Question 1
12 leading companies were brought into an issue in deciding about the allocation of land related properties by a Government agency. The companies were numbered from 2 to 12. Among these the companies numbered 7,9 and 10 were three leading business rivals. It was decided to use two dice and bet on the numbers emerging out of such throws. It was seen that results range from 2 to 12. Out of such throws which of the following: 7 or 9 or 10 is likely to appear more often than the other numbers? (i.e will any of the companies will have advantage over the others with the method followed.)
a) 10 b) 9 c) 7 d) all are equally possible.
Answer : c) 7
Solution :
Let (x,y) represent any throw of a dice where x represents number of I dice and y represents number on II dice
When two dice are thrown possible occurences of 7 include : (1,6), (6,1), (2,5) (5,2) (3,4), (4,3)
When two dice are thrown possible occurences of 9 include : (3,6),(6,3) (4,5) (5,4)
When two dice are thrown possible occurences of 10 include : (5,5) , (5,5)
Sum 7 can appear in at least 6 combinations as against 9 which occurs in 4 combinations and 10 which occurs in 2 combinations. Hence company 7 should have more chances of winning the bid over the others.
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Question 2
Three school friends studying in a Higher Secondary School studying in standard 10, 11 and 12 repectively met in a leading five star hotel in a tea party. During discussions they had an argument about the chances of getting specific numbers by the throw of 2 dice. Which two numbers from 9,10 and 11 will have equal chances of appearing.
a) 9,10 b) 10,11 c) 9,11 d) none of the above
Answer : b) 10,11
Solution :
The solution to this question is very similar to that of the first question.
When two dice are thrown possible occurences of 9 include : (3,6),(6,3), (4,5) (5,4)
When two dice are thrown possible occurences of 10 include : (5,5), (5,5)
When two dice are thrown possible occurences of 11 include : (5,6), (6,5)
Therefore 9 can appear in 4 possible combinations, 10 can appear in 2 possible combinations and 11 can appear in 2 possible combinations. Therefore numbers 10 and 11 have equal chances of appearing during the throw of two dice.
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Question 3
Mrs. Kalyani Chatterjee of Bhubaneshwar went abroad and settled in USA about twenty years ago. She returned on an holiday trip to Bhubaneshwar and went to Konarak Sun Temple for sightseeing. To her surprise she met ten of her class mates at Konarak . What is the probability that at least one of the ten people she met has the same birth day as that of Mrs.Kalyani Chatterjee. Her birth day – 14th March.
a) 0.0028 b) 0.014 c) 0.128 d) none of these.
Answer : a) 0.0028.
Solution :
Probability of a friend having birthday on same date as that of Kalyani = 1/365
Probability of a friend not having birthday on same date as that of Kalyani = 1 - 1/365 = 364/365
Probability of 2 friends not having birthday on same date as that of Kalyani = (364/365).(364/365) = (364/365)2
Similarly Probability of n friends not having birthday on same date as that of Kalyani = (364/365)n
Now, Probability of n friends having birthday on same date as that of Kalyani = 1 - Probability of n friends not having birthday on same date as that of Kalyani
= 1 - (364/365)n
Substituting n = 10 in above formula we get,
1 - (364/365)10 = 0.0028
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Question 3
Three students of Victory Engineering College - Saravanan, Meenakshi and Mangesh were given a mathematics problem. If their chances of solving the problem are 60%, 50% and 40%, what is the probability the problem will be solved?
a) 0.33 b) 0.66 c)0.77 d) 0.88
Answer : d) 0.88
Solution :
For solving such problems one has to employ the below simple equation :
P(problem is solved) = 1 – P(problem is not solved)
In other words, P(problem is solved) = 1 – P(none of the students solve the problem) -> eq 1
But probability of no one solves the problem can be found using the below equation
P(none of the students solve the problem) = P(Saravanan being not able to solve) x P(Meenakshi being not able to solve) x P(Mangesh being not able to solve)
= [1 - P(Saravanan solving the problem)] x [1 - P(Meenakshi solving the problem)] x [1 - P(Mangesh solving the problem)]
P(none of the students solve the problem) = [1 - 0.6][1 - 0.5][1 - 0.4] = .4 x .5 x .6 = .12 -> eq 2
Substitute the value for P(none of the students solve the problem) from eq 2 in eq 1, we get :
P(problem is solved) = 1 – 0.12 = 0.88
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Question 4
The New College, Chennai conducted NCC shooting camp at St. Thomas Mount. Students were asked to participate in the shooting competition. Four students – Goutham, Malavika, Geetha and Vinya participated in the competition. If the probability of their hitting the target Apple correctly is 0.7, 0.6, 0.5 and 0.4 respectively, what is the probability that four shots aimed at the target will hit correctly?
a) 0.964 b) 0.624 c) 0.824 d)0.756
Answer : a) 0.964
Solution :
The target Apple will be brought down even if one of the four shots hit the target.
The opposite of this is the situation when none of the shoots hit the target.
The probability of none of the shoot hits the target = P(Goutham missing the target) x P(Malavika missing the target) x P(Geetha missing the target) x P(Vinya missing the target)
= [1 - P(Goutham hitting the target)] x [1 - P(Malavika hitting the target)] x [1 - P(Geetha hitting the target)] x [1 - P(Vinya hitting the target)]
(1-0.7)(1-0.6)(1-0.5)(1-0.4)=0.3*0.4*0.5*0.6=0.036
So, the probability that at least one of the shoots hits the target Apple = 1 - probability of none of the shoot hits the target = 1 – 0.036 = 0.964.
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Question 5
In the tenth standard class of St.Michael’s School, Palayamkottai, 30% of the students were offered Tamil, 20% were offered Telugu and 10% were offered both. If a student is selected at random, what is the probability that he has offered Tamil or Telugu?
a) 2/5 b) 3/4 c) 3/5 d) 3/10
Answer : a) 2/5.
Solution :
P(Student offered Tamil) = Students offered Tamil / Total Students = 30/100 = 3/10,
P(Telugu) = Students offered Telugu / Total Students = 20/100 = 1/5
And P (Students offered both Tamil and Telugu) = P(Tamil Ω Telugu) = 10/100 = 1/10
Probability that a selected student at random has been offered Tamil or Telugu can be found by the below simple formula
P(Tamil or Telugu) = P (Tamil U Telugu) = P(Tamil) + P(Telugu) - P(Tamil Ω Telugu)
= (3/10 + 1/5 - 1/10) = 4/10 = 2/5
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Question 6
A company has three staff members working in communications department. They are Peter, Satish and Raju. The company has received a work of sending communication to its potential customers. The time the three take to complete the dispatch work together is 2 days less than Peter would have taken to do it alone, 10 days less than Satish to do the work alone and one-third of the time that Raju would have taken working alone. How many days will the three people take to do the dispatch of all the mails working together?
a) 2 days b) 4 days c) 5 days d) 3 days
Answer : a) 2 days.
Solution :
Let the number of letters to be dispatched be X.
If the three people together take ‘D’ days to complete the work then
Peter will take (D +2) days to complete it alone.
Satish will take (D + 10) days to complete it alone.
And Raju will take 3 D days to complete it alone.
Speed of Peter --- X / (D+2) letters per day
Speed of Satish -- X / (D + 10) letters per day
Speed of Raju -- X / 3D letters per day
All of them working together for one day will dispatch –
X/(D+2) + X/(D+10) + X/3D letters -> eq 1
Number of days it will take for them to complete the dispatch of all the X letters = X / Letters that can be dispatched by all of them in one day = X / [ X/D+2 + X/D+10 + X/3D]
But we had assumed the number of days to dispatch all X letters as D.
Therefore, D = X / [ (X/D+2) + (X/D+10) + (X / 3D)]
Or X/D = (X /D+2) + (X/D+10) + (X/3D)
Or 1/(D+2) + 1/(D+10) + 1/ 3D = 1 /D
Substituting the options one by one, we can find that D = 2 satisfies the equation.
i.e Substitute D = 2 in both sides of 1/(D+2) + 1/(D+10) + 1/ 3D = 1 /D.
Substituting D = 2 On LHS : 1/4 + 1/12 + 1/6 = 6/12 = 1/2
On RHS D = 2 we get 1/2
Therefore, LHS = RHS when D =2 and hence 2 is the right answer
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Question 7
Roja and Edward were working in a courier company. Roja takes 6 hours to pack 32 parcels while Edward takes 5 hours to pack 40 parcels. How long they will take to pack 330 parcels working together ?
a) 24 hours 45 minutes b) 23 hours c) 25 hours 15 minutes d) none of these.
Answer : c) 25 hours 15 minutes
Solution :
Speed of Roja per hour = 32/6 parcels
Speed of Edward per hour = 40/5 = 8 parcels
When both of them work together in one hour they will pack
32/6 + 8 = 13 1/3 parcels
For packing 330 parcels it will take = 330 / parcels that can be packed by them in one hour = 330 / 13 1/3 = 25 hours 15 minutes
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Question 8
An overseas software company entrusted the work to a company in India. The Indian company conducted campus recruitment drives and recruited men and women on a large scale. It gave training to newcomers in its training centre with advanced facilities at Mysore for over three weeks. The company conducted periodical tests to assess the progress in terms of knowledge and output of its employees in order to ensure uniform output being given by all the employees. Company’s 48 programmers wrote 48 lines of program in 36 minutes. How many programmers are required to write 192 lines program in 24 minutes.?
a) 24 b) 36 c) 288 d) 72
Answer : c) 288
Solution :
Programmers Lines Minutes
48 48 36
? 192 24
For such problems where efficiency of programmers are considered to be the same, one can employ the equation
P1 X M1 / L1 = P2 X M2 / L2 -> eq 1
Where P1,M1 and L1 are the number of programmers, minutes and lines of code respectively in I case and
P2,M2 and L2 are the number of programmers, minutes and lines of code respectively in II case.
From the question, we can find that P1 = 48, M1 = 36, L1 = 48, M2 = 24 and L2 = 192. P2 is to be found out.
Substituting the above values in equation 1 we get,
(48 x 36) / 48 = (P2 X 24)/192
Or, P2 = (48 x 36 x 192) / (48 x 24) = 288
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Question 9
Ramesh Khanna was five times old as his son Kishan Khanna eight years ago. Now he is three times as old as his son Kishan Khanna. Assume they both live in a country where people understand only binary numbers and they use symbol '0' for binary 1 and '1' for binary zero. Also they add 1 before any binary number. Express Ramesh Khanna’s present age in accordance to that world
a) 1001111 b) 1001111 c) 1100011 d) 1010000
Answer : a) 1001111
Solution :
Let Ramesh Khanna’s age be represented as ‘F’ eight years ago and the age of Kishan Khanna be represented as ‘S”.
F = 5 S
F + 8 = 3 (S + 8)
5S + 8 = 3S + 24
5S – 3 S = 24 – 8
2S = 16
S = 8
Now Ramesh Khanna’s age will be 5S + 8 = (5x8) + 8 = 48
Expressed as binary 110000 . However the people in their country use symbol 0 for binary one and 1 for binary zero. Replacing 1 with 0 and 0 with 1 we get 001111. Also those people usually add 1 before any binary number. Hence the answer becomes 1001111.
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Question 10
Bhavna is presently three times as old as her daughter Anushka. Ten years ago Bhavna was five times as old as her daughter Anushka was. After how many years, sum of their ages will be 100.
a) 10 b) 20 c)15 d) none of these.
Answer : a) 10
Solution :
Let Bhavna’s present age be represented as B
Let her daughter Anushka’s age be represented as A.
B = 3 A
Ten years ago,
B - 10 = 5(A – 10)
3A - 10 = 5A - 50
-10 + 50 = 5A – 3A = 2 A
40 = 2 A
A = 20.
B = 3 x A = 60.
Present sum of their ages = A + B = 80.
When 10 is added to both A and B, Sum of their ages will become 100. Hence, after 10 years, their sum of ages will be 100.
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Question 11
Monisha was seven times as old as her daughter Mercy eight years ago. Now three times Mercy’s age is her mother age. Before how many years the ratio of their ages (ratio of mother's age to Mercy's age) will be increased by 1 from the current ratio
a) 5 b) 4 c) 8 d) none of these.
Answer : b) 4
Solution :
Eight years ago -- let Monisha’s age be M and Mercy’s age be D.
M = 7 D
Now M + 8 = 3 (D + 8)
7 D + 8 = 3D + 24
7D - 3 D = 24 - 8 = 16
D = 4 = age of Mercy before 8 years
M = 28 = age of Mother before 8 years
Current age of Mercy = D + 8 = 12
Current age of Monisha = M + 8 = 36
Current ratio of mother's age to Mercy's age = 36/12 = 3.
Let before x years this ratio becomes 1 more than the current ratio. i.e x denotes the number of years before which the ratio becomes 3 + 1 = 4.
Then 36 - x / 12 - x = 4
36 - x = 48 - 4x
3x = 12
Or x = 4
i.e before 4 years, the ratio of their ages will be 4. Therefore answer = 4
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Question 12
“Hunterkey” a leading software company situate in China was started in the year 1988. The company had been progressing well since inception and the company has achieved a turn over of over 10 billion $. The management of the company has plans for achieving greater heights in terms of turnover and has been training its members in various skills such as technical skills, soft skills, programming skills etc. 60 programmers of the company write 60 lines of programs in 60 minutes totally. How long will it take for 84 programmers to write 84 lines of programs?
a) 84 min. b) 48 min c)60 min d) 72 min
Answer : c) 60 min.
Solution:
For problems like this where efficiency of the programmers/workers are assumed constant implicitly, you can apply the below formula
P1 X M1 / L1 = P2 X M2 / L2
Here, P1,M1 and L1 are number of programmers, number of minutes and number of lines respectively in case I
And, P2,M2 and L2 are number of programmers, number of minutes and number of lines respectively in case II.
Given P1 = 60, M1 = 60, L1 = 60, P2 = 84, L2 = 84.
Substituting in above formula we get.
60 x 60 / 60 = 84 x M2 / 84.
Simplifying we get, M2 = 60 minutes.
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Question 13
“Universal Software Inc.” USA is situated in California. The company was started in the year 1975 and has been progressing extremely well. It is aiming to reach the topmost position in the near future. During the year 2010, in a team, on any particular day, 48 programmers of the company were able to write 48 lines of software programs in 48 minutes. The company recruits 72 more programmers. Also the team management improves the throughput by making them to work for 72 more minutes every day. What will be the increase of the number of lines of code possible now ?
a) 300 lines b)252 lines c)48 lines d) 48 lines
Answer : b) 252 lines
Solution :
Using the same formula as in first question :
P1 X M1 / L1 = P2 X M2 / L2
Given, P1 = 48, M1 = 48 and L1 = 48
Given P2 = P1 + 72 = 48 + 72 = 120
Given M2 = M1 + 72 = 48 + 72 = 120
Substituting values in formula we get,
48 x 48 / 48 = 120 x 120 / L2
Simplifying we get L2 = 300
Increase in number of lines of code = L2 - L1 = 300 - 48 = 252
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Question 14
“Extremely fast solutions” a software programmer provider has been training its manpower in such a way that 36 programmers could write 36 software programs of similar nature in 36 hours. The company has received an order for getting 84 software programs of similar nature in 24 hours. How many additional programmers should the company employ for this project?
a) 90 b)42 c)66 d)44
Answer : a) 90
Solution :
Using the same formula as in first question (Note : M1 and M2 denote time in hours and not minutes. Left hand side and Right hand side should have same units and thats the deal.)
P1 X M1 / L1 = P2 X M2 / L2
Given P1 = 36, M1 = 36, L1 = 36.
Given L2 = 84, M2 = 24
Substituting we get,
36 x 36 / 36 = P2 x 24 / 84
Or P2 = 36 x 84 / 24 = 126
Number of additional programmers = P2 - P1 = 126 - 36 = 90.
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Question 15
Mohan bought a new bike. The number plate consists of 4 digits. How many 4-digit numbers can be formed for the number plate of Mohan’s bike from the digits 2, 4, 6, 7, 5 and 8 which are divisible by 5 with the condition that none of the digit is repeated?
a) 60 b) 30 c) 24 d) 120
Answer : a) 60
Solution :
Since the number on number plate is divisible by 5, the last digit should be 5 only. (0 is not allowed as per the question)
And the first digit can be filled by any of the remaining numbers (2, 4, 6, 7 and 8).
So, the number of ways of filling first digit without repetition = 5 ways
And the second digit can be filled by the remaining 4 numbers.
So, the number of ways of filling second digit without repetition = 4 ways
Therefore, number of ways of selecting 4-digit number is 5 x 4 x 3 x 1 = 60.
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Question 16
A family went to a studio to take a family photo. There are 5 members in that family namely a husband and his wife and their 3 children. In how many ways they can be made to stand (arranged) to take a photo in such a way that the husband is adjacent to his wife?
a) 80 ways b) 30 ways c) 120 ways d) 60 ways
Answer : b) 30 ways
Solution :
Number of ways of placing husband = 5 ways
Since husband must be adjacent to his wife, number of ways of placing wife = 1 way (near husband)
There are three spots remaining :
Any of the three children can be made to stand in first place. Therefore there are 3 ways to fill this spot.
Any of the remaining two children can be made to stand in the second spot. Therefore there are 2 ways to fill this spot.
The third spot can be filled in one way as there will be only one child remaining.
Therefore, the number of ways of arranging the five members = 5 x 1 x 3 x 2 x 1 = 30
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Question 17
Anand has to send 5 different greeting cards to his 5 friends. His five friends are located in Chennai, Mumbai, Bangalore, Delhi and Mysore. In how many ways he can send the greetings to his friends such that no friend is getting 2 greeting cards?
a) 80 ways b) 30 ways c) 120 ways d) 60 ways
Answer : c) 120 ways
Solution :
The number of ways of sending the first greeting card is 5 ways
Since no friend has to get two greeting cards,
The number of ways of sending the second greeting card is 4 ways.
The number of ways of sending the third, fourth and the fifth greeting card are 3 , 2 and 1 ways.
Hence the number of ways of sending all the 5 greeting cards = 5 x 4 x 3 x 2 x 1 = 120 ways
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Question 18
Four years ago, the average age of Deena and Prakash was 21 years. With Harish joining then now the average becomes 25 years. How old is Harish now?
a) 28 years b) 26 years c) 24 years d) 25 years.
Answer : d) 25 years.
Solution:
Let the present age of Deena, Prakash and Harish be A, B and C respectively.
4 years ago, the average age of Deena and Prakash was 21 years.
(i.e.) (A - 4 + B - 4) / 2 = 21
=> A + B = 50 -------- (1)
But now, the average age of Deena, Prakash and Harish is 25 years.
(i.e.) (A + B + C) / 3 = 25
=> A + B + C = 75 ------- (2)
(2) – (1) => C = 75 - 50 = 25.
Hence, the Present age of Harish is 25 years.
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Question 19
The present age of Harini’s father is 4 times the present age of Harini. And the age difference of Harini’s father and mother is 5 years. Five years ago, the sum of all the three person’s age was 70 years. What is the present age of Harini’s father?
a)44years b) 40years c) 50years d) 48years
Answer : b) 40
Solution:
Let the Present ages of Father, Mother and Harini be F, M and H respectively.
Present age of Father = 4 * Present age of Harini
i.e. F = 4H ------- (1)
Age difference between Harini’s Parent's = 5
i.e. F – M = 5 -----------(2)
Five years ago, the sum of ages of Father, Mother and Harini is 70 years
i.e. F – 5 + M – 5 + H – 5 = 70
=> F + M + H = 85 -------- (3)
(2) + (3) => 2F + H = 90 ----- (4)
Substitute (1) in (4), we get H = 10.
Then F = 4*10 = 40.
Therefore the present age of Harini’s father is 40 years.
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Question 20
Haritha told her friends that if she add ten times her age ten years from now to five times her age five years ago is same as the 20 times of her current age. How old Haritha will be ten years from now?
a)15 years b) 20 years c) 23 years d) 25 years
Answer : d) 25 years
Solution:
Let the age of Haritha be X.
Then, 10 times her age after 10 years + 5 times her age before 5 years = 20 times her present age.
i.e. 10 * (X + 10) + 5 * (X - 5) = 20 * X
15X + 75 = 20X
5X = 75
X = 15
Hence, Haritha’s Present age is 15 years.
So, after 10 years, Haritha’s age will be 25 years.
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Question 21
Three years ago, the ages of Geetha, Reena and Surya are in the ratio of 3:4:5. Three years hence, the sum of their ages is 78. What is the age of Reena at present?
a)18 years b) 23 years c) 28 years d) 20 years
Answer : b) 23 years
Solution:
Let the ages of Geetha, Reena and Surya 3 years ago be 3X, 4X and 5X respectively.
After 3 years, the sum of their ages is 78.
i.e. (3X + 3) + 3 + (4X + 3) + 3 + (5X + 3) + 3 = 78
12X = 60
X = 5.
Hence, Reena’s present age is 4X + 3 = 4(5) + 3 = 23 years.
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Question 22
Two countries - Germany and France are participating in a hockey game. Mr. Randor Guy, the famous astrologer in Germany is able to predict the winner of each match with great success. It is rumoured that in a match between 2 teams X and Y,Rando Guy picks X with the same probability as X's chances of winning. Let's assume such rumours to be true and that in a match between Germany and France, Germany the stronger team has a probability of 7/10 of winning the game. What is the probability that Randor Guy will correctly pick the winner of the Germany - France game?
a) 0.58 b) 0.68 c) 0.82 d) 0.42
Answer : a) 0.58
Solution :
Randor Guy predicts the winner in games with the same probability of winning. --> statement I
Germany has got a chance of winning the game with 7/10 probability. France has got a chance of winning the game with 3/10 probability.
Probability of Randor Guy picking the Germany correctly is 7/10 while picking France is 3/10 (as per statement I).
Probability of Randor Guy picking the winning team correctly = (Probability of Germany winning the game) AND (Probability of Randor Guy picking Germany correctly) OR (Probability of France winning the game) AND (Probability of Randor Guy picking France correctly).
Note : In probability related formulas generally AND translates to x and OR translates to +
Therefore, Probability of Randor Guy picking the winning team correctly = (Probability of Germany winning the game) * (Probability of Randor Guy picking Germany correctly) + (Probability of France winning the game) * (Probability of Randor Guy picking France correctly).
Probability of predicting winning team = (7/10 * 7/10) + (3/10 * 3/10)
=0.58
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Question 23
A box has 45 chocolates of different colours 20 red, 15 white and 10 black chocolates. If a chocolate is chosen at random, what is the probability of that being white chocolate?
a) 1/2 b) 1/3 c) 2/3 d) 3/4
Answer : b) 1/3
Solution :
Note : This is a simplest of the simple questions that you can expect on probability. This is a usual case in placement tests. You should be able to identify such simple questions amidst tough questions. At times, you could get mislead by thinking this is a tricky one.
Any of 45 chocolates could be chosen. So, totally there are 45 possibilities. Since there are only 15 white chocolates, only 15 possibilities can lead to a white chocolate. Hence P(white chocolate) = 15/45 or 1/3.
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Question 24
In an off campus placement programme, a software company recruiter interviewed 75 prospective candidates -- 10 from Civil engineering department, 5 from Bio-chemical department and rest from computer science department. If the software company finally issued offer letter to 17 candidates, what is the probability that all the selected candidates belonged to only computer science department?
a) 60C17/ 75C17 b) 65C17/75C17 c) 70C17/75C17 d) None of these.
Answer : a) 60C17/75C17
Solution :
17 out of 75 candidates selected in 75C17 ways. 17 out of 60 from computer science department can be selected in 60C17 ways.
So the required probability = Ways of selecting 17 computer science candidates / Ways of selecting 17 candidates from all departments
= 60C17 / 75C17
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Question 25
Chandrasekaran worked very sincerely in the software company which selected him in the campus recruitment. At the end of the first year he was given a salary increase of 5%. In the second year another friend from his college also joined the company and they spent more time together. As a result of this performance of Chandrasekaran came down in terms of the team leader. The team leader recommended a cut of 2.5% from the salary of Chandrasekaran. In the third year at the end of the first month he received Rs.34398 as salary and felt aggrieved. He started thinking about resigning from this company and try for other companies which pay a higher salary. What was the salary of Chandrasekaran at the beginning of first year?
a) Rs.32400 b) Rs.36800 c) Rs.22000 d) Rs.33600
Answer : d) Rs. 33600
Solution :
Let the salary at the beginning of first year be Rs.X
At the beginning of second year it will then be 5% more than his beginning salary. Therefore his second year salary = 105% of X = 105X/100
On this 2.5% reduction , his salary now will be 2.5% lesser than his second year salary = 97.5% x Second Year Salary = (97.5/100)(105X/100) – But this value is given as Rs.34398
Solving (97.5/100)(105X/100) = 34398, we get,
X = 34398 x 100 x 100 divided by 105 x 97.5
= Rs. 33600
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Question 26
Miss. Ranjitha bought a new wall clock which rings at every one hour. At 3 O’ clock the clock ticks 3 times and at 4 O’ clock the clock ticks 4 times. At 4 O’ clock the time between the first tick and the last tick is 21 seconds. How long does it tick at 12 O’ clock?
a) 66 seconds b) 55 seconds c) 77 seconds d)44 seconds
Answer : c) 77 seconds
Solution :
This is a simple problem. The gap between the first tick and the last tick at 4 O' clock is 3 gaps. Time taken = 21 seconds. That means between two gaps it takes 7 seconds.
At 12 O clock there will be 11 gaps. Total time for 11 gaps = 77 seconds.
Though the question could look confusing at first, this is actually a very simple questions. With practice, you should be ready to identify these kinds of questions which could potentially save a lot of your time during tests.
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Question 27
Anushka has a jewel chest containing Rings, Pins and Ear-rings. The chest contains 26 pieces. Anushka has 2 1/2 times as many rings as pins, and the number of pairs of ear rings is 4 less than the number of rings. How many earrings does Anushka have?
a) 12 b)8 c)6 d)10
Answer : a) 12
Solution :
Assume that there are R rings, P pins and E pair of ear-rings.
It is given that, she has 2 1/2 times as many rings as pins.
R = (5/2) * P or P = (2*R)/5
And, the number of pairs of earrings is 4 less than the number of rings.
E = R - 4 or R = E + 4
Also, there are in total 26 pieces.
R + P + 2*E = 26
R + (2*R)/5 + 2*E = 26
5*R + 2*R + 10*E = 130
7*R + 10*E = 130
7*(E + 4) + 10*E = 130
7*E + 28 + 10*E = 130
17*E = 102
E = 6
Hence, there are 6 pairs of Ear-rings i.e. total 12 Ear-rings
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