FREE 6th Grade Common Core Math Practice Test
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perimeter of the trapezoid below is 38. What is its area?
A. 198 cm\(^2\)
B. 162 cm\(^2\)
C. 99 cm\(^2\)
D. 81cm\(^2\)
6- Which of the following expressions has the greatest value?
A. \( 3^1+12\)
B. \( 3^3-3^2\)
C. \( 3^4-60\)
D. \( 3^5-218\)
7- Alfred has \(x\) apples. Alvin has 40 apples, which is 15 apples less than several apples Alfred owns. If Baron has \(\frac{1}{5}\) times as many apples as Alfred has. How many apples does Baron have?
A. 5
B. 11
C. 55
D. 275
8- In the following triangle, find \(α\).
A. \(100^\circ\)
B. \(90^\circ\)
C. \(60^\circ\)
D. \(30^\circ\)
9- The price of a laptop is decreased by \(15\%\) to $425. What is its original price?
A. $283
B. $430
C. $500
D. $550
10- Find the perimeter of the shape in the following figure? (all angles are right angles)
A. 21
B. 22
C. 24
D. 20
11- What are the values of mode and median in the following set of numbers?
\(1,3,3,6,6,5,4,3,1,1,2\)
A. Mode: 1, 2, Median: 2
B. Mode: 1, 3, Median: 3
C. Mode: 2, 3, Median: 2
D. Mode: 1, 3, Median: 2.5
12- Which expression equivalent to \(x × 92\)?
A. \((x×90)+2\)
B. \(x×9×2\)
C. \((x×90)+(x×2)\)
D. \((x×90)+2\)
13- The ratio of pens to pencils in a box is 3 to 5. If there are 96 pens and pencils in the box altogether, how many more pens should be put in the box to make the ratio of pens to pencils 1: 1?
A. 22
B. 23
C. 24
D. 25
14- If point A placed at \(-\frac{24}{3}\) on a number line, which of the following points has a distance equal to 5 from point A?
A. \(-13\)
B. \(-3\)
C. \(-2\)
D. A and B
15- Which of the following shows the numbers in increasing order?
A. \(\frac{3}{13}, \frac{4}{11}, \frac{5}{14}, \frac{2}{5}\)
B. \(\frac{3}{13}, \frac{5}{14}, \frac{4}{11}, \frac{2}{5}\)
C. \(\frac{3}{13}, \frac{5}{14}, \frac{2}{5}, \frac{4}{11}\)
D. \(\frac{5}{14}, \frac{3}{13}, \frac{2}{5}, \frac{4}{11}\)
16- If \(x=- 4\), which of the following equations is true?
A. \(x(3x-1)=50\)
B. \(5(11-x^2 )=-25\)
C. \(3(-2x+5)=49\)
D. \(x(-5x-19)=-3\)
17- What is the missing prime factor of number 450?
\(450=2^1×3^2×…\) _________
18- What is the perimeter of the following shape? (it’s a right triangle)
A. 14 cm
B. 18 cm
C. 24 cm
D. 32 cm
19- 65 is what percent of 50?
A. \(50 \%\)
B. \(77 \%\)
C. \(130 \%\)
D. \(140 \%\)
20- Which of the following expressions has a value of \(-23\)?
A. \(-10+(-8)+ \frac{5}{2}×(-2)\)
B. \(5×3+(-2)×18\)
C. \(-10+6×8÷(-4)\)
D. \((-3) × (-7) + 2\)
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Answers:
1- D
First, we need to find the GCF (Greatest Common Factor) of 143 and 55.
\(143=11×13\)
\(55=5×11→\) GFC\( = 11\)
Therefore, we need 11 boxes.
2- C
\(2205÷315=\frac{2205}{315}=\frac{441}{63}=\frac{147}{21}= 7\)
3- B
\(112=22+x \)
Subtract 22 from both sides of the equation. Then:
\(x=112-22=90\)
4- D
Distance that car B travels \(=1.2 ×\) distance that car A travels
=\(1.2×221.5=265.8 \) km
5- D
The perimeter of the trapezoid is 38.
Therefore, the missing side (height) is \(= 38 – 8 – 10 – 11 = 9\)
Area of the trapezoid: \(A = \frac{1}{2} h (b_1 + b_2) = \frac{1}{2}1 (9) (8 + 10) = 81\)
6- D
A. \(3^1+12=3+12=15\)
B. \(3^3-3^2=27-9=18\)
C. \(3^4-60=81-60=21\)
D. \(3^5-218=243-218=25\)
7- B
Alfred has \(x\) apple which is 15 apples more than number of apples Alvin owns. Therefore:
\(x-15=40→x=40+15=55\)
Alfred has 55 apples.
Let \(y\) be the number of apples that Baron has. Then: \(y=\frac{1}{5}×55=11\)
8- A
Complementary angles add up to 180 degrees.
\( β+150^\circ=180^\circ→β=180^\circ-150^\circ=30^\circ\)
The sum of all angles in a triangle is 180 degrees. Then:
\(α+β+50^\circ=180^\circ→α+30^\circ+50^\circ=180^\circ\)
\(→α+80^\circ=180^\circ→α=180^\circ-80^\circ=100^\circ\)
9- C
Let \(x\) be the original price.
If the price of a laptop is decreased by \(15\%\) to $425, then:
\(85 \% \space of \space x=425⇒ 0.85x=425 ⇒ x=425÷0.85=500\)
10- C
Let \(x\) and \(y\) be two sides of the shape. Then:
\(x+1=1+1+1→x=2\)
\(y+6+2=5+4→y+8=9→y=1\)
Then, the perimeter is:
\(1+5+1+4+1+2+1+6+2+1=24\)
11- B
First, put the numbers in order from least to greatest: \(1, 1, 1, 2, 3, 3, 3, 4, 5, 6, 6\)
The Mode of the set of numbers is: 1 and 3 (the most frequent numbers)
The median is: 3 (the number in the middle)
12- C
\(x×92=x×(90+2)=(x×90)+(x×2)\)
13- C
The ratio of pens to pencils is \(3: 5\). Therefore, there are 3 pens out of all 8 pens and pencils. To find the answer, first dived 96 by,8 then multiply the result by 3.
\(96÷8=12→12×3=36\)
There are 36 pens and 60 pencils \((96-36)\). Therefore, 24 more pens should be put in the box to make the ratio \(1: 1\)
14- D
If the value of point A is greater than the value of point B, then the distance between the two points on the number line is: the value of A- the value of B
A. \(-\frac{24}{3}-(-13)=-8+13=5=5\)
B. \(-3-(-\frac{24}{3})=-3+8=5=5\)
C. \(-2-(-\frac{24}{3})=-2+8=6≠5\)
15- B
\(\frac{3}{13}≅0.23, \frac{5}{14}≅0.357, \frac{4}{11}≅0.36, \frac{2}{5}=0.4\)
16- B
Plug the value of \(x\) in the equations. \(x = -4\), then:
A.\(x(3x-1)=50→-4(3(-4)-1)=-4(-12-1)=-4(-13)=52≠50\)
B. \(5(11-x^2 )=-25→5(11-(-4)^2 )= 5(11-16)=5(-5)=-25\)
C. \(3(-2x+5)=49→3(-2(-4)+5)=3(8+5)=39≠49\)
D. \(x(-5x-19)=-3→-4(-5(-4)-19=-4(20-19)=-4≠-3\)
17- 5
Let \(x\) be the missing prime factor of 450.
\(450= 2 × 3 × 3 × x ⇒ x =\frac{450}{18} ⇒ x = 25=5×5\)
18- C
Use the Pythagorean theorem to find the hypotenuse of the triangle.
\(a^2+b^2=c^2→6^2+8^2=c^2→36+64=c^2→100=c^2→c=10\)
The perimeter of the triangle is: \(6+8+10=24\)
19- C
Use the percent formula:
\(Part = \frac{percent}{100} × whole\)
\(65= \frac{percent}{100} × 50⇒ 65 = \frac{percent ×50}{100}⇒ 65=\frac{percent ×5}{10}\)
Multiply both sides by 10.
\(650 =percent ×5, \space divide \space both \space sides \space by \space 5.\)
130 = percent
The answer is \(130\%\)
20- A
Let’s check the options provided.
A. \(-10+(-8)+ (\frac{5}{2})×(-2)=-10+(-8)+(-5)=-10-13=-23\)
B. \(5×3+(-2)×18=15+(-38)=-21\)
C. \(-10+6×8÷(-4)=-10+48÷(-4)=-10-12=-22\)
D. \((-3)× (-7)+ 2=21+2=23\)
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