Quantitative Aptitude Quiz – 73 (Quadratic)

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Welcome To D2G’s. This Quiz deals with some quadratic equations and probability questions with proper explanations for your practice. The only way to learn mathematics is to do mathematics. Keep learning.

Directions (Q. Nos.1-5) In the following questions two Equations I and II are given. You have to solve both the equations and give answer.

a) If X > Y
b) If X >= Y
c) If X < Y
d) If X =< Y
e) If X = Y or the relationship can’t be established.

1. I. 3X^2 + 8X + 4 = 0
   II. 4Y^2 – 19Y + 12 = 0

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Answer c) X < Y
Explanation:
I. 3X^2 + 8X + 4 = 0
=>3X^2 + 6X + 2X + 4 = 0
=>3X(X+2) + 2(X+2) = 0
=>X = -2, -2/3
II. 4Y^2 – 19Y + 12 = 0
=>4Y^2 – 16Y – 3Y + 12 = 0
=>4Y(Y-4) – 3(Y-4) = 0
=>Y = 4, 3/4
Hence, Y > X.

2. I. 16X^2 + 20X + 6 = 0
   II. 10Y^2 + 38Y + 24 = 0

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Answer a) X > Y
Explanation:
I. 16X^2 + 20X + 6 = 0
Here, 16*6 = 96(12*8)
By changing sign, X = -12/16, -8/16
=>X = -3/4, -1/2
II. 10Y^2 + 38Y + 24 = 0
Here, 24*10 = 240(30*8)
By changing sign, Y = -30/10, -8/10
=>Y = -3, -4/5
Hence, X > Y.

3. I. 18X^2 + 18X + 4 = 0
   II. 12Y^2 + 29Y + 14 = 0

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Answer b) X >= Y
Explanation:
I. 18X^2 + 18X + 4 = 0
Here, 18*4 = 72(12*6)
By changing sign, X = -12/18, -6/18
=>X = -2/3, -1/3
II. 12Y^2 + 29Y + 14 = 0
Here, 12*14 = 168(21*8)
By changing sign, Y = -21/12, -8/12
=>Y = -7/4, -2/3
Hence, X >= Y.

4. I. 8X^2 + 6X = 5
   II. 12Y^2 – 22Y + 8 = 0

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Answer d) X =< Y
Explanation:
I. 8X^2 + 6X – 5 = 0
Here, 8*(-5) = -40{10*(-4)}
By changing sign, X = -10/8, 4/8
=>X = -5/4, 1/2
II. 12Y^2 – 22Y + 8 = 0
Here, 12*8 = 96{(-16)*(-6)}
By changing sign, Y = 16/12, 6/12
=>Y = 4/3, 1/2
Hence, X =< Y.

5. I. 17X^2 + 48X = 9
   II. 13Y^2 = 32Y – 12

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Answer c) X < Y
Explanation:
I. 17X^2 + 48X – 9 = 0
Here, 17*(-9) = -153{51*(-3)}
By changing sign, X = -51/17, 3/17
=>X = -3, 3/17
II. 13Y^2 – 32Y + 12 = 0
Here, 13*12 = 156{(-26)*(-6)}
By changing sign, Y = 26/13, 6/13
=>Y = 2, 6/13
Hence, X < Y.

6. In how many different ways can the letters of the word DESIGN be arranged, so that the vowels are at the two ends?
a) 48
b) 72
c) 36
d) 24
e) None of these

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Answer a) 48
Explanation:
Total letters = 6
Number of vowels(E, I) = 2
Total number of ways of arrangement, so that vowels are at the both ends = 2P2 * 4P4 = 2*1*4*3*2*1 = 48.

7. Out of 5 men and 3 women, a committee of 3 members is to be formed, so that it has 1 women and 2 men. In how many different ways can this be done?
a) 20
b) 10
c) 23
d) 30
e) None of these

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Answer d) 30
Explanation:
Required number of ways = 3C1 * 5C2 = 3*(5*4)/2*1 = 30.

Directions (Q. Nos.8-10): Study the given information carefully and answer the questions that follow.

A basket contains 4 red, 5 blue and 3 green marbles.

8. If two marbles are drawn at random, what is the probability that both are red?
a) 3/7
b) 1/11
c) 2/11
d) 1/6
e) None of these

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Answer b) 1/11
Explanation:
Total no. of possible outcomes = n(S) = 12C2 = 66
Total no. of favorable events = n(E) = 4C2 = 6
Required probability = 6/66 = 1/11.

9. If three marbles are picked at random, what is the probability that at least one is blue?
a) 7/12
b) 37/44
c) 5/12
d) 7/44
e) None of these

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Answer b) 37/44
Explanation:
Total no. of possible outcomes = 12C3 = 220
Number of events which do not contain blue marbles(3 marbles out of 7 marbles) = 7C3 = 35
Required probability = 1 – 35/220 = 37/44.

10. If three marbles are picked at random, what is the probability that at least one is blue?
a) 7/44
b) 7/12
c) 5/12
d) 1/44
e) None of these

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Answer d) 1/44
Explanation:
Total no. of ways of selection of 3 marbles out of 12 = n(S) = 220
Total no. of favorable events n(E) = 3C3 + 4C3 = 1 + 4 = 5
Required probability = 5/220 = 1/44.

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D2G’s Team..