Solution
Given• 7 times the number of coins that A has is equal to 5 times the number of coins B has
• 6 times the number of coins B has is equal to 11 times the number of coins C has.
To find• The minimum number of coins with A, B and C put together.
Approach and Working out Though it is a 700-level question, but it can be easily solved by applying the ‘process skill of inference’.
Let number of coins with A = x, with B = y, with C = z. (applying the process skill of ‘translation’)
Simplifying the ratios in terms of z alone
• \(7x = 5*(\frac{11z}{6})\)
o \(x = \frac{55z}{42} = \frac{11 * 5}{7 * 3 * 2}\)
• Using second equation y = \(\frac{11z}{6}\)
Using the above information, we have following inferences.
• Inference 1: since x =\(\frac{55z}{42}\) and there is no common factor between 55 and 42, thus z should be a multiple of 42. (since x has to be an integer)
• Inference 2: Since y = \(\frac{11z}{6}\) and there is no common factor between 11 and 6, thus z should be a multiple of 6. (since y has to be an integer)
• Inference 3: Since sum = x + y + z = \(\frac{55z}{42} + \frac{11z}{6} + z = \frac{29z}{7}\), thus z should be a multiple of 7
Using the inferences found above, we can have one more inference as follows.
• z should be a multiple of 7, 42 and 6.
• Thus z = k * LCM of (6,7,42)
• z (min) = 42
Now to evaluate the sum,
• Sum = \(\frac{29*z}{7} = \frac{29*42}{7} = 29*6 = 174\)
Correct Answer: Option EThus, the above question could be solved easily using the inferences found above. Essentially, we did the following.
• Applied the process skill of inference
• Used the conceptual knowledge of LCM
If we didn’t apply the process skills, the solution would have been longer and would have involved more complexity. Application of process skills help us solve the GMAT questions by reducing the complexity of the problems thereby improving both the efficiency as well as the accuracy.
To know more about process skills... visit this
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