9
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:

Nine different letters of alphabet are given, words with 5 letters are formed from these given letters. Then, how many such words can be formed which have at least one letter repeated ?

A) 43929 B) 59049
C) 15120 D) 0
 
Answer & Explanation Answer: A) 43929

Explanation:

Number of words with 5 letters from given 9 alphabets formed =

Number of words with 5 letters from given 9 alphabets formed such that no letter is repeated is = 

Number of words can be formed which have at least one letter repeated =  -  

= 59049 - 15120

= 43929

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6 65
Q:

In how many different ways can the letters of the word 'ABYSMAL' be arranged ?

A) 5040 B) 3650
C) 4150 D) 2520
 
Answer & Explanation Answer: D) 2520

Explanation:

Total number of letters in the word ABYSMAL are 7

Number of ways these 7 letters can be arranged are 7! ways

But the letter is repeated and this can be arranged in 2! ways

=> Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.

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8 173
Q:

In how many ways the letters of the word OLIVER be arranged so that the vowels in the word always occur in the dictionary order as we move from left to right ?

A) 186 B) 144
C) 136 D) 120
 
Answer & Explanation Answer: D) 120

Explanation:

In given word OLIVER there are 3 vowels E, I & O. These can be arranged in only one way as dictionary order E, I & O.

There are 6 letters in thegiven word.

First arrange 3 vowels.

This can be done in 6C3 ways and that too in only one way.(dictionary order E, I & O)

Remaining 3 letters can be placed in 3 places = 3! ways

Total number of possible ways of arranging letters of OLIVER = 3! x 6C3 ways

= 6x5x4 = 120 ways.

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7 231
Q:

Find the total numbers greater than 4000 that can be formed with digits 2, 3, 4, 5, 6 no digit being repeated in any number ?

A) 120 B) 256
C) 192 D) 244
 
Answer & Explanation Answer: C) 192

Explanation:

We are having with digits 2, 3, 4, 5 & 6 and numbers greater than 4000 are to be formed, no digit is repeated.

The number can be 4 digited but greater than 4000 or 5 digited.

Number of 4 digited numbers greater than 4000 are
4 or 5 or 6 can occupy thousand place => 3 x 4P3 = 3 x 24 = 72.

5 digited numbers = 5P5 = 5! = 120

So the total numbers greater than 4000 = 72 + 120 = 192

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9 228
Q:

In how many ways the letters of the word NUMERICAL can be arranged so that the consonants always occupy the even places ?

A) 5! B) 6!
C) 4! D) Can't determine
 
Answer & Explanation Answer: D) Can't determine

Explanation:

NUMERICAL has 9 positions in which 2, 4, 6, 8 are even positions.

And it contains 5 consonents i.e, N, M, R, C & L. Hence this cannot be done as 5 letters cannot be placed in 4 positions.

Therefore, Can't be determined.

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6 218