math f,4 問題－kfsdgte的部落格

標題:

math f,4 問題

發問:

題1THe perimeter of a rectangular playground is 324m.The length is incresed by 6m and the width by 3m. The area is then increased by 1/9 of its former value. Find the original dimensions of the playground.題2A two-digit number is less than three times the product of its by 8, and the digit in the tenth place... 顯示更多題1 THe perimeter of a rectangular playground is 324m.The length is incresed by 6m and the width by 3m. The area is then increased by 1/9 of its former value. Find the original dimensions of the playground. 題2 A two-digit number is less than three times the product of its by 8, and the digit in the tenth place exceeds the digit in the unit place by 2. Find the number.

最佳解答:

1) Let X & Y be the length & width of playground respectively, 2(X + Y) = 324 (X + Y) = 162 X = 162 - Y -- (i) (X + 6)*(Y + 3) = XY(1+1/9) XY + 6Y + 3X + 18 = 10/9*XY 9XY + 54Y + 27X + 162 = 10XY 54Y + 27X - XY + 162 = 0 -- (ii) sub (i) into (ii), 54Y + 27(162-Y) - (162-Y)Y + 162 = 0 54Y + 4374 - 27Y - 162Y + Y^2 + 162= 0 Y^2 - 135Y + 4536 = 0 (Y-72)(Y-63) = 0 Y=72 or Y=63 X = 162 - Y X = 90 or 99 Original dimension = 90*72 or 99*63 2) Let X be the ten digit place, Y be the unit digit 10X + Y = original no. 10X + Y = 3(XY) - 8 10X + Y = 3XY - 8 10X + Y - 3XY + 8 = 0 -- (i) X -Y = 2 X = 2 + Y -- (ii) sub (ii) into (i), 10(2+Y) + Y - 3(2+Y)Y + 8 = 0 20 + 10Y + Y - 6Y - 3Y^2 + 8 = 0 -3Y^2 + 5Y + 28 = 0 (Y-4)(-3Y-7) = 0 Y = 4 or -7/3 (rejected) Y = 4 and X = 6 Hence, the no. is 64

其他解答:

題1 Let the original length be x m. so, the original width = (162 - x) m. Let the original area be y m^2. y = x(162-x) -----(1) So, the new area: 10/9y = (x+6)(162-x+3) y = 9/10(x+6)(165-x) -----(2) Then, we combine (1) and (2): x(162-x) = 9/10(x+6)(165-x) 10(162x-x^2) = (9x+54)(165-x) 1620x-10x^2 = 1485x-9x^2+8910-54x x^2-1674x+10395=0 Then, you an find out the length and the width. 題2 Sorry, I don't really understand your question....SORRY!!!