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math41to50 (Math problems ** )

Mathematical problems — Problems 41 to 50

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

Problem 41:

If it was 2 hours later than it is right now, it would be 1/2 as long to noon as it would be if it was 1 hour earlier.  What is the current time?

Problem 42:

Dan is Tom’s father.  Five years from now, Dan will be six times as old as Tom was five years ago.  Also, five years from now, Dan will be three times as old as Tom will be.  How old was Dan when Tom was born?

Problem 43:

To date, Michael has run 36 half-marathon races.  Michael is currently 66 years old.  He runs 4 half-marathons per year.  How old will Michael be when the number of half-marathons run will equal his age?

Problem 44:

You are drawing a card from a standard deck of playing cards.  What is the probability that you will draw either an ace or a club?

Problem 45:

Ruth usually cuts a pie into 8 equal pieces.  She desires, however, to cut a particular pie into pieces that are 1 and 1/3 larger than her usual pieces.  How many pieces of pie will she cut the pie into?

Problem 46:

Can you create each of the numbers 0 through 20 using four 4’s and the operations of addition, subtraction, multiplication, and division only?

Problem 47:

A right triangle with integer sides a, b, and c exists such that a + c = 49.  What is the triangle’s area?

Problem 48:

What is the difference between an 80% decrease and a 60% decrease followed by a 20% decrease?

Problem 49:

A store advertises a 60% decrease in cost on all items in the store.  What does the price tag have to read so that an item originally costing $100 would still cost $100 after the 60% discount?

Problem 50:

There are 40 socks in a drawer, and 65% of the socks are blue and 35% of the socks are black.  You are blind.  How many socks do you have to remove from the drawer before you are certain that you have a matching pair?

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Solution to Problem 41:

  • x = current time
  • 12 – (x + 2) = 1/2 (12 – (x – 1))
  • 10 – x = 1/2 (13 – x)
  • 20 – 2x = 13 – x
  • x = 7 (o’clock)

Solution to Problem 42:

  • D = Dan’s current age and T = Tom’s current age
  • D + 5 = 6 (T – 5)
  • Also, D + 5 = 3 (T + 5)
  • Thus, 6 (T – 5) = 3 (T + 5)
  • 6T – 30 = 3T + 15
  • 3T = 45
  • T = 15, so Tom is currently 15 years old
  • Since D + 5 = 6 (T – 5), then D + 5 = 6 (15 – 5)
  • D = 60 – 5 = 55, so Dan’s current age is 55.
  • Since Tom is 15, the Dan was 40 when Tom was born.

Solution to Problem 43:

  • Let x = the number of years that will elapse from now until the years in which the number of races equals Michael’s age in years
  • 36 + 4x = 66 + x
  • Thus, 3x = 30
  • So, x = 10
  • 36 + 4(10) = 76
  • So 10 years from now (at the end of the year) Michael will be 76 years old and he will have run 76 half-marathons.

Solution to Problem 44:

  • Since there are 4 aces and 13 clubs in a deck, there are 17 cards that qualify.  However, the ace of clubs is in both categories, so the number of qualified cards is 17 – 1 = 16.  Thus, 16 ÷ 52 = 4 ÷ 13 = 0.3077 or approximately 31%.

Solution to Problem 45:

  • 1/8 x 4/3 = 1/6.  Thus, she will have 6 pieces of pie, equally sized.

Solution to Problem 46:

  • 0 = (4 / 4) – (4 / 4) = ((4 – 4) / 4 ) • 4
  • 1 = (4 • 4) / (4 • 4) = (4 + 4) / (4 + 4)
  • 2 = (4 / 4) + (4 / 4) = (4 • 4) / (4 + 4)
  • 3 = (4 + 4 + 4) / 4
  • 4 = ((4 – 4) / 4) + 4
  • 5 = ((4 • 4) + 4) / 4
  • 6 = ((4 + 4) / 4) + 4
  • 7 = (4 + 4) – (4 / 4)
  • 8 = (4 + 4) • (4 / 4)
  • 9 = (4 + 4) + (4 / 4)
  • 10 =
  • 11 =
  • 12 = (4 – (4 / 4)) • 4
  • 13 =
  • 14 =
  • 15 = (4 • 4) – (4 / 4)
  • 16 = (4 • 4) • (4 / 4)
  • 17 = (4 • 4) + (4 / 4)
  • 18 =
  • 19 =
  • 20 = ((4 / 4) + 4) • 4

Solution to Problem 47:

  • We know that a2 + b2 = c2 and a + c = 49 and that A = 1/2 (ab).  However, there are too many unknowns to solve the problem
  • There are only 17 “Pythagorean Triples” with integer sides where c < 100.  They are: (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41), (11, 60, 61), (12, 35, 37), (13, 84, 85), (15, 20, 25), (16, 63, 65), (20, 21, 29), (28, 45, 53), (33, 56, 65), (36, 77, 85), (39, 80, 89), (48, 55, 73), and (65, 72, 97).
  • Only 2 of these triangles have a + c = 49  (in bold print above).
  • Another approach is this:  Knowing that a has to be 24 or less because it cannot be larger than c and and a + c must equal 49, one can look at all the possible combinations of a and c  that meet these criteria.  The only combinations are the two bold groupings shown above.
  • The area of both of these triangles = 210 square units.

Solution to Problem 48:

  • 80% decrease:  x – 0.8x = 0.2x
  • 60% decrease followed by 20% decrease:  ( x – 0.6x ) – ( 0.2 ( x – 0.6x )) = 0.4x – 0.2x + 0.12x = 0.32x
  • Thus for $100, the difference between the two is $12 ( $32 – $20 ).

Solution to Problem 49:

  • x – 0.6x = 100
  • 0.4x = 100
  • x = $250

Solution to Problem 50:

  • Only three socks need to be taken out of the drawer.  At least two of the sock have to be the same color.  Being blind, however, you would still not know which two are black or which two are blue.

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