💡First, I need to determine the relative speed between the train and the man since they are moving in opposite directions. I'll add their speeds together.Next, I'll convert the relative speed from kilometers per hour to meters per second to match the unit of the train's length.Finally, I'll calculate the time it takes for the train to cross the man by dividing the length of the train by the relative speed.

💡Okay, so I have this problem where the ratio of A:B:C is 3:2:5, and I need to find the value of (5A + 3B) divided by (3C - 2A). Hmm, ratios can sometimes be tricky, but I think I can handle this by breaking it down step by step.First, I remember that when we have a ratio like A:B:C = 3:2:5, it means that A is to B as 3 is to 2, and B is to C as 2 is to 5. So, I can express A, B, and C in terms of a common variable to make the calculations easier. Let me choose a variable, say, k, which will represent the common multiplier for the ratios.So, if I let A = 3k, B = 2k, and C = 5k, that should maintain the given ratio. That makes sense because if k is 1, then A:B:C would be 3:2:5, which matches the problem statement. If k is any other number, the ratios will still hold because each term is scaled by the same factor.Now, I need to substitute these expressions into the given equation: (5A + 3B) ÷ (3C - 2A). Let me write that out with the substitutions:(5*(3k) + 3*(2k)) ÷ (3*(5k) - 2*(3k))Let me compute the numerator first: 5*(3k) is 15k, and 3*(2k) is 6k. So, adding those together, 15k + 6k equals 21k.Now, the denominator: 3*(5k) is 15k, and 2*(3k) is 6k. So, subtracting those, 15k - 6k equals 9k.So now, the expression simplifies to 21k ÷ 9k. Wait, both the numerator and the denominator have a k term. Since k is not zero (because we can't divide by zero), I can safely cancel out the k terms. That leaves me with 21 ÷ 9.Now, simplifying 21 divided by 9. Both numbers are divisible by 3. So, dividing numerator and denominator by 3, I get 7/3. That's the simplified form of the fraction.Let me double-check my steps to make sure I didn't make a mistake. I expressed A, B, and C in terms of k, substituted them into the expression, simplified both the numerator and the denominator, canceled out the k, and then simplified the resulting fraction. Everything seems to add up.Wait, just to be thorough, let me plug in a specific value for k to see if the ratio holds. Let's say k = 2. Then A = 6, B = 4, and C = 10. Plugging these into the original expression:Numerator: 5*6 + 3*4 = 30 + 12 = 42Denominator: 3*10 - 2*6 = 30 - 12 = 18So, 42 ÷ 18 simplifies to 7/3. Yep, that's the same result as before. So, it seems my answer is correct.I think I'm confident that the value of (5A + 3B) ÷ (3C - 2A) is 7/3 when A:B:C is 3:2:5.

💡Alright, I've got this problem here about two functions intersecting. Let me try to understand what it's asking and figure out how to solve it step by step.First, the problem mentions a linear function ( y_1 = x - a + 2 ) and an inverse proportion function ( y_2 = frac{k}{x} ) where ( k neq 0 ). It says these two functions intersect, which means they have at least one point in common.**Part (1): Determine whether ( y_2 ) passes through the point ( (k, 1) ).**Okay, so I need to check if the point ( (k, 1) ) lies on the graph of ( y_2 ). To do this, I can substitute ( x = k ) and ( y = 1 ) into the equation of ( y_2 ) and see if it holds true.Let's plug in ( x = k ) into ( y_2 ):[ y_2 = frac{k}{k} ]Simplifying that, we get:[ y_2 = 1 ]So, when ( x = k ), ( y_2 = 1 ). That means the point ( (k, 1) ) does lie on the graph of ( y_2 ). Therefore, the answer to part (1) is yes.**Part (2): If the graph of ( y_1 ) passes through the point ( (k, 1) ), and ( 2a + k = 5 ).**Alright, now we have additional information. The linear function ( y_1 ) also passes through ( (k, 1) ), and there's a relationship between ( a ) and ( k ) given by ( 2a + k = 5 ).**① Find the expression of ( y_2 ).**To find ( y_2 ), I need to determine the value of ( k ). Since ( y_1 ) passes through ( (k, 1) ), I can substitute ( x = k ) and ( y = 1 ) into the equation of ( y_1 ):[ 1 = k - a + 2 ]Simplifying this equation:[ 1 = k - a + 2 ]Subtract 2 from both sides:[ -1 = k - a ]So, we have:[ k - a = -1 ]Now, we also know from the problem that:[ 2a + k = 5 ]Now, we have a system of two equations:1. ( k - a = -1 )2. ( 2a + k = 5 )Let me solve this system. I can use substitution or elimination. Let's try elimination.From the first equation, I can express ( k ) in terms of ( a ):[ k = a - 1 ]Now, substitute this expression for ( k ) into the second equation:[ 2a + (a - 1) = 5 ]Simplify:[ 2a + a - 1 = 5 ][ 3a - 1 = 5 ]Add 1 to both sides:[ 3a = 6 ]Divide both sides by 3:[ a = 2 ]Now, substitute ( a = 2 ) back into the expression for ( k ):[ k = 2 - 1 = 1 ]So, ( k = 1 ). Therefore, the expression for ( y_2 ) is:[ y_2 = frac{1}{x} ]**② Compare the values of ( y_1 ) and ( y_2 ) when ( x > 0 ).**Now that we have ( a = 2 ) and ( k = 1 ), let's write down the expressions for both functions:- ( y_1 = x - 2 + 2 = x )- ( y_2 = frac{1}{x} )So, ( y_1 = x ) and ( y_2 = frac{1}{x} ). We need to compare these two functions for ( x > 0 ).Let's analyze the behavior of both functions:1. **When ( 0 < x < 1 ):** - ( y_1 = x ) will be between 0 and 1. - ( y_2 = frac{1}{x} ) will be greater than 1. - Therefore, ( y_1 < y_2 ) in this interval.2. **When ( x = 1 ):** - ( y_1 = 1 ) - ( y_2 = 1 ) - So, ( y_1 = y_2 ) at ( x = 1 ).3. **When ( x > 1 ):** - ( y_1 = x ) will be greater than 1. - ( y_2 = frac{1}{x} ) will be between 0 and 1. - Therefore, ( y_1 > y_2 ) in this interval.To summarize:- For ( 0 < x < 1 ), ( y_1 < y_2 ).- At ( x = 1 ), ( y_1 = y_2 ).- For ( x > 1 ), ( y_1 > y_2 ).This makes sense because the linear function ( y_1 = x ) increases steadily, while the inverse proportion function ( y_2 = frac{1}{x} ) decreases as ( x ) increases.I think that covers both parts of the problem. I should double-check my calculations to make sure I didn't make any mistakes.For part (1), substituting ( x = k ) into ( y_2 ) gives ( y = 1 ), which matches the point ( (k, 1) ). So that's correct.For part (2), solving the system of equations gave me ( a = 2 ) and ( k = 1 ), which seems right. Plugging these back into the original functions confirms that ( y_1 ) and ( y_2 ) pass through ( (1, 1) ) and intersect there.Comparing the functions for ( x > 0 ) also seems logical based on their behaviors. The linear function will eventually overtake the inverse proportion function as ( x ) increases, and vice versa for smaller ( x ).Overall, I think my solutions are correct.

💡First, I need to understand the operation defined as ( n heartsuit m = kn^3m^2 ) where ( k = 3 ). This means that for any two numbers ( n ) and ( m ), the operation ( heartsuit ) multiplies ( n ) cubed by ( m ) squared and then by the constant ( k ), which is 3 in this case.Next, I need to compute ( 3 heartsuit 5 ). Using the definition of the operation:[3 heartsuit 5 = 3 times 3^3 times 5^2]Calculating each part step by step:[3^3 = 27 quad text{and} quad 5^2 = 25]So,[3 heartsuit 5 = 3 times 27 times 25 = 2025]Then, I need to compute ( 5 heartsuit 3 ):[5 heartsuit 3 = 3 times 5^3 times 3^2]Again, calculating each part:[5^3 = 125 quad text{and} quad 3^2 = 9]So,[5 heartsuit 3 = 3 times 125 times 9 = 3375]Now, I need to find the ratio of ( 3 heartsuit 5 ) to ( 5 heartsuit 3 ):[frac{3 heartsuit 5}{5 heartsuit 3} = frac{2025}{3375}]Simplifying this fraction by dividing both numerator and denominator by 675:[frac{2025 div 675}{3375 div 675} = frac{3}{5}]Therefore, the value of ( frac{3heartsuit 5}{5heartsuit 3} ) is ( frac{3}{5} ).

💡Okay, so I have this problem here: For positive real numbers ( x ) and ( y ) satisfying ( x + y = 1 ), I need to find the minimum value of the expression ( frac{1 + y}{x} + frac{1}{y} ). The options are A: ( 3 + sqrt{2} ), B: ( 2 + 2sqrt{2} ), C: 5, and D: ( frac{11}{2} ).Alright, let me try to figure this out step by step. First, since ( x ) and ( y ) are positive real numbers and their sum is 1, I can express one variable in terms of the other. Maybe that will help simplify the expression.So, if ( x + y = 1 ), then ( x = 1 - y ). That means I can substitute ( x ) in the expression with ( 1 - y ). Let me try that.Substituting ( x = 1 - y ) into the expression:[frac{1 + y}{x} + frac{1}{y} = frac{1 + y}{1 - y} + frac{1}{y}]Hmm, that looks a bit complicated. Maybe I can combine these two fractions into one to make it easier to handle. Let me find a common denominator.The denominators are ( 1 - y ) and ( y ), so the common denominator would be ( y(1 - y) ). Let me rewrite each term with this common denominator:[frac{(1 + y) cdot y}{y(1 - y)} + frac{1 cdot (1 - y)}{y(1 - y)} = frac{y(1 + y) + (1 - y)}{y(1 - y)}]Now, let's expand the numerator:First term: ( y(1 + y) = y + y^2 )Second term: ( 1 - y )So, adding them together:[y + y^2 + 1 - y = y^2 + 1]Wait, the ( y ) and ( -y ) cancel out, so the numerator simplifies to ( y^2 + 1 ). So now, the expression becomes:[frac{y^2 + 1}{y(1 - y)}]Hmm, that's a bit simpler. So, now I have:[frac{y^2 + 1}{y(1 - y)}]I need to find the minimum value of this expression for ( y ) in the interval ( (0, 1) ) since ( x ) and ( y ) are positive and their sum is 1.Maybe I can take the derivative of this function with respect to ( y ) and find its critical points. That should help me find the minimum.Let me denote the function as:[f(y) = frac{y^2 + 1}{y(1 - y)}]First, let's simplify ( f(y) ) a bit more to make differentiation easier. Let's split the fraction:[f(y) = frac{y^2}{y(1 - y)} + frac{1}{y(1 - y)} = frac{y}{1 - y} + frac{1}{y(1 - y)}]Wait, that might not necessarily make it easier. Maybe I should just proceed with the quotient rule.The quotient rule states that if ( f(y) = frac{u(y)}{v(y)} ), then:[f'(y) = frac{u'(y)v(y) - u(y)v'(y)}{[v(y)]^2}]So, let me set ( u(y) = y^2 + 1 ) and ( v(y) = y(1 - y) = y - y^2 ).First, compute ( u'(y) ):[u'(y) = 2y]Next, compute ( v(y) ):[v(y) = y - y^2]Compute ( v'(y) ):[v'(y) = 1 - 2y]Now, plug these into the quotient rule:[f'(y) = frac{(2y)(y - y^2) - (y^2 + 1)(1 - 2y)}{(y - y^2)^2}]Let me compute the numerator step by step.First term: ( (2y)(y - y^2) = 2y^2 - 2y^3 )Second term: ( (y^2 + 1)(1 - 2y) ). Let's expand this:Multiply ( y^2 ) by ( 1 - 2y ): ( y^2 - 2y^3 )Multiply 1 by ( 1 - 2y ): ( 1 - 2y )So, adding them together: ( y^2 - 2y^3 + 1 - 2y )Therefore, the second term is ( y^2 - 2y^3 + 1 - 2y )Now, the numerator is:First term - Second term:[(2y^2 - 2y^3) - (y^2 - 2y^3 + 1 - 2y) = 2y^2 - 2y^3 - y^2 + 2y^3 - 1 + 2y]Simplify term by term:- ( 2y^2 - y^2 = y^2 )- ( -2y^3 + 2y^3 = 0 )- ( -1 )- ( +2y )So, the numerator simplifies to:[y^2 + 2y - 1]Therefore, the derivative ( f'(y) ) is:[f'(y) = frac{y^2 + 2y - 1}{(y - y^2)^2}]We need to find the critical points by setting ( f'(y) = 0 ). Since the denominator is always positive for ( y in (0,1) ), the critical points occur when the numerator is zero:[y^2 + 2y - 1 = 0]Let's solve this quadratic equation for ( y ):[y = frac{-2 pm sqrt{(2)^2 - 4(1)(-1)}}{2(1)} = frac{-2 pm sqrt{4 + 4}}{2} = frac{-2 pm sqrt{8}}{2} = frac{-2 pm 2sqrt{2}}{2} = -1 pm sqrt{2}]Since ( y ) must be positive and less than 1, we discard the negative solution:[y = -1 + sqrt{2} approx -1 + 1.414 approx 0.414]So, ( y = sqrt{2} - 1 ) is the critical point. Let's compute the corresponding ( x ):Since ( x + y = 1 ), ( x = 1 - y = 1 - (sqrt{2} - 1) = 2 - sqrt{2} ).Now, let's verify if this critical point gives a minimum. We can check the second derivative or analyze the behavior of the first derivative around this point.Alternatively, since the function tends to infinity as ( y ) approaches 0 or 1, and there's only one critical point in the interval ( (0,1) ), this critical point must correspond to a minimum.Now, let's compute the value of the expression at ( y = sqrt{2} - 1 ):First, compute ( x = 2 - sqrt{2} ).Compute ( frac{1 + y}{x} + frac{1}{y} ):Let me compute each term separately.Compute ( 1 + y = 1 + (sqrt{2} - 1) = sqrt{2} ).So, ( frac{1 + y}{x} = frac{sqrt{2}}{2 - sqrt{2}} ).Let me rationalize the denominator:Multiply numerator and denominator by ( 2 + sqrt{2} ):[frac{sqrt{2}(2 + sqrt{2})}{(2 - sqrt{2})(2 + sqrt{2})} = frac{2sqrt{2} + 2}{4 - 2} = frac{2sqrt{2} + 2}{2} = sqrt{2} + 1]Next, compute ( frac{1}{y} = frac{1}{sqrt{2} - 1} ).Again, rationalize the denominator:Multiply numerator and denominator by ( sqrt{2} + 1 ):[frac{sqrt{2} + 1}{(sqrt{2} - 1)(sqrt{2} + 1)} = frac{sqrt{2} + 1}{2 - 1} = sqrt{2} + 1]Therefore, the expression ( frac{1 + y}{x} + frac{1}{y} ) becomes:[(sqrt{2} + 1) + (sqrt{2} + 1) = 2sqrt{2} + 2]So, the minimum value is ( 2 + 2sqrt{2} ), which corresponds to option B.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, when I substituted ( x = 1 - y ), that was correct. Then, I combined the fractions correctly, leading to ( frac{y^2 + 1}{y(1 - y)} ). Taking the derivative using the quotient rule, I got the numerator as ( y^2 + 2y - 1 ), which led to the critical point ( y = sqrt{2} - 1 ). Plugging back into the expression, I correctly rationalized the denominators and found the sum to be ( 2 + 2sqrt{2} ). That seems correct.Alternatively, maybe I can try using the AM-GM inequality to see if I can get the same result without calculus. Let me try that approach.Given ( x + y = 1 ), and we need to minimize ( frac{1 + y}{x} + frac{1}{y} ).Let me rewrite the expression:[frac{1 + y}{x} + frac{1}{y} = frac{1 + y}{x} + frac{1}{y}]Since ( x + y = 1 ), ( 1 = x + y ), so ( 1 + y = x + 2y ). Therefore, the expression becomes:[frac{x + 2y}{x} + frac{1}{y} = 1 + frac{2y}{x} + frac{1}{y}]So, the expression simplifies to:[1 + frac{2y}{x} + frac{1}{y}]Now, let me focus on the terms ( frac{2y}{x} ) and ( frac{1}{y} ). Maybe I can apply the AM-GM inequality to these two terms.Recall that the AM-GM inequality states that for non-negative real numbers ( a ) and ( b ), the arithmetic mean is greater than or equal to the geometric mean:[frac{a + b}{2} geq sqrt{ab}]Equality holds when ( a = b ).So, let me set ( a = frac{2y}{x} ) and ( b = frac{1}{y} ). Then:[frac{frac{2y}{x} + frac{1}{y}}{2} geq sqrt{frac{2y}{x} cdot frac{1}{y}} = sqrt{frac{2}{x}}]Wait, that doesn't seem directly helpful because it still involves ( x ). Maybe I need to adjust my approach.Alternatively, perhaps I can consider the entire expression ( frac{2y}{x} + frac{1}{y} ) and try to apply AM-GM to it.Let me denote ( A = frac{2y}{x} ) and ( B = frac{1}{y} ). Then, the expression is ( A + B ).But I need to relate ( A ) and ( B ) in a way that allows me to apply AM-GM. Maybe I can express ( A ) and ( B ) in terms of a single variable.Since ( x + y = 1 ), ( x = 1 - y ). So, ( A = frac{2y}{1 - y} ) and ( B = frac{1}{y} ).Alternatively, maybe I can use substitution to express everything in terms of ( y ) and then apply calculus or another inequality.Wait, I already did the calculus approach and got ( 2 + 2sqrt{2} ). Maybe I can see if that's consistent with AM-GM.Let me try to write the expression ( frac{2y}{x} + frac{1}{y} ) as a sum of terms that can be bounded below by AM-GM.Let me consider ( frac{2y}{x} ) as two separate terms: ( frac{y}{x} + frac{y}{x} ). Then, the expression becomes:[frac{y}{x} + frac{y}{x} + frac{1}{y}]Now, I have three terms: ( frac{y}{x} ), ( frac{y}{x} ), and ( frac{1}{y} ). Let's apply AM-GM to these three terms.The AM-GM inequality for three terms states that:[frac{frac{y}{x} + frac{y}{x} + frac{1}{y}}{3} geq sqrt[3]{frac{y}{x} cdot frac{y}{x} cdot frac{1}{y}} = sqrt[3]{frac{y^2}{x^2 y}} = sqrt[3]{frac{y}{x^2}}]Hmm, that doesn't seem to directly give me a constant lower bound. Maybe this approach isn't the best.Alternatively, perhaps I can use the method of Lagrange multipliers, but that might be overkill for this problem.Wait, another idea: since ( x + y = 1 ), maybe I can express ( x = 1 - y ) and substitute into the expression, then find the minimum using calculus as I did before. But I already did that and got the correct answer.Alternatively, maybe I can use substitution to express everything in terms of a single variable and then complete the square or something like that.Let me try expressing the original expression in terms of ( y ):[frac{1 + y}{x} + frac{1}{y} = frac{1 + y}{1 - y} + frac{1}{y}]Let me denote ( t = y ), so ( t in (0,1) ). Then, the expression becomes:[frac{1 + t}{1 - t} + frac{1}{t}]Let me combine these two terms into one fraction:[frac{(1 + t) cdot t + (1 - t)}{t(1 - t)} = frac{t + t^2 + 1 - t}{t(1 - t)} = frac{t^2 + 1}{t(1 - t)}]So, the expression simplifies to:[frac{t^2 + 1}{t(1 - t)}]Now, let me denote this as ( f(t) = frac{t^2 + 1}{t(1 - t)} ). To find the minimum, I can take the derivative as I did before, but maybe there's a smarter way.Alternatively, let me consider the function ( f(t) ) and see if I can write it in a form that allows me to apply some inequality.Let me try to write ( f(t) ) as:[f(t) = frac{t^2 + 1}{t(1 - t)} = frac{t^2 + 1}{t - t^2}]Hmm, not sure if that helps. Maybe I can perform polynomial division or something.Alternatively, let me consider that ( t^2 + 1 = t^2 + 1 ), and ( t(1 - t) = t - t^2 ). Maybe I can write ( t^2 + 1 = (t - t^2) + (1 + t^2 - t + t^2) ). Wait, that seems convoluted.Alternatively, maybe I can express ( t^2 + 1 ) as ( (t^2 - 2t + 1) + 2t ), which is ( (t - 1)^2 + 2t ). Then, the expression becomes:[frac{(t - 1)^2 + 2t}{t(1 - t)} = frac{(1 - t)^2 + 2t}{t(1 - t)} = frac{(1 - t)^2}{t(1 - t)} + frac{2t}{t(1 - t)} = frac{1 - t}{t} + frac{2}{1 - t}]So, ( f(t) = frac{1 - t}{t} + frac{2}{1 - t} ).Now, let me denote ( a = frac{1 - t}{t} ) and ( b = frac{2}{1 - t} ). So, ( f(t) = a + b ).I can try to apply AM-GM to ( a ) and ( b ). Let me see:The AM-GM inequality states that ( frac{a + b}{2} geq sqrt{ab} ).So,[frac{frac{1 - t}{t} + frac{2}{1 - t}}{2} geq sqrt{frac{1 - t}{t} cdot frac{2}{1 - t}} = sqrt{frac{2}{t}}]Hmm, that gives me a lower bound involving ( t ), which isn't helpful because I need a constant lower bound.Alternatively, maybe I can consider the two terms ( frac{1 - t}{t} ) and ( frac{2}{1 - t} ) and see if I can find a substitution that allows me to apply AM-GM more effectively.Let me set ( u = frac{1 - t}{t} ). Then, ( u = frac{1}{t} - 1 ), so ( t = frac{1}{u + 1} ).Also, ( 1 - t = 1 - frac{1}{u + 1} = frac{u}{u + 1} ).So, the second term ( frac{2}{1 - t} = frac{2(u + 1)}{u} ).Therefore, the expression ( f(t) = u + frac{2(u + 1)}{u} ).Simplify:[f(t) = u + frac{2u + 2}{u} = u + 2 + frac{2}{u}]So, ( f(t) = u + 2 + frac{2}{u} ).Now, this looks more manageable. Let me denote ( g(u) = u + frac{2}{u} ). Then, ( f(t) = g(u) + 2 ).Now, I can apply AM-GM to ( g(u) ). The AM-GM inequality for two positive numbers ( u ) and ( frac{2}{u} ) states that:[frac{u + frac{2}{u}}{2} geq sqrt{u cdot frac{2}{u}} = sqrt{2}]Multiplying both sides by 2:[u + frac{2}{u} geq 2sqrt{2}]Therefore,[g(u) geq 2sqrt{2}]So,[f(t) = g(u) + 2 geq 2sqrt{2} + 2]Thus, the minimum value of ( f(t) ) is ( 2 + 2sqrt{2} ), which occurs when ( u = frac{2}{u} ), i.e., when ( u = sqrt{2} ).Recall that ( u = frac{1 - t}{t} ), so setting ( u = sqrt{2} ):[sqrt{2} = frac{1 - t}{t} implies sqrt{2} t = 1 - t implies t(sqrt{2} + 1) = 1 implies t = frac{1}{sqrt{2} + 1}]Rationalizing the denominator:[t = frac{1}{sqrt{2} + 1} cdot frac{sqrt{2} - 1}{sqrt{2} - 1} = frac{sqrt{2} - 1}{2 - 1} = sqrt{2} - 1]So, ( y = t = sqrt{2} - 1 ), which matches the critical point I found earlier. Therefore, the minimum value is indeed ( 2 + 2sqrt{2} ).Just to make sure, let me plug ( y = sqrt{2} - 1 ) back into the original expression:[frac{1 + y}{x} + frac{1}{y} = frac{1 + (sqrt{2} - 1)}{1 - (sqrt{2} - 1)} + frac{1}{sqrt{2} - 1}]Simplify:[frac{sqrt{2}}{2 - sqrt{2}} + frac{1}{sqrt{2} - 1}]Rationalizing both terms:First term:[frac{sqrt{2}}{2 - sqrt{2}} cdot frac{2 + sqrt{2}}{2 + sqrt{2}} = frac{2sqrt{2} + 2}{4 - 2} = frac{2sqrt{2} + 2}{2} = sqrt{2} + 1]Second term:[frac{1}{sqrt{2} - 1} cdot frac{sqrt{2} + 1}{sqrt{2} + 1} = frac{sqrt{2} + 1}{2 - 1} = sqrt{2} + 1]Adding them together:[(sqrt{2} + 1) + (sqrt{2} + 1) = 2sqrt{2} + 2]Yes, that confirms the minimum value is ( 2 + 2sqrt{2} ), which is option B.I think I've thoroughly checked this problem using both calculus and the AM-GM inequality, and both methods lead me to the same answer. So, I'm confident that the correct answer is B: ( 2 + 2sqrt{2} ).

💡First, I need to determine the total increase in the price of the desk. The price started at 15 and ended at 65, so the increase is 65 - 15 = 50.Next, I know that the auctioneer raises the price by 5 for each new bid. To find out how many bids were made in total, I divide the total price increase by the amount raised per bid: 50 / 5 = 10 bids.Each person involved in the bidding war placed 5 bids. To find out how many people were involved, I divide the total number of bids by the number of bids each person made: 10 bids / 5 bids per person = 2 people.Therefore, there were 2 people involved in the bidding war.