Full Length STAAR Grade 8 Practice Test

The correct answer is \(7.32\)
The weight of \(12.2\) meters of this rope is:
\(12.2 \ × \ 600\) g \(= 7320\) g
\(1\) kg \(= 1000\) g, therefore, \(7320\) g \(÷ 1000 = 7.32\) kg

The correct answer is \(600\)
The ratio of boys to girls is \(3:7\).
Therefore, there are \(3\) boys out of \(10\) students.
To find the answer, first divide the number of boys by \(3\), then multiply the result by \(10\).
\(180 \ ÷ \ 3 = 60 ⇒\)
\(60 \ × \ 10 = 600\)

The correct answer is \(38\)
the population is increased by \(15\%\) and \(20\%\).
\(15\%\) increase changes the population to \(115\%\) of original population.
For the second increase, multiply the result by \(120\%\).
\((1.15) \ × \ (1.20) = 1.38 = 138\%\)
\(38\) percent of the population is increased after two years.

A linear equation is a relationship between two variables, \(x\) and \(y\), that can be put in the form \(y = m \ x \ + \ b\).
A non-proportional linear relationship takes on the form \(y = m \ x \ + \ b\), where \(b ≠ 0\) and its graph is a line that does not cross through the origin.

The correct answer is \(5, \ 10\)
The perimeter of the rectangle is:
\(2 \ x \ + \ 2 \ y=30→\)
\(x \ + \ y=15→\)
\(x=15 \ - \ y\)
The area of the rectangle is:
\(x \ × \ y=50→\)
\((15 \ - \ y) \ (y)=50→\)
\(y^2 \ - \ 15 \ y \ + \ 50=0\)
Solve the quadratic equation by factoring method.
\((y \ - \ 5) \ (y \ - \ 10)=0→\)
\(y=5\)
(Unacceptable, because \(y\) must be greater than \(5\)) or \(y=10\)
If \(y=10 →\)
\(x \ × \ y=50→\)
\(x \ × \ 10=50→\)
\(x=5\)

The correct answer is \(120 \ x\ + \ 22.000 \ ≤ \ 40.000\)
Let \(x\) be the number of new shoes the team can purchase. Therefore, the team can purchase \(240 \ x\).
The team had \($40,000\) and spent \($22,000\).
Now the team can spend on new shoes \($18,000\) at most.
Now, write the inequality: \(120 \ x\ + \ 22.000 \ ≤ \ 40.000\)

The correct answer is \(10\) cm
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem: \(a^2 \ + \ b^2 = c^2\)
\(6^2 \ + \ 8^2 = c^2 ⇒\)
\(100 = c^2 ⇒ c = 10\)

The correct answer is \(25.5\)
\(3 \ x \ - \ 5=8.5→\)
\(3 \ x=8.5 \ + \ 5=13.5→\)
\(x=\frac{13.5}{3}=4.5 \)
Then; \(5 \ x \ + \ 3=5 \ (4.5) \ + \ 3=22.5 \ + \ 3=25.5\)

The correct answer is \($1800\)
Use simple interest formula:
\(I=prt\)
(\(I =\) interest, \(p =\) principal, \(r =\) rate, \(t =\) time)
\(I=(8000) \ (0.045) \ (5)=1800\)

The correct answer is \(210\)
Let \(x\) be the number of soft drinks for \(252\) guests.
Write the proportion and solve for \(x\).
\(\frac{10 \ soft \ drinks}{12 guests}= \frac{x}{252 \ guests}\)
\(x = \frac{252 \ × \ 10}{12} ⇒x=210\)

The correct answer is \(600\) ml
\(4\%\) of the volume of the solution is alcohol.
Let \(x\) be the volume of the solution.
Then: \(4\%\) of \(x = 24\) ml \(⇒ 0.04 \ x = 24 ⇒\)
\(x = 24 \ ÷ \ 0.04 = 600\)

The correct answer is \(40\) ft.\(^2\)
Use the area of rectangle formula \((s = a \ × \ b)\).
To find area of the shaded region subtract smaller rectangle from bigger rectangle.
S\(_{1} \ –\) S\(_{2} = (10\) ft \(× \ 8\) ft) \(– \ (5\) ft \(× \ 8\) ft) \(⇒\) S\(_{1} \ –\) S\(_{2} = 40\) ft.\(^2\)

The correct answer is \(30\%\)
Use the formula for Percent of Change
\(\frac{New \ Value-Old \ Value}{Old \ Value} \ × \ 100\%\)
\(\frac{28 \ - \ 40}{40} \ × \ 100\% = \ – \ 30\%\)
(negative sign here means that the new price is less than old price).

The correct answer is \($1000\)
Use simple interest formula:
\(I=prt\)
(\(I =\) interest, \(p =\) principal, \(r =\) rate, \(t =\) time)
\(I=(5000) \ (0.05) \ (4)=1000\)

The correct answer is \(8\)
Use formula of rectangle prism volume.
\(V =\) (length) (width) (height) \(⇒ 2000 = (25) \ (10)\) (height) \(⇒\)
height \(= 2000 \ ÷ \ 250 = 8\) feet

The correct answer is \(20\%\)
Use this formula: Percent of Change
\(\frac{New \ Value-Old \ Value}{Old \\ Value} \ × \ 100\%\)
\(\frac{16000 \ - \ 20000}{20000} \ × \ 100\% = 20\%\) and
\(\frac{12800 \ - \ 16000}{16000} \ × \ 100\% = 20\%\)

The correct answer is \(20 \ \pi\)
To find the area of the shaded region subtract smaller circle from bigger circle.
S \(_{bigger} \ –\) S \(_{smaller} = π \ (r _{bigger} )^2 \ – \ π \ (r _{smaller} )^2 ⇒\)
S \(_{bigger} \ –\) S \(_{smaller} = π \ (6)^2 \ – \ π \ (4)^2 ⇒\)
\(36 \ π \ – \ 16 \ π = 20 \ π\)

The correct answer is \(18\)
\(a=6⇒\) area of the triangle is \(=\frac{1}{2} \ (6 \ × \ 6)=\frac{36}{2}=18\) cm \(^2\)

The correct answer is \($70\)
\($9 \ × \ 10=$90\)
Petrol use:
\(10 \ × \ 2=20\) liters
Petrol cost:
\(20 \ × \ $1=$20\)
Money earned:
\($90 \ - \ $20=$70\)

The correct answer is \(20\)
Five years ago, Amy was three times as old as Mike.
Mike is \(10\) years now.
Therefore, \(5\) years ago Mike was \(5\) years.
Five years ago, Amy was:
A \(=3 \ × \ 5=15 \)
Now Amy is \(20\) years old:
\(15 \ + \ 5 = 20\)

The correct answer is \(22\)
\(\begin{cases}\frac{ - \ x}{2} \ + \ \frac{y}{4}=1\\ \frac{- \ 5 \ y}{6} \ + \ 2 \ x =4 \end{cases} \mapsto\) Multiply the top equation by \(4\).
Then:
\(\begin{cases}- \ 2 \ x \ + \ y=4\\ \frac{- \ 5 \ y}{6} \ + \ 2 \ x =4 \end{cases} \mapsto\) Add two equations.
\(\frac{1}{6} \ y=8→y=48 \) , plug in the value of \(y\) into the first equation \(→x=22\)

The correct answer is \(4.8\)
Two triangles \(\triangle\)BAE and \(\triangle\)BCD are similar. Then:
\(\frac{AE}{CD}=\frac{AB}{BC}→\)
\(\frac{4}{6}=\frac{x}{12}→\)
\(48 \ - \ 4 \ x=6 \ x→\)
\(10 \ x=48→\)
\(x=4.8\)

The correct answer is \(10\)
\(\frac{2}{5} \ × \ 25=\frac{50}{5}=10\)

The correct answer is \(y=x\)
The slop of line A is: \(m=\frac{y_{2} \ - \ y_{1}}{x_{2} \ - \ x_{1}}=\frac{3 \ - \ 2}{4 \ - \ 3}=1\)
Parallel lines have the same slope and only choice D \((y=x)\) has slope of \(1\).

The correct answer is \(5\)
\(x\) is directly proportional to the square of \(y\). Then:
\(x=c \ y^2\)
\(12=c \ (2)^2→\)
\(12=4 \ c→\)
\(c=\frac{12}{4}=3\)
The relationship between \(x\) and \(y\) is:
\(x=3 \ y^2\)
\(x=75\)
\(75=3 \ y^2→\)
\(y^2=\frac{75}{3}=25→y=5\)

The correct answer is \(54\)
The amount of money that jack earns for one hour:
\(\frac{$616}{44}=$14\)
\(\frac{$826 \ - \ $616}{1.5 \ × \ $14}=10\)
Number of total hours is:
\(44 \ + \ 10=54\)

The correct answer is \(a= c\)
Let’s find the mean (average), mode and median of the number of cities for each type of pollution.
Number of cities for each type of pollution: \(6, \ 3, \ 4, \ 9, \ 8\)
average (mean) \(= \frac{sum \ of \ terms}{number \ of \ terms}= \frac{6 \ + \ 3 \ + \ 4 \ + \ 9 \ + \ 8}{5}=\frac{30}{5}=6 \)
Median is the number in the middle. To find median, first list numbers in order from smallest to largest.
\(3, \ 4, \ 6, \ 8, \ 9\)
Median of the data is \(6\).
Mode is the number which appears most often in a set of numbers.
Therefore, there is no mode in the set of numbers.
Median \(=\) Mean, then, \(a= c\)

The correct answer is \(60\%, \ 40\%, \ 90\%\)
Percent of cities in the type of pollution A: \(\frac{6}{10} \ × \ 100=60\%\)
Percent of cities in the type of pollution C: \(\frac{4}{10} \ × \ 100=40\%\)
Percent of cities in the type of pollution E: \(\frac{9}{10} \ × \ 100=90\%\)

The correct answer is \(2\)
Let the number of cities should be added to type of pollutions B be \(x\). Then:
\(\frac{x \ + \ 3}{8}=0.625→\)
\(x \ +\ 3=8 \ × \ 0.625→\)
\(x \ + \ 3=5→\)
\(x=2\)

The correct answer is \(\frac{1}{2}\)
AB \(=12\) And AC \(=5\)
BC \(=\sqrt{12^2 \ + \ 5^2}=\sqrt{144\ +\ 25}=\sqrt{169}=13\)
Perimeter \(=5 \ + \ 12\ + \ 13=30 \)
Area \(=\frac{5 \ × \ 12}{2}=5 \ × \ 6=30\)
In this case, the ratio of the perimeter of the triangle to its area is: \(\frac{30}{30}=1\)
If the sides AB and AC become twice longer, then:
AB \(=24\) And AC \(=10\)
BC \(=\sqrt{24^2 \ + \ 10^2}=\sqrt{576 \ + \ 100}=\sqrt{676}=26\)
Perimeter \(=26 \ + \ 24 \ + \ 10=60\)
Area \(=\frac{10 \ × \ 24}{2}=10 \ × \ 12=120\)
In this case the ratio of the perimeter of the triangle to its area is: \(\frac{60}{120}=\frac{1}{2}\)

The correct answer is \(21\)
The capacity of a red box is \(20\%\) bigger than the capacity of a blue box and it can hold \(30\) books.
Therefore, we want to find a number that \(20\%\) bigger than that number is \(30\).
Let \(x\) be that number. Then:
\(1.20 \ × \ x=30\), Divide both sides of the equation by \(1.2\). Then:
\(x=\frac{30}{1.20}=25\)
Number of books in \(30\%\) of red box \(= \frac{30}{100} \ × \ 30=9→\)
\(30 \ - \ 9=21\)

The correct answer is \(- \ 5\)
The smallest number is \(- \ 15\).
To find the largest possible value of one of the other five integers, we need to choose the smallest possible integers for four of them. Let \(x\) be the largest number. Then:
\(- \ 70=(- \ 15) \ + \ (- \ 14) \ + \ (- \ 13) \ + \ (- \ 12) \ + \ (- \ 11) \ + \ x→\)
\(- \ 70=- \ 65 \ +\ x→\)
\(x=- \ 70 \ + \ 65=- \ 5\)

The correct answer is \(- \ 5\)
\(α=180^° \ - \ 112^°=68^°\)
\(β=180^° \ -\ 135^°=45^°\)
\(x \ + \ α \ + \ β=180^°→\)
\(x=180^° \ - \ 68^° \ - \ 45^°=67^°\)

The correct answer is \(f(x)=\sqrt{x} \ + \ 4\)
A. \(f(x)=x^2 \ - \ 5\)      if \(x=1→f(1)=(1)^2 \ - \ 5=1 \ - \ 5=- \ 4≠5\)
B. \(f(x)=x^2 \ - \ 1\)      if \(x=1→f(1)=(1)^2 \ - \ 1=1 \ - \ 1=0≠5\)
C. \(f(x)=\sqrt{x \ + \ 2}\)    if \(x=1→f(1)=\sqrt{1 \ + \ 2}=\sqrt{3}≠5\)
D. \(f(x)=\sqrt{x} \ + \ 4\)   if \(x=1→f(1)=\sqrt{1} \ + \ 4=5\)

The correct answer is \($810\)
Let \(x\) be all expenses, then \(\frac{22}{100} \ x=$660→\)
\(x=\frac{100 \ × \ $660}{22}=$3000\)
He spent for his rent: \(\frac{27}{100} \ × \ $3000=$810\)

The correct answer is \(\frac{100 \ x \ + \ 800}{x}\)
The amount of money for \(x\) bookshelf is: \(100 \ x\)
Then, the total cost of all bookshelves is equal to: \(100 \ x \ + \ 800\)
The total cost, in dollar, per bookshelf is:
\(\frac{Total \ cost}{number \ of \ items}=\frac{100 \ x \ + \ 800}{x}\)

The correct answer is \(0\)
\(\sqrt{x}=4→x=16\)
then; \(\sqrt{x} \ - \ 7=\sqrt{16} \ - \ 7=4 \ - \ 7=- \ 3\) and \(\sqrt{x \ - \ 7}=\sqrt{16 \ - \ 7}=\sqrt{9}=3\)
Then: \((\sqrt{x \ - \ 7}) \ + \ (\sqrt{x \ - \ 7})=3 \ + \ (-\ 3)=0\)

The correct answer is \(25\)
The angles on a straight line add up to \(180\) degrees. Then:
\(x \ + \ 25 \ + \ y \ + \ 2 \ x \ + \ y=180\)
Then, \(3 \ x \ + \ 2 \ y=180 \ - \ 25→\)
\(3 \ (35) \ + \ 2y=155→\)
\(2 \ y=155 \ - \ 105=50→\)
\(y=25\)

The correct answer is \(37\)
Square root of \(16\) is \(\sqrt{16}=4\ < \ 6\)
Square root of \(25\) is \(\sqrt{25}=5 \ < \ 6\)
Square root of \(37\) is \(\sqrt{37}=\sqrt{36 \ + \ 1} \ >\ \sqrt{36}=6\)
Square root of \(49\) is \(\sqrt{49}=7 \ > \ 6\)
Since, \(\sqrt{37} \ < \ \sqrt{49}\), then the answer is C.

The correct answer is \(11\)
\(|- \ 12 \ -\ 5| \ - \ |- \ 8 \ + \ 2|=|- \ 17| \ - \ |- \ 6|=17 \ - \ 6=11\)