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mathproblems91to100 ( ** )

Mathematical problems — Problems 91 to 100

The ten problems are presented first, followed by the solutions to the ten problems.  NOTE: Problems 91 through 96 are ‘mixture’ problems.  The following formula is very helpful in solving ‘mixture’ problems:  (quantity A  x  quality A) + (quantity B  x  quality B) =  quantity (A+B) x  quality C, where A<C<B or B<C<A.

Problem 91:

Here is an interesting mixture problem.  Cashews sell for \$3 per pound and walnuts sell for \$4 per pound,  How many pounds of each of these two nuts should be mixed together to make a 5-pound box of mixed nuts that costs \$3.60 per pound?

Problem 92:

A truck driver wants to mix up some 20% antifreeze.  He currently has 50 gallons of 10% antifreeze.  How much 40% antifreeze does he need to add to his 50 gallons of 10% in order to end up with a 20% mixture?

Problem 93:

A farmer has 100 gallons of 20% antifreeze.  He wants to raise the percentage of the antifreeze up to 35%.  How many gallons of pure (100%) antifreeze must he add to the original 100 gallons to raise the concentration to the desired level?

Problem 94:

How much water must be added to 8 ounces of 40% alcohol to produce a mixture of 7% alcohol?

Problem 95:

A farmer needs a 9% insecticide mixture.  He has 5% and 15% insecticide mixtures.  a) What is the mixture ratio to get 9%?  b) If he needs 40 gallons of the 9% mixture, how much of the 5% and the 15% mixtures should he use?

Problem 96:

A farmer has a 32-quart vat that contains 32 quarts of 20% weed-killer solution.  How much of the current solution has to be removed and replaced by pure 100% weed-killing solution to achieve a final concentration of 35%?

Problem 97:

If 2 more than 3 times your age equals 32 more than half your age, how old are you?

Problem 98:

The father of a set of triplets is 25 years older than the triplets.  If the sum of the ages of all the triplets and the father is 15 greater than the age of the father, then how old are the triplets and how old is the father?

Problem 99:

Max is three times as old as Don.  If the sum of their ages is 16 more than twice Don’s age plus 4, the how old are Max and Don?

Problem 100:

Joe loves to solve puzzles.  He has attempted 180 puzzles and has successfully solved 90% of them.  How many more puzzles must Joe solve successfully, without a failure, to achieve a 95% success rate?

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Solution to Problem 91:

• The ‘mixture formula’ is  (quantity A  x  quality A) + (quantity B  x  quality B) =  quantity (A+B) x  quality C, where A<C<B or B<C<A.
• Let c = cashew pounds and w = walnut pounds,
• Thus,  (c • 3) + (w • 4) = (5 • 3.6) with (c + w) = 5.
• Since w = 5 – c, we can substitute and arrive at  3c + 4(5 – c) = 18.
• Thus, 3c + 20 – 4c =  18, and then  c = 2.  Thus, we would add 2 pounds of cashews to 3 pounds of walnuts to achieve the desired result,

Solution to Problem 92:

• (quantity A  x  quality A) + (quantity B  x  quality B) =  quantity (A+B) x  quality C
• Let x = the amount of 40% solution needed
• Thus,  50 (0.1) + x (0.4) = (50 + x) 0.2
• Simplifying yields  5 + 0.4 (x) = 10 + 0.2 (x)
• Thus, 0.2 (x) = 5, so x = 25.  The truck driver needs to add 25 gallons of 40% antifreeze to the 50 gallons of 10% antifreeze.

Solution to Problem 93:

• (100 gallons)(0.2) + (x gallons)(1) = (100 + x gallons)(0.35)
• 20 + x = 35 + 0.35x
• 0.65x = 15, so x = 23.077.  So the farmer needs to add just over 23 gallons of pure antifreeze to achieve his goal.

Solution to Problem 94:

• 8 ounces of 40% alcohol would contain 3.2 ounces of pure alcohol  ( 0.4 x 8 = 3.2 ).
• A 7% alcohol mixture that contains 3.2 ounces of pure alcohol would require 46.7 total ounces of water  ( 3.2 ÷ 0.7 = 45.7 )
• 45.7 ounces minus 8 ounces = 37.7 ounces, thus one would need to add 37.3 ounces of water to the 40% mixture to achieve a 7% mixture.

Solution to Problem 95:

• Using the mixture formula, we can say that  x (0.05) + y (0.15) = ( x + y )(0.09)
• This equation simplifies to  0.04 • x = 0.06 • y, and thus  3x / 2y = 1
• Thus, the mixture ratio for a 9% mixture is 3 parts 5% and 2 parts 15%
• 40 gallons total would thus require 24 gallons of 5% (3/5 of 40) and 16 gallons of 15% (2/5 of 40).

Solution to Problem 96:

• Let x = the amount to be removed from the vat.
• Then,  (32 – x)(0.2) + x( 1 ) = 32(0.35)
• So,  6.4 – 0.2x + x = 11.2
• And,  0.8x = 4.8
• Thus, x = 6, so 6 quarts must be removed from the vat and replaced by 6 quarts of pure weed-killer solution to achieve a 35% solution.

Solution to Problem 97:

• If x equals your age, then:  3x + 2 = ½ • x + 32
• So, 2½ • x = 30, and x = 12.  You are 12 years old.

Solution to Problem 98:

• Let x = the age of the three triplets.  The father’s age would then be x + 25.
• Thus, x + x + x + (x + 25) = (x + 25) + 15.
• Solving for x,  4x + 25 = x + 40, and then 3x = 15, and x = 5.  The triplets are 5 years old and the father is 30.

Solution to Problem 99:

• Let x = Max’s age and y = Don’s age.
• Thus, x = 3y and (x + y) – 16 = 2y + 4.
• Substituting and simplifying yields  (3y + y) – 16 = 2y + 4  and thus 4y – 16 = 2y + 4.
• Thus, 2y = 20 and y = 10.  So, Don is 10 and Max is 30.

Solution to Problem 100:

• First we calculate how many puzzles Joe has solved successfully.  The equation would be  x ÷ 180 = 0.90.  Solving for x yields 162.
• We now let x = the number of puzzles Joe must solve successfully in the future to achieve a 95% success rate.
• Thus, (162 + x) ÷ (180 + x) = 0.95.  And, 162 + x = 0.95 (180 + x).  Simplification yields 162 + x = 171 + 0.95x.
• Further simplification yields  0.05x = 9, and x thus equals 180.
• Remarkably, Joe must solve another 180 puzzles successfully, without any misses, in order to achieve a 95% success rate.