A bag of sweets contains only mints, sherberts and toffees - OCR - GCSE Maths - Question 6 - 2019 - Paper 4

The ratio of sherberts to toffees is given as 7 : 5. This means if we let the number of sherberts be 7y, then the number of toffees will be 5y, where y is another common multiplier.

From the values of sherberts, we have two expressions: 3x = 7y. To find a common expression for sherberts, we can equate these two:

$3x = 7y$

To solve for x and y in terms of a common variable, let’s find a common multiple for 3 and 7, which is 21. Therefore, let x = 7k and y = 9k for some integer k.

Now substituting back, we have:

• Mints = 2x = 2(7k) = 14k
• Sherberts = 3x = 3(7k) = 21k
• Toffees = 5y = 5(9k) = 45k

Total sweets = Mints + Sherberts + Toffees = 14k + 21k + 45k = 80k.

The fraction of the sweets that are sherberts is given by:

$\text{Fraction of sherberts} = \frac{\text{Number of sherberts}}{\text{Total number of sweets}} = \frac{21k}{80k} = \frac{21}{80}$

Thus, the fraction of the sweets that are sherberts is ( \frac{21}{80} ).