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30 Questions MCQ Test UP PGT Mock Test Series 2024 - UP PGT Math Mock Test - 1

UP PGT Math Mock Test - 1 for UPTET 2024 is part of UP PGT Mock Test Series 2024 preparation. The UP PGT Math Mock Test - 1 questions and answers have been prepared according to the UPTET exam syllabus.The UP PGT Math Mock Test - 1 MCQs are made for UPTET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for UP PGT Math Mock Test - 1 below.
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UP PGT Math Mock Test - 1 - Question 1

Consider the following statements:

I: If A = {x: x is an even natural number} and B = {y: y is a natural number}, A subset B.

II: Number of subsets for the given set A = {5, 6, 7, 8) is 15.

III: Number of proper subsets for the given set A = {5, 6, 7, 8) is 15.

Which of the following statement(s) is/are correct?

Detailed Solution for UP PGT Math Mock Test - 1 - Question 1

Concept:

The null set is a subset of every set. (ϕ ⊆ A)

Every set is a subset of itself. (A ⊆ A)

The number of subsets of a set with n elements is 2n.

The number of proper subsets of a given set is 2n - 1

Calculation:

Statement I: If A = {x: x is an even natural number} and B = {y: y is a natural number}, A subset B.

A = {2, 4, 6, 8, 10, 12, ...} and A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, ...}.

It is clear that all the elements of set A are included in set B.

So, set A is the subset of set B.

Statement I is correct.

Statement II:  Number of subsets for the given set A = {5, 6, 7, 8) is 15.

Given: A = {5, 6, 7, 8}

The number of elements in the set is 4

We know that,

The formula to calculate the number of subsets of a given set is 2n

 = 2= 16

Number of subsets is 16

Statement II is incorrect.

Statement III: Number of proper subsets for the given set A = {5, 6, 7, 8) is 15.

The formula to calculate the number of proper subsets of a given set is 2n - 1

 = 2- 1

 = 16 - 1 = 15

The number of proper subsets is 15.

Statement III is correct.

∴ Statements I and III are correct.

UP PGT Math Mock Test - 1 - Question 2

The eccentricity of the ellipse 9x2 + 5y2 – 30y = 0 is:

Detailed Solution for UP PGT Math Mock Test - 1 - Question 2

9x2 + 5y2 - 30y = 0
9x2 + 5(y−3)2 = 45
We can write it as : [(x-0)2]/5 + [(y-3)2]/9 = 1
Compare it with x2/a2 + y2/b2 = 1
e = [(b2 - a2)/b2]½
e = [(9-5)/9]1/2
e = (4/9)½
e = ⅔

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UP PGT Math Mock Test - 1 - Question 3

The G.M. between the numbers: 56 and 14 is:

Detailed Solution for UP PGT Math Mock Test - 1 - Question 3

Geometric mean of two numbers a and b is √ab
As two numbers are 14 and 56
Geometric mean is √14×56
= ± √2×7×2×2×2×7
= ±(2×2×7)
= ±28

UP PGT Math Mock Test - 1 - Question 4

Detailed Solution for UP PGT Math Mock Test - 1 - Question 4

Area of standard ellipse is given by :πab.

UP PGT Math Mock Test - 1 - Question 5

Detailed Solution for UP PGT Math Mock Test - 1 - Question 5


Apply , C1 → C1 - C2, C2 → C2 - C3,

Because here row 1 and 2 are identical

UP PGT Math Mock Test - 1 - Question 6

The argument of the complex number -i

Detailed Solution for UP PGT Math Mock Test - 1 - Question 6

z = -i
Comlex number z is of the form x + iy
x = 0, y = -1
arg(z) = π − tan−1|y/x|
⇒ π − tan−1|-1/0|
= π - (π/2)
⇒  π - π/2
⇒ π/2

UP PGT Math Mock Test - 1 - Question 7

If a,b,c are integers not all equal andw is a cube root of unity (ω ≠ 1) then the minimum value of |a + bω+ cω2| is

Detailed Solution for UP PGT Math Mock Test - 1 - Question 7




When a = b = 1, c = 2, it gives minimum value (since a,b,c not all equal)

UP PGT Math Mock Test - 1 - Question 8

Let f be a real valued function defined on (0, 1) ∪ (2, 4) such that f ‘ (x) = 0 for every x, then

Detailed Solution for UP PGT Math Mock Test - 1 - Question 8

f ‘ (x) = 0 ⇒ f (x)is constant in (0 , 1)and also in (2, 4). But this does not mean that f (x) has the same value in both the intervals . However , if f (c) = f (d) , where c ∈ (0 , 1) and d ∈ (2, 4) then f (x) assumes the same value at all x ∈ (0 ,1) U (2, 4) and hence f is a constant function.

UP PGT Math Mock Test - 1 - Question 9

If the sum of the roots of the equation ax2 + bx + c = 0 is equal to sum of the squares of their reciprocals, then bc2 , ca2 , ab2 are in

Detailed Solution for UP PGT Math Mock Test - 1 - Question 9


UP PGT Math Mock Test - 1 - Question 10

If A, B, C are angles of a triangle ABC, then

is less than or equal to 

Detailed Solution for UP PGT Math Mock Test - 1 - Question 10

ANSWER :- b

Solution :- sin(A+B+C) = cos(A+B+C)/2 = 0

Hence, sinA/2 sinB/2 sinC/2 ≤ 1/8

UP PGT Math Mock Test - 1 - Question 11

A coin is tossed n times. The probability of getting head at least once is greater than 0.8. Then the least value of n is

Detailed Solution for UP PGT Math Mock Test - 1 - Question 11

Probability of not getting a head = 1/2
Probability of getting at least one head in n tosses is > 0.8 (given)
⇒ 1 -P (getting no head) > 0.8


Putting values for n, we see that the least value of n satisfying (1) is n=3.

UP PGT Math Mock Test - 1 - Question 12

Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation

Detailed Solution for UP PGT Math Mock Test - 1 - Question 12

Let the two numbers be α and β. Given that 

∴ Required equation is x2 – 18x + 16 = 0 

UP PGT Math Mock Test - 1 - Question 13

Nidhi has 6 friends. In how many ways can she invite one or more of them to a party at her home?

Detailed Solution for UP PGT Math Mock Test - 1 - Question 13

She has 6 friends and he wants to invite one or more. That is the same as saying he wants to invite at least 1 of his friends.
 
So, the number of ways he could do this is:
Invite only one friend
Invite any two friends
Invite any three friends
Invite any four friends
Invite any five friends
Invite all six friends
This can be thought of in terms of combinations. Inviting  r  friends out of  n  is same as choosing  r  friends out of  n . So, we can write the possibilities as:
6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 
= 6 + 15 + 20 + 15 + 6 + 1 
= 63

UP PGT Math Mock Test - 1 - Question 14

Two matrices A and B are multiplicative inverse of each other only if

Detailed Solution for UP PGT Math Mock Test - 1 - Question 14

If AB = BA = I , then A and B are inverse of each other. i.e. A is invers of B and B is inverse of A.

UP PGT Math Mock Test - 1 - Question 15

Detailed Solution for UP PGT Math Mock Test - 1 - Question 15

UP PGT Math Mock Test - 1 - Question 16

A fair die is tossed eight times. The probability that a third six is observed on the eight throw is

Detailed Solution for UP PGT Math Mock Test - 1 - Question 16

Required probability 

UP PGT Math Mock Test - 1 - Question 17

The coefficient of y in the expansion of (y² + c/y)5 is 

Detailed Solution for UP PGT Math Mock Test - 1 - Question 17

Given, binomial expression is (y² + c / y)5 
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r 
= 5Cr × y10-3r × Cr 
Now, 10 – 3r = 1 
⇒ 3r = 9 
⇒ r = 3 
So, the coefficient of y = 5C3 × c³ = 10c³

UP PGT Math Mock Test - 1 - Question 18

If the roots of the cubic equation ax3 + bx2 + cx + d = 0 are in G.P., then

Detailed Solution for UP PGT Math Mock Test - 1 - Question 18

Let
If α, β,and γ are the three roots then we must have
ax3 + bx2 + cx + d = a(x−α)(x−β)(x−γ)
Comparing coefficients of various power of x on both sides leads to
α + β + γ = −b/a −−−−(1)
αβ + βγ + γα = c/a −−−−(2)
αβγ= −d/a−−−(3)
On substituting β = αr and γ = αr2 we get
α(1+r+r2) = −b/a−−−(4)
α2(r+r2 + r3) = c/a−−(5)
α3/r3 = -d/a−−−−−(6)
Now, dividing (5) by (4) to αr=− c/b and substituting this in(6)weget
(−c/b)3 = -d/a
⇒ c3/b3 = -d/a
∴ c3a = b3d

UP PGT Math Mock Test - 1 - Question 19

For any two non-zero complex numbers z1 and z2 if then the difference of amplitudes of z1 and z2 is      

Detailed Solution for UP PGT Math Mock Test - 1 - Question 19







UP PGT Math Mock Test - 1 - Question 20

Detailed Solution for UP PGT Math Mock Test - 1 - Question 20

UP PGT Math Mock Test - 1 - Question 21

Express the shaded area in the form of an integral.

Detailed Solution for UP PGT Math Mock Test - 1 - Question 21

As the curve goes from c to d and the equation is x = f(y)
So the shaded area is ∫(c to d)f(y)dy

UP PGT Math Mock Test - 1 - Question 22

The equation of a circle which is concentric to the given circle x2 + y2 - 4x - 6y - 3 = 0 and which touches the Y-axis is:

Detailed Solution for UP PGT Math Mock Test - 1 - Question 22

UP PGT Math Mock Test - 1 - Question 23

If n (P) = 5, n(Q) = 12 and n(P U Q) = 14 then n(P ∩ Q) =

Detailed Solution for UP PGT Math Mock Test - 1 - Question 23

n (P) = 5, n(Q) = 12 and n(PUQ) = 14
n(PUQ) = n(P) + n(Q) - n(P∩Q) 
14 = 5 + 12 - n(P∩Q)
n(P∩Q) = 3

UP PGT Math Mock Test - 1 - Question 24

Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral is:

Detailed Solution for UP PGT Math Mock Test - 1 - Question 24

Hexagon has 6 vertices. Three vertices are chosen.
∴ n(S) = Total number of triangles formed


Let E : Triangle is equilateral
Out of these 20 triangles, only two triangles ACE and BDF are equilateral and each of their side is where ‘a’ is the side of regular hexagon.

UP PGT Math Mock Test - 1 - Question 25

If n is a +ve integer, then the binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are

Detailed Solution for UP PGT Math Mock Test - 1 - Question 25

(x+a)n = nC0 xn + nC1 x(n-1) a1 + nC2 x(n-2) a2 + ..........+ nC(n-1) xa(n-1) + nCn  an
Now, nC0 = nCn, nC1 = nCn-1,    nC2 = nCn-2,........
therefore, nCr = nCn-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.

UP PGT Math Mock Test - 1 - Question 26

 is equal to 

Detailed Solution for UP PGT Math Mock Test - 1 - Question 26

Suppose that the Reqd. Limit L = lim x→0 (sinx − x)/x3.
Substiture x = 3y , so that, as x→0 , y→0.
∴ L = lim y→0 sin3y − 3y/(3y)3,
= lim y→0 (3siny − 4sin3y)− 3y)/27y3,
= lim y→0 {3(siny − y)/27y3) − (4sin3y/27y3)},
⇒ L = lim y→0 1/9 * (siny − y)/y3) − 4/27* (siny/y)3...(∗)
.Note that, here,
= lim y→0(siny − y)/y3)
= lim x→0 (sinx−x)/x3) = L.
Therefore, (∗) ⇒ L = 1/9*L − 4/27
or,  8/9L = −4/27
Hence, L = −4/27*9/8
=−1/6

UP PGT Math Mock Test - 1 - Question 27

A batsman scores runs in 10 innings 38,70,48,34,42,55,63,46,54 and 44, then the mean deviation is

Detailed Solution for UP PGT Math Mock Test - 1 - Question 27

Arranging the given data in ascending order,
we have 34, 38, 42, 44, 46, 48, 54, 55, 63, 70
Here, Median M = (46+48)/2
=47
(∵ n = 10, median is the mean of 5th and 6th items)
∴ Mean deviation = ∑|xi−M|/n
=∑|xi−47|/10
= (13+9+5+3+1+1+7+8+16+23)/10
=8.6

UP PGT Math Mock Test - 1 - Question 28

Let R be the relation on N defined as x R y if x + 2 y = 8. The domain of R is

Detailed Solution for UP PGT Math Mock Test - 1 - Question 28

As x R y if x + 2y = 8, therefore, domain of the relation R is given by x = 8 – 2y ∈ N.
When y = 1, 
⇒ x = 6 ,when y = 2, 
⇒ x = 4, when y = 3, 
⇒ x = 2.
therefore domain is {2, 4, 6}.

UP PGT Math Mock Test - 1 - Question 29

In a triangle ABC if A = π/4 and tanB tanC = K, then K must satisfy. 

Detailed Solution for UP PGT Math Mock Test - 1 - Question 29

In a ΔABC, we know that
tanA + tanB + tanC = tanA tanB tanC
∴ tanB + tanC = tanA(tanBtanC – 1) 


⇒ tan2 B - (K - 1) tan B + K = 0
For real values of tan B, Disc.
(K – 1)2 – 4K > 0 
K2 – 6K + 1 >

UP PGT Math Mock Test - 1 - Question 30

The three arithmetic mean between – 2 and 10 are:

Detailed Solution for UP PGT Math Mock Test - 1 - Question 30

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