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# mathproblems101to110 ( ** )

### Mathematical problems — Problems 101 to 110

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

Problem 101:

Gracie has twice as many nickels as she has quarters.  She also has 5 less than three times the number of dimes as she has quarters.  The total amount of money she has is \$6.00.  How many of each coin does she have?

Problem 102:

Ralph spent \$20.40 at a concession stand.  He bought three hot dogs, two serving of French fries, and one large drink.  One hot dog costs \$1.80 more than one serving of fries.  A drink costs \$1.20 less than a hot dog.  How much did each item cost?  [Ignore sales tax]

Problem 103:

A bag is to contain bolts that cost \$5 per pound, nails that cost \$3 per pound, screws that cost \$2 per pound, and washers that cost \$6 per pound.  There should be an equal amount, by weight, of the bolts and the washers.  There should be twice as many nails as washers, and twice as many screws as nails.   How much of each type of hardware should be included if the bag is to cost \$4.25?

Problem 104:

Jake is 6 years older than Jack.  In seven years, the sum of Jake’s and Jack’s ages with equal 52.  How old are each of these young men?

Problem 105:

Stan is four times as old as Steve.  In 10 years, Stan will be twice as old as Steve.  What are the ages of Stan and Steve presently?

Problem 106:

A right triangle with sides ‘a’ (opposite), ‘b’ (adjacent), and ‘c’ (hypotenuse) has side lengths that are all integers.  Twice the length of ‘a’ plus twice the length of ‘b’ is 4 less than 3 times the length of ‘c’.   Also, twice the length of ‘a’ plus the length of ‘b’ equals 40.  What are the lengths of the three sides?

Problem 107:

James has \$100,000 to invest.  He wants to put some of this money in an account that earns 8 percent interest and the rest in a riskier account that promises to earn at the rate of 12%.  He needs a yearly earnings of \$4,500.  How much should he put in each of the two accounts?

Problem 108:

The square root of a certain son’s age is three less that the square root of his father’s age.  Three times the son’s current age is 9 less than the father’s current age.  How old are the father and the son?

Problem 109:

A cash drawer contains one-dollar bills, five-dollar bills, ten-dollar bills, and twenty-dollar bills.  There are two more \$10’s than \$5’s, eight less than twice as many \$20’s than \$10’s, and ten more than twice as many \$1’s than \$20’s.  The total amount of money in the drawer is \$650.  How many of each type of bills are there?

Problem 110:

Today, Al’s age is 4 years more than twice Ed’s age.  Three years ago, the sum of their ages was 31.  How old are Al and Ed?

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Solution to Problem 101:

• Let q = quarter and d = dime and n = nickel.
• 2q = n and (3q – 5) = d
• Converting dollars to cents yields  5n + 10d + 25q = 600
• Substitution yields 5(2q) + 10(3q – 5) + 25 q = 600
• Simplifying, 10q + 30q – 50 + 25q = 600
• Further simplifying yields  65q – 650.  Thus, Gracie has 10 quarters.
• That means she has 20 nickels and 25 dimes also.
• 10 quarters plus 20 nickels plus 25 dimes does indeed equal \$6.00

Solution to Problem 102:

• Let h = the cost of a hot dog, f = the cost of an order of French fries, and d = the cost of a drink.
• We know that   3h + 2f + d = 2040  (converting all dollar amounts into cents)
• We also know that  h = f + 180 and that h = d + 120.  Thus,  f = h – 180  and d = h – 120
• Placing all terms of the first equation in terms of h leads to   3h + 2(h – 180) + (h – 120) = 2040
• Thus, 3h + 2h – 360 + h – 120 = 2040
• And, 6h = 2520
• h = 420, or \$4.20.  Thus the hot dogs cost \$4.20, the fries cost \$2.40, and the drinks cost \$3.00.

Solution to Problem 103:

• Let b = weight of the bolts, n = the weight of the nails, s = the weight of the screws, and w = the weight of the washers.
• Thus, b = w,  2w = n,  2n = s, and  5b + 3n + 2s + 6w = 4.25
• Stating all variables in terms of  b  yields:  5b + 3(2b) + 2(4b) + 6b = 4.25
• Simplifying yields:  25b = 4.25, and thus  b = 0.17.
• Thus, the weights are 0.17 pounds of bolts, 0.17 pounds of nails, 0.34 pounds of screws, and 0.68 pounds of washers.

Solution to Problem 104:

• Let x = Jake’s current age, and let y = Jack’s current age.
• Thus, x – 6 = y, and (x + 7) + (y + 7) = 52.
• Substitution yields  (x + 7) + ((x – 6) + 7) = 52.
• Simplifying yields  x + 7 + x + 1 + 52, and thus 2x = 44.
• Solving for x yields 22.  Thus, Jake is 22 years old and Jack is 16 years old.

Solution to Problem 105:

• Let x = Stan’s age, and y = Steve’s age.
• We know that  x = 4y  and that (x + 10) = 2 (y + 10).
• Substituting the first equation into the second equation yields  (4y + 10) = 2 (y + 10).
• Simplification yields  4y + 10 = 2y + 20,  and thus 2y = 10.  Thus, y = 5.
• Stan’s age is 20 and Steve’s age is 5.

Solution to Problem 106:

• We know that a² + b² = c²; we also know that 2a + 2b = 3c – 4; we also know that 2a + b = 40.
• From the last equation shown above, we derive that b = 40 – 2a
• Substituting that value of b into the second equation listed above yields  2a + 2(40 – 2a) = 3c – 4
• Thus, 2a + 80 – 4a = 3c – 4
• Solving for c yields  3c = 2a – 4a + 80 + 4, or c = (-2a + 84) ÷ 3
• Substituting the derived values for b and c into the first equation listed above yields  a² + (40 – 2a)² = ((-2a + 84) ÷ 3)²
• This simplifies to  a² + 4a² – 160a + 1600 = (4a² – 336a +7056) ÷ 9  and further to  9a² + 36a² – 1440a +14400 = 4a² – 336a + 7056
• This simplifies to 41a² – 1104a + 7344 = 0
• This polynomial factors to  (41a – 612)(a – 12) = 0
• Thus, a = 12,  b = 40 – 2(12) = 16, and c = (-2(12) + 84) ÷ 3 = 20.

Solution to Problem 107:

• Let x = the amount to be put in the 8% account.  The amount for the 12% account would then be \$50,000 – x
• Thus, (0.08) x + (50,000 – x)(0.12) = 4500
• So, 4500 = (0.08) x + 6000 – (0.12) x
• So, 0.04 x = 1500, and x = \$37,500
• And, \$50,000 – \$37,500 = \$12,500.  Thus the amount for the 8% interest account equals \$37,500 and the amount for the 12% account equals \$12,500.

Solution to Problem 108:.

• Let x = the son’s current age and y = the father’s current age.
• Thus, (√x) + 3 = (√y)  and  3x + 9 = y.
• Substituting for y yields  (√x) + 3 = (√(3x + 9))
• Squaring both sides of the equation yields  x + 6√x + 9 = 3x + 9
• Rearranging yields  6√x = 2x
• Squaring both sides yields 36x = 4x², and thus x = 9
• The son’s age is 9 and the father’s age is 36.

Solution to Problem 109:

• Let a = the number of \$1’s, b = the number of \$5’s, c = the number of \$10’s, and d = the number of \$20’s
• Thus, b + 2 = c,  2c – 8 = d,  2d + 10 = a, and a + 5b + 10c + 20d = 650.
• Solving each of the above equations in terms of d yields and substituting into the longer equation yields:
• (2d +10) + 5((d + 4) ÷ 2) + 10((d + 8) ÷ 2) + 20d = 650
• Simplifying yields 2d + 10 + (5/2)d + 10 + 5d + 40 + 20 d = 650
• Further simplification yields  29½ • d = 590, and thus d = 20
• With d = 20, it can be found that the number of \$1’s = 50, \$5’s = 12, \$10’s = 14, and \$20’s = 20.  The total equals \$650.

Solution to Problem 110:

• Let  x = Al’s current age and let  y = Ed’s current age
• We know that  2y + 4 = x  and that  (y – 3) + (x – 3) = 31
• Substituting yields  (y – 3) + ((2y + 4) – 3) = 31
• Thus,  y – 3 + 2y + 1 = 31  and then  3y – 2 = 31
• Thus, 3y = 33 and  y = 11.   Ed’s age is 11 and Al’s age is 26.