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math141to150 ( ** )

Mathematical problems — Problems 141 to 150

The ten problems are presented first, followed by the solutions to the ten problems.

Problem 141:

How could you divide $80 among three people such that the second person would have twice as much money as the first person, and the third person would have $5 less than the second person?

Problem 142:

The sum of two consecutive odd numbers is 552.  What are the numbers?

Problem 143:

Julie has $50, which is 8 more than twice what John has.  How much money does John have?

Problem 144:

A bucket and water in the bucket weigh 50 pounds.  When half the water is emptied from the bucket, the bucket and the remaining water weigh 27 pounds.  How much does the bucket weigh?

Problem 145:

Tom can paint a given house by himself in 14 days.  Dick can paint the same house by himself in just 10 days.  Harry, the fastest painter of the three, can paint the same house in 8 days.  How long would it take the three of them, working together, to paint the house?

Problem 146:

A large pool is being filled by 2 hoses.  One of the hoses, by itself, can fill the pool in 10 hours, and the other hose, by itself, can fill the pool in 14 hours.  How long will it take both hoses, running at the same time, to fill the pool?

Problem 147:

You have a total of 40 coins, made up of just nickels and quarters.  If you double the number of quarters and add 12, that will equal the same number as if you multiply the number of nickels by 4 and subtract 4.  How much money do you have in quarters?

Problem 148:

Art can run a marathon (26.2 miles) in 2.5 hours.  Bart can run a marathon in 3 hours.  Bart is planning on running in an upcoming marathon, and Art has a proposal.  Art wants to start running at some distance behind the actual starting line of the marathon at the same time that Bart starts the normal marathon.  Art wants to start far enough behind that he will cross the finish line at the very same time that Bart crosses the line.  How far back from the starting point should Art begin his run?

Problem 149:

A fence is enclosing a rectangular piece of ground that has an area of 6400 square feet.  The length of the fence is equal to 3 times the width of the fence minus 40 feet.  What are the length and width of the fence, and how many total feet of fencing were required to enclose the area?

Problem 150:

What is the length of any side of a square that fits into a circle whose radius is 5 inches?

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Solution to Problem 141:

  • Let x = the amount of money for the first person, y =  the amount for the second person, and z = the amount for the third person.
  • We know that x + y + z = 80, and that y = 2x, and that z = 2x – 5.
  • Thus,  x + 2x + (2x – 5) = 80.  Simplification yields  3x – 5 = 80, and thus x = 17.
  • Thus, the first person get $17, the second person gets $34, and the third person gets $29.  The total is $80.

Solution to Problem 142:

  • Let x and y represent the numbers.  We know that y is 2 greater than x.  Thus,  x + (x + 2) = 552.
  • Then, 2x + 2 = 552, and x = 275.  Thus, the 2 odd numbers are 275 and 277.

Solution to Problem 143:

  • Let x = the amount of John’s money.  Thus,  50 = 2x + 8.  Solving for x yields 21.  Thus John has $21.

Solution to Problem 144:

  • Let b = the weight of the bucket and w = the weight of the initial amount of water.  Thus, b + w = 50.
  • We also know that b + ½w = 27.
  • Solving for b in both equations yields b = 50 – w and b = 27 – ½w.  Thus, 50 – w = 27 – ½w.
  • Solving for w yields ½w = 23, and then w = 46.  Since b + w = 50, the bucket must weigh 4 pounds.

Solution to Problem 145:

  • Let h = 1 painted house.  Thus, (1 / 14 days) + (1 / 10 days) + (1 / 8 days) = 1 house combined.
  • Finding a common denominator yields  (80 / 1120) + (112 / 1120) + (140/1120) = 332/1120 = 3.37.
  • Thus, the three working together could paint the house in 3.37 days.

Solution to Problem 146:

  • The following equation can be written:  (1 pool / 10 hours) + (1 pool / 14 hours) = 1 pool /  x hours.
  • Using a common denominator gives 14/140 + 10/140 = 1/x.  Thus, 24/140 = 1/x.  Thus x = 5.83 hours.

Solution to Problem 147:

  • Let n = the number of nickels and q = the number of quarters.
  • We know that n + q = 40, and we also know that 2q + 12 = 4n – 4.
  • Substituting 40 – n for q in the second equation yields  2(40 – n) + 12 = 4n – 4
  • Simplification yields 80 – 2n + 12 = 4n – 4, and then 96 = 6n.  Thus n, or the number of nickels, equals 16.
  • That means there are 24 quarters, or $6 in quarters.

Solution to Problem 148:

  • Dealing with distances and speed, we know that distance = rate x time, and thus time = distance / rate.
  • Since Art and Bart are going to run the race in the same time, we know that d / r for Art = d / r for Bart.
  • Letting x = Art’s distance behind the starting line, we can write the following equation:
    • (26.2 + x) / (26.2 / 2.5) = 26.2 / (26.2 / 3)
  • Simplifying this equation leads to  2.5 (26.2 + x) = 26.2 ( 3 )
  • Then 26.2 + x = 78.6 ÷ 2.5 = 31.44
  • Thus, x = 5.24 miles.  Thus, Art should start 5.24 miles behind the normal starting line at the same time that Bart starts his race.
  • NOTE: The solution to this problem assumes that Art can run the additional distance at the same speed he runs the marathon distance.

Solution to Problem 149:

  • Let w = width of the field and l = length of the field.  We know that 3w – 40 = l
  • Since w x l = area, we know that A = w (3w – 40).  Thus, 3w² – 40w = 6400, and thus 3w² – 40w – 6400 = 0
  • Factoring the polynomial yields (3w – 160)(w + 40) = 0.  Thus, w = 53.333 feet.  The length would then be 120 feet.
  • The total length of fencing to enclose the field would be 106.666 + 240 = 346.666 feet.

Solution to Problem 150:

  • If a line is drawn from the center of the circle to a corner of the square, and then another line is drawn from the center of the circle to an adjacent corner of the square, a right triangle is formed with sides A and B equal to 5 inches.
  • Thus, knowing that A² + B² = C² is applicable to a right triangle, we have 5² + 5² = C², with C equaling a radius of the circle.
  • Thus, C² =  25 + 25 = 50, and C= 7.07

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