Victory for the ACT Student Text 15e

M ATH T EST M ECHANICS • 193

ANATOMY

DIRECTIONS: Solve each item and choose the correct answer choice. Calculator use is permitted; however, some items are best solved without the use of a calculator. NOTE: All of the following should be assumed, unless otherwise stated. ͳǤ ŽŽ—•–”ƒ–‹˜‡ ϐ‹‰—”‡• ƒ”‡ ‡…‡••ƒ”‹Ž› drawn to scale. 2. The word average indicates arithmetic mean. 3. The word line indicates a straight line. ͶǤ ‡‘‡–”‹…ϐ‹‰—”‡• Ž‹‡ ‹ ƒ ’Žƒ‡Ǥ Paulo bicycled 6 3 1 miles on Wednesday 2 miles on Thursday. How many miles did he bicycle during those two days? A. 14 15 2 B. 14 4 1 C. 14 8 3 D. 14 15 11 E. 14 15 13 2. The average weight of 5 packages in a shipment is 8.7 pounds. The weights of 4 of the packages are 6.3 pounds, 7.5 pounds, 8.9 pounds, and 9.6 pounds. Šƒ– ‹• –Š‡ ™‡‹‰Š– ‘ˆ –Š‡ ϐ‹ˆ–Š ’ƒ…ƒ‰‡ǫ F. 8.5 G. 8.8 H. 9.4 J. 11.2 K. 12.3 1. and 8 5

The Cambridge Edge For most Mathematics items, answer choices are arranged in order from least to greatest or vice versa. This answer choice arrangement makes it easier for you to ϐ‹† ›‘—”…Š‘‹…‡ ‹ the list. It also sets up an important test-taking strategy of starting with the middle choice when applying the “test- the-test” strategy, which you’ll learn about later. The Cambridge Edge Answer choices are carefully constructed. The choices on the ACT test usually correspond to errors in thinking—not errors in arithmetic. You are not likely to ϐ‹† ƒ ’”‘„Ž‡™‹–Š answer choices like the following: A. $183.27 B. $183.28 C. $183.29 D. $183.30 E. $183.31 This feature is important because it is the basis for a couple of time- saving strategies that you’ll learn shortly.

You really do not need the directions at all. You can use a calculator, and there will be further discussion of this in class. Ž•‘ǡ ϔ‹‰—”‡• ƒ”‡ ‘– ‡…‡••ƒ”‹Ž› †”ƒ™ –‘ •…ƒŽ‡Ǥ ‘™‡˜‡”ǡ ‹ ’”ƒ…–‹…‡ǡ –Š‡ ϔ‹‰—”‡• ƒ”‡ ƒŽ‘•– always drawn to scale. Again, this issue will be discussed in detail later during the Math Lesson.

1. (D) This item is solved by addition : 6 3 1 8 5 2 3 19 5 42 + = +

95 = +

126

15

15

221

=

15

11

=

14

15

2. (J) Š‹• ‹–‡ ‹• ƒ „‹– ‘”‡ †‹ˆϔ‹…—Ž–Ǥ ‹”•–ǡ •‡– up an equation for the average:

x

. 5 6 3 7 5 8 9 9 6 + + + + . . .

=

.8 7

Then, solve for the missing weight:

. + + + + =

x

( . 6 3 7 5 8 9 9 6 . . )

( . ) ( ) 8 7 5

. 43 5 32 3 11 2 = − = . .

x