Question

Difficulty: MediumDecimals and Scientific Notation

The mass of a sample in a laboratory experiment is given by the expression M=0.000375×10nM = 0.000375 \times 10^n grams, where nn is an integer. When MM is written in standard scientific notation as a×10ka \times 10^k, where 1a<101 \le a < 10 and kk is an integer, the exponent kk is equal to 2-2. What is the value of nn?

  1. A
    6-6
  2. B
    2-2
  3. 22Answer
  4. D
    44
  5. E
    66

Answer

The value of nn is 22.
Writing 0.0003750.000375 as 3.75×1043.75 \times 10^{-4} allows the mass MM to be rewritten as 3.75×104+n3.75 \times 10^{-4 + n}. Matching this with the standard scientific notation form a×10ka \times 10^k (where a=3.75a = 3.75 and k=2k = -2) gives the equation 4+n=2-4 + n = -2. Solving for nn yields n=2n = 2.

Step-by-Step Solution

1
Convert the decimal decimal coefficient to scientific notation.
0.000375=3.75×1040.000375 = 3.75 \times 10^{-4}
Moving the decimal point 4 places to the right puts the leading number in the required range 1a<101 \le a < 10.
2
Substitute this conversion back into the expression for MM and combine powers of 10.
M=(3.75×104)×10n=3.75×10n4M = (3.75 \times 10^{-4}) \times 10^n = 3.75 \times 10^{n - 4}
By exponent rules, 10a×10b=10a+b10^a \times 10^b = 10^{a+b}.
3
Equate the exponent of 10 in the simplified expression to the given value of kk.
n4=2    n=2n - 4 = -2 \implies n = 2
The scientific notation requires a×10ka \times 10^k where k=2k = -2, so n4n - 4 must equal 2-2.

Key Concept

Converting numbers between standard decimal form and scientific notation using place value and exponent properties
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