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math81to90 ( ** )

Mathematical problems — Problems 81 to 90

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

Problem 81:

Find the sum of the 16 terms in the arithmetic sequence that starts with 15 and ends with 60.  What is the difference between the terms?

Problem 82:

If the first term of an arithmetic sequence is 3 and the 50th term is 297, then what is the common difference between the terms?

Problem 83:

Three points on a graph lie on the perimeter, or circumference, of a circle.  The three points, points A, B, and C, are (4, 20), (16, 14), and (4, -14).  What are the coordinates of the center of the circle?  What is the diameter of the circle?  [I conceived of, designed, and solved this problem myself]

Problem 84:

A man drives his van to work each day, carrying a certain number of riders with him.  It costs $12 per day for gasoline, which is equally divided between the riders and the driver himself.  Two new riders join the carpool, and the driver can reduce the cost by $1 per day per rider.  How many are now riding to work in the van?

Problem 85:

Using only eight 8’s and addition signs, can you create an addition problem with the sum equal to 1000?

Problem 86:

The number of boys in a school is ten less than twice the number of girls.  Also, six less than half the total number of students in the school have cars.  If 202 students have cars, then how many girls and how many boys are in the school?

Problem 87:

If you add 5 to a given positive integer and then subtract 4 from the same integer, and then multiply the sum by the difference, a product of 36 is obtained.  What is the original integer?

Problem 88:

The area of a triangle is 20 square inches.  The base of the triangle is 6 inches longer than the height of the triangle.  How many inches in length are the base and the height?

Problem 89:

A man bought two car lots for $50,000 apiece.  There were a total of 7 cars on the two car lots.  The individual cars on one lot cost $15,000 more than the cars on the other lot.  What did the man pay for the individual cars on each lot?

Problem 90:

A local business needs $10,000 in interest annually to fund its charitable-giving program.  The business now has $60,000 invested, with one third of the money in a low-risk stock earning 5% interest, and the rest in a diversified fund earning 10% interest.  This division of the funds is mandated by the business’s board of directors. How much more money does the business need to reach its charitable goal, and how much should be invested in each of the two investments?

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Solution to Problem 81:

  • Sn = n/2 (a1 + an)
  • Sn = 16/2 (15 + 60)
  • Sn = 8 • 75 = 600  (the sum of the numbers)
  • an = a1 + d(n – 1)
  • 60 = 15 + d(16 – 1)
  • 45 = 15d
  • d = 3  (the common difference)

Solution to Problem 82:

  • an = a1 + d(n – 1)
  • 297 = 3 + d(50 – 1)
  • 294 = 49d
  • d = 6  (the common difference is 6)

Solution to Problem 83:

  • The solution must be found in steps:
    • Step one is to determine the midpoint of a line between two of the three points.
      • Let’s choose points A (4, 20) and B (16, 14).
      • The midpoint formula is:   midpoint = ((x1 + x2) / 2, (y1 + y2) / 2)
      • Thus, midpointAB = ((4 + 16) / 2, (20 + 14) / 2) = (10, 17)
    • Step two is to determine the slope of the line running between points A and B.
      • m = (y2 – y1) / (x2 – x1), thus mAB = (20 – 14) / (4 – 16)
      • mAB = 6 / -12 = – 1/2
    • Step three
      • By definition, a line perpendicular to line AB and running threw the midpoint of AB would pass through the center of the circle with which we are dealing. A line perpendicular to AB would have a slope of +2 since the slope of AB is – 1/2.  Knowing the slope ( +2 )and a point on a line (10, 17) that passes through the center of the circle, we can know use the point-slope equation to write an equation this line
      • The point-slope equation is: ( y – y1) = m ( x – x1) ,  so (y – 17) = +2 (x – 10)
      • Thus, y – 17 = 2x – 20, and  y = 2x – 3
    • We will now repeat steps one through three using points B and C, to determine the equation of another line that runs through the center of the circle.
    • Step four is to determine the midpoint of line BC
      • We will now work with and B (16, 14) and C (4, -14).
      • The midpoint formula is:   midpoint = ((x1 + x2) / 2, (y1 + y2) / 2)
      • Thus, midpointBC = ((16 + 4) / 2, (14 + -14) / 2) = (10, 0)
    • Step five is to determine the slope of the line running between points B and C.
      • m = (y2 – y1) / (x2 – x1), thus mBC = (-14 – 14) / (4 – 16)
      • mBC = -28 / -12 =  7/3
    • Step 6
      • By definition, a line perpendicular to line BC and running threw the midpoint of BC would pass through the center of the circle with which we are dealing. A line perpendicular to BC would have a slope of – 3/7 since the slope of BC is 7/3.  Knowing the slope ( -3/7 )and a point on a line (10, 0) that passes through the center of the circle, we can know use the point-slope equation to write an equation this line
      • The point-slope equation is: ( y – y1) = m ( x – x1) ,  so (y – 0) = -3/7 (x – 10)
      • Thus, y = -3/7 x + 30/7
    • Step 7
      • We now have the equations of TWO lines that pass through a common point (the center of the circle).  We can use these two line equations to eliminate one of the variables (y) and obtain the value of the other variable (x).
      • Hence, since  y = 2x – 3  and  y = -3/7 x + 30/7 , we know that  2x – 3 =  -3/7 x + 30/7
      • Multiplying the various terms by a common factor of 7 yields: 14x – 21 = -3x + 30
      • Thus,  17x = 51, and x = 3.  We now know that the x coordinate of the center of the circle is 3.
    • Step 8
      • Since the x coordinate of the center of the circle is 3, we can plug that value into one of the other line equations to determine y.
      • y = 2x – 3
      • y = 2 (3) – 3
      • y = 3.  Hence, the coordinates of the center of the circle are (3, 3).
    • Step 9
      • We now know a point on the circumference of the circle, as well as the center point of the circle.  We can determine the distance between these two points using the 2-point distance formula.
      • The distance between two points (x1, y1) and (x2, y2) equation is:  d = √ (x2 – x1)2 + (y2 – y1) 2
      • Thus,  d = √ (16 – 3)2 + (14 – 3) 2
      • Thus,  d = √ (13)2 + (11) 2
      • Thus,  d = √ 169 + 121  = √ 290 = 17.03 or 17
      • The diameter of the circle would be twice that amount, or 34.

Solution to Problem 84:

  • If x = the number of riders and y represents the cost per rider, then 12 / x = y
  • With the addition of 2 rides and dropping the cost, we get:  12 / (x + 2) = y – 1
  • Solving the second equation for y yields:  y = (12 / (x + 2)) + 1
  • Setting the y from equation 1 and equation 3 as equals yields: 12 / x =  (12 / (x + 2)) + 1
  • Simplifying yields:  12x + 24 = 12x + x2 + 2x
  • Thus:  x2 + 2x – 24 = 0
  • And:  ( x – 4 )( x + 6 ) = 0
  • Thus, the original number of riders was 4 and the current number is 6.  The original riders were paying $4 and the current riders pay $3.
  • This problem can also be solved by starting with x • y = 12 and ( x + 2)( y – 1 ) = 12.  See if you can reach the answer with this starting point.

Solution to Problem 85:

  • 888 + 88 + 8 + 8 + 8 =  1000

Solution to Problem 86:

  • b = boys,  g = girls
  • b = 2g – 10  and 1/2 (b + g) – 6 = 202
  • Substituting yields: (1/2 ((2g – 10) + g)) – 6 = 202
  • Thus, (1/2 (3g – 10)) – 6 = 202
  • And, (3/2 g – 5) – 6 = 202
  • 3/2 g = 213, so 3g = 426
  • g (the number of girls) = 142
  • Thus, the number of boys = 2(142) – 10 = 274

Solution to Problem 87:

  • (x + 5)(x – 4) = 36
  • x² – 4x + 5x – 20 = 36
  • x² + x – 56 = 0
  • (x – 7)(x + 8) = 0
  • x = 7    (-8 is also a viable answer but it is not a positive integer)

Solution to Problem 88:

  • The area of a triangle = ½ (b • h)
  • ½ (b)(b – 6) = 20
  • ½ (b² – 6b) = 20
  • b² – 6b = 40
  • b² – 6b – 40 = 0
  • (b – 10)(b + 4) = 0
  • The base of the triangle = 10 inches and the height = 4 inches

Solution to Problem 89:

  • We will assign ‘x’ to the number of cars on the first lot, and thus the other lot will have 7 – x cars.
  • So, ($50,000 ÷ x) = (($50,000 ÷ (7 – x )) + $15,000
  • Multiplying all factors by ((x) (7 – x)) yields (7 – x)($50,000) = (x)($50,000) + (x)(7 – x)($15,000)
  • Dividing all factors of the equation by 5,000 yields (7 – x)(10) = (x)(10) + (x)(7 – x)(3)
  • Simplification yields 70 – 10x = 10x +21x – 3x2
  • Thus,  3x2 – 41x +70 = 0
  • Factoring the polynomial yields  (3x – 35)(x – 2) = 0
  • Thus, x = 2, which means that there are 2 cars on the first lot, costing $25,000 each, and 5 cars on the other lot, costing $10,000 each.  The more expensive cars are $15,000 more than the less expensive cars.

Solution to Problem 90:

  • If x equals the amount of money needed in the low-risk stock fund, then 2x would represent the amount in the diversified fund
  • Then,  0.05(x) + 0.1(2x) = 10,000
  • Simplifying yields 0.05x + 0.2x = 10,000
  • And then 0.25x = 10,000
  • So, x = 40,000.  This is the amount needed in the first fund.
  • This amount in the second fund is double what is in the first fund, so there is $80,000 needed in the second fund
  • Thus, the total amount of money needed in both funds is $120,000.
  • The business currently has $60,000 invested, so they need twice that amount, or $120,000.

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