CBSE Class 6 Mathematics Revision Worksheet Set 10

Question 1. Define the following : i) point ii) line iii) plane iv) ray v) intersecting line
Answer: i) Point: A point is a mark of position. It has no length, breadth, or thickness, and is usually denoted by a capital letter.
ii) Line: A line is a straight path that extends indefinitely in both directions without any endpoints.
iii) Plane: A flat surface that extends indefinitely in all directions in two dimensions is called a plane.
iv) Ray: A ray is a portion of a line that starts at one endpoint and extends indefinitely in one direction.
v) Intersecting line: Two or more lines that meet or cross each other at a common point are called intersecting lines.
In simple words: Basic geometric terms describe fundamental shapes and paths with specific endpoints or extensions.

Exam Tip: Always remember that a line has no endpoints, a ray has one endpoint, and a line segment has two endpoints.

Question 2. Draw any circle and mark : a) its centre b) a diameter c) a sector d) an arc
Answer: a) Centre: The fixed point inside the circle from which all points on the boundary are equidistant.
b) Diameter: A straight line segment passing through the centre whose endpoints lie on the circle.
c) Sector: The region enclosed by two radii and an arc.
d) An arc: Any connected piece of the boundary of a circle.
In simple words: Different parts of a circle serve specific geometric functions related to its perimeter and area.

Exam Tip: Practice drawing a neat circle using a compass and label each requested component clearly with distinct arrows or shading.

Question 3. Draw a rough sketch of a quadrilateral ABCD. State, a) two pairs of opposite sides. b) two pairs of opposite angles. c) two pairs of adjacent sides. d) two pairs of adjacent angles.
Answer: a) Two pairs of opposite sides: \(AB\) and \(DC\), \(AD\) and \(BC\)
b) Two pairs of opposite angles: \(\angle A\) and \(\angle C\), \(\angle B\) and \(\angle D\)
c) Two pairs of adjacent sides: \(AB\) and \(BC\), \(BC\) and \(CD\)
d) Two pairs of adjacent angles: \(\angle A\) and \(\angle B\), \(\angle B\) and \(\angle C\)
In simple words: Opposite sides and angles do not share a common vertex or side, whereas adjacent ones share a common vertex or side.

Exam Tip: Label vertices in cyclic order (clockwise or counter-clockwise) to correctly identify opposite and adjacent elements.

Question 4. Construct two line segments of lengths 4.1 cm and 3.3 cm. Construct a segment whose length is equal to the sum of lengths of these segment.
Answer: Steps of construction:
1. Draw a line of sufficient length and mark a point \(A\) on it.
2. Using a ruler and compass, measure a length of \(4.1\text{ cm}\) and cut off segment \(AB = 4.1\text{ cm}\).
3. From point \(B\), using a compass, measure and cut off segment \(BC = 3.3\text{ cm}\).
4. The total length of segment \(AC = AB + BC = 4.1\text{ cm} + 3.3\text{ cm} = 7.4\text{ cm}\).
In simple words: To find the sum of two line segments, place them end-to-end along a straight line.

Exam Tip: Always use a sharp pencil and a graduated ruler for accurate geometric constructions.

Question 5. Using a protector, construct angles of the following measures : \(50^\circ\), \(95^\circ\), \(100^\circ\)
Answer: Steps of construction:
1. Draw a base ray with an endpoint as the vertex of the angle.
2. Place the center of the protractor on the vertex and align the baseline of the protractor with the ray.
3. Mark the required degree point (\(50^\circ\), \(95^\circ\), and \(100^\circ\) respectively) on the paper.
4. Remove the protractor and draw a ray from the vertex through the marked point to complete the angle.
In simple words: Use a protractor by aligning its center with the vertex and reading the correct scale.

Exam Tip: Check whether the angle is acute or obtuse before drawing to ensure you read the correct inner or outer scale on the protractor.

Question 6. Define the following : Scalene triangle, Isosceles triangle, Equilateral triangle, Acute angled triangle, Obtuse angled triangle, Right angled triangle.
Answer: - Scalene triangle: A triangle with all three sides of different lengths.
- Isosceles triangle: A triangle with two sides of equal length.
- Equilateral triangle: A triangle with all three sides of equal length.
- Acute angled triangle: A triangle in which all three interior angles are acute (less than \(90^\circ\)).
- Obtuse angled triangle: A triangle in which one of the interior angles is obtuse (greater than \(90^\circ\)).
- Right angled triangle: A triangle in which one of the interior angles is a right angle (\(90^\circ\)).
In simple words: Triangles are classified either by comparing their side lengths or by the measures of their interior angles.

Exam Tip: Memorize both side-based and angle-based classifications clearly as they are frequently tested in geometry.

Question 7. Find the cost of fencing a square park of side 200 m at the rate of Rs 25 per metre.
Answer: Side of the square park = 200 m
Perimeter of the square park = \(4 \times \text{side}\)
Perimeter = \(4 \times 200 = 800\text{ m}\)
Rate of fencing = Rs 25 per metre
Total cost of fencing = Perimeter \(\times\) Rate
Total cost = \(800 \times 25 = \text{Rs } 20,000\)
In simple words: Fencing requires calculating the total boundary length (perimeter) and multiplying it by the cost per unit length.

Exam Tip: For boundary problems like fencing, always calculate the perimeter rather than the area.

Question 8. Amrita runs around a square field of side 80 m. Babita runs around a rectangular field whose length is 90 m and breadth is 55 m. Who covers the larger distance and by how much?
Answer: Side of square field = 80 m
Distance covered by Amrita (Perimeter of square) = \(4 \times 80 = 320\text{ m}\)
Length of rectangular field = 90 m
Breadth of rectangular field = 55 m
Distance covered by Babita (Perimeter of rectangle) = \(2 \times (\text{length} + \text{breadth})\)
Distance covered by Babita = \(2 \times (90 + 55) = 2 \times 145 = 290\text{ m}\)
Comparing distances: \(320\text{ m} > 290\text{ m}\), so Amrita covers the larger distance.
Difference in distance = \(320 - 290 = 30\text{ m}\)
In simple words: Find the perimeter for both runners and compare their totals to find the difference.

Exam Tip: Write formula statements clearly before substituting values to avoid calculation errors.

Question 9. Find the length of a rectangle given that its perimeter is 880m and breadth is 88m.
Answer: Perimeter of the rectangle = 880 m
Breadth of the rectangle = 88 m
Perimeter = \(2 \times (\text{length} + \text{breadth})\)
\(880 = 2 \times (\text{length} + 88)\)
\(\text{length} + 88 = \frac{880}{2} = 440\)
Length = \(440 - 88 = 352\text{ m}\)
In simple words: Divide the perimeter by 2 and subtract the given breadth to find the length.

Exam Tip: Double-check your result by recalculating the perimeter: \(2 \times (352 + 88) = 2 \times 440 = 880\text{ m}\).

Question 10. The perimeter of a triangle is 64 cm. Two of its sides measures 15 cm and 24 cm resp. Find the length of the third side of the triangle.
Answer: Perimeter of the triangle = 64 cm
First side = 15 cm
Second side = 24 cm
Perimeter = Sum of all three sides
\(64 = 15 + 24 + \text{third side}\)
\(64 = 39 + \text{third side}\)
Third side = \(64 - 39 = 25\text{ cm}\)
In simple words: Subtract the sum of the two known sides from the total perimeter to find the remaining side.

Exam Tip: Verify that the sum of any two sides is greater than the third side: \(15 + 24 = 39\) (not greater than 25? Wait, \(15 + 25 = 40 > 24\), triangle inequality holds).

Question 11. A hall is in the form of a rectangular region is 14m x 12m. How many slabs 8cmx6cm are needed to cover the floor of the hall?
Answer: Dimensions of the hall = \(14\text{ m} \times 12\text{ m}\)
Convert dimensions to centimeters: length = \(14 \times 100 = 1400\text{ cm}\), breadth = \(12 \times 100 = 1200\text{ cm}\)
Area of the hall = \(1400 \times 1200 = 1,680,000\text{ sq. cm}\)
Dimensions of one slab = \(8\text{ cm} \times 6\text{ cm}\)
Area of one slab = \(8 \times 6 = 48\text{ sq. cm}\)
Number of slabs needed = \(\frac{\text{Area of the hall}}{\text{Area of one slab}}\)
Number of slabs = \(\frac{1680000}{48} = 35,000\)
In simple words: Divide the total area of the floor by the area of a single slab, ensuring both use the same unit of measurement.

Exam Tip: Always convert units to be consistent (meters to centimeters) before calculating areas.

Question 12. A square field is of side 400m .Find its area & express it in hectares.
Answer: Side of the square field = 400 m
Area of the square = \(\text{side} \times \text{side}\)
Area = \(400 \times 400 = 160,000\text{ sq. m}\)
Conversion to hectares: \(1\text{ hectare} = 10,000\text{ sq. m}\)
Area in hectares = \(\frac{160000}{10000} = 16\text{ hectares}\)
In simple words: Find the area in square meters and divide by 10,000 to convert into hectares.

Exam Tip: Remember the standard conversion factor: \(1\text{ hectare} = 10,000\text{ m}^2\).

Question 13. The area of a rectangle is 150 sq. cm. If one of its side is 10cm long, how long is its other side? Also, find the perimeter of the rectangle.
Answer: Area of the rectangle = \(150\text{ sq. cm}\)
Given side (length) = 10 cm
Area = \(\text{length} \times \text{breadth}\)
\(150 = 10 \times \text{breadth}\)
Breadth = \(\frac{150}{10} = 15\text{ cm}\)
Perimeter of the rectangle = \(2 \times (\text{length} + \text{breadth})\)
Perimeter = \(2 \times (10 + 15) = 2 \times 25 = 50\text{ cm}\)
In simple words: Divide the area by the known side to find the other side, then use both sides to calculate the perimeter.

Exam Tip: Keep units clear: area is in square centimeters while perimeter is in centimeters.

Question 14. A worker is paid Rs 168 for 6 days. What wages would he get for 28 days?
Answer: Wages for 6 days = Rs 168
Wages for 1 day = \(\frac{168}{6} = \text{Rs } 28\)
Wages for 28 days = \(28 \times 28 = \text{Rs } 784\)
In simple words: Use the unitary method by finding the wage for a single day first, then multiply by the required number of days.

Exam Tip: Always show the intermediate step for 1 unit when solving unitary method problems.

Question 15. Using a ruler & a compass, construct angles of the following measures : \(30^\circ\), \(45^\circ\), \(75^\circ\), \(90^\circ\), \(105^\circ\), \(15^\circ\)
Answer: Steps of construction:
1. \(90^\circ\): Construct a perpendicular on a line using a compass.
2. \(60^\circ\): Construct an equilateral triangle angle using an arc from the vertex.
3. \(30^\circ\): Bisect the \(60^\circ\) angle.
4. \(15^\circ\): Bisect the \(30^\circ\) angle.
5. \(45^\circ\): Bisect the \(90^\circ\) angle.
6. \(75^\circ\): Construct the angle midway between \(60^\circ\) and \(90^\circ\) (bisecting the angle between them).
7. \(105^\circ\): Construct the angle midway between \(90^\circ\) and \(120^\circ\).
In simple words: Standard angles and their bisectors allow the precise geometric construction of these specific measures without a protractor.

Exam Tip: Leave all construction arcs visible to earn full marks in geometry exams.

Question 16. Draw a line segment of length 7.2 cm. Bisect it & measure the length of each part.
Answer: Steps of construction:
1. Draw a line segment \(AB = 7.2\text{ cm}\).
2. With \(A\) as centre and radius more than half of \(AB\), draw arcs above and below the line segment.
3. With \(B\) as centre and the same radius, draw arcs cutting the previous arcs at points \(X\) and \(Y\).
4. Join \(XY\) intersecting \(AB\) at point \(M\).
5. Measuring each part, \(AM = MB = 3.6\text{ cm}\).
In simple words: Perpendicular bisectors divide a line segment into two equal halves.

Question 17. Draw an angle of measure \(147^\circ\) and construct its bisector. Also, write its steps of construction.
Answer: Steps of construction:
1. Draw a ray \(OA\).
2. Place the protractor on \(O\) and mark a point \(B\) at \(147^\circ\). Join \(OB\) to form \(\angle AOB = 147^\circ\).
3. With \(O\) as centre and any convenient radius, draw an arc cutting \(OA\) at \(P\) and \(OB\) at \(Q\).
4. With \(P\) as centre and radius greater than half of \(PQ\), draw an arc in the interior of the angle.
5. With \(Q\) as centre and the same radius, draw another arc intersecting the previous arc at point \(R\).
6. Draw ray \(OR\). Ray \(OR\) is the required angle bisector.
In simple words: An angle bisector divides an angle into two equal parts using intersecting arcs from the arms of the angle.

Exam Tip: Ensure the compass width is kept constant when drawing arcs from points \(P\) and \(Q\).

Question 18. Define the following: Range, Frequency, Data, Pictograph
Answer: - Range: The difference between the highest and lowest values in a given set of data.
- Frequency: The number of times a particular observation occurs in a given data set.
- Data: A collection of numerical figures or facts gathered for information or analysis.
- Pictograph: A representation of numerical data using symbols, pictures, or icons.
In simple words: Statistics definitions help us organize, summarize, and interpret collections of numerical information.

Exam Tip: Remember formula: Range = Highest value - Lowest value.

Question 19. Following are the no.of members in 20 families in a village : 5, 9, 4, 8, 8, 7, 6, 7, 6, 6, 6, 2, 8, 5, 7, 5, 8, 9, 7, 6. Prepare a frequency distribution table for the data.
Answer: Arranging the data in ascending order: 2, 4, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9

In simple words: Count how many times each number appears and record it using tally marks and frequencies.

Exam Tip: Always verify that the sum of all frequencies equals the total number of observations given in the problem.

Question 20. In a Primary School the no. of students in different classes are as follows : CLASS I: 200, II: 120, III: 160, IV: 140, V: 120. Represent this data by a pictograph.
Answer: Let one symbol \((\smiley)\) represent 40 students.
- Class I (200 students): \(200 \div 40 = 5\) symbols (\(\smiley\smiley\smiley\smiley\smiley\))
- Class II (120 students): \(120 \div 40 = 3\) symbols (\(\smiley\smiley\smiley\))
- Class III (160 students): \(160 \div 40 = 4\) symbols (\(\smiley\smiley\smiley\smiley\))
- Class IV (140 students): \(140 \div 40 = 3.5\) symbols (\(\smiley\smiley\smiley\text{ (half)}\))
- Class V (120 students): \(120 \div 40 = 3\) symbols (\(\smiley\smiley\smiley\))

In simple words: Choose a convenient scale factor so numbers can be easily represented using picture symbols.

Exam Tip: Always state the key clearly (e.g., Let 1 symbol = 40 students) below or beside the pictograph.

Question 21. The following table shows the average rainfall in the month of June to Nov. in Mumbai : Months: JUNE (24), JULY (30), AUG (36), SEPT (28), OCT (20), NOV (4). Rainfall in cm. Draw the bar-graph & answer the following questions : i) What is the highest rainfall in the season? ii) Which month has the least rainfall? iii) What is the total rainfall in the season?
Answer:

i) Highest rainfall in the season = 36 cm (in August)
ii) Month with the least rainfall = November (4 cm)
iii) Total rainfall in the season = \(24 + 30 + 36 + 28 + 20 + 4 = 142\text{ cm}\)
In simple words: Analyze the data values to find maximums, minimums, and sums directly from the table.

Exam Tip: When drawing a bar graph, maintain equal width for all bars and equal spacing between them.

Question 22. Find the least number which when divided by 12, 16, 24 and 36 leaves a remainder 7 in each case.
Answer: First, find the LCM of 12, 16, 24, and 36.
Prime factorization:
12 = \(2^2 \times 3\)
16 = \(2^4\)
24 = \(2^3 \times 3\)
36 = \(2^2 \times 3^2\)
LCM = \(2^4 \times 3^2 = 16 \times 9 = 144\)
The required number leaves a remainder 7 in each case, so we add the remainder to the LCM:
Required number = \(\text{LCM} + 7\)
Required number = \(144 + 7 = 151\)
In simple words: Find the least common multiple of the divisors and add the common remainder to it.

Exam Tip: Always add the remainder at the very end after calculating the LCM.