Appearance
question:8% of people in a village in Sri Lanka died by bombardment, 15% of the remainder left the village on account of fear. The population is now reduced to 3553. How much was the population in the beginning?
answer:Let's denote the initial population of the village as P. According to the given information, 8% of the population died due to bombardment. Therefore, the population after the bombardment is 92% of the initial population (since 100% - 8% = 92%). So, after the bombardment, the population is: 0.92 * P Then, 15% of the remaining population left the village due to fear. This means that 85% of the population after the bombardment stayed in the village (since 100% - 15% = 85%). So, the population after people left due to fear is: 0.85 * (0.92 * P) According to the problem, this final population is 3553. So we can set up the equation: 0.85 * (0.92 * P) = 3553 Now, we can solve for P: P = 3553 / (0.85 * 0.92) P = 3553 / 0.782 P ≈ 4547.31 Since the population cannot be a fraction, we round to the nearest whole number: P ≈ 4547 Therefore, the initial population of the village was approximately boxed{4547} people.
question:Sandy gets 4 marks for each correct sum and loses 3 marks for each incorrect sum. Sandy attempts 40 sums within a 45minute time limit and obtains 72 marks. How many sums did Sandy get correct?
answer:Let's denote the number of correct sums as ( c ) and the number of incorrect sums as ( i ). From the information given, we have two equations: 1. The total number of sums attempted is 40: [ c + i = 40 ] 2. The total marks obtained by Sandy is 72, with 4 marks for each correct sum and a loss of 3 marks for each incorrect sum: [ 4c - 3i = 72 ] We can solve these two equations simultaneously to find the value of ( c ) (the number of correct sums). From the first equation, we can express ( i ) in terms of ( c ): [ i = 40 - c ] Now, we substitute ( i ) in the second equation: [ 4c - 3(40 - c) = 72 ] Expanding the equation: [ 4c - 120 + 3c = 72 ] Combining like terms: [ 7c = 72 + 120 ] [ 7c = 192 ] Dividing both sides by 7 to solve for ( c ): [ c = frac{192}{7} ] [ c = 27.4285714 ] Since the number of correct sums must be a whole number, we round down to the nearest whole number, which is 27. Therefore, Sandy got boxed{27} sums correct.
question:f the sum of the squares of nonnegative real numbers x, y, and z is 52, and xy + yz + zx = 27, then what is the sum of x, y, and z?
answer:Given the equations: 1. x^2 + y^2 + z^2 = 52 2. xy + yz + zx = 27 We are to find the sum x+y+z. Start by using the identity (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx). Plugging in the values from the problem, we have: [ (x+y+z)^2 = 52 + 2 times 27 = 52 + 54 = 106 ] Taking the square root of 106, we obtain: [ x+y+z = pm sqrt{106} ] Since x, y, z are nonnegative, their sum must also be nonnegative, so we discard the negative value. Thus: [ x+y+z = sqrt{106} ] Conclusion with boxed answer: [ boxed{sqrt{106}} ]
question:The plane is partitioned into regions by a finite number of lines no three of which are concurrent. Two regions are called neighbors if the intersection of their boundaries is a segment, or halfline or a line (a point is not a segment). An integer is to be assigned to each region in such a way that: i) the product of the integers assigned to any two neighbors is less than their sum; ii) for each of the given lines, and each of the halfplanes determined by it, the sum of the integers, assigned to all of the regions lying on this halfplane equal to zero. Prove that this is possible if and only if not all of the lines are parallel.
answer:1. **Coloring the Regions**: We start by coloring the regions formed by the lines in a checkerboard pattern, such that no two adjacent regions share the same color. Let's call the colors white and black. 2. **Defining Good Points**: A "good point" is defined as the intersection of two lines. Since no three lines are concurrent, each intersection point is shared by exactly four regions. 3. **Assigning Integers to Regions**: - For a white region, count the number of good points on its boundary and assign this number to the region. - For a black region, count the number of good points on its boundary, multiply this number by (-1), and assign this product to the region. 4. **Checking the First Condition**: - Consider two neighboring regions, one white and one black. Let (a) be the integer assigned to the white region and (-b) be the integer assigned to the black region. - The product of the integers assigned to these two regions is (a cdot (-b) = -ab). - The sum of the integers assigned to these two regions is (a + (-b) = a - b). - Since (a) and (b) are both positive (as they count the number of good points), we have (-ab < a - b), satisfying the first condition. 5. **Checking the Second Condition**: - For each line, consider the half-planes determined by it. Each half-plane contains a certain number of regions. - The sum of the integers assigned to all regions in a half-plane must be zero. - This is ensured by the checkerboard coloring and the assignment of integers. Each good point contributes equally to the sum of the integers in both half-planes, but with opposite signs due to the coloring. Thus, the contributions cancel out, resulting in a sum of zero. 6. **Special Case of Parallel Lines**: - If all lines are parallel, there are no good points, and thus each region would be assigned the integer 0. - This would violate the first condition, as the product of the integers assigned to any two neighbors would be 0, which is not less than their sum (which would also be 0). 7. **Conclusion**: The assignment of integers to regions is possible if and only if not all of the lines are parallel. (blacksquare)