Probability and Statistics - Online Test
Q1. Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations.
I. 8x2 + 30x + 28 = 0
II. 9y2 + 11y+2 =0
I. 8x2 + 30x + 28 = 0
II. 9y2 + 11y+2 =0
Answer : Option C
Explaination / Solution:
I. 8x2 + 30x + 28 = 0
x = (-7/4, -2)
II. 9y2 + 11y+2 =0
y = (-1, -2/9)
So x<y
Q2. Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations.
I. 5x2- 24x +16 = 0
II. 5y2 + 29y + 20 = 0
I. 5x2- 24x +16 = 0
II. 5y2 + 29y + 20 = 0
Answer : Option A
Explaination / Solution:
I. 5x2- 24x +16 = 0
x = 4/5, 4
II. 5y2 + 29y + 20 = 0
y = (-4/5, -5)
So X>Y
Q3. Direction: In the following question, there are two equations. Solve the equations and answer accordingly:
I. 6x2 + 46x +60 = 0
II. 4y2+ 29y + 45= 0
II. 4y2+ 29y + 45= 0
Answer : Option E
Explaination / Solution:
I. 6x2 + 46x +60 = 0
3x2 + 23x + 30 = 0
3x2 + 18x + 5x + 30 = 0
3x(x+6)+5(x+6)=0
(x+6)(3x+5)=0
x = -5/3, -6
II. 4y2 + 29y + 45= 0
4y2 + 20y + 9y + 45= 0
4y(y+5)+9(y+5)=0
(4y+9)(y+5)=0
y = -5, -9/4
So Relationship cannot be established
Q4. Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations.
I. 8x – y = 40
II. 4x + y = 32
I. 8x – y = 40
II. 4x + y = 32
Answer : Option C
Explaination / Solution:
After solving the both equation got values of x = 6 and y = 8
So X
Q5. Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations.
I. 4x2 - 24x + 32 = 0
II. y2 = (12)2 – 140
I. 4x2 - 24x + 32 = 0
II. y2 = (12)2 – 140
Answer : Option B
Explaination / Solution:
I. 4x2 - 24x + 32 = 0
x = (4, 2)
II. y = (+2,-2)
So X ≥Y
Q6.
Directions: Study the given information carefully and answer the questions that follow—
A store contains 5 red, 4 blue, 5 green shirts.
If two shirts are picked at random, what is the probability that both are blue?
Answer : Option C
Explaination / Solution:
Probabilities if both are blue
4C2/14C2 = 6/91
Q7.
Directions: Study the given information carefully and answer the questions that follow—
A store contains 5 red, 4 blue, 5 green shirts.
If three shirts are picked at random, what is the probability that two are blue and one is red?
Answer : Option A
Explaination / Solution:
Probability if two are Blue and one are Red
[(4C2*5C1)/14C3] = (6*5)/364 -->15/182
Q8.
Directions: Study the given information carefully and answer the questions that follow—
A store contains 5 red, 4 blue, 5 green shirts.
If two shirts are picked at random, what is the probability that at least one is green?
Answer : Option B
Explaination / Solution:
Probability if at least one is Green
[1-(9C2/14C2)] = 55/91
Q9.
Directions: Study the given information carefully and answer the questions that follow—
A store contains 5 red, 4 blue, 5 green shirts.
If two shirts are picked at random, what is the probability that either both are red or both are green?
Answer : Option D
Explaination / Solution:
Probabilities if both either are Red or either are green
(5C2 + 5C2)/ 14C2 = (10+10)/91 --> 20/91
Q10.
Directions: Study the given information carefully and answer the questions that follow—
A store contains 5 red, 4 blue, 5 green shirts.
If two shirts are picked at random, what is the probability that at most one is blue?
Answer : Option B
Explaination / Solution:
Probabilities if at most one is blue =
[(4C0*10C2+ 4C1*10C1)/14C2] = (1*45 + 4*10)/( 91) = 85/91
