Free Online ATI TEAS Test Practice with Answers 10

7. 93 m is equal to _______ km.

A. 0.093
B. 9.3
C. 93,000
D. 9,300

Below is a chart of the metric prefixes most commonly used from smallest (milli) to largest (kilo).

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The Rules:
For every step you move to the right, you divide your number by that power of ten.
For every step you move to the left, you multiply your number by that power of ten.

Correct Answer: A

Answer Explanation:

Step 1: Determine how you are moving.
We are moving three steps to the right, so we will divide
by 1,000.
Step 2: Set up your problem and solve.
All we do here is set up the problem we created from step
one and solve it.
93 m ÷ 1,000 km/m = 0.093 km

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8. Solve ​\( 3\frac{5}{8} \)​ × ​\( 2\frac{2}{5} \)

A. ​\( 6\frac{1}{4} \)
B. ​\( \frac{10}{87} \)
C. ​\( 8\frac{7}{10} \)
D. ​\( 7\frac{9}{16} \)

Correct Answer: C

Answer Explanation:

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9. On a recent road trip, Orlando drove an unknown number of miles during his first week. He then drove fifteen more miles in his second week and twenty-five fewer miles in his third week when compared to the first week. Write an expression that represents the total number of miles that Orlando drove in the
3-week road trip.

A. 3x – 40
B. 3x + 10
C. 3x + 40
D. 3x – 10

Correct Answer: D

Answer Explanation:

Step 1: Translate given information into algebra
expression.
Number of miles driven first week → x
We are also given that Orlando drove fifteen more miles in
his second week in comparison to the number he drove
the first week.
Number of miles driven second week → x + 15
Finally, we are told that he drove twenty-five fewer miles
in his third week when compared to the first week.
Number of miles driven third week → x – 25
Step 2: Determine the total.
Now that we have all individual expressions, we can find
the final sum.
We are asked to write an expression that represents the
total number of miles that Orlando drove in the 3-week
road trip. To find the total, we need to find the expression
that adds all expressions and simplify.
Step 3: Create equivalent fractions using the LCM as
the new denominator.
x + x + 15 + x – 25
x + x + 15 + x – 25
x + x + x + 15 – 25
3x – 10

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