I would go with D.

If you are confused why Statement (2) is sufficient, please watch the session "Problems on which traditional algebra doesn't work" from 'Thursdays with Ron':

https://vimeo.com/16699881

ikaplan wrote:I would go with D.

If you are confused why Statement (2) is sufficient, please watch the session "Problems on which traditional algebra doesn't work" from 'Thursdays with Ron':

https://vimeo.com/16699881

Statement 2 can be solved by using the equation 1.3A+1.5B=14.2
or 13A+15B=142
To buy max no of items, make B=0,therefore A+B<11
To buy min no of items, make A=0,therefore A+B>9
Hence A+B=10
Solving further A:B=4:6::2:3

Nice, Tricky problem!

Statement 1:
Average = (Total Revenue)/(Total number of candies)
1.42 = 1.3x + 1.5y/(x + y)
1.42x + 1.42y = 1.3x + 1.5y

The ratio of (x/y) can be obtained from this.
Thus, Statement 1 is SUFFICIENT.

Statement 2:
Total Revenue = 1.3x + 1.5y
14.2 = 1.3x + 1.5y
Prima facie, it seems that this equation cannot be solved for (x/y).
But there is a catch: x and y have to hold integral value. It might be possible that (x,y) hold just one possible integral pair of values for the above condition to be satisfied. And, we must check for that.

Let's multiply both sides by 10 to make it look better.
142 = 13x + 15y
For both x and y to be integers, the only possible solution is x=4 and y=6 (plug-in values for x to find this).
Thus, Statement (2) is also SUFFICIENT.

(D) is the answer.

In the first hourof the bake sale student sold either candy A which sold for £1.3 or Candy B which sold for £1.5. What is the ratio of the Candy A sold to Candy B during the first half of the sale.

a)The average price of the candy sold during the first hour was £1.42
b)Total price of all the goods sold was £14.2.

Shouldn't the question be 'What is the ratio of the Candy A sold to Candy B during the first hour of the sale ?' ?

Shouldn't the second statement be b)Total price of all the goods sold during the first hour was £14.2 ?

I agree with neegandham the incomplete sentence makes you wonder and go for E