Problem Of The Week

MATH CHALLENGE OF THE WEEK

This is the solution to the bonus math challenge that was given during last Thursday's EXPLORATIONS EVENT:

PART ONE = 22 Times. The hands of a clock lie directly opposite each other 11 times in each twelve hour interval or 22 times each day.

PART TWO = 44 Times. The hands of a clock are perpendicular to each other 22 times in each twelve hour interval or 44 times each day.

THIS WEEK'S CHALLENGE:

INCREASE THE AVERAGE

The average of five numbers is 18. Let the first number be increased by 1, the second number by 2, the third number by 3, the fourth number by 4 and the fifth number by 5. What is the average of the set of increased numbers. Remember to prove your answer by showing all of your work.


LAST WEEK'S CHALLENGE:

YOU CAN'T BUY IT IF YOU DON'T HAVE ENOUGH MONEY!

Alice and Betty each want to buy the same kind of ruler. But Alice is 22 cents short and Betty is 3 cents short. When they combine their money, they still do not have enough money. What is the MOST the ruler could cost. Remember to prove your answer by showing all of your work.

LAST WEEK'S ANSWER:

Suppose the ruler costs R cents. Then Alice has (R - 22) cents, Betty has (R - 3) cents, and together they have (2R - 25) cents. But this sum is STILL not enough to pay for this ruler.

Given                                                       STEP 1   2R - 25 < R
Subtract R from both sides of step 1   STEP 2     R - 25 < 0
Add 25 to both sides of step 2              STEP 3     R < 25

Since R is less than 25 cents, the most R could be is 24 cents.
 
Last Modified: Saturday, May. 18, 2013
 
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