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

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 cancelled 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 of the area of two circles. (\(S_1\): the area of bigger circle.\( S_2\): the area of the smaller circle). Use the area of 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 laptop and \(C\) be the price of 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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