the value of one part in the given ratio. Since Vasim's share corresponds to 5 parts and is equal to Rs. 1500, I can calculate the value of one part by dividing Rs. 1500 by 5.Next, using the value of one part, I can find Faruk's share, which corresponds to 3 parts, by multiplying the value of one part by 3.Similarly, I can find Ranjith's share, which corresponds to 6 parts, by multiplying the value of one part by 6.Finally, to find the difference between Faruk's and Ranjith's shares, I subtract Faruk's share from Ranjith's share.

answer：Okay, so I have this problem here: I need to find the number of terms in the expansion of ((sqrt{x} - frac{1}{x})^{12}) that contain (x) raised to a positive integer power. The options are A: 1, B: 3, C: 2, D: 4. Hmm, let me try to figure this out step by step.First, I remember that when you expand a binomial like ((a + b)^n), each term is given by the binomial theorem, which is (T_{r+1} = C(n, r) cdot a^{n-r} cdot b^r), where (C(n, r)) is the combination of n things taken r at a time. So, in this case, (a) is (sqrt{x}) and (b) is (-frac{1}{x}), right?So, applying the binomial theorem, the general term (T_{r+1}) in the expansion of ((sqrt{x} - frac{1}{x})^{12}) would be:[T_{r+1} = C(12, r) cdot (sqrt{x})^{12 - r} cdot left(-frac{1}{x}right)^r]Let me simplify this term. First, let's deal with the exponents. The term ((sqrt{x})^{12 - r}) can be written as (x^{frac{12 - r}{2}}) because the square root of x is (x^{1/2}), and raising that to the power of (12 - r) gives (x^{frac{12 - r}{2}}).Next, the term (left(-frac{1}{x}right)^r) can be broken down into ((-1)^r cdot x^{-r}) because (frac{1}{x}) is (x^{-1}), and raising that to the power of r gives (x^{-r}).Putting it all together, the general term becomes:[T_{r+1} = C(12, r) cdot (-1)^r cdot x^{frac{12 - r}{2}} cdot x^{-r}]Now, when multiplying the x terms, we add their exponents. So, (frac{12 - r}{2} - r) is the exponent of x in the term. Let me compute that:[frac{12 - r}{2} - r = frac{12 - r - 2r}{2} = frac{12 - 3r}{2}]So, the exponent of x in each term is (frac{12 - 3r}{2}). The problem asks for the number of terms where this exponent is a positive integer. So, I need to find all integer values of r (from 0 to 12, since it's a binomial expansion) such that (frac{12 - 3r}{2}) is a positive integer.Let me denote the exponent as k, so:[k = frac{12 - 3r}{2}]Since k must be a positive integer, both numerator and denominator must result in an integer, and k > 0.First, let's ensure that (frac{12 - 3r}{2}) is an integer. For this fraction to be an integer, the numerator (12 - 3r) must be even because the denominator is 2. So, (12 - 3r) must be divisible by 2.Let me see: 12 is even, and 3r is either even or odd depending on r. Since 3 is odd, 3r is even if r is even, and odd if r is odd. Therefore, 12 - 3r is even if r is even because even minus even is even, and even minus odd is odd. So, to have (12 - 3r) even, r must be even.So, r must be even. That's one condition.Next, since k must be positive, we have:[frac{12 - 3r}{2} > 0]Multiply both sides by 2:[12 - 3r > 0]Which simplifies to:[12 > 3r implies 4 > r implies r < 4]So, r must be less than 4. But r is an integer between 0 and 12, inclusive. Also, from the previous condition, r must be even. So, the possible even values of r less than 4 are r = 0, 2.Wait, hold on, r can be 0, 2, 4, 6, etc., but since r must be less than 4, r can only be 0, 2.Wait, but let me check if r=4 is allowed. If r=4, then the exponent would be:[frac{12 - 3*4}{2} = frac{12 - 12}{2} = 0]But 0 is not a positive integer, so r=4 is excluded. Similarly, higher even r's would give negative exponents, which are not positive integers.So, only r=0 and r=2 satisfy both conditions: r is even and the exponent is a positive integer.Therefore, there are 2 such terms.Wait, but let me double-check. Let's compute the exponent for r=0:For r=0:[frac{12 - 0}{2} = 6]Which is a positive integer.For r=2:[frac{12 - 6}{2} = 3]Which is also a positive integer.For r=4:[frac{12 - 12}{2} = 0]Not positive.For r=6:[frac{12 - 18}{2} = -3]Negative, so not positive.Similarly, higher r's will give more negative exponents.So, indeed, only r=0 and r=2 give exponents that are positive integers.Therefore, the number of such terms is 2.Wait, but let me think again. Is there a possibility that for some other r, the exponent could be a positive integer? Maybe I missed something.Let me try plugging in r=1:[frac{12 - 3*1}{2} = frac{9}{2} = 4.5]Not an integer.r=3:[frac{12 - 9}{2} = frac{3}{2} = 1.5]Not an integer.r=4:As before, 0.r=5:[frac{12 - 15}{2} = frac{-3}{2} = -1.5]Negative and not integer.r=6:-3, as above.So, no, only r=0 and r=2 give exponents that are positive integers.Therefore, the answer should be 2, which is option C.Wait, but let me check once more. Maybe I made a mistake in the exponent calculation.The exponent is (frac{12 - 3r}{2}). For it to be a positive integer, both numerator and denominator must satisfy that.So, 12 - 3r must be even and positive.We saw that 12 - 3r is even only when r is even because 3r is even only when r is even.So, r must be even.Then, 12 - 3r > 0 => r < 4.So, r can be 0, 2.Thus, two terms.Yes, that seems correct.Alternatively, maybe I can think of it differently. Let's list all possible exponents for r from 0 to 12 and see which ones are positive integers.But that might take too long, but let me try a few more.r=0: exponent=6r=1: 4.5r=2:3r=3:1.5r=4:0r=5:-1.5r=6:-3r=7:-4.5r=8:-6r=9:-7.5r=10:-9r=11:-10.5r=12:-12So, only r=0 and r=2 give positive integer exponents. So, that's two terms.Therefore, the answer is 2, which is option C.I think that's solid. I don't see any mistakes in my reasoning.**Final Answer**boxed{C}

answer：Okay, so I have this problem about subsets of the set ( S_n = {1, 2, ldots, n} ). The capacity of a subset is the sum of its elements, and subsets are classified as odd or even based on whether their capacity is odd or even. There are three parts to this problem, and I need to tackle them one by one.Starting with part 1: Prove that the number of odd subsets of ( S_n ) is equal to the number of even subsets. Hmm, so I need to show that for any ( n ), the number of subsets with an odd sum is equal to the number with an even sum. Let me think about small cases first to get an intuition. For ( n = 1 ), the subsets are the empty set (sum 0, even) and the set itself (sum 1, odd). So, one even and one odd subset. They are equal in number.For ( n = 2 ), the subsets are:- Empty set: sum 0 (even)- {1}: sum 1 (odd)- {2}: sum 2 (even)- {1,2}: sum 3 (odd)So, two even subsets and two odd subsets. Still equal.For ( n = 3 ), the subsets are:- Empty set: 0 (even)- {1}: 1 (odd)- {2}: 2 (even)- {3}: 3 (odd)- {1,2}: 3 (odd)- {1,3}: 4 (even)- {2,3}: 5 (odd)- {1,2,3}: 6 (even)Counting them: even subsets are empty set, {2}, {1,3}, {1,2,3} – that's four. Odd subsets are {1}, {3}, {1,2}, {2,3} – also four. So, equal again.Seems like for ( n = 1, 2, 3 ), the number of odd and even subsets are equal. Maybe this is a general pattern.How can I prove this for any ( n )? Maybe using induction.**Base case:** ( n = 1 ). As above, one even and one odd subset. So, it holds.**Inductive step:** Assume that for some ( k geq 1 ), the number of odd and even subsets of ( S_k ) are equal. Now, consider ( S_{k+1} = S_k cup {k+1} ).The subsets of ( S_{k+1} ) can be divided into two categories: those that do not contain ( k+1 ) and those that do. The number of subsets not containing ( k+1 ) is equal to the number of subsets of ( S_k ), which by the inductive hypothesis has equal numbers of odd and even subsets.Now, consider subsets that do contain ( k+1 ). For each subset ( X ) of ( S_k ), we can form a subset ( X cup {k+1} ). The capacity of this new subset is the capacity of ( X ) plus ( k+1 ).If ( k+1 ) is odd, then adding it to a subset will flip the parity of the capacity. So, if ( X ) was odd, ( X cup {k+1} ) becomes even, and vice versa. Therefore, the number of odd subsets in ( S_{k+1} ) will be equal to the number of even subsets in ( S_k ), and the number of even subsets in ( S_{k+1} ) will be equal to the number of odd subsets in ( S_k ). Since by inductive hypothesis, these are equal, the total number of odd and even subsets in ( S_{k+1} ) will also be equal.If ( k+1 ) is even, adding it to a subset does not change the parity of the capacity. So, the number of odd subsets in ( S_{k+1} ) will be twice the number of odd subsets in ( S_k ), and similarly for even subsets. But since in ( S_k ), the number of odd and even subsets are equal, doubling them preserves the equality.Therefore, by induction, for all ( n geq 1 ), the number of odd and even subsets of ( S_n ) are equal. That proves part 1.Moving on to part 2: Prove that for ( n geq 3 ), the total capacity of all odd subsets of ( S_n ) is equal to the total capacity of all even subsets of ( S_n ).Hmm, this is a bit trickier. I need to show that if I sum up the capacities of all odd subsets, it equals the sum of capacities of all even subsets.Again, let's test with small ( n ). For ( n = 3 ), as above, the odd subsets are {1}, {3}, {1,2}, {2,3} with capacities 1, 3, 3, 5. Summing these: 1 + 3 + 3 + 5 = 12.The even subsets are empty set (0), {2}, {1,3}, {1,2,3} with capacities 0, 2, 4, 6. Summing these: 0 + 2 + 4 + 6 = 12.So, for ( n = 3 ), the total capacities are equal.For ( n = 4 ), let's compute:Subsets of ( S_4 ):- Odd subsets: {1}, {3}, {1,2}, {2,3}, {1,4}, {3,4}, {1,2,3}, {2,3,4}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}... Wait, actually, listing all subsets would take too long. Maybe there's a smarter way.Alternatively, perhaps using generating functions or some combinatorial argument.Wait, another idea: For each element ( i ) in ( S_n ), how many times does it appear in all subsets? It appears in half of the subsets, which is ( 2^{n-1} ) times. So, the total capacity of all subsets is ( (1 + 2 + ldots + n) times 2^{n-1} ).But since the total capacity is split equally between odd and even subsets, as per part 2, each should have half of that total.Wait, but part 2 is about the total capacity of odd subsets equaling that of even subsets. So, if the total capacity is ( T ), then ( T_{odd} = T_{even} = T / 2 ).But how do I show that ( T_{odd} = T_{even} )?Alternatively, perhaps using a similar induction approach as in part 1.Base case: ( n = 3 ). As above, total capacities are equal, 12 each.Assume for ( n = k geq 3 ), the total capacities of odd and even subsets are equal. Now, consider ( n = k + 1 ).Again, the subsets of ( S_{k+1} ) can be divided into those that do not contain ( k+1 ) and those that do.Let ( T_k ) be the total capacity of all subsets of ( S_k ). Then, the total capacity of all subsets of ( S_{k+1} ) is ( T_k + (k+1) times 2^k ), since each subset of ( S_k ) can be combined with ( k+1 ), adding ( k+1 ) to each of the ( 2^k ) subsets.By the induction hypothesis, the total capacity of odd subsets ( A_k = B_k ), so ( T_k = 2A_k ).Now, when adding ( k+1 ), if ( k+1 ) is odd, then adding it flips the parity of the subset's capacity. So, the total capacity contributed by subsets containing ( k+1 ) is ( (k+1) times 2^k ). Since the number of odd and even subsets are equal, the total capacity added to odd subsets is equal to the total capacity added to even subsets. Therefore, the total capacities remain equal.If ( k+1 ) is even, adding it doesn't change the parity. So, the total capacity added to odd subsets is ( (k+1) times A_k ) and similarly for even subsets. But since ( A_k = B_k ), the total capacities remain equal.Therefore, by induction, for all ( n geq 3 ), the total capacities of odd and even subsets are equal. That proves part 2.Now, part 3: For ( n geq 3 ), find the total capacity of all odd subsets of ( S_n ).From part 2, we know that the total capacity of all odd subsets equals the total capacity of all even subsets. Therefore, the total capacity of all subsets is twice the total capacity of all odd subsets.The total capacity of all subsets is the sum of capacities of all subsets of ( S_n ). As I thought earlier, each element ( i ) appears in ( 2^{n-1} ) subsets. Therefore, the total capacity is ( (1 + 2 + ldots + n) times 2^{n-1} ).The sum ( 1 + 2 + ldots + n ) is ( frac{n(n+1)}{2} ). Therefore, the total capacity is ( frac{n(n+1)}{2} times 2^{n-1} = n(n+1)2^{n-2} ).Since the total capacity is twice the total capacity of all odd subsets, we have:( 2 times text{Total capacity of odd subsets} = n(n+1)2^{n-2} )Therefore, the total capacity of all odd subsets is ( frac{n(n+1)2^{n-2}}{2} = n(n+1)2^{n-3} ).So, the answer is ( n(n+1)2^{n-3} ).**Final Answer**The total capacity of all odd subsets of ( S_n ) for ( n geq 3 ) is boxed{n(n+1)2^{n-3}}.

answer：Okay, so I've got this problem about the function ( f(x) = x^2 - 2ax + 1 ). It has two parts. Let me try to work through each step carefully.First, part (1) asks me to find sets ( A ) and ( B ). Starting with set ( A ): It's the range of ( a ) such that ( f(x) ) has no real roots. Hmm, okay. For a quadratic equation ( ax^2 + bx + c = 0 ), the discriminant ( D ) is ( b^2 - 4ac ). If ( D < 0 ), the quadratic has no real roots. So, applying this to ( f(x) = x^2 - 2ax + 1 ). Here, ( a = 1 ), ( b = -2a ), and ( c = 1 ). So the discriminant ( D ) would be ( (-2a)^2 - 4*1*1 ). Let me compute that:( D = 4a^2 - 4 ).We need ( D < 0 ), so:( 4a^2 - 4 < 0 )Divide both sides by 4:( a^2 - 1 < 0 )Which simplifies to:( a^2 < 1 )Taking square roots:( |a| < 1 )So, ( a ) is between -1 and 1. Therefore, set ( A ) is ( (-1, 1) ).Alright, that seems straightforward. Now, moving on to set ( B ). It's the range of ( a ) such that ( f(x) ) is not monotonic on the interval ( (m, m+3) ). Wait, what does it mean for a function to be monotonic on an interval? A function is monotonic if it is entirely non-increasing or non-decreasing on that interval. For quadratics, which are parabolas, they have a vertex which is either a minimum or maximum. So, if the interval ( (m, m+3) ) contains the vertex, then the function will change direction there, making it non-monotonic on that interval.The function ( f(x) = x^2 - 2ax + 1 ) is a parabola opening upwards (since the coefficient of ( x^2 ) is positive). Its vertex is at ( x = frac{-b}{2a} ). Wait, in this case, the function is ( x^2 - 2ax + 1 ), so ( a ) in the standard quadratic form is 1, ( b ) is -2a. So, the vertex is at ( x = frac{2a}{2*1} = a ).So, the vertex is at ( x = a ). Therefore, for the function to not be monotonic on ( (m, m+3) ), the vertex ( x = a ) must lie within that interval. So, ( a ) must satisfy ( m < a < m + 3 ).Hence, set ( B ) is ( (m, m + 3) ).Wait, hold on. Is that correct? Let me think again. If the vertex is inside the interval ( (m, m+3) ), then the function will have a minimum at ( x = a ), so it will decrease before ( a ) and increase after ( a ) within that interval, making it non-monotonic. If the vertex is outside the interval, then the function is either entirely increasing or decreasing on that interval, hence monotonic.Therefore, yes, for ( f(x) ) to not be monotonic on ( (m, m+3) ), ( a ) must lie within ( (m, m+3) ). So, set ( B ) is indeed ( (m, m + 3) ).So, part (1) is done. ( A = (-1, 1) ) and ( B = (m, m + 3) ).Moving on to part (2): It says that if ( x in A ) is a sufficient but not necessary condition for ( x in B ), find the range of values for ( m ).Hmm. Let's parse this. "If ( x in A ) is a sufficient but not necessary condition for ( x in B )." So, in logical terms, ( x in A ) implies ( x in B ), but ( x in B ) does not necessarily imply ( x in A ). In set terms, this means that ( A ) is a subset of ( B ), but ( B ) is not a subset of ( A ). So, ( A subset B ).Therefore, every element of ( A ) is in ( B ), but there are elements in ( B ) that are not in ( A ).Given that ( A = (-1, 1) ) and ( B = (m, m + 3) ), we need ( (-1, 1) subset (m, m + 3) ).So, for ( (-1, 1) ) to be a subset of ( (m, m + 3) ), the interval ( (m, m + 3) ) must contain all elements of ( (-1, 1) ). That means:1. The left endpoint of ( B ), which is ( m ), must be less than or equal to the left endpoint of ( A ), which is -1. So, ( m leq -1 ).2. The right endpoint of ( B ), which is ( m + 3 ), must be greater than or equal to the right endpoint of ( A ), which is 1. So, ( m + 3 geq 1 ).Therefore, we have two inequalities:( m leq -1 )and( m + 3 geq 1 )Solving the second inequality:( m + 3 geq 1 )Subtract 3 from both sides:( m geq -2 )So, combining both inequalities:( -2 leq m leq -1 )Therefore, the range of values for ( m ) is from -2 to -1, inclusive.Wait, let me double-check. If ( m = -2 ), then ( B = (-2, 1) ). Since ( A = (-1, 1) ), then ( A ) is entirely within ( B ). Similarly, if ( m = -1 ), then ( B = (-1, 2) ), which also contains ( A = (-1, 1) ). For values of ( m ) between -2 and -1, ( B ) will still contain ( A ).But the problem says that ( x in A ) is sufficient but not necessary for ( x in B ). So, ( A ) must be a proper subset of ( B ). That is, ( B ) must contain some elements not in ( A ).Looking at our earlier conclusion, when ( m = -2 ), ( B = (-2, 1) ). Since ( A = (-1, 1) ), ( B ) includes numbers less than -1, which are not in ( A ). Similarly, when ( m = -1 ), ( B = (-1, 2) ), which includes numbers greater than 1, not in ( A ). So, in both cases, ( B ) contains elements not in ( A ), so ( A ) is a proper subset of ( B ).Therefore, the range of ( m ) is from -2 to -1, inclusive.I think that makes sense. Let me visualize it. If ( m ) is -2, ( B ) spans from -2 to 1, which covers ( A ) from -1 to 1. If ( m ) is -1, ( B ) spans from -1 to 2, which also covers ( A ). For any ( m ) in between, ( B ) will still cover ( A ) and extend beyond on one side or the other. So yes, the range is correct.I don't see any mistakes in my reasoning, so I think that's the answer.**Final Answer**The range of values for ( m ) is boxed{[-2, -1]}.

answer：Alright, so I've got this problem about an ant crawling along the wire frame of a cube. The ant never turns back, and the question is whether it's possible for the ant to have visited one vertex 25 times while each of the remaining vertices has been visited 20 times. Hmm, interesting.First, I need to visualize the cube and its wire frame. A cube has 8 vertices and 12 edges. Each vertex is connected to three others. So, the ant is moving along these edges, never turning back, meaning it can't immediately go back to the vertex it just came from.Now, the ant is visiting one vertex 25 times and the others 20 times each. That means the total number of visits is 25 + 7*20 = 25 + 140 = 165 visits. Since the ant starts at some vertex, each move takes it to a new vertex, so the number of moves is one less than the number of visits. So, the ant makes 164 moves.But wait, in graph theory terms, this is like an Eulerian trail or something similar, but with multiple visits allowed. An Eulerian trail visits every edge exactly once, but here the ant can visit vertices multiple times. So, it's more like a walk on the graph.Now, I remember that in a cube, which is a bipartite graph, the vertices can be divided into two sets such that no two vertices within the same set are adjacent. In a cube, these sets are often referred to as "black" and "white" vertices, like a checkerboard pattern. So, if the ant starts on a black vertex, it alternates between black and white vertices with each move.Given that, the number of times the ant visits black and white vertices should differ by at most one, right? Because it alternates between them. So, if the ant makes an even number of moves, it ends on the same color it started on. If it makes an odd number of moves, it ends on the opposite color.In this case, the ant makes 164 moves, which is even. So, it should end on the same color it started on. Let's say it starts on a black vertex. Then, it should have visited black vertices one more time than white vertices, or vice versa?Wait, no. If it starts on a black vertex and makes an even number of moves, it ends on a black vertex. So, the number of visits to black vertices would be one more than the number of visits to white vertices.But according to the problem, one vertex is visited 25 times, and the others are visited 20 times. So, let's see: if one vertex is visited 25 times, and the others 20 times, then depending on the color of that one vertex, the total visits to black and white vertices would be different.Let me calculate the total visits. There are 8 vertices: one visited 25 times, and seven visited 20 times each. So, total visits are 25 + 7*20 = 25 + 140 = 165 visits.Now, if the cube is bipartite, with 4 black and 4 white vertices, then the total visits to black vertices plus the total visits to white vertices equals 165.If the ant starts on a black vertex and ends on a black vertex after 164 moves, then the number of visits to black vertices should be one more than the number of visits to white vertices. So, if B is the number of visits to black vertices and W is the number of visits to white vertices, then B = W + 1, and B + W = 165.Solving these equations: B = W + 1, so W = B - 1. Then, B + (B - 1) = 165 => 2B - 1 = 165 => 2B = 166 => B = 83, and W = 82.So, the ant should have visited black vertices 83 times and white vertices 82 times.Now, in the problem, one vertex is visited 25 times, and the others 20 times. Let's see if this fits into the B and W counts.If the vertex visited 25 times is a black vertex, then the total visits to black vertices would be 25 + sum of visits to other black vertices. Since there are 4 black vertices, one is visited 25 times, and the other three are visited 20 times each. So, total black visits would be 25 + 3*20 = 25 + 60 = 85.But earlier, we found that B should be 83. So, 85 is 2 more than 83. That's a problem.Alternatively, if the vertex visited 25 times is a white vertex, then total white visits would be 25 + 3*20 = 85, and black visits would be 4*20 = 80. But earlier, we found that W should be 82 and B should be 83. So, 80 and 85 don't match 82 and 83.Wait, that doesn't add up. So, there's a contradiction here. The total visits to black and white vertices don't match what's expected based on the ant's path.Therefore, it's impossible for the ant to have visited one vertex 25 times and each of the remaining vertices 20 times.I think that's the crux of it. The bipartite nature of the cube imposes a constraint on the number of visits to each set of vertices, and the given visit counts don't satisfy that constraint.

answer：Okay, so I have this problem where a plot of land is represented on a map, and the scale is 1 cm equals 1 mile in reality. The plot's dimensions on the map are 20 cm by 25 cm. I need to find out how large this plot is in acres, given that one square mile equals 640 acres.Alright, first, I think I need to figure out the area of the plot on the map. Since it's a rectangle, the area should be length multiplied by width. So, 20 cm times 25 cm. Let me calculate that: 20 times 25 is 500. So, the area on the map is 500 square centimeters.Now, the tricky part is converting this area from square centimeters on the map to actual square miles in reality. The scale is 1 cm equals 1 mile, so does that mean 1 square cm equals 1 square mile? That seems straightforward, but I want to make sure I'm not missing anything. So, if 1 cm corresponds to 1 mile, then 1 cm squared would correspond to 1 mile squared. Therefore, 500 square centimeters on the map would correspond to 500 square miles in reality.Wait, that seems like a lot. 500 square miles is a huge area. Let me double-check. If each cm is a mile, then each dimension is 20 miles by 25 miles, so the area is indeed 20 times 25, which is 500 square miles. Okay, that checks out.Now, I need to convert square miles to acres. The problem states that one square mile equals 640 acres. So, if I have 500 square miles, I can multiply that by 640 to get the area in acres. Let me do that: 500 times 640. Hmm, 500 times 600 is 300,000, and 500 times 40 is 20,000. So, adding those together, 300,000 plus 20,000 equals 320,000 acres.Wait, that's 320,000 acres. That seems enormous. Is that right? Let me think again. If 1 square mile is 640 acres, then 500 square miles would be 500 times 640. Yes, that's 320,000 acres. Okay, that seems correct.I guess the key steps here are:1. Calculate the area on the map: 20 cm times 25 cm equals 500 cm².2. Convert the area from cm² to square miles using the scale: 1 cm² equals 1 square mile, so 500 cm² equals 500 square miles.3. Convert square miles to acres: 500 times 640 equals 320,000 acres.I don't see any mistakes in this reasoning. So, the actual plot of land is 320,000 acres.