If there are x cars and y yellow cars, clearly the new car is yellow, since 1/9 > 1/10
y = x/10
y+1 = (x+1)/9
There are now 81 cars, with 9 yellow.
How many cars are now in the car park?
y = x/10
y+1 = (x+1)/9
There are now 81 cars, with 9 yellow.
count=0
for i=1..2000
  with student[i] do {
    read   name,sex,score
    pct = score*5/2
    sub45 = false
    if pct < 45 then
      if sex = "F" then
      count++
      print name
    }
  }
}
print "There were ",count," females scoring below 45%"
Given that one-tenth of the cars are yellow initially, we can express this as 1/10. Let's represent the total number of cars in the car park as 'C'. Therefore, the number of yellow cars before the additional car arrives is (1/10)C.
After the extra car arrives, one-ninth of the cars become yellow. So, the new fraction of yellow cars is 1/9. We can represent this as (1/9)(C + 1), where 'C + 1' represents the total number of cars after the arrival of the extra car.
According to the given information, the fractions of yellow cars before and after the extra car's arrival should be equal. Therefore, we can set up the equation:
(1/10)C = (1/9)(C + 1)
To solve this equation, we can simplify and cross-multiply:
9C = 10(C + 1)
9C = 10C + 10
C - 10C = 10
-C = 10
C = -10
Since negative car quantities don't make sense in this context, we know that the equation cannot be solved with the given information. Therefore, we cannot determine the total number of cars in the car park based on the information provided.