AWES PGT Math Mock Test - 6 - AWES TGT/PGT MCQ

The correct answer is Bal Gangadhar Tilak.

Key Points

• Bal Gangadhar Tilak gave a slogan 'Freedom is our birthright and we must have it.
• Lokmanya Bal Gangadhar Tilak roused the Indian people and injected fresh life into our liberation cause with his electrifying phrase, "Swaraj is my birth-right and I shall have it."
• Lokmanya Tilak was born in 1856, barely one year before the First War of Indian Independence, into a disappointed and sullen India, and following the defeat of the 1857 insurrection, into a swirl of sorrow and hopelessness.

Important Points

• Bal Gangadhar Tilak worked valiantly to aid the country's independence from British domination.
• Bal Gangadhar Tilak, a member of the Lal-Bal-Pal troika, was dubbed the "Father of Indian Unrest" by British colonial rulers.
• Lokmanya Tilak founded and published two newspapers: the Marathi Kesari and the English The Mahratta. To criticize the colonial overlords, he used his pen as a weapon.

Renu Sud Karnad has been appointed as the chairperson and independent director of PayU Payments Private Ltd. Her extensive experience, including her tenure at HDFC Bank, highlights PayU's strategy of leveraging seasoned leadership to navigate the evolving fintech landscape.

G. Sathiyan made history by winning the men's singles trophy at the WTT Feeder Beirut 2024 event, becoming the first Indian to achieve this feat in a World Table Tennis Feeder Series tournament. This victory showcased his exceptional skills and talent in table tennis on the international stage.

The United Nations Environment Programme’s (UNEP) Production Gap report has been released in anticipation of the COP28 climate conference.
It warns that global fossil fuel production over the next seven years is projected to exceed more than double the levels required to stay within the climate goals outlined in the 2015 Paris climate agreement, including the crucial target of limiting global warming to 1.5 degrees Celsius.

The Indian Navy concluded the bi-annual 'Prasthan' exercise off the Mumbai coast, working with defence, state, and civil agencies to enhance preparedness and coordination in addressing contingencies related to oil production platforms.

The National Education Policy (NEP) was approved by the Union Cabinet in July 2020 and aims to implement various improvements to the Indian educational system from the school level to the college level.

Key Points

• According to NEP 2020, Children from 3 to 18 years old will have access to a free and compulsory high-quality education.
• This implies that all children between the ages of 3 and 18 will be entitled to receive a quality education from the government, regardless of their socioeconomic status.
• Everyone has access to get-quality education for all children from the age of 3-18 years instead of 6-14 years as per the RTE 2009.

 Therefore, the Arrangement of free and compulsory education for 3-18-year-old students are the changes required to be made in RTE 2009 as per NEP 2020.

Additional Information

• Right to Education Act (RTE) provided free and compulsory education to children between the age group of 6-14 years in 2009 and enforced it as a fundamental right under Article 21-A.
• India became one of 135 countries to make education a fundamental right of every child when the Act came into force on 1 April 2010.

► n(Persons travelling by car) = 20 = n(C)
► n(Persons travelling by bus) = 50 = n(B)
► n(Persons travelling by both car and bus) = 10 = n(C ∩ B)
∴ n(C ∪ B) = n(C) + n(B) - n(C ∩ B)
= 20 + 50 - 10
= 60 

Step-by-step Explanation:
Since we have given that,
Percentage of families own a cell phone = 25%
Percentage of families own a scooter = 15%
Percentage of families neither own = 65%
So, it becomes,
P(C ∪ S)^' = 1 - P(C ∪ S)
0.65 = 1 - P(C ∪ S)
1 - 0.65 = P(C ∪ S)
P(C ∪ S) = 0.35
So, it becomes, 
P(C ∪ S) = P(C) + P(S) - P(C ∩ S)
0.35 = 0.25 + 0.15 - x
0.35 = 0.40 - x
x = 0.05
Thus, 0.05 x Total number of families in town = 1500
Total number of families in the town =
1500 / 0.05 = 30,000
Hence, the total number of families in the town is 30,000.

U = Set of all teachers in a school
B = Set of all Mathematics teachers in a school
B’ = U – B = Set of all Non-Mathematics teachers in a school

Since f(2) = f(3) = 1, then inverse does not exists
Inverse exist for (C) f^-1 = {(3, 1), (2, 3), (1, 2)}.

n! < [(n+1)/2]ⁿ
For(n=1) 1! < [(1+1)/2]¹
= 1 < (2/2)¹
1 < 1 (False)
 
For(n=2) 2! < [(2+1)/2]²
= 2 < (3/2)²
1 < 6/4 
1 < 3/2
1 < 1.5 (True)
Therefore smallest number will be 2.

 Number 1 is odd. As any number is divisible by 1: 1
Do prime factorisation of 3600: 2*2*2*2*3*3*5*5
Select all odd numbers from above: 3,3,5,5
Try every possible products of these:
Single number: 3, 5
Two numbers: 3*3, 3*5, 5*5: 9,15,25
Three numbers: 3*3*5, 3*5*5: 45, 75
All four: 3*3*5*5: 225
The odd divisors are: 1,3,5,9,15,25,45,75,225
There are 9 odd divisors of 3600.

There are 8 letters in the word TRIANGLE.
2 alphabets are fixed, remaining are 6 alphabets
So, number of arrangements = ⁶P₆
= 6!
= 720

⇒ αS = α + 2α² + 3α³ + ...+(n -1)α^n-1 +nαⁿ

On subtracting, we get

Since, α is nth root of unity, therfore
αⁿ = 1 

z =| z | -3+ 2i ⇒ | z |²= (| z | -3)² + 4 = | z |² -6 | z | +9 + 4Þ | z | 13/6
=
Thus,  z = - 5/6 + 2i

(a + ib)² = 5 + 12i
use (a + b)² = a² + b² + 2*a*b
=> a² + b²*i² + 2*a*b*i = 5 + 12i
i² = -1
⇒ a² - b² + 2*a*b*i = 5 + 12i
equate the real and complex coefficients
⇒ a² - b² = 5 and ab = 6
a = 6/b
substitute in a² - b² = 5
⇒ 36/b² - b² = 5
⇒ 36 - b⁴ = 5b²
⇒ b⁴ + 5b² - 36 = 0
⇒ b⁴ + 9b² - 4b² - 36 = 0
⇒ b²(b² + 9) - 4(b² + 9) = 0
⇒ (b² - 4)(b² + 9) = 0
⇒ b² = 4 and b² = -9
b is a real number, so we eliminate b² = -9
b² = 4
⇒ b = 2 and b = -2
a = 3 and a = -3
The required number can be 3 + 2i and -3 -2i

The ratio of the coefficients of x^p and x^q in the expansion of (1+x)^p+q is

Given, G = (5√5 + 11)^{2n + 1} , 0 < G < 1 and R = (5√5 + 11)^{2n + 1}

d = 100
S_n = 21700
n = 14
S_n = n/2[2a + (n-1)d]
21700 = 14/2[2a + (14-1)100]
= 21700 = 7[2a + 1300]
= 3100 = 2a + 1300
⇒ 3100 - 1300 = 2a
⇒ 1800 = 2a
⇒ a = 900

1
2 3
4 5 6
7 8 9 10
..........
Series for starting number:
1, 2, 4, 7, ...
Let  Tn = an² + bn + c
a+b+c=1                         ......(1)
4a+2b+c=2                   ......(2)
9a+3b+c=4                   .......(3)
From eq (1), (2), (3), we get:
a= 1/2, b= −1/2, c=1
∴ T_n = n²/2 − n/2 + 1
∴ T₅₀ = (50)²/2 − 50/2 + 1 = 1226
Now, elements in 50th group:
1226, 1227, 1228,.....(50 terms)
∴ Sn = 50/2[2×1226+(50−1)×1]
= 62525

aα² + bα + c = 0 and bα² + cα + a = 0 ⇒ α = 1 ∴ a + b + c = 0

A minor, Mij, of the element aij is the determinant of the matrix obtained by deleting the ith row and jth column.

Let the centre of hyperbola be (α,β)
As y=5 line has the foci, it also has the major axis.
∴ [(x−α)²]/a² − [(y−β)²]/b² = 1
Midpoint of foci = centre of hyperbola
∴ α=1,β=5
Given, e= 5/4
We know that foci is given by (α±ae,β)
∴ α+ae=6
⇒1+(5/4a)=6
⇒ a=4
Using b² = a²(e² − 1)
⇒ b²=16((25/16)−1)=9
∴ Equation of hyperbola ⇒ [(x−1)²]/16−[(y−5)²]/9=1

Pt of intersection of x-y-1=0 and 3x-4y-6=0 is (-2,-3) other asymptote will be in the form of 4x + 3y + λ =0 and it should pass through (-2,-3). Thus λ = -17

sin³10° + sin³50° - sin³70° 
= 1/4 [(3 sin 10° - sin 30°) + (3 sin 50° - sin 150°) - (3 sin 70° - sin 210°)]

Therefore, the D.C.’s of vector AB are given by: 

If a line has the direction ratios – 18, 12, – 4, then its direction cosines are given by: