Top 10 7th Grade Common Core Math Practice Questions

B. 8

C. 18

D. 45

4- Right triangle ABC has two legs of lengths 6 cm (AB) and 8 cm (AC). What is the length of the third side (BC)?

A.  4 cm

B. 6 cm

C. 8 cm

D. 10 cm

5- The marked price of a computer is D dollar. Its price decreased by \(20\%\) in January and later increased by \(10\%\) in February. What is the final price of the computer in D dollar?

A. 0.80 D

B. 0.88 D

C. 0.90 D

D. 1.20 D

6- \([6 × (–24) + 8] – (–4) + [4 × 5] ÷ 2 = \)?

A. \(-122\)

B. \(-112\)

C. \(-102\)

D. \(-92\)

7- The area of a circle is \(64 π\). What is the circumference of the circle?

A. \(8 π\)

B. \(16 π\)

C. \(32 π\)

D. \(64 π\)

8- A $40 shirt now selling for $28 is discounted by what percent?

A. \(20\%\)

B. \(30\%\)

C. \(40\%\)

D. \(60\%\)

9- From last year, the price of gasoline has increased from $1.25 per gallon to $1.75 per gallon. The new price is what percent of the original price?

A. \(72\%\)

B. \(120\%\)

C. \(140\%\)

D. \(160\%\)

10- If \(40\%\) of a class are girls, and \(25\%\) of girls play tennis, what fraction of the class play tennis?

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A. \(10\%\)

B. \(15\%\)

C. \(20\%\)

D. \(40\%\)

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Answers:

2- 60
Jason needs an \(75\%\) average to pass for five exams. Therefore, the sum of 5 exams must be at lease \(5 × 75 = 375\)
The sum of 4 exams is:
68 + 72 + 85 + 90 = 315
The minimum score Jason can earn on his fifth and final test to pass is:
375 – 315 = 60

3- A
It’s needed to have a ratio to find value of \(x\).
\(\frac{45}{40}=\frac{2x+4}{16}⇒ 40(2x+4)=45×16 ⇒ x=7\)

4- D
Use Pythagorean Theorem: \(a^2 + b^2 = c^2\)
\(62 + 82 = c^2 ⇒ 100 = c^2 ⇒ c = 10\)

5- B
To find the discount, multiply the number by (\(100\% –\) rate of discount).
Therefore, for the first discount we get:
(D) (\(100\% – 20\%) =\) (D) (0.80) = 0.80 D
For increase of \(10%\):
(0.80 D) (\(100\% + 10\%) =\) (0.85 D) (1.10) = 0.88 D \(= 88\%\) of D

6- A
Use PEMDAS (order of operation):
\([6 × (– 24) + 8] – (– 4) + [4 × 5] ÷ 2 = [– 144 + 8] – (– 4) + [20] ÷ 2 = [– 144 + 8] – (– 4) + 10 =
[– 136] – (– 4) + 10 = [– 136] + 4 + 10 = – 122\)

7- B
Use the formula of areas of circles.
Area \(= πr^2 ⇒ 64 π = πr^2 ⇒ 64 = r^2 ⇒ r = 8\)
Radius of the circle is 8. Now, use the circumference formula:
Circumference \(= 2πr = 2π (8) = 16 π\)

8- B
Use the formula for Percent of Change
\(\frac{New \space Value \ – \ Old \space Value}{Old \space Value}× 100\%\)
\(\frac{28-40}{40}× 100\% = – 30\% \)
(negative sign here means that the new price is less than old price).

9- C

The question is this: 1.75 is what percent of 1.25?
Use percent formula:

part \(= \frac{percent}{100}×\) whole
\(\frac{percent}{100}× 1.25 ⇒ 1.75 = \frac{percent ×1.25}{100}⇒175 =\) percent \(×1.25 ⇒\) percent \(= \frac{175}{1.25}= 140\)

10- A
Let \(x\) be the amount of students in the class.
\(40\%\) of \(x =\) girls
\(25\%\) of girls = tennis \space player
Input \(40\%\) of a class instead of girls in second formula. Therefore, \(25\%\) of \(40\%\) of a class = tennis player
tennis player \(= 10\%\)

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