Aptitude Equation ProblemsPage 1
1. | In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. X² + 13X + 42 = 0
2. Y² + 19Y + 90 = 0 | (a)X < Y | (b)X > Y | (c)X ≤ Y | (d)X = Y |
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Answer is: BSolve:-
X² + 13X + 42 = 0
X² + 7X + 6X + 42 = 0
X(X + 7) + 6(X + 7) = 0
X = -7, -6
Again-
Y² + 19Y + 90 = 0
Y² + 10Y + 9Y + 90 = 0
Y(Y + 10) + 9(Y + 10) = 0
Y = -10,-9
So, X is greater than Y.
2. | In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. X² + 15X + 56 = 0
2. Y² - 23Y + 132 = 0 | (a)X < Y | (b)X > Y | (c)X ≤ Y | (d)X = Y |
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Answer is: ASolve:-
X² + 15X + 56 = 0
X² + 8X + 7X + 56 = 0
X(X + 8) + 7(X + 8) = 0
X = -8, -7
Again
Y² - 23Y + 132 = 0
Y² - 12Y -11Y + 132 = 0
Y(Y - 12) -11(Y -12) = 0
Y = 12, 11
So, Y > X, is true.
3. | In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. X² + 7X + 12 = 0
2. Y² + 6Y + 8 = 0 | (a)X < Y | (b)X > Y | (c)X ≤ Y | (d)X = Y |
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Answer is: CSolve:-
X² + 7X + 12 = 0
X² + 4X + 3X+ 12 = 0
X(X + 4) + 3(X + 4) = 0
(X + 4)(X + 3) = 0
X = -4, -3
Again
Y² + 6Y + 8 = 0
Y² + 4Y + 2Y + 8 = 0
Y(Y + 4) + 2(Y + 4) = 0
(Y + 4)(Y + 2) = 0
Y = -4, -2
So, X ≤ Y is right.
4. | In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. X² - 22X + 120 = 0
2. Y² - 26Y + 168 = 0
| (a)X < Y | (b)X > Y | (c)X ≤ Y | (d)X = Y |
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Answer is: CSolve:-
X² - 22X + 120 = 0
X² - 12X -10X + 120 = 0
X(X - 12) -10(X - 12) = 0
(X - 12)(X - 10) = 0
X = 12, 10
Again
Y² - 26Y + 168 = 0
Y² - 14Y - 12Y + 168 = 0
Y(Y - 14) - 12(Y - 14) = 0
(Y -14)(Y - 12) = 0
Y = 14, 12
So, X ≤ Y is right.
5. | In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. X² + 12X + 32 = 0
2. Y² + 17Y + 72 = 0 | (a)X < Y | (b)X > Y | (c)X ≤ Y | (d)X ≥ Y |
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Answer is: DSolve:-
X² + 12X + 32 = 0
X² + 8X + 4X+ 32 = 0
X(X + 8) + 4(X + 8) = 0
(X + 8)(X + 4) = 0
X = -8, -4
Again
Y² + 17Y + 72 = 0
Y² + 9Y + 8Y + 72 = 0
Y(Y + 9) + 8(Y + 9) = 0
(Y + 9)(Y + 8) = 0
Y = -9, -8
So, X ≥ Y is true.
6. | In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. X² - 4 = 0
2. Y² + 6Y + 9 = 0 | (a)X < Y | (b)X > Y | (c)X ≤ Y | (d)X ≥ Y |
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Answer is: BSolve:-
X² - 4 = 0
X = ± 2
Again
Y² + 6Y + 9 = 0
Y² + 3Y + 3Y + 9 = 0
Y(Y + 3) + 3(Y + 3) = 0
(Y + 3)(Y + 3) = 0
Y = -3, -3
So, X > y is true.
7. | In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. X² - 7X + 12 = 0
2. Y² + Y - 12 = 0
| (a)X < Y | (b)X > Y | (c)X ≤ Y | (d)X ≥ Y |
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Answer is: DSolve:-
X² - 7X + 12 = 0
X² - 4X -3X + 12 = 0
X(X - 4) - 3(X - 4) = 0
(X - 4)(X - 3) = 0
X = 4, 3
Y² + Y - 12 = 0
Y² + 4Y - 3Y - 12 = 0
Y(Y + 4) - 3(Y + 4) = 0
(Y + 4)(Y - 3) = 0
Y = -4, 3
So, X ≥ Y is right.
8. | In each of the equations, two equations numbered 1 and 2 are given. You have to solve both the equations-
1. X² = 729
2. Y = √729
| (a)X < Y | (b)X > Y | (c)X ≤ Y | (d)X = Y |
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Answer is: CSolve:-
X² = 729
X = ± √729 = ± 27
Again
Y = √729
Y = 27
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