Let's say the cost of a chocolate was c dollars.
The cost of a sweet was c - $1.70 dollars.
The total cost of chocolates that John bought is 2c.
The total cost of sweets that John bought is 7c - 7*$1.70 = 7c - $11.90.
The total cost of chocolates and sweets that John bought is 2c + (7c - $11.90) = 2c + 7c - $11.90 = 9c - $11.90.
The total cost of chocolates and sweets is equal to $299.60, so we have 9c - $11.90 = $299.60.
Adding $11.90 to both sides, we get 9c = $299.60 + $11.90 = $311.50.
Dividing both sides by 9, we get c = $311.50 / 9 = $34.61.
The total cost of sweets that John bought is 7c - $11.90 = 7*$34.61 - $11.90 = $242.27 - $11.90 = $230.37.
The total number of chocolates and sweets that John bought is 2 + 7 = <<2+7=9>>9.
The total number of chocolates and sweets that John bought is also the sum of the ratios, which is 2 + 7 = <<2+7=9>>9.
Since 9 chocolates and sweets cost $299.60, the cost of each chocolate and sweet is $299.60 / 9 = $33.29.
The cost of all the chocolates and sweets that John bought is $33.29 * 9 = $299.61. Answer: \boxed{9}.
The ratio of the number of chocolates to the number of sweets that John bought was 2: 7. The cost of a chocolate was $1.70 more than the cost of a sweet. He paid $299.60 for the chocolates and sweets. If the total cost of the sweets was $92.40 more than the total cost of chocolates, how many chocolates and sweets did John buy altogether?
1 answer