3."Diophantus passed 1/6 of his life in childhood, 1/12 in youth and 1/7 more as a bachelor. Five years after his marriage, was born a son who died 4 years before his father, at 1/2 his father's final age." How old was Diophantus when he died? : Let x = his final age : a lengthy but simple equation : Final age = child + youth + bach + marriage + son's age + 4 x= x+ x+ x+5+ x+4 : Common denominator of 12*7 = 84 x= x+ x= : x+9 x- x+ x+5+ x=9 : x=9 : Multiply both sides by 84 9x = 84(9) : x = 84 is his final age : Check: 14 + 7 + 12 + 5 + 42 + 4 = 84 x+4 i n a chemical laboratory one carboy contains 12 gallons of acids and 18 gallons of water. ----------------How many gallons must be drawn from each carboy and combined to form a solution that is 7 gallons acid and 7 gallons water? ---Acid Equation: 12A + 9B = 7 gallons Water Equatio: 18A + 3B = 7 gallons -------------------------------------Solve by elimination: Multiply thru 2nd equation by 3 to get: 54A + 9B = 21 ---------------------Subtract the 1st equation from that and solve for A: 42A = 14 . one carboy contains 12 gallons of acids and 18 gallons of water. How many gallons must be drawn from each carboy and combined to form a solution that is 7 gallons acid and 7 gallons water? In a chemical laboratory A. the alcohol comprising 25%.12. B. the alcohol comprising 25%. How much of the mixture must be drawn off and replaced by alcohol so that the tank contain a mixture of which 50% is alcohol?asoline tank of a car contains 50 liters of gasoline and alcohol. The The gasoline tank of a car contains 50 liters of gasoline and alcohol. Another carboy contains 9 gallons of acid and 3 gallons of water. How much of the mixture must be drawn off and replaced by alcohol so that the tank contain a mixture of which 50% is alcohol? 15. Another carboy contains 9 gallons of acid and 3 gallons of water. $3 is the price 27. let blend one price be $x Let blend 2 price be y 12x+10y=54 8x+15y=61 2(12x+10y) . how long will it take B to do it? 6/8=50/b B= 66.. E₁ can solve 1/x of a problem. Find the price per lb in each grade. The time required for two examinees to solve the same problem differs by two minutes. A tobacco dealer mixed 12 lbs of one grade of tobacco with 10 lbs of another grade to obtain a blend worth $54.7 29. E₂ solves 60/(x − 2) problems .3(8x+15y)= 2*54 . $3 for the secon blend plug the value of y in equation 12x+10y=24 x=2 $2. In one hour. How long will it take for the slower problem solver to solve the problem? So in 1 min. He then made a second blend worth $61 by mixing 8 lbs of the first grade and 15 lbs of the second grade. E₁ solves 60/x problems Similarly. in one hour. Together they can solve 32 problems in one hour.A = 1/3 1/3 1st carboy gives 4 gals acid and 6 gals water Substitute into 18A + 3B = 7 and solve for B: 18(1/3) + 3B = 7 3B = 1 B = 1/3 1/3 2nd carboy give 3 gals acid and 1 gal of water ============================== 1/3 of 1st carboy = 10 gals 1/3 of 2nd carboy = 4 gals 16. If A's rate of doing work is to B's is 6:8 and A does a piece of work in 50 days...3*61 24x+20y-24x-45y= -75 -25y=-75 y=3. Together they solve : 60/x + 60/(x − 2) = 32 Solving for x. So x = 5 31. A cyclist traveled a certain distance at her usual speed. peed__________time__________distance Usual________________r_____________t_____________d Faster______________r+2___________t-1____________d Slower______________r-2___________t+2____________d EQUATIONS - - - Use the second and third equations of the simplified system and the Elimination Method to get values for r and for t. but x = 3/4 is not possible since E₂ solves (3/4 − 2) = −1. If her speed had been 2 mi/hr slower. she would have traveled the distance in 1 hr less time. Find the distance traveled and her usual speed. If her speed had been 2 mi/hr faster.25 min to solve a problem. and getting x = 3/4 or 5. You can then use them in any of the equations to find value for d. she would have taken 2 hr longer. You can put this solution on YOUR website! system of linear equations . If her speed had been 2 mi/hr faster. If her speed had been 2 mi/hr slower.5x -18 = 0 2x^2 + x .5) 8/ x .5x x^2 + 0. Find the original speed of the boat.5) 36x + 18 . the boat arrives 10 minutes earlier then if the original speed had been maintained.5) = 1/6 MULTIPLY BOTH SIDES BY x *(x+0.5) = 10/60 = 1/6 6/ x . Let the original speed be x km / hour Time taken at original speed = 8 / x hours Time taken at increased speed after 2km = 2/ x + 6/ (x+0. Find the distance traveled and her usual speed. she would have taken 2 hr longer. By doing this.6 *x*6 = x ( x+0.6 / ( x+0. A boat going across a lake 8 km wide proceeds 2 km at a certain speed and then completes the trip at a speed of 0. she would have traveled the distance in 1 hr less time.6/ (x+0. Normal speed: mph Distance: miles 33 .5) .2/x .36 = 0 2x^2 -8x +9x -36 = 0 2x ( x-4) +9( x -4) = 0 .A cyclist traveled a certain distance at her usual speed.5) *6 36( x+0.5 km/hr faster.36x = x^2 + 0. 1N + 0. we get.30 .65 0.1N + 0.5N + 0.1N + 0.( 2x +9) ( x-4) = 0 x = 4km/ hour ANSWER CHECK 8/4 = 2 hours 2/4 = 0.5 hours 6/4.80 N=4 Q=N+1=4+1=5 Total Number of Quarters = 5 .5 = 4/3 hours = 1 hour 20 minutes ADD = 1 hour 50 minutes Difference = 10 minutes 43.1 (N/2) = 1.25Q + 0.25(N+1) + 0.5(N+1) + 0.65 0. How many quarters are there? Total Number of Nickels = N Total Number of Dimes = D Total Number of Quarters = Q Relations: D=N/2 Q=N+1 Value of Quarters * Q + Value of Nickels * N + Value of Dimes * D = 1.05N + 0. [0.65 taking two as common denominator.05N + 0.7N = 2.5 +0.1N] / 2 = 1. dimes and quarters.5 0.0.1D = 1.5N + 0.65 0.1N = 3. There are twice as many nickels as dimes and one more quarter than nickels. coin bank contains $1.1N = 3.30 0.65 in nickels. 2 = 14 2n = 16 n=8 . a toy savings bank contains $17. the number of dimes exceeds twice the number of nickels by 3 and the number of quarters is 4 less than 5 times the number of nickels.2 .145/25 + n/5 + 2d/5 Drop the d and q into the third equation and solve for n: n + d + q = 14 n + n . Solve the first one for q: 145 = 5n + 10d + 25q 25q = 145 .2d/5 Drop that q into the second equation and solve for d: n=2+d+q n = 2 + d + 145/25 .n/5 .2d/5 = 14 n + n . There are 2 more dimes and quarters combined. I'm going to assume that "There are 2 more dimes and quarters combined" means there are 2 more NICKELS THAN dimes and quarters combined. which you can solve. 145 = 5n + 10d + 25q n=2+d+q n + d + q = 14 Now you have 3 equations in 3 variables. how many quarters are in the bank? 5x+20x+30+125x-100=1730 150x+30 =1830 150x=1830-30 150x=1800 /150 x=12 nickels 15 2x+3 dimes = 27 5x-4 = 56 quarters 45.2d/5 d = n . Dimes and Quarters.145/25 + n/5 + 2d/5 + 145/25 . She has 14 coins in Nickels.45 in coins.30 consisting of nickels. dimes and quarters.5n .n/5 . How many nickels does she have.44. Mary Rose has $1.n/5 .2 .10d q = 145/25 .