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

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##### Out of 300 students in a school, 95 play cricket only, 120 play football only, 80 play volleyball only and 5 play no games. If one student is chosen at random, find the probability that he plays either cricket or volleyball-
 1) 5/12 2) 7/12 3) 11/12 4) 3/11 5) None of these
Explanation :
Total number of trials = 300 (Since there are 300 students all together).
Number of times a cricket player is chosen = 95 (Since 95 students play cricket).
Number of times a football player is chosen = 120.
Number of times a volleyball player is chosen = 80.
Number of times a student is chosen who plays no games = 5.
The probability of getting a player who plays either cricket or volleyball
= (95+80)/300 = 175/300 = 7/12
=

##### Directions: Study the following information carefully and answer the given questions.In a bag X there are some identical green balls and rest other are blue balls (blue balls greaterthan the green balls).(i) Probability of getting 1 green and 1 blue ball without replacement is 40/87.(ii) Probability of getting both green is A.(iii) If there is another bag (Y) in which some blue balls and rest others are green.(iv) If a number of blue balls taken out from Y is 4 more than the green balls taken out from bag Y and added to bag X. Then the probability of getting both green from bag X is 13/123.Find the number of blue balls taken out from bag Y.
 1) 2 2) 4 3) 5 4) 6
Explanation :
Bag X:
Two balls taken from bag x, total probability,
According to the question,
40/87 = (Probability of getting 1st green and second blue * Probability of getting 1st blue and second green)
Take green = 10 and blue = 20
According to the question,
40/87 = 10 * 20 * 2/(30 * 29)
40/87 = 40/87
Let us take number of green balls taken out from bag Y be a
Number of blue balls taken out from bag Y = a + 4
According to the question,
(10 + a)C2/(30 + 2a + 4)C2 = 13/123
(10 + a) * (9 + a)/[(34 + 2a) * (33 + 2a)] = 13/123
Simplify the equation we get, a = 4
Number of blue balls taken out from bag Y = a + 4 = 4 + 4 = 8
##### Directions: Study the following information carefully and answer the givenquestions.In a bag X there are some identical green balls and rest other are blue balls (blue balls greaterthan the green balls).(i) Probability of getting 1 green and 1 blue ball without replacement is 40/87.(ii) Probability of getting both green is A.(iii) If there is another bag (Y) in which some blue balls and rest others are green.(iv) If a number of blue balls taken out from Y is 4 more than the green balls taken out frombag Y and added to bag X. Then the probability of getting both green from bag X is 13/123.What should be total number of balls in bag X ?
 1) 10 2) 20 3) 29 4) 30
Explanation :
Bag X:
Two balls taken from bag x, total probability,
According to the question,
40/87 = (Probability of getting 1st green and second blue * Probability of getting 1st blue and
second green)
Take green = 10 and blue = 20
According to the question,
40/87 = 10 * 20 * 2/(30 * 29)
40/87 = 40/87
This satisfies the given condition.
So, total number  balls in box X should be 30.
##### Directions : Study the following information carefully and answer the given questions.In a bag X there are some identical green balls and rest other are blue balls (blue balls greater than the green balls).(i) Probability of getting 1 green and 1 blue ball without replacement is 40/87.(ii) Probability of getting both green is A.(iii) If there is another bag (Y) in which some blue balls and rest others are green.(iv) If a number of blue balls taken out from Y is 4 more than the green balls taken out from bag Y and added to bag X. Then the probability of getting both green from bag X is 13/123.What should be the value of A ?
 1) 2/29 2) 3/29 3) 4/29 4) 5/29
Explanation :
Bag X:
Two balls taken from bag x, total probability,
According to the question,
40/87 = (Probability of getting 1st green and second blue * Probability of getting 1st blue and second green)
Take green = 10 and blue = 20
According to the question,
40/87 = 10 * 20 * 2/(30 * 29)      or  40/87 = 40/87
This satisfies the given condition.
Probability of both green ball (A) = 10C2/30C2
= (10 * 9)/(29 * 30) = 3/29
##### Out of 15 students studying in a class, 7 are from Maharastra, 5 are from Karnataka and 3 are from Goa. Four students are to be selected at random. What are the chances that at least one is from Karnataka ?
 1) 1/15 2) 11/13 3) 12/13 4) 3/5 5) None of these
Explanation : Given , total numbers of students are 15.
##### A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue ?
 1) 10/21 2) 11/21 3) 2/7 4) 5/7 5) None of these
Explanation : In the bag,
##### In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green ?
 1) 1/2 2) 1/3 3) 4/9 4) 8/21
Explanation :
Total number of balls = n(S) = (8 + 7 + 6) = 21.
Let E = event that the ball drawn is neither red nor green = event that the ball drawn is blue.
So, n(E) = 7.
Therefore, P(E) = n(E) / n(S) =  7 / 21=1 /3 .
##### What is the probability of getting a sum 9 from two throws of a dice ?
 1) 1/6 2) 1/9 3) 1/8 4) 1/12 5) None of these
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)} = 4.
P(E) = n(E)/n(S) = 4/36 = 1/9 .

##### In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize ?
 1) 1/5 2) 2/5 3) 2/7 4) 3/5 5) None of these
Explanation :

Thus, P (getting a prize) =10 /(10 + 25) =10 /35 = 2 / 7

##### 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