# Full-Length 7th Grade IAR Math Practice Test-Answers and Explanations

Did you take the 7th 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.

## 7th Grade IAR Math Practice Test Answers and Explanations

1- **Choice C is correct**

If the score of Mia was \(90\), then the score of Ava is \(30\). Since the score of Emma was one and a half as that of Ava, therefore, the score of Emma is \(1.5×30=45\).

2- **Choice A is correct**

Write the ratio and solve for \(x\). \(\frac{60}{50}=\frac{5x+2}{10}⇒ 12=5x+2 ⇒12-2=5x⇒ x=\frac{10}{5}=2\)

3- **Choice B is correct**

Let \(x\) be the number of students in the class. \(40\%\) of \(x\) \(=\) girls, \(25\%\) of girls \(=\) tennis player,

Find \(25\%\) of \(40\%\). Then: \(25\%\) of \(40\%\)\(=0.25×0.40=0.1=10\%\) or \(\frac{10}{100}=\frac{1}{10}\)

4- **Choice C is correct**

Use the information provided in the question to draw the shape.

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

\(30^2+40^2=c^2⇒ 900+1,600= c^2⇒2,500= c^2⇒c=50\)

5- **Choice A is correct**

Write a proportion and solve for \(x\). \(\frac{12 \space Cans}{$ 7.40}=\frac{30 \space Cans}{x}, x= \frac{7.40×30}{12}⇒x=$18.5\)

6- **Choice D is correct**

Use the volume of square pyramid formula.

\(V= \frac{1}{3} a^2 h ⇒V=\frac{1}{3} (12\space m)^2×20 \space m ⇒ V=960\space m^3\)

7- **Choice C is correct**

Let x be the number of soft drinks for \(240\) guests. Write a proportional ratio to find \(x\). \(\frac{6 \space soft \space drinks}{8 \space guests} =\frac{x}{240 \space guests}\),

\(x=\frac{240×6}{8}⇒x=180\)

8- **Choice B is correct**

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

\(\frac{1.75-1.4}{1.4}×100\%=25\%\)

9- **The answer is: -99**

Use PEMDAS (order of operation): \([8×(-14)+15]-(10)+[4×6]÷3=[-122+15]-(10)+8=-97-10+8=-99\)

10- **Choice D is correct**

Simplify. \(5x^2 y(2xy^3)^4=5x^2 y(16x^4 y^{12 })=80x^6 y^{13}\)

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11- **Choice C is correct**

The distance between Jason and Joe is \(14\) miles. Jason running at \(6\) miles per hour and Joe is running at the speed of \(8\) miles per hour. Therefore, every hour the distance is \(2\) miles less.

\(14÷2=7\)

12- **Choice A is correct**

Let \(x\) be the integer. Then: \(5x-9=101\), Add \(9\) both sides: \(5x=110\), Divide both sides by \(5\): \(x=22\)

13- **Choice D is correct**

Two and half times of \(18,000\) is \(45,000\). One-fifth of them canceled their tickets.

One sixth of \(45,000\) equals \(9,000(\frac{1}{5} ×45,000=9,000)\).

\(36,000(45,000-9,000=36,000)\) fans are attending this week.

14- **Choice C is correct**

Write the numbers in order: \(25,12,13,18,22,36,22\)

Since we have \(7\) numbers (\(7\) is odd), then the median is the number in the middle, which is \(22\).

15- **Choice D is correct**

The question is: \(615\) is what percent of \(820\)?

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

\(615=\frac{percent}{100}×820 ⇒ 615=\frac{percent ×820}{100}⇒61,500=percent×820 \)⇒

percent\(=\frac{61,500}{820}=75, 615\) is \(75\%\) of \(820\). Therefore, the discount is: \(100\%-75\%=25\%\)

16- **The answer is \(22 \frac{1}{3}\) miles.**

Robert runs \(4 \frac{1}{3}\) miles on Saturday and \(2(4 \frac{1}{3} )\) miles on Monday and Wednesday.

Robert wants to run a total of \(35\) miles this week. Therefore, subtract \(4 \frac{1}{3}+2(4 \frac{1}{3} )\) from \(35\).

\(35-(4 \frac{1}{3}+2(4 \frac{1}{3} ))=35-12\frac{2}{3}=22 \frac{1}{3}\) miles

17- **Choice B is correct**

To find the area of the shaded region, find the difference between the area of the two circles. (\(S_1\): the area of the bigger circle.\( S_2\): the area of the smaller circle). Use the area of the circle formula. \(S=πr^2\)

\(S_1- S_2=π(6cm)^2- π(4cm)^2⇒S_1- S_2=36π \space cm^2-16π \space cm^2 ⇒ S_1- S_2 =20π \space cm^2\)

18- **Choice A is correct**

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

\(12^2+5^2=c^2⇒ 144+25= c^2 ⇒ c^2=169 ⇒c=13\)

19- **Choice A is correct**

Let \(L\) be the price of a laptop and \(C\) be the price of a computer. \(4(L) =7(C)\) and \(L = $240 + C\)

Therefore, \(4($240 + C) =7C ⇒ $960 + 4C = 7C ⇒ C=$320\)

20- **The answer is 70.**

Jason needs an \(75\%\) average to pass five exams. Therefore, the sum of \(5\) exams must be at least \(5×75=375\), The sum of \(4\) exams is \(62+73+82+88=305\).

The minimum score Jason can earn on his fifth and final test to pass is: \(375-305=70\)

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21- **Choice B is correct**

Let \(x\) be the original price. If the price of a laptop is decreased by \(15\%\) to \($425\), then:

\(85\%\ of x=425 ⇒ 0.85x=425 ⇒ x=425÷0.85=500\)

22- **Choice C is correct**

The weight of \(12\) meters of this rope is: \(12×450 \space g=5,400\space g\)

\(1\space kg=1,000 \space g\), therefore, \(5,400 \space g÷1,000=5.4\space kg\)

23- **Choice D is correct**

Only option D is correct. Other options don’t work in the equation. \((4x-2)x=42\)

24- **Choice C is correct**

Compare each score: In Algebra Joe scored \(24\) out of \(32\) in Algebra that it means \(75\%\) of total mark. \(\frac{24}{32}= \frac{x}{100}⇒x=75\)

Joe scored \(28\) out of \(40\) in science that it means \(70\%\) of total mark. \(\frac{28}{40}=\frac{x}{100}⇒x=70\)

Joe scored \(72\) out of \(90\) in mathematic that it means \(80\%\) of total mark. \(\frac{72}{90}=\frac{x}{100} ⇒x=80\)

Therefore, his score in mathematics is higher than his other scores.

**25-Choice B is correct**

To find the discount, multiply the number by (\(100\%\)\(-\)rate of discount).

Therefore, for the first discount we get: \((D)(100\%-25\%)=(D)(0.75)=0.75\)

For increase of \(15\%\): \((0.75D)(100\%+15\%)=(0.75D)(1.15)=0.8625 \)

\(D=86.25\%\) of \(D\)

**26-Choice B is correct**

Write the numbers in order: \(42,21,15,28,43,34,26\), since we have \(7\) numbers (\(7\) is odd), then the median is the number in the middle, which is \(28\).

**27-Choice C is correct **

The average speed of John is \(210÷7=30\) \(km\), and the average speed of Alice is: \(160÷5=32\) \(km\), Write the ratio and simplify. \(30\)∶ \(32 ⇒ 15∶16\)

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

\(\frac{42-56}{56}×100\%=-25\%\) (negative sign here means that the new price is less than old price).

**29-Choice C is correct**

Use the formula of areas of circles.\(Area=πr^2 ⇒ 121π= πr^2 ⇒ 121= r^2⇒ r=11\) Radius of the circle is \(11\). Now, use the circumference formula: Circumference\(=2πr=2π(11)=22π\)

**30-Choice B is correct**

Let x be the number of balls. Then: \(\frac{1}{2}x+\frac{1}{5}x+\frac{1}{10} x+12=x\)

\((\frac{1}{2}+\frac{1}{5}+\frac{1}{10})x+12=x\), \(\frac{8}{10}x+12=x,x=60\), In the bag of small balls \(\frac{1}{5}\) are white, then: \(\frac{60}{5}=12\), There are \(12\) white balls in the bag.

**31-Choice A is correct**

William ate \(\frac{4}{5}\) of \(10\) parts of his pizza that it means \(8\) parts out of \(10\) parts

(\(\frac{4}{5}\) of \(10\) parts \(=x ⇒ x=8\)) and left \(2\) parts.

Ella ate \(\frac{1}{2}\) of \(10\) parts of her pizza that it means \(5\) parts out of \(10\) parts (\(\frac{1}{2}\) of \(10\) parts \(= x\) ⇒ \(x=5\)) and left \(5\) parts. Therefore, they ate \((5+2)\) parts out of \((10+10)\) parts of their pizza and left \((5+2)\) parts out of \((10 + 10)\) parts of their pizza. It means: \(\frac{7}{20}\), After simplification we have: \(\frac{7}{20}\)

**32-Choices D is correct.**

The failing rate is \(14\) out of \(50=\)\(\frac{14}{50}\), Change the fraction to percent: \(\frac{14}{50} ×100\%=28\%\)

\(28\) percent of students failed. Therefore, \(72\) percent of students passed the exam.

**33-Choice C is correct **

\(x\%\) of \(50\) is \(6.2\), then: \(0.50x=6.2 ⇒x=6.2÷0.50=12.4\)

**34-The answer is 56**

Use the area of the square formula. \(S=a^2 ⇒ 196= a^2 ⇒ a=14\) One side of the square is \(14\) feet. Use the perimeter of the square formula. \(P=4a ⇒ P=4(14) ⇒ P=56\)

**35-Choice B is correct.**

Input the points instead of \(x\) and \(y\) in the formula. Only option B works in the equation.

\(6x-14=4y, 6(2)-14=4(-\frac{1}{2})⇒-2=-2\)

**36- Choice B is correct**The sum of supplement angles is \(180\). Let \(x\) be that angle. Therefore, \(x+4x=180\)

\(5x=180\), divide both sides by \(5\): \(x=36\)

**37- Choice B is correct**

Use simple interest formula: \(I=prt\) (\(I=\)interest,\(p=\)principal,\(r=\) rate,\(t=\)time)

\(I=(16,000)(0.035)(3)=1,680\)

**38- Choice B is correct.**

Total number of way is \(6×6=36\), favorable cases is \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\). Thus probability that sum of two tice get \(7\) is \(\frac{6}{36}=\frac{1}{6}\)

**39- The answer is 168**To find the number of possible outfit combinations, multiply the number of options for each factor: \(3×8×7=168\)

**40- Choice B is correct.**\(7\%\) of the volume of the solution is alcohol. Let \(x\) be the volume of the solution.

Then: \(7\%\) of \(x=35\) \(ml\) ⇒ \(0.07 x=35\) ⇒ \(x=35 ÷ 0.07=500\)

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