CBSE Class 8 Mathematics Practice Worksheet Set 02

Question 1. Various modes of transport used by 1260 students in a given school are given below:

Represent the above data by a pie chart.
Answer: To represent the data in a pie chart, we first calculate the central angle for each category using the formula: Central Angle = \(\left(\frac{\text{Component Value}}{\text{Total Value}}\right) \times 360^\circ\)
Total number of students = 1260
1. School Bus: \(\left(\frac{350}{1260}\right) \times 360^\circ = 100^\circ\)
2. Private Bus: \(\left(\frac{245}{1260}\right) \times 360^\circ = 70^\circ\)
3. Bicycle: \(\left(\frac{210}{1260}\right) \times 360^\circ = 60^\circ\)
4. Rickshaw: \(\left(\frac{175}{1260}\right) \times 360^\circ = 50^\circ\)
5. On foot: \(\left(\frac{280}{1260}\right) \times 360^\circ = 80^\circ\)
Verification of total angles: \(100^\circ + 70^\circ + 60^\circ + 50^\circ + 80^\circ = 360^\circ\)
In simple words: Find the fraction of each transport mode out of 1260 and multiply by 360 degrees to get the slice angle for the pie chart.

Exam Tip: Always double check that the sum of all calculated central angles equals exactly 360 degrees before drawing the pie chart.

Question 2. The following table gives the number of different fruits kept in a hamper:

Represent the above data by a pie chart.
Answer: Total number of fruits = \(26 + 30 + 21 + 5 + 8 = 90\)
Central Angle = \(\left(\frac{\text{Number of each fruit}}{90}\right) \times 360^\circ = \text{Number of each fruit} \times 4^\circ\)
1. Mangoes: \(26 \times 4^\circ = 104^\circ\)
2. Apples: \(30 \times 4^\circ = 120^\circ\)
3. Oranges: \(21 \times 4^\circ = 84^\circ\)
4. Coconuts: \(5 \times 4^\circ = 20^\circ\)
5. Pineapples: \(8 \times 4^\circ = 32^\circ\)
Verification of total angles: \(104^\circ + 120^\circ + 84^\circ + 20^\circ + 32^\circ = 360^\circ\)
In simple words: First find the total sum of all fruits (90), then multiply each fruit's count by 4 degrees to find its sector angle.

Exam Tip: Simplify the fraction \(\frac{360^\circ}{\text{Total}}\) first to make your angle calculations much quicker and less prone to arithmetic errors.

Question 3. The angles of a quadrilateral are in the ratio 1:2:3:4. Find the measure of all the angles.
Answer: Let the common ratio be \(x\).
The four angles of the quadrilateral are \(1x\), \(2x\), \(3x\), and \(4x\).
We know that the sum of all interior angles of a quadrilateral is \(360^\circ\).
\(1x + 2x + 3x + 4x = 360^\circ\)
\(10x = 360^\circ\)
\(x = \frac{360^\circ}{10} = 36^\circ\)
Now, substitute the value of \(x\) to find each angle:
First angle = \(1 \times 36^\circ = 36^\circ\)
Second angle = \(2 \times 36^\circ = 72^\circ\)
Third angle = \(3 \times 36^\circ = 108^\circ\)
Fourth angle = \(4 \times 36^\circ = 144^\circ\)
In simple words: Assume the angles are \(x, 2x, 3x, 4x\), add them up to equal \(360^\circ\), solve for \(x\), and then multiply to find each angle.

Exam Tip: Always state the angle sum property of a quadrilateral clearly at the beginning of your solution to secure full steps marks.

Question 4. Three angles of a quadrilateral are \(45^\circ\), \(162^\circ\), \(35^\circ\), find the fourth angle.
Answer: Let the fourth angle of the quadrilateral be \(x\).
The sum of all four interior angles of a quadrilateral is always \(360^\circ\).
\(45^\circ + 162^\circ + 35^\circ + x = 360^\circ\)
\(242^\circ + x = 360^\circ\)
\(x = 360^\circ - 242^\circ\)
\(x = 118^\circ\)
Therefore, the fourth angle is \(118^\circ\).
In simple words: Add the three given angles together and subtract their sum from \(360^\circ\) to find the remaining fourth angle.

Exam Tip: Double-check your addition of the three given angles before performing the final subtraction.

Question 5. On a graph paper plot the following points taking the axes X'OX and Y'OY: A(0, 6), B(-4, -3), C(3, 0) and D(3, -5). Join AB, BC, CD and DA. What figure do you obtain.
Answer:
1. Plot the given points on the Cartesian plane:
- Point \(A(0, 6)\) lies on the positive y-axis.
- Point \(B(-4, -3)\) lies in the third quadrant.
- Point \(C(3, 0)\) lies on the positive x-axis.
- Point \(D(3, -5)\) lies in the fourth quadrant.
2. Join the points in order: segment AB, segment BC, segment CD, and segment DA.
3. Observing the joined vertices and sides on the graph, the enclosed geometric shape formed is an irregular quadrilateral.
In simple words: Plot the four points on graph paper and connect them in order. The resulting closed four-sided shape is an irregular quadrilateral.

Exam Tip: Clearly label the origin, axes arrows, and coordinates of all plotted points on your graph paper.