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# math1to9 ( ** )

### Mathematical problems — Problems 1 to 10

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

Problem 1:

Starting from the same point, Jim walks north and Joe walks east.  After a period of time, Jim has walked 1 mile less than twice as far as Joe.  At that moment, the two men are 17 miles apart.  How far have both of them walked?

Problem 2:

When a couple married, the wife was 3/4 the husband’s age.  Twelve years later, the wife was 5/6 the husband’s age.  How old was the wife when they married?

Problem 3:

You have hired a man to work for you for seven days.  His pay is to be 1/7 (one-seventh) of a bar of gold per day, and he requires that he actually receive the 1/7 payment, in hand, every day.  What is the fewest number of cuts of the bar that are necessary to meet his demands?

Problem 4:

Metal train tracks expand as they warm.  Gaps are purposely left between track lengths to allow for this expansion.  Inadvertently, one mile of track was laid without including any expansion gaps.  Upon warming, the one-mile section buckled upward exactly in the middle, forming a perfect V-shape.  How high did the track rise?

Problem 5:

A stable boy was sent to count the number of persons and horses in a nearby paddock.  Being an imp, he instead counted the total number of heads and legs, both human and equine.  There were 74 heads and 196 legs.  Can the number of persons and humans be determined?

Problem 6:

Diophantus was a Greek mathematician.  His boyhood was 1/6 of his life.  He married after 1/7 more of his life.  His beard grew during the next 1/12 of his life, and his son was born 5 years after that.  The son lived to 1/2 of his father’s total age, and Diophantus died 4 years after his son died.  How long did Diophantus live?

Problem 7:

If you subtract 1 from a certain number and then take the square root of that difference, you will get the same result as subtracting the number from 7.  What is the number (or numbers)?

Problem 8:

A rectangular lot is 50 feet longer than it is wide.  The diagonal distance of the lot is 50 feet longer than the longest side.  What are the dimensions of the lot?

Problem 9:

If you were to add 4 years to Don’s current age, take the square root of that sum, and then multiply that root by 5, that product would equal Tom’s age 10 years ago.  Don is 10 years older than Tom.  What are the current ages of the two men?

Problem 10:

What temperature is the same in both degrees Fahrenheit and degrees Centigrade?

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Solution to Problem 1:

• Joe’s distance = x
• Jim’s distance = 2x – 1
• Using the Pythagorean theorem, x2 + (2x-1)2 = 172
• x2 + 4x2 – 4x + 1 = 289
• 5x2 – 4x – 288 = 0
• (5x + 36) (x – 8) = 0
• x = 8
• Joe’s distance = 8 miles and Jim’s distance = 15 miles

Solution to Problem 2:

• x = the husband’s age at marriage
• (3/4) x + 12 = 5/6 (x + 12)
• (3/4) x + 12 = (5/6) x + 10
• (1/12) x = 2
• x = 24
• At the marriage, the husband was 24 and the wife was 3/4 of 24, or 18

Solution to Problem 3:

• Only 2 cuts are needed, as follows:
• Divide the bar into 7 equal pieces by drawing parallel lines at 1/7th, 2/7th, 3/7th, etc.  Cut the bar at the 1/7 and 3/7 lines.
• On day one, give him the 1/7 piece.
• On day two, take back the 1/7 piece and give him the 2/7 piece.
• On day three, have him keep the 2/7 piece and give him back the 1/7 piece.
• On day four, take back both the smaller pieces and give him the 4/7 piece.
• On day five, have him keep the 4/7 piece and give him back the 1/7 piece.
• On day six, have him keep the 4/7 piece, take back the 1/7 piece, and give him the 2/7 piece.
• One day seven, have him keep the 4/7 piece, the 2/7 piece, and give him back the 1/7 piece.  He is now in possession of the entire bar of gold.

Solution to Problem 4:

• The buckling formed a right triangle, with the base equal to 2640 feet, the hypotenuse equal to 2640.5 feet.  We need to solve for the height.
• Using the Pythagorean Theorem, h2  + 26402 = 2640.52
• h2 + 6,969.600 = 6,972,240.25
• h = 51.38 feet

Solution to Problem 5:

• Yes, the persons and horses can be determined.  Let H = the number of horses and P = the number of persons.
• Since horses and persons have one head each, then H + P = 74
• Since horses have 4 legs and persons have 2 legs, then 4H + 2P = 196
• Substituting P = 74 – H into the above equation gives 4H + 2(74 – H) = 196
• So, 4H + 148 – 2H = 196
• Thus, 2H = 48, and H = 24
• Since the number of horses is 24, then the number of persons is 74 – 24, or 50
• 24 horses and 50 persons yields 74 heads and 196 legs.

Solution to Problem 6:

• Let x = the lifespan of Diophantus
• 1/6 x + 1/7 x + 1/12 x + 5 + 1/2 x + 4 = x
• Using the common denominator, 14/84 x + 12/84 x + 7/84 x + 42/84 x + 9 = x
• Thus, 75/84 x + 9 = x
• So 9/84 x = 9
• Thus, x = 84.  Diophantus lived 84 years.

Solution to Problem 7:

• √(x – 1) = 7 – x
• Squaring both sides, x – 1 = (7 – x)2
• x – 1 = 49 – 14x + x2
• x2 – 15x + 50 = 0
• (x – 5)(x – 10) = 0
• x = 5 or 10  (the calculation for 5 is straight-forward; with 10, we need to remember that the square root of 9 is 3 and -3.

Solution to Problem 8:

• The dimensions mentioned describe a right triangle, so the Pythagorean theory applies
• If the shortest side is represented by x, then x2 + (x + 50)2 = (x +100)2
• Thus, x2 + x2 + 100x + 2500 = x2 + 200x + 10000
• So, x2 – 100x – 7500 = 0
• Factoring this polynomial yields (x + 50)(x – 150) = 0
• So, the short side of the lot is 150 feet, the longer side is 200 feet, and the diagonal length is 250 feet.

Solution to Problem 9:

• 5 ( √ (d + 4) = t – 10    with d = Don’s age and t = Tom’s age
• d = t + 10, or t = d – 10
• Substituting yields,  5 ( √ (d + 4) = (d – 10) – 10
• Squaring both sides of the equation, 25 ( d + 4 ) = ( d – 20 )2
• So, 25d + 100 = d2 – 40d + 400
• Thus, d2 – 65d + 300 = 0
• Factoring the polynomial yields (d – 5)(d – 60) = 0
• Hence, Don’s age is 60 and Tom is 50.

Solution to Problem 10:

• C° = 5/9 (F° – 32)
• If C = F, then  C = 5/9 (C – 32)
• Thus,  9/5 C = C – 32
• So, 4/5 C = -32
• And, 4C = -160
• C = -40°
• Thus, -40° C = -40° F