Download 1. 2. 3. Quiz Questions There are 60 students going

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Common Core Standards : A-CED.1
Quiz
Quiz Questions
1. There are 60 students going on a field trip to the chocolate factory. The students are
from three different classes. Mrs. Hooper's class has 24 students and Mr. Gomez's class
has 18 students. Which of the equalities correctly describes the students and could be
used to solve for how many students are from Mr. Anderson's class? (Let A = the number
of students in Mr. Anderson's class.)
(A)
(B)
(C)
(D)
A + 18 = 24
A + A + A = 60
60 – 18 = A – 24
24 + 18 + A = 60
Explanation:
The relation Hooper's class + Gomez's class + Anderson's class = 60 students going on a field
trip becomes an equation by changing the written descriptions into numbers and variables. Mrs.
Hooper's class has 24 students and Mr. Gomez's class has 18 students giving 24 + 18 + Anderson's
class = 60 students going on a field trip. The number of students in Anderson's class is the unknown
and must be represented by a variable likeA for Anderson. That means 24 + 18 + A = 60.
2. There are 60 students going on a field trip to the chocolate factory. The students are
from three different classes. Mrs. Hooper's class has 24 students and Mr. Gomez's class
has 18 students. How many students are from Mr. Anderson's class?
(A) 16 students
(B) 18 students
(C) 20 students
(D) 22 students
Explanation:
The equation 24 + 18 + A = 60 (where A = students in Anderson's class) needs to be solved to
isolate A. First, simplify 24 + 18 to 42 to get 42 + A = 60. Then, subtract 42 from both sides to
get A = 60 – 42. This gives A = 18. There are 18 students from Anderson's class.
3. There are six chaperones going on a field trip. There are two buses for the trip. The
chaperones divide so there is the same number of chaperones on each bus. Which of the
equations could be utilized to find the number of chaperones on the first bus? (Let c =
number of chaperones on first bus.)
(A)
(B)
(C) 6 × 2 = c
(D) 6 × c = 2
Explanation:
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The relation
becomes an
equation by changing the written descriptions into numbers and variables. There are six chaperones
on the trip giving
buses, giving
. There are two
. The number of chaperones on the first
bus is an unknown represented by the variable c. This results in the equation
.
4. There are six chaperones going on a field trip. There are two buses for the trip. The
chaperones divide so there is the same number of chaperones on each bus. How many
chaperones are there on the first bus?
(A) 2
(B) 3
(C) 6
(D) 12
Explanation:
The equation = c (where c = number of chaperones on first bus) needs to be solved for the
variable c. This gives 3 = c or c = 3. There are 3 chaperones on the first bus.
5. A total of 66 people attended a field trip to a chocolate factory for a tour. A maximum of
15 people are allowed to tour at one time. Which equation correctly describes how many
tour groups to organize? (Let g = the number of groups.)
(A)
(B)
(C)
(D)
Explanation:
The relation
becomes an
equation by changing the written descriptions into numbers and variables. The 66 and 15 become
numbers directly giving
. The number of tour groups is an unknown so
we need to make a variable g = number of tour groups. The resulting equation is
.
6. A total of 66 people attended a field trip to a chocolate factory for a tour. A maximum of
15 people are allowed to tour at one time. What is the minimum number of tour groups that
can be formed?
(A) 4
(B) 5
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(C) 15
(D) 66
Explanation:
The equation
(where g = number of tour groups) needs to be solved for the variable g.
Multiplying each side by g to remove it from the denominator gives 66 ≤ 15 ×g. Dividing each side by
15 then gives
which becomes 4.4 ≤ g. Reversing this gives g ≥ 4.4. The actual number of
tour groups formed must be a whole number since there cannot be fractions of tours. The smallest
whole number that is greater than or equal to 4.4 is 5. A minimum of 5 tour groups must be formed.
7. A heart shaped chocolate box is composed of one square and two half circles. The
total number of chocolates in the box is calculated by adding the area of a square given
by 4x2 and the area of a circle approximated by 3x2. The company plans to add a small
additional box for a promotional campaign containing one row (2x) of chocolates. If the
total combined heart shape and small box contain 69 chocolates, which of these
equations could be utilized to solve for the number of chocolates in the small box (2x)?
(A) 4x2 + 3x2 + 2x = 69
(B) 4x2 – 3x2 + 2x = 69
(C) 4x2 + 3x2 – 2x = 69
(D) 4x2 – 3x2– 2x = 69
Explanation:
Begin with the relation: combine heart shape and small box = 69 chocolates total. The key words
"and, "adding," and "add" all indicate summing the three relations for numbers of chocolates. This
gives 4x2 + 3x2 + 2x. Combining these gives the equation 4x2 + 3x2 + 2x = 69.
8. A heart shaped chocolate box is composed of one square and two half circles. The
total number of chocolates in the box is calculated by adding the area of a square given
by 4x2 and the area of a circle approximated by 3x2. The company plans to add a small
additional box for a promotional campaign containing one row (2x) of chocolates. If the
total combined heart shape and small box contain 69 chocolates, how many chocolates
are in the small box (2x)?
(A)
(B)
(C)
(D)
3
4
5
6
Explanation:
The equation 4x2 + 3x2 + 2x = 69 can be solved for 2x, the number of chocolates in the small box.
This gives 7x2 + 2x – 69 = 0 which is factored to (7x + 23)(x –​
3) = 0. The solutions are
and x = 3. These give 2x = -6.572 and 2x = 6, respectively. The negative
solution does not make sense because you cannot have a negative number of chocolates in a box.
So, the correct solution must be 2x = 6. There are 6 chocolates in the small box.
9. On the day of the class field trip, the chocolate factory produced three times as many
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plain chocolate bars as crispy bars. They produced 50 more nutty bars than crispy bars.
The ratio of plain chocolate bars produced to nutty bars produced was 2 to 1. Which of the
equations below could be utilized to solve for the number of crispy bars produced on the
day of the field trip?
(A) 3c + 2c = 50
(B)
(C) 2 × 3c = 1 × (c + 50)
(D)
Explanation:
The key word "ratio" indicates division such that
. Then, you
need number of plain chocolate bars in terms of crispy bars, which is 3c from the key word "times."
And, you need number of nutty bars in terms of crispy bars, which is c + 50 from the key word
"more." Combining these gives the equation
.
10. On the day of the class field trip, the chocolate factory produced three times as many
plain chocolate bars as crispy bars. They produced 50 more nutty bars than crispy bars.
The ratio of plain chocolate bars produced to nutty bars produced is 2 to 1. How many
crispy bars were produced?
(A) 50
(B) 100
(C) 125
(D) 175
Explanation:
The equation
equates the 2 to 1 ratio with 3c (the number of plain chocolate bars) to c +
50 (the number of nutty bars). To solve for c, the number of crispy bars, first multiply both sides
by c + 50 to get 3c = 2c + 100. Then, subtract 2c from both sides to get c = 100. So, 100 crispy bars
were produced.
11. A large box of 144 chocolates has a width that is three times the height of the box and
a length that is twice the width of the boxes. Each chocolate rests in a cube that is 1 in × 1
in × 1 in. Which equation could be utilized to calculate the height of the box in inches?
(A) 2 × 6h + 2 × 3h + 2 × h = 144
(B) h × h × h = 144
(C) (2 × 3h) × 3h × h = 144
(D) 2 × 3h + 3h + h = 144
Explanation:
The initial relation for this problem requires knowing that the volume of the box will be length × width
× height. Since each chocolate has a volume of 1 in3, the total volume of the box with 144
chocolates will be 144 in3. So, the basic relationship is length × width × height = 144. We do not
know length, but we do know that it is twice the width. And, we know the width is three times the
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height. The key words "twice" and "times" both indicate multiplication so we have width = 3 × height =
3h and length = 2 × width = 2 × 3h. And the height is, of course, just h. So, the overall equation
becomes (2 × 3h) × 3h × h = 144.
12. A large box of 144 chocolates has a width that is three times the height of the box and
a length that is twice the width of the boxes. Each chocolate rests in a cube that is 1 in × 1
in × 1 in. What is the height of the box in inches?
(A) 1 in
(B) 2 in
(C) 6 in
(D) 12 in
Explanation:
Given the relation length × width × height = volume, the resulting equation is (2 × 3h) × 3h × h = 144.
This simplifies to 18h3 = 144. First, each side is divided by 18 to get h3 = 8. Then, taking the cubed
root of both sides gives h = 2. The height of the box is 2 in.
13. The candy company's total revenue this year was 1.1 times the revenue last year. If
they experience the same growth every year, what equation describes how many years
will it take before the revenue is more than double?
(A)
(B)
(C) 1.1y ≥ 2
(D) 1.1y ≥ 2
Explanation:
The basic relation here is for year revenue to be at least initial revenue. That is yearly revenue ≥ 2
× initial revenue. But there is no information given about the total initial revenue or the total year
revenue. Instead, we know that the first year the revenue will be 1.1 × initial revenue. The second
year, it will be 1.1 × first year revenue, or 1.1 × 1.1 × initial revenue, or 1.12 × initial revenue. For any
given year, the revenue is 1.1y ×initial revenue, where y is the number of years. Going back into the
inequality gives 1.1y× initial revenue ≥ 2 × initial revenue. Canceling the initial revenue from both
sides gives the equation 1.1y ≥ 2.
14. The candy company's total revenue this year was 1.1 times the sales last year. If they
experience the same growth every year, how many years will it take before the revenue is
more than double?
(A) 8
(B) 7
(C) 3
(D) 2
Explanation:
For any given year, the revenue will be 1.1y times the initial revenue. More than double means
1.1y ≥ 2. Taking the log10 of both sides gives y × log101.1 ≥ log102. Using a calculator for the base
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10 log gives log101.1 = 0.0414 and log102 = 0.301. The equation becomes 0.0414y ≥ 0.301.
Dividing both sides by 0.0414 gives y ≥ 7.27. Looking at whole numbers of years, the first year with
more than double the current revenue is 8 years.
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