Number and Operations

Section 4: Accurately Solves Problems

NECAP: M(N&O) – X – 4
Vermont: MX: 4

N&O – 35 Accurately solves problems: The intent of this GLE is to ensure that students
solve problems at various Depth of Knowledge levels (See NECAP Mathematics Test
Specifications) by performing accurate calculations (without the use of calculator,
manipulatives, or other tools).

Note: As the Depth of Knowledge levels increase the computational demand does not necessarily increase.
An attempt is made in the NECAP items to keep the level of computation required at a reasonable level and
focus on assessing concepts. Also note, two of the three testing sessions of the NECAP assessment allow
the use of calculators. Items where a calculator would take away from the construct being measured (e.g.,
accurately solves problems) will appear on the session that does not allow calculators, manipulatives, or
other tools.

Example 35.1 – (Grade 3) Accurately solves problems involving … addition and
subtraction of decimals (in the context of money):

Answer: $5.09; $3.50 + $1.59 = $3.00 + $1.00 + $1.00 + $0.09 = $5.09

Example 35.2 – (Grade 4) Accurately solves problems involving … addition and
subtraction of decimals :

Answer: $4.00; $1.50 + $0.50+ $0.50+ $0.50+ $0.50+ $0.50 = $4.00

Example 35.3 – (Grade 6) Accurately solves problems involving multiple
operations of decimals :

Carolyn and Kim sold 55 cups of lemonade on Monday.
• A cup of lemonade cost $0.10 to make.
• Each cup of lemonade is sold for $0.25.

a. How much did it cost to make 55 cups of lemonade? Show or explain your
work.

b. How much money did Carolyn and Kim collect? Show or explain your work.

c. How much profit did Carolyn and Kim collect for selling 55 cups of lemonade?
Show or explain your work.

Item Source: Adapted from
2002 – 6^th grade NHIEAP

Answer:
a. $5.50; 55 × $0.10 = $5.50
b. $13.75; Each group of 4 cups sells for $1.00. 55 ÷ 4 =13 remainder 3. So, selling 55 cups will make
$13.75.
c. $8.25; (55 × $0.25) – (55 × $0.10) = $8.25
d. 67 cups; There are 6 groups of $0.15 in each $1.00 with $0.10 left over. So, in $10.00 there are 60 groups
of $0.15 with $1.00 left over. Therefore, 66 cups would be $0.10 less than $10.00 profit. So, they need to sell
67 cups to make at least $10.00 profit.

N&O – 36 In and out of context: In and out of context means that some problems will
be cast in a context (see Examples 35.1 – 35.3), and others will not be in a context.

Example 36.1 – Non-contextual example:

What is the least common multiple of 12 and 9?

Answer: 36

N&O – 37 Concept of multiplication: Solving problems that involve the concept of
multiplication means that the problems must demand an understanding of multiplication,
not just the application of an algorithm.

Example 37.1:

Answer: 3 rows of 10 chairs, 5 rows of 6 chairs, 6 rows of 5 chairs

N&O – 38 Factor: An integer b is a factor of a given integer if the product of b and some
other integer is the given integer (e.g., given the integer 12, 4 is a factor of 12
since 4·3=12 ). A factor is also called a divisor since it divides the given integer evenly
(when the given integer is divided by the factor the remainder is 0).

Example 38.1 – Determine whole number factors:

Determine all the whole number factors of 36.

Answer: 1, 2, 3, 4, 6, 9, 12, 18, and 36 are all factors of 36 because each number divides into 36 a whole
number of times with no remainder.

Answer: The integer factors of are –36, –18, –12, –9, –6, –4, –3, –2, –1, 1, 2, 3, 4, 6, 9, 12, 18, 36, because
each divides into 36 an integral number of times with no remainder.

At grades K – 6 students will be expected to only generate the whole number factors for a
given whole number. For 36, students would only have to generate 1, 2, 3, 4, 6, 9, 12, 18,
and 36.

N&O – 39 Multiples: A multiple is a number that is the product of a given number and
an integer.

Example 39.1: List five multiples of 3.

Sample Answer: 3, 6, 9, 12, and 15 (3 is the given number in this example and each multiple is produced by
multiplying 3 by 1, 2, 3, 4, and 5, respectively.)

Example 39.2: List all the multiples of 3.

Answer: The set of multiples of 3 is {…, –15, –12, –9, –6, –3, 0, 3, 6, 9, 12, 15, …}.

Given number Integer Product Multiple

N&0 – 9 Integer: An integer is a number in the set {…, –3, –2, –1, 0, 1, 2, 3, …}.

N&O – 40 Prime numbers: A prime number is a whole number greater than 1 that is
only divisible by 1 and itself. (Its only factors are 1 and itself.)

Example 40.1:

5 is a prime number because its only factors are 1 and 5.
11 is a prime number because its only factors are 1 and 11.
12 is NOT a prime number because its factors are 1, 2, 3, 4, 6, and 12.

N&O – 41 Composite number: A composite number is a number that is not prime (has
factors other than 1 and itself).

Example 41.1:

12 is a composite number since its whole number factors are 1, 2, 3, 4, 6, and 12.
2 is NOT a composite number since its only factors are 1 and itself.

N&O – 42 Greatest Common Factor ( GCF ): The greatest common factor of two or
more positive integers is the largest factor they have in common.

Example 42.1: What is the greatest common factor (GCF) of 24, 36, and 60?

Answer: Though there are many common factors of 24, 36, and 60, the greatest common factor (GCF) of
24, 36, and 60 is 12.

N&O – 43 Least Common Multiple (LCM): The least common multiple of two or more
positive integers is the smallest positive multiple that they have in common.

Example 43.1: What is the least common multiple (LCM) of 9, 12, and 18?

Answer: The least common multiple of 9, 12, and 18 is 36.

N&O – 44 Proportional reasoning: Solving problems involving proportional reasoning
means to use proportional reasoning in problem solving situations that may involve
ratios, proportions, rates, slope , scale, similarity, percents, probability, and others. It is
assumed that throughout instruction students have sufficient opportunities to connect
each of these situations to proportional reasoning. (e.g., Students should realize that
proportional relationships are described by linear functions of the form y = kx).