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OTET Exam  >  OTET Tests  >  OAVS TGT Math Mock Test - 5 - OTET MCQ

OAVS TGT Math Mock Test - 5 - OTET MCQ


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30 Questions MCQ Test - OAVS TGT Math Mock Test - 5

OAVS TGT Math Mock Test - 5 for OTET 2024 is part of OTET preparation. The OAVS TGT Math Mock Test - 5 questions and answers have been prepared according to the OTET exam syllabus.The OAVS TGT Math Mock Test - 5 MCQs are made for OTET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for OAVS TGT Math Mock Test - 5 below.
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OAVS TGT Math Mock Test - 5 - Question 1
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Expansion and simplification of 8(3h - 4) + 5(h - 2) gives

Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 1

Now, 8 (3h - 4) + 5 (h - 2)

= 24h - 32 + 5h - 10, the expansion

= 24h + 5h - 32 - 10

= 29h - 42, the simplification

OAVS TGT Math Mock Test - 5 - Question 2
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A square of side 7cmis inscribed in a circle. The area enclosed between the circle and the square is

Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 2


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OAVS TGT Math Mock Test - 5 - Question 3
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Direction: In the following questions, a statement of Assertion is given followed by a corresponding statement of Reason just below it. Of the statements, mark the correct answer as

Assertion : A diameter of a circle is the longest chord of the circle and all diameters have equal length.

Reason : Length of a diameter = radius.

Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 3

Length of diameter = 2 × radius.

So, Reason is false.

Also, chord is a line connecting two points on a circle. The farthest points are the ones which are collinear to the centre of the circle. Points collinear to the centre of the circle are joined by the diameter. Hence diameter is the longest chord. Assertion is correct but Reason is false.

OAVS TGT Math Mock Test - 5 - Question 4
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In the given figure, AB = EF, BC = DE, AB ⊥ BD and EF ⊥ CE. Which of the following criterion is true for ΔABD ≅ ΔEFC?
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 4

In triangle ABD and FEC:
► AB = FE ( given )
► ∠FEC = ∠ABD ( 90degree)
► BC = DE
CD is common part coming in both triangles.
► BC + CD = CD + DE
► BD = CE
Therefore, triangle ABD is congruent to triangle FEC by SAS rule of congruence.

OAVS TGT Math Mock Test - 5 - Question 5
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The point (0, 3) lies on :
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 5
This is because y is also called X=0 and x=0,y=3 so it will lie on positive y axis(+ve y axis)
OAVS TGT Math Mock Test - 5 - Question 6
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What is the value of p if x-2 is a factor of x2 – 6x + p ?
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 6

P(x) = ×2 - 6×+p
p (2)= (2)2-6 (2) +p
4- 12 + p = 0
-8 + p = 0
p = 8 
 

OAVS TGT Math Mock Test - 5 - Question 7
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The degree of the polynomial x4 – 3x3 + 2x2 – 5x + 3 is:
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 7

The degree refers to the highest power of the polynomial. In this polynomial X has highest power 4.So the degree of polynomial is 4.

OAVS TGT Math Mock Test - 5 - Question 8
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The age of 18 students of a class is reported below. Their modal age is 10, 17, 14, 10, 11, 12, 12, 13, 17, 13, 14, 14, 15, 16, 17, 15, 17, 16​
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 8

Mode is the highest number of times an observation occurs, which means the number that appears most frequently in a set. So modal age is the age which occurs maximum number of times. It is evident that 17 occurs most of the times. So, modal age is 17 years.

OAVS TGT Math Mock Test - 5 - Question 9
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The diagonals of rhombus are 12 cm and 16 cm. The length of the side of rhombus is:
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 9

We know that, the diagonals of a rhombus are perpendicular bisector of each other.
Given,  AC = 16 cm and BD = 12 cm        [let]
∴ AO = 8cm, SO = 6cm
and  ∠AOB = 90°
In right angled ∠AOB,

OAVS TGT Math Mock Test - 5 - Question 10
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If A = 2n + 13, B = n + 7, where n is a natural number then HCF of A and B is:      
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 10

A = 2n + 13
B = n + 7

1) If the value of n = 1, then

A = (2*1) + 13
= 2 + 13
= 15

B = 1 + 7
= 8

Now, A = 15 and B = 8
HCF of 15 and 8 is 1.

2) If the value of n =2 

A = (2*2) + 13
= 4 + 13 
= 17

B = 2 + 7 
= 9

A = 17 and B = 9
HCF of 17 and 9 is 1.

3) If the value of n 3

A = (2*3) + 13
= 6 + 13
= 19

B = 3 + 7
= 10

A = 19 and B = 10
HCF of 19 and 10 is 1

4) If the value of n = 4
A = (2*4) + 13
= 8 +13 
= 21

B = 4 + 7
= 11

A = 21 and B = 11

HCF of 21 and 11 is 1

5) If the value of n = 5
A = (2*5) + 13
= 10 + 13
= 23

B = 5 + 7
= 12

A = 23 and B = 12
HCF of 23 and 12 is 1

6) If the value of n = 6
A = (2*6) + 13
= 12 + 13
= 25

B = 6 + 7
= 13

A = 25 and B = 13
HCF of 25 and 13 is 1

7) If the value of n = 7
A = (2*7) + 13
14 + 13
= 27

B = 7 + 7
= 14

A = 27 and B = 14
HCF of 27 and 14 is 1

8) If the value of n = 8
A = (2*8) + 13
16 + 13
= 29

B = 8 + 7
= 15

A = 29 and B = 15
HCF of 29 and 15 is 1

9) If the value of n = 9
A = (2*9) + 13
18 + 13
= 31

B = 9 + 7
= 16

A = 31 and B = 16
HCF of 31 and 16 is 1

10) If the value of n = 10 
A = (2*10) + 13
20 + 13
= 33

B = 10 + 7
= 17

A = 33 and B = 17
HCF of 33 and 17 is 1

We have seen that whatever the value n has, the HCF of A and B (separate values they hold in each situation) is always 1. It means we can take any value of n and the HCF will be 1.

OAVS TGT Math Mock Test - 5 - Question 11
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The surface area of a cube of side 27 cm is :
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 11

The surface area of a cube of side 27 cm can be found using the formula:Surface area of a cube = 6a^2, where a is the length of the side of the cube.Given:
Length of side of the cube = 27 cmSubstituting the value into the formula:Surface area of the cube = 6 × 27^2
Surface area of the cube = 6 × 729
Surface area of the cube = 4374 cm^2Therefore, the surface area of the cube of side 27 cm is 4374 cm^2, which corresponds to option C

OAVS TGT Math Mock Test - 5 - Question 12
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A garden roller has a circumference of 4 m. The number of revolutions it makes in moving 40 metres are:
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 12

We have circumference which is equal to the boundary of the circle. So we have one revolution equal to circumference of the circle which is equal to 4
Distance covered by the roller=circumference of the roller*number of revolutions
⇒ number of revolutions = distance covered by the roller/circumference of the roller
= 40/4 = 10

OAVS TGT Math Mock Test - 5 - Question 13
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Read the following text and answer the following questions on the basis of the same: Amit has a packet of Candies. It consists of 288 candies. He arranges the candies in a way that first row contains 3 candies, second row has 5 and third row has 7 and so on.

Q. Find the difference in the candies placed in the 10th and 15th row.
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 13
a15 − a10 = a + 14d − a − 9d = 5d = 5(2) = 10
OAVS TGT Math Mock Test - 5 - Question 14
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Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

Assertion (A): Sum of all 11 terms of an A.P whose middle most term is 30 is 330.

Reason (R): Sum of first n terms of an A.P is given by l is the middle term
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 14
Middle most term =

a6 = a + 5d = 30

= 11 × 30 = 330

Also, Sum of first n terms of an A.P is given by is last term(Here we can’t apply this formula as no last term is mentioned in the statement.)

Therefore, A is true but R is false.

OAVS TGT Math Mock Test - 5 - Question 15
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The area of a circle is 38.5 sq. cm. Its circumference is
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 15

Therefore, the correct answer is Option 1: 22 cm.

OAVS TGT Math Mock Test - 5 - Question 16
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What is the probability of getting one head and one tail when a coin is tossed 2 times?
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 16

Assume that is an equal chance of the coin landing on heads or tail (the coin is fair, not biased).

Probability of coin landing on head and then on tails=1/2 x 1/2=1/4

Probability of coin landing on tails and then on heads= 1/2 x 1/2 =1/4

Therefore, probability of getting one head and one tail in two coin tosses=1/4+1/4=1/2

OAVS TGT Math Mock Test - 5 - Question 17
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Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

Assertion (A): The roots of the quadratic equation x2 + 2x + 2 = 0 are imaginary.

Reason (B): If discriminant D = b2 – 4ac < 0 then the roots of the quadratic equation ax2 + bx + c = 0 are imaginary.
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 17
Let us apply the reason on the equation,

x2 + 2x + 2 = 0

D = b2 – 4ac = (2)2 – 4(1)(2) = 4 – 8 = – 4

is unreal.

No real value is possible in this case.

Therefore, Both A and R are true and R is the correct explanation for A.

OAVS TGT Math Mock Test - 5 - Question 18
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The nature of the roots of the equation x2 – 5x + 7 = 0 is –
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 18

Given equation is x2-5x+7=0
We have discriminant as b2-4ac=(-5)2-4*1*7= -3
And x = , Since we do not have any real number which is a root of a negative number, the roots are not real.

OAVS TGT Math Mock Test - 5 - Question 19
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If α, β are roots of the equation x2 + 5x + 5 = 0, then equation whose roots are α + 1 and β + 1 is
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 19

OAVS TGT Math Mock Test - 5 - Question 20
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The number   is
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 20





Since √2 and √5 both are irrational number therefore  is an irrational number.

OAVS TGT Math Mock Test - 5 - Question 21
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The angle of elevation of the top of a tower from two points P and Q at distances of ‘a’ and ‘b’ respectively from the base and in the same straight line with it are complementary. The height of the tower is
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 21

Let TW be the tower of height hh meters and P and Q are two points such that PW = a and QW = b.



OAVS TGT Math Mock Test - 5 - Question 22
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The curved surface area of a right circular cone whose slant height is 14 cm and base radius is 21 cm is​
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 22

L = 14 cm
R = 21 cm
C S A of Cone = πRL
22/7 * 21 * 14 cm
22 * 21 * 2 cm
462 * 2 cm
924 cm2

OAVS TGT Math Mock Test - 5 - Question 23
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If the two zeroes of the quadratic polynomial 7x2 – 15x – k are reciprocals of each other, the value of k is:​
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 23

OAVS TGT Math Mock Test - 5 - Question 24
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If a - b = 3 and a2 + b2 = 29, find the value of ab.
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 24

2ab = (a2 + b2) - (a - b)2
= 29 - 9 = 20
⇒ ab = 10.

OAVS TGT Math Mock Test - 5 - Question 25
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In a triangle ABC if ∠A = 53° and ∠C = 44° then the value of ∠B is:
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 25
B=180-(A+C) B=180-53+44 B=180-97 B=83
OAVS TGT Math Mock Test - 5 - Question 26
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In an AP, if a = 3.5, d = 0, n = 101, then an will be
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 26

a101 = 3.5 + 0 (100) = 3.5

OAVS TGT Math Mock Test - 5 - Question 27
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The product of three consecutive integers is divisible by
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 27

Let the three consecutive positive integers be n, n+1 and n+2.

Whenever a number is divided by 3, the remainder obtained is either 0,1 or 2.

Therefore, n=3p or 3p+1 or 3p+2, where p is some integer.

If n=3p, then n is divisible by 3.

If n=3p+1, then n+2=3p+1+2=3p+3=3(p+1) is divisible by 3.

If n=3p+2, then n+1=3p+2+1=3p+3=3(p+1) is divisible by 3.

So, we can say that one of the numbers among n,n+1 and n+2 is always divisible by 3 that is:

n(n+1)(n+2) is divisible by 3.

Similarly, whenever a number is divided by 2, the remainder obtained is either 0 or 1.

Therefore, n=2q or 2q+1, where q is some integer.

If n=2q, then n and n+2=2q+2=2(q+1) is divisible by 2.

If n=2q+1, then n+1=2q+1+1=2q+2=2(q+1) is divisible by 2.

So, we can say that one of the numbers among n, n+1 and n+2 is always divisible by 2.

Since, n(n+1)(n+2) is divisible by 2 and 3.

Hence, n(n+1)(n+2) is divisible by 6.

OAVS TGT Math Mock Test - 5 - Question 28
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If cos (40° + A) = sin 30°, the value of A is:​
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 28

cos(θ)=sin(90-θ)
so 40+A+30=90
A=20

OAVS TGT Math Mock Test - 5 - Question 29
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Three unbiased coins are tossed. What is the probability of getting at most two heads?
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 29

OAVS TGT Math Mock Test - 5 - Question 30
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Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.​
Detailed Solution for OAVS TGT Math Mock Test - 5 - Question 30

Let x and (16 - x) are two parts of 16 where (16 - x) is longer and x is smaller .
A/C to question,
2 × square of longer = square of smaller + 164
⇒ 2 × (16 - x)² = x² + 164
⇒ 2 × (256 + x² - 32x ) = x² + 164
⇒ 512 + 2x² - 64x = x² + 164
⇒ x² - 64x + 512 - 164 = 0
⇒ x² - 64x + 348 = 0
⇒x² - 58x - 6x + 348 = 0
⇒ x(x - 58) - 6(x - 58) = 0
⇒(x - 6)(x - 58) = 0
⇒ x = 6 and 58

But x ≠ 58 because x < 16
so, x = 6 and 16 - x = 10

Hence, answer is 6 and 10.

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