SOLUTION: Hi 60% of the buses in a city are owned by A and the rest by B. When B bought 80 new buses, the percentage of its buses increased to 52% of the total of the number of buses. How m

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Hi 60% of the buses in a city are owned by A and the rest by B. When B bought 80 new buses, the percentage of its buses increased to 52% of the total of the number of buses. How m      Log On

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Question 1118530: Hi
60% of the buses in a city are owned by A and the rest by B. When B bought 80 new buses, the percentage of its buses increased to 52% of the total of the number of buses. How many buses did
B have after the increase.
thanks
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


60% = 3/5; so A owned 3/5 of the buses and B owned 2/5.

Let the number owned by A be 3x and the number owned by B be 2x; the total number of buses is 5x.

When B adds 80 buses, it now has (2x+80) buses, which is 52% of the total, and the total number of buses is now 5x+80:

%282x%2B80%29%2F%285x%2B80%29+=+0.52+=+13%2F25
25%282x%2B80%29+=+13%285x%2B80%29
50x%2B2000+=+65x%2B1040
15x+=+960
x+=+64

Originally the number of buses A had was 3x = 3*64 = 192, and the number B had was 2x = 2*64 = 128; the total number of buses was 5x = 320.

After adding 80 new buses, the number B has is 208.

CHECK: B now has 208 buses, and the total number is 192+208 = 400. 208/400 = 52/100 = 52%.