A man shared his lottery winnings of $15,000 with his wife and two

Question

A man shared his lottery winnings of $15,000 with his wife and two children. How much did his wife receive?

(1) He gave his children $2,000 each.
(2) The wife received $1,000 less than her husband and three thousand more than each of the children.

chetan2u · Answer

The important point is shared WITH, which means it was divide amongst 4- man, his wife and TWO chidren..

(1) He gave his children $2,000 each. 
so he and his wife took 15000-2*2000 = $11,000
Insuff

(2) The wife received $1,000 less than her husband and three thousand more than each of the children. 
W = M-1000 = C+3000.....
 W+M+2C=15,000.....
W+(W+1000)+2(W-3000) = 15000... 
the equation has ONLY one variable 
so Suff
4W-5000=15000...........W=5000
Suff

OptimusPrepJanielle · Answer

Total lottery = $15,000 sahred with wife and 2 children.
Share of wife = ?

Statement 1: He gave his children $2,000 each
Amount remaining = $11,000
No information about the sharing with wife.
INSUFFICIENT

Statement 2: Wife received $1000 less than husband and $3000 more than children.
Assume amount received by wife = x

Amount with husband = x + 1000
Amount with children = x - 3000

We know that total = 15000
Hence we can solve x + (x + 1000) + 2*(x - 3000) = 15000 to get the value of x
SUFFICIENT

Correct Option : B

MathRevolution · Answer

In the original condition, there are 4 variables (h:husband, w:wife x: son, y: son) and 1 equation (h+w+x+y=15,000). In order to match the number of variables and the number of equations, we need 3 equations. Since there are one condition 1) and one condition 2), we lack 2 equations. Therefore, there is high chance that E is likely the answer. 
However, the condition 1) has 2 equations and the condition 2) has 3 equations. So, the condition 2) is sufficient and the correct answer is B.

- 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.

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A man shared his lottery winnings of $15,000 with his wife and two

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