Numerical Reasoning Test 2 Nlng Job

Let the weight of one bag of wheat be \(w\) pounds, one bag of corn be \(c\) pounds, and one bag of oats be \(o\) pounds.

From the first statement, the two sides have equal weights:\[3w + 2c + o = w + c + 5o\]Rearranging this equation gives:\[2w + c - 4o = 0\]which can be written as:\[2w = 4o - c\]
From the second statement, two bags of corn and one bag of oats weigh 26 pounds. Since one bag of corn weighs 10 pounds, substitute \(c = 10\):\[2(10) + o = 26\]\[20 + o = 26\]\[o = 6\]
Substitute \(o = 6\) and \(c = 10\) into \(2w = 4o - c\):\[2w = 4(6) - 10 = 24 - 10 = 14\]\[w = 7\]
This shows that one bag of wheat weighs 7 pounds.

To determine which Production Team made the most motors without defects, the actual number of defect-free motors produced by each team must be calculated, not just the defect percentages.

From the diagram, the total number of motors produced by each team is:
Team A = 4,500  
Team B = 2,500  
Team C = 3,500  
Team D = 4,000  
Team E = 3,000  

The number of motors without defects is calculated using:
(100% − defect percentage) × total production

Team A: 75% × 4,500 = 3,375  
Team B: 94% × 2,500 = 2,350  
Team C: 91% × 3,500 = 3,185  
Team D: 85% × 4,000 = 3,400  
Team E: 87% × 3,000 = 2,610  

Thus, Team D produce the most cars without defects.

\[ \text{Cost of machine} = 50{,}000\] \[ \text{Profit in March} = 7{,}876\] \[ \text{Monthly increase} = 2\% \Rightarrow r = 1.02\] \[ a = 7{,}876\] \[ S_n = a \frac{r^n - 1}{r - 1}\] \[ 7{,}876 \times \frac{1.02^n - 1}{0.02} \ge 50{,}000\] \[ 7{,}876(1.02^n - 1) \ge 1{,}000\] \[ 1.02^n - 1 \ge \frac{1{,}000}{7{,}876}\] \[ 1.02^n \ge 1.127\] \[ \log(1.02^n) \ge \log(1.127)\] \[ n\log(1.02) \ge \log(1.127)\] \[ n \ge \frac{\log(1.127)}{\log(1.02)}\] \[ n \ge 6.06\] \[ n = 7\]

Counting from March:

1. March
2. April
3. May
4. June
5. July
6. August
7. September