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Q:

In a Plane there are 37 straight lines, of which 13 pass through the point A and 11 pass through the point B. Besides, no three lines pass through one point, no lines passes through both points A and B , and no two are parallel. Find the number of points of intersection of the straight lines.

A) 525 B) 535
C) 545 D) 555

Answer:   B) 535

Explanation:

The number of points of intersection of 37 lines is . But 13 straight lines out of the given 37 straight lines pass through the same point A. Therefore instead of getting  points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting  points, we get only one point B.

Hence the number of intersection points of the lines is  = 535

  

Q:

Out of seven consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed ?

A) 25200 B) 25000
C) 25225 D) 24752
 
Answer & Explanation Answer: A) 25200

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

=(   ) = ( ) = 210.

Number of groups, each having 3 consonants and 2 vowels =210

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves = 5! = (5  4 3  2 1)=120.

Required number of words = (210 120) = 25200.

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Q:

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated ?

A) 15 B) 20
C) 5 D) 10
 
Answer & Explanation Answer: B) 20

Explanation:

Since each number to be divisible by 5, we must have 5 0r 0 at the units place. But in given digits we have only 5.
So, there is one way of doing it.
Tens place can be filled by any of the remaining 5 numbers.So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1  5  4) = 20.

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Q:

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there ?

A) 205 B) 194
C) 209 D) 159
 
Answer & Explanation Answer: C) 209

Explanation:

We may have (1 boy and 3 girls)or(2boys and 2 girls)or(3 boys and 1 girl)or(4 boys).
Required number of ways = () + () + () + ()

= () +  +  + 

= (24+90+80+15)

= 209.

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Q:

In a bag, there are 8 red, 7 blue and 6 green flowers. One of the flower is picked up randomly. What is the probability that it is neither red nor green ?

   

A) Option A B) Option B
C) Option C D) Option D
 
Answer & Explanation Answer: A) Option A

Explanation:

Total number of flowers = (8+7+6) = 21.
Let E = event that the flower drawn is neither red nor green.
= event taht the flower drawn is blue.
 n(E)= 7
 P(E)=  

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Q:

A box contains 5 green, 4 yellow and 5 white pearls. Four pearls are drawn at random. What is the probability that they are not of the same colour ?

A)  B)  C)  D) 

A) Option A B) Option B
C) Option C D) Option D
 
Answer & Explanation Answer: D) Option D

Explanation:

Let S be the sample space. Then,
n(s) = number of ways of drawing 4 pearls out of 14
=  ways =  = 1001
Let E be the event of drawing 4 pearls of the same colour.
Then, E = event of drawing (4 pearls out of 5) or (4 pearls out of 4) or (4 pearls out of 5)

 +  +  = 5+1+5 =11

 P(E) = 

  Required probability =  

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