💡First, I need to determine the total amount of work required to paint two houses. If eight people can paint two houses in twelve hours, I can calculate the total work in person-hours.Next, I'll use the concept of inverse proportionality between the number of people and the time taken to complete the work. This means that if fewer people are working, the time required will increase proportionally.Finally, I'll apply this relationship to find out how many hours it would take for six people to paint the same two houses by setting up and solving the appropriate equation.

💡Alright, so I've got this problem here with five statements labeled A through E, and I need to figure out which one is correct. Let me take it step by step.Starting with statement A: "If ( x < -1 ), then ( x^2 > x )." Hmm, okay. If x is less than -1, that means x is a negative number, right? And when you square a negative number, it becomes positive. So, for example, if x is -2, then ( x^2 ) would be 4. Comparing 4 and -2, clearly 4 is greater than -2. So, in this case, ( x^2 > x ) holds true. Maybe statement A is correct, but I should check the others to be sure.Moving on to statement B: "If ( x^2 > 1 ), then ( x > 0 )." Hmm, this seems a bit tricky. If ( x^2 > 1 ), that means x is either greater than 1 or less than -1. So, for example, if x is 2, ( x^2 = 4 ), which is greater than 1, and x is positive. But what if x is -2? Then ( x^2 = 4 ) as well, which is still greater than 1, but x is negative. So, statement B isn't always true because x could be negative. Therefore, statement B is false.Next is statement C: "If ( x^2 > x + 1 ), then ( x > 0 )." Let's test this. Suppose x is 2. Then ( x^2 = 4 ) and ( x + 1 = 3 ). So, 4 > 3, which is true, and x is positive. But what if x is -1? Then ( x^2 = 1 ) and ( x + 1 = 0 ). So, 1 > 0, which is true, but x is negative. This means statement C isn't always true either. So, statement C is false.Statement D: "If ( x^2 > x + 1 ), then ( x < 0 )." Using the same example as statement C, if x is 2, ( x^2 = 4 ) and ( x + 1 = 3 ), so 4 > 3 is true, but x is positive, not negative. This shows that statement D is also false because the condition can be met with a positive x.Finally, statement E: "If ( x < 0 ), then ( x^2 < x )." Let's test this. If x is -1, ( x^2 = 1 ) and x is -1. So, 1 is not less than -1; in fact, it's greater. Therefore, statement E is false.After going through all the statements, only statement A holds true in all cases where ( x < -1 ). The other statements can be disproven with counterexamples. So, I'm pretty confident that statement A is the correct one.**Final Answer**boxed{A}

💡Alright, let me try to work through this problem step by step. So, we have a set ( S ) composed of some positive integers not greater than 15. The key condition is that any two disjoint subsets of ( S ) must have unequal sums. Our goal is to find the maximum possible sum of the elements in ( S ) that satisfies this property.First, I need to understand the problem clearly. We need a set where if you take any two disjoint subsets, their sums are different. That means there can't be two different groups of numbers from ( S ) that don't share any elements and yet add up to the same total. I think the first thing to consider is the size of the set ( S ). If ( S ) is too large, it might be difficult to satisfy the condition because there are more subsets to consider, increasing the chances that two disjoint subsets could have the same sum. Let me think about the number of subsets. For a set with ( n ) elements, the number of subsets is ( 2^n ). But since we are only considering non-empty subsets, it's ( 2^n - 1 ). However, we are specifically looking at disjoint subsets, so the number of possible pairs of disjoint subsets is more complicated. Maybe I should think about the maximum number of elements ( S ) can have without violating the condition.I recall something about the pigeonhole principle, which states that if you have more pigeons than pigeonholes, at least two pigeons must share a hole. Maybe I can apply this principle here. If the number of possible subset sums is less than the number of subsets, then by the pigeonhole principle, there must be two subsets with the same sum.Let me try to calculate the number of possible subset sums. For a set ( S ) with elements ( a_1, a_2, ldots, a_n ), the maximum possible subset sum is the sum of all elements, which is ( sum_{i=1}^{n} a_i ). The minimum subset sum is 1 (if 1 is in the set). So, the number of possible subset sums is roughly ( sum_{i=1}^{n} a_i ).But wait, the number of possible subset sums isn't exactly the same as the range from 1 to the total sum because some sums might not be achievable. However, for the sake of estimation, maybe I can consider the total sum as an upper bound for the number of possible subset sums.Now, the number of non-empty subsets of ( S ) is ( 2^n - 1 ). So, if ( 2^n - 1 ) is greater than the total sum of ( S ), then by the pigeonhole principle, there must be two subsets with the same sum. Therefore, to satisfy the condition that all subset sums are unique, we need ( 2^n - 1 leq sum_{i=1}^{n} a_i ).This seems like a useful inequality. Let me write it down:[2^n - 1 leq sum_{i=1}^{n} a_i]But in our case, the elements of ( S ) are not arbitrary; they are positive integers not greater than 15. So, the maximum possible sum of ( S ) is ( 15 + 14 + ldots + 1 = 120 ). However, we need to find the maximum sum under the given condition.Wait, maybe I'm getting ahead of myself. Let's think about smaller sets first.If ( n = 1 ), then ( S ) has only one element, so there are no two disjoint subsets to compare. The condition is trivially satisfied. The maximum sum is 15.If ( n = 2 ), then ( S ) has two elements. The possible subset sums are the two elements and their sum. Since the elements are distinct, their sums will also be distinct. So, the condition is satisfied. The maximum sum would be 15 + 14 = 29.For ( n = 3 ), the number of non-empty subsets is 7. The maximum possible sum is 15 + 14 + 13 = 42. The number of possible subset sums is 7, and the range of sums is from 13 to 42, which is 30 possible sums. So, it's possible to have unique sums.Wait, but actually, the number of subset sums is not necessarily equal to the number of subsets because different subsets can have the same sum. But in our case, we need all subset sums to be unique.So, for ( n = 3 ), we need all 7 non-empty subsets to have distinct sums. Is this possible?Let me try with the largest numbers: 15, 14, 13.The subsets are:- {15}, sum = 15- {14}, sum = 14- {13}, sum = 13- {15,14}, sum = 29- {15,13}, sum = 28- {14,13}, sum = 27- {15,14,13}, sum = 42All these sums are unique. So, yes, ( n = 3 ) is possible with a maximum sum of 42.Moving on to ( n = 4 ). The number of non-empty subsets is 15. The maximum possible sum is 15 + 14 + 13 + 12 = 54. The number of possible subset sums is 15, and the range of sums is from 12 to 54, which is 43 possible sums. So, it's still possible to have unique sums.Let me try constructing such a set. Starting with the largest numbers: 15, 14, 13, 12.Check the subset sums:- Single elements: 12, 13, 14, 15- Pairs: 12+13=25, 12+14=26, 12+15=27, 13+14=27, 13+15=28, 14+15=29Wait, here we have a problem. The pair {12,15} sums to 27, and the pair {13,14} also sums to 27. So, these two disjoint subsets have the same sum, which violates the condition.Therefore, the set {12,13,14,15} doesn't satisfy the condition. I need to adjust the set to avoid such conflicts.Maybe replace 12 with a smaller number. Let's try 11.Set: {15,14,13,11}Check the subset sums:- Single elements: 11,13,14,15- Pairs: 11+13=24, 11+14=25, 11+15=26, 13+14=27, 13+15=28, 14+15=29- Triplets: 11+13+14=38, 11+13+15=39, 11+14+15=40, 13+14+15=42- Full set: 11+13+14+15=53Now, let's check for duplicate sums. The single elements are all unique. The pair sums are 24,25,26,27,28,29, which are all unique. The triplet sums are 38,39,40,42, which are unique. The full set sum is 53, which is unique. Also, check if any pair sum equals a triplet sum or the full set sum. 24,25,26,27,28,29 are all less than 38, so no overlap. Similarly, the triplet sums are all less than 53. So, no duplicates here.Therefore, the set {11,13,14,15} satisfies the condition with a sum of 11+13+14+15=53.Wait, but can we get a higher sum? Let's see. Maybe replace 11 with 12, but as we saw earlier, that causes a conflict. What if we replace 11 with 10? Then the set would be {10,13,14,15}. Let's check:Single elements: 10,13,14,15Pairs: 10+13=23, 10+14=24, 10+15=25, 13+14=27, 13+15=28, 14+15=29Triplets: 10+13+14=37, 10+13+15=38, 10+14+15=39, 13+14+15=42Full set: 10+13+14+15=52All sums are unique. The total sum is 52, which is less than 53. So, {11,13,14,15} is better.Alternatively, what if we replace 13 with 12? Set {11,12,14,15}:Single elements: 11,12,14,15Pairs: 11+12=23, 11+14=25, 11+15=26, 12+14=26, 12+15=27, 14+15=29Oops, here we have {11,15} and {12,14} both summing to 26. That's a conflict. So, this set doesn't work.So, {11,13,14,15} seems to be the best for ( n = 4 ) with a sum of 53.Now, let's try ( n = 5 ). The number of non-empty subsets is 31. The maximum possible sum is 15 + 14 + 13 + 12 + 11 = 65. The range of possible sums is from 11 to 65, which is 55 possible sums. Since 31 < 55, it's theoretically possible to have unique subset sums.But constructing such a set is more challenging. Let's try to build it step by step.Start with the largest numbers: 15,14,13,12,11.Check for conflicting subset sums. For example:- 15 + 12 = 27- 14 + 13 = 27So, these two disjoint subsets have the same sum, which violates the condition. Therefore, we need to adjust the set.Maybe replace 12 with a smaller number. Let's try 10.Set: {15,14,13,11,10}Check for conflicts:- 15 + 10 = 25- 14 + 11 = 25Conflict again. So, {15,14,13,11,10} doesn't work.What if we replace 11 with 9? Set {15,14,13,10,9}.Check:- 15 + 9 = 24- 14 + 10 = 24Conflict again.Hmm, maybe replace 10 with 8. Set {15,14,13,11,8}.Check for conflicts:- 15 + 8 = 23- 14 + 9 = 23 (but 9 isn't in the set)- 13 + 10 = 23 (10 isn't in the set)- 11 + 12 = 23 (12 isn't in the set)Wait, actually, in the set {15,14,13,11,8}, the only pairs are:15+14=29, 15+13=28, 15+11=26, 15+8=23,14+13=27, 14+11=25, 14+8=22,13+11=24, 13+8=21,11+8=19.All these pair sums are unique. Now, check triplet sums:15+14+13=42, 15+14+11=40, 15+14+8=37,15+13+11=39, 15+13+8=36,15+11+8=34,14+13+11=38, 14+13+8=35,14+11+8=33,13+11+8=32.All triplet sums are unique. Now, check if any pair sum equals a triplet sum:Pair sums: 23,25,22,24,21,19,29,28,27, etc.Triplet sums: 42,40,37,39,36,34,38,35,33,32.No overlaps here. Now, check the full set sum: 15+14+13+11+8=61.Also, check if any subset sum equals another. For example, is there a single element equal to a pair sum? 15 is larger than any pair sum except 15+14=29, but 15 is less than 29. Similarly, 14 is less than 25, etc. So, no single element equals a pair sum.Similarly, check if any pair sum equals a triplet sum. As above, no overlaps.Also, check if any triplet sum equals the full set sum. 61 is unique.Therefore, the set {15,14,13,11,8} satisfies the condition with a total sum of 61.Is this the maximum possible? Let's see if we can get a higher sum by adjusting the elements.Suppose we try to replace 8 with 9. Set {15,14,13,11,9}.Check for conflicts:- 15 + 9 = 24- 14 + 10 = 24 (but 10 isn't in the set)- 13 + 11 = 24Ah, here we have {13,11} and {15,9} both summing to 24. That's a conflict. So, this set doesn't work.What about replacing 8 with 10? Set {15,14,13,11,10}.As before, 15 +10=25 and 14 +11=25. Conflict.Alternatively, replace 11 with 12. Set {15,14,13,12,8}.Check conflicts:- 15 +12=27- 14 +13=27Conflict again.So, it seems that replacing 8 with a higher number causes conflicts, while keeping 8 allows the set to satisfy the condition. Therefore, {15,14,13,11,8} with a sum of 61 is a valid set.Is there a way to get a higher sum? Let's try another approach. Maybe instead of replacing 12 with 8, replace a different element.For example, replace 13 with 12. Set {15,14,12,11,8}.Check for conflicts:- 15 +12=27- 14 +13=27 (but 13 isn't in the set)- 14 +11=25- 15 +11=26- 14 +8=22- 15 +8=23- 12 +11=23- 12 +8=20- 11 +8=19Wait, here we have {15,8} and {12,11} both summing to 23. Conflict again.So, replacing 13 with 12 causes a conflict. What if we replace 14 with 12? Set {15,13,12,11,8}.Check conflicts:- 15 +12=27- 13 +14=27 (but 14 isn't in the set)- 15 +11=26- 13 +13=26 (but we only have one 13)- 15 +8=23- 13 +10=23 (10 isn't in the set)- 12 +11=23- 12 +8=20- 11 +8=19Here, {15,8} and {12,11} both sum to 23. Conflict again.It seems challenging to replace any element without causing a conflict. Therefore, the set {15,14,13,11,8} with a sum of 61 might be the maximum possible.Wait, let me try another combination. Maybe include 15,14,12,10,9.Set: {15,14,12,10,9}Check conflicts:- 15 +9=24- 14 +10=24Conflict again.Alternatively, {15,14,12,11,7}.Check:- 15 +7=22- 14 +8=22 (but 8 isn't in the set)- 12 +10=22 (10 isn't in the set)- 11 +11=22 (only one 11)- 15 +11=26- 14 +12=26Conflict here: {15,11} and {14,12} both sum to 26.Hmm, still conflicts.What if we try {15,14,13,10,8}?Check:- 15 +10=25- 14 +11=25 (11 isn't in the set)- 13 +12=25 (12 isn't in the set)- 15 +8=23- 14 +9=23 (9 isn't in the set)- 13 +10=23- 10 +13=23Wait, {15,8} and {13,10} both sum to 23. Conflict again.It seems that no matter how I adjust the set, replacing elements to include higher numbers often leads to conflicts. Therefore, the set {15,14,13,11,8} with a sum of 61 might indeed be the maximum possible.To confirm, let's check all subset sums:Single elements: 8,11,13,14,15Pairs:- 8+11=19- 8+13=21- 8+14=22- 8+15=23- 11+13=24- 11+14=25- 11+15=26- 13+14=27- 13+15=28- 14+15=29Triplets:- 8+11+13=32- 8+11+14=33- 8+11+15=34- 8+13+14=35- 8+13+15=36- 8+14+15=37- 11+13+14=38- 11+13+15=39- 11+14+15=40- 13+14+15=42Four elements:- 8+11+13+14=46- 8+11+13+15=47- 8+11+14+15=48- 8+13+14+15=50- 11+13+14+15=53Full set: 8+11+13+14+15=61Now, let's list all these sums:Single: 8,11,13,14,15Pairs:19,21,22,23,24,25,26,27,28,29Triplets:32,33,34,35,36,37,38,39,40,42Four elements:46,47,48,50,53Full set:61Looking through these, all sums are unique. There are no duplicates, which means the set {8,11,13,14,15} satisfies the condition.Is there a way to include a higher number instead of 8 to increase the total sum? Let's try replacing 8 with 9.Set: {9,11,13,14,15}Check subset sums:Single elements:9,11,13,14,15Pairs:- 9+11=20- 9+13=22- 9+14=23- 9+15=24- 11+13=24- 11+14=25- 11+15=26- 13+14=27- 13+15=28- 14+15=29Here, {9,15} and {11,13} both sum to 24. Conflict.So, replacing 8 with 9 causes a conflict. Similarly, replacing 8 with 10 causes conflicts as we saw earlier.Therefore, it seems that 8 is the smallest number we can include without causing conflicts when paired with the larger numbers. Thus, the set {8,11,13,14,15} with a total sum of 61 is the maximum possible under the given conditions.I think this is the optimal solution. I don't see a way to include a higher number without violating the subset sum condition. Therefore, the maximum sum of the elements in ( S ) that satisfies the property is 61.

💡First, I'll define the variables for Somu's current age as ( S ) and his father's current age as ( F ).From the first statement, Somu's age is one-fourth of his father's age. This gives the equation:[S = frac{1}{4}F]From the second statement, 12 years ago, Somu was one-seventh of his father's age. This gives the equation:[S - 12 = frac{1}{7}(F - 12)]Next, I'll express ( F ) in terms of ( S ) using the first equation:[F = 4S]Substituting ( F = 4S ) into the second equation:[S - 12 = frac{1}{7}(4S - 12)]Multiplying both sides by 7 to eliminate the fraction:[7(S - 12) = 4S - 12]Expanding and simplifying:[7S - 84 = 4S - 12]Subtracting ( 4S ) from both sides:[3S - 84 = -12]Adding 84 to both sides:[3S = 72]Finally, dividing both sides by 3 to solve for ( S ):[S = 24]Therefore, Somu's present age is (boxed{24}) years.

💡Okay, so I have this problem where I need to find the area of a triangular region enclosed by three lines: y = x + 2, y = -3x + 9, and y = 2. Hmm, let me think about how to approach this.First, I remember that to find the area of a triangle, I need to know the coordinates of its vertices. Since the triangle is formed by the intersection of these three lines, I should find the points where each pair of lines intersect. That means I need to solve for the intersection points of y = x + 2 and y = -3x + 9, y = x + 2 and y = 2, and y = -3x + 9 and y = 2.Let me start with the first pair: y = x + 2 and y = -3x + 9. To find their intersection, I can set the two equations equal to each other because at the intersection point, the y-values will be the same. So:x + 2 = -3x + 9Now, I'll solve for x:x + 3x = 9 - 2 4x = 7 x = 7/4Okay, so x is 7/4. Now, I'll plug this back into one of the original equations to find y. Let's use y = x + 2:y = (7/4) + 2 y = 7/4 + 8/4 y = 15/4So, the first intersection point is (7/4, 15/4).Next, I'll find the intersection of y = x + 2 and y = 2. Setting them equal:x + 2 = 2 x = 0So, the second intersection point is (0, 2).Now, the third intersection point is between y = -3x + 9 and y = 2. Setting them equal:-3x + 9 = 2 -3x = 2 - 9 -3x = -7 x = 7/3Plugging this back into y = 2, the third intersection point is (7/3, 2).Alright, so now I have the three vertices of the triangle: (7/4, 15/4), (0, 2), and (7/3, 2). To find the area of the triangle, I can use the formula for the area of a triangle given its vertices. The formula is:Area = (1/2) * |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))|Let me label the points for clarity:- Point A: (7/4, 15/4)- Point B: (0, 2)- Point C: (7/3, 2)Plugging these into the formula:Area = (1/2) * |( (7/4)(2 - 2) + 0(2 - 15/4) + (7/3)(15/4 - 2) )|Simplifying each term:- The first term: (7/4)(0) = 0- The second term: 0*(something) = 0- The third term: (7/3)(15/4 - 8/4) = (7/3)(7/4) = 49/12So, Area = (1/2) * |0 + 0 + 49/12| = (1/2) * (49/12) = 49/24Hmm, that seems a bit small. Let me double-check my calculations.Alternatively, I can use the base and height method. Looking at the points, I notice that two of the points are on the line y = 2: (0, 2) and (7/3, 2). The distance between these two points is the base of the triangle.Calculating the base:Base = |7/3 - 0| = 7/3Now, the height of the triangle is the vertical distance from the third point (7/4, 15/4) to the line y = 2.Calculating the height:Height = |15/4 - 2| = |15/4 - 8/4| = 7/4Now, using the area formula:Area = (1/2) * base * height = (1/2) * (7/3) * (7/4) = (1/2) * 49/12 = 49/24Okay, so both methods give me the same result. That makes me more confident that the area is indeed 49/24.Wait, just to be thorough, let me visualize the triangle. The line y = 2 is a horizontal line. The other two lines intersect above this line, forming the apex of the triangle. The base is along y = 2 from x = 0 to x = 7/3, which is a bit over 2 units. The height is from y = 2 up to y = 15/4, which is 3.75, so the height is about 1.75 units. Multiplying half of the base (which is about 1.1667) by the height (1.75) should give me an area around 2.04, which is approximately 49/24 (since 49 divided by 24 is about 2.04). That seems reasonable.I think I've covered all the steps and double-checked my work. I don't see any mistakes in my calculations, so I feel confident that the area of the triangular region is 49/24.

💡Okay, so I need to find the terminating decimal expansion of 13 over 200, which is written as 13/200. Hmm, I remember that terminating decimals are those that end after a certain number of decimal places, unlike repeating decimals which go on forever. First, I think I should try to convert the fraction into a decimal by dividing 13 by 200. But I'm not sure if I should do long division or if there's a quicker way. Maybe I can simplify the fraction first? Let me see. The denominator is 200, which I know is 2 to the power of 3 multiplied by 5 squared, right? So, 200 equals 2^3 times 5^2. For a fraction to have a terminating decimal, the denominator after simplifying should only have prime factors of 2 and/or 5. Since 200 is already 2^3 times 5^2, that means 13/200 should indeed be a terminating decimal. Wait, so maybe I can convert the denominator to a power of 10 to make it easier to write as a decimal. Powers of 10 are 10, 100, 1000, etc., which are 10^1, 10^2, 10^3, and so on. To get from 200 to 1000, I need to multiply by 5 because 200 times 5 is 1000. But if I multiply the denominator by 5, I have to do the same to the numerator to keep the fraction equivalent. So, multiplying both numerator and denominator by 5 gives me (13 times 5) over (200 times 5), which is 65 over 1000. Wait, 65 divided by 1000 is 0.065. But hold on, that doesn't seem right because 13 divided by 200 should be less than 0.1, right? Because 13 is less than 200, so the decimal should be less than 1. Wait, maybe I made a mistake in my multiplication. Let me check again. 13 times 5 is 65, and 200 times 5 is 1000. So 65/1000 is indeed 0.065. Hmm, but I thought 13/200 was 0.052. Did I do something wrong here?Let me try another approach. Maybe I should adjust the denominator to a power of 10 by multiplying both numerator and denominator by a suitable number. The denominator is 200, which is 2^3 times 5^2. To make it a power of 10, I need to have equal numbers of 2s and 5s. Since there are three 2s and two 5s, I need to multiply by 2 to make the number of 5s equal to the number of 2s. So, multiplying numerator and denominator by 2 gives me (13 times 2) over (200 times 2), which is 26 over 400. But 400 is still not a power of 10. Wait, 400 is 2^4 times 5^2. Hmm, maybe I need to multiply by 5 again to get more 5s. Let's see, multiplying numerator and denominator by 5 gives me (26 times 5) over (400 times 5), which is 130 over 2000. 2000 is 2^4 times 5^3. Still not a power of 10. This seems like it's not working. Maybe I should go back to the original fraction and try dividing 13 by 200 using long division. Let me set up the long division problem: 13 divided by 200. Since 200 is larger than 13, I know that the decimal will start with 0. So, I write 0. and then proceed. I add a decimal point and some zeros to 13, making it 13.000. Now, I divide 130 by 200. 200 goes into 130 zero times, so I write 0. Then I bring down the next 0, making it 1300. Now, how many times does 200 go into 1300? Let's see, 200 times 6 is 1200, and 200 times 7 is 1400, which is too much. So, it goes 6 times. 6 times 200 is 1200. Subtracting that from 1300 gives me 100. Bring down the next 0, making it 1000. How many times does 200 go into 1000? Exactly 5 times because 200 times 5 is 1000. Subtracting that leaves 0. So, the division is complete. Putting it all together, I have 0.065. Wait, that's what I got earlier, but I thought it should be 0.052. Maybe I was confused earlier. Let me check with another method. Alternatively, I can express 200 as 100 times 2. So, 13 divided by 200 is the same as 13 divided by 100 divided by 2. 13 divided by 100 is 0.13, and then dividing that by 2 gives me 0.065. So, that confirms it. Wait, but earlier I thought it was 0.052. Maybe I made a mistake in my initial thought. Let me verify by multiplying 0.065 by 200. 0.065 times 200 is 13, which is correct. So, 0.065 is indeed the correct decimal expansion. I think I confused myself earlier when I thought it was 0.052, but actually, 0.065 is the right answer. So, the terminating decimal expansion of 13/200 is 0.065.