answer:Let's denote the width of the grassy plot as ( w ) meters. The area of the grassy plot is ( 110 times w ) square meters. Since there is a gravel path 2.5 m wide all around the grassy plot, the length of the plot including the path will be ( 110 + 2 times 2.5 ) and the width will be ( w + 2 times 2.5 ). So, the dimensions of the plot including the path are ( 110 + 5 ) by ( w + 5 ). The area of the plot including the path is ( (110 + 5) times (w + 5) ) square meters. The area of the path alone is the area of the plot including the path minus the area of the grassy plot, which is: [ (115 times (w + 5)) - (110 times w) ] Given that the cost of gravelling the path is 680 at 80 paise per square meter, we can set up the equation: [ 0.80 times [(115 times (w + 5)) - (110 times w)] = 680 ] Now, let's solve for ( w ): [ 0.80 times [115w + 575 - 110w] = 680 ] [ 0.80 times [5w + 575] = 680 ] [ 4w + 460 = 680 times 1.25 ] (since 0.80 is the same as dividing by 1.25) [ 4w + 460 = 850 ] [ 4w = 850 - 460 ] [ 4w = 390 ] [ w = frac{390}{4} ] [ w = 97.5 ] So, the width of the grassy plot is boxed{97.5} meters.

answer:Bruce's score is given directly: [ text{Bruce's score} = 4 ] Michael scored 3 times more than Bruce, so his score is: [ text{Michael's score} = 4 times 3 = 12 ] Combining both scores to find the total: [ text{Total score} = text{Bruce's score} + text{Michael's score} = 4 + 12 = 16 ] Therefore, the total number of goals scored by Bruce and Michael is boxed{16}.

answer:Solution: Since begin{cases} a^3 - a = 0 frac{a}{1-a} neq 0 end{cases}, we solve and find a = -1. Therefore, the correct choice is: boxed{A}. This is solved by setting the real part to 0 and ensuring the imaginary part is not 0 to find the value of the real number a. This question examines the basic concept of complex numbers, specifically the condition for a complex number to be a pure imaginary number. It is a fundamental question.

answer:First, we rearrange the equation: [ 108x^2 - 35x - 77 = 0. ] Since x = frac{3}{4} is a solution, one of the factors of the quadratic must be 4x - 3. We need to find another factor, such that when multiplied by 4x - 3, the original equation 108x^2 - 35x - 77 = 0 is reconstructed. Assuming another factor in the form (ax - b), we expect: [ (4x - 3)(ax - b) = 108x^2 - 35x - 77. ] Expanding and comparing coefficients with the original quadratic: [ 4ax^2 - (3a + 4b)x + 3b = 108x^2 - 35x - 77. ] Setting the coefficients equal: 1. 4a = 108 Rightarrow a = 27 2. 3a + 4b = 35 Rightarrow 3(27) + 4b = 35 Rightarrow 81 + 4b = 35 Rightarrow 4b = -46 Rightarrow b = -frac{46}{4} = -frac{23}{2} Thus, the other factor is 27x + frac{23}{2}. Solving for x in 27x + frac{23}{2} = 0 results in: [ 27x = -frac{23}{2} Rightarrow x = -frac{23}{54}. ] Thus, the other solution is x = boxed{-frac{23}{54}}.