Did you take the TABE 11 & 12 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.

## TABE 11 & 12 Math Practice Test Answers and Explanations

**Part 1: Mathematics Computation**

**1- Choice C is correct**\(14 × 6 = 84\)

**2- Choice C is correct**\(748 + 37 = 785\)

**3- Choice B is correct**\(1,147 + 687 = 1,834\)

**4- Choice B is correct**\(378 – 132 = 246\)

**5- Choice C is correct**\(1,983 – 362 = 1,621\)

**6- Choice C is correct**\(198 + 2,295 = 2,493\)

**7- Choice B is correct**\(7,000 ÷ 200 = 35\)

**8- Choice A is correct**\(2,357 – 1,649 = 708\)

**9- Choice B is correct**\(78.32 +15.73 = 94.05\)

**10- Choice B is correct**\(3.6+6.1=9.7\)

**11- Choice A is correct**\(32.17 – 27.65 = 4.52\)

**12- Choice C is correct**\(\cfrac{\begin{align} 5.8 \\ × 3.4 \end{align}}{23.12} \)

**13- Choice C is correct**\(17.8÷100=0.178\)

**14- Choice B is correct**\(450÷6=75\)

**15- Choice B is correct**\(9ab – 2ab=7ab\)

**16- Choice C is correct**\(\frac{3}{5}+\frac{4}{5}=\frac{7}{5}\)

**17- Choice D is correct**\(14 –(–6)=14+6=20\)

**18- Choice B is correct**\(13.5÷4.5=3\)

**19- Choice D is correct**\(148 ÷ 4 = 37\)

**20- Choice B is correct**\(2,768 ÷ 8 = 346\)

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**21- Choice B is correct**\(-3+9-4= 2\)

**22- Choice B is correct**\(\frac{5}{7}-\frac{2}{7}=\frac{3}{7}\)

**23- Choice D is correct**\(\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\)

**24- Choice A is correct**\(\frac{3}{7}×\frac{2}{5}=\frac{6}{35}\)

**25- Choice B is correct**\(7^3× 7^4=7^7\)

**26- Choice D is correct**\(3 \frac{1}{5}×5 \frac{2}{3}= \frac{16}{5}×\frac{17}{3}=\frac{272}{15}=18 \frac{2}{15}\)

**27- Choice C is correct**\(3 \frac{2}{5}+5 \frac{1}{2}=8 \frac{9}{10}\)

**28- Choice B is correct**\(5 \frac{3}{4}-3 \frac{1}{4}=5+\frac{3}{4}-3-\frac{1}{4}=2 \frac{2}{4}=2 \frac{1}{2}\)

**29- Choice B is correct**\(4 \frac{3}{5}-\frac{1}{2}=4 \frac{1}{10}\)

**30- Choice C is correct**\(\frac{3}{7}÷\frac{2}{5}=\frac{15}{14}\)

**31- Choice B is correct**\(3^3× 3^4=3^{3+4}= 3^7\)

**32- Choice A is correct**\(3 \frac{2}{7}÷2 \frac{2}{5}=\frac{23}{7} ÷\frac{12}{5}= \frac{115}{84}=1 \frac{31}{84}\)

**33- Choice C is correct**\(15\%\) of \(60\) \(= 0.15 × 60 = 9\)

**34- Choice B is correct**\(20\%\) of \(20\) \(= 0.2 × 20 = 4\)

**35- Choice C is correct**\(25 \%\) of \(40 = 10\)

**36- Choice A is correct**\(6\%\) of

*__*\(= 36, \frac{6}{100} × x = 36, \frac{6x}{100}=36, x=\frac{3,600}{6}=600\)

**37- Choice D is correct**\(7x+4x=11x\)

**38- Choice B is correct\(8×(-3)=-24\)**

**39- Choice C is correct**\(9ab-ab=8ab\)

**40- Choice A is correct**\(73.50÷3.75=19.6\)

**Part 2: Applied Mathematics**

**41- Choice B is correct**\(3^5=3×3×3×3×3=243\)

**42- Choice B is correct**A. \(\frac{1}{5}=0.2\)

B. \(\frac{2}{3}=0.67\)

C. \(\frac{4}{7}=0.57\)

D. \(\frac{1}{2}=0.5\)

**43- Choice B is correct**\(\frac{11+ 23+17+27+7}{5}=\frac{85}{5}=17\)

**44- Choice A is correct**Use PEMDAS (order of operation):

\(-9-3×(–3)+[-3+11×(-3)]÷2=-9+9+[-3-33]÷2=0+[-36]÷2=-36÷2=-18\)

**45- Choice C is correct**\(160÷20=8\)

**46- Choice B is correct**\(x + y = N → x = 5 → 5 + y = N → y = N –5 → 5y = 5(N – 5)\)

**47- Choice D is correct**\($15×10=$150\) → Petrol use: \(10×3=30\) liters → Petrol cost: \(30×$2=$60\)

Money earned: \($150-$60=$90\)

**48- Choice A is correct**\(2x+3=7.2→2x=7.2-3=4.2→x=\frac{4.2}{2}=2.1\)

Then; \(3x-5=3(2.1)-5=6.3-5=1.3\)

**49- Choice C is correct**First draw an isosceles triangle. Remember that two sides of the triangle are equal.

Let put a for the legs. Then:

\(a=8\) ⇒ area of the triangle is \(=\frac{1}{2}(8×8)=\frac{64}{2}=32 \space cm^2\)

**50- Choice C is correct**Write the numbers in order: 3, 5, 11, 15, 19, 21, 25

Since we have 7 numbers (7 is odd), then the median is the number in the middle, which is 15.

**51- Choice D is correct**To find the discount, multiply the number by \((100\% – \)rate of discount)\).

Therefore, for the first discount we get: \((420)(100\% –17\%)=(420)(0.83)\)

For the next 13% discount: \((420)(0.83)(0.87)\)

**52- Choice C is correct**Plug in each pair of numbers in the equation: \(2x-3y=5\)

A. \((-2,1): ⇒2 (-2) -3 (1) = -7\)

B. \((–1,3): ⇒2 (–1) -3 (3) = -11\)

C. \(( 4,1): ⇒2 (4) -3 (1) = 5\)

D. \((2,2): ⇒2 (3) -3 (2) = 0\)

Choice C is correct.

**53- Choice D is correct**average \(=\frac{sum \space of \space terms}{number \space of \space terms} ⇒ 25=\frac{21+18+17+x}{4} ⇒100=56+x ⇒ x=44\)

**54- Choice A is correct**Use this formula: Percent of Change: \(\frac{New \space Value-Old \space Value}{Old \space Value}×100\%\)

\(\frac{25500-30000}{30000}×100\%=15\% \space and \space \frac{21678-25500}{25500}×100\%=15\%\)

**55- Choice C is correct**Let \(x\) be the original price. If the price of the sofa is decreased by \(16\%\) to $378, then: \(84\%\) of \(x=378 ⇒ 0.84x=378 ⇒ x=378÷0.84=450\)

**56- Choice C is correct**The sum of supplement angles is 180. Let \(x\) be that angle. Therefore, \(x+7x=180\)

\(8x=180\), divide both sides by 8: \(x=22.5\)

**57- Choice D is correct**The average speed of john is: \(120÷4=30\) km

The average speed of Alice is: \(180÷5=36\) km

Write the ratio and simplify. \(30 ∶ 36 ⇒ 5 ∶ 6\)

**58- Choice D is correct**Use Pythagorean Theorem: \(a^2+b^2=c^2\)

\(12^2+16^2=c^2 ⇒ 144+256=c^2 ⇒ 400=c^2⇒c=20\)

**59- Choice B is correct**The percent of girls playing tennis is: \(60\%×30\%=0.60×0.30=0.18=18\%\)

**60- Choice B is correct**Let x be the number. Write the equation and solve for \(x\). \((30-x)÷x=4\),

Multiply both sides by \(x\).

\((30- x)=4x\), then add x both sides. \(30=5x\), now divide both sides by 5. \(x=6\)

**61- Choice D is correct**The area of the floor is: \(9 \space cm × 25 \space cm = 225 \space cm^2\)

The number of tiles needed \(= 225 ÷ 5 = 45\)

**62- Choices A is correct**Area of the circle is less than 16π. Use the formula of areas of circles.

Area\(=πr^2 ⇒ 16 π> πr^2⇒ 16> r^2⇒ r<4\)

Radius of the circle is less than 4. Let’s put 4 for the radius. Now, use the circumference formula:

Circumference\(=2πr=2π (4)=8π\)

Since the radius of the circle is less than 4. Then, the circumference of the circle must be less than \(8π\).

Only choice A is less than \(8π\).

**63- Choice A is correct**Solving Systems of Equations by Elimination

\(\begin{cases}2x+3y=-7 \\x-2y=7\end{cases}\)

Multiply the second equation by \(-2\), then add it to the first equation.

\(\begin{cases}2x+3y= -7\\-2(x-2y=7)\end{cases}\)

**⇒**\(\begin{cases}2x+3y= -7\\-2x+4y=-14\end{cases}\)

⇒ \(7y=-21 ⇒ y=-3\)

**64- Choice C is correct**If the length of the box is 18, then the width of the box is one third of it, 6, and the height of the box is 2 (one third of the width). The volume of the box is: \(V=lwh=(18)(6)(2)=216\)

**65- Choice D is correct**

Write the equation and solve for B: \(0.75A=0.25B\), divide both sides by \(0.25\), then:

\(\frac{0.75}{0.25}A=B\), therefore: \(B=3A\), and \(B\) is 3 times of \(A\) or it’s \(300\%\) of \(A\).

**66- Choice A is correct**To find the number of possible outfit combinations, multiply number of options for each factor: \(4×5×3=60\)

**67- Choice C is correct**Use simple interest formula: I = prt (I = interest, p = principal, r = rate, t = time)

\(I=(10,000)(0.030)(5)=1,500\)

**68- Choice D is correct**Use percent formula: part\(=\frac{percent}{100}×whole\)

\(125=\frac{percent}{100}×50 ⇒ 125=\frac{percent ×50}{100} ⇒ 125= \frac{percent ×5}{10}\), multiply both sides by \(10\).

\(1,250=percent×5\), divide both sides by 5. \(250=percent\)

**69- Choice B is correct**The perimeter of the trapezoid is 54.

Therefore, the missing side (height) is \(=46-14-9-11=12\)

Area of a trapezoid: \(A=\frac{1}{2} h (b_1+ b_2)=\frac{1}{2}(12)(9+11)=120\)

**70- Choice D is correct**Add the first 5 numbers. 60 + 55 + 45 + 70 + 65 = 295

To find the distance traveled in the next 5 hours, multiply the average by number of hours.

Distance = Average × Rate = 65 × 5 = 325, Add both numbers. 325 + 295 = 620

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*TABE 11 & 12*

*Math*

**71- Choice C is correct**Use distance formula: Distance = Rate × time ⇒ \(360 = 48 × T\), divide both sides by 48. \(\frac{360}{48}= T\) ⇒ \(T = 7.5\) hours.

Change hours to minutes for the decimal part. 0.5 hours = 0.5 × 60 = 30 minutes.

**72- Choice D is correct**Let x be the number. Write the equation and solve for \(x\).

\(\frac{2}{3}×24=\frac{2}{5}. x ⇒ \frac{2×24}{3}= \frac{2x}{5}\), use cross multiplication to solve for \(x\).

\(10×24=2x×3 ⇒240=6x ⇒ x=40\)

**73- Choice C is correct**To find the discount, multiply the number by \((100\% – rate \space of \space discount)\).

Therefore, for the first discount we get: \((D) (100\% – 15\%) = (D) (0.85) = 0.85 D\)

For increase of \(8\%\): \((0.85 D) (100\% + 8\%) = (0.85 D) (1.08) = 0.918 D = 91.8\%\) of \(D\)

**74- Choice C is correct**Use the formula for Percent of Change: \(\frac{New \space Value-Old \space Value}{Old \space Value}× 100\%\)

\(\frac{32-50}{50}× 100\% = –36\%\) (negative sign here means that the new price is less than old price).

**75- Choice A is correct**Th ratio of boy to girls is 3 : 5. Therefore, there are 3 boys out of 8 students. To find the answer, first divide the total number of students by 8, then multiply the result by 3.

400 ÷ 8 = 50 ⇒ 50 × 3 = 150

**76- Choice B is correct**The question is this: 1.89 is what percent of 1.48?

Use percent formula: part \(= \frac{percent}{100} × whole\)

\(1.89 = \frac{percent}{100} × 1.48 ⇒ 1.89 = \frac{percent ×1.48}{100} ⇒189 = percent ×1.48 ⇒ percent = \frac{189}{1.48}= 127.7\)

**77- Choice D is correct**The question is this: 514.5 is what percent of 735?

Use percent formula: part \(= \frac{percent}{100} × whole\)

\(514.5 = \frac{percent}{100}× 735 ⇒ 514.5 =\frac{percent ×735}{100} ⇒51,450 = percent ×735 ⇒\)

\(percent = \frac{51,450}{735} = 70\)

\(514.5\) is \(70 \%\) of \(735\). Therefore, the discount is: \(100\% – 70\% = 30\%\)

**78- Choice A is correct**If the score of Mia was 50, therefore the score of Ava is 25. Since, the score of Emma was half as that of Ava, therefore, the score of Emma is 12.5.

**79- Choice C is correct**If 14 balls are removed from the bag at random, there will be one ball in the bag.

The probability of choosing a red ball is 1 out of 15. Therefore, the probability of not choosing a red ball is 14 out of 15 and the probability of having not a red ball after removing 14 balls is the same.

**80- Choice D is correct**Let \(x\) be the smallest number. Then, these are the numbers:

\(x, x+1, x+2, x+3, x+4\)

average \(=\frac{sum \space of \space terms}{number \space of \space terms}\)⇒ \(47 = \frac{x+(x+1)+(x+2)+(x+3)+(x+4)}{5}\)⇒\(47=\frac{5x+10}{5} ⇒ 235 = 5x+10 ⇒225 = 5x ⇒ x=45\)

**81- Choice C is correct**The equation of a line is in the form of \(y=mx+b\), where m is the slope of the line and \(b\) is the \(y\)-intercept of the line.

Two points (2, 3) and (1, 1) are on line A. Therefore, the slope of the line A is:

slope of line A\(=\frac{y_2- y_1}{x_2 – x_1}=\frac{1-3}{1-2} =\frac{-2}{-1}=2\)

The slope of line A is 2. Thus, the formula of the line A is: \(y=mx+b=2x+b\), choose a point and plug in the values of \(x\) and \(y\) in the equation to solve for \(b\).

Let’s choose point (1, 1). Then: \(y=2x+b→1=2+b→b=1-2=-1\)

The equation of line A is: \(y=2x-1\)

Now, let’s review the choices provided:

A. \((1,2) →y=2x-1→2=2-1=1\) This is not true.

B. \((3,6) →y=2x-1→6=2(3)-1=5\) This is not true.

C. \((3,5) →y=2x-1→5=2(3)-1=5\) This is true !

D. \((1,-2) →y=2x-1→-2=2-1=1\) This is not true.

**82- Choice C is correct**The weight of 15.5 meters of this rope is: 15.5 × 500 g = 7,750 g

1 kg = 1,000 g, therefore, 7,750 g ÷ 1000 = 7.75 kg

**83- Choice C is correct**5% of the volume of the solution is alcohol. Let \(x\) be the volume of the solution.

Then: \(5\%\) of \(x = 45\) ml ⇒ \(0.05 x = 45 ⇒ x = 45 ÷ 0.05 = 900\)

**84- Choice D is correct**average \(= \frac{sum \space of \space terms}{number \space of \space terms}\)

The sum of the weight of all girls is: 16 × 50 = 800 kg

The sum of the weight of all boys is: 24 × 65 = 1560 kg

The sum of the weight of all students is: 1560 + 800 = 2360 kg

average \(= \frac{2,360}{40} = 98.33\)

**85- Choice C is correct**Let \(x\) be the original price. If the price of a laptop is decreased by \(20\%\) to $860, then:

\(80\%\) of \(x=860⇒ 0.80x=860 ⇒ x=860÷0.80=1,075\)

**86- Choices C is correct**Some of prime numbers are: 2, 3, 5, 7, 11, 13

Find the product of two consecutive prime numbers: 2 × 3 = 6 (not in the options)

3 × 5 = 15 (not in the options) 5 × 7 = 35 (bingo!) Choice C is correct.

**87- Choice B is correct**Surface Area of a cylinder = 2πr (r + h),

The radius of the cylinder is 4 inches and its height is 10 inches. π is 3.14. Then:

Surface Area of a cylinder = 2 (3.14) (4) (4 + 10) = 351.68

**88- Choice D is correct**Use the information provided in the question to draw the shape.

Use Pythagorean Theorem: \(a^2 + b^2 = c^2\)

\(80^2 + 60^2 = c^2 ⇒ 6400 + 3600 = c^2 ⇒ 10,000 = c^2 ⇒ c = 100\)

**89- Choice A is correct**Let \(x\) be the number of years. Therefore, $1,600 per year equals \(1,600x\).

starting from $18,000 annual salary means you should add that amount to \(1,600x\).

Income more than that is: \(I > 1,600x + 18,000\)

**90- Choice A is correct**For each option, choose a point in the solution part and check it on both inequalities.

\(y≤x+4\)

\(2x+y≤-4\)

A. Point \((–4, –4)\) is in the solution section. Let’s check the point in both inequalities.

\(–4 ≤ – 4 + 4\), It works

\(2 (–4) + (–4) ≤ –4 ⇒ – 12 ≤ – 4,\) it works (this point works in both)

B. Let’s choose this point \((0, 0)\)

\(0 ≤ 0 + 4\), It works

\(2 (0) + (0) ≤ –4\), That’s not true!

C. Let’s choose this point \((–5, 0)\)

\(0 ≤ –5 + 4\), That’s not true!

D. Let’s choose this point \((0, 5)\)

\(5 ≤ 0 + 4\), That’s not true!