Question

Hi guys I am here to answer the question which I ask which is The 50 cars used by afirm, 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

ANSWER
NUMBER OF UNIVERSAL SET n(U) =50
NUMBER OF CAR WITH FAULTY BRAKES n(FB) =10
NUMBER OF CAR WITH FAULTY TYRES =15 n(FG)
THE NUMBER CARS WITH FAULTY BUT GOOD TYRES =2
HOW MAY CARS HAD GOOD BRAKES AND GOOD TYRES? n(fbUft) '
BY USING FORMULA
n(U) =n(FT U FB) +n(FT U FB)'
BUT WEDON'T HAVE THE NUMBER OF FAULTY BRAKE UNON NUMBER OF FAULTY TYRE, n(FB U FT) BUT WE HAVE NUMBER OF CAR WITH FAULTY BRAKES BUT GOOD TYRES =2 SO WE CAN GET n(FT U FB)
We take number of faulty brakes minus x

10-x=2
10-x=2
X=10-2
X=8
So n(FT U FB) =10-8+8+15-8
=17
From formula
N(U) =n(FtUFb) +(FTUFB) '
50 =17 +n(FTUFB)'
N(FTUFB) =50-17
=33 ANS



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Answer 1

Explanation:

This is not that hard....just use a Venn diagram to illustrate the info given...then cars with good brakes and good tires = 50 - 2 - 8 - 7 = 33