Quantitative aptitude Question for all banking Bank PO,Clerk,IBPS PO,Railway,SSC,PSUs,IAS,OAS ,C-SAT Exams

Mathmetics

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

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Q531.

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

1)	1/2	2)	1/3
3)	3/4	4)	4/5
5)	None of thesae

Answer : 3/4
Explanation :

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1),

(4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n(E) = 27.

So, P(E) = n(E) / n(S) =27 /36 = 3/4

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Q532.

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

1)	1/24	2)	21/46
3)	1/48	4)	24/117
5)	None of these

Answer : 21/46
Explanation :

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = ²⁵C₃= Number ways of selecting 3 students out of 25

= (25 x 24 x 23) / (3 x 2 x 1) = 2300.

n(E) = (¹⁰C₁ x ¹⁵C₂)

= 10 x (15 x 14) /(2 x 1) = 1050.

So, P(E) = n(E) / n(S) =1050 / 2300 =21/46

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Q533.

Three unbiased coins are tossed. What is the probability of getting at most two heads ?

1)	1/4	2)	3/4
3)	7/8	4)	1/36
5)	None of these

Answer : 7/8
Explanation :

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = Event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

So, P(E) = n(E) / n(S) =7 /8 .

View Answer

Q534.

What is the probability of getting a sum 9 from two throws of a dice ?

1)	1/6	2)	1/8
3)	1/9	4)	1/12
5)	None of these

Answer : 1/9
Explanation :

In two throws of a die, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

So,P(E) = n(E) / n(S) = 4 /36 =1/9.

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Q535.

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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Q536.

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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Q537.

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

View Answer

Q538.

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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Q539.

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together ?

1)	100800	2)	120960
3)	240150	4)	4989600
5)	None of thesae

Answer : 120960
Explanation :

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

Number of ways of arranging these letters = 8! / (2!)(2!)= 10080.

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters = 4! /2! = 12.

So, required number of words = (10080 x 12) = 120960.

View Answer

Q540.

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women ?

1)	42	2)	48
3)	63	4)	90
5)	None of these

Answer : 63
Explanation : The required number of ways = (⁷C₅ x³C₂) = (⁷C₂ x ³C₁) = (7 x 6 )/(2 x 1) x 3 = 63.

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Mathmetics

Q531.

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

Q532.

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

Q533.

Three unbiased coins are tossed. What is the probability of getting at most two heads ?

Q534.

What is the probability of getting a sum 9 from two throws of a dice ?

Q535.

How many words with or without meaning, can be formed by using all the letters of the word, 'DELHI' using each letter exactly once ?

Q536.

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

Q537.

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 ?

Q538.

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed ?

Q539.

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together ?

Q540.

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women ?

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