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Number of Questions: 10
Time: 10 Minutes
Category: Aptitude
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Question 1 of 30
1. Question
In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?
Correct
The word ‘OPTICAL’ has 7 letters. It has the vowels ‘O’,’I’,’A’ in it and these 3 vowels should always come together. Hence these three vowels can be grouped and considered as a single letter. That is, PTCL(OIA).
Hence we can assume total letters as 5 and all these letters are different.
Number of ways to arrange these letters
=5!=5×4×3×2×1=120=5!=5×4×3×2×1=120All the 3 vowels (OIA) are different
Number of ways to arrange these vowels among themselves
=3!=3×2×1=6=3!=3×2×1=6Hence, required number of ways
=120×6=720Incorrect
The word ‘OPTICAL’ has 7 letters. It has the vowels ‘O’,’I’,’A’ in it and these 3 vowels should always come together. Hence these three vowels can be grouped and considered as a single letter. That is, PTCL(OIA).
Hence we can assume total letters as 5 and all these letters are different.
Number of ways to arrange these letters
=5!=5×4×3×2×1=120=5!=5×4×3×2×1=120All the 3 vowels (OIA) are different
Number of ways to arrange these vowels among themselves
=3!=3×2×1=6=3!=3×2×1=6Hence, required number of ways
=120×6=720 
Question 2 of 30
2. Question
In how many different ways can the letters of the word ‘LEADING’ be arranged such that the vowels should always come together?
Correct
The word ‘LEADING’ has 7 letters. It has the vowels ‘E’,’A’,’I’ in it and these 3 vowels should always come together. Hence these 3 vowels can be grouped and considered as a single letter. that is, LDNG(EAI).
Hence we can assume total letters as 5 and all these letters are different. Number of ways to arrange these letters
=5!=5×4×3×2×1=120=5!=5×4×3×2×1=120In the 3 vowels (EAI), all the vowels are different. Number of ways to arrange these vowels among themselves
=3!=3×2×1=6=3!=3×2×1=6Hence, required number of ways
=120×6=720Incorrect
The word ‘LEADING’ has 7 letters. It has the vowels ‘E’,’A’,’I’ in it and these 3 vowels should always come together. Hence these 3 vowels can be grouped and considered as a single letter. that is, LDNG(EAI).
Hence we can assume total letters as 5 and all these letters are different. Number of ways to arrange these letters
=5!=5×4×3×2×1=120=5!=5×4×3×2×1=120In the 3 vowels (EAI), all the vowels are different. Number of ways to arrange these vowels among themselves
=3!=3×2×1=6=3!=3×2×1=6Hence, required number of ways
=120×6=720 
Question 3 of 30
3. Question
A coin is tossed 3 times. Find out the number of possible outcomes.
Correct
Explanation:
When a coin is tossed once, there are two possible outcomes: Head(H) and Tale(T)
Hence, when a coin is tossed 3 times, the number of possible outcomes
=2×2×2=8=2×2×2=8(The possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT )
Incorrect
Explanation:
When a coin is tossed once, there are two possible outcomes: Head(H) and Tale(T)
Hence, when a coin is tossed 3 times, the number of possible outcomes
=2×2×2=8=2×2×2=8(The possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT )

Question 4 of 30
4. Question
In how many different ways can the letters of the word ‘DETAIL’ be arranged such that the vowels must occupy only the odd positions?
Correct
Explanation:
The word ‘DETAIL’ has 6 letters which has 3 vowels (EAI) and 3 consonants(DTL)
The 3 vowels(EAI) must occupy only the odd positions. Let’s mark the positions as (1) (2) (3) (4) (5) (6). Now, the 3 vowels should only occupy the 3 positions marked as (1),(3) and (5) in any order.
Hence, number of ways to arrange these vowels
= ^{3}P_{3} =3!=3×2×1=6=3!=3×2×1=6Now we have 3 consonants(DTL) which can be arranged in the remaining 3 positions in any order. Hence, number of ways to arrange these consonants
= ^{3}P_{3}=3!=3×2×1=6=3!=3×2×1=6Total number of ways
= number of ways to arrange the vowels × number of ways to arrange the consonants
=6×6=36Incorrect
Explanation:
The word ‘DETAIL’ has 6 letters which has 3 vowels (EAI) and 3 consonants(DTL)
The 3 vowels(EAI) must occupy only the odd positions. Let’s mark the positions as (1) (2) (3) (4) (5) (6). Now, the 3 vowels should only occupy the 3 positions marked as (1),(3) and (5) in any order.
Hence, number of ways to arrange these vowels
= ^{3}P_{3} =3!=3×2×1=6=3!=3×2×1=6Now we have 3 consonants(DTL) which can be arranged in the remaining 3 positions in any order. Hence, number of ways to arrange these consonants
= ^{3}P_{3}=3!=3×2×1=6=3!=3×2×1=6Total number of ways
= number of ways to arrange the vowels × number of ways to arrange the consonants
=6×6=36 
Question 5 of 30
5. Question
In how many different ways can the letters of the word ‘JUDGE’ be arranged such that the vowels always come together?
Correct
Explanation:
The word ‘JUDGE’ has 5 letters. It has 2 vowels (UE) and these 2 vowels should always come together. Hence these 2 vowels can be grouped and considered as a single letter. That is, JDG(UE).
Hence we can assume total letters as 4 and all these letters are different. Number of ways to arrange these letters
=4!=4×3×2×1=24=4!=4×3×2×1=24In the 2 vowels (UE), all the vowels are different. Number of ways to arrange these vowels among themselves
=2!=2×1=2=2!=2×1=2Total number of ways =24×2=48
Incorrect
Explanation:
The word ‘JUDGE’ has 5 letters. It has 2 vowels (UE) and these 2 vowels should always come together. Hence these 2 vowels can be grouped and considered as a single letter. That is, JDG(UE).
Hence we can assume total letters as 4 and all these letters are different. Number of ways to arrange these letters
=4!=4×3×2×1=24=4!=4×3×2×1=24In the 2 vowels (UE), all the vowels are different. Number of ways to arrange these vowels among themselves
=2!=2×1=2=2!=2×1=2Total number of ways =24×2=48

Question 6 of 30
6. Question
What is the probability of getting a number less than 4 when a die is rolled?
Correct
Explanation :
Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces)
i.e., n(S) = 6E = Getting a number less than 4 = {1, 2, 3}
Hence, n(E) = 3P(E) = = =
Incorrect
Explanation :
Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces)
i.e., n(S) = 6E = Getting a number less than 4 = {1, 2, 3}
Hence, n(E) = 3P(E) = = =

Question 7 of 30
7. Question
A dice is thrown. What is the probability that the number shown in the dice is divisible by 3?
Correct
Explanation :
Total number of outcomes possible when a die is rolled, n(S) = 6 (∵ 1 or 2 or 3 or 4 or 5 or 6)
E = Event that the number shown in the dice is divisible by 3 = {3, 6}
Hence, n(E) = 2
P(E) = = =
Incorrect
Explanation :
Total number of outcomes possible when a die is rolled, n(S) = 6 (∵ 1 or 2 or 3 or 4 or 5 or 6)
E = Event that the number shown in the dice is divisible by 3 = {3, 6}
Hence, n(E) = 2
P(E) = = =

Question 8 of 30
8. Question
John draws a card from a pack of cards. What is the probability that the card drawn is a card of black suit?
Correct
Explanation :
Total number of cards, n(S) = 52
Total number of black cards, n(E) = 26
Incorrect
Explanation :
Total number of cards, n(S) = 52
Total number of black cards, n(E) = 26

Question 9 of 30
9. Question
What is the probability of selecting a prime number from 1,2,3,… 10 ?
Correct
Explanation :
Total count of numbers, n(S) = 10
Prime numbers in the given range are 2,3,5 and 7
Hence, total count of prime numbers in the given range, n(E) = 4P(E) = = =
Incorrect
Explanation :
Total count of numbers, n(S) = 10
Prime numbers in the given range are 2,3,5 and 7
Hence, total count of prime numbers in the given range, n(E) = 4P(E) = = =

Question 10 of 30
10. Question
A letter is randomly taken from English alphabets. What is the probability that the letter selected is not a vowel?
Correct
Explanation :
Total number of alphabets, n(S) = 26
Total number of characters which are not vowels, n(E) = 21
P(E) = = \frac { 21 }{ 26 }
Incorrect
Explanation :
Total number of alphabets, n(S) = 26
Total number of characters which are not vowels, n(E) = 21
P(E) = = \frac { 21 }{ 26 }

Question 11 of 30
11. Question
In how many ways, a committee of 5 members can be selected from 6 men and 5 ladies, consisting of 3 men and 2 ladies?
Correct
(3 men out 6) and (2 ladies out of 5) are to be chosen.
Required number of ways = (^{6}c_{3}x^{5}c_{2}) = [6x5x4/3x2x1] x [5×4/2×1] = 200.
Incorrect
(3 men out 6) and (2 ladies out of 5) are to be chosen.
Required number of ways = (^{6}c_{3}x^{5}c_{2}) = [6x5x4/3x2x1] x [5×4/2×1] = 200.

Question 12 of 30
12. Question
In how many ways can a cricket eleven be chosen out of a batch of 15 players ?
Correct
Required number of ways = ^{15}c_{11 }= ^{15}c_{(1511)} = ^{11}c_{4}
= 15x14x13x12/4x3x2x1 = 1365.
Incorrect
Required number of ways = ^{15}c_{11 }= ^{15}c_{(1511)} = ^{11}c_{4}
= 15x14x13x12/4x3x2x1 = 1365.

Question 13 of 30
13. Question
How many words can be formed from the letters of the word ‘DIRECTOR’ So that the vowels are always together?
Correct
In the given word, we treat the vowels IEO as one letter.
Thus, we have DRCTR (IEO).
This group has 6 letters of which R occurs 2 times and others are different.
Number of ways of arranging these letters = 6!/2! = 360.
Now 3 vowels can be arranged among themselves in 3! = 6 ways.
Required number of ways = (360×6) = 2160.
Incorrect
In the given word, we treat the vowels IEO as one letter.
Thus, we have DRCTR (IEO).
This group has 6 letters of which R occurs 2 times and others are different.
Number of ways of arranging these letters = 6!/2! = 360.
Now 3 vowels can be arranged among themselves in 3! = 6 ways.
Required number of ways = (360×6) = 2160.

Question 14 of 30
14. Question
How many words can be formed from the letters of the word ‘EXTRA’ so that the vowels are never together?
Correct
The given word contains 5 different letters.
Taking the vowels EA together, we treat them as one letter.
Then, the letters to be arranged are XTR (EA).
These letters can be arranged in 4! = 24 ways.
The vowels EA may be arranged amongst themselves in 2! = 2 ways.
Number of words, each having vowels together = (24×2) = 48 ways.
Total number of words formed by using all the letters of the given words
= 5! = (5x4x3x2x1) = 120.
Incorrect
The given word contains 5 different letters.
Taking the vowels EA together, we treat them as one letter.
Then, the letters to be arranged are XTR (EA).
These letters can be arranged in 4! = 24 ways.
The vowels EA may be arranged amongst themselves in 2! = 2 ways.
Number of words, each having vowels together = (24×2) = 48 ways.
Total number of words formed by using all the letters of the given words
= 5! = (5x4x3x2x1) = 120.

Question 15 of 30
15. Question
How many words can be formed by using all letters of the word ‘DAUGHTER’ so that the vowels always come together?
Correct
Given word contains 8 different letters. When the vowels AUE are always together, we may suppose them to form an entity, treated as one letter.
Then, the letters to be arranged are DGNTR (AUE).
Then 6 letters to be arranged in ^{6}p_{6} = 6! = 720 ways.
The vowels in the group (AUE) may be arranged in 3! = 6 ways.
Required number of words = (720×6) = 4320.
Incorrect
Given word contains 8 different letters. When the vowels AUE are always together, we may suppose them to form an entity, treated as one letter.
Then, the letters to be arranged are DGNTR (AUE).
Then 6 letters to be arranged in ^{6}p_{6} = 6! = 720 ways.
The vowels in the group (AUE) may be arranged in 3! = 6 ways.
Required number of words = (720×6) = 4320.

Question 16 of 30
16. Question
How many words can be formed by using all letters of the word “BIHAR”
Correct
The word BIHAR contains 5 different letters.
Required number of words = ^{5}p_{5} = 5! = (5x4x3x2x1) = 120.
Incorrect
The word BIHAR contains 5 different letters.
Required number of words = ^{5}p_{5} = 5! = (5x4x3x2x1) = 120.

Question 17 of 30
17. Question
Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?
Correct
We have n(s)=52c2=(52*51)/(2*1)=1326.
Let A=event of getting both black cards
B=event of getting both queens
aÇb=event of getting queen of black cards
n(A)=^{26}c2=(26*25)/(2*1)=325,
n(b)=^{4}c2=(4*3)/(2*1)=6 and
n(aÇb)=2c2=1
p(A)=n(A)/n(S)=325/1326;
p(B)=n(B)/n(S)=6/1326 and
p(aÇb)=n(aÇb)/n(s)=1/1326
p(aÈb)=p(a)+p(b)p(aÇb)=(325+61/1326)=330/1326=55/221
Incorrect
We have n(s)=52c2=(52*51)/(2*1)=1326.
Let A=event of getting both black cards
B=event of getting both queens
aÇb=event of getting queen of black cards
n(A)=^{26}c2=(26*25)/(2*1)=325,
n(b)=^{4}c2=(4*3)/(2*1)=6 and
n(aÇb)=2c2=1
p(A)=n(A)/n(S)=325/1326;
p(B)=n(B)/n(S)=6/1326 and
p(aÇb)=n(aÇb)/n(s)=1/1326
p(aÈb)=p(a)+p(b)p(aÇb)=(325+61/1326)=330/1326=55/221

Question 18 of 30
18. Question
.Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6
Correct
Clearly n(s)=6*6=36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.Then
e={(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),
(6,6)}
n(e)=14.
Hence p(e)=n(e)/n(s)=14/36=7/18
Incorrect
Clearly n(s)=6*6=36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.Then
e={(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),
(6,6)}
n(e)=14.
Hence p(e)=n(e)/n(s)=14/36=7/18

Question 19 of 30
19. Question
A bag contains 6 white and 4 black balls .2 balls are drawn at random. find the probability that they are of same colour.
Correct
let s be the sample space
Then n(s)=no of ways of drawing 2 balls out of (6+4)=10c2=(10*9)/(2*1)=45
Let e=event of getting both balls of same colour
Then n(e)=no of ways(2 balls out of six) or(2 balls out of 4)
=(^{6}c2+^{4}c2)=(6*5)/(2*1)+(4*3)/(2*1)=15+6=21
p(e)=n(e)/n(s)=21/45=7/15
Incorrect
let s be the sample space
Then n(s)=no of ways of drawing 2 balls out of (6+4)=10c2=(10*9)/(2*1)=45
Let e=event of getting both balls of same colour
Then n(e)=no of ways(2 balls out of six) or(2 balls out of 4)
=(^{6}c2+^{4}c2)=(6*5)/(2*1)+(4*3)/(2*1)=15+6=21
p(e)=n(e)/n(s)=21/45=7/15

Question 20 of 30
20. Question
A boy has nine trousers and 12 shirts. In how many different ways can he select a trouser and a shirt?
Correct
The boy can select one trouser in nine ways.
The boy can select one shirt in 12 ways.The number of ways in which he can select one trouser and one shirt is 9 * 12 = 108 ways.Incorrect
The boy can select one trouser in nine ways.
The boy can select one shirt in 12 ways.The number of ways in which he can select one trouser and one shirt is 9 * 12 = 108 ways. 
Question 21 of 30
21. Question
How many three letter words are formed using the letters of the word TIME?
Correct
The number of letters in the given word is four.
The number of three letter words that can be formed using these four letters is ⁴P₃ = 4 * 3 * 2 = 24.Incorrect
The number of letters in the given word is four.
The number of three letter words that can be formed using these four letters is ⁴P₃ = 4 * 3 * 2 = 24. 
Question 22 of 30
22. Question
Using all the letters of the word “THURSDAY”, how many different words can be formed?
Correct
Total number of letters = 8
Using these letters the number of 8 letters words formed is ⁸P₈ = 8!.Incorrect
Total number of letters = 8
Using these letters the number of 8 letters words formed is ⁸P₈ = 8!. 
Question 23 of 30
23. Question
The number of arrangements that can be made with the letters of the word MEADOWS so that the vowels occupy the even places?
Correct
The word MEADOWS has 7 letters of which 3 are vowels.
VVVAs the vowels have to occupy even places, they can be arranged in the 3 even places in 3! i.e., 6 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.Hence the total ways are 24 * 6 = 144.Incorrect
The word MEADOWS has 7 letters of which 3 are vowels.
VVVAs the vowels have to occupy even places, they can be arranged in the 3 even places in 3! i.e., 6 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.Hence the total ways are 24 * 6 = 144. 
Question 24 of 30
24. Question
The probability that a number selected at random from the first 50 natural numbers is a composite number is .
Correct
.
The number of exhaustive events = ⁵⁰C₁ = 50.
We have 15 primes from 1 to 50.Number of favourable cases are 34.Required probability = 34/50 = 17/25.Incorrect
.
The number of exhaustive events = ⁵⁰C₁ = 50.
We have 15 primes from 1 to 50.Number of favourable cases are 34.Required probability = 34/50 = 17/25. 
Question 25 of 30
25. Question
If four coins are tossed, the probability of getting two heads and two tails is .
Correct
Since four coins are tossed, sample space = 2^{4}
Getting two heads and two tails can happen in six ways.n(E) = six waysp(E) = 6/2^{4} = 3/8Incorrect
Since four coins are tossed, sample space = 2^{4}
Getting two heads and two tails can happen in six ways.n(E) = six waysp(E) = 6/2^{4} = 3/8 
Question 26 of 30
26. Question
If six persons sit in a row, then the probability that three particular persons are always together is –
Correct
Six persons can be arranged in a row in 6! ways. Treat the three persons to sit together as one unit then there four persons and they can be arranged in 4! ways. Again three persons can be arranged among them selves in 3! ways. Favourable outcomes = 3!4! Required probability = 3!4!/6! = 1/5
Incorrect
Six persons can be arranged in a row in 6! ways. Treat the three persons to sit together as one unit then there four persons and they can be arranged in 4! ways. Again three persons can be arranged among them selves in 3! ways. Favourable outcomes = 3!4! Required probability = 3!4!/6! = 1/5

Question 27 of 30
27. Question
A basket has 5 apples and 4 oranges. Three fruits are picked at random. The probability that at least 2 apples are picked is .
Correct
Total fruits = 9
Since there must be at least two apples,(⁵C₂ * ⁴C₁)/⁹C₃ + ⁵C₃/⁹C₃ = 25/42.Incorrect
Total fruits = 9
Since there must be at least two apples,(⁵C₂ * ⁴C₁)/⁹C₃ + ⁵C₃/⁹C₃ = 25/42. 
Question 28 of 30
28. Question
Three 6 faced dice are thrown together. The probability that all the three show the same number on them is .
Correct
It all 3 numbers have to be same basically we want triplets. 111, 222, 333, 444, 555 and 666. Those are six in number. Further the three dice can fall in 6 * 6 * 6 = 216 ways.
Hence the probability is 6/216 = 1/36Incorrect
It all 3 numbers have to be same basically we want triplets. 111, 222, 333, 444, 555 and 666. Those are six in number. Further the three dice can fall in 6 * 6 * 6 = 216 ways.
Hence the probability is 6/216 = 1/36 
Question 29 of 30
29. Question
A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that all the four bulbs are defective.
Correct
Out of nine, five are good and four are defective. Required probability = ⁴C₄/⁹C₄ = 1/126
Incorrect
Out of nine, five are good and four are defective. Required probability = ⁴C₄/⁹C₄ = 1/126

Question 30 of 30
30. Question
A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that exactly three bulbs are good.
Correct
Required probability = (⁵C₃ . ⁴C₁)/⁹C₄ = (10 * 4)/126 = 20/63
Incorrect
Required probability = (⁵C₃ . ⁴C₁)/⁹C₄ = (10 * 4)/126 = 20/63