OAR Math Practice Test Questions
Preparing students for the OAR Math test? Try these free OAR Math Practice questions. Reviewing practice questions is the best way to brush up on your student’s Math skills. Here, we walk you through solving 10 common OAR Math practice problems covering the most important math concepts on the OAR Math test. For additional educational resources, visit the U.S. Department of Education website.
These OAR Math practice questions are designed to be similar to those found on the real OAR Math test. They will assess your students’ level of preparation and will give them a better idea of what to study on their exams. For additional educational resources, visit the U.S. Department of Education website.
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C. 14
D. 16
6- If \(40\%\) of a number is 4, what is the number?
A. 4
B. 8
C. 10
D. 12
7- The average of five numbers is 24. If a sixth number 42 is added, then, which of the following is the new average?
A. 25
B. 26
C. 27
D. 42
8- The ratio of boys and girls in a class is 4:7. If there are 44 students in the class, how many more boys should be enrolled to make the ratio 1:1?
A. 8
B. 10
C. 12
D. 14
9-What is the slope of the line: \(4x-2y=6\):___________
10- A football team had $20,000 to spend on supplies. The team spent $14,000 on new balls. New sports shoes cost $120 each. Which of the following inequalities represent the number of new shoes the team can purchase.
A. \(120x+14,000 \leq 20,000\)
B. \(20x+14,000 \geq 20,000\)
C. \(14,000x+120 \leq 20,000\)
D. \(14,000x+12,0 \geq 20,000\)
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Answers:
1- 60
Jason needs a score 75 average to pass for five exams. Therefore, the sum of 5 exams must be at lease 5 \(\times \) 75 = 375
The sum of 4 exams is:
68 + 72 + 85 + 90 = 315.
The minimum score Jason can earn on his fifth and final test to pass is:
\(375 – 315 = 60\)
2- B
\(Probability = \frac{number \space of \space desired \space outcomes}{number \space of \space total \space outcomes} = \frac{18}{12+18+18+24} = \frac{18}{72} = \frac{1}{4}\)
3- B
The area of the square is 595.36. Therefore, the side of the square is the square root of the area.
\(\sqrt{595.36}=24.4\)
Four times the size of the square is the perimeter:
\(4 {\times} 24.4 = 97.6\)
4- A
The width of the rectangle is twice its length. Let \(x\) be the length. Then, width=2\(x\)
Perimeter of the rectangle is \(2 (width + length) = 2(2x+x)=60 {\Rightarrow} 6x=60 {\Rightarrow} x=10 \)
Length of the rectangle is 10 meters.
5- D
average \(= \frac{sum \space of \space terms}{number \space of \space terms} {\Rightarrow} (average \space of \space 6 \space numbers) \space 12 = \frac{sum \space of \space terms}{6} ⇒sum \space of \space 6 \space numbers\space is \)
\(12 {\times} 6 = 72\)
\((average \space of \space 4 \space numbers) \space 10 = \frac{sum \space of \space terms}{4}{\Rightarrow} sum \space of \space 4 \space numbers \space is \space 10 {\times} 4 = 40\)
sum of 6 numbers – sum of 4 numbers = sum of 2 numbers
\(72 – 40 = 32\)
average of 2 numbers = \(\frac{32}{2} = 16 \)
6- C
Let \(x\) be the number. Write the equation and solve for \(x\).
\(40\% \space of \space x=4{\Rightarrow} 0.40 \space x=4 {\Rightarrow} x=4 {\div}0.40=10\)
7- C
First, find the sum of five numbers.
average \(=\frac{ sum \space of \space terms }{ number \space of \space terms } ⇒ 24 = \frac{ sum \space of \space 5 \space numbers }{5}\)
\( ⇒ sum \space of \space 5 \space numbers = 24 × 5 = 120\)
The sum of 5 numbers is 120. If a sixth number that is 42 is added to these numbers, then the sum of 6 numbers is 162.
120 + 42 = 162
average \(==\frac{ sum \space of \space terms }{ number \space of \space terms } = \frac{162}{6}=27\)
8- C
The ratio of boys to girls is 4:7.
Therefore, there are 4 boys out of 11 students.
To find the answer, first, divide the total number of students by 11, then multiply the result by 4.
\(44 {\div} 11 = 4 {\Rightarrow} 4 {\times} 4 = 16\)
There are 16 boys and 28 \((44 – 16)\) girls. So, 12 more boys should be enrolled to make the ratio 1:1
9- 2
Solve for y.
\(4x-2y=6 {\Rightarrow} -2y=6-4x {\Rightarrow} y=2x-3\)
The slope of the line is 2.
10- A
Let \(x\) be the number of new shoes the team can purchase. Therefore, the team can purchase 120 \(x\).
The team had $20,000 and spent $14000. Now the team can spend on new shoes $6000 at most.
Now, write the inequality:
\(120x+14,000 {\leq}20,000\)
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