OAVS TGT Math Mock Test - 4 - OTET MCQ

In triangle OAB

OA = OB (radius of a circle)

∠OAB = ∠OBA

∠OBA = 40º (angles opposite to equal sides are equal)

Using the angle sum property

∠AOB + ∠OBA + ∠BAO = 180º

Substituting the values

∠AOB + 40º + 40º = 180º

By further calculation

∠AOB + 80º = 180º

∠AOB = 180º - 80º

∠AOB = 100º

As the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle

∠AOB = 2 ∠ACB

Substituting the values

100º = 2 ∠ACB

Dividing both sides by 2

∠ACB = 50º

Therefore, ∠ACB is equal to 50º.

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

Solution: From equation (i) x = 14 - 4y ....(iii)

Substitute the value of x in equation (ii)

⇒ 7 (14 - 4y) - 3y = 5

⇒ 98 - 28y - 3y = 5

⇒ 98 - 31y = 5

⇒ 93 = 31y

⇒ y = 93/31

⇒ y = 3

Now substitute value of y in equation (iii)

⇒ 7x - 3 (3) = 5

⇒ 7x - 3 (3) = 5

⇒ 7x = 14

⇒ x = 14/7 = 2

So the solution is x = 2 and y = 3

 Given area of circle=154
      ⇒area of circle=πr²
                             =22/7 ×7×7.
                             =154 
so radius of the circle=7cm
 perimeter of circle=2πr
                           =2 ×22/7×7
                           =44cm
⇒perimeter of the circle=44 cm.

Assume that √7 be a rational number.
i.e. √7 = p/q, where p and q are co prime.
⇒ 7 = p²/q² 
⇒ 7q² = p²    ...(1)
⇒ p² is divisible by 7, i.e. p is divisible by 7
⇒ For any positive integer c, it can be said that p = 7c, p² = 49c²

Equation (1) can be written as: 7q² = 49c²
⇒ q² = 7c²
This gives that q is divisible by 7.

Since p and q have a common factor 7 which is a contradiction to the assumption that they are co-prime.
Therefore, √7 is an irrational number.

Number of Total outcomes = 52
Number of aces and Number of kings = 4 + 4 = 8
Number of cards except ace and king = 52 – 8 = 44
Required Probability = 44/52 = 11/13

Median=L + 
=40 + 
=40 + 6 = 46

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.

The least number of acute angles that a triangle can have is 2.

As we cannot have more than one right angle or obtuse angle, we have only two or three acute angles in a triangle.

Further, if one angle is acute, sum of other two angles is more than 90⁰ and we cannot have two right angles or obtuse angles.

a₂₁ -a₇ = {a + (21 - 1)d}

- {a + (7 - 1)d} = 84

a + 20d- a - 6d = 84

14d = 84

d = 18/14 = 6

d = 6

So, A is incorrect but R is correct

Product of Zeroes

⇒ k² – 4k + 4 = 0

⇒ (k – 2)² = 0 ⇒ k = 2

The probability of a sure event is always 1.

The reason is also true as n(n+1) is divisible by 2 always for all natural number n.

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

Since, HCF of 56 and 72, by Euclid’s divsion lemma,
72 = 56 × 1 + 16 ……….(i)
56 = 16 ×3 + 8 ……….(ii)
16 = 8× 2 + 0 ……….(iii) 
∴ HCF of 56 and 72 is 8. 
∴ 8 = 56 – 16× 3
8 = 56 – (72 – 56 ×1) ×3
[From eq. (i) : 16 = 72 – 56× 1]
8 = 56 – 3 ×72 + 56× 3
8 = 56 × 4 + (–3) × 72 
∴ x = 4,y = −3

The first 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.

Mean = Sum of Observations/Total No of Observations

Mean = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 ÷ 10

Mean = 55/10

Mean = 5.5

l = 2 x 5 = 10 cm , b = 5 cm

and h = 5 cm

Total surface area = 2(lb + bh + lh)

= 2(10 x 5 + 5 + 5 x 10 x 5)

= 2 x 125 = 250 cm²

Pentagon (5 sided polygon).

Assertion: Consider the given polynomial x² + 4x + 5
as its discriminant is negative b² -4ac = 16 - 20 = -4 so it dont have real roots ,
∴ x² + 4x + 5 has no zeroes.
∴ Assertion is false.
Reason: Clearly, Reason is true.
Since Assertion (A) is false but reason (R) is true.

A prime number is a natural number greater than 1 that is not a product of two smaller natural numbers. So 10 prime numbers are: 2,3,5,7,11,13,17,19,23,29
Since the number of terms are even ,
Median = 
 10/2=5th term=11
10/2 +1=6th term=13
Median = 

Given AB is the tower.

P and Q are the points at distance of 4m and 9m respectively.

From fig, PB = 4m, QB = 9m.

Let angle of elevation from P be α and angle of elevation from Q be β.

Given that α and β are supplementary. Thus, α + β = 90

In triangle ABP,

tan α = AB/BP – (i)

In triangle ABQ,

tan β = AB/BQ

tan (90 – α) = AB/BQ (Since, α + β = 90)

cot α = AB/BQ

1/tan α = AB/BQ

So, tan α = BQ/AB – (ii)

From (i) and (ii)

AB/BP = BQ/AB

AB^2 = BQ x BP

AB^2 = 4 x 9

AB^2 = 36

Therefore, AB = 6.

Hence, height of tower is 6m.

P = 640/1000 = 0.64