# Full-Length 6th Grade IAR Math Practice Test-Answers and Explanations

Did you take the 6th Grade IAR 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.

## 6th Grade IAR Math Practice Test Answers and Explanations

1- **Choice B is correct**

Plug in the value of \(x\) and \(y\) and use order of operations rule. \(x=2\) and \(y=-3\)

\(5(4x-3y)-7y^2=5(4(2)-3(-3))-7(-3)^2=5(8+9)-7(9)=5(17)-63=85-63=22\)

**2- Choice C is correct**

For one hour he earns \($18\), then for \(t\) hours, he earns \($18t\). If he wants to earn at least \($78\), therefore, the number of working hours multiplied by \(18\) must be equal to \(78\) or more than \(78\). \(18t≥78\)

**3- Choice B is correct**\((108-(3×9))÷9=9^3÷81=9\)

**4- Choice B is correct**The ratio of boys 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\(3\).

\(240÷8=30 ⇒ 30×3=90\)

**5- Choice A is correct**

Probability \(=\frac{number \space of \space desired \space outcomes}{number \space of \space total \space outcomes} = \frac{9}{9+15+14+16}=\frac{9}{54}=\frac{1}{6}=0.16\)

**6- Choice D is correct**Let’s compare each fraction: \(\frac{2}{3}<\frac{3}{4}<\frac{7}{9}<\frac{4}{5}\)

Only choice D provides the right order.

**7- Choice B is correct**Let \(y\) be the width of the rectangle. Then; \(14×y=84→y=\frac{84}{14}=6\)

**8- Choice B is correct**\(4×\frac{5}{16}=\frac{20}{16}=1.25\)

A. \(1.25>2\)

B. \(1<1.25<2\) This is the answer!

C. \(\frac{3}{8}=1.25\)

D. \(1.25=2^2\)

**9- Choice B is correct**In any rectangle, The measure of the sum of all the angles equals \(180^\circ\).

**10- Choice C is correct**\(\frac{824}{17}=48.5\)

**11- The answer is **\(7^2\).

\(588=2^2×3^1×7^2\)

**12- Choice B is correct**The area of the trapezoid is: 𝑎𝑟𝑒𝑎\(=\frac{(𝑏𝑎𝑠𝑒 \space 1+𝑏𝑎𝑠𝑒 \space 2)}{2}

×ℎ𝑒𝑖𝑔ℎ𝑡=\frac{12+10}{2}𝑥=𝐴→11𝑥=𝐴→𝑥=\frac{A}{11}\)

**13- Choice B is correct**\(\frac{72}{8}=9, \frac{648}{72}=9, \frac{5832}{648}=9\), Therefore, the factor is \(9\).

**14- Choice C is correct**Simplify each option provided.

A. \(13-(3×6)+(7×(-6))=13-18+(-42)=-5-42=-47\)

B. \((\frac{25}{400})+(\frac{7}{50})=\frac{25}{400}+\frac{56}{400}=\frac{81}{400}\)

C. \(((22×\frac{30}{6})-(7×\frac{144}{12}))×\frac{18}{2}=(110-84)×9=26×9=234\) (this is the answer)

D. \((\frac{6}{24}+\frac{12}{33})-50=(\frac{1}{4}+\frac{1}{3})-50=(\frac{3}{12}+\frac{4}{12})-50=\frac{7}{12}-50=\frac{-593}{15}\)

**15- Choice D is correct**To find the discount, multiply the number \((100\%- rate \space of \space discount)\)

Therefore; \(450(100\%-16\%)=450(1-0.16)=450-(450×0.16)\)

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**16- Choice A is correct**\(1,400\) out of \(11,900\) equals to \(\frac{1,400}{11,900}=\frac{200}{1,700}=\frac{2}{17}\)

**17- Choice C is correct**The opposite of Nicolas’s integer is \(25\). So, the integer is \(-25\). The absolute value of \(25\) is also \(25\).

**18- Choice B is correct**Volume of a box = length\( ×\) width \(×\) height \(= 7 × 4 × 12 = 336\)

**19- Choice C is correct**\(1\) yard \(= 3\) feet, Therefore: \(33,759 \space ft×\frac{1 \space yd}{3 \space ft}=11,253 \space yd\)

**20- Choice B is correct**\(16\%\) of the volume of the solution is alcohol. Let \(x\) be the volume of the solution.

Then: \(16\%\) of \(x=38\) \(ml\) ⇒ \(0.16x=38 ⇒ x=38÷0.16=237.5\)

**21- Choice C is correct**\((-2)(9x-8)=(-2)(9x)+(-2)(-8)=-18x+16\)

**22- Choice D is correct**\(1 pt = 16\) fluid ounces. \(576 ÷ 16 = 36 \)

Then: \(576\) fluid ounces \(= 36 pt\)

**23- Choice D is correct**\(1 kg = 1,000 g\) and \(1 g = 1,000 mg\),

\(120 kg = 120 × 1,000 g = 120 × 1,000 × 1,000 = 120,000,000 mg\)

**24- Choice C is correct**The diameter of a circle is twice the radius. Radius of the circle is \(\frac{14}{2}=7\).

Area of a circle \(= πr^2=π(7)^2=49π=49×3.14=153.86≅153.9\)

**25- Choice B is correct**Average (mean)\(=\frac{sum \space of \space terms}{number \space of \space terms}=\frac{15+17+12+16+21+23}{6}=\frac{104}{6}=17.33\)

**26- Choice C is correct**Prime factorizing of \(18=2×3×3\), Prime factorizing of \(24=2×2×2×3\)

\(LCM\) \(= 2×2×2×3×3=72\)

**27- Choice B is correct**The coordinate plane has two axes. The vertical line is called the \(y\)-axis and the horizontal is called the \(x\)-axis. The points on the coordinate plane are addressed using the form \((x,y)\). Point \(A\) is one unit on the left side of the \(x\)-axis, therefore its \(x\) value is \(4\) and it is two units up, therefore its \(y\)-axis is \(2\). The coordinate of the point is: \((4, 2)\)

**28- Choice B is correct**\(α\) and \(β\) are supplementary angles. The sum of supplementary angles is \(180\) degrees.

\(α+β=180^\circ→α=180^\circ -β=180^\circ-125^\circ=55^\circ\), Then, \(\frac{α}{β}=\frac{55}{125}=\frac{11}{25}\)

**29- Choice C is correct**The opposite number of any number \(x\) is a number that if added to \(x\), the result is \(0\). Then:

\(7+(-7)=0\) and \(4+(-4)=0\)

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**30- Choice C is correct**\(16=-129+x\), First, subtract \(129\) from both sides of the equation. Then:

\(16+129=-129+x+129 →145=x\)

**31- Choice D is correct**\(-5<2x+7≤3\)→ (add (\(-7\)) all sides) \((-7)+(-5)<2x+7+(-7)≤3+(-7) →-12<2x≤-4→\) (divide all sides by \(2\)), \(-6<x≤-2\)

In inequality \(-6<x≤-2\), \(x\) is fewer or equal to \(-2\) and more than \(-6\). Only choice D represents the same inequality on the number line.

**32- Choice B is correct**A. Number of books sold in April is: \(690\)

The number of books sold in July is: \(1150→ \frac{690}{1150}≠2\)

B. Number of books sold in July is: \(1150\)

Half the number of books sold in May is: \(\frac{1150}{2} =575→690>575\) (it’s correct)

C. Number of books sold in June is: \(375\)

Half the number of books sold in April is: \(\frac{690}{3}=230→240<375\)

D. \(690+375=1,065>1150\)

**33- Choice B is correct**\(51∶18=17∶6, 17×3=51\) and \(8×3=18\)

**34- Choice A is correct**There are \(6\) integers from \(9\) to \(15\). Set of numbers that are not composite between \(9\) and \(15\) is: \([{11,13}]\), \(2\) integers are not composite. Probability of not selecting a composite number is: Probability\(= \frac{number \space of \space desired \space outcomes}{number \space of \space total \space outcomes}= \frac{2}{6}=\frac{1}{3}\)

**35- Choice A is correct**Number of biology book:\(20\), Total number of books; \(20+59+36=115\)

The ratio of the number of biology books to the total number of books is: \(\frac{20}{115}=\frac{4}{23}\)

**36- Choice B is correct**A. \(\frac{5}{8}<0.8→ \frac{5}{8}=0.6\). Therefore, this inequality is not correct.

B. \(23\%<\frac{2}{3} →23\%= 0.23, \frac{2}{3}=0.66\),Therefore \(0.23<0.66\) .

C. \(8<\frac{18}{3} →\frac{18}{3}=6\). Therefore, this inequality is not correct.

D. \(0.6>\frac{7}{9}→ \frac{7}{9}=0.77→0.77>0.6\), this inequality is not correct.

**37- Choice A is correct**\(38÷2=19\)

**38- The answer is 0.38**\(38\) percent \(= 0.38\)

**39- Choice B is correct**Let’s write the inequality for each statement.

A. \(\frac{x}{19}<9\)

B. \(\frac{9}{x}≤19\) This is the inequality provided in the question.

C. \(\frac{x}{9}≤19\)

D. \(\frac{x}{9}<19\)

**40- Choice C is correct**Since \(E\) is the midpoint of \(AB\), then the area of all triangles \(DAE, DEF, CFE\), and \(CBE\) are equal. The total area of \(ABCD\) is \(180\). So, the area of each triangle is \(180 ÷ 4 = 45\)

The area of the triangle \(ADE\) is: \(45\)

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