Maths Mock Test- 3 - Class 9 MCQ

To calculate the volume of sand needed to fill the rectangular sandbox, we can use the formula:Volume = Length × Width × DepthGiven:
Length = 2 m
Width = 5 m
Depth = 10 cm = 0.1 mSubstituting the values into the formula:Volume = 2 m × 5 m × 0.1 m
Volume = 1 cubic meterTherefore, the correct answer is A: 1 cubic meter.

• In a coordinate plane, the quadrants are defined as follows:
  
  
  • I Quadrant: (x > 0, y > 0)
  
  
  • II Quadrant: (x < 0, y > 0)
  
  
  • III Quadrant: (x < 0, y < 0)
  
  
  • IV Quadrant: (x > 0, y < 0)
  
   
• The point B(-3, 4) has x-coordinate -3 and y-coordinate 4.
• Since x is negative and y is positive, it lies in the II Quadrant.

The length of the longest rod that can fit in a cubical vessel of side 10 cm is equal to the length of the diagonal of the cube. Using the Pythagorean theorem, the length of the diagonal of the cube can be found as:Diagonal = √(length² + breadth² + height²)
Diagonal = √(10² + 10² + 10²)
Diagonal = √300
Diagonal = 10√3 cmTherefore, the length of the longest rod that can fit in a cubical vessel of side 10 cm is 10√3 cm, which corresponds to option C.

- A rational number is a number that can be expressed as a fraction where both the numerator and denominator are integers.
- √196 is a rational number because it simplifies to 14/1, which is a fraction where both the numerator and denominator are integers.
- Options B and C are irrational numbers because they cannot be expressed as fractions.
- Option A is a repeating decimal, which can be rational if it eventually settles into a repeating pattern, but without further information, it is not clear if this is the case.

Required answer,

Note:
We have taken integral value only, not the fractional.
For example 146/5 = 29.2 but we have taken 29 and so on.

2ACB=AOB
ACB=40
CAB+ACB+ABC=180
CAB=180-70
CAB=110
since,OA=OB(radius)
OAB=OBA

AOB+OAB+OBA=180
2OBA=100
OBA=50

CAO=CAB-OBA=110-5O
CAO=60^o

The point D(3, -6) lies in the IV Quadrant.
Explanation: In the Cartesian plane, the X-axis and Y-axis divide the plane into four quadrants. The I Quadrant is where both x and y are positive. The II Quadrant is where x is negative and y is positive. The III Quadrant is where both x and y are negative. The IV Quadrant is where x is positive and y is negative.
The point D(3, -6) has a positive x-coordinate and a negative y-coordinate. Therefore, it lies in the IV Quadrant.

Let a = 3x and b = 5 (3x)³ - (5)³ - 3×3x×5(3x- 5) = 27x³ - 125 -45x ( 3x -5) = 27x³ -125 - 135x² + 225x = 27x³ - 135x² + 225x - 125

X+ 1/x = 7 then,
cubing both side
= (x +1/x)³ = (7)³
= x³ + 1/x³ + 3×x×1/x×(x+1/x) = 343
= x³ +1/x³ +3(7) = 343
= x³ +1/x³ +21 =343
= x³ +1/x³ =343 - 21
= x³ +1/x³ = 322