1 2 3 4 5 There were 414 tickets purchased for a major leagu

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There were 414 tickets purchased for a major league baseball game. The general admission tickets cost $6.50 and the upper reserved tickets cost $8.00. The total amount of money spent was $2890.50. How many of each kind of ticket were purchased? How many general admission tickets were purchased? How many upper reserved tickets were purchased?

Solution

1. Let the number of general admission tickets and upper reserved tickets purchased be x and y respectively. Then x + y = 414...(1) and 6.5x +8y = 2890.5 or, 65x +80y = 28905 or, 13x + 16y = 5781...(2) From the 1st equation , we have y = 414 -x. On substituting this value of y in the 2nd equation, we have 13x + 16(414-x) = 5781 or, 13x -16x + 6624 = 5781 or, 3x = 6624-5781 = 843 so that x = 843/3 = 281. Then y = 414-x = 414-281 = 133. Thus, 281general admission tickets were purchased and 133 upper reserved tickets were purchased.

2. Let x liters of 35% solution be mixed with y liters of 90 % solution to get 55 liters of 65 % solution. Then x + y = 55...(1) and x*0.35 + y*90 = 55*65 or, 0.35x + 0.90y = 35.75 or, 35x+90y = 3575 or, 7x + 18y = 715...(2) From the 1st equation, we have y = 55 -x . On substituting this value of y in the 2nd equation, we have 7x + 18(55-x) = 715 or, 7x - 18x + 990 = 715 or, 11x = 990 - 715 = 275 so that x = 275/11 = 25 . Then y = 55 -x = 55 -25 = 30.Thus, there needs to be 25 liters of 35% solution and 30 liters of 90% solution in the mixture.

.x + y + z = 8...(1), x - z = 5...(2), and x + y = 6. ...(3). From the 2nd and the 3rd equation, we have z = x -5 and y = 6 -x. On substituting these values of y and z in the 1st equation, we have x + 6 -x + x - 5 = 8 or x +1 = 8 so that x = 8-1 = 7. Then y = 6-x = 6 - 7 = -1 and z = x - 5 = 7 -5 = 2. We can verify the result by substituting these values of x, y, z in the 1st equation.Thus, there is one solution. The solution is x = 7, y = -1 and z = 2.

4. Let the capacities of the Pumps A, B and C be x, y, and z gal per hour respectively. Then, x + y + z = 3500 ...(1), x + y = 2100...(2) and x + z = 2300...(3) From the 2nd and the 3rd equation, we have y = 2100 -x and z = 2300 -x. On substituting these values of y and z in the 1st equation, we have x + 2100 -x + 2300 -x = 3500 or -x +4600 = 3500 so that x = 4400 - 3500 = 900. Then y = 2100 -x = 2100 -900 = 1200, and z= 2300 -x = 2300- 900= 1400 1200.We can verify the result by substituting these values of x, y, z in the 1st equation.Thus, thepumping capacities of pump A is 900 gal per hour, that of pump B is 1200 gal per hour and that of pump C is 1400 gal per hour.

5. Let x , y and z pounds of the 1st , 2nd and the 3rd fertizer be mixed to get 200 lbs of the final mixture of fertilizer. Then x + y + z = 200...(1) , 0.2x + 0.1y + 0z = 0.115*200 or, 2x + y = 230...(2), 0.3x + 0.2y+ 0.4z = 0.26*200 or, 3x + 2y + 4z = 520...(3) and 0.5x +0.7y +0.6z = 200* 0.625 or, 5x + 7y + 6z = 1250 ...(4)From the 2nd equation, we have y = 230 -2x. On substituting this value of y in the 1st, 3rd and the 4th equations, we get x -z = 30...(5), 4z -x = 60...(6) and 9x -6z = 360....(7). On adding the 5thand the 6th equations, we get 3z 90 so that xz = 30. Then x -30 = 30 so that x = 60. Also then y = 230 -2x = 230 -120 = 110. The solution is x = 60, y = 110 and z = 30.The answer is one must mix 60 lbs of 1st mixture, 110lbs of 2nd mixture and 30lbs of the 3rd mixture to obtain the desired reult.