Each new car produced in a certain factory includes at least one of th

Let a,b and c are number of cars having only one option, let p,a and r are number of cars having exactly 2 options and x number of cars have exactly 3 options.
68= (a+b+c)+2(p+q+r)+3x -----(1)
50= (a+b+c) +(p+q+r)+ x -----(2)
Now (1)-(2)
18= (p+q+r) +2x
Option 1
P+q+r=2, x=8 fits
Option 2
P+q+r= 4, x=7 fits
Option 3
P+q+r =6, x = 6 fits
Option E is the answer

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Solution:

We can use the formula:

Total = n(A) + n(B) + n(C) - n(Exactly two sets) - 2 * n(All three sets) + n(No set)

Since n(No set) = 0, we have:

Total = n(A) + n(B) + n(C) - n(Exactly two sets) - 2 * n(All three sets)

Substituting the numbers, we have:

50 = 25 + 13 + 30 - n(Exactly two sets) - 2 * n(All three sets)

50 = 68 - n(Exactly two sets) - 2 * n(All three sets)

n(Exactly two sets) + 2 * n(All three sets) = 18 → Eq. 1

Now, let’s analyze each statement.

I. There were 2 cars that had exactly two of the options and there were 8 cars that had all three options.

n(Exactly two sets) = 2 and n(All three sets) = 8

Let’s substitute this into Eq. 1 and see if it works:

2 + 2 * 8 = 18 ?

18 = 18 ? → Yes

Statement I could be true.

II. There were 4 cars that had exactly two of the options and there were 7 cars that had all three options.

n(Exactly two sets) = 4 and n(All three sets) = 7

Let’s substitute this into Eq. 1 and see if it works:

4 + 2 * 7 = 18 ?

18 = 18 ? → Yes

Statement II could be true.

II. There were 6 cars that had exactly two of the options and there were 6 cars that had all three options.

n(Exactly two sets) = 6 and n(All three sets) = 6

Let’s substitute this into Eq. 1 and see if it works:

6 + 2 * 6 = 18 ?

18 = 18 ? → Yes

Statement III could be true.

Answer: E