ACT Sample Test #1 Problems 1 to 30

1. The weights of quintuplets at birth was:

1 pound, 8 ounces

2 pound, 2 ounces

2 pound, 4 ounces

1 pound, 10 ounces

1 pound, 14 ounces

If there are 16 ounces in a pound, what is the average weight of the babies?

A. 1 pound, 4 ounces

B. 1 pound, 8 ounces

C. 1 pound, 14 ounces

D. 2 pounds, 2 ounces

E. 2 pounds, 6 ounces

1. Answer: C

The total weight of the quintuplets was 7 pounds, 38 ounces. If the problem is changed to ounces, the total weight is (7x16)+38=150 ounces. Divide by 5 to find the average: 150/5=30 ounces = 1 pound, 14 ounces.

2. A cafe offers a buffet with a meat dish, a vegetable, a bread, and a dessert. If there are five choices of a meat dish, four choices of a vegetable, three choices of a bread, and two choices of a dessert, how many different meals is it possible to order?

F. 5

G. 2⁴

H. 4⁴

I. 4·3·2·1

J. 5·4·3·2

2. Answer: J

Use the multiplication principal of counting. If there are 5 different choices of meats and four choices of a vegetable, there are 5x4 choices of a meat/vegetable combination. And if each of the meat/vegetable combinations can be mixed with a bread, there are (5x4)x3 of the meat/vegetable/bread combinations. Lastly, if there a two different desserts and each meat/vegetable/bread combination is tried with a different dessert, then there are [(5x4)x3]x2 meat/vegetable/bread/dessert combinations.

3. Given two points, (a, b) and (2, -3), which of the following represents the slope of the straight line containing these two points?

A. b-(-3) / a-2

B. a-b / 2-(-3)

C. a-2 / b-(-3)

D. a+2 / b+(-3)

E. b-2/ a-(-3)

3. Answer: A

The values of y are b and -3, and the values of x are a and 2. Thus, starting at point (a, b) the slope of a line is the

(change in y)/(change in x) = (b-(-3))/(a-2).

4. Find the value of x if

5x + 3y = 19

3x + 5y = 5

F. -2

G. 1

H. 3

I. 5

J. 8

4. Answer: I

Solve for x by multiplying the first equation by 5 and the second equation by negative 3.

5x + 3y = 19 Þ 25x + 15y = 95

3x + 5y = 5 Þ -9x - 15y = -15

Adding the two equations together yields

16x = 80

Þx = 5

5. ABCD is a square and P, Q, R, and S are the midpoints of the sides. If the side of the square is 4 inches, what is the area of the shaded region?

A. 4

B. 6

C. 8

D. 10

E. 12

5. Answer: E

Partition the shaded area into equal triangles. Since the area of the square is 16 (s²=4²), each triangle is 2 square inches and 6x2=12.

Note: To picture the area of one of the triangles examine CRQ: A=(1/2)bh=(1/2)(2)(2) = 2.

6. Sara's insurance company raised her auto rates by 12%. If she had been paying $240 per month before, how much is she paying now?

F. $28.80

G. $252.00

H. $268.80

I. $288.00

J. $360

6. Answer: H

If her rates were raised by 12% then the increase is 12% of $240

= .12 x 240

= $28.80

Added to her original premium, her new premium is now $268.80.

7. Three concentric circles having radii of 1, 2, and 3 units are pictured below. Region A is the inside circle, Region B is the ring between the smaller circle and the middle circle, and Region C is the ring between the middle circle and the larger circle. If a bug lands on one of the regions, what is the probability that it will land on Region C?

A. 1/3

B. 4/9

C. 1/2

D. 5/9

E. 2/3

7. Answer: D

The total area of regions A and B can be found by finding the area of the middle circle using A=πr² = π 2² or 4π . The area of the large circle is π 3² or 9π. Therefore, the area of the shaded region is 9π-4π=5π. The probability that the bug will land on the shaded area is (5π)/(9π) or 5/9.

8. If there are 16 ounces to a pound, how many pounds, to the nearest tenth of a pound, are there in 485 ounces?

F. 30.3

G. 30.5

H. 33.0

I. 77.6

J. 7760

8. Answer: F

Divide 485 by 16.

9. In the figure below AB = 4, CD = 9, and AF = 12. If all angles are right angles, what is the perimeter of the figure?

A. 25

B. 37

C. 50

D. 432

E. not enough information

9. Answer: C

Note that AB + CD = FE and AF = BC + DE. Thus the perimeter is found by adding the sum of the lengths of the sides.

A. Only the numbers in the pictures were added.

B. Partial answer

C. Correct

D. 4x9x12

E. It helps to redraw the picture.

10. The average of the interior angles of a polygon is 90 degrees. How many sides does the polygon have?

F. 3

G. 4

H. 5

I. 6

J. 8

10. Answer: G

A quadrilateral has 360 degrees. Thus, the average number of degrees in each interior angle is 90 degrees.

11. Simplify the expression

A. 1/12

B. 7/12

C. 42/12

D. 7/60

E. 7/72

11. Answer: E

The numerator is

1/4 + 1/3 = 3/12 + 4/12 = 7/12

Now 7/12 ¸ 6 = 7/12 x 1/6 = 7/72.

12. Find the radius of a circle that has a circumference of 54p.

F. 3Ö3

G. 27

H. 3pÖ3

I. 3Ö6

J. 27p

12. Answer: G

The formula for the circumference of the circle is C = 2pr. Substituting for C we have 54p = 2pr. Thus r = 54p/2p = 27.

13. Let Line 1 and Line 2 be parallel, and angles with degree measure (z+40) and (z+80) as shown. Find z.

A. -15

B. 15

C. 30

D. 45

E. 60

13. Answer: C

Since Line 1 is parallel to Line 2 (z+80) and (z+40) are supplementary angles. Therefore

(z+40)° + (z+80)° = 180°

2z + 120° = 180°

2z = 60°

z = 30°

14. What is the slope of the equation

15x - 3y + 8 = 0?

F. -15

G. -5

H. 5/3

I. 8/3

J. 5

14. Answer: J

Solution:

15x - 3y + 8 = 0

-3y = -15x - 8

3y = 15x + 8

y = 5x + (8/3)

The slope/intercept form of a line is y=mx+b where m is the slope. In this case the slope is 5.

15. If the length of a rectangle is 6 times the width and if the area is 54, what is the length of the rectangle?

A. 2

B. 3

C. 4

D. 9

E. 18

15. Answer: E

Let w = the width of the rectangle

Then 6w = the length

Using the formula for the area of a rectangle, A=lw, we have

54 = 6w · w

9 = w · w

3 = w

therefore, l = 6w = 6 · 3 = 18

16. If sin(a) = 3/4 and tan(a) = (3Ö7)/7, then cos(a) = ?

F. 3/7

G. 4/7

H. (Ö7)/12

I. (Ö7)/4

J. (4Ö7)/3

16. Answer: I

Since sin(a)/cos(a) = tan(a)

cos(a) = sin(a) / tan(a)

= (3/4)¸(3Ö7)/7)

= 3/4 x 7/(3Ö7)

= 7/(4Ö7)

= (7×Ö7)/(4×Ö7×Ö7)

= (7×Ö7)/28

= (Ö7)/4

17. If the length of side RQ is 20 meters and angle PRQ = 60 degrees then the length of side RP is

A. 5 meters

B. 10 meters

C. 20 meters

D. 5√3 meters

E. 10√3 meters

17. Answer: B

Solution: This is a 30-60-90 right triangle with sides in proportion 1: √3:2 where 1 is the side opposite the 30 degree angle, √3 is the side opposite the 60 degree angle, and 2 is the hypotenuse. Since the hypotenuse in the problem is 20, the sides are in the ratio of

10 : 10√3 : 20.

18. Find the slope of a line that passes through the points (-3,2) and (5,6).

F. -3

G. 1/5

H. 1/2

I. 2

J. 5

18. Answer: H

Solution:

Slope = rise/run = (y₂-y₁)/(x₂-x₁)

= (6-2)/(5- (-3))

= 4/8 = 1/2

19. |-5 · 2| · |-2| =

A. -20

B. -9

C. 9

D. 20

E. 50

19. Answer: D

|-5 · 2| · |-2|

= |-10| · |-2|

= 10 · 2

= 20

20. Find the sum of 6x² y²z + 3xy and 6xy² + 3x²y²z

F. 9x² y²z + 9xy²

G. 12x² y²z + 6x²y²z

H. 9x² y²z + 3xy + 6xy²

I. 12x² y²z + 3xy + 6xy²

J. 18x² y²z

20. Answer: H

Group like terms. Remember that like terms have the same variables and the same exponents.

(6x² y²z + 3xy) + (6xy² + 3x²y²z)

= (6x² y²z + 3x²y²z) +3xy + 6xy²

= 9x² y²z + 3xy + 6xy²

21. If 3x - 18 = 3 then what is the value of x² + x - 5x?

A. -91

B. 7

C. 21

D. 56

E. 91

21. Answer: C

Solve and plug in.

3x - 18 = 3

3x -18 +18 = 3 + 18

3x = 21

x = 7

x² + x - 5x

= 7² + 7 - 5(7)

= 56 - 35

= 21

22. If the measure of each interior angle of a regular polygon is 120°, what kind of polygon is it?

F. Triangle

G. Square

H. Pentagon

I. Hexagon

J. Octagon

22. Answer: H

The number of degrees in the interior angles of a polygon with n sides is 180(n-2). In this problem let n be the number of sides of the polygon (which is also the number of vertices). Then the number of degrees in the interior angles is 120n. Now you have an equivalent relationship:

180(n-2) = 120n

180n - 360 =120n

60n = 360

n = 6

If there are six sides you have a hexagon.

Extra: Test the cases.

Polygon	Number of sides	Number of degrees in interior angles	Average number degrees each interior angle
Triangle	3	180	180/3 = 60
Square	4	180(4-2) = 360	360/4 = 90
Pentagon	5	180(5-2) = 540	540/5 = 108
Hexagon	6	180(6-2) = 720	720/6 = 120
Octagon	8	180(8-2) = 1080	1080/8 = 135

23. Let t, u, v ≠ 0. Then

23. Answer: A

Solution: Treat as three separate problems.

t⁴/t⁹ = t^4-9 = t^-5 = 1/t⁵

u²/u^-2 = u^2-(-2) = u⁴

v/v^-9 = v¹/v^-9 =v^1-(-9) = v¹⁰

24. Which expression represents the area of the shaded region given that the larger circle has radius 4 and the smaller circle has radius 2.

F. π(4)² - 2²

G. π (4-2)²

H. π (4² - 2²)

I. 2π (4-2)

J. 2π (4-2)²

24. Answer: H

Area of larger circle: π r² = π 4²

Area of smaller circle: π r² = π 2²

Difference: π 4² - π2² = π (4² - 2²)

25. 4 1/3 - 8 5/6 =

A. - 10 1/3

B. -5 1/3

C. -4 1/2

D. -3 5/6

E. 5 1/3

25. Answer: C

Subtract:

- 8 5/6 + 4 1/3

= -(8 5/6 - 4 1/3)

= - (8 5/6 - 4 2/6)

= - 4 3/6

= - 4 1/2

26. The decimal expansion of 152/1111 is 0.02430243… where the digits 0243 repeat. The 45th digit to the right of the decimal point is

F. 0

G. 1

H. 2

I. 3

J. 4

26. Answer: F

The decimal expansion of 152/1111 is 0.02430243… where the digits 0243 repeat. Therefore, 3 is the 4th, 8th, 12th, … 4x digit, where x is a positive integer. Also, 0 is the 1st, 5th, 9th, 13th, and the 4x+1 digit. Since 45 is equal to 4·11+1, the 45th digit is 0.

Shortcut: Since the pattern consists of four numbers, divide 45 by 11. The remainder signals where in the pattern the 45th term will fall.

45¸4 = 11, remainder 1 Þ the 45th digit is in the first place in the pattern.

27. = ?

A.

B.

C.

D.

E.

27. Answer: C

= =

Note that the -3 and the -2 in the numerator and the denominator cancel.

28. There are yellow, red, and white marbles in a jar. If 1/4 are yellow, 2/5 are red, and 21 are white, how many marbles are in the jar?

F. 7

G. 39

H. 60

I. 70

J. 210

28. Answer: H

Let y represent the fraction of white marbles.

Then 1/4 + 2/5 + y = 1

Þ 5/20 + 8/20 + y = 20/20

Þy = 7/20

Let t represent the total number of marbles. Then, since there are 21 white marbles,

(7/20)t = 21

Þt = (20/7)x21

Þ t = 60

29. For all x ≠ -5,

A. x + 4

B. x - 5

C. x - 9

D. 2x + 5

E. - (⁴/₅)x - 9

29. Answer: C

Factor x²- 4x - 45 to get (x - 9)(x + 5). Therefore,

Now cancel the (x+5) terms so that the expression is simplified to (x - 9).

Shortcut: Rather than factoring test the answer choices to see which one is a factor.

30. The area of a triangle with sides of length x, y, and z is 12 units. What is the area of a triangle with sides of length 2x, 2y, and 2z?

F. (√3/4)12

G. 18

H. 24

I. 48

J. 144

30. Answer: I

If the dimensions of a polygon are doubled the area is quadrupled. See, for example, the figure below. Let H represent the height of triangle ABC. We won't present a formal proof here* but triangle ABC is similar to triangle A'BC', triangle ABP is similar to triangle A'BP' and the height in triangle A'BC' is double the height of triangle ABC. Therefore

Area of ABC = (1/2) bh = (1/2)zH

Area of A'BC' = (1/2)bh = (1/2)(2z)(2H)

= 4(1/2)zH = 4 (Area of ABC) = 4(12) = 48.

Also, the midpoints of a triangle form four similar triangles. This makes one triangle equal to one-fourth of the total area.

*For a more formal proof see Heron's formula for the area of a triangle in the math review.