Top 10 7th Grade MAP Math Practice Questions
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ts. William and his sister Ella ordered two pizzas. William ate \(\frac{1}{4}\) of his pizza and Ella ate \(\frac{1}{2}\) of her pizza. What part of the two pizzas was left?
A. \(\frac{1}{2}\)
B. \(\frac{1}{3}\)
C. \(\frac{3}{8}\)
D. \(\frac{5}{8}\)
3- Simplify \(6x^2 y^3 (2x^2 y)^3\).
A. \( 12x^4 y^6\)
B. \( 12x^8 y^6\)
C. \( 48x^4 y^6\)
D. \( 48x^8 y^6\)
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 dollars?
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?
A. \(10 \%\)
B. \(15 \%\)
C. \(20 \%\)
D. \(40 \%\)
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Answers:
1- C
In this case, to find the area of the shaded region, subtract the area of two circles. (S1: the area of the big circle. S2: the area of a little circle)
Use the area of a circle formula.
S = \(π\)r\(^2\)
\(S_{1} – S_{2}= π( 5 + 3 cm)^2 – π(5 cm)^2\) ⇒ \(S_{1} – S_{2} = π64 cm^2 – π25 cm^2\) ⇒ \(S_{1} – S_{2} = 39π cm^2\)
2- D
William ate \(\frac{1}{4}\) of 8 parts of his pizza which means 2 parts out of 8 parts ( \(\frac{1}{4}\) of 8 parts = x ⇒ x = 2) and left 6 parts.
Ella ate \(\frac{1}{2}\) of 8 parts of her pizza which means 4 parts out of 8 parts ( \(\frac{1}{2}\) of 8 parts = x ⇒ x = 4) and left 4 parts.
Therefore, they ate \((4 + 2)\) parts out of\( (8 + 8)\) parts of their pizza and left \((6 + 4)\) parts out of \((8 + 8)\) parts of their pizza that it means: \(\frac{10}{16}\)
After simplification we have: \(\frac{5}{8}\)
3- D
Simplify.
\(6x^2 y^3 (2x^2 y)^3= 6x^2 y^3 (8x^6 y^3 ) = 48x^8 y^6\)
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 \(10 \%\):
\((0.80 D) (100\% + 10\%) = (0.85 D) (1.10) = 0.88 D = 88\% D\)
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 =\( πr^2 ⇒ 64 π = πr^2 ⇒ 64 = r^2 ⇒ r = 8\)
The 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
\(\frac{New Value-Old Value}{Old Value}× 100 \%\)
\(\frac{28-40}{40}× 100 \% = – 30 \% \)
(The negative sign here means that the new price is less than the old price).
9- C
The question is this: 1.75 is what percent of 1.25?
Use the 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 number of students in the class.
\(40 \%\) of x = girls
\(25 \%\) of girls = tennis player
Input \(40\%\) of a class instead of girls in the second formula. Therefore, \(25\%\) of \(40\%\) of a class = tennis player
tennis player = \(10\%\)
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