1/2 elim. after Round 1, 1/3 after Rd. 2, 1/4 after Rd. 3,

[lisaz0224]

Several Freshmen tried out for the school track team.

After round one, 1/2 of them were eliminated.
After round two, 1/3 of the remainder were eliminated
After round three 1/4 of the remainder were eliminated
After round four 1/5 of the remainder were eliminated
After round five 1/6 of the remainder were eliminated.
In the end, there were 10 people on the team. How many were there in the beginning?

How do you solve this problem?? Please explain the steps!!

 

[Deleted member 4993]

Re: Math Question

Several Freshmen tried out for the school track team.

After round one, 1/2 of them were eliminated.
After round two, 1/3 of the remainder were eliminated
After round three 1/4 of the remainder were eliminated
After round four 1/5 of the remainder were eliminated
After round five 1/6 of the remainder were eliminated.
In the end, there were 10 people on the team. How many were there in the beginning?
How do you solve this problem?? Please explain the steps!!

Start naming a variable (preferably the value of the parameter they are asking)

Let the number of people in the begining = P


After round one, 1/2 of them were eliminated. - P/2 left

After round two, 1/3 of the remainder were eliminated - P/2 * 2/3 = P/3 left

and continue....

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[Loren]

Re: Math Question

Maybe you would want to experiment by starting at the end and working backwards. For instance....
"After round five 1/6 of the remainder were eliminated.
In the end, there were 10 people on the team."

10 people remained after 1/6 of the previous number were eliminated means that 10 is 5/6 of the preceding number (1/6 were eliminated so the rest (5/6) remained). 10=(5/6)x Solve for x to get the number of people in round five.

 

[stapel]

How do you solve this problem?? Please explain the steps!!

Here's how my twelve-year-old suggests doing it:

You end up with 10, which is 5/6 of what there had been before that Round. Ten divided by five is two; six parts of two each is twelve.

The twelve is 4/5 of what was before that Round. Twelve divided by four is three; five parts of three is fifteen.

The fifteen is 3/4 of what was before that Round. Fifteen divided by three is five; four parts of five is twenty.

The twenty is 2/3 of what was before that Round. Twenty divided by two is ten; three parts of ten is thirty.

The thirty is 1/2 of what was before that Round. Thirty divided by one is thirty; two parts of thirty is sixty.

Hope that helps!

Eliz.

 

[Deleted member 4993]

Re:

Here's how my twelve-year-old suggests doing it:
Eliz.

Just the other day - he was eleven - time is just flying......

 

[mmm4444bot]

(Who writes these questions?)

... In the end, there were 10 people on the team. How many were there in the beginning?

There were not any people on the team in the beginning.

:twisted:

 

[chivox]

The word "of" in a story problem like this usually means to multiply. That is indeed the case here.

(1/2) * (2/3) * (3/4) * (4/5) * (5/6) * p = 10

(( cancel and solve for p ))

(1/6) p = 10

p = 60

That is, if 1/3 "of" the remaining people were eliminated after the second round, you would multiply 2/3 (1 - 1/3) by the number of people who were there at the end of the previous round, which we know was 1/2 (1 - 1/2) of the original number, for which we substitute the variable p. We have to subtract the number given in the story problem from 1 for each round, because we are given the number of students remaining at the end (10), while the story problem gives us the fraction of students who were eliminated in each round... tricky.

Since multiplication is a commutative operation, it doesn't matter what order we use, and the problem is set up so all but one of the numerators and denominators cancel out.

-Paul