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Equations for 3 overlapping sets?

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Equations for 3 overlapping sets?

by makhija1 » Fri Jun 21, 2013 8:22 pm
Q1. In a group of 68 students, each student is registered for at least one of three classes - History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

Q2. At a certain dealership, every car on the lot has at least one of three modest options: windows, brakes and radio. 40 cars have windows, 30 cars have brakes, and 50 cars have a radio. 21 cars have B&R, 13 cars have B&W and 17 cars have W&R. If 11 cars have all 3 options, what is the total number of cars on the lot?

For Q1, we are applying the following equation for 3 overlapping sets:

Total = H + M + E - (HM + ME + HE) - 2(HME)
68 = 25 + 25 + 34 - Sum - 6
Sum = 10

While for Q2, we are applying a different equation for 3 overlapping sets:

Total = B + W + R - (BR + BW + WR) + BWR
Total = 40 + 30 + 50 -21 - 13 - 17 + 11
Total = 80

Please explain the different scenarios/equations and when they must be applied...

Thanks,
Adi
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by GMATGuruNY » Sat Jun 22, 2013 6:45 pm
makhija1 wrote:At a certain dealership, every car on the lot has at least one of three modest options: windows, brakes and radio. 40 cars have windows, 30 cars have brakes, and 50 cars have a radio. 21 cars have B&R, 13 cars have B&W and 17 cars have W&R. If 11 cars have all 3 options, what is the total number of cars on the lot?
The formula used for the first problem can also be applied here:

T = W + B + R - (WB+ WR + BR) - 2(WBR)

The big idea with overlapping group problems is to SUBTRACT THE OVERLAPS.
When we add together every car in W, every car in B, and every car in R:
Those in exactly 2 of the groups (WB+ WR + BR) are counted twice, so they need to be subtracted from the total ONCE.
Those in all 3 groups (WBR) are counted 3 times, so they need to be subtracted from the total TWICE.
By subtracting the overlaps, we ensure that no car is overcounted.

Values in the formula:
W = 40.
B = 30.
R = 50.
Of the 13 cars that have windows and brakes, 11 have all 3 options, so cars with ONLY windows and brakes = WB = 13-11 = 2.
Of the 17 cars that have windows and a radio, 11 have all 3 options, so cars with ONLY windows and a radio = WR = 17-11 = 6.
Of the 21 cars that have brakes and a radio, 11 have all 3 options, so cars with ONLY brakes and a radio = BR = 21-11 = 10.
WBR = 11.

Thus:
T = 40+30+50 - (2+6+10) - 2(11) = 80.
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