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555-605 Level|   Word Problems|            
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sagnik242
Bunuel
If Pei ordered a total of 63 bottles of Cola, Root Beer and Ginger Ale for a part. How many Cola bottles did she order?

Given: {Cola} + {Beer} + {Ale} = 63.
Question: {Cola} = ?

(1) The no of bottles of root beer Pei order was 80% of bottles of ginger ale that she ordered --> {Beer} = 0.8{Ale} --> 5*{Beer} = 4*{Ale} ({Beer} is a multiple of 4 and {Ale} is a multiple of 5). If {Beer} = 4 and {Ale} = 5, then {Cola} = 54 but if {Beer} = 8 and {Ale} = 10, then {Cola} = 45. Not sufficient.

(2) The no of bottles of cola Pei order was 75% of total no. of bottles of ginger ale and root beer that she ordered --> {Cola} = 0.75*({Beer} + {Ale}) --> {Beer} + {Ale} = 4/3*{Cola} --> {Cola} + 4/3*{Cola} = 63. We can solve for C. Sufficient.

Answer: B.

Question : How did you get from 0.8 Ale to 5 Beer , and then to the 4 ale, beer ? Also, how did you get from 0,75 beer ale to 4/3 cola?

Hi sagnik242,

let me try to help.
Beer = 0.8 Ale
Beer = \(\frac{8}{10}\) Ale
10 Beer = 8 Ale (multiply the entire expression with 10)
5 Beer = 4 Ale (simplify by dividing the entire expression with 2)

same goes with 0,75 beer ale to 4/3 cola. hope that was helpful :D
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This question could be very tricky if total number of bottles bought was between 10 and 17. A could be sufficient.


When I solved A, I didn't use any equation. I tried to think logically that Number of cola bottle could be any number if I different number of Ginger ale bottles.

But after looking the official solution, I realized that my thought process for A could be dangerous. I have to actually form equation in this type of situation.
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If Pei ordered a total of 63 bottles of Cola, Root Beer and Ginger Ale for a part. How many Cola bottles did she order?

(1) The no of bottles of root beer Pei order was 80% of bottles of ginger ale that she ordered
(2) The no of bottles of cola Pei order was 75% of total no. of bottles of ginger ale and root beer that she ordered

Take into consideration the av.time taken by participants, I have doubt that the first statement was fully tested. Or I do it too slowly :-)

C + RB + GA = 63

Statement 1
RB = 8/10 of GA = 4/5 of GA

C + 4/5GA + GA = 63
5C + 9GA = 315
C = 9(35-GA)/5

Hence, C is a multiple of 9, yet it can by any number within aforementioned restrictions.
The quibble here is if those numbers were primes, the probability that we can get correct answer is very high.

Statement 2
is indeed sufficient, we get sum of RB + GA, and, thus, we can deduct it from total and get the result.
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