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PS problems .. Help to understand and solve..

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PS problems .. Help to understand and solve..

by Immy » Wed May 27, 2009 1:26 pm
Hey guys,

I seem to be having some problems with my exponent laws, if you dont mind helping me with this and some other questions, that would be great.

1. 5^21 x 4^11 = 2 x 10^n what is the value of n ?

2. If n is a positive interger and the product of all the intergers from 1 to n , inclusive, is a multiple of 00-, what is the least possible value of n ?

10
11
12
13
14

3. A certain office supply store stocks 2 sizes of selt-stick notepads, each in 4 colors; blue,green,yellow, or pin. The store packs the notepads in packages that contain either 3 notepads of the same size and the same color o4 3 notepads of the same size and of 3 different colors. If the order in which the colors are packed is not considered, how many different packages of the types described above are possible ?

6
8
16
24
32

When a certain tree was first planted, it was 4 feet tall, and the height of the tree increased by a contsant amount of each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller that it was at the end of the 4th year. By how many feet did the height of the three increase each year.

3/10
2/5
1/2
2/3
6/5

Thanks so much..

Immy
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by joostinshu » Wed May 27, 2009 5:30 pm
I am sitting for the exam on Friday so I will take a shot at your first question


1. 5^21 x 4^11 = 2 x 10^n what is the value of n ?

On a problem like this first look to see where you can break down numbers to make each side look similar.

- on the right side of the equation 10 breaks down to (2^n) (5^n) and since we have a 2 on that side by itself which is essentiall 2^1 on the right side of the equation we are left with 2^(n+1) * 5^n

- on the left side we do the exact same thing and realize that 4 is the same as 2^2. Remembering our exponent rules we realized that 4^11 becomes 2^(2*11) or 2^11.

Finally we know that when the bases are the same on both sides of the equation we can drop the exponents and set them equal, thus

22 = n + 1 ------ or N=21

or we can drop the exponents on the 5 and find that

N=21

I hope this helps and is also correct!
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by joostinshu » Thu May 28, 2009 4:06 am
3. A certain office supply store stocks 2 sizes of selt-stick notepads, each in 4 colors; blue,green,yellow, or pin. The store packs the notepads in packages that contain either 3 notepads of the same size and the same color o4 3 notepads of the same size and of 3 different colors. If the order in which the colors are packed is not considered, how many different packages of the types described above are possible ?

6
8
16
24
32

First -- Break this down into what you know:

Colors - 4 of them (blue, green, yellow, pink)
Sizes - 2 of them (large, small)
Types of packages - 4 (small with same color, small with different colors, large with same color, large with different colors)

Really we are determining how many different small packages and large packages we can create.

Now this becomes pretty simple especially with the same color - we have 4 colors and 2 size packages so there are 8 possibilities with the same color arrangement.

With the mixed colors you need to calculate how many ways you can pick 3 different colors from 4 possible colors or 4! / 3!1! = 4 and then multiple that by the number of packages to get another 8 possibilities.

The correct answer is 16.
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by joostinshu » Thu May 28, 2009 4:27 am
When a certain tree was first planted, it was 4 feet tall, and the height of the tree increased by a contsant amount of each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller that it was at the end of the 4th year. By how many feet did the height of the three increase each year.

3/10
2/5
1/2
2/3
6/5

This is a progression problem, meaning the tree grows the same amount each year so you can quickly notice the tree's height will equal the following

Year 1 - (4+x)
Year 2 - (4+2x)
Year 3 - (4+3x)
Year 4 - (4+4x)
Year 5 - (4+5x)
Year 6 - (4+6x)

Now we are told that in year 6, the tree is 1/5 taller than it was in year 4 so we have to set this up:
Year 6 1/5th more than year 4
(4+6x) = 6/5(4+4x)

Now we solve for x.

4+6x = 24/5 + 24/5x
6/5x = 4/5
6x = 20/5
6x = 4
x = 4/6
x = 2/3

2/3 is the answer.
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