jaydevsachdeva wrote:

Answer is 62. Solved by both equations:

Hi, for all that are confused as to why there are 2 different answers from the given 2 equations,

There are two formulas for 3 overlapping sets:

Total=A+B+C−(sum of 2−group overlaps)+(all three)+Neither.

Total=A+B+C−(sum of EXACTLY 2−group overlaps)−2∗(all three)+Neither

The answer is supposed to be 62 for both of them.

For 1st Equation, as it is a sum of 2 group overlaps, it will also include the common(all three) 4 part with it.

Total=A+B+C−(sum of 2−group overlaps)+(all three)+Neither

Total= 20+30+40-((5+4)+(9+4)+(6+4))+4+0=62

=90-32+4

=62

For 2nd Equation, as it is a sum of 2 Exactly group overlaps, it will not include the common 4 part with it.

Total=A+B+C−(sum of EXACTLY 2−group overlaps)−2∗(all three)+Neither

=20+30+40-(5+9+6)-2*4+0

=90-20-8

=62

I hope I am right, and I hope everyone got it.

I completely agree with you in regards to this particular problem. See Below:

Reading through the history of this problem I can see why there is confusion. I understand applying Venn Diagrams and the formulas associated with such. My confusion initially resulted from the GMAT Club Math Book's formula explanation: Pay attention to the description on what it means to sum the intersections - it states to add "g" to every intersection (d, e, and f)...then you subtract from 90, then you add "g" again.

Total = A+B+C - (sum of 2-group overlaps) + (All 3) + Neither

"In the formula above, Sum of 2-group overlaps = AnB+AnC+BnC, where AnB means intersection of A and B (sections d, and g), AnC means intersection of A and C (sections e and g), and BnC means intersection of B and C (sections f and g).

Now, when we subtract AnB (d and g), AnC (e and g), and BnC (f and g) from A+B+C, we are subtracting sections "d, e, and f" ONCE BUT section "g" THREE TIMES (and we need to subtract section "g" only twice), therefore we should add only section "g", which is intersection of A, B, and C (AnBnC) again to get: Total = A+B+C - (sum of 2-group overlaps) + (all three) + Neither Due to this, I initially obtained 62 as well because I did as it states: I summed AnB+AnC+BnC by adding d and g, e and g, and f and g...THAT is the key...although I now understand the concept, not just memorizing, I did not when I started learning this concept. It did not mention to only sum d, e, and f.....But that is what I had to do to obtain 74. Thus, by adding (5+4)+(6+4)+(9+4) we get 32....subtracting it from 90 = 58, + 4 = 62. Now, in the Mathbook, the second formula is explained well in that you only add what which is common for A and C, which is "e", for example, not "e and g".

I love that mathbook, it has been my favorite prep for formulas and methods, I am eternally grateful....

Going back through I realize now that although it may say to specifically sum d+g, e+g, f+g, I know that when the question asks for "both A and B" I know not to add the "all 3" in it as well because anything shared amongst all 3 will be included in those 2. I may have read to much into the Mathbook but by understanding the concept it just makes sense now. Bunuel did a great job explaining, thank you again.