Question

A company received two shipments of ball bearnings. In the first shipment, I percent of the ball bearnings were defective. In the second shipment, which was twice as large as the first, 4.5 percent of the ball bearings were defective. If the company received a total of 100 defective ball bearings, how many ball bearings were in the first shipment?

• A. 2000
• B. 1000
• C. 990
• D. 3000

Answer 1

The Correct Option is B

Let's call the number of ball bearings in the first shipment "x" and the percentage of defective ball bearings in the first shipment "I."
So, in the first shipment, there were \((\frac{1}{100})\times\) x defective ball bearings. 
In the second shipment, which was twice as large as the first, there were 2x ball bearings, and 4.5 percent of them were defective. 
So, in the second shipment, there were \((\frac{4.5}{100})\times2\)x defective ball bearings.
The total number of defective ball bearings is given as 100: 
\((\frac{I}{100})\times x+(\frac{4.5}{100})\times2x=100\) 
Now, let's solve for x: 
\((\frac{I}{100})\times x+(\frac{9}{100})\times x=100\)
\((\frac{I}{100}+\frac{9}{100})\times x=100\)
\((\frac{(I+9)}{100})\times x=100\)
Now, we can solve for x: 
\(x=\frac{(100\times100)}{(I+9)}\)
We are given that the total number of defective ball bearings is 100, so: 
\((\frac{1}{100})\times x+(\frac{4.5}{100})\times2x=100\)
\((\frac{1}{100})\times x+(\frac{9}{100})\times x=100\)
\((\frac{1}{100}+\frac{9}{100})\times x=100\)
\((\frac{(I+9)}{100})\times x=100\)
\(\frac{(I+9)}{100}=\frac{100}{x}\) 
\(x=100\times\frac{100}{(I+9)}\)
Now, we know that x must be a positive integer, so we need to find a value of I for which x is a positive integer. 
Let's try I = 1: 
\(x=\frac{100\times100}{(I+9)}\)
\(x=\frac{10000}{10}\)
\(x=1000\)
So, when I = 1, x = 1000, which is a positive integer. 
\(\therefore\) the number of ball bearings in the first shipment was 1000.
The correct option is (B): 1000