answer：Alright, so I have this problem where I need to use the digits 1 through 7 to form a 7-digit number that's a multiple of 11. Hmm, okay, let me think about how to approach this.First, I remember that a number is divisible by 11 if the difference between the sum of its digits in the odd positions and the sum of its digits in the even positions is a multiple of 11, including zero. So, if I denote the sum of the digits in the odd positions as ( a ) and the sum in the even positions as ( b ), then ( a - b ) should be a multiple of 11.Given the digits 1, 2, 3, 4, 5, 6, 7, the total sum is ( 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 ). So, ( a + b = 28 ). Since ( a - b ) must be a multiple of 11, let's consider the possible differences. The possible multiples of 11 that make sense here are 0, 11, and -11. But since ( a ) and ( b ) are both positive and their sum is 28, the difference can't be too large. Let's see:If ( a - b = 11 ), then adding this to ( a + b = 28 ) gives ( 2a = 39 ), which would make ( a = 19.5 ). That's not possible because ( a ) has to be an integer. Similarly, if ( a - b = -11 ), then ( 2a = 17 ), which also isn't an integer. So, the only feasible option is ( a - b = 0 ), meaning ( a = b = 14 ).Okay, so both the sum of the digits in the odd positions and the sum in the even positions must be 14. Now, I need to figure out how many ways I can split the digits 1 through 7 into two groups of four and three digits, respectively, each summing to 14.Let me list all possible combinations of three digits from 1 to 7 that add up to 14:1. 1, 6, 7: ( 1 + 6 + 7 = 14 )2. 2, 5, 7: ( 2 + 5 + 7 = 14 )3. 3, 4, 7: ( 3 + 4 + 7 = 14 )4. 3, 5, 6: ( 3 + 5 + 6 = 14 )So, there are four valid combinations for the three digits that sum to 14. Each of these combinations will occupy the odd positions, and the remaining four digits will occupy the even positions.Now, for each combination, I need to calculate how many ways I can arrange the digits. First, for the three digits in the odd positions: there are 4 odd positions in a 7-digit number (positions 1, 3, 5, 7). Wait, no, actually, in a 7-digit number, the positions are 1 to 7. So, the odd positions are 1, 3, 5, 7 – that's four positions. But I only have three digits to place here. Hmm, that doesn't seem right. Wait, no, actually, the number of digits is 7, so if I'm splitting them into odd and even positions, I need to clarify how many digits go into each.Wait, in a 7-digit number, there are four odd positions (1st, 3rd, 5th, 7th) and three even positions (2nd, 4th, 6th). So, I need to place four digits in the odd positions and three digits in the even positions. But earlier, I thought that both sums ( a ) and ( b ) are 14. But if ( a ) is the sum of four digits and ( b ) is the sum of three digits, and both are 14, that doesn't add up because ( 14 + 14 = 28 ), which is the total sum. So, that's correct.But wait, the problem is that I initially thought of ( a ) as the sum of three digits, but actually, since there are four odd positions, ( a ) is the sum of four digits, and ( b ) is the sum of three digits. So, I need to find all sets of four digits that sum to 14, and the remaining three digits will also sum to 14.Wait, that changes things. Earlier, I was looking for three digits that sum to 14, but actually, I need four digits that sum to 14. Let me correct that.So, let's find all combinations of four digits from 1 to 7 that add up to 14.Let me list them:1. 1, 2, 4, 7: ( 1 + 2 + 4 + 7 = 14 )2. 1, 2, 5, 6: ( 1 + 2 + 5 + 6 = 14 )3. 1, 3, 4, 6: ( 1 + 3 + 4 + 6 = 14 )4. 2, 3, 4, 5: ( 2 + 3 + 4 + 5 = 14 )Are there more? Let me check:- 1, 3, 5, 5: Not possible, duplicates.- 1, 4, 5, 4: Duplicates.- 2, 2, 5, 5: Duplicates.- 3, 3, 4, 4: Duplicates.So, I think these four combinations are the only ones.Wait, let me double-check:- 1, 2, 3, 8: But 8 is not in our digits.- 1, 2, 4, 7: Already listed.- 1, 2, 5, 6: Already listed.- 1, 3, 4, 6: Already listed.- 2, 3, 4, 5: Already listed.Yes, that's all. So, there are four combinations of four digits that sum to 14.Now, for each of these four combinations, the remaining three digits will automatically sum to 14 as well because the total is 28.So, for each of these four combinations, I can arrange the four digits in the odd positions and the three digits in the even positions.Now, let's calculate the number of permutations for each case.For the four digits in the odd positions: There are 4 positions, and we have 4 distinct digits, so the number of ways to arrange them is ( 4! = 24 ).For the three digits in the even positions: There are 3 positions, and we have 3 distinct digits, so the number of ways to arrange them is ( 3! = 6 ).Therefore, for each combination, the total number of arrangements is ( 24 times 6 = 144 ).Since there are four such combinations, the total number of 7-digit numbers that are multiples of 11 is ( 4 times 144 = 576 ).Wait, but hold on a second. I think I might have made a mistake earlier. Initially, I thought of three digits summing to 14, but then realized it's four digits. But in the initial problem, the user mentioned arranging the seven digits into a 7-digit number, so all digits must be used exactly once. So, my approach seems correct.But let me double-check the combinations of four digits summing to 14:1. 1, 2, 4, 7: Correct.2. 1, 2, 5, 6: Correct.3. 1, 3, 4, 6: Correct.4. 2, 3, 4, 5: Correct.Yes, that's all. So, four combinations.Therefore, the total number of such numbers is 4 (combinations) multiplied by 24 (arrangements of four digits) multiplied by 6 (arrangements of three digits), which is 4 × 24 × 6 = 576.Wait, but let me think again. Is there a possibility that some of these combinations might result in the same number? For example, if two different combinations of digits lead to the same arrangement when considering the positions. But since all digits are unique and we're permuting all positions, I don't think so. Each combination leads to unique permutations.Also, I should confirm that the difference ( a - b = 0 ) is the only possibility. Earlier, I considered ( a - b = 11 ) and ( a - b = -11 ), but those led to non-integer values for ( a ) and ( b ). So, indeed, the only feasible solution is ( a = b = 14 ).Therefore, I'm confident that the total number of such 7-digit numbers is 576.

answer：First, I need to determine the percentage of votes the winning candidate received. Let's denote this percentage as ( P% ).Since the total number of votes polled is 400, the winning candidate received ( P% ) of 400 votes, which is ( frac{P}{100} times 400 ).The losing candidate received the remaining percentage of votes, which is ( (100 - P)% ), or ( frac{100 - P}{100} times 400 ).The winning candidate won by a 160 vote majority. This means the difference between the votes received by the winning candidate and the losing candidate is 160 votes.Setting up the equation:[frac{P}{100} times 400 - frac{100 - P}{100} times 400 = 160]Simplifying the equation:[4P - 4(100 - P) = 160][4P - 400 + 4P = 160][8P - 400 = 160][8P = 560][P = 70]Therefore, the winning candidate received 70% of the votes.

answer：First, I need to determine the total number of students in the class by adding the number of boys and girls.Next, I will divide the total number of students by the number of members required per group to find out how many groups can be formed.Finally, I will present the final answer in a clear and concise manner.

answer：First, I recognize that (sin 150^circ) can be simplified using the identity for sine of supplementary angles: (sin(180^circ - theta) = sin theta). Therefore, (sin 150^circ = sin 30^circ).Next, I substitute (sin 30^circ) into the original expression:[cos 15^circ cos 30^circ - sin 15^circ sin 30^circ]I recall the cosine difference identity:[cos(A - B) = cos A cos B + sin A sin B]However, the expression I have is (cos A cos B - sin A sin B), which resembles the cosine of a sum:[cos(A + B) = cos A cos B - sin A sin B]Applying this identity to my expression with (A = 15^circ) and (B = 30^circ), I get:[cos(15^circ + 30^circ) = cos 45^circ]Finally, I know that (cos 45^circ = frac{sqrt{2}}{2}), so the value of the original expression is (frac{sqrt{2}}{2}).

answer：First, I recognize that the problem involves calculating the rate of simple interest. I recall the formula for simple interest:[ text{Simple Interest (SI)} = frac{text{Principal (P)} times text{Rate of Interest (R)} times text{Time (T)}}{100} ]Given:- Simple Interest (SI) = Rs. 7200- Principal (P) = Rs. 20000- Time (T) = 3 yearsI need to find the Rate of Interest (R). To do this, I'll rearrange the formula to solve for R:[ R = frac{text{SI} times 100}{text{P} times text{T}} ]Plugging in the given values:[ R = frac{7200 times 100}{20000 times 3} ][ R = frac{720000}{60000} ][ R = 12 ]Therefore, the rate of interest is 12% per annum.

answer：Okay, so I have this triangle ABC, and the problem says that sin B times sin C equals cos squared of half A. I need to figure out what kind of triangle ABC is based on this condition. The options are isosceles, right, equilateral, or isosceles right triangle.First, let me write down what I know. In any triangle, the sum of the angles is 180 degrees, or pi radians. So, A + B + C = pi. That might come in handy later.The given equation is sin B * sin C = cos²(A/2). Hmm, I remember that cos²(theta) can be written using a double-angle identity. Specifically, cos²(theta) = (1 + cos 2theta)/2. So, if I let theta be A/2, then cos²(A/2) becomes (1 + cos A)/2. That might simplify things.So, substituting that into the equation, I get:sin B * sin C = (1 + cos A)/2.Now, I need to relate this to the angles of the triangle. Since A + B + C = pi, I can express A as pi - B - C. So, cos A would be cos(pi - B - C). I remember that cos(pi - x) = -cos x, so cos(pi - B - C) = -cos(B + C). Therefore, cos A = -cos(B + C).Let me substitute that back into the equation:sin B * sin C = (1 - cos(B + C))/2.Hmm, okay. Now, I need to figure out what cos(B + C) is. Using the cosine addition formula, cos(B + C) = cos B cos C - sin B sin C. So, substituting that in:sin B * sin C = (1 - (cos B cos C - sin B sin C))/2.Let me simplify the right side:sin B * sin C = (1 - cos B cos C + sin B sin C)/2.Multiply both sides by 2 to eliminate the denominator:2 sin B sin C = 1 - cos B cos C + sin B sin C.Subtract sin B sin C from both sides:sin B sin C = 1 - cos B cos C.Hmm, interesting. So, sin B sin C + cos B cos C = 1.Wait, that looks familiar. Isn't that the formula for cos(B - C)? Because cos(B - C) = cos B cos C + sin B sin C. So, yes, that's exactly it.So, cos(B - C) = 1.Now, when does cosine of an angle equal 1? That happens when the angle is 0 radians, or a multiple of 2pi, but since we're dealing with angles in a triangle, which are between 0 and pi, the only possibility is that B - C = 0. So, B = C.Therefore, angles B and C are equal. In a triangle, if two angles are equal, then the sides opposite those angles are equal as well. So, triangle ABC is isosceles with sides opposite B and C being equal.Wait, but let me double-check. Could there be another possibility? If cos(B - C) = 1, then B - C must be 0 modulo 2pi, but since B and C are both between 0 and pi, the only solution is B = C. So, yes, it must be an isosceles triangle.But let me think again. The original equation was sin B sin C = cos²(A/2). If B = C, then let's see what happens. Let me denote B = C = x. Then, since A + 2x = pi, A = pi - 2x.So, cos²(A/2) = cos²((pi - 2x)/2) = cos²(pi/2 - x) = sin²x. Because cos(pi/2 - x) = sin x.So, the equation becomes sin x * sin x = sin²x, which is true. So, that checks out.Alternatively, if I consider whether the triangle could be right-angled or equilateral. If it's a right triangle, let's say angle A is 90 degrees. Then, A = pi/2, so cos²(A/2) = cos²(pi/4) = (sqrt(2)/2)^2 = 1/2.Then, sin B sin C = 1/2. But in a right-angled triangle, B + C = pi/2, so C = pi/2 - B. Then, sin C = sin(pi/2 - B) = cos B. So, sin B sin C = sin B cos B = (1/2) sin 2B. For this to equal 1/2, sin 2B must be 1, so 2B = pi/2, so B = pi/4. Therefore, angles would be pi/2, pi/4, pi/4. So, it's an isosceles right triangle. So, in that case, it's both isosceles and right-angled.But wait, in the earlier reasoning, we concluded that B = C, which would make it isosceles. So, is it necessarily a right triangle? Or is it just isosceles?Wait, let me see. If angle A is pi/2, then B and C are pi/4 each, so it's an isosceles right triangle. But if angle A is not pi/2, can we still have B = C?Yes, for example, if angle A is 60 degrees, then B and C would be 60 degrees each, making it equilateral. But wait, in that case, sin B sin C = sin 60 * sin 60 = (sqrt(3)/2)^2 = 3/4. And cos²(A/2) = cos²(30) = (sqrt(3)/2)^2 = 3/4. So, that also satisfies the equation. So, an equilateral triangle would also satisfy the condition.But wait, in the earlier reasoning, we concluded that B = C, which is true for both isosceles and equilateral triangles. So, does that mean the triangle could be either isosceles or equilateral? But equilateral is a special case of isosceles where all sides are equal.But the options are: A: Isosceles triangle, B: Right triangle, C: Equilateral triangle, D: Isosceles right triangle.So, if the triangle could be either isosceles or equilateral, but equilateral is a subset of isosceles, then the most specific answer would be isosceles triangle. But wait, in the case where it's equilateral, it's also isosceles, but the problem might be expecting the most specific answer.Wait, but in the case where A is 60 degrees, making the triangle equilateral, it's also isosceles. So, the answer could be either A or C. But in the options, C is equilateral triangle, which is a more specific case. So, is the triangle necessarily equilateral?Wait, let's test another case. Suppose angle A is 90 degrees, making it a right triangle. Then, as we saw earlier, angles B and C are 45 degrees each, so it's an isosceles right triangle. So, in that case, it's both isosceles and right-angled.But in the case where angle A is 60 degrees, it's equilateral. So, depending on angle A, the triangle can be either isosceles right-angled or equilateral. So, does that mean the answer is both A and D? But the options are separate.Wait, but in the problem, the condition is given, and we have to choose the correct option. So, perhaps the triangle must be isosceles, but it could also be right-angled or equilateral. But since the options are separate, and isosceles is a broader category, perhaps the answer is A: Isosceles triangle.But wait, in the case where it's equilateral, it's also isosceles, so A would still be correct. But if the triangle is equilateral, it's also isosceles, but the problem might be expecting the most specific answer. However, since the condition is satisfied by both isosceles and equilateral triangles, but equilateral is a subset, the answer is A: Isosceles triangle.Wait, but earlier, when I considered the case where angle A is 90 degrees, the triangle is isosceles right-angled, which is also a specific case. So, does that mean the answer could be D? But in that case, it's also isosceles.Hmm, this is confusing. Let me try another approach.We have sin B sin C = cos²(A/2). We transformed it to cos(B - C) = 1, which implies B = C. So, the triangle must be isosceles with B = C. So, regardless of whether it's right-angled or equilateral, it's isosceles. So, the answer must be A: Isosceles triangle.But wait, in the case where it's equilateral, all angles are 60 degrees, so B = C = 60 degrees, so it's isosceles. Similarly, in the right-angled case, B = C = 45 degrees, so it's isosceles. So, in all cases, it's isosceles. Therefore, the answer is A: Isosceles triangle.But wait, the original problem didn't specify whether it's necessarily right-angled or equilateral, just that it's isosceles. So, the answer is A.Wait, but in the initial reasoning, I concluded that B = C, so it's isosceles. So, the answer is A.But let me check again. If B = C, then the triangle is isosceles. So, regardless of whether it's right-angled or equilateral, it's isosceles. So, the answer is A.Wait, but in the case where it's equilateral, it's also isosceles, but the problem might be expecting the most specific answer. However, since the condition is satisfied by both isosceles and equilateral triangles, but equilateral is a subset, the answer is A: Isosceles triangle.Wait, but in the case where it's equilateral, it's also isosceles, so A would still be correct. But if the triangle is equilateral, it's also isosceles, but the problem might be expecting the most specific answer. However, since the condition is satisfied by both isosceles and equilateral triangles, but equilateral is a subset, the answer is A: Isosceles triangle.Wait, but in the case where it's equilateral, it's also isosceles, so A would still be correct. But if the triangle is equilateral, it's also isosceles, but the problem might be expecting the most specific answer. However, since the condition is satisfied by both isosceles and equilateral triangles, but equilateral is a subset, the answer is A: Isosceles triangle.Wait, but in the case where it's equilateral, it's also isosceles, so A would still be correct. But if the triangle is equilateral, it's also isosceles, but the problem might be expecting the most specific answer. However, since the condition is satisfied by both isosceles and equilateral triangles, but equilateral is a subset, the answer is A: Isosceles triangle.Wait, I think I'm going in circles here. Let me summarize:From the given condition, we derived that B = C, so the triangle is isosceles. Therefore, the answer is A: Isosceles triangle.Yes, that makes sense. So, the answer is A.