Math Question -- Algebra?

Question

A car dealer has 22 vehicles on his lot. If 8 of the vehicles are vans and 6 of the vehicles are red, and 10 vehicles are neither vans nor red, how many red vans does he have on his lot?

Answer 1

I think the answer is 2, but I'm not sure. If 10 vehicles are neither red nor vans, then 12 vehicles are either a van or red (since it must total 22). If he has 8 vans and 6 red vehicles, that totals 14, so the overlap of vans and red must be 2, in order for the totals to add up correctly.

Answer 2

Working backwards, 10 vehicles are neither vans nor red, so 22 - 10 = 12 vehicles that are either vans or red.

Probability of one of the 12 vehicles being vans is 8/12 = 2/3

Probability of one of the 12 vehicles being red is 6/12 = 1/2

Probability of one of the 12 vehicles being BOTH read AND a van is the product of the previous two probabilities

(2/3)(1/2) = 2/6 = 1/3. So, there is a one in three chance that a vehicle on his lot is a red van.

Since there are 22 vehicles on the lot,
22(1/3) = 7 red vans, rounded off.

P.S. It is understandable that sales are slow here. Who on earth would buy a large red van?

Answer 3

There are 22 vehicles on the lot. 10 vehicles are neither red nor vans, so 22-10 = 12 vehicles or either red or vans.
Let A=red
B=van
n(A or B)= n(A)+n(B)-n(A and B)
12=8+6-n(A and B) --- (1)
so there are 2 red vans in his lot. Equation (1) is satisfied for red vans = 2.

Answer 4

2 Vans must be red because only 4 of the 14 non-vans are red. Since there are 6 red vehicles, 2 of them must be red vans.

Answer 5

(10 + 6 +8) - 22
= 2

2 red vans