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A bag contains red, white, and blue marbles of which 2/3 of the marbles are blue. If there are 1/3 as many red marbles as blue, and there are 20 white marbles in the bag, how many marbles are in the bag?


A. 60
B. 100
C. 120
D. 140
E. 180


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If a and b are different prime numbers greater than 2, and n = a x b, then how many positive factors, not including 1 and n, does n have?

A. 0
B. 1
C. 2
D. 3
E. 4

2007-09-30 11:31:46 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

q1
20=(1-2/3-1/3*2/3)x
=x/9
x=9*20
=180(E)

q2
n=a*b=1*n
ans=2(C)

2007-09-30 11:47:16 · answer #1 · answered by Mugen is Strong 7 · 0 0

To solve the first question, simply set up a system of equations. Let R, W, and B be the numbers of red, white, and blue marbles, respectively.

The total number of marbles is T = R+W+B, and we know that 2/3 the marbles are blue:
B/T = 2/3
B/(R+W+B) = 2/3

We also know that there are 1/3 as many red marbles as blue marbles:
R/B = 1/3

We also know that there are 20 white marbles:
W = 20

So simply solve the system of equations:
"W = 20
R/B = 1/3
B/(R+W+B) = 2/3"
for the three variables.

You should end up with
W = 20
R = 40
B = 120

Thus, the total number of marbles is
T = R+W+B
T = 20+40+120
T = 180

To solve the second problem, consider what factors a prime number has: only 1 and itself, by definition. Now, when you multiply two numbers together, their product has all the factors that the two numbers had, and all multiplicative combinations of one number

For example, 6 has factors 1, 2, 3, 6 while 5 has factors 1, 5. 6*5 = 30, so 30 has the following factors:
1*1 = 1
1*2 = 2
1*3 = 3
1*6 = 6
5*1 = 5
5*2 = 10
5*3 = 15
5*6 = 30

Notice I just multiplied all pairs of factors from the two original numbers, and also notice that you can't use a factor more than once unles that factor is present in both numbers. For example, 5*5 = 25 is NOT a factor of 30, because there is only one factor of 5 present in the starting numbers.

So, prime A has factors 1, A, and prime B has factors 1, B. Thus, N = A*B has these factors:
1*1 = 1
1*A = A
B*1 = B
B*A = B*A = N

Thus, N has only four factors, 1, A, B, and N. The question states we cannot include 1 or N, so only factors A and B remain. That's two factors.

2007-09-30 11:38:57 · answer #2 · answered by lithiumdeuteride 7 · 0 0

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