CBSE Class 5 Mathematics Tenths And Hundredths Worksheet Set 05

Question 1. Arrange in ascending order : 0.002, 0.02, 0.2, 2.0.
Answer: Ascending order means arranging numbers from the smallest to the largest.
Comparing the decimal numbers by their place values:
0.002 = 0 thousandths
0.02 = 20 thousandths
0.2 = 200 thousandths
2.0 = 2000 thousandths
Therefore, the ascending order is: \(0.002, 0.02, 0.2, 2.0\)
In simple words: Compare decimals from left to right, matching place values, and list them from smallest to largest.

Exam Tip: Make all decimals have the same number of decimal places by adding trailing zeros (0.002, 0.020, 0.200, 2.000) to avoid comparison errors.

Question 2. Arrange in descending order 21.111, 12.121, 22.121, 22.111
Answer: Descending order means arranging numbers from the largest to the smallest.
Comparing the whole number parts first: 22 is greater than 21, and 21 is greater than 12.
Comparing the numbers starting with 22: 22.121 is greater than 22.111.
Comparing the numbers starting with 21 and 12: 21.111 comes next, followed by 12.121.
Therefore, the descending order is: \(22.121, 22.111, 21.111, 12.121\)
In simple words: Order the numbers from the biggest value to the smallest value.

Exam Tip: Always compare the whole number part first before looking at the decimal places.

Question 3. Add : a) 463.23+ 55.555+ 444.5+ 6.503
Answer: Align the decimal points vertically and add zeros to make all numbers have the same number of decimal digits:
\(463.230 + 55.555 + 444.500 + 6.503\)
Sum of thousandths: \(0 + 5 + 0 + 3 = 8\)
Sum of hundredths: \(3 + 5 + 0 + 0 = 8\)
Sum of tenths: \(2 + 5 + 5 + 5 = 17\) (write 7, carry 1)
Sum of ones: \(3 + 5 + 4 + 6 + 1 = 19\) (write 9, carry 1)
Sum of tens: \(6 + 5 + 4 + 1 = 16\) (write 6, carry 1)
Sum of hundreds: \(4 + 4 + 1 = 9\)
Total sum = 969.788
In simple words: Line up the decimal points and add column by column from right to left.

Exam Tip: Pad numbers with trailing zeros so every number has an equal number of decimal places before adding.

Question 3. Add : b) 228.18 + 72.65 + 115.38
Answer: Align the decimal points vertically:
\(228.18 + 72.65 + 115.38\)
Sum of hundredths: \(8 + 5 + 8 = 21\) (write 1, carry 2)
Sum of tenths: \(1 + 6 + 3 + 2 = 12\) (write 2, carry 1)
Sum of ones: \(8 + 2 + 5 + 1 = 16\) (write 6, carry 1)
Sum of tens: \(2 + 7 + 1 + 1 = 11\) (write 1, carry 1)
Sum of hundreds: \(2 + 1 + 1 = 4\)
Total sum = 416.21
In simple words: Align decimals and add each place value carefully with carrying over.

Exam Tip: Double-check your column addition and carry-overs to avoid simple arithmetic mistakes.

Question 4. Evaluate : a) 126.84 – 151.6 + 218.931
Answer: Group the positive numbers together and then subtract the negative number:
Positive numbers: \(126.840 + 218.931 = 345.771\)
Now subtract 151.600:
\(345.771 - 151.600 = 194.171\)
Total result = 194.171
In simple words: First add all positive numbers together, then subtract the negative number.

Exam Tip: Use the BODMAS/PEMDAS rule by performing addition before subtraction, or group positive terms first.

Question 4. Evaluate : b) 29.6 + 15.2 – 6.9
Answer: First, add the positive numbers:
\(29.6 + 15.2 = 44.8\)
Next, subtract 6.9 from the sum:
\(44.8 - 6.9 = 37.9\)
Total result = 37.9
In simple words: Add the first two numbers and then subtract the third number.

Exam Tip: Perform calculations step-by-step from left to right when operators have the same priority.

Question 5. Draw the net of a cuboid in three different ways.S
Answer: A cuboid has 6 rectangular faces. Its net can be drawn in multiple ways by unfolding the faces along its edges. One standard net is a cross shape with flaps, another is a straight strip of 4 rectangles with top and bottom attached to the second rectangle, and a third is an L-interlocking arrangement.
In simple words: A cuboid net is a flat 2D shape that can be folded up to make a 3D rectangular box.

Exam Tip: Ensure the net has exactly 6 rectangular faces with opposite faces being identical in dimensions.

Question 6. If the distance between two towns is 90km. Find the distance between them in the map if 1cm = 12km.
Answer: Actual distance between two towns = 90 km
Map scale: \(1\text{ cm} = 12\text{ km}\)
Distance on the map = \(\frac{\text{Actual distance}}{\text{Scale}}\)
Distance on the map = \(\frac{90}{12} = 7.5\text{ cm}\)
In simple words: Divide the total actual distance by the scale value to find the distance on the map.

Exam Tip: Always check the units and divide actual distance by scale factor when converting to map distance.

Question 7. Draw the net of a dice in two different ways.
Answer: A dice is a cube having 6 square faces. Two different nets can be drawn:
1. The standard T-net or cross-net (a line of 4 squares with one square attached on either side of the second square).
2. A stair-step net (a row of 3 squares with offset squares attached above and below).
In simple words: A dice net consists of 6 connected squares that fold into a cube.

Exam Tip: Verify that folding the net mentally results in all 6 faces enclosing a complete cube without overlapping.

Question 8. Meena’s sister in America had sent her 12 USA dollars as a gift. She used Rs. 525 for books. How much money was left with her? (1 USA dollar = Rs. 62.50)
Answer: Total dollars received = 12 USA dollars
Value of 1 USA dollar = Rs. 62.50
Total money in rupees = \(12 \times 62.50 = \text{Rs. } 750.00\)
Money spent on books = Rs. 525.00
Money left = \(750.00 - 525.00 = \text{Rs. } 225.00\)
In simple words: Multiply the dollars by the conversion rate to get total rupees, then subtract the money spent.

Exam Tip: Convert foreign currency into local currency first using multiplication before performing subtraction.

Question 9. A brick has a length of 22cm, breadth of 6cm and height of 5cm. What is the volume of 5 bricks?
Answer: Length of one brick (\(l\)) = 22 cm
Breadth of one brick (\(b\)) = 6 cm
Height of one brick (\(h\)) = 5 cm
Volume of one brick = \(l \times b \times h = 22 \times 6 \times 5\)
\(22 \times 30 = 660\text{ cm}^3\)
Volume of 5 bricks = \(5 \times 660 = 3300\text{ cm}^3\)
In simple words: Find the volume of one brick by multiplying length, breadth, and height, then multiply by 5.

Exam Tip: Volume of a cuboid is \(l \times b \times h\) and its unit is cubic units (\(\text{cm}^3\)).

Question 10. Write in metres. a) 89 centimetre = ____________ metres.
Answer: Since \(100\text{ cm} = 1\text{ m}\), to convert centimetres to metres, divide by 100.
\(89\text{ cm} = \frac{89}{100} = 0.89\text{ metres}\)
In simple words: Divide centimetres by 100 to get metres.

Exam Tip: Moving the decimal point two places to the left converts cm to m.

Question 10. Write in metres. b) 15 metre 2 centrimetre = ____________ metres.
Answer: \(15\text{ m}\) is already in metres.
Convert \(2\text{ cm}\) into metres: \(\frac{2}{100} = 0.02\text{ m}\)
Total = \(15 + 0.02 = 15.02\text{ metres}\)
In simple words: Keep the metres as whole numbers and convert the centimeters into a decimal, then add them.

Exam Tip: Watch out for missing place values; 2 cm is 0.02 m, not 0.2 m.

Question 10. Write in metres. c) 1 centimetre = ____________ metre.
Answer: \(1\text{ cm} = \frac{1}{100}\text{ m} = 0.01\text{ metre}\)
In simple words: 1 centimeter is one-hundredth of a meter.

Exam Tip: Remember that centi means one-hundredth (\(\frac{1}{100}\)).

Question 11. How many pens can Radha buy in Rs. 120, if the cost of a pen is Rs. 13. How much money is left with her?
Answer: Total money available = Rs. 120
Cost of 1 pen = Rs. 13
Number of pens bought = \(120 \div 13\)
\(13 \times 9 = 117\)
So, Radha can buy 9 pens.
Money spent = \(9 \times 13 = \text{Rs. } 117\)
Money left = \(120 - 117 = \text{Rs. } 3\)
In simple words: Divide total money by the cost of one pen to find the quantity, and use the remainder for money left.

Exam Tip: The quotient gives the number of items bought, and the remainder gives the leftover money.

Question 12. If Dinesh grows 55cm more then his height will be 2 metre. What is his height today?
Answer: Target height = \(2\text{ m} = 200\text{ cm}\)
Growth needed = 55 cm
Current height = Target height - Growth needed
Current height = \(200\text{ cm} - 55\text{ cm} = 145\text{ cm}\) (or \(1.45\text{ m}\))
In simple words: Subtract 55 cm from 200 cm to find Dinesh's current height.

Exam Tip: Convert all units to the same system (centimetres) before performing subtraction.

Question 13. Temperature of Jaipur at 3 PM = \(42.8^{\circ}\text{C}\). How many degrees will the temperature need to rise for it to reach \(45^{\circ}\text{C}\) in Jaipur.
Answer: Current temperature = \(42.8^{\circ}\text{C}\)
Desired temperature = \(45.0^{\circ}\text{C}\)
Temperature rise needed = \(45.0 - 42.8 = 2.2^{\circ}\text{C}\)
In simple words: Subtract the current temperature from the target temperature to find the required rise.

Exam Tip: Align decimal points properly when subtracting decimal temperatures.

Question 14. Draw the three dimensional figures for the following nets.
Answer: 1. The first net (a semicircle sector folding into a cone with a circular base) represents a Cone.
2. The second net (a rectangle with two circles attached at opposite ends) represents a Cylinder.
In simple words: The first net folds into a cone, and the second net folds into a cylinder.

Exam Tip: Recognize standard circular 3D shapes and their flat curved nets.

Question 15. Cost of an egg is Rs. 5, find the cost of two and half dozen eggs?
Answer: Cost of 1 egg = Rs. 5
1 dozen = 12 eggs
Two and half dozen = \(2.5 \times 12 = 30\text{ eggs}\)
Total cost = \(30 \times 5 = \text{Rs. } 150\)
In simple words: Calculate total eggs by multiplying 2.5 by 12, then multiply by the cost per egg.

Exam Tip: Remember that a dozen equals 12 items.

Question 16. Find the volume of – a) A cuboid having length 22cm, breadth 12cm and height 8cm.
Answer: Length (\(l\)) = 22 cm, Breadth (\(b\)) = 12 cm, Height (\(h\)) = 8 cm
Volume of cuboid = \(l \times b \times h\)
\(= 22 \times 12 \times 8\)
\(= 22 \times 96 = 2112\text{ cm}^3\)
In simple words: Multiply length, breadth, and height to get the volume.

Exam Tip: Volume is always expressed in cubic units (\(\text{cm}^3\)).

Question 16. Find the volume of – b) A cube of side 15cm.
Answer: Side of cube (\(s\)) = 15 cm
Volume of cube = \(s^3 = s \times s \times s\)
\(= 15 \times 15 \times 15\)
\(= 225 \times 15 = 3375\text{ cm}^3\)
In simple words: Multiply the side length by itself three times.

Exam Tip: A cube has equal length, breadth, and height.

Question 17. The fencing of a square garden is 24m. Find the area of a garden.
Answer: Perimeter (fencing) of square garden = 24 m
Perimeter of square = \(4 \times \text{side}\)
Side = \(\frac{24}{4} = 6\text{ m}\)
Area of square = \(\text{side} \times \text{side} = 6 \times 6 = 36\text{ m}^2\)
In simple words: Divide the perimeter by 4 to get the side length, then square it to find the area.

Exam Tip: Perimeter is the total boundary length, whereas area is the surface space inside.

Question 18. How many square stamps of side 5cm can fit on to a page of length 90cm and breadth 25cm?
Answer: Area of one square stamp = \(5 \times 5 = 25\text{ cm}^2\)
Area of the page = \(\text{length} \times \text{breadth} = 90 \times 25 = 2250\text{ cm}^2\)
Number of stamps that can fit = \(\frac{\text{Area of page}}{\text{Area of one stamp}}\)
\(= \frac{2250}{25} = 90\text{ stamps}\)
In simple words: Divide the total area of the page by the area of one stamp.

Exam Tip: Dividing total area by unit area is the quickest way to find how many items fit without gaps.

Question 19. Draw four different rectangles of area 24 sq.cm. a) Find the perimeter of each rectangle.
Answer: Rectangles with area 24 \(\text{cm}^2\) have length (\(l\)) and breadth (\(b\)) such that \(l \times b = 24\).
Possible dimensions (length \(\times\) breadth) and their perimeters (\(2 \times (l + b)\)):
1. \(24\text{ cm} \times 1\text{ cm}\): Perimeter = \(2 \times (24 + 1) = 50\text{ cm}\)
2. \(12\text{ cm} \times 2\text{ cm}\): Perimeter = \(2 \times (12 + 2) = 28\text{ cm}\)
3. \(8\text{ cm} \times 3\text{ cm}\): Perimeter = \(2 \times (8 + 3) = 22\text{ cm}\)
4. \(6\text{ cm} \times 4\text{ cm}\): Perimeter = \(2 \times (6 + 4) = 20\text{ cm}\)
In simple words: Find pairs of factors of 24 and calculate the perimeter for each pair.

Exam Tip: Perimeter formula for a rectangle is \(2 \times (\text{length} + \text{breadth})\).

Question 19. Draw four different rectangles of area 24 sq.cm. b) Which rectangle has largest perimeter?
Answer: Comparing the perimeters (50 cm, 28 cm, 22 cm, 20 cm):
The rectangle with dimensions \(24\text{ cm} \times 1\text{ cm}\) has the largest perimeter of 50 cm.
In simple words: The longest and skinniest rectangle has the largest perimeter.

Exam Tip: For a fixed area, the rectangle with the most extreme factor pair has the largest perimeter.

Question 19. Draw four different rectangles of area 24 sq.cm. c) Find the difference between largest and smallest perimeter.
Answer: Largest perimeter = 50 cm
Smallest perimeter = 20 cm (\(6\text{ cm} \times 4\text{ cm}\))
Difference = \(50 - 20 = 30\text{ cm}\)
In simple words: Subtract the smallest perimeter from the largest perimeter.

Exam Tip: Always subtract the minimum value from the maximum value when finding differences.

Question 20. Draw four different rectangles having the same perimeter of 18cm. a) Find the area of each rectangle.
Answer: Since Perimeter = \(2 \times (l + b) = 18\), the sum of length and breadth is \(\frac{18}{2} = 9\text{ cm}\).
Possible pairs of integer dimensions (\(l, b\)) where \(l + b = 9\) and their areas (\(l \times b\)):
1. \(8\text{ cm} \times 1\text{ cm}\): Area = \(8 \times 1 = 8\text{ cm}^2\)
2. \(7\text{ cm} \times 2\text{ cm}\): Area = \(7 \times 2 = 14\text{ cm}^2\)
3. \(6\text{ cm} \times 3\text{ cm}\): Area = \(6 \times 3 = 18\text{ cm}^2\)
4. \(5\text{ cm} \times 4\text{ cm}\): Area = \(5 \times 4 = 20\text{ cm}^2\)
In simple words: Find pairs of numbers that add up to 9, then multiply them to find their areas.

Exam Tip: Half of the perimeter gives the sum of length and breadth.

Question 20. Draw four different rectangles having the same perimeter of 18cm. b) Find the difference between largest and smallest area.
Answer: Largest area = \(20\text{ cm}^2\)
Smallest area = \(8\text{ cm}^2\)
Difference = \(20 - 8 = 12\text{ cm}^2\)
In simple words: Subtract the smallest area from the largest area.

Exam Tip: For a fixed perimeter, dimensions that are closest together give the largest area.

Question 21. How many square cakes each of side 4cm can be arranged on a tray of length 80cm and breadth 60cm?
Answer: Area of one square cake = \(4 \times 4 = 16\text{ cm}^2\)
Area of the tray = \(80 \times 60 = 4800\text{ cm}^2\)
Number of cakes = \(\frac{\text{Area of tray}}{\text{Area of one cake}}\)
\(= \frac{4800}{16} = 300\text{ cakes}\)
In simple words: Divide the total area of the tray by the area of one cake.

Exam Tip: Alternatively, calculate how many cakes fit along the length (\(80 \div 4 = 20\)) and breadth (\(60 \div 4 = 15\)), then multiply them (\(20 \times 15 = 300\)).

Question 22. Draw a square of 16 sq.cm. Name it as A. Draw another square with the double size of A . Name it as B. a) Find the side of A and B.
Answer: Area of square A = \(16\text{ cm}^2\)
Side of square A = \(\sqrt{16} = 4\text{ cm}\)
Square B has double the area of A: Area of square B = \(2 \times 16 = 32\text{ cm}^2\)
Side of square B = \(\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}\text{ cm} \approx 5.66\text{ cm}\).
In simple words: Side of A is 4 cm because \(4 \times 4 = 16\). Side of B is the square root of 32.

Exam Tip: Doubling the area does not double the side length; side length increases by a factor of \(\sqrt{2}\).

Question 22. Draw a square of 16 sq.cm. Name it as A. Draw another square with the double size of A . Name it as B. b) Find the relation between the area of A and B.
Answer: Area of A = \(16\text{ cm}^2\)
Area of B = \(32\text{ cm}^2\)
Relation: The area of B is twice (2 times) the area of A (\(\text{Area}_B = 2 \times \text{Area}_A\)).
In simple words: B has twice the surface area of A.

Exam Tip: Read carefully whether "double size" refers to area or side length.

Question 22. Draw a square of 16 sq.cm. Name it as A. Draw another square with the double size of A . Name it as B. c) Find the relation between the perimeter of A and B.
Answer: Perimeter of A = \(4 \times 4 = 16\text{ cm}\)
Perimeter of B = \(4 \times 4\sqrt{2} = 16\sqrt{2}\text{ cm} \approx 22.63\text{ cm}\)
Relation: The perimeter of B is \(\sqrt{2}\) times the perimeter of A (\(\text{Perimeter}_B = \sqrt{2} \times \text{Perimeter}_A\)).
In simple words: The perimeter of B is scaled by a factor of \(\sqrt{2}\) compared to A.

Exam Tip: When area doubles, perimeter scales by the square root of 2, not by 2.