Quantitative Aptitude DI
Feb 15 2021
Here we are providing new series of Quantitative Aptitude Questions for upcoming exams, so the aspirants can practice it on a daily basis.
Study the following data carefully and answer the questions:
Bar graph given below shows the average number of Red and Blue balls and difference between Red and Blue balls in five different bags A, B, C, D and E.
Total balls in the bag = Red balls + Blue balls + Green balls
Note: Red balls are more than Blue balls in only bags A and E and total number of balls in any one can’t exceed 20.
Ratio of total balls in bag A to that in bag B is 2: 1 and probability of selecting 2 balls from bag B out of which one is green and other is blue is (1/7). Probability of selecting 2 balls from bag C in such a way that both are either Red or Blue is (4/45). Average of total number of balls in all the five bags together is 11.6 and probability of selecting 2 Green balls from bag E is (1/12).
1) If 3 red balls from bag A are painted Green, then probability of selecting 2 green balls from that bag is increased by what percent from the original probability?
a) 120%
b) 280%
c) 180%
d) 320%
e) None of these
2) If 3 more balls of colour red and blue colour are added to bag B and now the probability of selecting 2 balls out of which one is Blue and one is green becomes (1/11), then what will be the probability of selecting 3 red balls from that bag?
a) 2/33
b) 3/44
c) 4/55
d) 5/66
e) None of these
3) What will be the probability of selecting 2 balls of different colours from bag C?
a) 21/55
b) 31/45
c) 11/35
d) 1/15
e) None of these
4) If ‘x’ white balls are added to the bag D in such a way that probability of selecting 2 blue balls from the bag becomes (1/10), then what will be the probability of selecting ‘x/2’ red balls from bag D after adding ‘x’ white balls?
a) 1/131
b) 1/144
c) 2/121
d) 1/133
e) None of these
5) What will be the probability of selecting 3 balls from bag E in such a way that 2 balls are of same colour and one is of different colour?
a) 55/84
b) 45/74
c) 65/94
d) 53/88
e) None of these
Answers :
Directions (1-5) :
Table given below shows the red and blue balls in five bags:
Probability of selecting 2 balls from bag B out of which one is green and other is blue = (4 * b)/7 + bC2 = (1/7)
8b/(7 + b)(6 + b) = 1/7
56b = b2 + 13a + 42
b2 – 43b + 42 = 0
b = 1 or 42 (Not valid)
Total balls in bag B = 3 + 4 + 1 = 8
Total balls in bag A = 8 * (2/1) = 16
Total green balls in bag A = a = 16 – 8 – 3 = 5
Probability of selecting 2 balls from bag C in such a way that both are either Red or Blue = (2C2 + 3C2)/5 + cC2 = (4/45)
8/(5 + c)(4 + c) = 4/45
(5 + c)(4 + c) = 90
c = 5 and -14 (Not valid)
Total balls in bag C = 2 + 3 + 5 = 10
Probability of selecting 2 Green balls from bag E = eC2/6 + eC2 = (1/12)
e(e – 1)/(6 + e)(5 + e) = 1/12
11e2 – 23e – 30 = 0
e = 3
Total balls in bag E = 4 + 2 + 3 = 9
Total balls in bag D = (11.6 * 5) – (16 + 8 + 10 + 9) = 15
Green balls in bag D = 15 – 5 – 7 = 3
1) Answer: c)
Probability of selecting 2 green balls from bag A = 5C2/16C2 = 10/120
Number of green balls after painting = 5 + 3 = 8
Now, Probability of selecting 2 green balls from bag A = 8C2/16C2 = 28/120
Increase in the probability = (28/120) – (10/120) = (18/120)
Required percent = [(18/120)/(10/120)] * 100 = 180%
2) Answer: a)
Let ‘x’ blue balls are added to the bag B.
Probability of selecting 2 balls out of which one is Blue, and one is green =(4 + x)/11C2 = (1/11)
(4 + x)/5 = 1
x = 1
Hence, number of red balls added to bag B = 3 – x = 2
Probability of selecting 3 red balls from that bag = 5C3/11C3 = 10/165 = 2/33
3) Answer: b)
Probability of selecting one red and one blue ball from bag C = (2 * 3)/10C2
= 2/15
Probability of selecting one red and one green ball from bag C = (3 * 5)/10C2
= 1/3
Probability of selecting one blue and one green ball from bag C = (2 * 5)/10C2
= 2/9
Required probability = (2/15) + (1/3) + (2/9) = (6 + 15 + 10)/45 = 31/45
4) Answer: d)
Probability of selecting 2 blue balls from the bag = 7C2/15 + xC2 = (1/10)
42/(15 + x)(14 + x) = 1/10
x2 + 29x – 210 = 0
x = 6 and -35 (Not valid)
Probability of selecting ‘x/2 = 3’ red balls from bag D = 5C3/21C3 = 10/1330
= 1/133
5) Answer: a)
Probability of selecting 2 red and one blue balls from bag E = (4C2 * 2C1)/9C3
= 12/84
Probability of selecting 2 red and one green balls from bag E = (4C2 * 3C1)/9C3
= 18/84
Probability of selecting 2 blue and one red balls from bag E = (2C2 * 4C1)/9C3
= 4/84
Probability of selecting 2 blue and one green balls from bag E = (2C2 * 3C1)/9C3
= 3/84
Probability of selecting 2 green and one red balls from bag E = (3C2 * 4C1)/9C3
= 12/84
Probability of selecting 2 green and one blue balls from bag E = (3C2 * 2C1)/9C3
= 6/84
Required probability = (12 + 18 + 4 + 3 + 12 + 6)/84 = 55/84