Hello Readers,
Welcome To D2G’s. This Quiz contains some Quadratic equations and simple probability questions with proper explanation 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 relation can’t be established
b) If X > Y
c) If Y > X
d) If X >= Y, and
e) If Y >= X.
1. I. 14X^2 – 5X – 1 = 0
II. 2Y^2 + 3Y + 1 = 0
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Answer b) X > Y
Explanation:
I. 14X^2 – 5X – 1 = 0
=>14X^2 – 7X + 2X – 1 = 0
=>(7X + 1)(2X – 1) = 0
=>X = -1/7 or 1/2
II. 2Y^2 + 3Y + 1 = 0
=>2Y^2 + 2Y + Y + 1 = 0
=>(Y + 1)(2Y + 1) = 0
=>Y = -1 or -1/2
Hence, X > Y.
2. I. 2X + Y = 12
II. 3X + 5Y – 32 = 0
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Answer a) X = Y
Explanation:
I. 2X + Y = 12
=>10X + 5Y = 60
=>10X + 32 – 3X = 60
=>7X = 28
=>X = 4
Now, 2X + Y = 12
=>Y = 4
Hence, X = Y.
3. I. 2Y^2 + 45Y – 23 = 0
II. X^2 + 48X + 575 = 0
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Answer e) Y >= X
Explanation:
I. 2Y^2 + 45Y – 23 = 0
=>2Y^2 + 46Y – Y – 23 = 0
=>(2Y – 1)(Y + 23) = 0
=>Y = -23 or 1/2
II. X^2 + 23X + 25X + 575 = 0
=>(X + 23)(X + 25) = 0
=>X = -23 or -25
Hence, Y >= X.
4. I. 4Y^2 – 7Y + 3 = 0
II. X^2 – 2X + 1 = 0
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Answer d) X >= Y
Explanation:
I. 4Y^2 – 7Y + 3 = 0
=>4Y^2 – 4Y – 3Y + 3 = 0
=>4Y(Y – 1) – 3(Y – 1) = 0
=>(4Y – 3)(Y – 1) = 0
=>Y = 3/4 or 1
II. X^2 – 2X + 1 = 0
=>(X – 1)^2 = 0
=>X = 1
Hence, X >= Y.
5. I. 225X^2 + 105X – 44 = 0
II. 210Y^2 + 319Y + 121 = 0
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Answer d) X >= Y
Explanation:
I. 225X^2 + 105X – 44 = 0
=>225X^2 + 165X – 60X – 44 = 0
=>(15X + 11)(15X – 4) = 0
=>X = -11/15 or 4/15
II. 210Y^2 + 319Y + 121 = 0
=>210Y^2 + 165Y + 154Y + 121 = 0
=>(15Y + 11)(14Y + 11) = 0
=>Y = -11/15 or -11/14
Hence, X >= Y.
6. In a meeting between two countries, each country has 12 delegates. All the delegates of one country shake hands with all delegates of the other country. Find the number of handshakes possible?
a) 72
b) 144
c) 288
d) 234
e) None of these
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Answer b) 144
Explanation:
Total number of handshakes = 12 * 12 = 144.
7. There are 10 questions in a question paper. In how many ways, a student can solve these questions, if he solve one or more questions?
a) 1024
b) 1025
c) 1023
d) 1000
e) None of these
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Answer c) 1023
Explanation:
Required number of ways = 2^n – 1 = 2^10 – 1 = 1024 – 1 = 1023.
Directions (Q.8 – 10): Study the following information carefully to answer the questions that follow.
A box contains 2 blue caps, 4 red caps, 5 green caps and 1 yellow cap.
8. If two caps are picked at random, what is the probability that both are blue?
a) 1/6
b) 1/10
c) 1/45
d) 1/66
e) None of these
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Answer d) 1/66
Explanation:
Total number of caps = 2 + 4 + 5 + 1 = 12
Total number of outcomes = n(S) = 12C2 = 66
Favorable number of outcomes = n(E) = 2C2 = 1
Required probability = 1/66.
9. If one cap is picked at random, what is the probability that it is either blue or yellow?
a) 2/6
b) 1/4
c) 2/8
d) 6/11
e) None of these
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Answer b) 1/4
Explanation:
Total number of caps = 12
n(S) = 12C1 = 12
out of (2 blue + 1 yellow) caps, number of ways to pick one cap n(E) = 3C1 = 3
Required probability P(E) = n(E)/n(S) = 3/12 = 1/4.
10. if two caps are picked at random, what is the probability that atleast one is red?
a) 1/3
b) 16/21
c) 19/33
d) 7/19
e) None of these
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Answer c) 19/33
Explanation:
Required probability = 1 – 8C2/12C2 = 1 – 28/66 = 38/66 = 19/33.
Best Wishes,
D2G’s Team..
