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math21to30 (Math problems)

Mathematical problems — Problems 21 to 30


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

Problem 21:

Mary travels 30 miles to work.  She found that if she increases her normal speed by 10 mph, she arrives 6 minutes earlier to work.  What is her normal speed?

Problem 22:

Three positive integers exist such that c2 – a2 – b2 = 101 and a • b = 72.  What is the value of a + b + c ?

Problem 23:

If 86 in base 10 is represented by 321 in base x, what does 123 in base x equal in base 10?

Problem 24:

The square of a given number is 6 more than 5 times the number.  What is the number?

Problem 25:

The ‘ones’ digit of a 2 digit number is 4 less than the ‘tens’ digit.  Dividing the number by the sum of the digits yields 7.  What is the number?

Problem 26:

Divide 30 by half and add 10.  What is the answer?

Problem 27:

If you take 12 apples from a pile of 20 apples, how many apples do you have?

Problem 28:

Twelve years from now, Ralph will be 4 times as old as he was 15 years ago.  How old is Ralph now?

Problem 29:

If Tom is twice as old as Dick was 10 years ago, and if the sum of their ages 5 years ago was 90, how old is Tom now?

Problem 30:

Three times a certain positive integer is one less than the square of another positive integer.  The difference of the square of these two integers equals 9.  What are the integers?

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Solution to Problem 21:

  • r = rate or speed,  t = time
  • r1 x t1 = 30 miles,  r2 x t2 = 30 miles,  r1 x t1 = r2 x t2
  • r2 = r1 + 10,  t2 = t1 – 6 min = t1 – 1/10 hour
  • So, r1 x t1 = (r1 + 10) ( t1 – 1/10 )
  • rt = rt – r/10 +10t – 1
  • 10t – r/10 = 1, or 100t – r = 10
  • Since rt = 30, then t = 30/r
  • Thus,  100 (30/r) – r = 10
  • (3000/r) – r = 10
  • 3000 – r2 = 10r
  • r2 + 10r – 3000 = 0
  • (r + 60) (r – 50) = 0
  • r = 50 mph

Solution to Problem 22:

  • With three unknowns and only 2 equations, this problem cannot be solved mathematically, but it can be solved intuitively.
  • With a and b as integers, there are only 6 combinations where a • b = 72 ( 1 • 72, 2 • 36, 3 • 24, 4 • 18, 6 • 12, and 8 • 9 ).
  • With c2 = a2 + b2 + 101, we only need to find the a – b pair which forms an integer square root with 101.
  • Only the pairing of 4 and 18 works ( 212 = 42 + 182 + 101), or 441 = 441 .
  • Thus, a + b + c = 4 + 18 + 21 = 43

Solution to Problem 23:

  • 1 (x0) + 2 (x1) + 3 (x2) = 86
  • 3x2 + 2x + 1 = 0
  • (3x + 17)(x – 5) = 0
  • x = 5
  • 3(50) + 2(51) + 1(52) = y
  • 3 + 10 + 25 = 38, so 123 in base 5 equals 38 in base 10

Solution to Problem 24:

  • Thus, x2 – 6 = 5x
  • This yields a quadratic equation:  x2 – 5x – 6 + 0
  • Factoring yields (x – 6)(x + 1) = 0
  • Hence, the number is 6.  [Note: -1 is also a correct response]

Solution to Problem 25:

  • xy equals the number, with x = the ‘tens’ digit and y = the ‘ones’ digit
  • x = y + 4  and  (10x + y) / (x + y) = 7
  • Simplifying the second equation yields:  10x + y = 7x +7y
  • Thus, 3x = 6y
  • Substituting y + 4 for x yields 3(y + 4) = 6y
  • So, 3y + 12 = 6y
  • So, y = 4
  • Hence, the original number must be 84

Solution to Problem 26:

  • Did I trick you?  Notice that the problem says ‘by half’, not ‘in half’.  Dividing 30 by half yields 60, not 15.  The correct answer is 70.

Solution to Problem 27:

  • Did I ‘get’ you again?  If YOU take 12 apples from a pile, no matter how many apples are in the pile, YOU have 12 apples!  Did you say 8?

Solution to Problem 28:

  • R = Ralph’s current age
  • R + 12 = 4 (R-15)
  • R + 12 = 4R – 60
  • 3R = 72
  • R = 24, so Ralph is presently 24 years old.

Solution to Problem 29:

  • T = Tom’s age now;  D = Dick’s age now
  • T = 2(D – 10)  and (T – 5) + (D – 5) = 90
  • Substituting yields:  ((2(D – 10)) – 5) + (D – 5) = 90
  • So, 2D – 20 – 5 + D – 5 = 90
  • 3D – 30 = 90, so 3D = 120, and D = 40
  • So, T = 2(40 – 10), or T = 60
  • Tom is 60 and Dick is 40

Solution to Problem 30:

  • From the first statement:  3x + 1 = y2
  • From the second statement:  x2 – y2 = 9
  • Substituting from the 1st statement into the 2nd statement yields:  x2 – (3x + 1) = 9
  • Thus,  x2 – 3x – 1 = 9
  • So, x2 – 3x – 10 = 0
  • Factoring yields  (x – 5)(x + 2) = 0
  • So x = 5
  • If x = 5, then y = 4

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