FREE 7th Grade STAAR Math Practice Test

A. \(I > 2000 x + 24000\)

B. \(I > -2000 x + 24000\)

C. \(I < -2000 x + 24000\)

D. \(I < 2000 x – 24000\)

9- Which of the following graphs represents the compound inequality?

A. graph 1

B. graph 2

C. graph 3

D. graph 4

10- A football team had $20,000 to spend on supplies. The team spent $14,000 on new balls. New sports shoes cost $120 each. Which of the following inequalities represents how many new shoes the team can purchase?

A. \( 120x+14,000 ≤20,000 \)

B. \(120x+14,000 ≥20,000\)

C. \(14,000x+12,0 ≤20,000 \)

D. \( 14,000x+12,0 ≥20,000 \)

11- Two dice are thrown simultaneously, what is the probability of getting a sum of 6 or 9?

A. \(\frac{1}{3}\)

B. \(\frac{1}{4}\)

C. \(\frac{1}{6}\)

D. \(\frac{11}{36}\)

12- A swimming pool holds 2,000 cubic feet of water. The swimming pool is 25 feet long and 10 feet wide. How deep is the swimming pool? __________

13- Which graph corresponds to the following inequalities?
\(y≤ x + 4\)
\(2x + y ≤ – 4\)

A.

B.

C.

D.

14- A bank is offering \(4.5\%\) simple interest on a savings account. If you deposit $8,000, how much interest will you earn in five years?

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A. $360

B. $720

C. $1800

D. $3600

15- A card is drawn at random from a standard 52–card deck, what is the probability that the card is of Hearts? (The deck includes 13 of each suit clubs, diamonds, hearts, and spades)

A. \(\frac{1}{3}\)

B. \(\frac{1}{4}\)

C. \(\frac{1}{6}\)

D. \(\frac{1}{52}\)

16- How long does a 420–mile trip take moving at 50 miles per hour (mph)?

A. 4 hours

B. 6 hours and 24 minutes

C. 8 hours and 24 minutes

D. 8 hours and 30 minutes

17- 11 yards 6 feet and 4 inches equal to how many inches?

A. 388

B. 468

C. 472

D. 476

18- A shirt costing $200 is discounted \(15\%\). After a month, the shirt is discounted another \(15\%\). Which of the following expressions can be used to find the selling price of the shirt?

A. \((200) (0.70)\)

B. \((200) – 200 (0.30)\)

C. \((200) (0.15) – (200) (0.15)\)

D. \((200) (0.85) (0.85)\)

19- Which of the following points lies on the line \(2x + 4y = 10\)

A. \((2, 1)\)

B. \((–1, 3)\)

C. \((–2, 2)\)

D. \((2, 2)\)

20- \(5 + 8 × (–2) – [4 + 22 ×5] ÷ 6 = \)?

A. \(-30\)

B. \(-20\)

C. \(-10\)

D. 0

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Answers:

2- C
Simplify: \(5(x-2y)+(2-x)^2 = (5x-10y)+(4-4x+x^2) = x -10y +4 +x^2\)
When \(x=3\) and \(y=-2\),therefore:
\(x -10y +4 +x^2 =3+20+4+9 =36\)

3- C
\(average (mean) = \frac{sum \space of \space terms }{number \space of \space terms}⇒ 88 = \frac{sum \space of \space terms}{50}⇒ sum = 88 × 50 = 4400\)
The difference of 94 and 69 is 25. Therefore, 25 should be subtracted from the sum.
\(4400 – 25 = 4375\)
\(mean =\frac{sum \space of \space terms}{number \space of \space terms}⇒ mean = \frac{4375 }{50}= 87.5\)

4- D
If the length of the box is 27, then the box’s width is one-third of it, 9, and the height of the box is 3 (one-third of the width). The volume of the box is:
\(V = l.w.h = (27) (9) (3) = 729\)

5- C
Add the first 5 numbers. \(40 + 45 + 50 + 35 + 55 = 225\)
To find the distance traveled in the next 5 hours, multiply the average by number of hours.
\(Distance = Average × Rate = 50 × 5 = 250\)
Add both numbers.
\(250 + 225 = 475\)

6- 240
The ratio of boys to girls is 2:3. Therefore, there are 2 boys out of 5 students. To find the answer, first, divide the total number of students by 5, then multiply the result by 2.
\(600 ÷ 5 = 120 ⇒ 120 × 2 = 240\)

7- 130
The perimeter of the trapezoid is 54 cm.
Therefore, the missing side (high) is \(= 54 – 18 – 12 – 14 = 10\)
Area of a trapezoid:
\( A = \frac{1}{2}h (b_1 + b_2) = \frac{1}{2}(10) (12 + 14) = 130\)

8- A
Let \(x\) be the number of years. Therefore, $2,000 per year equals 2000\(x\).
starting from $24,000 annual salary means you should add that amount to 2000\(x\).
Income more than that is:
\(I > 2000x + 24000\)

9- D
Solve for \(x\).
\(-2≤2x-4<8 \)⇒ (add 4 all sides)\( -2+4≤2x-4+4<8+4 \)
\(⇒ 2≤2x<12 \)
⇒ (divide all sides by 2)\( 1≤x<6\)
\(x\) is between 1 and 6.

10- A
Let \(x\) be the number of new shoes the team can purchase. Therefore, the team can purchase 120 \(x\).
The team had $20,000 and spent $14000. Now the team can spend on new shoes $6000 at most.
Now, write the inequality:
\(120x+14.000 ≤20.000\)

11- B
For Sum 6: (1 & 5) and (5 & 1), (2 & 4) and (4 & 2), (3 & 3), so we have 5 options.
For sum 9: (3 & 6) and (6 & 3), (4 & 5) and (5 & 4), we have 4 options.
To get a sum of 6 or 9 for two dice: \(5+4=9\)
Since we have \(6 × 6 = 36\) total options, the probability of getting a sum of 6 and 9 is 9 out of 36 or \(\frac{1}{4}\).

12- 8
Use formula of rectangle prism volume.
\(V = (length) (width) (height) ⇒ 2000 = (25) (10) (height)\)
\( ⇒ height = 2000 ÷ 250 = 8\)

13- A
For each option, choose a point in the solution part and check it on both inequalities.
A. Point \((–4, –4)\) is in the solution section. Let’s check the point in both inequalities.
\(–4 ≤ – 4 + 4, \space It \space works\)
\(2 (–4) + (–4) ≤ –4 ⇒ – 12 ≤ – 4\) it works (this point works in both)
B. Let’s choose this point \((0, 0)\)
\(0 ≤ 0 + 4, \space It \space works\)
\(2 (0) + (0) ≤ –4, \space \space That’s \space not \space true!\)
C. Let’s choose this point \((–5, 0)\)
\(0 ≤ –5 + 4, \space That’s \space not \space true!\)
D. Let’s choose this point \((0, 5)\)
\(5 ≤ 0 + 4, \space That’s \space not \space true!\)

14- C
Use simple interest formula:
I=prt
(I = interest, p = principal, r = rate, t = time)
\(I=(8000)(0.045)(5)=1800\)

15- B
The probability of choosing a Hearts is \(\frac{13}{52}=\frac{1}{4}\)

16- C
Use distance formula:
\(Distance = Rate × time ⇒ 420 = 50 × T\)
divide both sides by 50.
\(\frac{420}{50} = T ⇒ T = 8.4 \space hours\)
Change hours to minutes for the decimal part.
\(0.4 \space hours = 0.4 × 60 = 24 \space minutes\)

17- C
\(11 × 36 + 6 × 12 + 4 = 472\)

18- D
To find the discount, multiply the number by (\(100\% – \)rate of discount).
Therefore, for the first discount we get:
\((200) (100\% – 15\%) = (200) (0.85) = 170\)
For the next \(15%\) discount: \((200) (0.85) (0.85)\)

19- B
Input \((-1, 3)\) in the \(2x + 4y = 10\) formula instead of \(x\) and y. So we have:
\( 2(-1) + 4(3) = 10\)
\(-2 + 12 = 10\)

20- A
Use PEMDAS (order of operation):
\(5 + 8 × (–2) – [4 + 22 ×5] ÷ 6 = 5 + 8 × (–2) – [4 + 110] ÷ 6 = 5 + 8 × (–2) – [114] ÷ 6 = 5 + (–16) – 19 = 5 + (–16) – 19 = –11 – 19 = –30\)

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