Mathmetics

ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams

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Q1.
A boy has 3 library tickets and 8 books of his interest in the library. Of these 8, he does not want to borrow Chemistry part II unless Chemistry part I is also borrowed. In how many ways can he choose the three books to be borrowed?

1) 26 2) 27
3) 41 4) 56
5)None of these
Answer : 41
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Q2.
In how many different ways can the letters of the word ‘PREVIOUS’ be arranged in such a way that the vowels always come together?
1) 2420 2) 2880
3) 3260 4) 4840
5)None of these
Answer : 2880
Explanation :
Treating the group of four vowels (E, I O and U) as one distinct letter then total number of letters will be 8 - 4 + 1 = 5 [i.e. P, R, V, S and (E, I, O, U)].
The five letters can be arranged in 5! = 120 ways The vowels among themselves can be arranged in 4! = 24 ways,
The letters of the word ‘PREVIOUS’ can be arranged in 120 × 24 = 2880 ways such that vowels always come together.

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Q3.
There are  5 men and 3 women to form a committee of 5-members where at least 3 men are there. How many way they can be selected ?
1) 42 2) 45
3) 46 4) 54
5)None of these
Answer : 46
Explanation :  Total ,5 men and 3 women i.e. 8 persons.
So, 3 men can selected from 5 men.
They can selected in (5C3  x 3 C2 ) or (5C4  x 3 C1 ) or (5C5  x 3 C0
Total numbers of way to select = [(5 x 4 )/2 x 3 ] +[(5 ) x 3 ] + [(1x 1) ] = 46

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Q4.
There are 5 professors from IIT Delhi and 3 professors from IIT Bombay who are sitting together to discuss a matter at a round table. In how many ways can they sit around the table so that no two professors from IIT Bombay are together ?
1) 600 2) 840
3) 1200 4) 1440
5)None of these
Answer : 1440
Explanation : Given, no professors of IIT-B together. Suppose, we first arrange 5 prof. IIT-D and after that we can arrange the IIT-B in space between IIT-D.
So, numbers of way of arranging  the IIT-Bin the circle = (5-1)! = 24 ways.
Numbers of way of arranging IIT-B will be by placing them in 5 space that are formed between the IIT-D.
This can be arranged in 5p3 = 60 ways.
therefore, total number of ways = 24 x 60 =1440
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Q5.
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw ?
1) 32 2) 48
3) 56 4) 64
5)None of these
Answer : 64
Explanation : Out of total 10 balls,


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Q6.
In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women ?
1) 266 2) 5040
3) 11760 4) 86400
5)None of these
Answer : 11760
Explanation : From 8 men and 10 women , the committee can be formed in following ways.

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Q7.
How many words with or without meaning, can be formed by using all the letters of the word, 'DELHI' using each letter exactly once ?
1) 60 2) 120
3) 180 4) 240
5)None of these
Answer : 120
Explanation :
The word 'DELHI' has 5 letters and all these letters are different.

Total words (with or without meaning) formed by using all these 5 letters using each letter exactly once

= Number of arrangements of 5 letters taken all at a time =  5P5 = 5! = 5 x 4 x 3 x 2 x 1 = 120
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Q8.
In how many ways can the letters of the word 'LEADER' be arranged ?
1) 60 2) 120
3) 240 4) 360
5)None of these
Answer : 360
Explanation :
The word 'LEADER' has 6 letters. 

But in these 6 letters, 'E' occurs 2 times and rest of the letters are different.

Hence,number of ways to arrange these letters = 6!2!=6×5×4×3×2×12×1 = 360
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Q9.
There are 8 men and 10 women and you need to form a committee of 5 men and 6 women. In how many ways can the committee be formed ?

1) 8750 2) 10420
3) 11760 4) 12420
5)None of these
Answer : 11760
Explanation :
We need to select 5 men from 8 men and 6 women from 10 women

Number of ways to do this  = 8C5 x 10C6

=  8C3 x 10C4 [Applied the formula nCr = nC(n - r) ]

= (8×7×63×2×1) (10×9×8×74×3×2×1)

= 56 x 210 = 11760
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Q10.
Out of 7 consonants and 4 vowels, how many words of 3  consonants and 2 vowels can be formed ?

1) 2100 2) 24400
3) 21300 4) 25200
5)None of these
Answer : 25200
Explanation :
Number of ways of selecting 3 consonants out of 7 = 7C3

Number of ways of selecting 2 vowels out of 4 = 4C2

Number of ways of selecting 3 consonants out of 7 and 2

vowels out of 4 = 7C3 x 4C2

=(7×6×5/3×2×1)×(4×3/2×1)=210

It means that we can have 210 groups where each group

contains total 5 letters(3 consonants and 2 vowels).

Number of ways of arranging 5 letters among themselves = 5! = 5 x 4 x 3 x 2 x 1 = 120

Hence, Required number of ways = 210 x 120 = 25200

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