In the first hour of 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
OA D
[spoiler]Doubt: Are one suppose to check even the equation one gets ? I didnt check the equation for option B since there were 2 variable and 1 equation ![/spoiler]
Veritas
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- ikaplan
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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
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
"Commitment is more than just wishing for the right conditions. Commitment is working with what you have."
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Statement 2 can be solved by using the equation 1.3A+1.5B=14.2ikaplan 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
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
- aneesh.kg
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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.
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.
Aneesh Bangia
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Shouldn't the question be 'What is the ratio of the Candy A sold to Candy B during the first hour of the sale ?' ?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 second statement be b)Total price of all the goods sold during the first hour was £14.2 ?
Anil Gandham
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