Hello Readers,
Welcome To D2G’s. This Quiz contains some Quadratic equations and simple probability questions with proper explanations for your practice. Keep Learning.
Directions(Q.1-5): In each question two equations are provided. On the basis of these you have to find out the relation between X and Y. Give answer
a) if X = Y or the relation can not be established.
b) if X > Y
c) if Y > X
d) if X >= Y
e) if Y >= X
1. I. 5X^2 – 8X + 3 = 0
II. 2Y^2 – 7Y + 5 = 0
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Answer e) Y >= X
Explanation:
I. 5X^2 – 8X + 3 = 0
=>5X^2 – 5X – 3X + 3 = 0
=>(5X – 3)(X-1) = 0
=>X = 1,3/5
II. 2Y^2 – 7Y + 5 = 0
=>Y = 1,5/2
Hence, Y >= X.
2. I. 12X^2 – X – 1 = 0
II. 6Y^2 – 5Y + 1 = 0
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Answer e) Y >= X
Explanation:
I. 12X^2 – X – 1 = 0
=>12X^2 + 3X – 4X – 1 = 0
=>(3X – 1)(4X +1) = 0
=>X = 1/3 or, -1/4
II. 6Y^2 – 5Y + 1 = 0
=>6Y^2 – 2Y – 3Y + 1 = 0
=>(2Y – 1)(3Y – 1) = 0
=>Y = 1/2,1/3
Hence, Y >= X.
3. I. 17X^2 + 15X – 2 = 0
II. 3Y^2 + 7Y + 4 = 0
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Answer d) X >= Y
Explanation:
I. 17X^2 + 15X – 2 = 0
=>17X^2 + 17X – 2X – 2 = 0
=>(17X – 2)(X + 1) = 0
=>X = 2/17 or, -1
II. 3Y^2 + 7Y + 4 = 0
=>(3Y + 4)(Y + 1) = 0
=>Y = -4/3 or, -1
Hence, X >= Y.
4. I. X^2 – 6X + 8 = 0
II. 3Y^2 – 10Y + 7 = 0
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Answer a) the relation can not be established
Explanation:
I. X^2 – 6X + 8 = 0
=>X^2 – 2X – 4X + 8 = 0
=>(X – 2)(X – 4) = 0
=>X = 2,4
II. 3Y^2 – 10Y + 7 = 0
=>Y = 1, 7/3
Hence, the relation can not be establised.
5. I. 8X^2 + 10X + 3 = 0
II. 5Y^2 – 8Y + 3 = 0
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Answer c) Y > X
Explanation:
I. 8X^2 + 6X + 4X + 3 = 0
=>X = -3/4 or, -1/2
II. 5Y^2 – 8Y + 3 = 0
=>5Y^2 – 3Y – 5Y + 3 = 0
=>Y = 3/5 or, 1
Hence, Y > X.
6. What is the probability of getting a sum ‘9’ from two throwns of a dice?
a) 1/6
b) 1/8
c) 1/9
d) 1/12
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Answer c) 1/9
Explanation:
In two throwns of dice, n(s) = 6*6 = 36
Let E = even of getting sum ‘9’ = {(3,6), (6,3), (4,5), (5,4)}
Now, n(E) / n(S) = 4/36 = 1/9.
7. Two dice are tossed. Probability that total score is in prime number is
a) 1/6
b) 5/12
c) 1/2
d) 7/9
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Answer b) 5/12
Explanation:
n(S) = 6*6 = 36
E = {(1,1), (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (4,1), (4,3), (5,2), (5,6), (6,1), (6,5)}
So, n(E) = 15
Then, P(E) = n(E) / n(S) = 15/36 = 5/12.
8. In how many ways can we arrange letters of the word ‘MATTER’?
a) 73
b) 144
c) 360
d) 720
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Answer c) 360
Explanation:
The word contains 1M, 1A, 2T, 1E, 1R.
Required number of ways = 6! / {(1!)(1!)(2!)(1!)(1!)} = 720/2 = 360.
9. In a lottery, there are 5 prizes and 10 blanks in tickets. A ticket is drawn at random. What is the probability of getting a prize?
a) 1/7
b) 2/7
c) 1/3
d) 2/3
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Answer c) 1/3
Explanation:
n(S) = 5 + 10 = 15
n(E) = 5
Then, P(E) = n(E) / n(S) = 5/15 = 1/3.
10. A glass jar contains 1 red, 3 green, 2 blue, and 4 yellow marbles. If a single marble is chosen at random from the jar, what is the probability that it is yellow or green?
a) 3/25
b) 7/10
c) 1/5
d) 9/10
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Answer b) 7/10
Explanation:
n(S) = 10
Now, probability of getting a yellow marble = 4/10
Probability of getting a green marble = 3/10
Since, the events are mutually exclusive
P(yellow or green) = P(yellow) + P(green) = 4/10 + 3/10 = 7/10.
Best Wishes,
D2G Team..
