# The Preparation Of Semester II Exam (Mathematics VII) By Mrssriendangs

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Quizzes Created: 2 | Total Attempts: 253
Questions: 15 | Attempts: 101  Settings  • 1.

### Given A = { the factors of 4 }, the all subsets of A is . .

• A.

{2,4}

• B.

{1,2,4}

• C.

{1},{2},{4},{1,2}{1,4}{2,4}{1,2,4}

• D.

{ },{1},{2},{4},{1,2},{1,4},{2,4}{1,2,4}

D. { },{1},{2},{4},{1,2},{1,4},{2,4}{1,2,4}
Explanation
The given answer { },{1},{2},{4},{1,2},{1,4},{2,4}{1,2,4} represents all the subsets of the factors of 4. The empty set {} is included as a subset. The subsets containing only one factor are {1}, {2}, and {4}. The subsets containing two factors are {1,2}, {1,4}, and {2,4}. The subset containing all the factors is {1,2,4}. Therefore, the given answer correctly lists all the subsets of the factors of 4.

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• 2.

### Let K = {4, 6, 8, 9, 10, 12, 14} K will be expressed by the set builder, so set P is ….

• A.

{x/4≤ x < 15, x is even numbers}

• B.

{x/4≤ x ≤14, x is composite numbers}

• C.

{x/4 < x ≤ 14, x is natural numbers}

• D.

{x/4 < x ≤ 14, x is whole numbers}

B. {x/4≤ x ≤14, x is composite numbers}
Explanation
The set K is defined as {4, 6, 8, 9, 10, 12, 14}, which consists of even numbers between 4 and 14. The correct answer is {x/4≤ x ≤14, x is composite numbers}, which means that the set P will consist of composite numbers between 4 and 14. Since 4, 9, and 14 are the only composite numbers in the set K, the answer is correct.

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• 3.

### Given      U = {1, 2, 3, …, 6, 7}        A = { 1, 2, 4, 6}      B = {  3, 5, 7}      The Venn diagram describing the relationship between these sets are ....

A.
Explanation
The Venn diagram describing the relationship between sets A and B is such that set A is represented by a circle containing elements 1, 2, 4, and 6, and set B is represented by a circle containing elements 3, 5, and 7. The circles overlap to show that there are no common elements between sets A and B.

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• 4.

### Let A={composite numbers between 6 and 15}       B= {odd numbers between 2 and 15}       The elements of set A – B are ….

• A.

{ 8, 10, 12, 14}

• B.

{9}

• C.

{3, 5, 7, 11, 13}

• D.

{ 6, 8, 10,12,14,15}

A.  { 8, 10, 12, 14}
Explanation
The correct answer is { 8, 10, 12, 14}. This is because set A contains composite numbers between 6 and 15, which are 8, 9, 10, 12, and 14. Set B contains odd numbers between 2 and 15, which are 3, 5, 7, 9, 11, 13, and 15. The elements of set A - B are the elements that are in set A but not in set B. Therefore, the elements { 8, 10, 12, 14} are the elements of set A - B.

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• 5.

### The ratio of an angle and its supplementary is 2 : 3 so the measure of the angle is ….

• A.

1440

• B.

1080

• C.

720

• D.

360

C. 720
Explanation
The ratio of an angle and its supplementary is 2:3. This means that the measure of the angle is two-thirds of its supplementary angle. Since the supplementary angle is the angle that, when added to the given angle, equals 180 degrees, we can set up the equation (2/3)x + x = 180, where x represents the measure of the angle. Solving this equation, we find that x = 720. Therefore, the measure of the angle is 720 degrees.

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• 6.

### Look at the figure! If  angle 1 = 1250 so the sum of angle  4 + 5 + 6 = ….

• A.

1250

• B.

1800

• C.

2250

• D.

3050

D. 3050
Explanation
The sum of angles 4, 5, and 6 can be found by subtracting angle 1 from the sum of all angles in the figure. Since the sum of all angles in a figure is 360 degrees, the sum of angles 4, 5, and 6 would be 360 - 1250, which equals 3050.

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• 7.

### Look at the figure! ABCD is a Rectangle angle BOC =1100 The measure of  angle ABO is …

• A.

1100

• B.

700

• C.

550

• D.

350

C. 550
Explanation
In a rectangle, opposite angles are equal. Therefore, angle BOC is equal to angle AOD. Since angle BOC is given as 1100, angle AOD is also 1100. Angle ABO is formed by the intersection of lines AB and BO. Since angle AOD is 1100, and angle BOC is 1100, the sum of angles AOD and BOC is 2200. Since the sum of angles in a triangle is 1800, angle ABO can be found by subtracting the sum of angles AOD and BOC from 1800. Therefore, angle ABO is 1800 - 2200 = 550.

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• 8.

A.
• 9.

### The area of a rectangle is 216 cm2. the ratio of the length and the width of the rectangle is      3 : 2. The length and the width of the rectangle are ….

• A.

L = 9 cm and w = 6 cm

• B.

L = 12 cm and w = 8 cm

• C.

L = 18 cm and w = 12 cm

• D.

L = 21 cm and w = 14 cm

C. L = 18 cm and w = 12 cm
Explanation
The area of a rectangle is calculated by multiplying its length and width. In this question, the ratio of the length and width is given as 3:2. Let's assume the length is 3x and the width is 2x. According to the given information, the area of the rectangle is 216 cm2. So, we can set up the equation 3x * 2x = 216. Solving this equation, we find that x = 6. Therefore, the length is 3 * 6 = 18 cm and the width is 2 * 6 = 12 cm, which matches the given answer.

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• 10.

### Look at the figure!    PQRS is a square If the length of PR is 8 cm. The area of a square PQRS is ….

• A.
• B.

32  cm2

• C.
• D.
B. 32  cm2
Explanation
The area of a square is calculated by multiplying the length of one side by itself. In this case, the length of PR is given as 8 cm. Since all sides of a square are equal in length, the length of each side of the square is also 8 cm. Therefore, the area of the square PQRS is 8 cm multiplied by 8 cm, which equals 64 cm^2.

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• 11.

### Look at the figure! PQRS is parallelogram The measure of   PQR is ….

• A.

16

• B.

48

• C.

164

• D.

132

D. 132
Explanation
The measure of angle PQR in the parallelogram PQRS is 132 degrees.

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• 12.

### Look at the figure ! The length of AB is 18 cm and BC is 15 cm if the length of height DF is 6 cm The length of DE is ….

• A.

10 cm

• B.

8 cm

• C.

6 cm

• D.

5 cm

D. 5 cm
Explanation
Based on the given information, AB is 18 cm and BC is 15 cm. DF is the height of the triangle and its length is given as 6 cm. Since DE is perpendicular to AB, it can be inferred that DE is also perpendicular to BC. Therefore, triangle ABC is a right triangle. Using the Pythagorean theorem, we can find the length of DE. By applying the theorem, DE^2 = DF^2 + EF^2. Since DF is 6 cm and EF is the length of DE, we can substitute the values and solve for EF. Solving the equation, we find that EF is equal to 5 cm. Therefore, the length of DE is 5 cm.

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• 13.

### Look at the figure! PQRS is Rhombus. If PR = 18 cm and QS = 24 cm; The perimeter of rhombus PQRS is ….

• A.

60 cm

• B.

48 cm

• C.

36 cm

• D.

24 cm

A. 60 cm
Explanation
The perimeter of a rhombus is equal to four times the length of one side. Since PQRS is a rhombus, all sides are equal in length. Given that PR = 18 cm and QS = 24 cm, we can conclude that all sides of the rhombus are equal to 18 cm. Therefore, the perimeter of PQRS is 4 * 18 cm = 72 cm. However, the given options do not include 72 cm, so the closest option is 60 cm.

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• 14.

### The diagonals of rhombus are 24 cm and x cm, if the perimeter of rhombus is 80 cm so the area of the rhombus is ….

C.
Explanation
The diagonals of a rhombus bisect each other at right angles. Therefore, the length of each diagonal is half the length of the perimeter. In this case, the sum of the diagonals is 24 cm + x cm = 80 cm. So, x cm = 80 cm - 24 cm = 56 cm. Since the diagonals of a rhombus are perpendicular bisectors, they divide the rhombus into four congruent right-angled triangles. The area of each triangle can be calculated using the formula (base * height)/2, where the base is 24 cm/2 = 12 cm and the height is 56 cm/2 = 28 cm. Therefore, the area of each triangle is (12 cm * 28 cm)/2 = 168 cm². Since there are four triangles, the total area of the rhombus is 4 * 168 cm² = 672 cm².

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• 15.

### The figure ABCD is kite, if The measure of    angle BAD = 1300 and angle  BCD = 70 so the measure of  angle  ADC = …. Back to top