# Full-Length 8th Grade MCAS Math Practice Test-Answers and Explanations

Did you take the 8th Grade Math Practice Test? If so, then it’s time to review your results to see where you went wrong and what areas you need to improve.

## 8th Grade Math Practice Test Answers and Explanations

**1- Choice C is correct**Jason ate \(\frac{1}{2}\) of 8 parts of his pizza. It means 4 parts out of 8 parts (\(\frac{1}{2}\) of 8 parts is \(x ⇒ x=4\)) and left 4 parts. Eva ate \(\frac{3}{4}\) of 8 parts of her pizza. It means 6 parts out of 8 parts (\(\frac{3}{4}\) of 8 parts is \(x ⇒ x=6\)) and left 2 parts.

Therefore, they ate (4 + 6) parts out of (8 + 8) parts of their pizza and left (4 + 2) parts out of (8 + 8) parts of their pizza that equals to: \(\frac{6}{16}\)

After simplification, the answer is: \(\frac{3}{8}\)

**2- The answer is** \(5 \frac{7}{10}\) miles.

Robert runs \(3 \frac{1}{10}\) miles on Saturday and \(2×(3 \frac{1}{10} )\) miles on Monday and Wednesday.

Robert wants to run a total of 18 miles this week.

Therefore: \(3 \frac{1}{10}+2×(3 \frac{1}{10})\) should be subtracted from 18:

\(18-(3\frac{1}{10}+2(3 \frac{1}{10}))=15-9 \frac{3}{10}=5 \frac{7}{10}\) miles.

**3- Choice A is correct**Let \(x\) be the integer. Then: \(2x+20=68\). Subtract 20 both sides: \(2x=48\). Divide both sides by \(2\) ⇒ \(x=24\)

**4- The answer is: –37**Use PEMDAS (order of operation):

\([3×(–21)+(5×2)]–(–25)+[(–3)×6]÷2=[-63+10]+25+[-18]÷2=-53+25-9=-37\)

**5- The answer is 768 cm.**Write the proportion and solve for missing side.

\(\frac{Smaller \space triangle \space height}{Smaller \space triangle \space base}

=\frac{Bigger \space triangle \space height}{Bigger \space triangle \space base} ⇒ \frac{100 \space cm}{160 \space cm}=\frac{100+380 \space cm}{x}⇒ x=768 \space cm\)

**6- Choice C is correct.**Write the proportion and solve. \(\frac{3 \space ft}{2 \space ft}= \frac{x}{38 \space ft} ⇒ x=57 \space ft\)

**7- Choice D is correct.**The distance that Mike runs can be found by the following equation:

\(D_M= 5.5t+7.5\). The distance Julia runs can be found by \(D_J=8t\)

Julia catches Mike if they run the same distance. Therefore:

\(8t=5.5t+7.5⇒2.5t=7.5 ⇒t= \frac{7.5}{2.5}=3\) hours

**8- Choice C is correct**x is the number of all sales profit and \(2\%\) of it is:

\(2\%×x=0.02x\). Employer’s revenue: \(0.2x+7,000\)

**9- The answer is 60.**Jason needs an \(75\%\) average to pass the exams. Therefore, the sum of 5 exams must be at least \(5×75=375\). The sum of 4 exams is: \(68+72+85+90=315\).

The minimum score Jason can earn on final test to pass is: \(375–315=60\)

**10- Choice C is correct.**We can write: \(\frac{25}{100}=\frac{8}{x}⇒\frac{8×100}{25}=x⇒x=32\)

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**11- Choice B is correct.**Let \(x\) be the amount of angle and y be the amount of its supplement. The angle and its supplement are \(180^\circ\) in total \((x+y=180^\circ)\). we have: \(x=\frac{1}{5} y\)

\(x+y=\frac{1}{5} y+y =180^\circ ⇒ y=150^\circ\) and \(x=30^\circ\)

**12- Choice D is correct**\(y=-2x+1\) ⇒if \(x=0\) therefore \(y=1\) and if \(x=1\) therefore \(y=-1\). Hence answer d is correct.

**13- Choice D is correct**\(\begin{cases}\frac{-x}{2}+\frac{y}{4}=1\\\frac{-5y}{6}+2x=4\end{cases}\)

⇒ Multiply the top equation by 4. Then,

\(\begin{cases}-2x+y=4\\\frac{-5y}{6}+2x=4\end{cases}\)

⇒Add two equations.

\(\frac{1}{6} y=8⇒y=48\) , plug in the value of y into the first equation ⇒\(x=22\)

**14- Choice C is correct**\(\begin{cases}x+4y=10\\5x+10y=20\end{cases}\)

⇒Multiply the top equation by -5 then,

\(\begin{cases}-5x-20y=-50\\5x+10y=20\end{cases}\)

⇒Add two equations

\(-10y=-30→y=3\) , plug in the value of \(y\) into the first equation

\(x+4y=10⇒x+4(3)=10⇒x+12=10\)

Subtract 12 from both sides of the equation. Then: \(x+12=10→x=-2\)

**15- Choice B is correct.**\(\frac{21+18+16+x}{4} =20⇒\frac{55+x}{4}=20⇒55+x=80⇒x=25\)

**16- Choice B is correct.**Solve for \(x\).

\(5≤3x-1<11\)⇒ (add 1 all sides) \(5+1≤3x-1+1<11+1 ⇒ 6≤3x<12\) ⇒ (divide all sides by 3) \(2≤x<4 ⇒x\) is between 2 and 4.

**17- Choice D is correct.**Distance between two points is equal: \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} =\sqrt{(13-(-2)^2+(-2-6)^2}=\sqrt{15^2+(-8)^2}=\sqrt{225+64}=\sqrt{289}=17\)

**18- Choice B is correct**Distance between two points is equal:

\(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=\sqrt{(9-4)^2+(7-(-5)^2}=\sqrt{(5)^2+(-12)^2}=\sqrt{169}=13\)

**19- Choice D is correct**The slop of line A is: \(m=\frac{y_2-y_1}{x_2-x_1}=\frac{-10-8}{4-(-8)}=-\frac{3}{2}\)

Also \((y-y_1 )=m(x-x_1 )⇒y-8=-\frac{3}{2}(x+8)⇒y=-\frac{3}{2} x-4\)

**20- Choice C is correct**The value of \(y\) in the \(x\)-intercept of a line is zero. Then:

\(y=0→10x-4(0)=5→10x=5→x=\frac{1}{2}\). Then, \(x\)-intercept of the line is \(\frac{1}{2}\)

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**21- Choice C is correct**The total amount of money Giselle made as a carpenter can be modeled by \(20x\), and the total amount of money she made as a blacksmith can be modeled by \(25y\). Since these together add up to $690, we get the following equation:

\(20x+25y=690\).

We are also given that last week, Giselle worked as a carpenter and a blacksmith for a total of 30 hours. This can be expressed as:

\(x+y=30⇒y=30-x\)

Therefor \(20x+25(30-x)=690⇒x=12\) and \(y=18\)

**22- Choice D is correct**\(\begin{cases}3x+y=8\\-5x-2y=0\end{cases}\)

Multiply the top equation by 2 then,

\(\begin{cases}6x+2y=16\\-5x-2y=0\end{cases}\)

⇒ Add two equations

\(x=16\) , plug in the value of \(y\) into the first equation

\(3x+y=8→3(16)+y=8→y=-40\)

**23- Choice D is correct**Let \(x=\) the total miles of the ride.

Therefore, \(x-1=\) the additional miles of the ride. The correct equation takes $1.25 and adds it to $1.15 times the number of additional miles, \(x-1\). Translating, this becomes: \(y\)(the total cost)\(=1.25+1.15(x-1)\), which is the same equation as \(y=1.15(x-1)+1.25\).

**24- Choice D is correct.**Write as two points in terms of: (number of people, cost in$) (15,120) and (25,200). Find the equation of the line using: m\(=\frac{y_2–y_1}{x_2–x_1}\) and \(y=mx+b\)

Equation: \(y=8x\) plug in \(x=40\), \(y=8(40)=320\). A party of 40 people will cost $320.00.

**25- Choice A is correct**\(C=250+150h\). Assuming the initial meeting counts for the 1st hour, you would plug in \(h=25\) for a total cost of $4000.00.

**26- Choice D is correct**Let the number be \(x\). Then the other number\(=x+8\). Sum of two numbers \(=30\). According to question, \(x+x+8=30 ⇒2x+8=30⇒ 2x=22⇒ x=11\). Therefore, \(x+8=11+8=19\)

**27- Choice C is correct.**\(0.0000005823=5.823 × 10^{-7}\)

**28- Choice B is correct.**\(28,000,000,000=2.8×10^{10}\)

**29- Choice B is correct.**The area of greater circle is: \(A_g=πr^2=π .(45)^2=6361.7 \space mm^2\)

The area of smaller circle is: \(A_s=πr^2=π .(33)^2=3421.2 \space mm^2\)

Then area of colored part is \(A_c=A_g-A_s=6361.7-3421.2=2940.5 \space mm^2\)

**30- Choice C is correct.**When a point is reflected over y axes, the \((x)\) coordinate of that point changes to \((-x)\), while its y coordinate remains the same.

**31- Choice C is correct.**Use \(A_{reg}=l×w⇒117=l×9\). We can solve for the length by dividing.

\(l=\frac{117}{9}=13\)

**32- Choice B is correct.**To estimate area of the shaded region, subtract area of the circle from area of the square. The area of the square formula: \(S =a^2\) and the area of circle formula: \(S= πr^2\) Therefore \(S_{square} –S _{circle} = a^2 – πr^2 ⇒

S_{square} –S _{circle}= (14)^2 – π(\frac{14}{2})^2 ⇒S_{square} –S _{circle}= 42 \space cm^2\)

**33- Choice B is correct.**The amount of money for \(x\) bookshelf is: \(100x\)

Then, the total cost of all bookshelves is equal to: \(100x+800\)

The total cost, in dollar, per bookshelf is: \(\frac{Total \space cost}{number \space of \space items}=\frac{100x+800}{x}\)

**34- Choice D is correct.**The median of the numbers is the number in the middle. First write the numbers in order from least to greatest.

12, 32, 33, 45, 48, 65, 65, 72, 82, 95

Since the number of numbers is even, we need to find the average of two numbers in the middle. \(48 + 65 = 113→\frac{13}{2} = 56.5\)

**35- Choice B is correct.**Write the numbers in order: 1, 3, 5, 9, 11, 14, 15, 18, 26. The median is the number in the middle, which is 11.

**36- Choice D is correct**To solve absolute values equations, write two equations.\(|-3x+4|=26\) can equal positive 26, or negative 26.Therefore \(-3x+4=26⇒x=\frac{22}{-3}\) and \(-3x+4=-26⇒x=\frac{-26-4}{-3}=10\)

**37- The answer is D.**Number of ways of selecting 3 consonants from 7 is 7C3

Number of ways of selecting 2 vowels from 4 is 4C2

Number of ways of selecting 3 consonants from 7 and 2 vowels from 4 is “7C3” × “4C2″= \((\frac{7×6×5}{3×2×1})×(\frac{4×3}{2×1})

=210\)

It means we can have 210 groups where each group contains total 5 letters (3 consonants and 2 vowels). Number of ways of arranging 5 letters among themselves = 5! =120 Hence, required number of ways = 210 × 120 = 25,200

**38- Choice A is correct**The area of Trapezoid is \(\frac{(a+b).h}{2}⇒A=\frac{(10+15)9}{2}=112.5\)

**39- Choice A is correct**Volume of pyramid is equal to \(\frac{1}{3} (base \space edge)^2 (height)⇒

V=\frac{1}{3} ×(18)^2×24=2592 \space cm^3\)

**40- Choice D is correct**He can go in any of the 25 buses (25 ways). Since he cannot come back in the same bus, he can return in 24 ways. Total number of ways =25 × 24 = 600

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