Reasonable Rates

Hard Scientific Notation Problems

Scientific notation is a method used to express very large or very small numbers as a product of a coefficient (between 1 and 10) and a power of ten.These scientific notation practice problems are commonly tested on the ACT Math section and require strong understanding of exponents, multiplication, division, and number conversion.

When multiplying and dividing numbers in scientific notation, you must adjust the coefficient so that it stays between 1 and 10 while correctly updating the exponent. , $$\dfrac{3\times10^7}{5\times10^3}=0.6 \times 10^4=6\times 10^3.$$

Hard scientific notation problems | Practice Questions

1.What is 0.00024 expressed in scientific notation? 2

$ 2.4\times 10^{-3} $
$ 2.4\times 10^{-4} $
$ 2.4\times 10^{-5} $
$ 2.4\times 10^4 $

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B

2. What is 362 million expressed in scientific notation? 2

$ 3.62\times10^7 $
$ 3.62\times10^6 $
$3.62\times10^9$
$ 3.62\times10^8$

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D

3. What is the total weight in pounds expressed in scientific notation of 57 million bricks if each brick weighs 4 pounds? 1

$ 2.28\times 10^8 $
$1.14\times 10^9$
$ 2.28\times 10^9$
$ 4.56\times 10^9 $

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A

4. What is $5\cdot 0.00000063$ in scientific notation? 3

$3.15\times 10^{-5}$
$3.15\times 10^{-6} $
$ 3.15\times 10^{-7} $
$ 3.15\times 10^{-8} $

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B

5. What is 12% of $2.5\times 10^{115}$? 3

$3.0\times 10^{114}$
$3.0\times 10^{115}$
$3.0\times 10^{116}$
$ 6.0\times 10^{114} $

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A

6. What is $3\times10^7+8\times10^5$ expressed in scientific notation? 3

$3.8\times10^7$
$3.8\times10^5$
$3.08\times10^5$
$3.08\times10^7$

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D

7. If a light-year is $5.9\times10^{12}$ miles, about how many light-years away is a star that is $3.7\times10^{15}$ miles away? 3

$63 $
$627 $
$629 $
$ 6268 $

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B

8. What value of $x$ makes the following equation true: $2\!\times\!10^{11}\cdot4.7\!\times\!10^{3x+5}\!=\!94,000$? 3

$ -2 $
$ -3 $
$-4 $
$ -5$

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C

9. What is $\sqrt{1.6\times10^{1301}}$? 3

$4\times10^{1300} $
$ 4\times10^{649}$
$ 4\times10^{1299} $
$ 4\times10^{650} $

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D

10. If a water sample taken near a chemical plant has 8 parts per million of a toxin, how many milliliters of the toxin are in 20,000 liters of the water? 4

$16$
$ 160$
$400 $
$ 1600 $

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B

11. What is $5\times10^{3000}\cdot7\times10^{5000}$ expressed in scientific notation? 4

$ 3.5\times10^{8001}$
$3.5\times10^{8000} $
$3.5\times10^{8002}$
$3.5\times10^{7999}$

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A

Answer key: 1.B, 2.D, 3.A, 4.B, 5.A, 6.D, 7.B, 8.C, 9.D, 10.B, 11.A.

Solutions

1. (B) Move the decimal point 4 places to the right to place the first non-zero digit (2) to the left of the decimal: $0.00024 \rightarrow 2.4$. Since the original number is less than 1, the exponent is negative, resulting in $2.4\times 10^{-4}$.

2. (D) Write 362 million as $362 \times 10^6$. Adjust the coefficient to be between 1 and 10 by moving the decimal point two places to the left ($362 = 3.62 \times 10^2$). Finally, combine the exponents: $3.62 \times 10^2 \cdot 10^6 = 3.62 \times 10^8$.

3. (A) One million is $10^6$, so 57 million is $57\times 10^6=5.7\times 10^7$. Multiply by the weight of each brick (4): $5.7\times 10^7 \cdot 4 = 22.8\times 10^7$. Adjust the coefficient to be between 1 and 10 to obtain the final answer: $22.8 \times 10^7 = 2.28\times 10^8$.

4. (B) Multiply the values: $5 \cdot 0.00000063 = 0.00000315$. To convert to scientific notation, move the decimal point 6 places to the right to create a number between 1 and 10 ($3.15$). This gives $3.15 \times 10^{-6}$.

5. (A) Convert 12% to 0.12 and multiply: $0.12\cdot 2.5\times 10^{115}=0.3\times 10^{115}=3.0\times 10^{114}$ in scientific notation.

6. (D) Convert the second term to the same exponent as the first and add. \[3 \times10^7+0.08\times10^7=3.08\times10^7. \]

7. (B) Divide the distances, and then convert to an integer. $\dfrac{3.7\times10^{15}}{5.9\times10^{12}} \approx 0.627 \times 10^3 = 627$.

8.(C) Since the coefficient on the left is $9.4$ and $94,000 = 9.4 \times 10^4$, we can equate the exponents of 10: $11+3x+5=4 \implies 3x=-12 \implies x=-4$.

9. (D) Rewrite the expression to make the exponent even: $1.6 \times 10^{1301} = 16 \times 10^{1300}$. Then, calculate the square roots: $\sqrt{16}=4$ and $\sqrt{10^{1300}}=10^{650}$. The answer is $4\times10^{650}$.

10. (B) Express the concentration as a fraction, multiply by the number of liters, and finally convert from liters to milliliters. $\left(\dfrac{8}{1,000,000}\right)\cdot20,000=0.16$ liters; $0.16\times1000=160$ milliliters.

11. (A) Multiply the coefficients and the powers of 10, then adjust to scientific notation. $$ 5\times7=35=3.5\times10^1; \quad 10^{3000}\cdot10^{5000}=10^{8000};\quad 3.5\times10^1\cdot10^{8000}=3.5\times10^{8001}. $$

Difficulty Key

1 Easiest
2 – 4 Intermediate
5 Most Difficult

Why scientific notation practice problems Matter on the ACT

Scientific notation problems are frequently included in the ACT Math test because they measure your ability to work efficiently with very large and very small numbers. These questions are designed to test speed, accuracy, and understanding of exponent rules.

Students who master this topic can solve problems faster without converting numbers into standard form.

Key Rules to Remember

When solving scientific notation problems:

Multiply coefficients and add exponents
Divide coefficients and subtract exponents
Always normalize the final coefficient (between 1 and 10)
Be careful with negative exponents
Watch for unit conversions and word problems

Common Mistakes Students Make

Many students lose points due to simple errors such as:

Forgetting to adjust the decimal correctly
Incorrect exponent operations
Misreading large numbers like “million” or “billion”
Skipping normalization step

ACT Exam Strategy

On the ACT, scientific notation problems are usually designed to look complex but follow simple rules. The fastest approach is to:

Identify the powers of 10 first
Ignore full decimal expansion
Combine exponents directly
Simplify at the end only

Extra Practice Tip

Try timing yourself while solving scientific notation problems. Most ACT questions of this type should take less than 45 seconds.