ACT Math Practice

ACT Math Practice

9th - 12th Grade

30 Qs

quiz-placeholder

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ACT Math Practice

ACT Math Practice

Assessment

Quiz

Mathematics

9th - 12th Grade

Hard

CCSS
HSG.GPE.B.6, 5.NBT.B.7, 6.EE.A.2C

+24

Standards-aligned

Created by

Andrea McFall

Used 4+ times

FREE Resource

30 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

A car averages 27 miles per gallon. If gas costs $4.04 per gallon, which of the following is closest to how much the gas would cost for this car to travel 2,727 typical miles?

$44.44

$109.08

$118.80

$408.04

$444.40

Answer explanation

If you divide 2,727 miles by 27 miles per gallon you will get the number of gallons: = 101. Then, multiply the number of gallons by the cost per gallon: 101(4.04) = 408.04. This gives the cost of gas for this car to travel 2,727 typical miles.

Tags

CCSS.5.NBT.B.7

2.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

When x = 3 and y = 5, by how much does the value of 3x2 – 2exceed the value of 2x2 – 3y ?

4

14

16

20

50

Answer explanation

14 is the correct answer. When you use x = 3 and y = 5 in the given expressions, 3x2 – 2y = 3(3)2 – 2(5) = 27 – 10 = 17 and 2x2 – 3y = 2(3)2 – 3(5) = 18 – 15 = 3. Then subtract 3 from 17 to get 14.

Tags

CCSS.6.EE.A.2C

3.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

What is the value of x when 2x + 3 = 3x – 4 ?

-7

-1/5

1

1/5

7

Answer explanation

You can solve this problem by first subtracting 2x from each side of the equation to get 3 = x – 4. Then add 4 to each side, so x = 7.

4.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

What is the greatest common factor of 42, 126, and 210 ?

2

6

14

21

42

Answer explanation

42 is the correct answer since it is the largest number that is a factor of all three numbers given. You can find the greatest common factor by writing out the prime factorization of all three numbers and then taking each of the common prime factors to the lowest power that appears for that factor: 42 = 2 × 3 × 7; 126 = 2 × 32 × 7; and 210 = 2 × 3 × 5 × 7. So the greatest common factor is 2 × 3 × 7 = 42.

5.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Sales for a business were 3 million dollars more the second year than the first, and sales for the third year were double the sales for the second year. If sales for the third year were 38 million dollars, what were sales, in millions of dollars, for the first year?

16

17.5

20.5

22

35

Answer explanation

If x = sales for the first year, then x + 3 = sales for the second year. Since sales for the third year were double the sales for the second year, sales for the third year = 2(x + 3). Sales for the third year were 38, so 2(x + 3) = 38. To solve this equation, you could first divide each side by 2 to get x + 3 = 19. Then, by subtracting 3 from both sides, x = 16.

Tags

CCSS.7.EE.B.4A

6.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Media Image

In the figure below, ray EF was constructed starting from rays ED and EG. By using a compass D and G were marked equidistant from E on rays ED and EG. The compass was then used to locate a point F, distinct from E, so that F is equidistant from D and G. For all constructions defined by the above steps, the measures of ∠DEF and ∠GEF:

are equal

are NOT equal

sum to 30 degrees

sum to 45 degrees

sum to 60 degrees

Answer explanation

If you draw line segments DF and FG, you can show ΔDEF ≅ ΔGEF by SSS (side-side-side congruence). So, ∠DEF ≅ ∠GEF because corresponding parts of congruent triangles are congruent.

Tags

CCSS.HSG.CO.C.9

7.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Abandoned mines frequently fill with water. Before an abandoned mine can be reopened, the water must be pumped out. The size of pump required depends on the depth of the mine. If pumping out a mine that is D feet deep requires a pump that pumps a minimum of  D2/25 + 4D – 250 gallons per minute, pumping out a mine that is 150 feet deep would require a pump that pumps a minimum of how many gallons per minute?

362

500

800

1,250

1,750

Answer explanation

If you substitute D with 150 in the expression, you get 150^2/25+ 4(150) – 250 = 22,500/25 + 600 – 250 = 1,250.

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