 Ask Remember Me? annesheridan Posts: 2, Reputation: 1 New Member #1 Dec 12, 2005, 05:34 PM
Math problem
Kyle is a world class tennis player and goes to practice with some bags of tennis balls. Kyle has five bags containing a total of 50 tennis balls. The first and second bags contain a total of 21 tennis balls, the second and third bags contain 20 tennis balls, and the third and fourth contain 21 tennis balls and the fourth and fifth contain 27 tennis balls. What is the total number of tennis balls kyle has in the first and fifth bags? CaptainForest Posts: 3,645, Reputation: 393 Ultra Member #2 Dec 12, 2005, 11:51 PM
There are 5 Bags that have a total of 50 balls in them

B1, B2, B3, B4, B5

Information Given in Question:

B1 + B2 = 21
B3 + B2 = 20
B3 + B4 = 21
B5 + B4 = 27

Solution

We know that B1 + B2 + B3 + B4 + B5 = 50
We know that B2 + B3 = 20 and B4 + B5 = 27. Therefore, that leaves only 3 balls left in B1 (50-20-27)

Threfore, B1 = 3
B2 = 21-3 = 18
B3 = 20-18 =2
B4 = 21-2 = 19
B5 = 27-19 = 8

Therefore, in B1 + B5 there are 11 balls (3+8) lilfyre Posts: 508, Reputation: 98 Senior Member #3 Dec 13, 2005, 04:46 PM
Okay was he right?
I hate it when they never come back and say thank you. :mad:

How long did it take you to get the answer? :confused: I hate word problems :confused: every time my daughter walks toward me with one i want to run. (O: Originally Posted by CaptainForest
There are 5 Bags that have a total of 50 balls in them

B1, B2, B3, B4, B5

Information Given in Question:

B1 + B2 = 21
B3 + B2 = 20
B3 + B4 = 21
B5 + B4 = 27

Solution

We know that B1 + B2 + B3 + B4 + B5 = 50
We know that B2 + B3 = 20 and B4 + B5 = 27. Therefore, that leaves only 3 balls left in B1 (50-20-27)

Threfore, B1 = 3
B2 = 21-3 = 18
B3 = 20-18 =2
B4 = 21-2 = 19
B5 = 27-19 = 8

Therefore, in B1 + B5 there are 11 balls (3+8) CaptainForest Posts: 3,645, Reputation: 393 Ultra Member #4 Dec 13, 2005, 05:01 PM

I have always enjoyed Algebra annesheridan Posts: 2, Reputation: 1 New Member #5 Dec 17, 2005, 01:13 PM
Math problem
Thank You So Much!! I Can't Believe What A Hard Time I Had With A Fourth Grade Math Problem!! :d :d :d CaptainForest Posts: 3,645, Reputation: 393 Ultra Member #6 Dec 17, 2005, 01:15 PM
Your welcome. Sometimes our mind goes blank on easy questions, it happens to all of us at some point. s_cianci Posts: 5,472, Reputation: 760 Uber Member #7 Dec 18, 2005, 06:15 PM
Make a table comparing the number of balls in each bag. Start by assuming that each bag contains 10 balls, even though you know that's not the case, for a total of 50:

Bag 1: Bag 2: Bag 3: Bag 4: Bag 5:
10 10 10 10 10

Adjust this initial guess to reflect the distribution in each pair of bags as specified in the problem:

Bag 1: Bag 2: Bag 3: Bag 4: Bag 5:
11 10 10 11 16

Notice that this distribution results in 8 balls too many, a total of 58 instead of 50. So, refine this second guess while still keeping the distribution in each pair of bags as specified in the problem:

Bag 1: Bag 2: Bag 3: Bag 4: Bag 5:
10 11 9 12 15

Notice that alternately subtracting a ball from one bag then adding a ball to the next, then subtracting, etc. results in a total of 57 balls, which is one closer to the target amount of 50. Repeat this 7 more times, i.e. subtract 7 more from bags 1 , 3 and 5 and add 7 more to bags 2 and 4 and get the following result:

Bag 1: Bag 2: Bag 3: Bag 4: Bag 5:
3 18 2 19 8

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