Free Full Length DAT Quantitative Reasoning Practice Test

The correct answer is \(180^\circ\)
The sum of all angles in a quadrilateral is \(360\) degrees. 
Let \(x\) be the smallest angle in the quadrilateral.
Then the angles are:
\(x, 2 \ x, 3 \ x, 6 \ x\)
\(x \ + \ 2 \ x \ + \ 3 \ x \ + \ 6 \ x=360→12 \ x=360→x=30\)
The angles in the quadrilateral are: \(30^\circ, 60^\circ, 90^\circ\), and \(180^\circ\)

The correct answer is \($1,458\)
Use simple interest formula:
\(I=prt\)
(\(I =\) interest, \(p =\) principal, \(r =\) rate, \(t =\) time)
\(I=(18000) \ (0.027) \ (3)=1458\)

The correct answer is \(\frac{6 \ x \ + \ 3}{2 \ x^2 \ +  \ 2 \  x}\)
\((\frac{f}{g})(x) = \frac{f(x)}{g(x)}=\frac{6 \ x \ + \ 3}{2 \ x^2 \ + \ 2\   x}\)

The correct answer is \(\frac{6 \ \sqrt{π}}{π}\)
Formula for the area of a circle is:
A \(=π \ r^2\)
Using \(36\) for the area of the circle we have:
\(36=π \ r^2\)
Let’s solve for the radius \((r)\).
\(\frac{36}{π}=r^2→r=\sqrt{\frac{36}{π}}=\frac{6}{\sqrt{π}}=\frac{6}{\sqrt{π}} \ × \ \frac{\sqrt{π}}{\sqrt{π}}=\frac{6\ \sqrt{π}}{π}\)

The correct answer is \(12,000\)
Number of visiting fans: \(\frac{4 \ × \ 30000}{10}=12,000\)

The correct answer is \(y=8 \ x \ − \ 22\)
The equation of a line is: 
\(y=m \ x \ + \ b\), where m is the slope and is the \(y-\)intercept.
First find the slope:
\(m=\frac{y_{2} \ - \ y_{1}}{x_{2} \ - \ x_{1}}=\frac{18 \ -\ (- \ 6)}{5 \ - \ 2}=\frac{24}{3}=8\)
Then, we have: \(y=8 \ x \ + \ b\)
Choose one point and plug in the values of \(x\) and \(y\) in the equation to solve for \(b\).
Let’s choose the point \((2, - \ 6)\)
\(y=8 \ x \ + \ b→- \ 6=8 \ (2) \ + \ b→- \ 6=16 \ + \ b→b=- \ 22\)
The equation of the line is: \(y=8 \ x \ - \ 22\)

The correct answer is \(120\) degree
The angle \(x\) and \(35\) are complementary angles.
Therefore:
\(x \ + \ 60=180\)
\(180^\circ \ - \ 60^\circ=120^\circ\)

The correct answer is \(\frac{\sqrt{20}}{4}\)
sin \(A=\frac{6}{4}⇒\) 
Since sin \(θ=\frac{opposite}{hypotenuse}\), we have the following right triangle. Then:
\(c=\sqrt{6^2 \ - \ 4^2 }=\sqrt{36 \ - \ 16}=\sqrt{20}\)
cos \(=\frac{\sqrt{20}}{4}\)

The correct answer is \(90\)
Length of the rectangle is: \(\frac{3}{2} \ × \ 18=27\)
perimeter of rectangle is: \(2 \ × \ (27 \ + \ 18)=90\)

The correct answer is \(40\)
First, find the number.
Let \(x\) be the number.
Write the equation and solve for \( x\). 
\(120\%\) of a number is \(60\), then:
\(1.2 \ × \ x=60 ⇒ x=60 \ ÷ \ 1.2=50\)
\(80\%\) of \( 50\) is:
\(0.9 \ × \ 50=40\)

The correct answer is \(96,000\)
Three times of \(32,000\) is \(128,000\).
One sixth of them cancelled their tickets.
One sixth of \(128,000\) equals \(32,000 \ (\frac{1}{4} \ × \ 128000=32000)\). 
\(96,000 \ (128000 \ – \ 32000=96000)\) fans are attending this week

The correct answer is \(512\) cm\(^3\)
If the length of the box is \(32\), then the width of the box is one Fourth of it, \(8\), and the height of the box is \(2\) (one Fourth of the width). 
The volume of the box is:
V \(=\) (length) \(×\) (wdth) \(×\) (height) \(=(32) \ × \ (8) \ × \ (2)=512\) cm\(^3\)

The correct answer is \(\frac{5}{13}\)
tan⁡\(=\frac{opposite}{adjacet}\), and tan⁡\(x=\frac{5}{12}\), therefore, the opposite side of the angle \(x\) is \(5\) and the adjacent side is \(12\).
Let’s draw the triangle.
Using Pythagorean theorem, we have:
\(a^2 \ + \ b^2=c^2→5^2 \ + \ 12^2=c^2→25 \ + \ 144=c^2→c=13\)
sin \(x⁡=\frac{opposite}{hypotenuse}=\frac{5}{13}\)

The correct answer is \(8.6 \ × \ 10^5\)
\(860000=8.6 \ × \ 10^5\)

The correct answer is \(1\)
Plug in the value of each option in the inequality.
A. \(1 \ \ \ (1 \ - \ 3)^2 \ + \ 2 \ > \ 4 \ (1) \ - \ 5→6 \ > - \ 1 \)    Bingo!
B. \(6 \ \ \ (6 \ - \ 3)^2 \ + \ 2 \ > \ 4 \ (6) \ - \ 5→11 \ > \ 19\) No!
C. \(8 \ \ \ (8 \ - \ 3)^2 \ + \ 2 \ > \ 4 \ (8) \ - \ 5→27 \ > \ 27\) No!
D. \(3 \ \ \ (3 \ - \ 3)^2 \ + \ 2 \ > \ 4 \ (3) \ - \ 5→2 \ > \ 7\)    No!
E. \(4 \ \ \ (4 \ - \ 3)^2 \ + \ 3 \ > \ 4 \ (4) \ - \ 5→4 \ > \ 11\)   No!

The correct answer is \(x \ ≥ \ 3 \ ∪ \ x \ ≤ − \ 11\)
\(x \ + \ 4 \ ≥ \ 7→x \ ≥ \ 7 \ - \ 4→x \ ≥ \ 3\)
Or
\(x \ + \ 4 \ ≤ \ - \ 7→x \ ≤ \ - \ 7 \ - \ 4→x \ ≤ \ - \ 11\)
Then, solution is: \(x \ ≥ \ 3 \ ∪ \ x \ ≤ \ − \ 11\)

The correct answer is \(5 \ \sqrt{5}\)
Based on triangle similarity theorem: \(\frac{a}{a \ + \ b}=\frac{c}{4}→c=\frac{4 \ a}{a \ + \ b}=\frac{4 \ \sqrt{5}}{4 \ \sqrt{5}}=1→\)
area of shaded region is: \((\frac{c \ + \ 4}{2}) \ (b)=\frac{5}{2} \ × \ 2 \ \sqrt{5}=5 \ \sqrt{5}\)

The correct answer is \(82\%\)
the population is increased by \(30\%\) and \(40\%\). 
\(15\%\) increase changes the population to \(130\%\) of original population. 
For the second increase, multiply the result by \(140\%\).
\((1.30) \ × \ (1.40)=1.82=182\%\)
\(82\) percent of the population is increased after two years.

The correct answer is \(1 \ - \ \sqrt{5} \)
\(x_{1,2} = \frac{- \ b \pm \sqrt{b^2 \ - \ 4 \ a \ c}}{2 \ a }\)
\(a \ x^2 \ + \ b \ x \ + \ c=0\)
\( 2 \ x^2 \ -  \ 4 \ x \ – \ 8=0 ⇒\) then: \(a=1, \ b=- \ 2 \) and \(c=  – \ 4\) 
\(x =\frac{ 2 \ + \ \sqrt{- \ 2 ^2  \ - \ (4) .(1) .(- \ 4)} }{2 .1}=1 \ - \ \sqrt{5 } \)
\(x =\frac{2 \ - \ \sqrt{- \ 2^2  \ - \ (4) .(1) .(- \ 4)} }{2 .1}= 1 \ + \ \sqrt{5 }\)

The correct answer is \(x^{\frac{12}{9}}\)
\((x^2)^{\frac{6}{9}} = x^{2 \ × \ \frac{6}{9}} = x^{ \frac{12}{6}} =\)

The correct answer is \($3\)
Let \(x\) be the cost of one-kilogram orange, then:
\(4\ x \ + \ (3 \ × \ 6)=30→\)
\(4 \ x \ + \ 18= 30→\)
\(4 \ x=30 \ - \ 18→\)
\(4 \ x=12→x=\frac{12}{4}=$3\)

The correct answer is \(40\)
Let \(x\) be the length of AB, then:
\(30=\frac{x \ × \ 2}{2}→x=15\)
The length of AC \(=\sqrt{15^2 \ + \ 8^2}=\sqrt{289}=17\)
The perimeter of \(\triangle\)ABC \(=15 \ + \ 8 \ + \ 17=40\)

The correct answer is \(\frac{2 \ x}{5}\)
Simplify the expression.
\(\sqrt{\frac{x^2}{5} \ - \ \frac{x^2}{25}}=\sqrt{\frac{5 \ x^2}{25} \ - \ \frac{x^2}{25}}=\sqrt{\frac{4 \ x^2}{25}}=\sqrt{\frac{4}{25} \ x^2}=\)
\(\sqrt{\frac{4}{25}} \ × \ \sqrt{x^2}=\frac{2}{5} \ × \ x=\frac{2 \ x}{5}\)

The correct answer is \(\frac{y}{25}\)
Solve for \(x\).
\(5 \ \sqrt{ x}=\sqrt{y}\)
Square both sides of the equation:
\(( 5\ \sqrt{ x})^2=(\sqrt{y})^2\)
\(25  \ x=y\)
\(x=\frac{y}{25}\)

The correct answer is \(\frac{2}{9}\)
The probability of choosing a Hearts is \(\frac{16}{72} = \frac{2}{9}\)

The correct answer is \(57.81\)
average \(= \frac{sum \ of \ terms }{number \ of \ terms}\)
The sum of the weight of all girls is: \(23 \ × \ 50=1150\) kg
The sum of the weight of all boys is: \(25 \ × \ 65=1625\) kg
The sum of the weight of all students is: \(1150 \ + \ 1625=2775\) kg
average \(= \frac{2775}{48}=57.81\)

The correct answer is \(135 \ x \ + \ 18,000 \ ≤ \ 35,000\)
Let \(x\) be the number of shoes the team can purchase. Therefore, the team can purchase \(120 \ x\).
The team had \($35,000\) and spent \($18000\).
Now the team can spend on new shoes \($17000\) at most. 
Now, write the inequality:
\(135 \ x \ + \ 18,000 \ ≤ \ 35,000\)

The correct answer is \(- \ 9\)
Plug in the value of \(x\) and, 
\(x=4\) and \(y= 2\)
\(4 \ (x \ + \ 3 \ y) \ - \ (3 \ + \ x)^2=\)
\(4 \ (4 \ + \ 3 \ ( 2)) \ - \ (3 \ + \ 4)^2=\)
\(4 \ (4 \ + \ 6) \ - \ (7)^2 = 40 \ - \ 49=- \ 9\)

The correct answer is \(25\)
Let \(x\) be the smallest number.
Then, these are the numbers:
\(x, x \ + \ 1, x \ + \ 2, x \ + \ 3, x \ + \ 4\)
average \(= \frac{sum \ of \ terms}{number \ of \ terms} ⇒\)
\(27= \frac{x \ + \ (x \ + \ 1) \ + \ (x \ + \ 2) \ + \ (x \ + \ 3) \ + \ (x \ + \ 4)}{5}⇒\)
\(27=\frac{5 \ x \ + \ 10}{5} ⇒\)
\(135 = 5 \ x \ + \ 10 ⇒\)
\(125 = 5 \ x ⇒ x=25\)

The correct answer is \(16 \ x^6\)
\(y=(- \ 4 \ x^3)^2=(- \ 4)^2 \ (x^3)^2=16 \ x^6\)

The correct answer is \(8\) cm
Formula for the Surface area of a cylinder is:
SA \(=2 \ π \ r^2 \ + \ 2 \ π \ h→192 \ π=2 \ π \ r^2 \ + \ 2 \ π \ r \ (4)→r^2 \ + \ 4 \ r \ - \ 96=0 \)
Factorize and solve for \(r\).
\((r \ + \ 12) \ (r \ - \ 8)=0→r=8\) or \(= - \ 12\) (unacceptable)

The correct answer is \(27\%\)
The question is this: \(616.85\) is what percent of \(845\)?
Use percent formula:
part \(= \frac{percent}{100} \ ×\) whole 
\(616.85 = \frac{percent}{100} \ × \ 845 ⇒\)
\(616.85=\frac{percent \ × \ 845}{100} ⇒\)
\(61685=\) percent \(× \ 845 ⇒\) 
percent \(= \frac{61685}{845} =73 \)
\(616.85\) is \(73\%\) of \(845\).
Therefore, the discount is: \(100\% \ – \ 73\%=27\%\)

The correct answer is \(30\)
The area of rectangle is: \(6 \ × \ 3=18\) cm\(^2\)
The area of circle is: \(π \ r^2=π \ × \ (\frac{8}{2})^2=3 \ × \ 16=48\) cm\(^2\)
Difference of areas is: \(48 \ - \ 18=30\)

The correct answer is \(130\) miles
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem: \(a^2 \ + \ b^2=c^2\)
\(50^2 \ + \ 120^2=c^2⇒\)
\(2500 \ + \ 14400 = c^2⇒\)
\(16900 = c^2⇒ c = 130\)

The correct answer is \(− \ \frac{1}{3}\)
The equation of a line in slope intercept form is: \(y=m \ x \ + \ b\)
Solve for \(y\).
\(9 \ x \ - \ 3 \ y=24 ⇒\)
\(- \ 3 \ y=24 \ - \ 9 \ x ⇒\)
\(y=(24 \ - \ 9 \ x) \ ÷ \ (- \ 3) ⇒\)
\(y=3\ x \ - \ 8\)
The slope is \(3\).
The slope of the line perpendicular to this line is:
\(m_{1} \ × \ m_{2} = - \ 1 ⇒\)
\(3 \ × \ m_{2} = - \ 1 ⇒\)
\(m_{2} = - \ \frac{1}{3}\)

The correct answer is \(\frac{ x^3}{54} \ - \ 5\)
\(f(g(x))=4 \ × \ (\frac{x}{6})^3 \ - \ 5=\frac{4 \ x^3}{216} \ - \ 5 = \frac{x^3 }{54} \  - \ 5\)

The correct answer is \(125\)
The ratio of boy to girls is \(1:5\).
Therefore, there are \(1\) boys out of \(6\) students.
To find the answer, first divide the total number of students by \(6\), then multiply the result by \(1\). 
\(750 \ ÷ \ 6=125 ⇒ 125 \ × \ 1=125\)

The correct answer is \(\frac{1}{14}\)
Write the ratio of \(7\ a\) to \(4 \ b\).
\(\frac{7 \ a}{4 \ b}=\frac{1}{8}\)
Use cross multiplication and then simplify.
\(7 \ a \ × \ 8=4 \ b \ × \ 1→56 \ a=4 \ b→a=\frac{4 \ b}{56}=\frac{b}{14}\)
Now, find the ratio of \(a\) to \(b\).
\(\frac{a}{b}=\frac{\frac{b}{14}}{b}→\frac{b}{14} \ ÷ \ b=\frac{b}{14} \ × \ \frac{1}{b}=\frac{b}{14 \ b}=\frac{1}{14}\)

The correct answer is \(\frac{\sqrt{3}}{2}\)
The relationship among all sides of right triangle \(30^\circ \ - \ 60^\circ \ - \ 90^\circ\) is provided in the following triangle: 
Sine of \(60^\circ\) equals to:
\(\frac{opposite}{hypotenuse}=\frac{x\sqrt{3}}{2 \ x}=\frac{\sqrt{3}}{2}\)

The correct answer is \(31\)
Plug in the value of \(x\) in the equation and solve for \(y\).
\(4 \ y=\frac{4 \ x^2}{2} \ - \ 4→\)
\(4 \ y = \frac{ 4 \ (8)^2}{2} \ - \ 4→\)
\(4 \ y= \frac{4 \ (64)}{2} \ - \ 4→\)
\(4 \ y= 128 \ - \ 4=124\)
\(4 \ y = 124→y=31\)