Quantitative Aptitude DI - Sep 23 2020
Here we are providing new series of Quantitative Aptitude Questions for upcoming exams, so the aspirants can practice it on a daily basis.
Direction (1 - 5): Study the following data carefully and answer the questions:
There are total 3 rooms A, B and C and in each room, there is meeting is going on between employees of five different companies HCL, IBM, Infosys, TCS and Wipro.
bar graph given below shows the per cent probability of selecting one HCL employee, one IBM employee and one Infosys employees from three rooms A, B and C respectively.
Number of ways in which two employees from TCS can be selected from room A is 3 and probability of selecting two employees at random from the room out of which 1 is from Wipro and other is from TCS is (1/20).
Probability in which 1 employee from TCS can be selected from room B is (1/10) while the probability in which 2 employees from Wipro can be selected from the room is (3/95).
Number of ways in which 3 employees from TCS can be selected from room C is 20 and number of ways in which 2 employees from TCS and 2 employees from IBM together can be selected from room C is 150.
1) What is the ratio of probability of selecting one TCS employee and one HCL employees when two employees at random are selected from room A to the probability of selecting one Wipro employee from room B?
a) 2: 5
b) 4: 7
c) 3: 8
d) 4: 5
e) None of these
2) If probability of selecting employees from room A, room B and room C is 40%, 10% and 50% respectively and two employees are selected at random from any room, then what is the probability that both the employees are from Infosys?
a) (69/1700)
b) (169/5700)
c) (129/1900)
d) (119/3700)
e) None of these
3) If five employees are transferred from room A to room C, then probability of selecting two IBM employees from room C becomes (7/75), then find the percentage of IBM employees tranferred to room C while comparing all the transferred employees to room C?
a) 80%
b) 75%
c) 50%
d) 60%
e) None of these
4) If overall four employees are selected at random, what is the probability that 1 HCL and 1 Wipro employees are from room A, 1 IBM employee is from room B and remaining 1 Infosys is from room C?
a) 21/430
b) 3/820
c) 7/750
d) 51/700
e) None of these
5) Total number of ways in which 2 IBM employees, 1 TCS employee and 2 Wipro employees can be selected from room A is ‘P’ while the total number of ways in which 1 HCL employee and 4 IBM employees can be selected from room B is ‘Q’, then what is the ratio of value of ‘P’ to that of ‘Q’?
a) 36: 35
b) 26: 25
c) 16: 15
d) 6: 5
e) None of these
Answers :
Directions (1 - 5):
Room A:
Ratio of number of HCL, IBM and Infosys employees in room A = 32: 16: 20 = 8: 4: 5
Let HCL, IBM and Infosys employees in room A is 8x, 4x and 5x respectively and total employees are 25x.
Let TCS employees in room A = a
^{a}C_{2} = 3
a(a – 1)/2 = 3
a(a – 1) = 6
a^{2} – a – 6 = 0
a = 3 and -2 (Not valid)
TCS employees = a = 3
Wipro employees = 25x – (8x + 4x + 5x + 3) = (8x – 3)
Probability of selecting two employees at random from the room out of which 1 is from Wipro and other is from TCS =
[3 * (8x – 3)]/[^{25x}C_{2}] = 1/20
6(8x – 3)/25x(25x – 1) = 1/20
24(8x – 3) = 5x(25x – 1)
192x – 72 = 125x^{2} – 5x
125x^{2} – 197x + 72 = 0
x = 1
HCL employees = 8x = 8
IBM employees = 4x = 4
Infosys employees = 5x = 5
Wipro employees = (8x – 3) = 5
Total employees = 25x = 25
Room B:
Ratio of number of HCL, IBM and Infosys employees in room B = 25: 35: 10 = 5: 7: 2
Let HCL, IBM and Infosys employees in room B is 5y, 7y and 2y respectively and total employees are 20y.
Let TCS employees in room B = b
Wipro employees = 20y – (5y + 7y + 2y + b) = (6y – b)
Probability of selecting 1 employee from TCS = b/20y = (1/10)
b = 2y ……….. (1)
Wipro employees = (6y – b) = 4y
Probability of selecting 2 employees from Wipro = ^{4y}C_{2}/^{20y}C_{2} = (3/95)
2y(4y – 1)/10y(20y – 1) = 3/95
19(4y – 1) = 3(20y – 1)
76y – 19 = 60y – 3
16y = 16
y = 1
HCL employees = 5y = 5
IBM employees = 7y = 7
Infosys employees = 2y = 2
TCS employees = b = 2y = 2
Wipro employees = 4y = 4
Total employees = 20y = 20
Room C:
Ratio of number of HCL, IBM and Infosys employees in room C = 15: 25: 20 = 3: 5: 4
Let HCL, IBM and Infosys employees in room C is 3z, 5z and 4z respectively and total employees are 20z.
Let number of TCS employees are ‘c’.
Number of ways in which 3 employees from TCS can be selected = ^{c}C_{3} = 20
c(c – 1)(c – 2) = 120
c = 6
TCS employees = c = 6
Wipro employees = 20z – (3z + 5z + 4z + 6) = (8z – 6)
Number of ways in which 2 employees from TCS and 2 employees from IBM together can be selected from room C = ^{6}C_{2} * ^{5z}C_{2} = 150
5z(5z – 1) = (150 * 2)/(3 * 5)
5z^{2} – z – 4 = 0
z = 1 and -4/5 (Not valid)
HCL employees = 3z = 3
IBM employees = 5z = 5
Infosys employees = 4z = 4
Wipro employees = (8z – 6) = 2
Total employees = 20z = 20
Rooms |
HCL |
IBM |
Infosys |
TCS |
Wipro |
Total |
A |
8 |
4 |
5 |
3 |
5 |
25 |
B |
5 |
7 |
2 |
2 |
4 |
20 |
C |
3 |
5 |
4 |
6 |
2 |
20 |
1) Answer: A
Probability of selecting one TCS employee and one HCL employees when two employees at random are selected from room A
= (^{3}C_{1} * ^{8}C_{1})/^{25}C_{2} = (3 * 8)/(25 * 12) = (2/25)
Probability of selecting one Wipro employee from room B
= ^{4}C_{1}/^{20}C_{1} = 4/20 = (1/5)
Required ratio = (2/25): (1/5) = 2: 5
2) Answer: B
Probability of selecting two Infosys employees from room A = (4/10) * (^{5}C_{2}/^{25}C_{2}) = (1/75)
Probability of selecting two Infosys employees from room B = (1/10) * (^{2}C_{2}/^{20}C_{2}) = (1/1900)
Probability of selecting two Infosys employees from room C = (5/10) * (^{4}C_{2}/^{20}C_{2}) = (3/190)
Required probability = (1/75) + (1/1900) + (3/190) = (76 + 3 + 90)/5700 =
(169/5700)
3) Answer: D
Let ‘x’ IBM employees from room B are transferred to room C.
New IBM employees in room C = (5 + x)
New total employees in room C = 20 + 5 = 25
According to the question:
^{(5 + x)}C_{2}: ^{25}C_{2} = (7/75)
[(5 + x)(4 + x)]/[25 * 24] = 7/75
(5 + x)(4 + x) = 56
x^{2} + 9x + 20 – 24 = 0
x^{2} + 9x – 36 = 0
x = 3 and -12 (Not valid)
Required per cent = (3/5) * 100 = 60%
4) Answer: C
Probability of selecting 1 HCL and 1 Wipro employees from room A = (^{8}C_{1} * ^{5}C_{1})/^{25}C_{2} = (8 * 5)/(25 * 12) = (2/15)
Probability of selecting 1 IBM employee from room B = ^{7}C_{1}/^{20}C_{1} = (7/20)
Probability of selecting 1 Infosys employee from room B = ^{4}C_{1}/^{20}C_{1} = (4/20) = (1/5)
Required total probability
= (2/15) * (7/20) * (1/5) = 7/750
5) Answer: A
Total number of ways in which 2 IBM employees, 1 TCS employee and 2 Wipro employees can be selected from room A = P
P = ^{4}C_{2} * ^{3}C_{1} * ^{5}C_{2} = (2 * 3) * 3 * (5 * 2) = 6 * 3 * 10
P = 180
Total number of ways in which 1 HCL employee and 4 IBM employees can be selected from room B = Q
Q = ^{5}C_{1} * ^{7}C_{4} = 5 * [(7 * 6 * 5)/(3 * 2)] = 5 * 35 = 175
Required ratio = 180: 175 = 36: 35