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Class 5 Maths Chapter 8 HOTS Questions - Chapter 8 - Mapping your Way

Q1: Calculate the area of rectangle if one square represents an area of 1 cm2.
Class 5 Maths Chapter 8 HOTS Questions - Chapter 8 - Mapping your Way A)  2cm2
B)  9cm2       
C)  50cm2
D)  24cm2
Ans: 
(d)
Sol: 
There are 6  ×4 = 24 squares and each square has an area of = 1cm2. So, the area of rectangle
= 6 × 4 × 1 = 24cm2.

Q2: What is the area of rectangle if Length = 2m, Breadth = 5m
A)  12m2
B)  14m2         
C)  10m       
D)  18m2
Ans:
(c)
Sol: The area of rectangle is given by length × breadth. So the area is 2 × 5 = 10m2.

Q3: Each side of square is of 10m. What will be the area of the square?
A)  100m2
B)  90m       
C)  34m2                           
D)  115m2
Ans:
(a)
Sol: Area of square = side × side =10 × 10= 100m2.

Q4: The third side of the triangle if the perimeter of a triangle is 20m and two of its sides are 5m and 4m, is
A) 11 m
B) 10m
C) 12m
D) 5m
Ans:
(a)
Sol: Perimeter of a triangle =1st side + 2nd side + 3rd side So, 20 = 5 + 4 + 3rd side 3rd side
= 20 − 9
= 11m.

Q5: Find the missing side in the figure given below if the perimeter of the figure is 28m.
Class 5 Maths Chapter 8 HOTS Questions - Chapter 8 - Mapping your WayA) 12m
B) 3m
C) 9m
D) 6m
Ans: 
(c)
Sol:
Given figure is a rectangle. Also, given I = length = 8m Let b be the breadth of the rectangle. Perimeter of rectangle = 2(1 + b) = 28
⇒   l + b = 28 / 2
⇒   l + b = 14 ⇒ b = 14 ? 8 = 6m.

Q6: Find the area of shaded region. (Each square = 1m2)

Class 5 Maths Chapter 8 HOTS Questions - Chapter 8 - Mapping your Way A)  21m2                             
B)  35m2
C)  34m2
D) 13m2         
Ans:
(a)
Sol: 
Calculate no. of shaded boxes. It comes out to be 21. And each box has area of 1m2. So the area of shaded region becomes 21m2.

Q7:The length and breadth of a rectangle are (3a + 2) and (2a - 1). Which of the following represents its perimeter?
A) 2(5a - 1)
B) (5a + 1)
C) (5a - 1)                  
D) 2(5a + 1)
Ans: 
(d)
Sol:
Perimeter = 2[3a + 2 + 2a - 1] = 2[5a + 1]

Q8: The difference between the length and the breadth of a rectangle is 8 cm and the perimeter is 64 cm. Which of the following can be the length and breadth of this rectangle?
A) L = 12 cm, B = 4 cm
B) L = 20 cm, B = 8 cm
C) L = 20 cm, B = 12 cm
D) L = 12 cm, B = 8 cm
Ans: 
(c)
Sol: 
Length of rectangle = x cm Breadth of rectangle = (x - 8) cm
Perimeter of rectangle = 2(x + x - 8)
= 2(2x - 8) Given 2(2x - 8)
= 64
⇒  2x = 40  
⇒  x = 20 L = 20,
B = 20 - 8 = 12

Q9: The area of a triangle whose base is 12 cm and the Height twice the base, is
A) 144sq.cm
B) 288 sq. cm
C) 289 sq. cm
D) Cannot be determined
Ans: 
(a)
Sol: Base ofΔ = 12 cm
Height ofΔ = 2 × 12 = 24cm
Area ofΔ = 12 × 12 × 24
= 144sq.cm.

Q10: The area of a square is 100 sq. cm. If the sides this square are increased by 10% then what will be the area of new square?
A) 121 sq. cm  
B) 101 sq. cm
C) 81 sq. cm                    
D) 169 sq. cm
Ans:
(a)
Sol: Given area of square = 100 side × side = 100 = 10 × 10
⇒Side = 10 cm
Increased in side of square = 10% of 10 = 1
∴ New side = 10 + 1 = 11
∴ Area of new square =(11)2
=121sq.cm.

Q11: The breadth of a rectangle is increased by 2 units. Its perimeter is now increased by?
A) 2 units                        
B) 4 units        
C) 8 units                        
D) 16 units
Ans:
(b)
The initial perimeter (P1) of the rectangle is given by:
P1 = 2L + 2B
Now, the breadth is increased by 2 units, so the new breadth is B + 2.
The new perimeter (P2) of the rectangle with the increased breadth is given by:
P2 = 2L + 2(B + 2) = 2L + 2B + 4
The increase in perimeter (ΔP) is the difference between the new perimeter and the initial perimeter:
ΔP = P2 - P1 = (2L + 2B + 4) - (2L + 2B) = 4
So, the perimeter is increased by 4 units when the breadth of the rectangle is increased by 2 units.
The correct answer is (B) 4 units.

Q12: The area of a square is equal to the area of a rectangle of I = 8 cm and b = 2 cm. What is the side of the square?
A) 6 cm                          
B) 4 cm          
C) 3 cm                          
D) 8 cm
Ans:
(b)


Q13: How many small cubes of side 2 cm can be put in a cubical box of side 6 cm?
A) 9
B) 12
C) 27
D) 6
Ans:
(c)
Sol: Volume of the cubical box = (side length)3 = (6 cm)3 = 216 cubic cm.
Volume of one small cube = (side length)3 = (2 cm)3 = 8 cubic cm.
Now, divide the volume of the box by the volume of one small cube:
Number of small cubes that can fit = (Volume of box) / (Volume of one small cube) = 216 cubic cm / 8 cubic cm = 27.
So, you can fit 27 small cubes with a side length of 2 cm inside a cubical box with a side length of 6 cm.
The correct answer is (C) 27.

Q14: A cuboid measures 24m × 12m × 10m.how many cubes of side 3m can fit in the box?
A) 9
B) 16
C) 15
D) 24
Ans: 
(a)
Sol: 9 m

Q15: Area of a square is:
A) product of all sides
B) sum of all sides
C)  side × side
D)  2 × side
Ans: 
(c)

Q16: Calculate the area of rectangle given below if each square has an area of 4m2.
Class 5 Maths Chapter 8 HOTS Questions - Chapter 8 - Mapping your WayA)  80m2
B)  64m2
C)  40m2
D)  15m2
Ans:
(c)
Sol:
There are 2  x 5 = 10 squares and each square has an area of 4m2 So, the area of rectangle
= 2 × 5 × 4 = 40m2.

Q17: Which of these figures represents the area of 24m2 if given that each block Class 5 Maths Chapter 8 HOTS Questions - Chapter 8 - Mapping your Way  = 2m2 
(a) Class 5 Maths Chapter 8 HOTS Questions - Chapter 8 - Mapping your Way
(b) Class 5 Maths Chapter 8 HOTS Questions - Chapter 8 - Mapping your Way
(c) Class 5 Maths Chapter 8 HOTS Questions - Chapter 8 - Mapping your Way
(d) Class 5 Maths Chapter 8 HOTS Questions - Chapter 8 - Mapping your Way
Ans:
(c)
Sol: Given each block is of 2m2,so there should be 12 blocks in a figure to get an area of 24m2. In option (c), there are 2 × 6 = 12 blocks in figure.

Q18: Area of rectangle is 32m2. If the length is 8m then breadth = ________
A) 20m
B) 4m  
C) 10m
D) 12m
Ans:
(b)
Sol: Given, area = length × breadth
⇒ breadth = area ÷ length
So, the breadth = 32/8
= 4m.

Q19: Consider the following two statements.
Statement A: Perimeter is defined as the length of boundary line of a close geometrical figure.
Statement B: Holding capacity of a container is called volume. 

Which one of the following option is correct about the above statements?
A) Only A is correct.
B) Only B is correct
C) Both A and B are correct.  
D) Neither A nor B are correct.
Ans:
(c)

  • Statement A accurately defines what perimeter is, which is the length of the boundary of a closed geometric figure.
  • Statement B accurately defines what volume is, which is the holding capacity of a container or the amount of space enclosed by a three-dimensional object.
  • Both statements are correct in their respective definitions.

Q20: Consider the following two statements.
Statement A: If edge of a cube measures 25 m and I = 20 m, b = 20m, h = 30 m are the measures of a cuboid then cuboid has the larger volume.
Statement B: Dilip breaks a rod of 90 cm into four equal pieces. Taking three parts he makes a triangle by joining end to end. Perimeter of the triangle is 67.5 cm.
Which one of the following option is correct about the above statements?                
A) Statement A is false and B is correct.
B) Statement A is true and B is false.
C) Both statements are true.
D) Both statements are false.
Ans:
(a)
Statement A is false. In statement A cube has the larger volume.
Statement B: Each part = 90 / 4 = 2.25cm.
∴ Perimeter of triangle = 3 × (22.5)cm = 67.5cm.

The document Class 5 Maths Chapter 8 HOTS Questions - Chapter 8 - Mapping your Way is a part of the Class 5 Course Mathematics for Class 5: NCERT.
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