Jan 2019 (Mathcounts)

I will post seven new problems every weekend, one for each day of the week. The idea is that you work on your own during the week and I will post answers the following Saturday. For those of you who practiced with me last year, the process is similar.

For instructions on submitting weekly answers, click on the “Logistics” tab.
For previously posted problems, click on the “Archive” tab.
For previous prizes, click on the “News” tab.

Week of Jan 27 – Feb 2.

Click GoogleForm to submit answers by Sat, Feb 2.

This week we will focus on problems related to volume. You are allowed to use a calculator if “Calculator” is indicated.

Jan 27: Two cylinders are equal in volume. The radius of one is doubled, and the height of the other cylinder is increased to k times its original height. If the two new cylinders are equal in volume, what is the value of k?

Answer: k=4.

Jan 28: The surface area of a sphere, in square meters, and its volume, in cubic meters, are numerically equal. What is the length of the radius of the sphere?

Answer: 4\pi r^2 = (4/3) \pi r^3. So the radius r = 3.

Jan 29: Boynton’s sheet cake measures 18 × 24 inches and has a height of 4 inches. However, these measurements include a 3/4-inch thick layer of frosting on the top and sides. What is the volume of Boynton’s cake excluding the frosting? Express your answer to the nearest whole number. (Calculator)

Answer: (18-1.5) * (24-1.5) * (4-0.75) =  1206.5625 ~ 1206.6

Jan 30: A certain box of width w has a length that is twice its width, and its height is three times its width. What is the total volume of 24 of these boxes? Express your answer in terms of w.

Answer: 24\cdot (2w)\cdot (3w)\cdot w = 144w^3

Jan 31: For a particular rectangular solid with integer dimensions, the sum of its length, width and height is 50 cm. What is the absolute difference between the greatest possible volume and the least possible volume of the solid?

Answer: w+h+l=50. For the largest volume, w, h and l are similar in values. (17, 17, 16). For the smallest volume, w=h=1, l=48. The difference is 17*17*16-1*1*48= 4576.

Feb 1: A pyramid with a square base of side length 6 m has a height equal to the length of the diagonal of the base. What is the volume of the pyramid? Express your answer in simplest radical form.

Answer: 6\cdot 6 \cdot 6\sqrt{2} /3 = 4576

Feb 2: Connecting the centers of the four faces of a regular tetrahedron creates a smaller regular tetrahedron. What is the ratio of the volume of the smaller tetrahedron to the volume of the original one? Express your answer as a common fraction.

Answer: The new tetrahedron has side length 2/3 of the original side length. Therefore the volume is 8/27.


Week of Jan 20 – Jan 26.

Click GoogleForm to submit answers by Sat, Jan 26.

Jan 20: Last year Mr. John Q. Public received an inheritance. He paid 20% in federal taxes on the inheritance, and paid 10% of what he had left in state taxes. He paid a total of $10,500 for both taxes. How many dollars was the inheritance?

Answer: 20\%x + 10\% \cdot 80\% x = 10500. Therefore, x= 10500/0.28 = 73214.

Jan 21: In a certain year the price of gasoline rose by 20% during January, fell by 20% during February, rose by 25% during March, and fell by x% during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is x?

Answer: 1.2 \cdot 0.8 \cdot 1.25 \cdot (1-x\%) = 1. Therefore, x\%=17\%.

Jan 22: Heather compares the price of a new computer at two different stores. Store A offers 15% off the sticker price followed by a $90 rebate, and store B offers 25% off the same sticker price with no rebate. Heather saves $15 by buying the computer at store A instead of store B. What is the sticker price of the computer, in dollars?

Answer: 85\%x-90 +15 = 75\% x. Therefore, x = 750.

Jan 23: The angles of quadrilateral ABCD satisfy ∠A=2∠B=3∠C=4∠D. What is the degree measure of ∠A, rounded to the nearest whole number?

Answer: 360=∠A+∠B+∠C+∠D = ∠A+∠A/2+∠A/3+∠A/4. Therefore ∠A = 172.8 =173.

Jan 24: A bag initially contains red marbles and blue marbles only, with more blue than red. Red marbles are added to the bag until only 1/3 of the marbles in the bag are blue. Then yellow marbles are added to the bag until only 1/5 of the marbles in the bag are blue. Finally, the number of blue marbles in the bag is doubled. What fraction of the marbles now in the bag are blue?

Answer: After yellow marbles are added, the total number of marbles is 5B where B is the original number of blue marbles. When blue marbles are increased to 2B, the total is 6B. Therefore 1/3 of all marbles are now blue. 

Jan 25: The average value of all the pennies, nickels, dimes, and quarters in Paula’s purse is 20cents. If she had one more quarter, the average value would be 21 cents. How many dimes does she have in her purse?

Answer: Let x be the number of coins in the bag. We have 20x +25 = 21(x+1). Therefore, x =4. To make 80 cents with 4 coins, the number of quarters must be 3, the number of nickels must be 1 and the number of dimes must be 0. 

Jan 26: The mean of three numbers is 10 more than the least of the numbers and 15 less than the greatest. The median of the three numbers is 5. What is their sum?

Answer: Let x bet the mean of 3 numbers. Then the numbers are a+15, 5, a-10, and we have 2a+10 = 3a. Therefore, a=10 and their sum is 30. 


Week of Jan 13 – Jan 19.

Click GoogleForm to submit answers by Sat, Jan 19.

Jan 13: The sum of three numbers is 20. The first is four times the sum of the other two. The second is seven times the third. What is the product of all three?

Answer: 28 (a=16, b=7/2, c=1/2)

Jan 14: Patty has 20 coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have 70 cents more. How much are her coins worth?

Answer: Patty has 14 more nickels than dimes. So she has 17 nickels and 3 dimes for a total of $1.15. 

Jan 15: Some boys and girls are having a car wash to raise money for a class trip to China. Initially 40% of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then 30% of the group are girls. How many girls were initially in the group?

Answer: (g+b)*40% = g, (g+b)*30% = g-2. So g=8, b=12.

Jan 16: The area of a rectangle remains unchanged when it is made 2 1/2 inches longer and 2/3 inches narrower, or when it is made 2 1/2 inches shorter and 4/3 inches wider. Its area, in square inches, is:

Answer: (l+5/2)(w-2/3) = (l-5/2)(w+4/3) = l w.  So l=15/2 and w=8/3. The area is 20 square inches.

Jan 17: The number of geese in a flock increases so that the difference between the populations in year n+2 and year n is directly proportional to the population in year n+1. If the populations in the years 1994, 1995, and 1997 were 39, 60, and 123, respectively, then the population in 1996 was:

Answer: Let x be the population in 1996, and c be the coefficient for the proportion. We have (x-39) = 60c and (123-60) = xc. Therefore, x=84 and c=3/4.

Jan 18: Elmo makes N sandwiches for a fundraiser. For each sandwich he uses B globs of peanut butter at 4 cents per glob and J globs of jam at 5 cents per glob. The cost of the peanut butter and jam to make all the sandwiches is $2.53. Assume that B, J, and N are all positive integers with N>1. What is the cost of the jam Elmo uses to make the sandwiches?

Answer: (4B+5J)N=253=11*23. If N=11, then J=3 and B=2. If N=23, there is no solution. So the cost of jam is 15cents.

Jan 19: Mr. Earl E. Bird leaves his house for work at exactly 8:00 A.M. every morning. When he averages 40 miles per hour, he arrives at his workplace three minutes late. When he averages 60 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?

Answer: Let d be the distance that he travels, then (d/40 – d/60)*60 = 6. So d=12 miles. 12/(12/40-3/60) = 48 mph.  

Happy New Year!

Week of Jan 6 – Jan 12. The seven problems this week are related to arithmetic and geometric sequences. If a_1, a_2, a_3, \dots is an arithmetic sequence, then the neighboring terms have a common difference a_i-a_{i-1}; if it is a geometric sequence, then the neighboring terms have a common ratio a_i/a_{i-1}.

Click GoogleForm to submit answers by Sat, Jan 12.

Jan 6: In a given arithmetic sequence, the first term is 2, the last term is 29, and the sum of all the terms is 155. The common difference is:

Answer: 3. The average of the terms is 31/2, so there are 10 terms in total. 

Jan 7: If the sum of the first 10 terms and the sum of the first 100 terms of given arithmetic progression are 100 and 10, respectively, then what is the sum of the first 110 terms?

Answer: (a + (a+9d))*5 = 100, (a + (a+99d))*50 = 10. Therefore, d=-0.22 and 2a=21.98 We want (a+(a+109d))*55 = -110

Jan 8: The measures of the interior angles of a convex polygon of n sides are in arithmetic progression. If the common difference is 5∘ and the largest angle is 160∘, then n equals:

Answer: n=9.    (n-2)*180 = (160 + 160-5(n-1))*n/2. 

Jan 9: In the sequence 2001, 2002, 2003, …, each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is 2001+2002−2003=2000. What is the 2004th term in this sequence?

Answer: 0.  

Jan 10: A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?

Answer: Let the three numbers in the arithmetic progression be 9, 9+d, 9+2d. The three numbers in the geometric sequence are 9, 11+d, 29+2d. So (11+d)^2 = 9(29+2d). Therefore, d=-14 or d=10. The smallest possible value is 29+2(-14)=1.

Jan 11: The first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 2005th term of the sequence?

Answer: 2005, 133, 55, 250, 133, 55, 250… So the 2005th term is 250.

Jan 12: Suppose that u_n is a sequence of real numbers satisfying u_{n+2}=2u_{n+1}+u_n, and that u_3=9 and u_6=128. What is u_5?

Answer: u_6 = 2u_5+u4, u_5=2u_4+u_3. This translates to 128=2u_5+u_4, u_5=2u_4+9u_5=53.