Question

Sophia has a different bag only containing red grapes and green grapes.
The number of grapes in her bag is different, but 4/9 of the grapes are also red.
She picks out a red grape from her bag and eats it.
3/7 of the remaining grapes in her bag are red. How many of the remaining grapes in her bag are red and how many are green?​

Answer 1

Let's represent the total number of grapes in Sophia's bag as x. Then, let's say she has y red grapes and z green grapes. We know that:

y + z = x (because the total number of grapes is the sum of the red and green grapes)

4/9 x = y (because 4/9 of the grapes are red)

Now, let's look at what happens when Sophia eats a red grape. There are now x - 1 grapes left in her bag, and y - 1 of them are red. We also know that 3/7 of the remaining grapes (or x - 1) are red. Putting it all together:

3/7 (x - 1) = y - 1

3/7 (x - 1) = 4/9 x - 1

Multiplying both sides by 63 (the least common multiple of 7 and 9), we get:

27(x - 1) = 28x - 63

27x - 27 = 28x - 63

x = 36

Now that we know there are 36 grapes in Sophia's bag, we can use our earlier equations to find that she has 16 red grapes (4/9 of 36). When she eats one of them, she is left with 35 grapes, and we can solve for the number of remaining red grapes:

3/7 (35) = y - 1

y = 16

So there are still 16 red grapes in her bag, and the remaining 19 grapes must be green. Answer:

16 red grapes and 19 green grapes.