6th & 7th Problems for March & April 2012
1. The mass of a $20 gold piece is exactly twice the mass of a $10 gold piece. Which is worth
more, one kilogram of $10 gold pieces or one-half kilogram of $20 gold pieces? Prove it
mathematically.
2. Alicia, Bryan, and Charles own all the shares in a company. Together, Alicia and Bryan own 1/2
of the shares. Alicia and Charles together own 1/3 of the shares, and Bryan and Charles
together own 1/5 of the shares. What fraction of the shares does each person own? Prove it
mathematically.
3. Kal was traveling with his parents. He noticed a sign that read, "Blue Lake—111 miles;
Brinesville—173 miles." He continued his book, figuring that he had a while before his parents
stopped for lunch in Brinesville. Sometime later, his mother said, "Look, Brinesville is exactly
twice as far away as Blue Lake." How many miles had they driven while Kal was reading? Prove
it mathematically.
4. A merchant can place 8 large boxes or 10 small boxes into a carton for shipping. The boxes are
never mixed in a carton. In one shipment, he sent a total of 96 boxes. If there are more large
boxes than small boxes, how many cartons did he ship? Prove it mathematically.
5. A bottle and a cup exactly balance a coffeepot. The bottle balances a cup and a saucer. Three
saucers can be used to balance two coffeepots. How many cups will it take to balance a bottle?
6. To find my number, take 1 fewer than the square of the smallest two-digit prime number and
divide this result by the difference between the squares of the fourth and first prime numbers.
What is my number? Prove it mathematically.
Answers
1. A kilogram of $10 gold pieces.
The monetary value and mass of the two coins are in the same ratio: a $20 piece is worth twice as much and has
twice the mass of a $10 piece.
Thus, the monetary values of different masses of the two coins are in the same ratio as the masses. A onekilogram mass of $10 coins has twice the value as one-half kilogram of $20 coins.
Another approach is to let x equal the mass of a $10 gold piece in grams. Then, 2x equals the mass of a $20 gold
piece, and the number of each coin in the given masses is (1000 g)/(x g) = 1000/x for the $10 pieces and (500
g)/(2x g) = 250/x for the $20 pieces.
There are four times as many $10 pieces, each of which is worth one-half as much as a $20 piece, so one
kilogram (1000 g) of the $10 pieces is worth twice as much as one-half kilogram (500 g) of $20 pieces.
2. Alicia owns 19/60, Bryan owns 11/60, and Charles owns 1/60 of the shares. Since A + C = 1/3, then −A
− C = −1/3. Also, A + B = 1/2, so if we add corresponding parts of the last two equations, we find that B
− C = 1/2 + −1/3, so B − C = 1/6. We also know that B + C = 1/5, so if we add corresponding parts of
these last two equations, we find that 2B = 1/6 + 1/5 = 11/30, so B = 11/60. Thus, since B + C = 1/5, we
find that C = 1/5 − 11/60 = 1/60. Finally, since A + C = 1/3, then A = 1/3 − 1/60 = 19/60.
3. Forty-nine miles. Use a diagram similar to the one below. If Brinesville is 173 miles away and Blue
Lake is 111 miles away, then the two towns are 62 miles apart. To be twice as far from Brinesville as
Blue Lake means that the family must be 62 miles away from Blue Lake. The distance from the first
sign to Blue Lake was 111 miles, and 111 − 62 = 49.
4. One approach is to use a table to find the number of cartons, based on the possible numbers of small
boxes, and obtain the number of large boxes by finding the difference. The shaded sections of the table
show combinations that will not work because the numbers of the large boxes will not fit evenly into
cartons. He shipped more large boxes than small boxes, so he must have used 11 boxes (56 versus 16
larger boxes).
5. 75 cups.
Letting B = bottle, C = cup, S = saucer, and P = pot, write the three equations for the given conditions:
(1) B + C = P
(2) B = C + S
(3) 3S = 2P
Multiply equation (1) by 2 to get 2(B + C) = 2P, or 2B + 2C = 2P.
Substitute from equation (3) to get equation (4): 2B + 2C = 3S.
Solve equation (2) for S and multiply by 3 to get 3S = 3B - 3C.
Substitute for 3S in equation (4) to get 2B + 2C = 3B - 3C, and rearrange to solve for B = 5C.
8a = 9, b = 1, and c = 0.
We first know that b = 1, because b in the quotient times divisor a must equal a. (Put 1's in for b's.)
Then since a subtracted from the two-digit number 1c is 1, we know that c = 0 and a = 9.
6. 8/3 or 2.666. . . . The smallest two-digit prime number is 11. Its square is 121, and 1 fewer is 120. The
fourth and first prime numbers are 7 and 2, respectively. The difference between their squares, 49 and 4,
respectively, is 45. When 120 is divided by 45, the result is 8/3.