Chapter 14 Practice Test 4 With Answers - Mcgraw-Hill'S Psat/nmsqt Page 42
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45514
M
GRAW-HILL’S PSAT/NMSQT
C
s could be 3 or 4
to be covered. The floor is 12 Â 15 ¼ 180 square feet.
r ¼ 3
4(3) þ 5s , 30
Each tile is made up of a 6- by 6-inch square attached
Subtract 12:
5s , 18
to a 458-458-908 triangle. Convert the inches into feet.
1 foot
6 inches Â
¼ 0:5 foot
12 inches
Since s must be greater than r, there are no values
possible when r . 2.
The tile consists of a 0.5-ft by 0.5-ft square that has an
Therefore there are 6 possible integer pairs (r, s).
area of (0.5)(0.5) ¼ 0.25 square foot and a triangle
(Chapter
10
Lesson
3:
Numerical
Reasoning
1
with an area of
(0.5 ft)(0.5 ft) ¼ 0.125 square foot.
Problems)
2
The total area of the tile is 0.25 þ 0.125 ¼ 0.375
square foot. To find out how many tiles you need,
divide the total area of the floor by the area of each tile.
31. 1089 When solving this problem, remember
180 4 0.375 ¼ 480 tiles
that you cannot repeat digits.
(Chapter 11 Lesson 5: Areas and Perimeters)
The smallest possible three-digit integer: 102
The largest possible three-digit integer: 987
987 þ 102 ¼ 1,089
35. 25
If you have the time, you could solve this
Don’t forget that 0 is a digit, but it can’t be the first
problem by actually adding up all of the even integers
digit of a three-digit integer.
less than or equal to 50 and all of the odd integers less
(Chapter
10
Lesson
3:
Numerical
Reasoning
than 50 and then subtracting the two sums. But it is
Problems)
simpler to write out the equation as it is written
below. In doing so you will find that there are 25
even numbers and 25 odd numbers, and each even
32. 120
number is 1 more than one of the odd numbers
x ft
14 in
(e.g., 2 is 1 more than 1, 4 is 1 more than 3, 6 is 1
more than 5, etc.). Therefore the sum of the even
numbers is 25(1) ¼ 25 more than the sum of the
25 ft
10 in
odd numbers.
2 þ 4 þ 6þ8 þÁ Á Á þ 44 þ 46 þ 48 þ 50
21 þ 3 þ 5þ7 þÁ Á Á þ 43 þ 45 þ 47 þ 49
1 þ 1 þ 1þ1 þÁ Á Á þ 1 þ 1 þ 1 þ 1 ¼ 25
(Chapter
10
Lesson
3:
Numerical
Reasoning
The shorter side of the garden measures 25 feet in
length and is represented on the map as 10 inches.
Problems)
The longer side of the garden would be represented
by the 14-inch side on the map.
36. 4/27
Each student can write either 1, 2, or 3 on
10 inches
14 inches
the card. Using the fundamental counting principle,
Set up a proportion:
¼
25 feet
x feet
you can calculate how many different permutations
of cards are possible:
3 Â 3 Â 3 ¼ 27
Cross-multiply:
10x ¼ 350
Next, find out how many permutations would
Divide by 10:
x ¼ 35
provide a sum greater than 7. To get a sum larger
than 7, the cards must be either 3, 3, and 3 or 3, 3,
To find the perimeter, add up all the sides:
and 2. There is only one way to get 3, 3, and 3—all
35 þ 25 þ 35 þ 25 ¼ 120
three students write a 3 on the card. There are three
(Chapter 11 Lesson 5: Areas and Perimeters)
ways to get the 3, 3, 2 permutation. Student 1 could
(Chapter 8 Lesson 4: Ratios and Proportions)
write the two, student 2 could write the two, or
student 3 could write the two. There are four ways
to get a sum larger than 7, or 4 out of 27 possible per-
33. 200
The ratio is a “ratio of parts” with a sum of
mutations.
1 þ 2 þ 4 þ 6 ¼ 13. The second largest part, then, is
(Chapter 10 Lesson 5: Counting Problems)
4/13 of the whole. 4/13 of $650 ¼ $200
(Chapter 8 Lesson 4: Ratios and Proportions)
37. 9
The volume of the cube is 216, so each edge
3
34. 480
To answer this question you can find the
has length 6.
Volume ¼ x
¼ 216
area of each tile and then the area of the entire floor
Take cube root:
x ¼
6
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