Did you take the ALEKS 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.

## ALEKS Math Practice Test Answers and Explanations

**1- The answer is 38**\($7×8=$56\), Petrol use: \(8×1.5=12\) liters

Petrol cost: \(12×$1.5=$18\). Money earned: \($56-$18=$38\)

**2- The answer is 22.016**First, multiply the tenths place of 4.3 by 5.12. The result is 1.536. Next, multiply 4 by 5.12 which results in 20.48. The sum of these two numbers is: 1.536 + 20.48 = 22.016

**3- The answer is 340**Dividing 85 by \(25\%\), which is equivalent to 0.25, gives 340. Therefore, \(25\%\) of 340 is 85.

**4- The answer is** \(25\%\)

Use this formula: Percent of Change \(=\frac{New \space Value-Old \space Value}{Old \space Value} ×100\%\)

\(\frac{24,000-32,000}{32,000}×100\%=25\%\) and \(\frac{18,000-24,000}{24,000}×100\%=25\%\)

**5- The answer is 490**Let x be the original price. If the price of the sofa is decreased by \(13\%\) to $426.3, then: \(87\%\) of \(x=426.3 ⇒ 0.87x=426.3 ⇒ x=426.3÷0.87=490\)

**6- The answer is** \(15\%\)

The percent of girls playing tennis is: \(30\% × 50\% = 0.30 × 0.50 = 0.15= 15 \%\)

**7- The answer is** \(200\%\)

Write the equation and solve for B: \(0.80 A = 0.40 B\), divide both sides by 0.40, then: \(\frac{0.80}{0.40}A=B\), therefore: \(B = 2A\), and B is 2 times of A or it’s \(200\%\) of A.

**8- The answer is 6 hours and 30 minutes**

Use distance formula: Distance = Rate × time ⇒ 390 = 60 × T, divide both sides by 60⇒ \(\frac{390}{60}\) = T ⇒ T = 6.5 hours. Change hours to minutes for the decimal part. 0.5 hours = 0.5 × 60 = 30 minutes.

**9- The answer is **\(120\%\)

Use percent formula: \(part = \frac{percent}{100}×whole\) , \(84=\frac{percent}{100}×70 ⇒ 84=\frac{percent ×70}{100} ⇒\), multiply both sides by 100 ⇒ 8400 = percent × 70, divide both sides by 70.

120 = percent

**10- The answer is 6.832**To add decimal numbers, line them up and add from the right. \(1.78+3.045+2.007=6.832\)

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**11- The answer is 17**Use Pythagorean Theorem: \(a^2+b^2=c^2, 82 + 152 = c^2 ⇒ 64+225=c^2 ⇒ 289=c^2⇒c=17\)

**12- The answer is 40**The sum of supplement angles is 180. Let \(x\) be that angle. Therefore, \(x = \frac{2}{9}×180, x = 40\)

**13- The answer is 6:7**The average speed of john is: \(210÷7=30\) km

The average speed of Alice is: \(175÷5=35\) km

Write the ratio and simplify. \(30: 35 ⇒ 6: 7\)

**14- The answer is $40**\($8.5×10=$85\), Petrol use: \(10×3=30\) liters, Petrol cost: \(30×$1.5=$45\)

Money earned: \($85-$45=$40\)

**15- The answer is 9**Let \(x\) be the number. Write the equation and solve for \(x\).

\((45 – x) ÷ x = 4\)

Multiply both sides by \(x\).

\((45 – x) = 4x\), then add \(x\) both sides. \(45 = 5x\), now divide both sides by 5. \(x = 9\)

**16- The answer is 2000**If the length of the box is 40, then the width of the box is one-fourth of it, 10, and the height of the box is 5 (one second of the width). The volume of the box is: V = lwh = (40) (10) (5) = 2000

**17- The answer is 140**To find the number of possible outfit combinations, multiply the number of options for each factor: 4 × 5 × 7 = 140

**18- The answer is 72**The area of the trapezoid is: \(Are=\frac{1}{2} h(b_1+b_2 )=\frac{1}{2}(x)(24+16)=300→20x=300→x=15\)⇒ \(z=\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17\)

The perimeter of the trapezoid is: \(16+8+17+16+15=72\)

**19- The answer is** \(\frac{8}{25}\)

There are 25 integers from 5 to 30. Set of numbers that are not composite between 5 and 30 is: \({ 5,7,11,13,17,19,23,29}\)⇒ 8 integers are not composite. Probability of not selecting a composite number is:

\(Probability= \frac{number \space of \space desired \space outcomes}{number \space of \space total \space outcomes}= \frac{8}{25}\)

**20- The answer is 26 miles**Use the information provided in the question to draw the shape.

Use Pythagorean Theorem:

\(a^2+ b^2=c^2 , 24^2+ 10^2=c^2

⇒ 576+100= c^2 ⇒ 676=c^2 ⇒ c=26\)

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**21- The answers are 11 and 13.5**You can find the possible values of and b in \((ax+3)(bx-4)\) by using the given equation \(a+b=5\) and finding another equation that relates the variables a and b. Since \((ax+3)(bx-4)=-6x^2+2cx+10\), expand the left side of the equation to obtain

\(abx^2-4ax+3bx-12=-6x^2+2cx+10\)

Since ab is the coefficient of \(x^2\) on the left side of the equation and 10 is the coefficient of \(x^2\) on the right side of the equation, it must be true that \(ab=-6\)

The coefficient of \(x\) on the left side is \(3b-4a\) and the coefficient of \(x\) in the right side is \(2c\). Then: \(\frac{3b-4a}{2}=c, a+b=5\), then: \(a=5-b\)

Now, plug in the value of \(a\) in the equation \(ab=-6\). Then: \(ab=-6→(5-b)b=-6→5b-b^2=-6\)⇒ Add \(-5b+b^2\) both sides. Then: \(b^2-5b-6=0\)

Solve for \(b\) using the factoring method. \(b^2-5b-6=0→(b-6)(b+1)=0\)

Thus, either \(b=6\) and \(a =-1\), or \(b =-1\) and \(a =6\). If \(b =6\) and \(a =-1\), then

\(c=\frac{3b-4a}{2}=\frac{3(6)-4(-1)}{2}=11\). If \(b=-1\) and \(a =6\), then, \(c=\frac{3b-4a }{2}=\frac{3(-1)-4(6)}{2}=\frac{27}{2}=13.5→c=13.5\). Therefore, the two possible values for \(c\) are 11 and 13.5.

**22- The answer is **\(15x+2\)If \(f(x)=7x+2(1-x)\), then find \(f(3x)\) by substituting \(3x\) for every \(x\) in the function. This gives: \(f(3x)=7(3x)+2(1-(3x))\) ⇒ It simplifies to: \(f(3x)=7(3x)+2(1-(3x))=21x+2-6x=15x+2\)

**23- The answer is 17**The input value is 5. Then: \(x=-4⇒ f(x)=2x^2+4x+1→f(-4)=2(-4)^2+4(-4)+1=17\)

**24- The answer is **\(cosA=\frac{3}{5}\)

To solve for \(cosA\) first identify what is known.

The question states that ∆ABC is a right triangle whose \(n∠B=90\circ\) and \(sinC=\frac{3}{5}\).

It is important to recall that any triangle has a sum of interior angles that equals 180 degrees. Therefore, to calculate \(cosA\) use the complementary angles identify of a trigonometric function. \(cosA=cos(90-C)\), Then: \(cosA=sinC\)

For complementary angles, the sin of one angle is equal to the cos of the other angle. \(cosA=\frac{3}{5}\)

**25- The answer is **\(y=x^2-4x+1\)

To figure out what the equation of the graph is, first find the vertex. From the graph, we can determine that the vertex is at \((2,-3)\).

We can use vertex form to solve for the equation of this graph.

Recall vertex form, \(y=a(x-h)^2+k\), where h is the x coordinate of the vertex, and k is the y coordinate of the vertex. Plugging in our values, you get \(y=a(x-2)^2-3\)

To solve for a, we need to pick a point on the graph and plug it into the equation.

Let’s pick \((4,1)\), \(1=a(4-2)^2-3\), \(1=a(2)^2-3\), \(1=4a-3\), \(a=1\)⇒ Now the equation is : \(y=(x-2)^2-3\), Let’s expand this, \(y=(x^2-4x+4)-3\), \(y=x^2-4x+1\)

**26- The answer is** \(-\frac{2}{3}\)

Multiplying each side of \(\frac{7}{x}=\frac{21}{x+4} by x(x+4)\) gives \(7(x+4)=21x\), divide two side by 7.

\(x+4=3x\) or \(x=2\). Therefore, the value of \(-\frac{x}{3}=-\frac{2}{3}\).

**27- The answer is 12**It is given that \(g(6)=8\). Therefore, to find the value of \(f(g(6))\), then \(f(g(6))=f(8)=12\)

**28- The answer is** \(49\sqrt{3}\)

The area of the triangle is: \(\frac{1}{2}\) AD × BC and AD is perpendicular to BC. Triangle ADC is a \(30^\circ-60^\circ- 90^\circ\) right triangle. The relationship among all sides of the right triangle \(30^\circ-60^\circ- 90^\circ\) is provided in the following triangle: In this triangle, the opposite side of the 30° angle is half of the hypotenuse. And the opposite side of \(60^\circ\) is opposite of \(30^circ × \sqrt{3}\)

CD = 7, then AD = \(7 × \sqrt{3}\)

Area of the triangle ABC is: \(\frac{1}{2}\) AD×BC = \(\frac{1}{2}\) \(7\sqrt{3}×14=49\sqrt{3}\)

**29- The answer is 40**It is given that \(g(4)=10\). Therefore, to find the value of \(f(g(4))\), substitute 10 for \(g(4)\).

\(f(g(4))=f(10)=40\).

**30- The answer is** \(\sqrt{17}\)

The equation of a circle with center (h, k) and radius r is \((x-h)^2+(y-k)^2=r^2\). To put the equation \(x^2+y^2-6x+4y=4\) in this form, complete the square as follows:

\(x^2+y^2-6x+4y=4, (x^2-6x)+(y^2+4y)=4\)

\((x^2-6x+9)-9+(y^2+4y+4)-4=4, (x-6)^2+(y+4)^2=17\)

Therefore, the radius of the circle is \(\sqrt{17}\)