NVS TGT Math Mock Test - 2 - NVS TGT/PGT MCQ

G20

• Founded in 1999.
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President Droupadi Murmu is set to confer the President’s Colour to the Armed Forces Medical College (AFMC) on December 1 in a ceremony marking its platinum jubilee year.

Answer:
The value displayed if the formula "= "$55.00" 5" is entered into a cell is 60 (Option B).
Here is the detailed solution:
1. The formula "= "$55.00" 5" combines a text string "$55.00" with the number 5.
2. When a text string is combined with a number using the plus sign (+) operator, Excel tries to convert the text into a number.
3. In this case, "$55.00" cannot be converted into a number because it contains a dollar sign and double quotes.
4. As a result, Excel treats "$55.00" as text and concatenates it with the number 5.
5. The result of the formula is the text string "$55.00" followed by the number 5, which is " $55.00 5".
6. Since the resulting value is a text string, it cannot be mathematically calculated.
7. Therefore, the value displayed in the cell will be the text string itself, which is "$55.00 5" (Option C).
8. Option B, which states 60, is incorrect because the formula does not perform any mathematical operations on the text string and number.
9. The correct answer is Option C, which displays the text string "$55.00 5".

The term 'micro teaching' was coined by Dwight Allen of the Stanford University. This is an innovative teacher training technique and focuses on developing teaching skills. This is highly individualised training skill. Micro teaching can also be defined as a teaching technique especially used in teachers' pre-service education to train them systematically by allowing them to experiment main teacher behaviours. By the help of this technique, teacher candidates can experiment and learn each of the teaching skills by breaking them into smaller parts and without encountering chaotic environment of the crowded classes.
In the original process, a teacher is asked to prepare a short lesson (usually 20 minutes) for a small group of learners who may not have been his/her own students. This is then recorded on video. After the lesson, the teacher, teaching colleagues, a master teacher and the students together view the video recording and comment on what they see happening, referencing the teacher's learning objectives. Watching the video and getting comments from colleagues and students provide teachers with an often intense "under the microscope" view of their teaching.

Correct answer is A.

We have the sequence 5n²+2n+3
It will be in AP if it satisfies an-an-1=d where d is a constant.
Here, a_n=5n²+2n+3
And, a_n-1=5(n-1)²+2(n-1)+3=5(n²+1-2n)+2n-2+3
=5n²+5-10n+2n+1=5n²-8n+6
a_n-a_n-1=5n²+2n+3-(5n²-8n+6)
=5n²+2n+3-5n²+8n-6
=10n-3, which depends on a variable , so its not constant.

use factor theorem as x+2 is factor of

x³-2ax²+16 so put x = -2 and equate the equation to 0

so putting x = -2

(-2)³-2a(-2)²+16 =0

-8-8a+16=0

-8a = -8

a = 8/8= 1

So,

a = 1

Two coins are simultaneously tossed.
So sample space={HH,HT,TH,TT}
No. of favourable outcomes=getting at least one head={HH,HT,TH}
=3
Total number of outcomes=4
Probability of getting at least one head=No. of favourable outcomes/Total number of outcomes
=3/4

The solution of the linear equation is represented by the ordered pair (x,y)

Hence if x>0 , y>0 , then the solution will always lie in the first quadrant.

Median is the middle value of the observations. So Median cannot be obtained when the observations are not in order. So first we have to arrange in ascending order. 
59, 46, 30, 23, 27, 44, 52, 40 and 29
Gives- 23,27,29,30,40,44,46,52,59
Since the no. of observations is odd 

sum=n/2x(2a+(n-1)d )
       =50/2(2x3+ (50-1)3 )
       = 25 ( 6+49x3 )
       = 25x 153
       = 3825

11th term is 35 so a+10d = 35 => equation 1

13th term is 41 so a+12d = 41 => equation 2

solve both equations by elimination method

we get common difference as 3.

If a, b and c are in A.P., then 

we have, 

2 = (3p - 2)/2   and 2p + 2  = (2q + 4)/2

3p = 6  and 2p = q

p = 2

and q = 2p = 4

In ∆ BDC and ∆ BAC
Angle BAC = BDC
(angle made on same segment BC)
Since ABC is making right angle (90)
So,
In ∆ABC
ABC +BAC+ACB=180
(angle sum property of triangle)
90+42+ACB=180
ACB=180-132
ACB=48^o

It is given that,
∠POQ =110°
Since, the Iangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ ∠OPT = 90°
and ∠OQT = 90°
Now, in quadrilateral POQT.
∠POQ + ∠OQT + ∠PTQ + ∠OPT = 360°
(angle sum property of quadrilateral)
⇒ 110° + 90° + ∠PTQ + 90° = 360°
⇒ ∠PTQ + 290° = 360°
⇒ ∠PTQ = 70°
Hence, right option is (B).

Let the interior angles be x and 3x 

We know that exterior angle of triangle is equal to sum of interior opposite angles.

⇒ x+3x=80^∘

⇒ 4x=80^∘

⇒ x=20^∘

So the angles are 

x=20^∘

3x=3×20^∘

= 60∘

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°.    

Option A: If n is integer similarly n²- 1 is not divisible by 8.
Option B: If n is a natural number then it is not possible for n=1,2.
Option C: If n is odd the possible for n is 3,5,7...
Option D: If n is even then it is not possible for n=2.

Thus, the option C is correct.

The denominator 2⁶ x 5² is of the form 2^m x 5ⁿ, where m and n are non-negative integers. Hence, it is a terminating decimal expansion.

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