BPSC TGT Math & Science Mock Test - 8 - Bihar PGT/TGT/PRT MCQ

As per the given above sentence is given in Passive Voice.
Use structure “To + subject( object ) + auxiliary + past participle”
To resign was his obligation.
Therefore ,required answer will be option B .

(b) represents 1st form of verb used with simple present tense, (c) taken is 3rd form of verb and (d) represents a continuous form of tense. For simple past tense, the 2nd form of verb is used hence; the correct answer is 'took'.

 "The little boy asked the man the way to Athens" from Active to Passive Voice is:

A. The man was asked the way to Athens by the little boy.

This option correctly presents the sentence in the Passive Voice while maintaining its original meaning.

दिए गए विकल्पों में सही उत्तर विकल्प 1 ‘घड़ा’ है। अन्य सभी विकल्प इसके अनुचित उत्तर होंगे।

Key Points

• दिए गए विकल्पों में घन का अनेकार्थी शब्द 'घड़ा' नहीं है।
• घड़ा के अनेकार्थी शब्द हैं- घट, शरीर, हृदय, कम।
• घड़ा के पर्यायवाची शब्द हैं- गगरा, घट, कुभ, कलश।

(यदि -तो ) मिश्र वाक्य का योजक चिन्ह है

As safe ans secure have same meaning in the same way protect and guard have same meaning.

Initial area = L × B

New area = (100 + 20)% of L × (100 – 12.50)% of B = 1.05L × B

Hence, Option A is correct.

X × 32 + 8 – 40  = X × 30
2X = 32
X = 16
Hence, Option A is correct.

Let AB be the observation tower and h be its height.

Also, let the ship be at C when the angle of elevation is 30° and at D when the angle of elevation is 45°.

The time taken by the ship to travel from C to D is 10 minutes and we need to find out the time the ship will take to reach B from D.

∴ tan 30º = AB/CB = h / CB = 1/ √3

⇒ CB = √3 x h

∴ tan 45º = AB / DB = h / DB = 1

⇒ DB = h

⇒ CD = CB – DB = (√3h – h) = h(√3– 1)

Now, as h(√3 – 1) distance is covered in 10 minutes, a distance of h is covered in = 13.66 minutes = 13 minutes 40 seconds

At sunrise morning begins, similarly at sunset night begins.

HCF = 13
Let the numbers be 13x and 13y.
Where x and y are co-prime.
∴ LCM = 13 xy
∴ 13 xy = 455

∴ Numbers are 13 × 5 = 65 and 13 × 7 = 91

The Himalaya mountains separate the Indian subcontinent from the rest of Asia. They extend for about 2,500 kilometers (1,550 miles) and are so vast that they are composed of three different mountain belts. The northernmost belt, known as the Great Himalayas, has the highest average elevation and contains the highest mountain summit in the world, Mount Everest.

Plasma Membrane is the outermost boundary of the cell. It is also known as the cell membrane. It is the membrane found in all cells that separate the interior of the cell from the outside environment. A cell wall is attached to the plasma membrane on its outside surface in bacterial and plant cells.

Let AB be the tower and CD be the observer

tan30° = AE/CE

⇒ 1/√3= (AB – CD) / x = (50 – 1.5) / x = 48.5/x

x = 84 m

Cache size is 32 KB and cache block size is 32 B. So,

So, number of index bits needed = 9 ( since 29 = 512). Number of offset bits = 5 (since 25 = 32 B is the block size and assuming byte addressing). So, number of tag bits = 32 - 9 - 5 = 18 (as memory address is of 32 bits).
So, time for comparing the data = Time to compare the data + Time to select the block in set = 0.6 + 18/10 ns = 2.4 ns.
(Two comparisons of tag bits need to be done for each block in a set, but they can be carried out in parallel and the succeeding one multiplexed as the output).

-3 is root of both

2(9) -3p -15 = 0

p = 1

and

second equation has two equal roots so, 

b²-4ac = 0

p² = 4pq

p = 4q

q = 1/4

We obtain zeros of the polynomial when y=0. Observing the points we see that we have point (5,0) which means when y=0 , x= 5 this means x=5 is a solution of the polynomial.

We have,

∠C = 90°

AB + BC = 25 cm and AC = 5 cm

Let BC be x cm, then AB = (25 – x) cm

By using Pythagoras theorem,

AB² = BC² + AC²

⇒ (25 – x²) = x² + (5)²

⇒ 625 – 50x + x² = x² + 25

⇒ 50x = 600

⇒ x = 600 / 50 = 12

Hence, BC = 12 cm

Zeroes of f(x) = x² − 4x − 5 are obtained by solving:
x² − 4x − 5 = 0, which implies 
x² − 5x + x − 5 = 0 or, x(x − 5) + 1(x − 5) = 0 
which means x = 5 or x = −1 
Thus the assertion is correct. 
However, the reason is incorrect. 
The numbers given are not the zeroes  of x² − 4x + 7

As the denominator has factor 5³ × 2² and which is of the type 5^m × 2ⁿ, So this is a terminating decimal expansion.

We know that x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx).
If x + y + z = 0, then x³ + y³ + z³ – 3xyz = 0 or x³ + y³ + z³ = 3xyz.

= (1 / 2)² + (1 / 2)²

= 1 / 4 + 1 / 4

= 2 / 4 = 1 / 2

Given: x = a sec θ cos φ, y = b sec θ sin φ and z = c tan θ

Initial unpolarized intensity is 2a² 
the intensity of light transmitted by the first polarizered will be I_unpolarized/2​​=a²
option A is correct

Plane mirror always forms a virtual image of the real object for all the rays emitted by the object. So, when a point source of light is placed in front of a  plane mirror, all the reflected rays meet at a point when produced backward.
Option A is correct.

From lens displacement method, f_max = d/4

Empirical formula of mass CH = (12 + 1) = 13 
Molecular formula = Empirical formula x n