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Zorluk: Çok zorFractions, Decimals, and Percents Arithmetic

A cloud data center processes an incoming raw data stream through sequential filtering stages. In Stage 1, 38\frac{3}{8} of the incoming raw data volume is discarded as noise, and 0.200.20 of the remaining data is flagged for long-term archiving. In Stage 2, 45\frac{4}{5} of the data not flagged for long-term archiving is processed into active storage, while the rest is discarded. If the volume of data processed into active storage in Stage 2 is 5050 terabytes greater than the volume of data discarded in Stage 1, what was the initial volume of the raw data stream, in terabytes?

  1. A
    400
  2. B
    1,600
  3. 2,000Cevap
  4. D
    2,500
  5. E
    4,000

Cevap

2,000 terabytes
Let the initial data volume be XX. Discarded in Stage 1 is 38X\frac{3}{8}X, leaving 58X\frac{5}{8}X. Archiving takes 20%20\% (0.20=150.20 = \frac{1}{5}) of this remainder, which is 15×58X=18X\frac{1}{5} \times \frac{5}{8}X = \frac{1}{8}X. The data entering Stage 2 is 58X−18X=12X\frac{5}{8}X - \frac{1}{8}X = \frac{1}{2}X. Active storage in Stage 2 is 45×12X=25X\frac{4}{5} \times \frac{1}{2}X = \frac{2}{5}X. Subtracting Stage 1 discarded volume from Stage 2 active storage volume gives 25X−38X=1640X−1540X=140X\frac{2}{5}X - \frac{3}{8}X = \frac{16}{40}X - \frac{15}{40}X = \frac{1}{40}X. Setting 140X=50\frac{1}{40}X = 50 yields X=2,000X = 2,000 terabytes.

Adım Adım Çözüm

1
Define the variable and compute Stage 1 discarded volume and remaining volume.
Discarded in Stage 1 = 38X\frac{3}{8}X; Remaining after Stage 1 discard = X−38X=58XX - \frac{3}{8}X = \frac{5}{8}X.
Establishing quantities in terms of the total initial volume XX allows setting up a single-variable linear equation.
2
Calculate the volume flagged for archiving and the volume available for Stage 2.
Archived volume = 0.20×58X=15×58X=18X0.20 \times \frac{5}{8}X = \frac{1}{5} \times \frac{5}{8}X = \frac{1}{8}X. Available for Stage 2 = \frac{5}{8}X - \frac{1}{8}X = \frac{4}{8}X = \frac{1}{2}X$.
The 0.200.20 decimal must be converted to a fraction (15\frac{1}{5}) and applied to the remaining 58X\frac{5}{8}X base.
3
Calculate the volume processed into active storage in Stage 2.
Active storage volume = 45×12X=25X\frac{4}{5} \times \frac{1}{2}X = \frac{2}{5}X.
Stage 2 processes 45\frac{4}{5} of the data that entered Stage 2.
4
Set up the algebraic equation comparing active storage volume in Stage 2 to discarded volume in Stage 1.
25X−38X=50\frac{2}{5}X - \frac{3}{8}X = 50.
The problem states that active storage volume in Stage 2 is 5050 terabytes greater than the Stage 1 discarded volume.
5
Find a common denominator and solve for XX.
(1640−1540)X=50  ⟹  140X=50  ⟹  X=2,000\left(\frac{16}{40} - \frac{15}{40}\right)X = 50 \implies \frac{1}{40}X = 50 \implies X = 2,000.
Converting fractions to a denominator of 4040 yields 140X=50\frac{1}{40}X = 50, so multiplying by 4040 gives X=2,000X = 2,000 terabytes.

Anahtar Kavram

Multi-step arithmetic operations involving sequential fractions, decimals, and percent bases
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Konular

Fractions, Decimals, and Percents Arithmetic

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