GRADE 6 EOG REVIEW PACKET

Problem 41 :

Which expression represents the perimeter of the triangle?

6th-grade-eog-q41.png

A)  9k + 36    B) 10k + 25    C) 20k + 25    D) 24k + 36

Solution :

Perimeter = 3k + 5 + 2k + 16 + 15 + 4k

= 3k + 2k + 4k + 5 + 16 + 15

= 9k + 36

So, option A is correct.

Problem 42 :

The length of a rectangle is 6 units longer than the width, w. Which choice is a correct expression for the perimeter of the rectangle?

A)  2w + 6    B) 2w + 12     C) 4w + 6     D) 4w + 12

Solution :

Perimeter of the rectangle = 2(length + width)

Width be w, length = w + 6

Perimeter = 2(w + 6 + w)

= 2(2w + 6)

= 4w + 12

So, option D is correct.

Problem 43 :

Jane wants to visit her sister.

  • Her car travels x miles per gallon of gas.
  • She will travel 1,000 miles to her sister’s house.
  • Gas costs $3.50 per gallon.

Which expression shows how much Jane will spend for gas on the trip to her sister’s house?

Solution :

Distance covered = x miles

Cost of gas = 3.50 per gallon

Money spent = distance covered x cost per gallon

= 3.50

In the total distance of 1000 miles, he covered x miles.

= (x/1000)

Jane will spend = 3.50(x/1000)

So, option B is correct.

Problem 44 :

Suppose that a stove and a freezer together weigh at least 370 pounds. The weight of the stove is 170 pounds. Which inequality correctly describes these conditions for the weight of the freezer, f ?

A)  f ≥ 200    B) f > 200      C) f ≤ 200     D) f < 200

Solution :

Let s be the weight of stove

s + f ≥ 370

Wight of stove = 170

 ≥ 370 - s

 ≥ 370 - 170

 ≥ 200

So, option A is correct.

Problem 45 :

The Wilsons want to put outdoor carpet on their porch

6th-grade-eog-q45.png

How much carpet will be needed for their porch?

A) 42 ft2     B) 72 ft2    C) 108 ft2     D) 144 ft2

Solution :

6th-grade-eog-q45p1.png

The top part is rectangle and bottom part is square.

Area of the given shape = Area of square + area of rectangle

= 6 x 6 + 12 x 6

= 36 + 72

= 108

108 ft2 

So, option C is correct.

Problem 46 :

What is the volume of the right rectangular prism below?

6th-grade-eog-q46.png

Solution :

Volume of rectangular prism = 2 x 1   1/4 x   1   1/2

= 2 x 5/4 x 3/2

= 15/4

= 3  3/4 

So, option C is correct.

Problem 47 :

In the graph below, each grid square represents one square yard.

6th-grade-eog-q47.png

What is the area of the shaded figure?

A) 20 yd2     B) 30 yd2      C) 36 yd2       D) 40 yd2

Solution :

Area of triangle + are of parallelogram

= ( 1/2) x base x height + base x height

= (1/2) x 5 x 4 + (5 x 4)

= 10 + 20

= 30 yd2

So, option B is correct.

Problem 48 :

Abby is making a decoration. When folded, the decoration is a triangular pyramid made of four congruent equilateral triangles. Approximately, what is the surface area of Abby’s decoration?

6th-grade-eog-q48.png

A) 64 in2      B) 85 in2      C) 97 in2     D) 170 in2

Solution :

From the figure, height of the triangle = 6.06, base = 7

Area of 1 triangle = (1/2) x 7 x 6.06

= 21.21

Area of 4 triangles = 4 (21.21)

= 84.84

85 in2

So, option B is correct.

Problem 49 :

Katherine earned 84, 92, 84, 75, and 70 on her first 5 tests. What is the minimum grade Katherine needs to earn on the next test to have a mean of 84?

A)  81    B)  84      C) 95     D)  99

Solution :

Let x be the unknown score.

(84 + 92 + 84 + 75 + 70 + x)/6 = 84

405 + x = 84(6)

x = 504 - 405

x = 99

So, the answer is option D.

Problem 50 :

The weather station recorded the high temperature each day for 30 days. The graph of the temperature data is shown below.

6th-grade-eog-q50.png

In which interval is the median temperature?

A)   41–50     B) 51–60     C) 61–70    D) 71–80

Solution :

= (3 + 8 + 6 + 7 + 4 + 2)

= 30 (even)

then average of 15th and 16th term will be median.

So, median is 51-60.

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