Problem of the Week

__5/24/19__

**A Day at the Fairs**

It seems that a humble merchant visited three fairs. At the first fair, early in the morning, he doubled his money selling his products, but spent $30 in food and buying other items.

At midday at the second fair, he tripled his money and spent $54. At the third fair in the afternoon he quadrupled his money but spent $72.

Upon his return home to his wife and ten children, late that day, he counted the money he had in his bag; there was $48.

How much did the man gain or lose during the day?

**Solution**

The merchant gained $19 during the day. The equation to solve is:

48 = 4[3(2x-30) – 54] – 72

x = 29, representing how much money the merchant began the day with. 48 – 29 = $19 gain.

__5/17/19__

**Squares Rectangles and Rhombuses**

All squares are both rectangles and rhombuses. All rhombuses and rectangles are parallelograms. On a sheet of paper Josh draws 19 rectangles, 15 rhombuses, and 7 squares. How man parallelograms did Josh draw?

**Solution**

In a Venn diagram, the area where rectangles and rhombuses overlap represents squares. In this region, there are 7 squares, leaving 12 non-square rectangles and 8 non-square rhombuses for a total of 27 parallelograms.

__5/10/19__

**Fair Die**

A fair die is tossed four times. What is the probability that it lands with either 5 or 6 on top at least once?

**Solution**

65/81 or approximately .80247.

The number of possible ways to roll four dice is 6^{4}, or 1296: six choices for each of the four rolls.

There are four ways to roll a single die once and not get a 5 or 6 (that is, to get a 1, 2, 3, or 4), so the number of ways to roll a die four times and not get a 5 or a 6 is 4^{4}, or 256.

Then 1296 - 256, or 1040, ways (the rest of the possibilities) exist to roll the die four times and get a 5 or 6 at least once.

Therefore, the probability of rolling a die four times and getting a 5 or a 6 at least once is 1040/1296, or 65/81, which is approximately .80247.

__5/3/19__

*AD***and ***BD*

What is the sum of the distances *AD *and *BD *in the figure shown?

**Solution**

27. By the Pythagorean theorem, *BD*= 10. Let *E*be a point on *AB*with ∠*DEA*= 90°. Then *DE*= 8, and *AE*= 21 – 6 = 15. By the Pythagorean theorem, *AD*= 17. Hence, 10 + 17 = 27.

__4/26/19__

**Dog Show**

Of the animals entered in a dog show, the number of poodles is at least one-fifth of the number of beagles and at most one-sixth the number of collies. The number of dogs that are poodles or beagles is at least 23. Find the minimum number of collies entered in the show.

**Solution**

*c*≥ 24. Let *p*, *b*, and *c*represent poodles, beagles, and collies, respectively. Then

(1/6)*c*≥ *p*≥ (1/5)*b*

and

*p + b*≥ 23.

Since *p*≥ (1/5)*b*, 5*p*≥ *b*, and 6*p*≥ *b + p*≥ 23,

6*p*≥ 23 and *p*≥ 23/6.

But *p*must be a natural number, so *p*≥ 4. Since

(1/6)*c*≥ *p*≥ 4

we have (1/6)*c*≥ 4 and *c*≥ 24.

__4/12/19__

**Less than 2005°**

The sum of the interior angles of a polygon is less than 2005°. What is the largest possible number of sides of the polygon?

**Solution**

13. The sum of the interior angles of a polygon = 180°(*n*– 2) < 2005° →180°*n*< 2365° →*n*< 13.139. Since *n*must be an integer, the largest value for *n*is 13.

__4/1/19__

**One or Ten?**

Given that they are made of the same material, which is heavier: a ball with a radius of 10 inches or 10 balls each with a radius of 1 inch? For credit, show your answer and PROVE WHY it is correct.

**Solution**

The ball with a radius of 10 inches is heavier. Since the volume of a sphere is (4/3)π*r*^{3}, one ball with a radius of 10 inches would have a volume of (4/3)π⋅10^{3}= (4000/3)π in^{3}. Thus 10 balls of radius 1 inch would have an accumulated volume of only

10⋅[(4/3)π⋅1^{3}] = (40/3)π in^{3}.

__3/25/19__

**Tamika > Carlos?**

Tamika selects two different numbers at random from the set {8, 9, 10} and adds them. Carlos takes two different numbers at random from the set {3, 5, 6} and multiplies them. What is the probability that Tamika's result is greater than Carlos's result?

**Solution**

4/9. Tamika can get the numbers 8 + 9 = 17, 8 + 10 = 18, or 9 + 10 = 19. Carlos can get 3 × 5 = 15, 3 × 6 = 18, or 5 × 6 = 30. The possible ways to pair these numbers are (17, 15), (17, 18), (17, 30), (18, 15), (18, 18), (18, 30), (19, 15), (19, 18), and (19, 30). Four of these nine pairs show Tamika with a higher result, so the probability is 4/9.

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