Using a Venn diagram, we can fill in the given information:
[asy]
label("Faulty Brakes", (0,80));
label("Faulty Tyres", (80,80));
label("Good Brakes", (40,0));
label(scale(0.8)*"$10$",(20,50));
label(scale(0.8)*"$15$",(60,50));
label(scale(0.8)*"$2$",(20,70));
filldraw(circle((30,50),20),gray(0.7));
filldraw(circle((50,50),20),gray(0.7));
[/asy]
From the diagram, we can see that there were 8 cars with only faulty brakes (i.e. not faulty tyres), and $10-2=8$ cars with both faulty brakes and faulty tyres. Therefore, we have accounted for $10+8+8+15=41$ cars. Thus, the remaining $50-41=\boxed{9}$ cars must have had good brakes and good tyres.
The 50 cars used by a firm were inspected. 10 had faulty brakes and 15 had faulty tyres. There were 2 cars with faulty brakes but good tyres. How many cars had good brakes and good tyres?
2 answers
The answer is 33