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mathproblems111to120 ( ** )

Mathematical problems — Problems 111 to 120

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

Problem 111:

Three years ago, the sum of the ages of quintuplets and their older sister was 155.  The older sister is 5 years older than the quintuplets.  How old are the quintuplets and how old is the older sister?

Problem 112:

When a husband marries his wife, he is four times as old as his bride.  Eight years later, he is three times as old as his wife.  Twenty-four years after that, he is only twice her age.  How old were the husband and wife when they married?

Problem 113:

Fran is 30 years older than Fern.  In 18 years, Fran will be twice as old as Fern.  How old will Fran be when she is twice as old as Fern?

Problem 114:

Ruth’s current age is two years more than twice Steve’s current age.  Nine years ago, Ruth’s age was three years more that three times Steve’s age.  How old will Ruth and Steve be in five years?

Problem 115:

Two trains leave the same train station traveling in opposite directions.  The first train leaves at 3 p.m.  The second train leaves 30 minutes later, and travels at a speed averaging 15 miles per hour faster than the first train.  At 9 p.m. that evening the trains are 600 miles apart.  How fast are the two trains traveling?

Problem 116:

Al can bicycle two miles per hour less than twice as fast as Bill, so Al didn’t leave for a given destination until two hours after Bill left for the same destination.  If the total combined distance traveled for Al and Bill was 504 miles, and if Bill biked for 10 hours, then how fast can Al cycle?

Problem 117:

Sue leaves Smithfield at 1 p.m., heading south toward Santaquin, traveling at 45 mph.  Jane leaves Santaquin at 2 p.m., heading north toward Smithfield, traveling at 55 mph.  If Smithfield and Santaquin are 145 miles apart, at what time will Sue and Jane meet?

Problem 118:

Henry left for an out-of-town work assignment at 7 a.m. and drove at an average speed of 45 mph.  At  7:30 a.m., he realized that he had forgotten his briefcase, so his wife, Ruth, jumped in her car and chased after Henry.  She traveled at an average speed of 55 mph.  How long did it take Ruth to catch Henry, and how far did each of them travel?

Problem 119:

Two cars are approaching each other on the same straight highway, one car traveling west at 45 mph and the other car traveling east at 30 mph.  When the cars were 300 miles apart, a bee started flying back and forth between the cars at 60 mph, reversing direction each time it meets one of the cars.  How many miles has the bee flown when the cars pass each other on the highway?

Problem 120:

Ben left his home at 7 am on his bicycle, headed to a city park.  Ben cycles at 6 miles per hour.  Brad, Ben’s brother, leaves the home at 7:30 am, headed for the same park.  He catches up with Ben at 7:45 am.  How fast did Brad travel on his bicycle?

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Solution to Problem 111:

  • Let x = the age of the quintuplets and let  y = the older sister’s age.
  • Thus we know that:  5(x – 3) + (y – 3) = 155.  We also know that: x + 5 = y.
  • Substituting yields:  5(x – 3) + ((x + 5) – 3) = 155.
  • Thus, 5x – 15 + x + 2 = 155, and  6x – 13 = 155.
  • So, 6x = 168, and x = 28.  Thus, the quintuplets are 28 years old and the older sister is 33 years old.

Solution to Problem 112:

  • We can solve the problem using the ‘eight years later’ information, or the ‘twenty-four years after that’ information.
  • ‘Eight years later’ solution:
    • Let the husband’s age at marriage = x, and let the wife’s age at marriage = y.
    • We know that  4y = x  and we know that  3(y + 8) = x + 8.
    • Substitution yields  3(y + 8) = 4y + 8, and then 3y + 24 = 4y + 8
    • So,  y = 16.  Thus the wife was 16 years old when they married and the husband was 64.
  • ‘Twenty-four years after that’ solution:
    • We again know that 4y = x, and we know that 2(y + 32) = x + 32.
    • Substitution yields  2(y + 32) = 4y + 32, and thus 2y + 64 = 4y + 32.
    • So,  2y = 32, and y again = 16, the wife’s age at marriage.

Solution to Problem 113:

  • Let x = Fran’s current age, and let  y = Fern’s current age.
  • We know that  y + 30 = x  and we know that 2(y + 18) = x + 18.
  • Substitution yields  2(y + 18) = (y + 30) + 18.  Thus,  2y + 36 = y + 48.
  • Solving for y yields  y = 12.  So Fern is currently 42 and Fran is currently 12.
  • So Fern will be 60 when Fran is 30, and that is when she will be twice as old Fran.

Solution to Problem 114:

  • Let x = Ruth’s current age and y = Steve’s current age.
  • We know that  2y + 2 = x
  • We also know that 3(y – 9) + 3 = x – 9
  • Substitution yields  3(y – 9) + 3 = (2y + 2) – 9
  • Simplifying yields 3y – 27 + 3 = 2y – 7, and this leads to y = 17
  • Thus Steve is 17 and Ruth is 36.  In five years Steve will be 22 and Ruth will be 41.

Solution to Problem 115:

  • Let x = the speed of train #1.  By 9 p.m., it has traveled 6 hours at x mph.
  • By 9 p.m., the second train has traveled 5.5 hours at (x + 15) mph.
  • At 9 p.m., the trains are 600 miles apart, so we can write  6x + 5.5(x + 15) = 600.
  • Simplification yields  6x + 5.5x + 82.5 = 600, and thus 11.5x = 517.5
  • Thus, x = 45 mph, which is the speed of train #1.  Train #2 must be traveling at 60 mph.

Solution to Problem 116:

  • Let x = the speed at which Bill can cycle.  Thus, Al’s speed would be  2x – 2.
  • Using the rate x time = distance formula, we can write that  8(2x – 2) + 10x = 504.
  • Simplifying leads to  16x – 16 + 10x = 504, and that 26x = 520.
  • Thus, x = 20 mph, which is Bill’s speed.  Al’s speed must be 38 mph.

Solution to Problem 117:

  • From 1 p.m. until 2 p.m., Sue will have traveled for 1 hour at 45 mph.  Thus, she will have traveled 45 miles.  Thus, she is now only 100 miles away from Jane when Jane commences her travel.
  • Letting x = the time that it will take for the two to meet, we can write  (x hours)(45 mph) + (x hours)(55 mph) = 100 miles.
  • Thus, 45x + 55x = 100, and then 100x = 100.
  • Then, x = 1 hour.  Sue and Jane will meet up at 3 p.m., which is 2 p.m. plus 1 hour.

Solution to Problem 118:.

  • Henry is going to travel x hours at a speed of 45 mph.  Ruth is also going to travel (x – 0.5) hours at a speed of 55 mph.  The distance traveled by Henry and Ruth will be the same.  Thus,  45 • x = 55 • (x – 0.5).  So, 45x = 55x – 27.5.  Solving for x yields 2.75 hours.
  • Henry traveled for 2.75 hours (2 hours 45 minutes), so Ruth traveled for 2 hours and 15 minutes.  Each of them traveled 123.75 miles.

Solution to Problem 119:

  • The combined speed of the two cars is 75 mph (30 mph + 45 mph).  Since the cars started out 300 miles apart, they will travel for 4 hours before meeting.  This is calculated by  300 miles ÷ 75 mph = 4 hours.  Since the bee is traveling at 60 mph, it will have traveled 260 miles when the cars meet.  This is calculated by 60 mph • 4 hours = 240 miles.

Solution to Problem 120:

  • Let x = the speed at which Brad travels in mph.  Thus, (0.25 hour)(x mph) = (0.75 hour)(6 mph)
  • Thus, 0.25x = 4.5, and x = 18 mph.  Thus, Brad cycled at 18 mph.

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