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# math61to70 (Math problems ** )

### Mathematical problems — Problems 61 to 70

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

Problem 61:

The sum of a number and 12 is equal to twice the number.  What is the number?

Problem 62:

The sum of a number and 3 more than twice the number equals 36.  What is the number?

Problem 63:

When 5 is subtracted from two times a given number, the result is two more than the number.  What is the number?

Problem 64:

The sum of a number and 8 times its reciprocal is 6.  What is the number?

Problem 65:

The sum of two integers is 9.  The sum of their squares is 21 greater than the product of the two integers.  What are the numbers?

Problem 66:

The sum of two integers is 21.  The sum of the squares of these integers is 233.  What are the integers?

Problem 67:

The product of two integers is 494.  One of the integers is 7 greater than the other.  What are the integers?

Problem 68:

If y varies directly with x, and y = 20 when x = 5, then what is the value of y when x = 2 ?

Problem 69:

If the square of y varies directly with the cube of x, and if y is equal to 3 when x is equal to 2, then what is the value of x when y is 24?

Problem 70:

The volume of a sphere varies directly with the cube of its radius.  If a sphere with a radius of 3 yards has a volume of 113.1 cubic yards, then what is the volume of a sphere that has a radius of 5 yards?

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Solution to Problem 61:

• x + 12 = 2x
• So, x = 12

Solution to Problem 62:

• x + ( 2x + 3 ) = 36
• 3x + 3 = 36
• 3x = 33
• So, x = 11

Solution to Problem 63:

• 2x – 5 = x + 2
• 2x – x = 2 + 5
• x = 7

Solution to Problem 64:

• x + (8/x) = 6
• Multiplying all components by x yields:  x2 + 8 = 6x
• So, x2– 6x + 8 = 0
• Factoring yields: (x – 2)(x – 4) = 0
• So the two solutions are 2 and 4.

Solution to Problem 65:

• x + y = 9  and  (x2 + y2) – 21 = x • y
• Substituting (9 – x) for y into the second equation yields: ( x2 + (9 – x)2 ) – 21 = x (9 – x)
• Simplifying yields:  x2 + 81 – 18x + x2 – 21 = 9x – x2
• Further simplifying yields:  3x2 – 27x + 60 = 0
• Factoring yields:  (3x – 15)(x – 4) = 0
• Hence, x = 4 and the other integer must be 5

Solution to Problem 66:

• x + y = 21 and x2 + y2 = 233
• Substituting (21 – x) for y in the second equation yields:  x2 + ( 21 – x )2 = 233
• Simplification yields: x2 + 441 – 42x + x2 = 233
• Further simplification yields: 2x2 – 42x + 208 = 0
• Factoring yields: ( 2x – 26 )( x – 8 ) = 0
• Hence, the two integers are 8 and 13.

Solution to Problem 67:

• x • y = 494 and x + 7 = y
• Substituting ( x + 7 ) for y in the first equation yields:  x • ( x + 7 ) = 494
• Simplification yields:  x2 + 7x – 494 = 0
• Factoring yields:  ( x – 19 )( x + 26 ) = 0
• Hence, the integers are 19 and 26.

Solution to Problem 68:

• 20 / 5 = y / 2
• 5y = 40
• y = 8

Solution to Problem 69:

• 32 / 23 = 242 / x3
• 9 / 8 = 576 / x3
• 9 x3 = 4,608
• x3 = 512
• x = 8

Solution to Problem 70:

• V1 / r13  =  V2 / r23
• 113.1 / 33 = V2 / 53
• V2 = (113.1) (125) / 27 = 523.6 cubic yards