Data Sufficiency

To receive a driver license, sixteen year-olds at Culliver High School have to pass both a written and a practical driving test. Everyone has to take the tests, and no one failed both tests. If 30% of the 16 year-olds who passed the written test did not pass the practical, how many sixteen year-olds at Culliver High School received their driver license?

(1) There are 188 sixteen year-olds at Culliver High School.

(2) 20% of the sixteen year-olds who passed the practical test failed the written test.

To receive a driver license, sixteen year-olds at Culliver High School have to pass both a written and a practical driving test. Everyone has to take the tests, and no one failed both tests. If 30% of the 16 year-olds who passed the written test did not pass the practical, how many sixteen year-olds at Culliver High School received their driver license?

(1) There are 188 sixteen year-olds at Culliver High School.

(2) 20% of the sixteen year-olds who passed the practical test failed the written test.

Matrix! Let's call 'x' the number of students who passed the written exam. Given that .30x passed the written and did not pass the practical, and that nobody failed both, our initial matrix will look like this:


Statement 1 tell us that there are 188 sixteen-year-olds in all. Now our matrix looks like this:




That won't tell us how many passed both tests. Not Sufficient.

Statement 2: Let's call the number of students who passed the practical test 'y.' Now we know that .2y passed the practical and failed the written and that .8y passed both. So there are two ways we can designate the "Passed practical/Passed written" cell, .7x and .8y. We now know that .7x = .8y. So our matrix will look like this:




We now have a ratio of y to x, (y/x = 7/8) but no values. Not Sufficient.

Together, our matrix will look like this:


Summing the bottom row, we have: x + .2y = 188. Summing the right-most column we have: y + .3x = 188. Two linear equations/two unknowns, so we can solve for our unknowns. If we can solve for x, we can solve for .7x. Together, the statements are sufficient. Answer is C

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.

To receive a driver license, sixteen year-olds at Culliver High School have to pass both a written and a practical driving test. Everyone has to take the tests, and no one failed both tests. If 30% of the 16 year-olds who passed the written test did not pass the practical, how many sixteen year-olds at Culliver High School received their driver license?

(1) There are 188 sixteen year-olds at Culliver High School.

(2) 20% of the sixteen year-olds who passed the practical test failed the written test.

From the original condition we can obtain the below table. This is a typical 2by2 question that we commonly face in GMAT math test.


From the table, we have 3 variables (a,b,c) and 1 equation (30%=c/(a+b+c)) therefore we need 2 more equations to match the number of variables and equations. Since there is 1 each in 1) and 2), there is high probability that C is the answer. It turns out that C actually is the answer.

Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

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Can this problem be solved with Venn Diagrams ?

Have been trying to do this via Venn Diagrams but too no avail