Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q1. 

1)  26  2)  27 
3)  41  4)  56 
5)  None of these 
Answer : 41
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Q2. 

1)  2420  2)  2880 
3)  3260  4)  4840 
5)  None of these 
Answer : 2880
Explanation :
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 5members 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.
Explanation : Total ,5 men and 3 women i.e. 8 persons.
So, 3 men can selected from 5 men.
They can selected in (^{5}C_{3}_{ }x ^{3} C_{2} ) or (^{5}C_{4}_{ }x^{ 3} C_{1} ) or (^{5}C_{5}_{ }x ^{3 }C_{0} )
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 IITB together. Suppose, we first arrange 5 prof. IITD and after that we can arrange the IITB in space between IITD.
Explanation : Given, no professors of IITB together. Suppose, we first arrange 5 prof. IITD and after that we can arrange the IITB in space between IITD.
So, numbers of way of arranging the IITBin the circle = (51)! = 24 ways.
Numbers of way of arranging IITB will be by placing them in 5 space that are formed between the IITD.
This can be arranged in ^{5}p_{3} = 60 ways.
therefore, total number of ways = 24 x 60 =1440
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Q5. 

1)  32  2)  48 
3)  56  4)  64 
5)  None of these 
Answer : 64
Explanation : Out of total 10 balls,
Explanation : Out of total 10 balls,
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Q6. 

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.
Explanation : From 8 men and 10 women , the committee can be formed in following ways.
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Q7. 

1)  60  2)  120 
3)  180  4)  240 
5)  None of these 
Answer : 120
Explanation :
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. 

1)  60  2)  120 
3)  240  4)  360 
5)  None of these 
Answer : 360
Explanation :
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. 

1)  8750  2)  10420 
3)  11760  4)  12420 
5)  None of these 
Answer : 11760
Explanation :
Explanation :
We need to select 5 men from 8 men and 6 women from 10 women
Number of ways to do this = ^{8}C_{5} x ^{10}C_{6}
= ^{8}C_{3} x ^{10}C_{4} [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. 

1)  2100  2)  24400 
3)  21300  4)  25200 
5)  None of these 
Answer : 25200
Explanation :
Explanation :
Number of ways of selecting 3 consonants out of 7 = ^{7}C_{3}
_{}
Number of ways of selecting 2 vowels out of 4 = ^{4}C_{2}
_{}
Number of ways of selecting 3 consonants out of 7 and 2
vowels out of 4 = ^{7}C_{3} x ^{4}C_{2}
_{}
=(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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