In a bag of marbles, 3/5 of them are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red?

Since it is a question of fractions, you can assume any reasonable number and work with the numbers, it's easier.

Say we have 10 marbles, 3/5 of 10=6 are blue, and the rest (4) are red.

If we double the red marbles, we have 8, and the total becomes 8+6=14.
The fraction of red will then be 8/14=4/7.

Try the same approach using 20 marbles to start and see if you get the same answer.

If you would like to solve it using fractions, it is the same approach.

Blue: 3/5
Red: 1-3/5=2/5
Double reds: 4/5
Total : 3/5+4/5=7/5
Fraction of red: (4/5)/(7/5) = 4/7

In a bag are only red and white marbles. The probability of choosing a red is 3/5. The probability of choosing 2 reds without replacement is 3/10 How many white marbles are in the bag?

To find the fraction of red marbles after doubling the number of red marbles, we first need to determine the current fraction of red marbles in the bag.

Given that 3/5 of the marbles are blue, we can conclude that the remaining fraction (2/5) represents the red marbles. Therefore, currently, 2/5 of the marbles are red, and 3/5 are blue.

Next, we are doubling the number of red marbles while keeping the number of blue marbles the same. Doubling the amount of red marbles means multiplying it by 2.

Let's assume there are a total of 5 marbles in the bag initially. Out of these, 2 are red (2/5) and 3 are blue (3/5).

If we double the red marbles, we have 4 red marbles (2 * 2) and still 3 blue marbles. The total number of marbles in the bag is now 7 (4 red + 3 blue).

Therefore, after doubling the red marbles, the fraction of red marbles will be 4/7, since we have four red marbles out of a total of seven marbles.

Hence, 4/7 of the marbles will be red.