Decimals and Scientific Notation Practice Questions — GRE General Test

Question 1Question →

A laboratory mixture is created by combining two solutions containing a specific tracer compound. Solution X has a volume of $2.5 \times 10^3$ milliliters and a tracer concentration of $4.8 \times 10^{-7}$ grams per milliliter. Solution Y has a volume of $7.5 \times 10^3$ milliliters and a tracer concentration of $1.6 \times 10^{-6}$ grams per milliliter. What is the total mass, in grams, of the tracer compound present in the combined mixture?

1.32 \times 10^{-2}

2.4 \times 10^{-3}

1.32 \times 10^{-5}

2.4 \times 10^{-2}

1.32 \times 10^{-1}

Show answer & explanation

Answer: $1.32 \times 10^{-2}$

Answer

1.32 \times 10^{-2}

grams

The mass in Solution X is

(2.5 \times 10^3) \times (4.8 \times 10^{-7}) = 1.2 \times 10^{-3}

grams, and the mass in Solution Y is

(7.5 \times 10^3) \times (1.6 \times 10^{-6}) = 1.2 \times 10^{-2}

grams. Rewriting

1.2 \times 10^{-3}

as

0.12 \times 10^{-2}

allows direct addition:

(0.12 + 1.2) \times 10^{-2} = 1.32 \times 10^{-2}

grams.

Step-by-Step Solution

1

Calculate the mass of the tracer in Solution X

M_X = (2.5 \times 10^3) \times (4.8 \times 10^{-7}) = 12.0 \times 10^{-4} = 1.2 \times 10^{-3}

grams

Mass equals volume multiplied by concentration.

2

Calculate the mass of the tracer in Solution Y

M_Y = (7.5 \times 10^3) \times (1.6 \times 10^{-6}) = 12.0 \times 10^{-3} = 1.2 \times 10^{-2}

grams

Mass equals volume multiplied by concentration.

3

Sum the tracer masses and adjust to scientific notation

M_{\text{total}} = 1.2 \times 10^{-3} + 1.2 \times 10^{-2} = 0.12 \times 10^{-2} + 1.2 \times 10^{-2} = 1.32 \times 10^{-2}

grams

To add terms in scientific notation, convert them to have matching exponents before adding coefficients.

Key Concept

Arithmetic operations with Scientific Notation and Decimals

Estimated Time:1m 30s

Question 2Question →

Consider the expression $K = \frac{0.000072 \times (1.5 \times 10^{-4})}{3.6 \times 10^{-11}}$ . Which of the following values or expressions are equivalent to $K$ ? Select all that apply.

Select all that apply

3.0 \times 10^2

0.3 \times 10^3

30,000 \times 10^{-2}

3.0 \times 10^{-2}

3.0 \times 10^4

Show answer & explanation

Answer: $3.0 \times 10^2$ ; $0.3 \times 10^3$ ; $30,000 \times 10^{-2}$

Answer

The expressions equivalent to

K

are

3.0 \times 10^2

,

0.3 \times 10^3

, and

30,000 \times 10^{-2}

.

Evaluating the expression

K

gives

300

. First, rewrite

0.000072

as

7.2 \times 10^{-5}

. Multiplying by

1.5 \times 10^{-4}

gives

(7.2 \times 1.5) \times 10^{-9} = 10.8 \times 10^{-9}

. Next, dividing by

3.6 \times 10^{-11}

yields

(10.8 / 3.6) \times 10^{-9 - (-11)} = 3.0 \times 10^2 = 300

. Testing the choices shows that

3.0 \times 10^2 = 300

,

0.3 \times 10^3 = 300

, and

30,000 \times 10^{-2} = 300

are all equal to

300

.

Step-by-Step Solution

1

Convert decimal numbers in the numerator to scientific notation.

0.000072 = 7.2 \times 10^{-5}

.

Converting all terms to powers of 10 simplifies multiplication.

2

Multiply the terms in the numerator.

(7.2 \times 10^{-5}) \times (1.5 \times 10^{-4}) = (7.2 \times 1.5) \times 10^{-5 + (-4)} = 10.8 \times 10^{-9}

.

Coefficients multiply together and exponents add during multiplication of powers with the same base.

3

Divide the numerator by the denominator.

\frac{10.8 \times 10^{-9}}{3.6 \times 10^{-11}} = \left(\frac{10.8}{3.6}\right) \times 10^{-9 - (-11)} = 3.0 \times 10^2 = 300

.

Dividing coefficients gives

3.0

, and subtracting the exponent of the denominator

-11

from

-9

yields

-9 + 11 = 2

.

4

Verify each option against the value

300

.

3.0 \times 10^2 = 300

,

0.3 \times 10^3 = 300

, and

30,000 \times 10^{-2} = 300

are all equivalent to

300

.

Matching each option's evaluated value ensures all correct representations are selected.

Key Concept

Simplifying numerical expressions involving decimals and scientific notation rules.

Question 3Question →

If $(5.0 \times 10^{-4}) \times (4.0 \times 10^{7}) = 2.0 \times 10^n$ , what is the value of $n$ ?

Show answer & explanation

Answer: 4

Answer

The value of

n

is 4.

Multiplying the coefficients yields

5.0 \times 4.0 = 20.0

, and multiplying the powers of ten yields

10^{-4} \times 10^7 = 10^3

. Combining these gives

20.0 \times 10^3

. Rewriting

20.0 \times 10^3

into standard scientific notation gives

2.0 \times 10^4

. Therefore,

n = 4

.

Step-by-Step Solution

1

Multiply the numerical coefficients

5.0 \times 4.0 = 20.0

When multiplying numbers in scientific notation, separate the coefficients from the exponential terms.

2

Add the exponents of the base 10 terms

10^{-4} \times 10^{7} = 10^{3}

By exponent rules,

10^a \times 10^b = 10^{a+b}

.

3

Adjust the product to standard scientific notation form

20.0 \times 10^3 = 2.0 \times 10^4

Shift the decimal point one place to the left to obtain a coefficient

2.0

(

1 \le a < 10

), which increases the power of 10 by 1.

4

Determine the exponent value

n

n = 4

Comparing

2.0 \times 10^4

to

2.0 \times 10^n

gives

n = 4

.

Key Concept

Scientific notation multiplication and place value rules

Question 4Question →

Given $x = 3.6 \times 10^{-3}$ and $y = 1.2 \times 10^{-4}$ , which of the following statements are true? Select all that apply.

Select all that apply

x

is greater than

y

.

The ratio

\frac{x}{y}

is equal to

30

.

The sum

x + y

is equal to

4.8 \times 10^{-7}

.

The product

x \times y

is equal to

4.32 \times 10^{12}

.

The value of

y

expressed as a standard decimal is

0.0012

.

Show answer & explanation

Answer: $x$ is greater than $y$ .; The ratio $\frac{x}{y}$ is equal to $30$ .

Answer

The statement that

x

is greater than

y

and the statement that the ratio

\frac{x}{y}

is equal to

30

are both correct.

The statement asserting that

x

is greater than

y

is true because

3.6 \times 10^{-3} = 0.0036

is larger than

1.2 \times 10^{-4} = 0.00012

. The statement regarding the ratio

\frac{x}{y}

is true because dividing coefficients

\frac{3.6}{1.2} = 3

and subtracting exponents

-3 - (-4) = 1

yields

3 \times 10^1 = 30

.

Step-by-Step Solution

1

Convert both numbers to standard decimal notation to compare them.

x = 0.0036

and

y = 0.00012

.

Converting to standard form makes place value comparison direct.

2

Compare

x

and

y

.

0.0036 > 0.00012

, so

x > y

.

The thousandths digit of

x

(

3

) is greater than the thousandths digit of

y

(

0

).

3

Calculate the ratio

\frac{x}{y}

using properties of exponents.

\frac{3.6 \times 10^{-3}}{1.2 \times 10^{-4}} = \left(\frac{3.6}{1.2}\right) \times 10^{-3 - (-4)} = 3 \times 10^1 = 30

.

Divide the coefficients and subtract the exponents for scientific notation division.

Key Concept

Decimals and Scientific Notation Operations

Question 5Question →

What is the value of $\frac{4.8 \times 10^{-3}}{1.2 \times 10^{-7}}$ expressed in scientific notation?

4.0 \times 10^4

4.0 \times 10^{-10}

4.0 \times 10^{-4}

3.6 \times 10^4

4.0 \times 10^3

Show answer & explanation

Answer: $4.0 \times 10^4$

Answer

4.0 \times 10^4

Dividing

4.8 \times 10^{-3}

by

1.2 \times 10^{-7}

requires dividing the coefficient

4.8

by

1.2

to obtain

4.0

, and subtracting the exponent of the denominator from that of the numerator:

-3 - (-7) = 4

. This yields

4.0 \times 10^4

.

Step-by-Step Solution

1

Divide the numerical coefficients.

4.8 \div 1.2 = 4.0

When dividing numbers in scientific notation, divide the coefficients independently from the powers of ten.

2

Apply exponent rules to divide the powers of ten.

10^{-3} \div 10^{-7} = 10^{-3 - (-7)} = 10^4

Subtract the exponent of the denominator from the exponent of the numerator.

3

Combine the coefficient and power of ten into scientific notation.

4.0 \times 10^4

The coefficient

4.0

is between

1

and

10

, so the expression is in proper standard scientific notation form.

Key Concept

Division of numbers in scientific notation

Question 6Question →

Let $M = 3.2 \times 10^{-5}$ and $N = 8.0 \times 10^{-4}$ . Which of the following statements must be true? Select all that apply.

Select all that apply

N - M = 7.68 \times 10^{-4}

The ratio

\frac{N}{M}

is equal to

25

\sqrt{M \cdot N} = 1.6 \times 10^{-5}

M + N = 1.12 \times 10^{-3}

\frac{1}{M} = 3.125 \times 10^4

Show answer & explanation

Answer: $N - M = 7.68 \times 10^{-4}$ ; The ratio $\frac{N}{M}$ is equal to $25$ ; $\frac{1}{M} = 3.125 \times 10^4$

Answer

The correct statements are

N - M = 7.68 \times 10^{-4}

, the ratio

\frac{N}{M}

is equal to

25

, and

\frac{1}{M} = 3.125 \times 10^4

.

The statement

N - M = 7.68 \times 10^{-4}

is true because expressing both numbers with the exponent

10^{-4}

gives

(8.0 - 0.32) \times 10^{-4} = 7.68 \times 10^{-4}

. The statement that the ratio

\frac{N}{M}

is equal to

25

is true because

\frac{8.0}{3.2} \times 10^{-4 - (-5)} = 2.5 \times 10 = 25

. The statement

\frac{1}{M} = 3.125 \times 10^4

is true because

\frac{1}{3.2} \times 10^5 = 0.3125 \times 10^5 = 3.125 \times 10^4

.

Step-by-Step Solution

1

Evaluate

N - M

N - M = 7.68 \times 10^{-4}

Convert

M

to the exponent

10^{-4}

:

M = 0.32 \times 10^{-4}

. Then

N - M = (8.0 - 0.32) \times 10^{-4} = 7.68 \times 10^{-4}

.

2

Evaluate

\frac{N}{M}

\frac{N}{M} = 25

Divide coefficients and subtract exponents:

\frac{8.0}{3.2} \times 10^{-4 - (-5)} = 2.5 \times 10^1 = 25

.

3

Evaluate

\sqrt{M \cdot N}

\sqrt{M \cdot N} = 1.6 \times 10^{-4}

Compute

M \cdot N = 25.6 \times 10^{-9} = 2.56 \times 10^{-8}

. Take the square root:

\sqrt{2.56} \times \sqrt{10^{-8}} = 1.6 \times 10^{-4}

.

4

Evaluate

M + N

M + N = 8.32 \times 10^{-4}

Express with common powers of ten:

0.32 \times 10^{-4} + 8.0 \times 10^{-4} = 8.32 \times 10^{-4}

.

5

Evaluate

\frac{1}{M}

\frac{1}{M} = 3.125 \times 10^4

Compute

\frac{1}{3.2 \times 10^{-5}} = 0.3125 \times 10^5 = 3.125 \times 10^4

.

Key Concept

Operations with numbers in scientific notation and place value adjustments

Question 7Question →

A supercomputer processor operates at a standard speed of $1.25 \times 10^9$ calculations per second. During a maintenance diagnostic, the processor's operational speed is reduced by $96\%$ . Operating exclusively at this reduced speed, how many seconds will it take the processor to complete a workload of $1.8 \times 10^{13}$ calculations?

3.6 \times 10^4

1.5 \times 10^4

3.6 \times 10^5

1.44 \times 10^5

3.6 \times 10^6

Show answer & explanation

Answer: $3.6 \times 10^5$

Answer

3.6 \times 10^5

seconds

The correct answer is derived by first finding the reduced processing speed:

4\%

of

1.25 \times 10^9

equals

0.04 \times 1.25 \times 10^9 = 5.0 \times 10^7

calculations per second. Dividing the target workload of

1.8 \times 10^{13}

by

5.0 \times 10^7

yields

\frac{1.8}{5.0} \times 10^{13-7} = 0.36 \times 10^6 = 3.6 \times 10^5

seconds.

Step-by-Step Solution

1

Determine the reduced operational speed percentage.

Remaining speed percentage is

100\% - 96\% = 4\% = 0.04

.

A

96\%

reduction means the processor operates at

4\%

of its original capacity.

2

Calculate the reduced operational speed in scientific notation.

Reduced speed

= 0.04 \times (1.25 \times 10^9) = 0.05 \times 10^9 = 5.0 \times 10^7

calculations per second.

Multiplying the decimal coefficient

0.04

by

1.25

gives

0.05

, which adjusts to

5.0 \times 10^7

in scientific notation.

3

Divide total workload by reduced speed to compute time in seconds.

Time

= \frac{1.8 \times 10^{13}}{5.0 \times 10^7} = \left(\frac{1.8}{5.0}\right) \times 10^{13 - 7} = 0.36 \times 10^6

seconds.

Workload divided by rate gives total duration. Exponents are subtracted when dividing powers of ten.

4

Convert the final result to standard scientific notation.

0.36 \times 10^6 = 3.6 \times 10^5

seconds.

Moving the decimal point one place to the right requires decreasing the exponent of 10 by 1.

Key Concept

Decimals and Scientific Notation Operations

Estimated Time:2m 0s

Question 8Question →

Let $p = 0.00036 \times 10^{-2}$ and $q = 9.0 \times 10^{-7}$ . Which of the following statements are true? Select all that apply.

Select all that apply

p + q = 4.5 \times 10^{-6}

\frac{p}{q} = 4

\sqrt{p \times q} = 1.8 \times 10^{-6}

p - q = 2.7 \times 10^{-7}

\sqrt{p^2 + q^2} = p + q

Show answer & explanation

Answer: $p + q = 4.5 \times 10^{-6}$ ; $\frac{p}{q} = 4$ ; $\sqrt{p \times q} = 1.8 \times 10^{-6}$

Answer

The correct statements are the ones asserting

p + q = 4.5 \times 10^{-6}

,

\frac{p}{q} = 4

, and

\sqrt{p \times q} = 1.8 \times 10^{-6}

.

Converting

p

to

3.6 \times 10^{-6}

and

q

to

0.9 \times 10^{-6}

shows that their sum is

4.5 \times 10^{-6}

, their ratio is

4

, and the square root of their product

\sqrt{3.24 \times 10^{-12}}

is

1.8 \times 10^{-6}

.

Step-by-Step Solution

1

Convert

p

into standard scientific notation.

p = 0.00036 \times 10^{-2} = 3.6 \times 10^{-4} \times 10^{-2} = 3.6 \times 10^{-6}

.

Expressing numbers in consistent scientific notation enables direct arithmetic operations.

2

Align

q

to the same power of ten for comparison.

q = 9.0 \times 10^{-7} = 0.9 \times 10^{-6}

.

Writing terms with matching exponents allows straightforward addition and subtraction.

3

Test each statement using the simplified values.

p + q = (3.6 + 0.9) \times 10^{-6} = 4.5 \times 10^{-6}

(True);

\frac{p}{q} = \frac{3.6 \times 10^{-6}}{0.9 \times 10^{-6}} = 4

(True);

\sqrt{p \times q} = \sqrt{3.24 \times 10^{-12}} = 1.8 \times 10^{-6}

(True);

p - q = 2.7 \times 10^{-6} \neq 2.7 \times 10^{-7}

(False);

\sqrt{p^2 + q^2} \neq p + q

(False).

Evaluating each given equation determines all correct options.

Key Concept

Decimal arithmetic and scientific notation require aligning powers of ten for addition/subtraction and proper application of radical properties.

Question 9Question →

If $k = 4.5 \times 10^{-8}$ , what is the value of $\frac{(0.000009)^2 \times 400,000}{k}$ expressed in scientific notation?

7.2 \times 10^2

7.2 \times 10^4

7.2 \times 10^{-14}

7.2 \times 10^{10}

7.2 \times 10^{-2}

Show answer & explanation

Answer: $7.2 \times 10^2$

Answer

7.2 \times 10^2

Converting

0.000009

to

9 \times 10^{-6}

and squaring yields

81 \times 10^{-12}

. Multiplying this by

400,000 = 4 \times 10^5

produces

324 \times 10^{-7}

. Dividing by

4.5 \times 10^{-8}

gives

72 \times 10^1

, which in proper scientific notation (

a \times 10^n

with

1 \le a < 10

) is

7.2 \times 10^2

.

Step-by-Step Solution

1

Convert the components of the numerator into scientific notation

0.000009 = 9 \times 10^{-6}

and

400,000 = 4 \times 10^5

Converting all numbers to scientific notation simplifies subsequent exponent operations.

2

Square the first decimal term

(9 \times 10^{-6})^2 = 9^2 \times (10^{-6})^2 = 81 \times 10^{-12}

Applying the exponent to both the coefficient and the power of 10 gives

81 \times 10^{-12}

.

3

Multiply the terms in the numerator

(81 \times 10^{-12}) \times (4 \times 10^5) = (81 \times 4) \times 10^{-12 + 5} = 324 \times 10^{-7}

Multiply coefficients and add powers of 10.

4

Divide the numerator by the denominator

k = 4.5 \times 10^{-8}

\frac{324 \times 10^{-7}}{4.5 \times 10^{-8}} = \left(\frac{324}{4.5}\right) \times 10^{-7 - (-8)} = 72 \times 10^1

Divide coefficients and subtract the denominator exponent from the numerator exponent.

5

Convert the final result to standard scientific notation format

a \times 10^n

where

1 \le a < 10

72 \times 10^1 = 7.2 \times 10^2

Shift the decimal point one place to the left and increase the exponent of 10 by 1.

Key Concept

Decimals and Scientific Notation Operations

Estimated Time:2m 0s

Question 10Question →

A high-precision instrument measures the mass of a single micro-particle $P$ as $4.8 \times 10^{-8}$ grams and a single micro-particle $Q$ as $8.0 \times 10^{-9}$ grams. A sample consists of a combination of $P$ and $Q$ particles in a ratio of $3$ particles of $P$ for every $5$ particles of $Q$ . If the total mass of the sample is $9.2 \times 10^{-5}$ grams, what is the total number of particles in the sample?

Show answer & explanation

Answer: 4000

Answer

The total number of particles in the sample is 4,000.

Aligning powers of 10 shows that one P particle weighs

4.8 \times 10^{-8}

g and one Q particle weighs

0.8 \times 10^{-8}

g. A combined unit of 3 P particles and 5 Q particles has a mass of

3(4.8 \times 10^{-8}) + 5(0.8 \times 10^{-8}) = 18.4 \times 10^{-8} = 1.84 \times 10^{-7}

g. Dividing the total sample mass of

9.2 \times 10^{-5}

g by

1.84 \times 10^{-7}

g yields 500 units. Since each unit contains 8 particles (3 + 5), the total number of particles is

500 \times 8 = 4,000

.

Step-by-Step Solution

1

Convert the mass of particle Q so that it shares the same exponent (

10^{-8}

) as particle P.

Mass of single particle Q =

0.8 \times 10^{-8}

grams.

Aligning powers of 10 is necessary before performing addition of masses.

2

Find the combined mass of a fundamental ratio group consisting of 3 particles of P and 5 particles of Q.

Mass of one ratio group =

3(4.8 \times 10^{-8}) + 5(0.8 \times 10^{-8}) = 14.4 \times 10^{-8} + 4.0 \times 10^{-8} = 18.4 \times 10^{-8} = 1.84 \times 10^{-7}

grams.

Determines the mass contributed by each set of 8 particles.

3

Divide the total mass of the sample by the mass of a single ratio group.

Number of groups =

\frac{9.2 \times 10^{-5}}{1.84 \times 10^{-7}} = \frac{9.2}{1.84} \times 10^2 = 5 \times 10^2 = 500

groups.

Determines how many full ratio sets of particles make up the sample.

4

Multiply the number of groups by the total number of particles contained in each group (

3 + 5 = 8

).

Total particles =

500 \times 8 = 4000

.

Yields the total count of individual particles in the sample.

Key Concept

Operations with scientific notation, decimal place value alignment, and weighted proportional sums.

Question 11Question →

A short-pulse laser emits a single pulse lasting $8.4 \times 10^{-9}$ seconds. A high-speed optical sensor completes one measurement cycle every $1.4 \times 10^{-11}$ seconds. How many measurement cycles does the sensor complete during the duration of a single laser pulse?

Show answer & explanation

Answer: 600

Answer

600

Dividing the pulse duration (

8.4 \times 10^{-9}

seconds) by the sensor cycle time (

1.4 \times 10^{-11}

seconds) yields

\frac{8.4}{1.4} \times 10^{-9 - (-11)} = 6 \times 10^2 = 600

complete cycles.

Step-by-Step Solution

1

Set up the division expression for the total number of cycles.

\frac{8.4 \times 10^{-9}\text{ seconds}}{1.4 \times 10^{-11}\text{ seconds}}

To find how many cycle intervals fit within the total pulse duration.

2

Divide the decimal coefficients.

\frac{8.4}{1.4} = 6

Separating the numerical coefficients from the powers of ten.

3

Apply exponent rules to divide powers of 10.

10^{-9 - (-11)} = 10^{-9 + 11} = 10^2 = 100

Dividing powers with the same base requires subtracting the denominator exponent from the numerator exponent.

4

Combine results to find total cycles.

6 \times 100 = 600

Multiplying coefficient quotient by the simplified power of ten.

Key Concept

Division of Numbers in Scientific Notation and Exponent Rules

Question 12Question →

A chemical laboratory starts with a liquid solution having a total volume of $3.2 \times 10^{-4}$ cubic meters. The solution is partitioned equally into $400$ identical micro-vials. Subsequently, heat treatment causes the volume of liquid in each micro-vial to evaporate, reducing its volume by $75\%$ . What is the final volume of liquid remaining in a single micro-vial, expressed in scientific notation?

2.0 \times 10^{-7}

cubic meters

8.0 \times 10^{-7}

cubic meters

6.0 \times 10^{-7}

cubic meters

2.0 \times 10^{-5}

cubic meters

2.0 \times 10^{-6}

cubic meters

Show answer & explanation

Answer: $2.0 \times 10^{-7}$ cubic meters

Answer

2.0 \times 10^{-7}

cubic meters

Dividing the initial total volume of

3.2 \times 10^{-4}

cubic meters by

400

(

4 \times 10^2

) gives an initial volume of

8.0 \times 10^{-7}

cubic meters per micro-vial. Reducing this volume by

75\%

means

25\%

of the liquid remains. Multiplying

8.0 \times 10^{-7}

by

0.25

yields

2.0 \times 10^{-7}

cubic meters.

Step-by-Step Solution

1

Express the number of micro-vials in scientific notation.

400 = 4 \times 10^2

Converting into scientific notation simplifies division involving powers of 10.

2

Calculate the initial volume per micro-vial before evaporation.

\frac{3.2 \times 10^{-4}}{4 \times 10^2} = \left(\frac{3.2}{4}\right) \times 10^{-4 - 2} = 0.8 \times 10^{-6} = 8.0 \times 10^{-7}

cubic meters

Equal partitioning requires dividing the total volume by the total number of containers.

3

Determine the remaining fraction of liquid after a

75\%

volume reduction.

Remaining fraction

= 100\% - 75\% = 25\% = 0.25

A reduction by

75\%

leaves

25\%

of the liquid volume in the vial.

4

Multiply the volume per micro-vial by the remaining fraction.

8.0 \times 10^{-7} \times 0.25 = 2.0 \times 10^{-7}

cubic meters

Applying the remaining fraction yields the final volume in standard scientific notation format

a \times 10^n

where

1 \le a < 10

.

Key Concept

Operations with Decimals, Place Value, and Scientific Notation

Estimated Time:2m 0s

Question 13Question →

Consider the expression $N = \frac{0.00048 \times (2.5 \times 10^7)}{1.2 \times 10^{-2}}$ . Which of the following expressions are equal to $N$ ? Select all such values.

Select all that apply

1.0 \times 10^6

(4.0 \times 10^4) \times 25

1.0 \times 10^2

\frac{5.0 \times 10^{-3}}{5.0 \times 10^{-9}}

0.00012 \times 10^9

Show answer & explanation

Answer: $1.0 \times 10^6$ ; $(4.0 \times 10^4) \times 25$ ; $\frac{5.0 \times 10^{-3}}{5.0 \times 10^{-9}}$

Answer

The expressions equal to

N

are

1.0 \times 10^6

,

(4.0 \times 10^4) \times 25

, and

\frac{5.0 \times 10^{-3}}{5.0 \times 10^{-9}}

.

Evaluating

N

simplifies to

1.0 \times 10^6

. The expression stating

1.0 \times 10^6

is an exact scientific notation match. The expression

(4.0 \times 10^4) \times 25

computes to

40,000 \times 25 = 1,000,000

. The fraction

\frac{5.0 \times 10^{-3}}{5.0 \times 10^{-9}}

evaluates to

1.0 \times 10^{-3 - (-9)} = 1.0 \times 10^6

.

Step-by-Step Solution

1

Express all numbers in the stem's numerator in standard scientific notation

0.00048 = 4.8 \times 10^{-4}

and the numerator becomes

(4.8 \times 10^{-4}) \times (2.5 \times 10^7)

Standardizing numbers into scientific notation simplifies operations involving powers of 10

2

Multiply the coefficients and combine the powers of 10 in the numerator

(4.8 \times 2.5) \times 10^{-4 + 7} = 12 \times 10^3 = 1.2 \times 10^4

Coefficients are multiplied directly while exponents are added according to index laws

3

Divide the simplified numerator by the denominator

\frac{1.2 \times 10^4}{1.2 \times 10^{-2}} = \left(\frac{1.2}{1.2}\right) \times 10^{4 - (-2)} = 1.0 \times 10^6 = 1,000,000

Subtracting a negative exponent in the denominator is equivalent to adding its positive value

4

Evaluate each given option to determine equivalence to

1.0 \times 10^6

The expressions representing

1.0 \times 10^6

are

1.0 \times 10^6

,

(4.0 \times 10^4) \times 25

, and

\frac{5.0 \times 10^{-3}}{5.0 \times 10^{-9}}

Direct comparison verifies which options match the calculated value of

N

Key Concept

Operations with Decimals and Scientific Notation

Question 14Question →

Let $A = 0.000032 \times 10^{n}$ and $B = 8.0 \times 10^{n-3}$ , where $n$ is an integer. If $\frac{A^2}{B} = 1.28 \times 10^{-4}$ , what is the value of $n$ ?

-2

1

3

4

9

Show answer & explanation

Answer: $3$

Answer

The value of

n

is

3

.

Rewriting

0.000032

as

3.2 \times 10^{-5}

allows

A

to be expressed as

3.2 \times 10^{n-5}

. Squaring

A

yields

10.24 \times 10^{2n-10}

. Dividing by

B = 8.0 \times 10^{n-3}

gives

1.28 \times 10^{n-7}

. Equating

n - 7 = -4

directly leads to

n = 3

.

Step-by-Step Solution

1

Convert

A

to standard scientific notation in terms of

n

A = 3.2 \times 10^{-5} \times 10^n = 3.2 \times 10^{n-5}

The decimal

0.000032

equals

3.2 \times 10^{-5}

because the decimal point is shifted

5

places to the right.

2

Calculate

A^2

A^2 = (3.2 \times 10^{n-5})^2 = (3.2)^2 \times 10^{2(n-5)} = 10.24 \times 10^{2n-10}

Apply the power of a product rule

(ab)^k = a^k b^k

and exponent power rule

(10^p)^q = 10^{pq}

.

3

Divide

A^2

by

B

\frac{A^2}{B} = \frac{10.24 \times 10^{2n-10}}{8.0 \times 10^{n-3}} = \left(\frac{10.24}{8.0}\right) \times 10^{(2n-10) - (n-3)} = 1.28 \times 10^{n-7}

Divide coefficients

(10.24 / 8.0 = 1.28)

and subtract powers of ten exponents

((2n - 10) - (n - 3) = n - 7)

.

4

Equate the simplified expression to the given value and solve for

n

1.28 \times 10^{n-7} = 1.28 \times 10^{-4} \implies n - 7 = -4 \implies n = 3

Since the coefficients match (

1.28

), equate the exponents of

10

to solve for

n

.

Key Concept

Scientific Notation Operations and Exponent Laws

Estimated Time:2m 0s

Question 15Question →

An astronomical digital sensor captures $4.5 \times 10^7$ raw data packets during an observation run, where each packet has a size of $1.6 \times 10^{-4}$ megabytes. A compression algorithm reduces the total volume of these combined packets by $75\%$ . What is the total volume, in megabytes, of the compressed data expressed in scientific notation?

1.8 \times 10^3

5.4 \times 10^3

7.2 \times 10^3

1.8 \times 10^4

7.2 \times 10^{-3}

Show answer & explanation

Answer: $1.8 \times 10^3$

Answer

1.8 \times 10^3

megabytes

Multiplying

4.5 \times 10^7

by

1.6 \times 10^{-4}

gives

7.2 \times 10^3

megabytes. Decreasing this total by

75\%

retains

25\%

of the volume, which equals

0.25 \times 7.2 \times 10^3 = 1.8 \times 10^3

megabytes.

Step-by-Step Solution

1

Calculate the total uncompressed size by multiplying the packet count by individual packet size.

(4.5 \times 10^7) \times (1.6 \times 10^{-4}) = (4.5 \times 1.6) \times 10^{7 + (-4)} = 7.2 \times 10^3

megabytes.

Total volume is the product of quantity and unit size.

2

Determine the remaining fraction of data after a

75\%

size reduction.

100\% - 75\% = 25\% = 0.25

A

75\%

reduction leaves

25\%

of the original data.

3

Multiply the uncompressed volume by

0.25

and write the result in standard scientific notation.

0.25 \times (7.2 \times 10^3) = 1.8 \times 10^3

megabytes.

The coefficient

1.8

meets the condition

1 \leq a < 10

for scientific notation

a \times 10^n

.

Key Concept

Decimals and Scientific Notation

Question 16Question →

An automated micro-dispenser delivers liquid reagents in precise doses. Each dose has a volume of $3.6 \times 10^{-5}$ liters. If a reservoir containing $0.0162$ liters of reagent is emptied completely by delivering these equal doses, how many doses were delivered?

Show answer & explanation

Answer: 450

Answer

450

Dividing the total volume of

1.62 \times 10^{-2}

liters by the single dose volume of

3.6 \times 10^{-5}

liters yields

\frac{1.62}{3.6} \times 10^{-2 - (-5)} = 0.45 \times 10^3 = 450

doses.

Step-by-Step Solution

1

Express the total reservoir volume in scientific notation

0.0162 = 1.62 \times 10^{-2}

liters

Converting decimals into standard scientific notation simplifies multiplication and division operations.

2

Set up the division for the number of doses

\text{Number of doses} = \frac{1.62 \times 10^{-2}}{3.6 \times 10^{-5}}

The total volume divided by the volume per single dose yields the total dose count.

3

Compute the division of coefficients and exponent terms independently

\frac{1.62}{3.6} = 0.45

and

\frac{10^{-2}}{10^{-5}} = 10^{-2 - (-5)} = 10^3

Applying the exponent quotient rule

10^a / 10^b = 10^{a-b}

yields

-2 - (-5) = 3

.

4

Convert from scientific notation to a standard integer

0.45 \times 10^3 = 450

Multiplying

0.45

by

1,000

shifts the decimal point 3 places to the right.

Key Concept

Division of numbers in scientific notation and place value manipulation

Estimated Time:1m 30s

Question 17Question →

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

-6

-2

2

4

6

Show answer & explanation

Answer: $2$

Answer

The value of

n

is

2

.

Writing

0.000375

as

3.75 \times 10^{-4}

allows the mass

M

to be rewritten as

3.75 \times 10^{-4 + n}

. Matching this with the standard scientific notation form

a \times 10^k

(where

a = 3.75

and

k = -2

) gives the equation

-4 + n = -2

. Solving for

n

yields

n = 2

.

Step-by-Step Solution

1

Convert the decimal decimal coefficient to scientific notation.

0.000375 = 3.75 \times 10^{-4}

Moving the decimal point 4 places to the right puts the leading number in the required range

1 \le a < 10

.

2

Substitute this conversion back into the expression for

M

and combine powers of 10.

M = (3.75 \times 10^{-4}) \times 10^n = 3.75 \times 10^{n - 4}

By exponent rules,

10^a \times 10^b = 10^{a+b}

.

3

Equate the exponent of 10 in the simplified expression to the given value of

k

.

n - 4 = -2 \implies n = 2

The scientific notation requires

a \times 10^k

where

k = -2

, so

n - 4

must equal

-2

.

Key Concept

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

Question 18Question →

A high-precision optical sensor measures a time interval as $T = \frac{0.00064 \times (2.5 \times 10^8)}{1.6 \times 10^{-1}}$ nanoseconds. Which of the following values are equivalent to $T$ ? Select all such values.

Select all that apply

1.0 \times 10^6

100,000

\frac{4.0 \times 10^7}{40}

(2.5 \times 10^3) \times (4.0 \times 10^2)

1.0 \times 10^{-6}

Show answer & explanation

Answer: $1.0 \times 10^6$ ; $\frac{4.0 \times 10^7}{40}$ ; $(2.5 \times 10^3) \times (4.0 \times 10^2)$

Answer

The values equivalent to

T

are

1.0 \times 10^6

,

\frac{4.0 \times 10^7}{40}

, and

(2.5 \times 10^3) \times (4.0 \times 10^2)

.

Evaluating

T

gives

\frac{(6.4 \times 10^{-4}) \times (2.5 \times 10^8)}{1.6 \times 10^{-1}} = \frac{1.6 \times 10^5}{1.6 \times 10^{-1}} = 1.0 \times 10^6

. The expression representing

1.0 \times 10^6

directly matches

T

. The quotient

\frac{4.0 \times 10^7}{40}

simplifies to

\frac{4.0 \times 10^7}{4.0 \times 10^1} = 1.0 \times 10^6

, which matches

T

. The product

(2.5 \times 10^3) \times (4.0 \times 10^2)

simplifies to

10.0 \times 10^5 = 1.0 \times 10^6

, which also matches

T

.

Step-by-Step Solution

1

Convert the decimal

0.00064

into scientific notation

0.00064 = 6.4 \times 10^{-4}

Expressing all terms in scientific notation simplifies exponent operations.

2

Simplify the numerator of the expression for

T

(6.4 \times 10^{-4}) \times (2.5 \times 10^8) = (6.4 \times 2.5) \times 10^{-4 + 8} = 16.0 \times 10^4 = 1.6 \times 10^5

Multiply coefficients directly and add exponents for product of powers with equal base.

3

Divide the numerator by the denominator

1.6 \times 10^{-1}

T = \frac{1.6 \times 10^5}{1.6 \times 10^{-1}} = \left(\frac{1.6}{1.6}\right) \times 10^{5 - (-1)} = 1.0 \times 10^6 = 1,000,000

Subtract the denominator exponent from the numerator exponent:

5 - (-1) = 6

.

4

Evaluate the choices to check equivalence to

1.0 \times 10^6

1.0 \times 10^6

,

\frac{4.0 \times 10^7}{40} = 1.0 \times 10^6

, and

(2.5 \times 10^3) \times (4.0 \times 10^2) = 10.0 \times 10^5 = 1.0 \times 10^6

are all equal to

T

.

Matching each simplified expression to the calculated value of

T

determines the correct options.

Key Concept

Operations with Decimals and Exponents in Scientific Notation

Question 19Question →

If $A = 0.00048 \times 10^{-3}$ and $B = 1.2 \times 10^{-5}$ , what is the value of $\frac{A + B}{4 \times 10^{-8}}$ expressed in scientific notation?

3.12 \times 10^2

1.5 \times 10^3

4.2 \times 10^2

3.12 \times 10^{-4}

3.12 \times 10^1

Show answer & explanation

Answer: $3.12 \times 10^2$

Answer

3.12 \times 10^2

Converting

A = 0.00048 \times 10^{-3}

to powers of 10 gives

4.8 \times 10^{-7}

, which equals

0.048 \times 10^{-5}

. Adding

B = 1.2 \times 10^{-5}

yields

(0.048 + 1.2) \times 10^{-5} = 1.248 \times 10^{-5}

. Dividing by

4 \times 10^{-8}

gives

\frac{1.248}{4} \times 10^{-5 - (-8)} = 0.312 \times 10^3 = 3.12 \times 10^2

.

Step-by-Step Solution

1

Express

A

in standard scientific notation and then match its exponent to

B

's exponent.

A = 0.00048 \times 10^{-3} = 4.8 \times 10^{-7} = 0.048 \times 10^{-5}

To perform addition between numbers in scientific notation, their exponents must be equal.

2

Add

A

and

B

.

A + B = 0.048 \times 10^{-5} + 1.2 \times 10^{-5} = 1.248 \times 10^{-5}

Combine the coefficients once the powers of 10 match.

3

Divide

A + B

by

4 \times 10^{-8}

and express the final result in scientific notation.

\frac{1.248 \times 10^{-5}}{4 \times 10^{-8}} = \left(\frac{1.248}{4}\right) \times 10^{-5 - (-8)} = 0.312 \times 10^3 = 3.12 \times 10^2

Divide the coefficients and subtract the exponent in the denominator from the exponent in the numerator.

Key Concept

Decimals, Place Value, and Operations in Scientific Notation

Question 20Question →

A manufacturing facility produces precision metal components. Each component has a mass of $8.4 \times 10^{-4}$ kilograms. If a shipment container holds a batch of these components with a total mass of $1.05$ kilograms, how many components are in the container?

Show answer & explanation

Answer: 1250

Answer

The container holds 1,250 components.

To find the number of components, divide the total batch mass by the mass per component:

\frac{1.05 \text{ kg}}{8.4 \times 10^{-4} \text{ kg}}

. Expressing

\frac{1}{10^{-4}}

as

10^4

transforms the expression into

\frac{1.05}{8.4} \times 10^4

. Dividing

1.05

by

8.4

yields

0.125

. Finally,

0.125 \times 10^4 = 1250

.

Step-by-Step Solution

1

Set up the ratio of total mass to single component mass

Number of components =

\frac{1.05}{8.4 \times 10^{-4}}

Dividing total mass by individual component mass gives the total count.

2

Apply exponent rules to move the power of 10 to the numerator

\frac{1.05}{8.4} \times 10^4

Since

\frac{1}{10^{-4}} = 10^4

, shifting the negative exponent to the numerator changes its sign.

3

Divide the decimal coefficients

\frac{1.05}{8.4} = 0.125

Simplifying

\frac{105}{840}

reduces to

\frac{1}{8} = 0.125

.

4

Evaluate the product with the place value shift

0.125 \times 10,000 = 1250

Multiplying by

10^4

shifts the decimal point 4 places to the right.

Key Concept

Division with scientific notation and decimal place value adjustment

Estimated Time:1m 30s

Decimals and Scientific Notation

Answer

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