Maths Mock Test- 4 - Class 10 MCQ

Taking LCM

Let p(x) =ax3 + bx2 + cx + d

Given that, one of the zeroes of the cubic polynomial p(x) is zero.

Let α, β and γ are the zeroes of cubic polynomial p(x), where α = 0.

We know that,

Step-by-step explanation:

sum of two zeros at a time = c/a

                      ∴αβ + βy + yα = c/a

                  ∴0×β + βy + y×0 = c/a

                                         βy = c/a

  hence, product of 2 zeros = c/a

Let radius of circle = R 
∴ πR² = 25π
⇒ R = 5cm
∴ Distance between two parallel tangents 
= diameter = 2 x 5 = 10 cm.

Area of the circle = sum of area of two circles

Angles of ΔABC:

• ∠A = 50°
• ∠B = 70°
• ∠C = 60°

Angles of ΔDEF:

• ∠D = 60°
• ∠E = 70°
• ∠F = 50°

We can observe the following:

• ∠A = 50° corresponds to ∠F = 50°
• ∠B = 70° corresponds to ∠E = 70°
• ∠C = 60° corresponds to ∠D = 60°

So, the corresponding angles of ΔABC and ΔDEF are equal, meaning the two triangles are similar.

The correct answer is:

d) ΔFED.

Angle of elevation is 60
Base = 30m
Height of the tree = Perpendicular
So in the right triangle
Where base is given and we have to find perpendicular we have only tan θ
So, Tan θ = P/B
Tan 60 = P/30
√30 = P/30
P = 30√30

Prime factorisation of 102 = 2 x 3 x 17.
Prime factorisation of 85 = 5 x 17 = (2+3) x 17.
The two numbers are:
1. 2 x 17 = 34.
2. 3 x 17 =51

⇒ a = 0.737373...
⇒  100a = 73.737373
= 73 + a
⇒ a = 73/99 = p/q
⇒  p=73 and q = 99 are co-prime.
Here, q=3² X 11.

Observing both the AP’s we see that the common difference of both the AP’s is same ,so difference between their corresponding terms will be same ie,a₁-b₁=2.5-3.72=-1.22
a₂-b₂=2.51-3.73=-1.22
 So , a₁₀₀₀₀₅-b₁₀₀₀₀₅=-1.22

tan 1.tan 2.tan 3...tan (90 - 3 ).tan ( 90 - 2 ).tan ( 90 - 1) 
=tan 1.tan 2 .tan 3...cot 3.cot 2.cot 1 
=tan 1.cot 1.tan 2.cot 2.tan 3.cot 3 ... tan 89.cot 89 
1 x 1 x 1 x 1 x ... x 1 =1

Let p(x) =ax³ + bx² + cx + d
Given that, one of the zeroes of the cubic polynomial p(x) is zero.
Let α, β and γ are the zeroes of cubic polynomial p(x), where a = 0.
We know that,

The centroid is the intersection of the three medians while the incentre is the intersection of the three (internal) angle bisectors. In an equilateral triangle, each median is also an angle bisector (and vice versa), the centroid coincides with the incentre. In fact, the centroid, incentre, circumcentre and orthocentre of an equilateral triangle are coincide at the same point.

Given that, a =x³y² = x × x × x × y × y
and b = xy³ = x × y × y × y
∴ HCF of a and b = HCF (x³y²,xy³) = x × y × y = xy² 
[Since, HCF is the product of the smallest power of each common prime factor involved in the numbers]

Area of circle with radius R₁ = πR₁²
Area of circle with radius R₂ = πR₂²
Area of circle with radius R = πR²
Area of circle with radius R₁+Area of circle with radius R₂ =Area of circle with radius R
πR₁² + πR₂²=πR²
R₁²+R₂²=R²

The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is 19.404 cm^3.

For the largest right circular cone, the height of the cone and the diameter of the circular base of the cone should be equal to the edge length of the cube
Edge of the cube = 4.2 cm

i.e 2r = 4.2

r = 4.2/2 = 2.1 cm

h = 4.2 cm

Volume of the cone = 1/3 * pi * r² * h

=> 1/3 * 22/7 * 2.1 * 2.1 * 4.2

=> 19.404 cm³

Hence, the volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is 19.404 cm³.

A consistent system means the equations have at least one solution. Graphically, this translates to the lines representing the equations either intersecting at a single point or completely overlapping (coinciding)

Isosceles triangle.