Biology
General
Easy

Question

Match the types of WBC listed under Column - I with the shape of nucleus given under Column - II and select the correct option from codes given below.

  1. (a) rightwards arrow (iii), (b) rightwards arrow (iv), (c) rightwards arrow (i), (d) rightwards arrow (ii)    
  2. (a) rightwards arrow (v), (b) rightwards arrow (iii), (c) rightwards arrow (i), (d) rightwards arrow (iv)    
  3. (a) rightwards arrow (ii), (b) rightwards arrow (i), (c) rightwards arrow (v), (d) rightwards arrow (iii)    
  4. (a) rightwards arrow (iii), (b) rightwards arrow (iv), (c) rightwards arrow (ii), (d) rightwards arrow (i)    

The correct answer is: (a) rightwards arrow (iii), (b) rightwards arrow (iv), (c) rightwards arrow (ii), (d) rightwards arrow (i)

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The number of ways in which four letters can put in four addressed envelopes so that no letter goes into the envelope meant for it is :

Total number of ways of putting 4 letters in 4 envelopes = 4! =24
Now, number of ways of putting only 1 letter in the right envelope= blank to the power of 4 C subscript 1 cross times 2 =8
number of ways of putting only 2 letters in the right envelopes =blank to the power of 4 C subscript 2=6
number of ways of putting 3 letters in the right envelope= 0, as when we will put three letters in the right envelope the last will automatically go in the right place.
number of ways of putting 4 letters in the right envelope= 1
So, the number of ways in which four letters can put in four addressed envelopes so that no letter goes into the envelope meant for it =24-(8+6+0+1)=24-15=9

The number of ways in which four letters can put in four addressed envelopes so that no letter goes into the envelope meant for it is :

Maths-General
Total number of ways of putting 4 letters in 4 envelopes = 4! =24
Now, number of ways of putting only 1 letter in the right envelope= blank to the power of 4 C subscript 1 cross times 2 =8
number of ways of putting only 2 letters in the right envelopes =blank to the power of 4 C subscript 2=6
number of ways of putting 3 letters in the right envelope= 0, as when we will put three letters in the right envelope the last will automatically go in the right place.
number of ways of putting 4 letters in the right envelope= 1
So, the number of ways in which four letters can put in four addressed envelopes so that no letter goes into the envelope meant for it =24-(8+6+0+1)=24-15=9
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In the given figure of the heart which of the labelled part (1, 2, 3, 4, 5) carries oxygenated blood?

In the given figure of the heart which of the labelled part (1, 2, 3, 4, 5) carries oxygenated blood?

biologyGeneral
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The given figure shows an angiogram of the coronary blood vessel. Which one of the following statements correctly describes, what is being done?

The given figure shows an angiogram of the coronary blood vessel. Which one of the following statements correctly describes, what is being done?

biologyGeneral
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There are 3 letters and 3 addressed envelopes corresponding to them. The number of ways in which the letters be placed in the envelopes so that no letter is in the right envelope is:

Total number ways in which 3 letters can be put in 3 different envelopes= 3! = 6
now, numbers of ways when only 1 letter is put in the right envelope = blank cubed C subscript 1=3
number of ways when 2 letters are put in the right envelope = 0
as when we put 2 letters in the right envelope the third letter would be automatically in the right envelope
number of ways when 3 letters are put in the right envelope = 1
So, the no. of ways when no letter will be in the right envelope=6-(3+0+1) =6-4 =2

There are 3 letters and 3 addressed envelopes corresponding to them. The number of ways in which the letters be placed in the envelopes so that no letter is in the right envelope is:

Maths-General
Total number ways in which 3 letters can be put in 3 different envelopes= 3! = 6
now, numbers of ways when only 1 letter is put in the right envelope = blank cubed C subscript 1=3
number of ways when 2 letters are put in the right envelope = 0
as when we put 2 letters in the right envelope the third letter would be automatically in the right envelope
number of ways when 3 letters are put in the right envelope = 1
So, the no. of ways when no letter will be in the right envelope=6-(3+0+1) =6-4 =2
General
biology

Match Column - I with Column - II and select the correct option from the codes give below.

Match Column - I with Column - II and select the correct option from the codes give below.

biologyGeneral
General
Maths-

The sum of all the numbers formed by taking all the digits from 3,4,5,6,7 is:

The number of numbers having 3 in the unit place= 4! =24
The number of numbers having 4 in the unit place= 4! =24
The number of numbers having 5 in the unit place= 4! =24
The number of numbers having 6 in the unit place= 4! =24
The number of numbers having 7 in the unit place= 4! =24
So the sum of the digits in the unit place of all the numbers=3 cross times 24 plus 4 cross times 24 plus 5 cross times 24 plus 6 cross times 24 plus 7 cross times 24
=24(3+4+5+6+7)
=24cross times25=600
Similarly the sum of the digits of all the numbers in each of the other places=600
The required sum =600 cross times 10000 plus 600 cross times 1000 plus 600 cross times 100 plus 600 cross times 10 plus 600 cross times 1
=600(10000+1000+100+10+1)
=600(11111)
=6666600

The sum of all the numbers formed by taking all the digits from 3,4,5,6,7 is:

Maths-General
The number of numbers having 3 in the unit place= 4! =24
The number of numbers having 4 in the unit place= 4! =24
The number of numbers having 5 in the unit place= 4! =24
The number of numbers having 6 in the unit place= 4! =24
The number of numbers having 7 in the unit place= 4! =24
So the sum of the digits in the unit place of all the numbers=3 cross times 24 plus 4 cross times 24 plus 5 cross times 24 plus 6 cross times 24 plus 7 cross times 24
=24(3+4+5+6+7)
=24cross times25=600
Similarly the sum of the digits of all the numbers in each of the other places=600
The required sum =600 cross times 10000 plus 600 cross times 1000 plus 600 cross times 100 plus 600 cross times 10 plus 600 cross times 1
=600(10000+1000+100+10+1)
=600(11111)
=6666600
parallel
General
Maths-

The sum of all the numbers formed by taking all the digits from 2,3,4,5 is:

The number of numbers having 2 in the unit place= 3! =6
The number of numbers having 3 in the unit place= 3! =6
The number of numbers having 4 in the unit place= 3! =6
The number of numbers having 5 in the unit place= 3! =6
So the sum of the digits in the unit place of all the numbers=2 cross times 6 plus 3 cross times 6 plus 4 cross times 6 plus 5 cross times 6
=12+18+24+30
=84
Similarly the sum of the digits of all the numbers in each of the other places=84
The required sum =84 cross times 1000 plus 84 cross times 100 plus 84 cross times 10 plus 84
=84(1000+100+10+1)
=84cross times 1111
=93324

The sum of all the numbers formed by taking all the digits from 2,3,4,5 is:

Maths-General
The number of numbers having 2 in the unit place= 3! =6
The number of numbers having 3 in the unit place= 3! =6
The number of numbers having 4 in the unit place= 3! =6
The number of numbers having 5 in the unit place= 3! =6
So the sum of the digits in the unit place of all the numbers=2 cross times 6 plus 3 cross times 6 plus 4 cross times 6 plus 5 cross times 6
=12+18+24+30
=84
Similarly the sum of the digits of all the numbers in each of the other places=84
The required sum =84 cross times 1000 plus 84 cross times 100 plus 84 cross times 10 plus 84
=84(1000+100+10+1)
=84cross times 1111
=93324
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The number of constant mappings from A equals left curly bracket 1 comma 2 comma 3 comma 4 horizontal ellipsis horizontal ellipsis comma n right curly bracket to B equals left curly bracket a comma b right curly bracket is

The number of constant mappings from A equals left curly bracket 1 comma 2 comma 3 comma 4 horizontal ellipsis horizontal ellipsis comma n right curly bracket to B equals left curly bracket a comma b right curly bracket is 2.

The number of constant mappings from A equals left curly bracket 1 comma 2 comma 3 comma 4 horizontal ellipsis horizontal ellipsis comma n right curly bracket to B equals left curly bracket a comma b right curly bracket is

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The number of constant mappings from A equals left curly bracket 1 comma 2 comma 3 comma 4 horizontal ellipsis horizontal ellipsis comma n right curly bracket to B equals left curly bracket a comma b right curly bracket is 2.
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The number of many one functions from A equals left curly bracket 1 comma 2 comma 3 right curly bracket text  to  end text B equals left curly bracket a comma b comma c comma d right curly bracket is

Given, A{1, 2, 3} and B{a, b, c, d}

Element A can have an image in 3 ways, i.e., (1, 2, 3)

Similarly, B can have an image in 4 ways.

Thus, total number of ways =4×4×4=64

The number of many one functions from A equals left curly bracket 1 comma 2 comma 3 right curly bracket text  to  end text B equals left curly bracket a comma b comma c comma d right curly bracket is

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Given, A{1, 2, 3} and B{a, b, c, d}

Element A can have an image in 3 ways, i.e., (1, 2, 3)

Similarly, B can have an image in 4 ways.

Thus, total number of ways =4×4×4=64

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The number of into functions that can be defined from A equals left curly bracket x comma y comma z comma w comma t right curly bracket text  to  end text bold italic B equals left curly bracket bold italic alpha comma bold italic beta comma bold italic gamma right curly bracket is

 

The number of into functions that can be defined from A equals left curly bracket x comma y comma z comma w comma t right curly bracket text  to  end text bold italic B equals left curly bracket bold italic alpha comma bold italic beta comma bold italic gamma right curly bracket is

 

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A body moving with velocity v has momentum and kinetic energy numerically equal. What is the value of v?

A body moving with velocity v has momentum and kinetic energy numerically equal. What is the value of v?

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The graph between the resistive force F acting on a body and the distance covered by the body is shown in figure. The mass of the body is 25KG and initial velocity is 2 m/s. When the distance covered by the body is 4km, its kinetic energy would be

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The number of onto functions that can be defined from A={a,b,c,d,e} to {1,2} is

Given, A{abcd, e} and B{,2}

Element a can has an image in 2 ways, i.e., (1 or 2).

Similarly, a, bc, d and e can have an image in 5 ways.

Thus, total number of ways =2×2×2×2×2=32

The number of onto functions that can be defined from A={a,b,c,d,e} to {1,2} is

Maths-General

Given, A{abcd, e} and B{,2}

Element a can has an image in 2 ways, i.e., (1 or 2).

Similarly, a, bc, d and e can have an image in 5 ways.

Thus, total number of ways =2×2×2×2×2=32

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The momentum of a body increases by 20%. The percentage increase in its kinetic energy is

The momentum of a body increases by 20%. The percentage increase in its kinetic energy is

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Two spheres A and B  of masses M1 and  M2 respectively collide. A is at rest initially and B is moving with velocity V along X-axis. After collision  has a velocity V over 2 in a direction perpendicular to the original direction. The mass  moves after collision in the direction

Two spheres A and B  of masses M1 and  M2 respectively collide. A is at rest initially and B is moving with velocity V along X-axis. After collision  has a velocity V over 2 in a direction perpendicular to the original direction. The mass  moves after collision in the direction

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