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# math71to80 (Math problems)

### Mathematical problems — Problems 71 to 80

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

Problem 71:

Two lines are on a graph.  The first line passes through the point (-4, 13) and has a slope of -3/2.  The second line passes through the point (10, 8) and has a slope of 1/2.  At what point on the graph do these two lines intersect?  This problem can be solved mathematically without actually drawing the lines on graph paper.

Problem 72:

The sum of three consecutive integers is 45.  What are the integers?

Problem 73:

The sum of four consecutive integers is 38.  What is the largest of the four integers?

Problem 74:

The sum of 8 consecutive numbers equals 4.  What is the product of these same 8 consecutive numbers?

Problem 75:

The product of two consecutive odd numbers is 143.  What are the two numbers?

Problem 76:

The sum of 4 consecutive odd integers is 64.  What are the numbers?

Problem 77:

The difference between the cubes of two consecutive integers is 127.  What are the two integers?

Problem 78:

The sum of 5 consecutive even integers is 100.  What is the difference between the largest and smallest of these 5 integers?

Problem 79:

The sum of five consecutive multiples of 4 is 20.  What is the product of the two largest numbers?

Problem 80:

In the sequence of terms beginning with 7 and with a common difference of 5, what is the 10th term in that sequence?

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Solution to Problem 71:

• Using the point-slope equation for line 1 yields:  (y – 13) = -3/2 (x – (-4))
• Thus, y – 13 = (-3/2 • x) – 6
• And, y = (-3/2 • x) + 7
• Using the point-slope equation for line 2 yields:  (y – 8) = 1/2 (x – 10)
• Thus, y – 8 = (1/2 • x) – 5
• And, y = (1/2 • x) + 3
• Since the intersection point will be at the same value for y, these two equations can be set as equal to each other.
• Hence,  (-3/2 • x) + 7 = (1/2 • x) + 3
• Simplifying yields:  2x = 4
• Thus, x = 2.  With x = 2, then y = 4.  The point of intersection of the two points is thus (2, 4).

Solution to Problem 72:

• x + (x + 1) + (x + 2) = 45
• 3x + 3 = 45
• 3x = 42
• x = 14
• The integers must be 14, 15, 16

Solution to Problem 73:

• x + (x + 1) + (x + 2) + (x + 3) = 38
• 4x + 6 = 38
• 4x = 32, or x + 8
• If 8 is the smallest of the integers, then 11 must be the largest.

Solution to Problem 74:

• x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6) + (x + 7) = 4
• 8x + 28 = 4, or 8x = -24
• So, x = -3
• Since x = -3, then the other seven numbers are -2, -1, 0, 1, 2, 3, and 4.
• With zero as one of the numbers, the product has to be equal to 0 (zero)!

Solution to Problem 75:

• x • y = 143, with y being 2 greater than x
• Thus, x • (x + 2) = 143
• So, x2 + 2x – 143 = 0
• Factoring yields: (x – 11) (x + 13) = 0
• Thus, the solutions are 11 and 13 and -11 and -13.

Solution to Problem 76:

• x + (x + 2) + (x + 4) + (x + 6) = 64
• 4x + 12 = 64
• 4x = 52. so x = 13
• Thus, the integers are 13, 15, 17, and 19.

Solution to Problem 77:

• x3 – (x – 1)3 = 127
• x3 – (x – 1)(x2 – 2x +1) = 127
• x3 – (x3 – 2x2 + x – x2 + 2x – 1) = 127
• x3 – (x3 – 3x2 + 3x – 1) = 127
• 3x2 – 3x + 1 = 127
• 3x2 – 3x -126 = 0
• Factoring yields:  (3x – 21)(x + 6) = 0
• Thus, x = 7 and the two numbers must be 6 and 7.
• 343 – 216 = 127

Solution to Problem 78:

• x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = 100
• 5x + 20 = 100
• 5x = 80
• So, x = 16.  The numbers must be 16, 18, 20, 22, and 24.
• The difference between the largest and smallest of the numbers is 24 – 16 = 8
• In actuality, one doesn’t need to know the specific numbers to determine the difference between the largest and smallest of the 5 integers.  The difference between the largest and smallest of 5 consecutive even integers will always be 8, no matter what the numbers are.

Solution to Problem 79:

• x + (x + 4) + (x + 8) + (x + 12) + (x + 16) = 20
• 5x + 40 = 20
• 5x = -20
• x = -4
• Thus, the 5 multiples of 4 are -4, 0, 4, 8, and 12.  Their sum does equal 20.
• The largest two of the five numbers are 8 and 12.  Their product is 96.

Solution to Problem 70:

• Remember the equation,  an = a1 + d(n – 1)  , with an being the nth term of an arithmetic sequence, a1 being the first term, and d being the common difference between the terms.
• Thus, a10 = 7 + 5(10 – 1) = 7 + 45 = 52
• Double-checking yields, 5 – 12 – 17 – 22 – 27 – 32 – 37 – 42 – 47 – 52.