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Difficulty:
5%
(low)
Question Stats:
82% (01:05) correct
18%
(01:13)
wrong
based on 809
sessions
History
Date
Time
Result
Not Attempted Yet
At a refreshment stand, each can of soda sells for the same price and each sandwich sells for the same price. What is the total price for 2 sandwiches and 3 cans of soda at the stand?
(1) At the stand the total price for 1 sandwich and 1 can of soda if $3
(2) At the stand the total price for 3 sandwiches and 2 cans of soda is $8
(1) At the stand the total price for 1 sandwich and 1 can of soda if $3
(2) At the stand the total price for 3 sandwiches and 2 cans of soda is $8
24 Jan 2015, 23:24
Kudos
Bookmarks
here we go...
Let price of Soda = p
Let price of sandwich = m
2m + 3p = ??
St1: m + p = 3
m = 2, p = 1 -----> holds true
m = 1.5, p = 1.5 -----> holds true
Clearly not sufficient
St2: 3m + 2p = 8
m = 2, p = 1----> holds true
m = 1.5, p = 1.75 -----> holds true
Clearly not sufficient
Combining St1 and St2
m + p = 3
3m + 2p = 8
Solve---
m = 2, p = 1
Hence C is the winner here.
Let price of Soda = p
Let price of sandwich = m
2m + 3p = ??
St1: m + p = 3
m = 2, p = 1 -----> holds true
m = 1.5, p = 1.5 -----> holds true
Clearly not sufficient
St2: 3m + 2p = 8
m = 2, p = 1----> holds true
m = 1.5, p = 1.75 -----> holds true
Clearly not sufficient
Combining St1 and St2
m + p = 3
3m + 2p = 8
Solve---
m = 2, p = 1
Hence C is the winner here.
General Discussion
25 Jan 2015, 00:34
Kudos
Bookmarks
Hi ajca,
This DS question is built on "system" Algebra, without any of the factors/multiples shortcuts that are common on DS versions of these questions.
We're asked for the total price of 2 sandwiches and 3 cans of soda.
**Note: When translating the Facts into Algebra, I'm going to use....
W = # of sandwiches
C = # of cans of soda
The question asks us for 2W + 3C
Fact 1: 1 sandwich and 1 can of soda = $3
W + C = 3
This is 1 equation with 2 variables, so there's no way to figure out the answer to the question. Here's the proof:
IF....
W = 1
C = 2
Then the answer to the question is 2(1) + 3(2) = 8
IF.....
W = 2
C = 1
Then the answer to the question is 2(2) + 3(1) = 7
Fact 1 is INSUFFICIENT
Fact 2: 3 sandwiches and 2 cans of soda = $8
We have a similar situation to what we have in Fact 1: 1 equation and 2 variables.....
3W + 2C = 8
IF....
W = 2
C = 1
Then the answer to the question is 2(2) + 3(1) = 7
IF....
W = 1
C = 2.5
Then the answer to the question is 2(1) + 3(2.5) = 9.5
Fact 2 is INSUFFICIENT
Combined, we have...
W + C = 3
3W + 2C = 8
This is a "system" - 2 variables and 2 unique equations, which means that we CAN solve for the values of W and C (and then answer the question.
Combined, SUFFICIENT
Final Answer:
GMAT assassins aren't born, they're made,
Rich
This DS question is built on "system" Algebra, without any of the factors/multiples shortcuts that are common on DS versions of these questions.
We're asked for the total price of 2 sandwiches and 3 cans of soda.
**Note: When translating the Facts into Algebra, I'm going to use....
W = # of sandwiches
C = # of cans of soda
The question asks us for 2W + 3C
Fact 1: 1 sandwich and 1 can of soda = $3
W + C = 3
This is 1 equation with 2 variables, so there's no way to figure out the answer to the question. Here's the proof:
IF....
W = 1
C = 2
Then the answer to the question is 2(1) + 3(2) = 8
IF.....
W = 2
C = 1
Then the answer to the question is 2(2) + 3(1) = 7
Fact 1 is INSUFFICIENT
Fact 2: 3 sandwiches and 2 cans of soda = $8
We have a similar situation to what we have in Fact 1: 1 equation and 2 variables.....
3W + 2C = 8
IF....
W = 2
C = 1
Then the answer to the question is 2(2) + 3(1) = 7
IF....
W = 1
C = 2.5
Then the answer to the question is 2(1) + 3(2.5) = 9.5
Fact 2 is INSUFFICIENT
Combined, we have...
W + C = 3
3W + 2C = 8
This is a "system" - 2 variables and 2 unique equations, which means that we CAN solve for the values of W and C (and then answer the question.
Combined, SUFFICIENT
Final Answer:
GMAT assassins aren't born, they're made,
Rich
