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# mathproblems121to130 ( ** )

### Mathematical problems — Problems 121 to 130

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

Problem 121:

Brad left Provo one hour after Cal.  Brad caught up with Cal 200 miles from Provo.  Brad traveled at a speed that was 125% of Cal’s speed.  How fast, in miles per hour, did Brad and Cal travel?

Problem 122:

You have 10 coins in quarters and pennies.  You have two less than five times as many quarters as pennies.  How much money do you have?

Problem 123:

Jim is 7 miles from shore in a small paddle boat.  He needs to get to a store for supplies.  The store is 32 miles further along the coast and 6 miles inland.  Jim can paddle 2½ times as fast as he can walk.  He paddles to shore x miles along the shore line from his present position, and then continues walking directly to the store.  Jim spent the same amount of time rowing as he did walking.  How far down the shore did he land?

Problem 124:

A rectangle is to have a perimeter of 36 feet and the greatest area possible.  What are the dimensions of the rectangle that will yield the largest area?

Problem 125:

John is 50 years of age, and he runs 3 marathons every year.  He has just completed his 30th marathon.  How old will John be when the number of marathons run will equal his age in years?

Problem 126:

The sum of two numbers is 40, and the product of those same two numbers is 204.  What are the numbers?

Problem 127:

Jane and Anne are 12 years apart in age, with Jane being younger.  If you divide each of their two ages by 4, and add together those results, it equals 15.  How old are Jane and Anne?

Problem 128:

The difference between 2 positive numbers is 20.  If you square the larger number and subtract 10 times the smaller number from the square, you get 575.  What are the two numbers?

Problem 129:

One integer is four smaller than the other.  The sum of their reciprocals is 5/24.  What are the numbers?

Problem 130:

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Solution to Problem 121:

• Let db = Brad’s distance traveled, rb = Brad’s speed, and tb = Brad’s time of travel.
• Let dc = Cal’s distrance traveled, rc = Cal’s speed, and tc = Cal’s time of travel.
• We know that d = rt, and knowing that Brad and Cal traveled the same distance, we can say that rb • tb = rc • tc
• Thus,  1.25 rc • (tc – 1) = rc • tc (Brad traveled at 1.25 times Cal’s speed, and for one less hour)
• Simplifying yields,  1.25 rc tc – 1.25 rc = rc • tc
• And thus,  0.25 rc tc = 1.25 rc
• And,  0.25 tc = 1.25 , so tc = 5
• Thus, Cal traveled 200 miles in 5 hours, which yields a speed of 40 mph.
• Since Brad’s speed was 1.25% of Cal’s speed, he traveled at 50 mph.

Solution to Problem 122:

• Let p = the number of pennies  and q = the number of quarters.
• Thus, 5p – 2 = q and  p + q = 10.
• Substitution yields  5p – 2 = 10 – p.
• Solving for p yields  6p = 12, and p = 2
• If p = 2, then q = 8.  Two pennies plus 8 quarters equals \$2.02.

Solution to Problem 123:

• The problem involves two right triangles.  The perpendicular sides of the first triangle are 7 and x in length.  The perpendicular sides of the second triangle are 6 and (32 – x) in length.  The slanting sides of the two triangles form a straight line together.  We know the length of the slanting line in triangle 1 will be 2.5 times as long as the slanting line in triangle 2.  We will call those lines 2.5y and y.
• Using the right triangle formula, we can say that  7² + x² = (2.5y)²  and that  6² + (32 – x)² = y²
• Solving for y² in the first equation yields  y² = (7² + x²) ÷ 6.25.
• A new equation can be written by setting the two y² amounts equal to each other thusly:  (7² + x²) / 6.25 = 6² + (32 – x)²
• This equation simplifies to  5.25x² – 400x +6576 = 0
• Using the quadratic equation, we find that x = 24.  Thus, Jim came ashore 24 miles further down the coast from where he started.

Solution to Problem 124:

• As a general rule, a square will always give the largest area for a given perimeter.
• Thus, if the side of the square is x feet, then we know that 4x = 36 feet.  Thus, the side of the square is 9 feet long, and the area would be 81 square feet.
• As a comparison, the rectangle with 2 sides 10 feet in length and 2 other sides 8 feet in length would have a perimeter of 36 feet, but the area would be only 80 square feet.

Solution to Problem 125:

• Let y = years to come.  Let m = the number of marathons run.  Thus, 30 + 3y = m.
• Also, 50 + y would equal John’s age to come.  If the age equals the marathons, then 30 + 3y = 50 + y.
• Then, 2y = 20, and y = 10.  Thus, in 10 years, John will be 60 years old and he will have run 30 + 30 marathons, or 60.

Solution to Problem 126:

• Let x and y represent the two numbers.  We know that x + y = 40, and we know that x • y = 204.
• Solving for x in the second equation yields  x = 204 / y.
• Substitution yields  (204 / y) + y = 40.  Multiplying all factors by y yields  204 + y² = 40y.
• Rearranging yields the polynomial equation  y² – 40y +204 = 0
• Factoring yields  (y – 6)(y – 34) = 0.  Thus, the two numbers are 6 and 34.

Solution to Problem 127:

• Let j = Jane’s age and a = Anne’s age.  Thus, j + 12 = a.  Also, we know that  (j / 4) + (a / 4) = 15.
• Substitution yields  (j / 4) + ((j + 12) / 4) = 15, and multiplying all factors by 4 yields  4(j / 4) + (j + 12) = 60.
• Simplification yields  j + j + 12 = 60,  2j = 48, and thus  j = 24.  Thus, Jane is 24 years old and Anne is 36 years old.

Solution to Problem 128:.

• Let x = the larger number and y = the smaller number.  Thus, x – y = 20, and y = x – 20.
• We also know that x² – 10y = 575.  Substituting for y, we get  x² – 10(x – 20) = 575.
• Simplifying yields  x² –  10x + 200 = 575, and thus  x² – 10x – 375 = 0.
• Factoring yields, (x – 25)(x + 15) = 0, and thus x = 25.  If x = 25, then y = 5.

Solution to Problem 129:

• Let x = the larger integer and y the smaller integer.  We know that x – 4 = y.
• We also know that (1/x) + (1/y) = 5/24.  Substituting for y yields  (1/x) + (1/x – 4) = 5/24.
• Multiplying both sides of the equation by (x)(x-4) yields  (x-4) + (x) = 5/24 (x)(x – 4)
• Simplification yields  24(2x – 4) = 5x² – 20x  and then  48x – 96 = 5x² – 20x
• Simplification yields  5x² – 68x + 96 = 0.  Factoring the polynomial yields  (5x – 8)(x – 12) = 0.
• Thus, x = 12 and y = 8.

Solution to Problem 130:

• Let x