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Problem of the Week Archive
Ready to Compete – January 30, 2017
Problems
Chapter competitions officially start this week! Are you ready to compete? Let’s try a few of last
year’s chapter problems to warm up.
The mean of a set of five numbers is 3k. When a sixth number is added to the set, the mean
increases by k. What is the ratio of the sixth number to the sum of the first five numbers? Express
your answer as a common fraction.
[Sprint #22]
Let’s call the sum of the set of five numbers S and the sixth number N. We know the mean of the set of five numbers is 3k, so
we can write the equation S/5 = 3k or S = 15k. When we add the sixth number the mean increases by k, so we can write the
equations (S + N)/6 = 4k or S + N = 24k. Using S = 15k, we find that N = 9k. The ratio of the sixth number to the sum of the
set of five numbers is therefore N/S = 9k/15k = 3/5.
The large rectangle shown here is composed of four squares. If the area of the
large rectangle is 240 in2, what is its perimeter?
[Sprint #23]
Looking at the two smallest squares, we know they will all have equal sides of length x. This means the square directly above
them must have side length 2x and the largest square to the right will have side length 3x. This means the rectangle has width
3x and length 5x. The area is 240 in2 so we can write the equation (3x)(5x) = 240. Solving for x we get 15x2 = 240 → x2 =
16 → x = 4. The perimeter is therefore 2(3x) + 2(5x) = 2(3 × 4) + 2(5 × 4) = 2(12) + 2(20) = 24 + 40 = 64 inches.
On Monday, a worker began painting a fence at 8:00 a.m. At 10:00 a.m., two more workers showed
up and helped the first worker paint. The three workers finished the job at 11:30 a.m. On Tuesday,
the first worker began painting an identical fence at 8:00 a.m. The two other workers showed up
earlier this time, and the job was finished at 10:54 a.m. All three workers paint at the same constant
rate. How many minutes did the first worker paint alone on Tuesday?
[Sprint #27]
Since all the workers paint at a constant rate, let’s call this rate W. On the first day, one worker painted for 3.5 hours and the
other two workers painted for 1.5 hours. On the second day, one worker painted for 2 hours 54 minutes or 2.9 hours, and the
other two workers painted for 2.9 hours minus the amount of time the first worker was painting alone. We will call this time T.
Using these times and variables we can set up the following equation: 3.5W +1.5(2W) = 2.9W + (2.9 – T)2W. Dividing
through by W and solving for T, we get 3.5 + 1.5(2) = 2.9 + (2.9 – T)2 → 6.5 = 8.7 – 2T → 2T = 2.2 → T = 1.1 hours or
1.1 × 60 = 66 minutes.
Problem of the Week Archive
Ready to Compete – January 30, 2017
Problems
Chapter competitions officially start this week! Are you ready to compete? Let’s warm up with a few
problems from the 2016 Chapter Competition.
The mean of a set of five numbers is 3k. When a sixth number is added to the set, the mean
increases by k. What is the ratio of the sixth number to the sum of the first five numbers? Express
your answer as a common fraction.
[Sprint #22]
The large rectangle shown here is composed of four squares. If the area of the
large rectangle is 240 in2, what is its perimeter?
[Sprint #23]
On Monday, a worker began painting a fence at 8:00 a.m. At 10:00 a.m., two more workers showed
up and helped the first worker paint. The three workers finished the job at 11:30 a.m. On Tuesday,
the first worker began painting an identical fence at 8:00 a.m. The two other workers showed up
earlier this time, and the job was finished at 10:54 a.m. All three workers paint at the same constant
rate. How many minutes did the first worker paint alone on Tuesday?
[Sprint #27]