Online Free Mathematics Mock Test MCQ type Questions [28.10.2019]
Online Free Mock Test MCQ type Questions
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Solutions:
1. Answer :Option B
The perimeter of a rhombus = 4l
Diagonal 6cm = 3cm + 3cm = legs of right triangle within rhombus.
Diagonal 8cm = 4cm + 4cm = legs of right triangle within rhombus.
Hence perimeter of rhombus = 4×5 = 20cm
Diagonal 8cm = 4cm + 4cm = legs of right triangle within rhombus.
Hence perimeter of rhombus = 4×5 = 20cm
2. Answer :Option C
Side of square 
= 10 cm
Now, length of the rectangle = (10 + 2) cm = 12 cm
We know that, 2(L + B) = Perimeter of rectangle = 40
2 × 12 + 2B = 40, 2B = 40-24 = 16, B = 16/2 = 8 cm
Hence area of rectangle = (12 × 8) cm2 = 96 cm2
= 10 cmNow, length of the rectangle = (10 + 2) cm = 12 cm
We know that, 2(L + B) = Perimeter of rectangle = 40
2 × 12 + 2B = 40, 2B = 40-24 = 16, B = 16/2 = 8 cm
Hence area of rectangle = (12 × 8) cm2 = 96 cm2
3. Answer :Option B
Perimeter of trapezium = 58 cm
Sum of its non-parallel = 20 cm
Sum of parallel sides = (58 – 20) = 38 cm
Area of trapezium=
×d × (Sum of parallel sides)
152 =
×d × 38
By solving this,we get
∴d = 8 cm
Sum of its non-parallel = 20 cm
Sum of parallel sides = (58 – 20) = 38 cm
Area of trapezium=
×d × (Sum of parallel sides)152 =
×d × 38By solving this,we get
∴d = 8 cm
4. Answer :Option A
Given: Cuboidal box – length = 13 cm long,
Breadth = 11 cm
and height = 9 cm
Cubical box – side = 12 cm.
Let’s find the volume of the cuboidal box.
We know, volume of cuboid = length × breadth × height
So, volume of cuboidal box = 13 × 11 × 9 [as given] = 1287 cm3
We know, volume of cube = (side)3
So, volume of cubical box = (12)3 = 1728 cm3
Total volume = 1287 + 1728 = 3015 cm3
According to the question,
Tanu wants to pack 3060 cubes of side 1 cm into this box.
Volume of one such cube of side 1 cm = (1)3 = 1 cm3
Total volume of cubes which Tanu wants to pack = 1 × 3060 = 3060 cm3
Since, only 3015 cm3 volume is available, it won’t be able to accumulate 3060 cm3 volume of total cubes.
∴ 3060 – 3015 = 45 boxes left unpack.
Thus, Option A is correct.
Breadth = 11 cm
and height = 9 cm
Cubical box – side = 12 cm.
Let’s find the volume of the cuboidal box.
We know, volume of cuboid = length × breadth × height
So, volume of cuboidal box = 13 × 11 × 9 [as given] = 1287 cm3
We know, volume of cube = (side)3
So, volume of cubical box = (12)3 = 1728 cm3
Total volume = 1287 + 1728 = 3015 cm3
According to the question,
Tanu wants to pack 3060 cubes of side 1 cm into this box.
Volume of one such cube of side 1 cm = (1)3 = 1 cm3
Total volume of cubes which Tanu wants to pack = 1 × 3060 = 3060 cm3
Since, only 3015 cm3 volume is available, it won’t be able to accumulate 3060 cm3 volume of total cubes.
∴ 3060 – 3015 = 45 boxes left unpack.
Thus, Option A is correct.
5. Answer :Option D
Given: Perimeter of a square = 44 cm
Perimeter of a rectangle = Perimeter of a rectangle = 44 cm
We know that, Perimeter of square = 4 × side
⇒ 44 = 4 × side
⇒ side = 44/4 = 11 cm
Thus, area of square = (side)2 =(11)2 = 121 cm2 …(i)
According to the question,
The length of the rectangle is 5 cm more than the side of the square.
We have found that side of the square = 11 cm
⇒ length of the rectangle = 5 cm + side of the square
⇒ length of the rectangle = 5 cm + 11 cm = 16 cm …(ii)
∵, Perimeter of the rectangle = 44 cm
⇒ 2 (length + breadth) = 44
⇒ 2 (16 + breadth) = 44 [by (ii)] ⇒ 16 + breadth = 22
⇒ breadth = 22 – 16 = 6 cm
Length and breadth of the rectangle are 16 cm and 6 cm respectively.
Thus, Area of the rectangle = 16 × 6 = 96 cm2 …(iii)
By equations (i) & (ii), we get
Sum of the areas of square and rectangle = 121 + 96 = 217 cm2
Hence, option D is correct.
Perimeter of a rectangle = Perimeter of a rectangle = 44 cm
We know that, Perimeter of square = 4 × side
⇒ 44 = 4 × side
⇒ side = 44/4 = 11 cm
Thus, area of square = (side)2 =(11)2 = 121 cm2 …(i)
According to the question,
The length of the rectangle is 5 cm more than the side of the square.
We have found that side of the square = 11 cm
⇒ length of the rectangle = 5 cm + side of the square
⇒ length of the rectangle = 5 cm + 11 cm = 16 cm …(ii)
∵, Perimeter of the rectangle = 44 cm
⇒ 2 (length + breadth) = 44
⇒ 2 (16 + breadth) = 44 [by (ii)] ⇒ 16 + breadth = 22
⇒ breadth = 22 – 16 = 6 cm
Length and breadth of the rectangle are 16 cm and 6 cm respectively.
Thus, Area of the rectangle = 16 × 6 = 96 cm2 …(iii)
By equations (i) & (ii), we get
Sum of the areas of square and rectangle = 121 + 96 = 217 cm2
Hence, option D is correct.
6. Answer :Option B
Given that, length of the carton is 8 cm and breadth of the carton is 8 cm.
Let h be the depth (or height) of carton.
We know that, 1 litre = 1000 cm3
Also, Volume of carton = length × breadth × height
⇒ 1000 = 8 × 8 × h [By substituting values in the formula] ⇒ h = 1000/64 = 15.625 ≈ 16 cm
Hence, option B is correct.
Let h be the depth (or height) of carton.
We know that, 1 litre = 1000 cm3
Also, Volume of carton = length × breadth × height
⇒ 1000 = 8 × 8 × h [By substituting values in the formula] ⇒ h = 1000/64 = 15.625 ≈ 16 cm
Hence, option B is correct.
7. Answer :Option C
Area of the square room of side 10 m or 1000 cm
= 1000 × 1000 sq cm
Now, area of tile = (50)2 = 50 × 50 sq cm
∴ Number of tiles needed.
{(1000*1000)/50*50} = 400
Hence, 400 tiles will be needed.
= 1000 × 1000 sq cm
Now, area of tile = (50)2 = 50 × 50 sq cm
∴ Number of tiles needed.
{(1000*1000)/50*50} = 400
Hence, 400 tiles will be needed.
8. Answer :Option A
With option (2) one can form a dice.
(2+5) = 7
(1+6) =7
(3+4) = 7
Hence they will be the opposite faces of each other.
(2+5) = 7
(1+6) =7
(3+4) = 7
Hence they will be the opposite faces of each other.
9. Option D
Measurement of each sheet of paper is 22 cm and 28 cm.
If one sheet rolled out into a cylinder through its length thus, the height of the cylinder, h = 22 cm and the area of the base = 28 cm
We know that, a circumference of base = 2πr
∴ In that case,
If other sheet rolled out into a cylinder through its width thus, the height of the cylinder, h = 28 cm and the area of the base = 22 cm
In that case,
Therefore, required difference = (1372 – 1078) cm3 =294cm3
If one sheet rolled out into a cylinder through its length thus, the height of the cylinder, h = 22 cm and the area of the base = 28 cm
We know that, a circumference of base = 2πr
∴ In that case,
If other sheet rolled out into a cylinder through its width thus, the height of the cylinder, h = 28 cm and the area of the base = 22 cm
In that case,
Therefore, required difference = (1372 – 1078) cm3 =294cm3