x bags of marbles are to be created, each consisting of y red marbles,

x bags of marbles are to be created, each consisting of y red marbles, z blue marbles, and nothing else. If the marbles are to be drawn from a total of 84 red marbles and 30 blue marbles and all marbles must be placed in one of the bags, what is the maximum number of bags that can be created?

Finding maximum common divisor between 84 red marbles and 30 blue marbles

84 = 6* 14
30 = 6 * 5

Therefore, maximum number of bags = 6

x bags of marbles are to be created, each consisting of y red marbles, z blue marbles, and nothing else. If the marbles are to be drawn from a total of 84 red marbles and 30 blue marbles and all marbles must be placed in one of the bags, what is the maximum number of bags that can be created?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

Since there would be equal number of red marbles in each bag and equal number of blue ones, the number of marbles, an integer, of each color from a total of 84 red marbles and 30 blue marbles must be minimum and have a common factor.

Now,
84:30 = 14:5

Beyond 14 and 5 values are not possible since number of marbles would be an integer.
Thus, 6 is the maximum common factor for both 14 and 5.

This means that each of the 6 bags must have 14 red marbles and 5 blue marbles.

Answer D.

Imo. D

Total 84 red and 30 blue marbles (total=114) are available.

Maximum # of bags to be created (x to be determine), each bag contain y red and z blue marbles and all 114 shall be consumed to create bags.

The quickest way to solve the question by working from answer choices.
The largest no. is 7, so let's divide 84 red marbles and 30 blue marbles in such a way that no marble shall be left. 84/7 = 12 and 30/7 (not divisible by 7), hence two marbles left behind. So, 7 is incorrect
Let's try, 6, 84 and 30, both numbers are divisible by 6. So, we can form 6 bags, each containing 14 red and 5 blue marbles. Correct.

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

Each bag consists of y red and z blue marbles.

84 red marbles and 30 blue marbles
—> there are equal amounts of red and blue marbles in each bag.

We need to find the Greatest Common Divisor of this two numbers:
—>84= \(2^{2}*3*7\)
30= 2*3*5
—> GCD(84,30)=6

There are 14 red and 5 blue marbles in each bag.
—> The maximum number of bags is 6
The answer is D

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since we need to determiner max no of bags , solve seeing answer options
option E ; 7 wont be possible since 30 is not divisible by 7 and each bag has to contain equal of red and blue
option D ; 6 bags possible i.e each bag with 5 of red and 12 of blue marbles
IMO D


x bags of marbles are to be created, each consisting of y red marbles, z blue marbles, and nothing else. If the marbles are to be drawn from a total of 84 red marbles and 30 blue marbles and all marbles must be placed in one of the bags, what is the maximum number of bags that can be created?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

\(prime.f(84)=42*2=21*2*2=7*3*2^2\)
\(prime.f(30)=15*2=5*3*2\)
\(gcf(85,30)=3*2=6\)

Answer (D)

start from the greatest number;
7: 30 can't be divided by 7 therefore no
6: 30 can be divided by 6 which is 5 but and 84 devided by 6 is 14
bingo, there will be 5 blue and 14 red per bag and a total of 6 bags

Therefore, D