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13 Feb 2018, 22:42
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
92% (01:26) correct
8%
(01:16)
wrong
based on 64
sessions
History
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John, Peter, and Paul together have ten marbles. If each has at least one marble, how many marbles does each boy have?
(1) John has 5 more than Paul.
(2) Peter has half as many as John.
(1) John has 5 more than Paul.
(2) Peter has half as many as John.
13 Feb 2018, 23:55
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Bunuel
We have one equation with three variables (John + Peter + Paul = 10), we need to find two more.
We'll see which statements give us this information, a Logical approach.
(1) gives only one equation (John = Paul +5)
Insufficient.
(2) gives only one equation (Peter = John/2)
Insufficient.
Combined: we now have 3 distinct equations with 3 variables, sufficient.
(C) is our answer.
16 Feb 2018, 13:46
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Bunuel
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
This question and conditions do not make sense for the following reason.
Let J, T and P be the numbers of marbles that John, Peter and Paul have, respectively.
We have J ≥ 1, T ≥ 1 and P ≥ 1 from the original condition.
J + T + P = 10.
Conditions 1) & 2):
We have J = P + 5, and P = (1/2)J or J = 2P.
Then 2P = P + 5 and we have P = 5 and J = 10
Since J + T + P = 10, P = 5 and J = 10, we have T = -5, which is a negative number.
However, the basic concept related to this question is as follows.
Since we have 3 variables (J, T and P) and 1 equation, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first.
Conditions 1) & 2):
We have J = P + 5, and P = (1/2)J or J = 2P.
Then 2P = P + 5 and we have P = 5 and J = 10
That's why we can determine the value of T.
Therefore the answer would be C.
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
