Aptitude :: Equation ProblemsPage 2

9.

In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. 2X² + 11X + 14 = 0
2. 4Y² + 12y + 9 = 0
(a)X < Y
(b)X > Y
(c)X ≤ Y
(d)X = Y
Answer is: ASolve:-
2X² + 11X + 14 = 0
2X² + 7X + 4X + 14 = 0
X(2X + 7) + 2(2X +7) = 0
(X + 2)(X + 7/2) = 0
X = -2, -7/2
Again
4Y² + 12y + 9 = 0
4Y² + 6y + 6Y + 9 = 0
2Y(2Y + 3) + 3(2Y + 3) = 0
(2Y +3)(2Y + 3) = 0
Y = -3/2, -3/2
So, Y>X is right.

10.

In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. X² - X - 12 = 0
2. Y² + 5Y + 6 = 0
(a)X < Y
(b)X > Y
(c)X ≤ Y
(d)X ≥ Y
Answer is: DSolve:-
X² - X - 12 = 0
X² - 4X + 3X - 12 = 0
X(X - 4) + 3(X - 4) = 0
(X - 4)(X + 3) = 0
X = 4, -3
Again
Y² + 5Y + 6 = 0
Y² + 3Y + 2Y + 6 = 0
Y(Y + 3) + 2(Y + 3) = 0
(y + 3)(Y + 2) = 0
Y = -3, -2
So, X ≥ Y is right.

11.

In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations
1. X² - 8X + 15 = 0
2. Y² - 3Y + 2 = 0
(a)X < Y
(b)X > Y
(c)X ≤ Y
(d)X ≥ Y
Answer is: BSolve:-
X² - 8X + 15 = 0
X² - 5X - 3X + 15 = 0
X(X - 5) - 3(X - 5) = 0
(X - 5)(X - 3) = 0
X = 5, 3
Again
Y² - 3Y + 2 = 0
Y² - 2Y - y + 2 = 0
Y(Y -2) -1(Y -2) = 0
(Y -2)(Y - 1) = 0
Y= 2, 1
So, X is greater than Y.

12.

In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. X² - 32 = 112
2. Y² - √169 = 0
(a)X < Y
(b)X > Y
(c)X ≤ Y
(d)X = Y
Answer is: ASolve:-
X² - 32 = 112
X² = 144
X = ± 12
Again
Y² - √169 = 0
Y² = √169
Y = 13
So, Y > X is true.

13.

In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. X - √121 = 0
2. Y² - 121 = 0
(a)X < Y
(b)X > Y
(c)X ≤ Y
(d)X ≥ Y
Answer is: DSolve:- X - √121 = 0
X = 11 Again Y² - 121 = 0 Y = ± 11 This statement X ≥ Y is true.

14.

In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. X² - 16 = 0
2. Y² - 9Y + 20 = 0
(a)X < Y
(b)X > Y
(c)X ≤ Y
(d)X = Y
Answer is: CSolve:-
X² - 16 = 0
X = ± 4
Again
Y² - 9Y + 20 = 0
Y² - 5Y - 4Y + 20 = 0
Y(Y - 5) - 4(Y - 5) = 0
(Y - 5)(Y - 4) = 0
Y = 5, 4
So, X ≤ Y is true.

15.

In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. 5X² - 18X + 9 = 0
2. 20Y² - 13Y + 2 = 0
(a)X < Y
(b)X > Y
(c)X ≤ Y
(d)X = Y
Answer is: BSolve:-
5X² - 18X + 9 = 0
5X² - 15X - 3X + 9 = 0
5X(X - 3) - 3(X - 3) = 0
(X - 3)(5X - 3) = 0
X = 3, 3/5
Again
20Y² - 13Y + 2 = 0
20Y² - 8Y - 5Y + 2 = 0
4Y(5Y - 2) - 1(5Y - 2) = 0
(4Y - 1)(5Y - 2) = 0
Y = 1/4, 2/5
So, X > Y is right.

16.

In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. X³ - 878 = 453
2. Y² - 82 = 39
(a)X < Y
(b)X > Y
(c)X ≤ Y
(d)X ≥ Y
Answer is: DSolve:- X³ - 878 = 453
X³ = 453 + 878
X³ = 1331 X = 11 Again Y² - 82 = 39 Y² = 39 + 82 Y² = 121 Y = ± 11 X ≥ Y, this statement is right.

Comments

I think qst 12 bit 2 is wrong

Amit On 17.10.2016. 15:32

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