💡Okay, so I have this problem here about a sequence and its sums. Let me try to understand what's being asked. We are given that ( S_n ) is the sum of the first ( n ) terms of the sequence ( {a_n} ). We know that ( S_3 = 1 ) and ( S_4 = 11 ). Also, there's a recurrence relation given: ( a_{n+3} = 2a_n ) for all positive integers ( n ). The question is asking us to find ( S_{3n+1} ).Alright, let's break this down step by step.First, since ( S_3 = 1 ) and ( S_4 = 11 ), we can find ( a_4 ) by subtracting ( S_3 ) from ( S_4 ). That is, ( a_4 = S_4 - S_3 = 11 - 1 = 10 ). So, ( a_4 = 10 ).Next, the recurrence relation ( a_{n+3} = 2a_n ) tells us that every term three places ahead is twice the current term. So, for example, ( a_7 = 2a_4 ), ( a_{10} = 2a_7 = 4a_4 ), and so on. This seems like a geometric progression with a common ratio of 2, but it's every third term. Let me think about how this affects the sequence. If we look at terms ( a_1, a_4, a_7, a_{10}, ldots ), each term is double the previous one. Similarly, terms ( a_2, a_5, a_8, a_{11}, ldots ) would form another geometric sequence, and the same with ( a_3, a_6, a_9, a_{12}, ldots ).So, if I consider the terms ( a_4, a_7, a_{10}, ldots ), they form a geometric sequence where each term is 2 times the previous one. Since ( a_4 = 10 ), the next term ( a_7 = 2 times 10 = 20 ), then ( a_{10} = 40 ), and so on. So, in general, ( a_{3n+1} = 10 times 2^{n-1} ). Wait, let me check that.If ( n = 1 ), then ( a_{4} = 10 times 2^{0} = 10 ), which is correct. For ( n = 2 ), ( a_{7} = 10 times 2^{1} = 20 ), which is also correct. So yes, ( a_{3n+1} = 10 times 2^{n-1} ).Now, since ( S_{3n+1} ) is the sum of the first ( 3n+1 ) terms, we need to find a way to express this sum in terms of ( n ). Let's think about how the sequence is structured.The sequence ( {a_n} ) can be divided into three interleaved geometric sequences:1. The first sequence: ( a_1, a_4, a_7, ldots )2. The second sequence: ( a_2, a_5, a_8, ldots )3. The third sequence: ( a_3, a_6, a_9, ldots )Each of these sequences has a common ratio of 2. However, we don't know the first terms of the second and third sequences yet. We only know ( a_4 = 10 ) and ( S_3 = 1 ).Wait, ( S_3 = a_1 + a_2 + a_3 = 1 ). So, the sum of the first three terms is 1. But we don't know the individual terms ( a_1, a_2, a_3 ). Hmm, that complicates things.But maybe we can express ( S_{3n+1} ) in terms of these sequences. Let's consider how many terms each of the three interleaved sequences contributes to ( S_{3n+1} ).For ( S_{3n+1} ), we have:- The first sequence contributes ( n+1 ) terms: ( a_1, a_4, a_7, ldots, a_{3n+1} )- The second sequence contributes ( n ) terms: ( a_2, a_5, a_8, ldots, a_{3n-1} )- The third sequence contributes ( n ) terms: ( a_3, a_6, a_9, ldots, a_{3n} )So, ( S_{3n+1} = ) (sum of first ( n+1 ) terms of first sequence) + (sum of first ( n ) terms of second sequence) + (sum of first ( n ) terms of third sequence)But we don't know the first terms of the second and third sequences. However, we do know that each of these sequences is a geometric sequence with ratio 2.Let me denote:- Let ( b_k ) be the first sequence: ( b_1 = a_1, b_2 = a_4, b_3 = a_7, ldots )- Let ( c_k ) be the second sequence: ( c_1 = a_2, c_2 = a_5, c_3 = a_8, ldots )- Let ( d_k ) be the third sequence: ( d_1 = a_3, d_2 = a_6, d_3 = a_9, ldots )Each of these sequences ( b_k, c_k, d_k ) is a geometric sequence with ratio 2.We know that ( b_2 = a_4 = 10 ). Since ( b_2 = 2b_1 ), we have ( 10 = 2b_1 ), so ( b_1 = 5 ). Therefore, the first sequence ( b_k ) has first term 5 and ratio 2.Similarly, we can find the first terms of ( c_k ) and ( d_k ). But wait, we don't have enough information yet. We only know ( S_3 = 1 ), which is ( a_1 + a_2 + a_3 = 1 ). Since ( a_1 = b_1 = 5 ), then ( 5 + a_2 + a_3 = 1 ), so ( a_2 + a_3 = -4 ).But ( c_1 = a_2 ) and ( d_1 = a_3 ). So, ( c_1 + d_1 = -4 ). Hmm, that's interesting. So, the sum of the first terms of the second and third sequences is -4.But each of these sequences is geometric with ratio 2. So, ( c_k = c_1 times 2^{k-1} ) and ( d_k = d_1 times 2^{k-1} ).Now, let's try to express ( S_{3n+1} ) in terms of these sequences.As I thought earlier, ( S_{3n+1} = ) (sum of first ( n+1 ) terms of ( b_k )) + (sum of first ( n ) terms of ( c_k )) + (sum of first ( n ) terms of ( d_k )).Let me write that out:( S_{3n+1} = sum_{k=1}^{n+1} b_k + sum_{k=1}^{n} c_k + sum_{k=1}^{n} d_k )Since each of these is a geometric series, we can use the formula for the sum of a geometric series:Sum of first ( m ) terms of a geometric sequence with first term ( t ) and ratio ( r ) is ( t times frac{r^m - 1}{r - 1} ).So, let's compute each sum:1. Sum of first ( n+1 ) terms of ( b_k ):( sum_{k=1}^{n+1} b_k = b_1 times frac{2^{n+1} - 1}{2 - 1} = 5 times (2^{n+1} - 1) )2. Sum of first ( n ) terms of ( c_k ):( sum_{k=1}^{n} c_k = c_1 times frac{2^n - 1}{2 - 1} = c_1 times (2^n - 1) )3. Sum of first ( n ) terms of ( d_k ):( sum_{k=1}^{n} d_k = d_1 times frac{2^n - 1}{2 - 1} = d_1 times (2^n - 1) )So, putting it all together:( S_{3n+1} = 5(2^{n+1} - 1) + c_1(2^n - 1) + d_1(2^n - 1) )We can factor out ( (2^n - 1) ) from the last two terms:( S_{3n+1} = 5(2^{n+1} - 1) + (c_1 + d_1)(2^n - 1) )Earlier, we found that ( c_1 + d_1 = -4 ). So, substituting that in:( S_{3n+1} = 5(2^{n+1} - 1) + (-4)(2^n - 1) )Let's expand this:First, expand ( 5(2^{n+1} - 1) ):( 5 times 2^{n+1} - 5 times 1 = 5 times 2^{n+1} - 5 )Then, expand ( (-4)(2^n - 1) ):( -4 times 2^n + 4 times 1 = -4 times 2^n + 4 )Now, combine these two results:( 5 times 2^{n+1} - 5 - 4 times 2^n + 4 )Simplify the constants:( 5 times 2^{n+1} - 4 times 2^n - 1 )Note that ( 5 times 2^{n+1} = 5 times 2 times 2^n = 10 times 2^n ). So, substituting that:( 10 times 2^n - 4 times 2^n - 1 )Combine like terms:( (10 - 4) times 2^n - 1 = 6 times 2^n - 1 )Wait, that simplifies to ( 6 times 2^n - 1 ). But let me double-check my steps to make sure I didn't make a mistake.Starting from:( S_{3n+1} = 5(2^{n+1} - 1) + (-4)(2^n - 1) )Expanding:( 5 times 2^{n+1} - 5 - 4 times 2^n + 4 )Combine constants: ( -5 + 4 = -1 )Combine the terms with ( 2^n ):( 5 times 2^{n+1} - 4 times 2^n = 5 times 2 times 2^n - 4 times 2^n = 10 times 2^n - 4 times 2^n = 6 times 2^n )So, altogether: ( 6 times 2^n - 1 )Hmm, but wait, the initial problem statement says ( a_{n+3} = 2a_n ). So, each term three apart is double. So, the sequences ( b_k, c_k, d_k ) are each geometric with ratio 2, which we used correctly.But let me check if the sum ( S_{3n+1} ) is correctly broken down into the three sequences.Yes, because ( S_{3n+1} ) includes ( n+1 ) terms from the first sequence (since it's every third term starting at 1), and ( n ) terms from the second and third sequences.So, the calculation seems correct. Therefore, ( S_{3n+1} = 6 times 2^n - 1 ).But wait, let me test this with a small value of ( n ) to see if it makes sense.Let's take ( n = 1 ). Then, ( S_{3(1)+1} = S_4 = 11 ).Plugging ( n = 1 ) into our formula: ( 6 times 2^1 - 1 = 12 - 1 = 11 ). That matches.Now, let's take ( n = 2 ). Then, ( S_{3(2)+1} = S_7 ).We can compute ( S_7 ) step by step.We know ( S_3 = 1 ), ( S_4 = 11 ).Then, ( a_5 = 2a_2 ), ( a_6 = 2a_3 ), ( a_7 = 2a_4 = 20 ).So, ( S_5 = S_4 + a_5 = 11 + 2a_2 )( S_6 = S_5 + a_6 = 11 + 2a_2 + 2a_3 )( S_7 = S_6 + a_7 = 11 + 2a_2 + 2a_3 + 20 )But we know that ( a_2 + a_3 = -4 ), so ( 2a_2 + 2a_3 = 2(-4) = -8 ).Therefore, ( S_7 = 11 - 8 + 20 = 23 ).Now, using our formula for ( n = 2 ): ( 6 times 2^2 - 1 = 6 times 4 - 1 = 24 - 1 = 23 ). That matches as well.Let's try ( n = 3 ). Then, ( S_{10} ).Compute ( S_{10} ):We have:( a_8 = 2a_5 = 2(2a_2) = 4a_2 )( a_9 = 2a_6 = 2(2a_3) = 4a_3 )( a_{10} = 2a_7 = 2(20) = 40 )So, ( S_7 = 23 )( S_8 = S_7 + a_8 = 23 + 4a_2 )( S_9 = S_8 + a_9 = 23 + 4a_2 + 4a_3 )( S_{10} = S_9 + a_{10} = 23 + 4a_2 + 4a_3 + 40 )Again, ( a_2 + a_3 = -4 ), so ( 4a_2 + 4a_3 = 4(-4) = -16 )Thus, ( S_{10} = 23 - 16 + 40 = 47 )Using our formula for ( n = 3 ): ( 6 times 2^3 - 1 = 6 times 8 - 1 = 48 - 1 = 47 ). Perfect, that matches too.So, it seems our formula ( S_{3n+1} = 6 times 2^n - 1 ) is correct.But wait, in the initial problem statement, the user wrote:"Given that ( S_{3}=1 ), ( S_{4}=11 ), ( a_{n+3}=2a_{n}(n∈N^{∗}) ), find ( S_{3n+1}= )____."But in the assistant's initial solution, they arrived at ( 3 times 2^{n+1} - 1 ), which is equal to ( 6 times 2^n - 1 ), which is the same as what I got. So, both methods lead to the same result.Wait, but in the assistant's solution, they said:"The sequence ( {S_{3n}} ) is also a geometric sequence with first term ( S_3 = 1 ) and common ratio 2. Hence, ( S_{3n} = 2^{n} ). Therefore, ( S_{3n+1} = S_{3n} + a_{3n+1} = 2^{n} + 10 times 2^{n-1} = 2^{n} + 5 times 2^{n} = 6 times 2^{n} ). Wait, but that doesn't match the previous conclusion."Wait, hold on, the assistant's initial solution seems to have a miscalculation. Let me check.The assistant wrote:"The sequence ( {S_{3n}} ) is also a geometric sequence with first term ( S_3 = 1 ) and common ratio 2. Hence, ( S_{3n} = frac{2^{n} - 1}{2 - 1} = 2^{n} - 1 ). Therefore, ( S_{3n+1} = S_{3n} + a_{3n+1} = (2^{n} - 1) + 10 times 2^{n - 1} = 2^{n} - 1 + 5 times 2^{n} = 6 times 2^{n} - 1 )."Wait, actually, that's correct. So, the assistant's initial solution is correct, arriving at ( 6 times 2^n - 1 ), which is the same as my result.But in the initial problem statement, the user wrote the assistant's solution as:"Given that ( S_{3}=1 ) and ( S_{4}=11 ), we have ( a_{4}=S_{4}-S_{3}=10 ). Given that ( a_{n+3}=2a_{n}(n∈N^{∗}) ), we have ( a_{3n+1}=2a_{3n-2} ). The sequence ( {a_{3n-2}} ) is a geometric sequence with first term ( a_{4}=10 ) and common ratio 2. Therefore, ( a_{3n-2}=a_{4}×2^{n-2}=10×2^{n-2} ). The sequence ( {S_{3n}} ) is also a geometric sequence with first term ( S_{3}=1 ) and common ratio 2. Hence, ( S_{3n+1}=S_{3n}+a_{3n+1}= frac {2^{n}-1}{2-1}+10×2^{n-1}= boxed{3×2^{n+1}-1} )."Wait, hold on, in the assistant's solution, they wrote ( S_{3n} = frac{2^{n} - 1}{2 - 1} = 2^{n} - 1 ). Then, ( S_{3n+1} = S_{3n} + a_{3n+1} = (2^{n} - 1) + 10 times 2^{n - 1} ). Calculating that:( 2^{n} - 1 + 10 times 2^{n - 1} = 2^{n} - 1 + 5 times 2^{n} = 6 times 2^{n} - 1 ).But in the assistant's solution, they wrote ( 3 times 2^{n+1} - 1 ). Wait, ( 3 times 2^{n+1} = 3 times 2 times 2^{n} = 6 times 2^{n} ). So, ( 3 times 2^{n+1} - 1 = 6 times 2^{n} - 1 ). So, both expressions are equivalent.Therefore, the assistant's solution is correct, and my calculation also leads to the same result. So, the final answer is ( 6 times 2^{n} - 1 ), which can also be written as ( 3 times 2^{n+1} - 1 ).But to make sure, let's see if ( 6 times 2^{n} - 1 ) is the same as ( 3 times 2^{n+1} - 1 ).Yes, because ( 3 times 2^{n+1} = 3 times 2 times 2^{n} = 6 times 2^{n} ). So, both expressions are equivalent.Therefore, the answer is ( 3 times 2^{n+1} - 1 ), which is the same as ( 6 times 2^{n} - 1 ).So, in conclusion, ( S_{3n+1} = 3 times 2^{n+1} - 1 ).**Final Answer**The value of ( S_{3n+1} ) is boxed{3 times 2^{n+1} - 1}.

💡Okay, so I have this problem here: the function is given as f(x) = 3cos(πx + θ), and it's stated that f(x) equals f(2 - x) for any x. I need to find what f(1) equals, and the options are A: ±3, B: 0, C: 3, D: -3.Hmm, let's break this down. First, the function is a cosine function with some amplitude and phase shift. The general form is A*cos(Bx + C), right? In this case, A is 3, B is π, and C is θ. So, f(x) = 3cos(πx + θ).Now, the condition given is f(x) = f(2 - x) for any x. That means if I plug in any x into the function, it should give me the same result as plugging in (2 - x). So, for example, if x is 0, f(0) should equal f(2). If x is 1, f(1) should equal f(1), which is trivial, but for other values, it's non-trivial.This seems like a symmetry condition. If f(x) = f(2 - x), then the function is symmetric about the line x = 1. Because reflecting x over x = 1 gives 2 - x. So, the graph of f(x) is symmetric around x = 1.Now, for a cosine function, which is naturally symmetric about its peaks and troughs, this symmetry condition might impose some restrictions on the phase shift θ.Let me write down the condition: f(x) = f(2 - x). So,3cos(πx + θ) = 3cos(π(2 - x) + θ)Simplify the right side:3cos(2π - πx + θ)So, we have:cos(πx + θ) = cos(2π - πx + θ)Now, when is cos(A) equal to cos(B)? That's when A = B + 2πk or A = -B + 2πk for some integer k.So, either:1. πx + θ = 2π - πx + θ + 2πkor2. πx + θ = -(2π - πx + θ) + 2πkLet's consider the first case:πx + θ = 2π - πx + θ + 2πkSimplify:πx = 2π - πx + 2πkBring πx to the right:2πx = 2π + 2πkDivide both sides by 2π:x = 1 + kBut this has to hold for all x, which is only possible if k = 0 and the equation reduces to x = 1, but that's only true for x=1, not for all x. So, this case doesn't hold for all x, which is required.Now, let's consider the second case:πx + θ = -(2π - πx + θ) + 2πkSimplify the right side:-2π + πx - θ + 2πkSo, the equation becomes:πx + θ = -2π + πx - θ + 2πkSubtract πx from both sides:θ = -2π - θ + 2πkBring θ to the left:2θ = -2π + 2πkDivide both sides by 2:θ = -π + πkSo, θ is equal to -π + πk, where k is any integer.Since θ is a phase shift, it's defined modulo 2π. So, let's find θ within a 2π interval. Let's take k = 1:θ = -π + π(1) = 0k = 0:θ = -π + 0 = -πk = 2:θ = -π + 2π = πSo, θ can be 0, π, or -π, etc., but since cosine is periodic with period 2π, θ = 0 and θ = 2π are the same, θ = π and θ = -π are the same, etc.So, essentially, θ can be either 0 or π (mod 2π).Therefore, the function f(x) can be written as either:f(x) = 3cos(πx + 0) = 3cos(πx)orf(x) = 3cos(πx + π) = 3cos(πx + π)But cos(πx + π) is equal to -cos(πx), because cos(A + π) = -cos(A). So, f(x) can be either 3cos(πx) or -3cos(πx).So, now, we need to find f(1). Let's compute f(1) for both cases.Case 1: f(x) = 3cos(πx)f(1) = 3cos(π*1) = 3cos(π) = 3*(-1) = -3Case 2: f(x) = -3cos(πx)f(1) = -3cos(π*1) = -3*(-1) = 3So, depending on the value of θ, f(1) can be either 3 or -3. Therefore, f(1) equals ±3.Looking back at the options, A is ±3, which matches our conclusion.Wait, but let me double-check. Is there any other possibility? Because θ could be other angles, but we saw that θ has to be either 0 or π modulo 2π, so the function is either 3cos(πx) or -3cos(πx). Therefore, f(1) is either 3 or -3, so ±3 is the correct answer.Alternatively, another way to think about it is that since the function is symmetric about x=1, then x=1 must be either a maximum or a minimum point. For a cosine function, the maximum is at 0, and the minimum is at π. So, if the function is shifted such that x=1 corresponds to either a maximum or minimum, then f(1) would be either 3 or -3.Yes, that makes sense. So, f(1) is either the maximum or the minimum of the function, which are 3 and -3, respectively.Therefore, the answer is A: ±3.**Final Answer**boxed{A}

💡First, determine the number of non-technician workers by subtracting the number of technicians from the total number of workers.Next, calculate the total salary of the technicians by multiplying the number of technicians by their average salary.Then, calculate the total salary of the non-technicians by multiplying the number of non-technicians by their average salary.Add the total salaries of both groups to find the overall total salary of all workers.Finally, divide the overall total salary by the total number of workers to find the average salary of all workers.

💡Okay, so I have this problem where I need to find the arithmetic mean of the sums |a₁ - a₂| + |a₃ - a₄| + |a₅ - a₆| + |a₇ - a₈| + |a₉ - a₁₀| for all permutations of the integers 1 through 10. Hmm, that sounds a bit complicated, but let me break it down.First, I know that a permutation of the integers 1 through 10 is just any rearrangement of those numbers. So, for each permutation, I have to calculate the sum of the absolute differences between pairs of consecutive numbers: (a₁, a₂), (a₃, a₄), and so on up to (a₉, a₁₀). Then, I need to find the average of all these sums across all possible permutations.Since there are 10 numbers, the total number of permutations is 10 factorial, which is 10! = 3,628,800. That's a huge number, so calculating each permutation individually isn't feasible. I need a smarter way to approach this.I remember that when dealing with expected values or averages over all permutations, sometimes you can exploit symmetry or linearity of expectation. Maybe I can find the expected value of each |a_i - a_j| term and then sum them up.Let me think about just one pair, say |a₁ - a₂|. Since all permutations are equally likely, the distribution of a₁ and a₂ is the same as any other pair. So, the expected value of |a₁ - a₂| should be the same as the expected value of |a₃ - a₄|, and so on for all five pairs.Therefore, if I can find the expected value of |a - b| where a and b are two distinct randomly chosen numbers from 1 to 10, then I can multiply that by 5 to get the total expected sum.Okay, so let's focus on finding E[|a - b|] where a and b are distinct integers from 1 to 10. How do I compute that?Well, the expected value is the average of |a - b| over all possible pairs (a, b) where a ≠ b. Since there are 10 numbers, there are 10 × 9 = 90 possible ordered pairs (a, b). But since |a - b| is the same as |b - a|, we can consider unordered pairs, which would be 45. However, since in permutations, the order matters, maybe I should stick with ordered pairs.Wait, in permutations, each pair (a, b) is equally likely, so whether I consider ordered or unordered pairs, the average should be the same because each unordered pair contributes twice to the ordered pairs. So, maybe it doesn't matter. Let me confirm.If I compute the average over ordered pairs, I have 90 pairs, each with |a - b|. If I compute the average over unordered pairs, I have 45 pairs, each with |a - b|, but each unordered pair corresponds to two ordered pairs with the same |a - b|. So, the average over ordered pairs is the same as the average over unordered pairs.Therefore, I can compute the average over unordered pairs, which might be simpler.So, how many unordered pairs are there? For 10 numbers, it's C(10, 2) = 45.Now, for each possible difference d = 1, 2, ..., 9, how many pairs have |a - b| = d?For d = 1: The pairs are (1,2), (2,3), ..., (9,10). So, there are 9 such pairs.For d = 2: The pairs are (1,3), (2,4), ..., (8,10). So, 8 pairs.Similarly, for d = k, the number of pairs is 10 - k.So, in general, for each d from 1 to 9, there are (10 - d) pairs with |a - b| = d.Therefore, the total sum of |a - b| over all unordered pairs is the sum from d=1 to 9 of d × (10 - d).Let me compute that:Sum = 1×9 + 2×8 + 3×7 + 4×6 + 5×5 + 6×4 + 7×3 + 8×2 + 9×1Let me calculate each term:1×9 = 92×8 = 163×7 = 214×6 = 245×5 = 256×4 = 247×3 = 218×2 = 169×1 = 9Now, adding them up:9 + 16 = 2525 + 21 = 4646 + 24 = 7070 + 25 = 9595 + 24 = 119119 + 21 = 140140 + 16 = 156156 + 9 = 165So, the total sum is 165.Since there are 45 unordered pairs, the average |a - b| is 165 / 45.Calculating that: 165 ÷ 45 = 3.666... which is 11/3.Wait, 165 ÷ 45 simplifies to 11/3 because 45 × 3 = 135, and 165 - 135 = 30, so 30/45 = 2/3, so total is 3 + 2/3 = 11/3.So, the expected value E[|a - b|] is 11/3.But wait, earlier I considered unordered pairs, but in permutations, the pairs are ordered. Does that affect the expectation?Let me think. If I consider ordered pairs, each unordered pair contributes two ordered pairs, but with the same |a - b|. So, the average over ordered pairs would still be the same as the average over unordered pairs because each |a - b| is counted twice, but the total sum would be double, and the number of pairs would also be double. So, the average remains the same.Therefore, E[|a - b|] = 11/3.Since there are five such pairs in each permutation, the expected total sum is 5 × (11/3) = 55/3.So, the arithmetic mean of the sums is 55/3.Wait, let me double-check my calculations to make sure I didn't make a mistake.First, the total number of unordered pairs is 45, and the total sum of |a - b| is 165. 165 divided by 45 is indeed 11/3. Then, multiplying by 5 gives 55/3. That seems correct.Alternatively, I can think about the expected value for each pair and then sum them up. Since each pair is independent in terms of expectation, the linearity holds, so adding them up is valid.Therefore, I think the answer is 55/3.**Final Answer**The arithmetic mean is boxed{dfrac{55}{3}}.

💡First, I'll define the variables for sugar (S), flour (F), and baking soda (B).From the given ratios:1. Sugar to flour is 5:2, so S/F = 5/2.2. Flour to baking soda is 10:1, so F/B = 10/1.Using the second ratio, I can express B in terms of F:B = F/10.Next, if there were 60 more pounds of baking soda, the new ratio of flour to baking soda would be 8:1:F/(B + 60) = 8/1.Substituting B = F/10 into this equation:F/(F/10 + 60) = 8/1.Solving for F:F = 8(F/10 + 60)F = 8F/10 + 48010F = 8F + 48002F = 4800F = 2400.Now, using the first ratio to find S:S/2400 = 5/2S = 2400 * (5/2)S = 6000.Therefore, there are 6000 pounds of sugar stored in the room.

💡Let's denote the total number of students as ( n ).Given that 20% of the students are juniors, the number of juniors is:[ 0.2n ]The remaining 80% are seniors, so the number of seniors is:[ 0.8n ]The average score on the test was 86, so the total score for all students is:[ 86n ]The average score of the seniors was 85, so the total score contributed by the seniors is:[ 85 times 0.8n = 68n ]The total score contributed by the juniors is the difference between the total score and the seniors' score:[ 86n - 68n = 18n ]Since all juniors scored the same, the score each junior received is:[ frac{18n}{0.2n} = 90 ]Therefore, each junior received a score of 90 on the test.(boxed{textbf{(B)} 90})