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# math11to20 (Math problems ** )

### Mathematical problems — Problems 11 to 20

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

Problem 11:

A student needs to calculate his GPA (grade-point average).  He received an A in a 4 credit-hour class, an A in a 1 credit-hour class, a B in a 5 credit-hour class, a C in a 3 credit-hour class, and a B in a 3 credit-hour class.  What is his GPA?

Problem 12:

You approach a checkout counter with merchandise costing \$75.45.  You have a coupon for 25% off all merchandise, and the current sales tax rate is 8%.  The checkout clerk says that, to simplify matters, she’ll just subtract the 8% tax from the 25% coupon, and charge you 17% for the purchase.  Should you accept the offer?

Problem 13:

Steven and Mark are raising goats.  Steven says to Mark, “If you let me have one of your goats, then I would have twice as many goats as you would then have.”  Mark then counters by saying, “But if you let me have one of your goats, then we would both have the same number of goats.”  How many goats does each of the boys have now?

Problem 14:

Four years ago, Jane was twice as old as Sam.  Four years from now, Sam will be 3/4 of Jane’s age.  How old are Jane and Sam now?

Problem 15:

A vendor had a basket of bananas.  His first customer bought half of the bananas plus half of a banana.  His second customer bought half of what he had left plus half a banana.  His third customer bought half of what was left plus half a banana.  This left the vendor with no bananas.  He had not cut or split a banana.  How many bananas were in the basket initially?

Problem 16:

A red car is traveling east on a long, straight road at 25 miles per hour.  A blue car is traveling west on the same road at 20 miles per hour, approaching the red car.  When the two cars are exactly 180 miles apart, a bee passes the red car, heading east at 30 miles per hour.  When the bee reaches the blue car, it immediately turns around and heads back toward the red car.  The bee continues flying back and forth between the cars until the cars pass each other.  How many miles has the bee flown when the cars pass each other?  (Assume that the bee can reverse direction without slowing down.)

Problem 17:

A right triangle, with legs a and b and hypotenuse c, exists such that a + c = 49 and c – b = 2 .  What is the area of the triangle?

Problem 18:

A group of boys and girls took a test.  Two-thirds of the boys and three-fourths of the girls passed the test.  If an equal number of boys and girls passed the test, what percentage of the entire group passed?

Problem 19:

A man bought some plants for his garden.  On the first day he planted half of the plants plus 10 more.  On the second day he planted half of the remaining plants plus 5 more.  On the third day he planted half of the remaining plants.  This left him with 5 plants.  How many plants did he originally buy?

Problem 20:

Two pipes are used to fill a pool.  Pipe A, by itself, can fill the pool in 4 hours.  Pipe B, by itself, can fill the pool in 6 hours.  How long would it take both pipes, running at the same time, to fill the pool?

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Solution to Problem 11:

• A = 4, B = 3, C = 2
• So, ( (4 x 4) + (4 x 1) + (3 x 5) + (2 x 3) + (3 x 3) ) / 16 credits = GPA
• GPA =  50 / 16  =  3.125

Solution to Problem 12:

• \$75.45 – ( 0.25 x \$75.45 ) = \$56.59
• \$56.45 + (0.08 x \$56.45 ) = \$61.12      [This would be the price if you don’t accept the clerk’s offer]
• \$75.45 – ( 0.17 x \$75.45 ) = \$62.62      [This would be the price if you do accept the clerk’s offer.  Don’t accept the offer! ]

Solution to Problem 13:

• s = the number of goats Steven presently has, and m = the number of goats Mark currently has.
• So, s + 1 = 2 (m – 1)  according to Steven’s plan
• And, s – 1 = m + 1  according to Mark’s plan
• Solving the second equation for s yields:   s = m + 2
• Substituting that value of s into the first equation yields:   (m + 2) + 1 = 2 (m – 1)
• Solving for m yields:  m + 3 = 2m – 2
• Thus m =  5.   If Mark has 5 goats then Steven has 7

Solution to Problem 14:

• J = Jane’s current age and S = Sam’s current age
• From the first part of the problem, we can state that:  J – 4 = 2 (S – 4)
• Thus, J-4 = 2S – 8
• So,  J = 2S – 4
• From the second part of the problem, we can state that:  3/4 (J + 4) = S + 4
• Substituting 2S – 4 for J in this equation, we get:  3/4 ((2S – 4) + 4) = S + 4
• Thus, 3/4 (2S) = S + 4
• 3/2 (S) = S + 4
• 1/2 (S) = 4, hence  S = 8  (Sam’s age)
• Since J = 2S – 4   then Jane’s age = 2 (8) – 4 = 12  (Jane’s age)
• Jane is 12 and Sam is 8

Solution to Problem 15:

• Beginning number of bananas =  x
• Customer # 1 bought: 1/2 (x) + 1/2
• Bananas left after customer # 1 =  x – (1/2 (x) + 1/2) = 1/2 (x) – 1/2
• Customer # 2 bought:  1/2 (1/2 (x) – 1/2) + 1/2 = 1/4 (x) + 1/4
• Bananas left after customer # 2 = (1/2 (x) – 1/2) – (1/4 (x) + 1/4) = 1/4 (x) – 3/4
• Customer # 3 bought:  1/2(1/4 (x) – 3/4) + 1/2 = 1/8 (x) + 1/8
• Bananas left after customer # 3 = (1/4 (x) – 3/4) – (1/8 (x) + 1/8) = 1/8 (x) – 7/8
• Thus,  1/8 (x) – 7/8 = 0
• So, 1/8 (x) = 7/8
• x = 7 bananas

Solution to Problem 16:

• First, calculate the time needed for the two cars to meet:
• (25 mph x t hours) + (20 mph x t hours) = 180 miles
• 25t  + 20t = 180
• 45t = 180
• t = 4 hours  (time elapsed when the cars meet)
• Since the bee has been flying at 30 mph during those 4 hours, it will have traveled 30 mph x 4 hours = 120 miles

Solution to Problem 17:

• a2 + b2 = c2
• a + c = 49 , so  a = 49 – c
• c – b = 2 , so  b = c – 2
• Substituting, we get  (49 – c)2 + (c – 2)2 =  c2
• 2401 – 98c + c2 + c2 – 4c + 4 = c2
• 2c2 – 102c + 2405 = c2
• c2 – 102c + 2405 = 0
• (c – 37)(c – 65) = 0
• c = 37 or 65
• 65 is ruled out, so c = 37
• Thus, b = 35 and a = 12
• Area of the triangle = (a x b) ÷ 2 = (12 x 35) ÷ 2 = 210 square units

Solution to Problem 18:

• Solution # 1
• b = total boys and g = total girls, so b + g = total number = 100%
• 2/3 (b) = 3/4 (g), so b = 9/8 (g)
• 9/8 (g) + g = 100%, so 17/8 (g) = 100, and thus g = 47.06%
• b = 100% – 47.06% = 52.94%
• Thus,  (2/3 (52.94) + 3/4 (47.06)) / 100 =  (35.29 + 35.29)/ 100 = 70.6%
• Solution # 2
• x =  # of boys that passed =  # of girls that passed
• b =  total boys,   g =  total girls
• Thus, 2/3 (b) = x  and 3/4 (g) = x
• So, b = 3/2 (x)  and g = 4/3 (x)
• Thus, (x + x) / (3/2 (x)) + (4/3 (x))  =  % passed
• So, 2x /  (17/6 (x)) = % passed
• Thus, 12/17 = % passed = 70.6 %

Solution to Problem 19:

• Original number of plants = p
• Day one:    number planted = 1/2 (p) + 10 ,   number remaining = p – (1/2 (p) + 10) = 1/2 (p) – 10
• Day two:    number planted = 1/2 (1/2 (p) – 10) + 5 = 1/4 (p) ,   number remaining = (1/2 (p) – 10) – 1/4 (p) = 1/4 (p) – 10
• Day three:     number planted = 1/2 (1/4 (p) – 10) = 1/8 (p) – 5 ,   number remaining = (1/4 (p) – 10) – (1/8 (p) – 5) = 1/8 (p) – 5
• Thus:     1/8 (p) – 5 = 5 ,  so 1/8 (p) = 10  , and p = 80 original plants

Solution to Problem 20:

• (1 pool / 4 hours) + (1 pool / 6 hours) = (1 pool / x hours)
• 1/4 + 1/6 = 1/x
• 3/12 + 2/12 = 1/2
• 5/12 = 1/x
• x = 12/5 = 2.4 hours = 2 hours 24 minutes