OAVS TGT Math Mock Test - 3 - OTET MCQ

If we throw a die once, then possible outcomes (s), are
S = { 1, 2, 3, 4, 5, 6 }
⇒    n(E) = 6
(i) Let E be the favourable outcomes of getting an even number, then
E = { 2, 4, 6 }
⇒ n(S) = 3

The correct option is Option A.

All equilateral triangles have the same angles by the congruence rule (SSS)

So, TTS <—> PQR

Factors of a² + ab + bc + ca

• Step 1: Factor out common terms

Given expression: a² + ab + bc + ca

= a(a + b) + c(b + a)

= a(a + b) + c(a + b)

• Step 2: Factor by grouping

= (a + c)(a + b)

Therefore, the factors of a² + ab + bc + ca are (a + c)(a + b).

Let the angle be x
Then the complement angle will be 90^∘ −x

Given x is 14^∘ more than 90^∘ −x
⟹ x = 14^∘ + 90^∘ −x
⟹ 2x = 104^∘
⟹ x = 52^∘
The measure of the angle is 52^∘

Bcs (ABC) is a full triangle ABC triangle divide into four equal partsTherefore, 64÷4=16 cm²

In a parallelogram ABCD, ∠D = 115° (Given)

Since, ∠A and ∠D are adjacent angles of parallelogram. 

We know, Adjacent angles of a parallelogram are supplementary. 

∠A + ∠D = 180° 

∠A = 180° – 115°  = 65° 

Measure of ∠A is 65°.

Correct Answer :- a

Explanation : Multiple of 3 : 3,6,9,12,15,18,21,24

Multiple of 5 : 5,10,15,20,25

Number of possible outcomes (multiple of 3 and 5) = {15} = 1

Number of Total outcomes = 25

∴ Required Probability = 1/25

5²ⁿ −2²ⁿ is of the form a²ⁿ − b²ⁿ which is divisible by both (a + b) and (a – b).
So, 5²ⁿ − 2²ⁿ is divisible by both 7, 3.

Upstream = (x-y) km/hr

Downstream =(x+y)km/hr

20(x+y)=2

X+Y=10.....(1)

4(x-y)=2

x-y=2....(2)

Solve both equations

x+y=10

x-y=2

2x=12

X=6

Substitute x=6in i

6+y=10

Y=4

Speed of rowing in still water is 6km/hr.

the speed of still water is 4km/hr

So option B is correct answer. 

Let three spheres are S₁, S₂ & S₃
having radii r₁ = 3cm, r₂ = 4cm & r₃ = 5cm respectively.
Let the radius of new Big sphere S is R.
A/Q,
Volume of new Sphere  S = Sum of volumes three Spheres S₁, S₂ & S₃

Distance increased every day is the common difference for the AP which is 0.5
Initial distance is 5 km , so a=5
To find the distance at 10th day, so n=10
l=a+(n-1)d
 =5+(10-1)0.5
 =5+4.5=9.5km

A non-leap year has 365 days

A year has 52 weeks. Hence there will be 52 Sundays for sure.

52 weeks = 52 x 7 = 364 days .

365– 364 = 1 day extra.

In a non-leap year there will be 52 Sundays and 1day will be left.

This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, friday, Saturday, Sunday.

Of these total 7 outcomes, the favourable outcomes are 1.

Hence the probability of getting 53 sundays = 1/7.

Let α,β,γ are the zeroes of the given polynomial. Given: α = 2
 Since αβγ = -d/a 

we can’t have values of 3 and – 5

As we will get forms in the expression which has no solution.

So, the assertion is correct.

D = b² – 4ac > 0 then the roots of the quadratic equation ax² + bx + c = 0 are real and unequal. The reason perfectly explains the roots of the equation whose D > 0. But it doesn’t explains the assertion.

Therefore, Both A and R are true and R is not correct explanation for A.

Since diagonals of a rhombus bisect each other at right angle .

∴ In △AOB , we have 

∠OAB + ∠x + 90° = 180° 

∠x = 180° -  90° - 35° [∵ ∠OAB = 35°]

= 55° 

Also, ∠DAO = ∠BAO = 35°  

∴ ∠y + ∠DAO + ∠BAO + ∠x = 180°  

⇒ ∠y + 35° + 35° + 55° =  180°  

⇒ ∠y = 180° - 125° = 55°

Hence the values of x and y are x =  55°, y =  55°.    

We have mean = 
Mean = 
Now each value is increased by 5
Mean = 
+5=65+5=70
Hence the mean is 70

Let us consider the sum of two angles as ∠A+∠B=116∘ and the difference can be written as ∠A−∠B=24^o

We know that the sum of all the angles in a triangle is 180^o.

So we can write it as

∠A+∠B+∠C=180^o

By substituting ∠A+∠B=116^o in the above equation
116^o + ∠C = 180^o

On further calculation

∠C=180^o − 116^o

By subtraction

∠C=64^o

It is given that ∠A−∠B=24^o

It can be written as ∠A=24^o+∠B

Now by substituting ∠A=24^o+∠Bin∠A+∠B=116^o

∠A+∠B=116^o

24^o+∠B+∠B=116^o

On further calculation

24^o+2∠B=116^o

By subtraction

2∠B=116^o−24^o

2∠B=92^o

By division

∠B=292​
∠B=46∘
By substituting ∠B=46∘ in ∠A=24∘+∠B

We get

∠A=24^o+46^o
By addition
∠A=70^o
Therefore, ∠A=70^o,∠B=46^o and ∠C=64^o

Let ABCD is a square circumscribing a circle of radius a cm.
The side of square ABCD = Diameter of circle
⇒AB=2a
Therefore, perimeter of square AB=4×AB=4×2a=8 cm

Mean of 5 nos.= 28

Therefore, sum of the 5 nos. = 28 * 5 = 140 

The the no. exclude by 'x'.

New sum of 4 nos.= 140 - x

New mean = 28- 2 = 26

Therefore, (140 - x ) / 4 = 26

=> 140 - x = 104

=> x = 36

A linear equation with one variable consists of numbers or constants and multiplies of a variable. The standard form of such an equation is ax+b=0, where a and b are constants and x is the variable.
Hence, 2x+7=8  is a linear equation in one variable.

Clarification: The graph of x²+2x+1 does not cut the x-axis, because it has imaginary roots.
x²+2x+2=0
D = b²-4ac = 2²-4(1)(2) = 4 - 8 = -4
as D<0 so it does not touch x-axis

A circle divides the plane on which it lies into three parts:
(i) Inside the circle, which is also called the interior of the circle.
(ii) The circle
(iii) Outside the circle, this is also called the exterior of the circle.

The circle and its interior make up the circular region.
 

The given AP is x – 7, x – 2, x + 3 …
So a= x-7, d=x - 2- x + 7 = x + 3 - x +2 = 5
l = a + (n - 1)d
=x - 7 + (15 - 1)5
=x - 7 + 70 = x + 63

∵ Probability of any event cannot be more than 1.
∴ 1.5 can not be the probability of any event.
∴ (a) is the answer.

Given:

Radius = 10.5 cm

Height = 6 m

Weight of 1 cm³ = 5 grams

Formula used:

Volume of cylinder = πr²h

1 m = 100 cm

⇒ 6 m = 600 cm

1 kg = 1000 gram

Calculation:

According to the question

Volume of cylinder = πr2h

⇒ (3.14 × 10.5 × 10.5 × 600) cm³

⇒ 207711 cm³

Now, Weight of 1 cm3​ = 5 grams

So, Weight of 207,711 cm³ = 5 × 207711

⇒ 1,038,555 grams

⇒ 1038.555 kg ~ 1038.5 kg