Liu has a bag only containing red grapes and green grapes - OCR - GCSE Maths - Question 11 - 2018 - Paper 1

Let the total number of grapes in Sophia's bag be $y$. We know that:

rac{4}{9}y are red grapes. After she eats one red grape, the number of remaining grapes is $y - 1$.

Out of the remaining grapes, rac{3}{7} are red. Thus, we can write:

rac{3}{7}(y - 1) are red grapes left.

Since originally there were rac{4}{9}y red grapes, after eating one, the equation becomes:

rac{4}{9}y - 1 = rac{3}{7}(y - 1)

To solve for $y$, we multiply through by 63 (the LCM of 9 and 7) to eliminate the fractions:

$28y - 63 = 27(y - 1)$ $28y - 63 = 27y - 27$ $y = 36$

So, Sophia has 36 grapes in total. The number of red grapes initially is:

rac{4}{9} imes 36 = 16

Now, after eating one, the remaining red grapes are:

16 - 1 = 15 red grapes.

The remaining grapes in total are $y - 1 = 36 - 1 = 35$, and the number of green grapes is:

Total remaining - Remaining red = 35 - 15 = 20.

Final counts:

• Red grapes: 15
• Green grapes: 20.