Maths Mock Test- 3 - Class 10 MCQ

2x² + 3 + 2√6x + x² = 3x² - 5x
3 + 2√6x + 5x= 0

Which is not a quadratic equation.

(a) The given equation is x² - 4x + 3√2 = 0.
On comparing with ax² + bx + c = 0, we get
a = 1, b = -4 and c = 3√2
The discriminant of x² - 4x + 3√2 = 0 is
D = b² - 4ac
= (-4)² - 4(1)(3√2) = 16 - 12√2 = 16 - 12 x (1.41)
= 16 - 16.92 = -0.92
⇒ b² - 4ac < 0
(b) The given equation is x² + 4x - 3√2 = 0
On comparing the equation with ax² + bx + c = 0, we get
a = 1, b = 4 and c = -3√2
Then, D = b² - 4ac = (-4)² - 4(1)(-3√2)
= 16 + 12√2 > 0
Hence, the equation has real roots.
(c) Given equation is x² - 4x - 3√2 = 0
On comparing the equation with ax² + bx + c = 0, we get
a = 1, b = -4 and c = -3√2
Then, D = b² - 4ac = (-4)² - 4(1) (-3√2)
= 16 + 12√2 > 0
Hence, the equation has real roots.
(d) Given equation is 3x^​2 + 4√3x + 4 = 0.
On comparing the equation with ax² + bx + c = 0, we get
a = 3, b = 4√3 and c = 4
Then, D = b² - 4ac = (4√3)² - 4(3)(4) = 48 - 48 = 0
Hence, the equation has real roots.
Hence, x² - 4x + 3√2 = 0 has no real roots.

Since HCF(p,q,r)*LCM(p,q,r) is not equal to pq/r, neither it is equal to qr/p and neither is p,q,r. So the correct answer is D . Also, HCF(p,q,r)*LCM(p,q,r) is not equal to p*q*r . This condition only holds for two numbers.

The correct option is Option D.

Given : D, E and F are tge mid-point of the sides BC, CA and AB respectively of Δ ABC.

To prove : Δ DEF is congruent to traingle

Proof : 

Since E and F are midpoints of AC and AB. 

BC II FE and FE = ½ BC = BD (By mid point theorem)

BD II FE and BD = FE 

Similarly, BF II DE and BF = DE

Hence, BDEF is a parallelogram (A pair of opposite sides are equal and parallel)

Similarly, we can prove that FDCE and AFDE are also parallelograms. 

Now, BDEF is a parallelogram so it's diagonal FD divides it into two traingles of equal areas. 

Therefore, ar(Δ BDF) = ar(Δ DEF).......... (i)

In parallelogram AFDE, 

ar(Δ AFE) = ar(Δ DEF)   (EF is a diagonal)......... (ii)

In parallelogram FDCE, 

ar(Δ CDE) = ar(Δ DEF)   (DE is a diagonal)...........(iii)

From (i), (ii) and (iii)

ar(Δ BDF) = ar(Δ AFE) = ar(Δ CDE) = ar(Δ DEF)..........(iv)

If area of traingles are equal then they are congruent.

Hence, Δ DEF is congruent to triangle Δ BDF = Δ AFE = Δ CDE.

In ΔCAB and ΔADB
Angle B is common and Angle A=Angle D
So the triangles are similar

a=cb/p
Now applying pythagoras theorem in 
Δ ABC
H² = P²+B²
BC²=AC²+AB²

we know sin(90 - a) = cos(a) 

cos(90 - a) = sin(a)

sin(a) = 1/cosec(a)

sec(a) = 1/cos(a)

cos40 = cos(90-50) = sin50

cosec40 = cosec(90-50) = sec50

so our expression becomes

sin50/sin50 + sec50/sec50 - 4cos50 / sin40

= 1 + 1 - 4(1)   since cos50 = sin40

= -2

We will factorize 24 to 6x4, take out the 4 from root to make both common

For first A.P
2+5+8+11+......+98
a=2,an​=98,d=3
an​=a+(n−1)d
98=2+(n−1)3
98=2+3n−3
3n=99
n=33
Number of term =33
For first A.P
3+8+13+18+......+198
a=3,an​=198,d=5
an​=a+(n−1)d
198=3+(n−1)5
198=3+5n−5
5n=200
n=40
No of terms =40
Common terms=40−33=7

l=a+(n-1)d
(p+q)th term is m
m=a+(p+q-1)d
m=a+pd+qd-d  ….1
(p-q)th term is n
n=a+(p-q-1)d
n=a+pd-qd-d   ….2
Adding 1 and 2
m+n=2a+2pd-2d
m+n=2(a+pd-d)
½(m+n=a+(p-1)d
So pth term is ½(m+n)

The area should be 144π,
 ,Area of the circle= πr²
144π = πr²
r = 12cm
Circumference=2πr = 24πcm

Volume of water displaced = 150 x 8 = 1200 m³ 
⇒ 90 x 40 x h = 1200
⇒ 

Construct DG || BF
In triangle ADG,
AE = ED ( given)
EF || DG ( by construction)
a line passing through the mid point of any side of a triangle, parallel to the other side, bisects the third side
So AF = FG    ...(1)
Now in triangle CBF
BD = DC ( given)
DG||BF ( by construction)
a line passing through the mid point of any side of a triangle, parallel to the other side, bisects the third side
So, FG = GC …(2)
Now, by (1) & (2)
AF = FG = GC
 AF : FC = 1:2

Here are the two solutions of each of the given equations.x−y+1=03x+2y−12=0


The triangle formed by given two linear equations is shaded in the graph. The triangle formed by given lines is ΔAEF Therefore, the vertices of triangle AEF are A ( –1, 0), E (2, 3) and F (4, 0).

Given: a₁ = 5,a₂ = 3,b₁ = −15,b₂ = −9,c₁ = 8 and 

Therefore, the the pair of given linear equations has infinitely many solutions.

Given: x−y = 2…..(i)
And x+y = 4 ………(ii)
Adding eq. (i) and (ii) for elimination of y, we get
2x = 6 ⇒ x = 3
Putting the values of x in eq. (i), we get
3−y = 2 ⇒ y = 1
∴x = α = 3 and y = β = 1

Since both these points lie on a straight line i.e x axis, distance will be the difference between the respective y coordinates

(0,-3)   (0,7)

7-(-3) = 7+3 = 10