Mathematical problems — Problems 51 to 60
The ten problems are presented first, followed by the solutions to the ten problems.
Problem 51:
The sum of the squares of two consecutive integers is equal to 145. What are the two integers?
Problem 52:
A jar contains 7 amoebas. Each amoeba replicates and doubles in size with each replication every minute. The amoebas filled the jar in 40 minutes. How long did it take to fill the jar halffull?
Problem 53:
A rectangular lawn measures 60′ by 80′. A sidewalk of uniform width replaces the outer border of the lawn. The remaining lawn measures 2400 square feet. How wide is the sidewalk?
Problem 54:
You have 48 cents in American coins in your pocket, comprised of 6 coins. One coin falls through a hole in your pocket. What are the odds that the coin was a dime, and what is the probability that the coin was a dime?
Problem 55:
The odds that a horse named ‘Mastodon’ will lose a horse race are given as 23 to 2. What is the probability that Mastodon will win the race?
Problem 56:
At full throttle in stationary water, Paul’s boat travels at 30 miles per hour. Paul travels up a river, against the current, at full throttle for one hour. He then returns to his starting point, with the current, at full throttle. He covers the return distance in just onehalf hour. What is the speed of the river’s current, and how far did he travel up the river?
Problem 57:
A horse is tied to a 20foot rope, and there is a bale of hay that is 35 feet away from the horse. The horse is still able to eat the bale of hay. How is this possible?
Problem 58:
Look at the triangle of letters shown below. The triangle is pointing upward. Can you reverse the triangle so that it is pointing downwards, only moving three of the letters?
A B C D E F G H I J
Problem 59:
A man went to a hardware store to make a purchase. He was informed that his purchase would cost $3 for 600, which meant that each part cost $1. How much would he have to pay for 601?
Problem 60:
What is special about the following sequence of numbers? 8, 5, 4, 9, 1, 7, 6, 10, 3, 2, 0
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Solution to Problem 51:
 x^{2} + (x + 1)^{2} = 145
 x^{2} + (x^{2} + 2x + 1) = 145
 2x^{2} + 2x – 144 = 0
 Factoring the polynomial yields: (2x + 18) (x – 8) = 0
 The only integer solution to this equation is 8. The numbers are 8 and 9.
 8^{2} + 9^{2} = 145
Solution to Problem 52:
 Logically, it took 39 minutes to fill the jar halffull. If the jar was full in 40 minutes and the amoebas double in volume each minute, then the jar must have been halffull one minute before it was completely full.
 Mathematically, the same result is achieved thusly:
 Volume at time 0 = 2^{0} • v = v
 Volume at time 1 minute = 2^{1} • v
 Volume at time x minute = 2^{x} • v
 If x = time at which the jar was 1/2 full, then:
 2 ( 2^{x} ) = 2^{40}
 2^{1} = 2^{40} / 2^{x}
 2^{1} = 2^{40x}
 x = 39
Solution to Problem 53:
 Sidewalk width = w
 Total area of the sidewalk = (60′ • 80′) – 2400 = 2400 ft^{2}
 Sidewalk area = 80w + 80w + ((60 – 2w) w) + ((60 – 2w) w) = 2400
 280w – 4w^{2} + 2400
 w2 – 70w + 600 = 0
 (w – 10)(w – 60) = 0
 w = 10 feet
Solution to Problem 54:
 The combination of 6 American coins that equals 48¢ is one quarter, two dimes, and three pennies. Assuming that each coin is as likely to slide through the hole as another, the probability of the coin being a dime is 2/(2+4) or 2/6 = 33%. The odds that the coin is a dime are 2 to 4.
Solution to Problem 55:
 We would first rewrite the odds of winning as 2 to 23. To determine the probability of winning we would write the fraction 2 / ( 2+ 23) = 2 / 25 = 0.08 = 8%.
Solution to Problem 56:

 d_{1} = ( r_{1} ) ( t_{1} ) [upriver] and d_{2} = (r_{2} ) ( t_{2} ) [downriver]
 Since d_{1} = d_{2}, then ( r_{1} ) ( t_{1} ) = (r_{2} ) ( t_{2} )
 If the river current speed = c , then (30 – c) x 1 = (30 + c) x 0.5
 Thus, 30 – c = 15 + c/2
 And, 15 = 3c / 2
 Thus, c = 10 mph (the speed of the river’s current)
 Thus, with d = rt, we have d = (30 – 10) x 1, so the distance = 20 miles
Solution to Problem 57:
 The horse is tied to the rope, but the other end of the rope is not tied to anything. The horse can easily walk to the bale of hay, dragging the rope along behind.
Solution to Problem 58:
 The triangle direction can be reversed by moving only the letters A, G, and J, as shown below.
A B C J D E F H I G
Solution to Problem 59:
 The cost for 601 would also be $3. The man is buying individual number stickers in order to place his house number on the exterior of his house. Each number costs $1. Thus, the three numbers 6, 0, and 0 cost the same as the three numbers 6, 0, and 1.
Solution to Problem 60:
 The numbers are in alphabetical order!
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