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E
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Difficulty:
15%
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Question Stats:
83% (02:07) correct
17%
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wrong
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Ann bought five different kinds of fruit: apples, oranges, pears, mangoes, and bananas. If the number of apples that Ann bought was twice the number of oranges and if the number of pears that Ann bought was the same as the number of apples and oranges combined, what fraction of the total number of pieces of fruit that Ann bought were pears?
(1) Ann bought a total of 18 pieces of fruit.
(2) Ann bought 5 bananas.
(1) Ann bought a total of 18 pieces of fruit.
(2) Ann bought 5 bananas.
26 Oct 2014, 06:39
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Ann bought five different kinds of fruit: apples, oranges, pears, mangoes, and bananas. If the number of apples that Ann bought was twice the number of oranges and if the number of pears that Ann bought was the same as the number of apples and oranges combined, what fraction of the total number of pieces of fruit that Ann bought were pears?
(1) Ann bought a total of 18 pieces of fruit.
(2) Ann bought 5 bananas
Let us assume 'x' be the number of oranges bought by Ann.
A (apples) -> 2x,O(oranges)->x,P(Pears)->3x,M(Mangos)->Not known,B(Bananas)-> Not Known
Option (1)
Even if we know the total number of pieces of fruit (18 i.e given),it will be required to find the number of quantities of anyone of Apples,Oranges or Pears (as we can derive 'x' based on the value).It is not possible to calculate 3x/18 (as we do not know the value of 'x')
Not Sufficient
Option (2)
In this option as well,we have been given the value of (bananas) B=5 that makes it impossible to evaluate the fraction of total number of pieces of pears.First off all we do not know the total number of quantities of fruit and secondly number of quantities of anyone of Apples,Oranges or pears is unknown.
Not Sufficient.
Option (1) + (2).
Even if we know the total number of pieces of fruit (option 1) and quantity of bananas (B=5)(Option 2) available,it will be still required to find the number of quantities of anyone of Apples,Oranges or Pears (as we can derive 'x' based on the value).It is not possible to calculate 3x/18 (as we do not know the value of 'x').
Not Sufficient.
Hence Option 'E' .
(1) Ann bought a total of 18 pieces of fruit.
(2) Ann bought 5 bananas
Let us assume 'x' be the number of oranges bought by Ann.
A (apples) -> 2x,O(oranges)->x,P(Pears)->3x,M(Mangos)->Not known,B(Bananas)-> Not Known
Option (1)
Even if we know the total number of pieces of fruit (18 i.e given),it will be required to find the number of quantities of anyone of Apples,Oranges or Pears (as we can derive 'x' based on the value).It is not possible to calculate 3x/18 (as we do not know the value of 'x')
Not Sufficient
Option (2)
In this option as well,we have been given the value of (bananas) B=5 that makes it impossible to evaluate the fraction of total number of pieces of pears.First off all we do not know the total number of quantities of fruit and secondly number of quantities of anyone of Apples,Oranges or pears is unknown.
Not Sufficient.
Option (1) + (2).
Even if we know the total number of pieces of fruit (option 1) and quantity of bananas (B=5)(Option 2) available,it will be still required to find the number of quantities of anyone of Apples,Oranges or Pears (as we can derive 'x' based on the value).It is not possible to calculate 3x/18 (as we do not know the value of 'x').
Not Sufficient.
Hence Option 'E' .
03 Dec 2015, 05:59
Kudos
Bookmarks
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
Ann bought five different kinds of fruit: apples, oranges, pears, mangoes, and bananas. If the number of apples that Ann bought was twice the number of oranges and if the number of pears that Ann bought was the same as the number of apples and oranges combined, what fraction of the total number of pieces of fruit that Ann bought were pears?
(1) Ann bought a total of 18 pieces of fruit.
(2) Ann bought 5 bananas.
We have 5 variables (a,o,p,m,b) and 2 equations (a=2o, p=a+o) in the original condition, but only 2 equations are given by the conditions, so there is high chance (E) will be the answer.
Looking at the conditions together, a+o+p+m+b=18, b=5, but a=2o, p=a+o=3o still does not give the number of p, so is insufficient, making the answer (E).
For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
Ann bought five different kinds of fruit: apples, oranges, pears, mangoes, and bananas. If the number of apples that Ann bought was twice the number of oranges and if the number of pears that Ann bought was the same as the number of apples and oranges combined, what fraction of the total number of pieces of fruit that Ann bought were pears?
(1) Ann bought a total of 18 pieces of fruit.
(2) Ann bought 5 bananas.
We have 5 variables (a,o,p,m,b) and 2 equations (a=2o, p=a+o) in the original condition, but only 2 equations are given by the conditions, so there is high chance (E) will be the answer.
Looking at the conditions together, a+o+p+m+b=18, b=5, but a=2o, p=a+o=3o still does not give the number of p, so is insufficient, making the answer (E).
For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
