# Six Algebra Word Problems : Ratio and Proportion, Length of Time Together and Alone, Length of Diagonal and Period of Swing

50. Foxes and rabbits. The ratio of foxes to rabbits in the Deerfield Forest Preserve is 2 to 9. If there are 35 fewer foxes than rabbits, then how many of each are there?

Page 379 Problem 40, 42

40. Envelope stuffing. Every week, Linda must stuff 100 envelopes. She can do the job by herself in 6 hours. If Laura helps, they get the job done in 5 Â½ hours. How long would it take Laura to do the job by herself?

42. Draining the vat. With only the small valve open, all of the liquid can be drained form large vat in 4 hours. With only the large value open, all of the can be drained from the same vat in 2 hours. How long would it take to drain the vat with both valves open?

Page 507 Problem 98,100

98. Side of a sign. Find the length of the side of a square sign whose area is 80 cubic feet.

100. Diagonal of a sign. What is the length of the diagonal of a rectangular billboard whose sides are 5 meters and 12 meters?

Page 534 Problem 96

96. Time to swing. The period T (time in seconds for one complete cycle) of a simple pendulum is related to the length L (in feet) of the pendulum by the formula 8T^2 = (3.14^2 or pie) L. If a child is on a swing with a 10 foot chain, then low long does it take to complete one cycle of the swing?

Page 542 Problem 76

76. Diagonal of a square. The diagonal of a square is 2 meter longer than a side. Find the length of a side.

If an object is given an initial velocity of vo feet per second form a height of so feet, then is heights S after t seconds is given by that formula S = -16t2+ vot+ so

13. (x-2)2=

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Page 364 Problem 34, 44

34.

44.

Page 371 Problem 50

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#### Solution Summary

Six Algebra Word Problems involving Ratio and Proportion, Length of Time Together and Alone, Length of Diagonal and Period of Swing are solved. Simplifying algebraic expressions is also shown. The solution is detailed and well presented.