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SSAT Middle Level Math : Numbers and Operations

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All SSAT Middle Level Math Resources

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Example Questions

Example Question #6 : Distributive Property

Simplify the expression:

8x - 7 (3x+ 6) - 5

Possible Answers:

-13 x + 1

-13 x - 1

-13 x - 47

-13 x + 37

-13 x - 37

Correct answer:

-13 x - 47

Explanation:

Distribute, then collect like terms:

8x - 7 (3x+ 6) - 5

$=8x - 7 \cdot 3x - 7 \cdot 6 - 5$

=8x - 21x - 42 - 5

$= \left ( 8 - 21 \right ) x - 42 - 5$

= -13 x - 47

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Example Question #7 : Distributive Property

Diana is thirty-three years older than her son Colin, who is three times as old as her niece Sharon. If is Diana's age, how old is Sharon?

Possible Answers:

$\frac{1}{3}D - 33$

$\frac{1}{3}D + 11$

$\frac{1}{3}D - 11$

3D - 99

3D-33

Correct answer:

$\frac{1}{3}D - 11$

Explanation:

Colin's age is thirty-three years less than Diana's age of , so Colin is D -33 years old; Sharon is one-third of this, or $\frac{1}{3 }\left (D -33 \right )$ . Using distribution, this can be rewritten as

$\frac{1}{3 }\left (D -33 \right ) = \frac{1}{3 } \cdot D - \frac{1}{3 } \cdot 33 = \frac{1}{3}D - 11$ .

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Example Question #8 : Distributive Property

Nina is twenty-one years younger than her mother Caroline, who is one-third as old as their neighbor Mr. Hutchinson. If is Nina's age, how old is Mr. Hutchinson?

Possible Answers:

$\frac{1}{3}N - 7$

$\frac{1}{3}N - 21$

3N + 63

3N + 21

3N + 7

Correct answer:

3N + 63

Explanation:

Caroline is twenty-one years older than Nina, so her age is N + 21 . Mr. Hutchinson is three times as old as Caroline, so he is $3\left (N + 21 \right )$ . Using distribution, this can be rewritten as

$3\left (N + 21 \right ) = 3 \cdot N + 3 \cdot 21= 3N + 63$ .

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Example Question #8 : Distributive Property

Which of the following expressions is equivalent to x(2-k) ?

Possible Answers:

x(k+2)

2x-kx

2x-2k

2(k-x)

Correct answer:

2x-kx

Explanation:

When distributing, the number outside the parentheses is multiplied by both of the numbers inside without changing any signs.

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Example Question #1 : Find Greatest Common Factor

Use the distributive property to express the sum 40+36 as the multiple of a sum of two whole numbers with no common factor.

Possible Answers:

None of these

3(10+9)

9(10+4)

4(10+9)

10(4+9)

Correct answer:

4(10+9)

Explanation:

The distributive property can be used to rewrite an expression. When we use this property we will identify and pull out the greatest common factor of each of the addends. Then we can create a quantity that represents the sum of two whole numbers with no common factor multiplied by their greatest common factor.

40+36

In this case, the greatest common factor shared by each number is:

$\text{GCF}=4$

After we reduce each addend by the greatest common factor we can rewrite the expression:

4(10+9)

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Example Question #1 : Find Greatest Common Factor And Least Common Multiple: Ccss.Math.Content.6.Ns.B.4

Use the distributive property to express the sum 16+28 as the multiple of a sum of two whole numbers with no common factor.

Possible Answers:

7(4+4)

7(4+7)

5(4+7)

4(4+7)

4(4+4)

Correct answer:

4(4+7)

Explanation:

16+28

In this case, the greatest common factor shared by each number is:

$\text{GCF}=4$

After we reduce each addend by the greatest common factor we can rewrite the expression:

4(4+7)

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Example Question #2 : Find Greatest Common Factor

Use the distributive property to express the sum 28+32 as the multiple of a sum of two whole numbers with no common factor.

Possible Answers:

5(7+8)

4(7+8)

None of these

8(7+5)

3(7+8)

Correct answer:

4(7+8)

Explanation:

28+32

In this case, the greatest common factor shared by each number is:

$\text{GCF}=4$

After we reduce each addend by the greatest common factor we can rewrite the expression:

4(7+8)

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Example Question #3 : Find Greatest Common Factor

Use the distributive property to express the sum 20+36 as the multiple of a sum of two whole numbers with no common factor.

Possible Answers:

5(5+9)

5(4+9)

4(5+9)

9(5+9)

9(5+4)

Correct answer:

4(5+9)

Explanation:

20+36

In this case, the greatest common factor shared by each number is:

$\text{GCF}=4$

After we reduce each addend by the greatest common factor we can rewrite the expression:

4(5+9)

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Example Question #2 : Find Greatest Common Factor

Use the distributive property to express the sum 20+25 as the multiple of a sum of two whole numbers with no common factor.

Possible Answers:

5(5+5)

5(4+5)

3(4+5)

5(4+5)

None of these

Correct answer:

5(4+5)

Explanation:

20+25

In this case, the greatest common factor shared by each number is:

$\text{GCF}=5$

After we reduce each addend by the greatest common factor we can rewrite the expression:

5(4+5)

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Example Question #1 : Find Greatest Common Factor

Use the distributive property to express the sum 55+100 as the multiple of a sum of two whole numbers with no common factor.

Possible Answers:

5(11+20)

20(11+5)

8(11+20)

5(10+20)

55(10+4)

Correct answer:

5(11+20)

Explanation:

55+100

In this case, the greatest common factor shared by each number is:

$\text{GCF}=5$

After we reduce each addend by the greatest common factor we can rewrite the expression:

5(11+20)

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SSAT Middle Level Math : Numbers and Operations

Study concepts, example questions & explanations for SSAT Middle Level Math

All SSAT Middle Level Math Resources

Example Questions

Example Question #6 : Distributive Property

Example Question #7 : Distributive Property

Example Question #8 : Distributive Property

Example Question #8 : Distributive Property

Example Question #1 : Find Greatest Common Factor

Example Question #1 : Find Greatest Common Factor And Least Common Multiple: Ccss.Math.Content.6.Ns.B.4

Example Question #2 : Find Greatest Common Factor

Example Question #3 : Find Greatest Common Factor

Example Question #2 : Find Greatest Common Factor

Example Question #1 : Find Greatest Common Factor

All SSAT Middle Level Math Resources

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