Bihar STET Paper 1 Mathematics Mock Test - 2 - Bihar PGT/TGT/PRT MCQ

Let AB and CD be two perpendicular diameters of a circle with centre O.

Now, ∠ABC = 90º

[Angle in semicircle is a right angle]

Similarly ∠ACD = ∠ADC

= ∠BAD = 90º ...(1)

In ΔAOB and ΔAOD, we have

AO = AO (Common)

∠AOB = ∠AOD (Each 90º, given)

BO = OD (Radii of circle)

ΔAOB ≅ ∠AOD(By SAS congruence)

AB = AD (By C.P.C.T.)

Similarly, we have,

AD = DC , DC = BC; BC = AB

Hence, AB = BC = CD = DA ...(2)

Also, it is given that diagonals of ABCD intersect at

90º ...(3)

By (1), (2) and (3) ABCD is a square.

Let the angle = x
Supplement of angle= 180-x

ATQ

180-x = x
180 = 2x

x = 90°

And its supplement will be 180-90=90°

x² + 7x + 10 = (x + 2)(x + 5)

So, the value of x² + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0

Therefore, the zeroes of x² + 7x + 10 are –2 and –5.

Sum of zeroes = -7 = –(Coefficient of x) / (Coefficient of x²)

Product of zeroes = 10 = Constant term / Coefficient of x²

Given: p(x) = 3x²−px−2 = 0
∴ p(2) = 3(2)²−p(2)−2 = 0
⇒ 12−2p−2 = 0
⇒ −2p = −10
⇒ p = 5

60°+ QSR=180°[co interior angles]

QSR=120°

since EF||QS

RFE=QSR (corresponding angles) 

So,RFE=120°

Number of possible outcomes (queen of diamonds) = 1
Number of Total outcomes = 52
Required Probability = 1/52

Assertion, a_n = 7 - 4n

d = a_{n + 1 -an}

= 7 - 4 (n + 1) - 7 (-4n)

= 7 - 4n - 4 - 7 + 4n = - 4

Given length of common chord AB =12 cm

Let the radius of the circle with centre O is OA = 10 cm

Radius of circle with centre P is AP = 8 cm

From the figure, OP⊥AB

⇒AC = CB

∴AC = 6 cm   (Since AB=12 cm)

In ΔACP, 

AP² = PC²+AC²    [By Pythagoras theorem]

⇒ 8² = PC²+6²  

⇒ PC² = 64–36 = 28

 PC = 2√7​ cm

Consider ΔACO,

AO² = OC²+AC²   [By Pythagoras theorem]

⇒102 = OC2+62  

⇒OC2 = 100−36 = 64

⇒OC = 8 cm
From the figure, OP = OC+PC = 8+2√7​ cm.
Hence, the distance between the centres is (8+2√7​) cm.

Internal angles of a regular Hexagon = 120°

Alpha(a) and beta(b) are roots of x^2 + px + 1

This implies that sum of roots= a+b = -p/1=-p

And the product of roots = ab = 1/1=1

Similarly ,

Gamma(c) and delta(d) are roots of x^2 + qx + 1

So c+d=-q and cd =1.

The above results can be obtained once we know that any quadratic equation has two roots and hence can be written as (x-p)(x-q)=0 where a and b are the roots .

So x^2 -(p+q)x + pq =0

Comparing this with the general form of quadratic equation :ax^2 + bx + c= 0 we get

Sum of the roots =p+q= -b/a

And product of the roots = pq = c/a}
 

RHS=(a-c)(b-c)(a+d)(b+d)

=(c^2-ac-bc +ab)(d^2 +bd +ad + ab)

We know ab=1

So RHS= (c^2-ac-bc +1)(d^2 +bd +ad + 1)

= (c^2)(d^2) +(a+b)c^2(d) + c^2 -(d^2)c(a+b) -(a+b)^2(cd) -(a+b)c + d^2 + bd + ad + 1

= 1 + ac+bc + c^2 - da-db - (a^2 + b^2 + 2(1)) -ac -bc + bd + ad +1

Cancelling off all the common terms,

We get c^2 +d^2-a^2-b^2

= c^2+d^2-a^2-b^2
=c^2 -a^2 + d^2 -b^2 + 2(1) -2(1){ ab=cd =1}
=c^2 + d^2 + 2cd - a^2 - b^2 - 2ab
LHS=(c+d)^2-(a+b)^2
Therefore,
But we know -p= a+b and -q = c+d
LHS= q^2-p^2

360^o - 130^o =230^o

Therefore, angle ABC =1/2reflex angle AOC

Angle ABC =1/2 230^o

So, angle ABC =115^o

Given , t_n = 2n+3
t1=2×1+3
t1=5
Sn=n/2(t1+tn)
=n/2 (5+2n+3)
=n/2(2n+8)
=n(n+4)

A line segment has two distinct endpoints. There is exactly one point that divides the segment into two equal parts. Hence a line segment has one and only one midpoint.

In right triangle ABC,

Hence, the the height of the pole is 10 m.

A+b =180(linear pair)

2x+3x=180

5x=180

x=180/5

x=36

a=2x=2*36=72

b=3x=3*36=108

b=d (vertical opposite angles are equal)

d=f (alternative interior angles are equal)

f=h (vertically opposite angles are equal)

So, h= 108

Number of total coins = 100 + 50 + 20 + 10 = 180
Number of coins except five rupee coins = 180 – 10 = 170
∴ Required Probability = 170/180 = 17/18

Sample space - {HH, HT, TH, TT}
Number of total possible outcomes = 4
Number of favourable outcome (both heads) = 1
∴  Probability of getting both head = 1/4

LCM of 50 and 48 = 1200
∴ 1200 sec = 20 min
Hence at 12.20 pm they will beep again for the first time.

Age of learner is proportional to the cognitive ability of the learner, thus teaching aids should be selected on the basis of the age. Use of any kind of teaching aid can only be helpful in serving the purpose, if it runs parallel to the cognitive ability of the learner.
The use of teaching aids facilitates the objective of teaching by assisting teachers in differentiating instruction. Hence, they must be chosen keeping in view what the teacher intends to teach the students.
The teaching aids also must be carefully chosen to match the intellectual level of the students. In case of students with intellectual disabilities, different kind of teaching aids are to be used, instead of those used for normal intellectual abilities.

If a student passes a comment, then the teacher must ignore the comment and continue taking class. 

Teaching is not affected by socio-economic background of students and teachers. Teaching is impartial.

Sarita's teacher is following learning as a social activity. Group work creates more opportunities for critical thinking and can promote student learning and achievement. Student group work enhances skills like communication and other professional development skills. All of these skills are critical to successful teamwork in the classroom.
Hence, the correct answer is, 'Learning as a social activity.'

Hence, the data is inadequate.

Four friends W, X, Y and Z are students of Class 10th.

i) W and X are good in Hindi but poor in English.

ii) W and Y are good in Science but poor in Mathematics.

iii) Y and Z are good in English but poor in Social Studies.

iv) Z and X are good in Mathematics as well as in Science.

Thus we can see Y and Z pairs of friends are good, both in English and Science

Hence, the correct answer is "Y and Z".