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| | | | | 1 . | Directions (Q. 1-5): Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between ‘x’ and ‘y’ and give answer.
(1) if x > y
(2) if x < y
(3) if x $\geq$ y
(4) if x $\leq$ y
(5) if x = y or no relation can be established between ‘x’ and ‘y’.
$Q.$ I. $6x^ 2 - 19x + 15 = 0 $
II.$ 10y^ 2 - 29y + 21 = 0$ | | A). x > y | | B). x < y | | C). x $\geq$ y | | D). x $\leq$ y | | Answer & Explanation | Answer : Option C | | | | Explanation : | | | $6x^ 2 - 19x + 15 = 0 $
$3x(2x - 3) - 5(2x - 3) = 0$
$ (3x - 5) (2x - 3) = 0$
$x$ = $5\over 3$ , $3\over 2$
$10y 2 - 15y - 14y + 21 = 0$
$ 5y(2y - 3) - 7(2y - 3) = 0$
$(5y - 7) (2y - 3) = 0$
y = $7\over5$ , $3\over 2$
$x \geq y$ | | | |
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| 2 . | I. 12 $x^ 2 $ + 11x - 56 = 0
II. 4 $y^ 2 $ - 15y + 14 = 0 | | A). x > y | | B). x < y | | C). x $\geq$ y | | D). x $\leq$ y | | Answer & Explanation | Answer : Option D | | | | Explanation : | | | I. $12 x^ 2 + 32x - 21x - 56 $ = 0
or, 4x(3x + 8) - 7(3x + 8) = 0
or, (4x - 7) (3x + 8) = 0
X = $7\over 4$ , $-8\over 3$
II. $ 4y^ 2 - 8y - 7y + 14 $= 0
or, 4y(y - 2) - 7(y - 2) = 0
or, (4y - 7) (y - 2) = 0
y = 2, $7\over 4$
$x \leq y$ | | | |
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| 3 . | I. $3x 2 + 13x + 12 $= 0
II.$ y^ 2 + 9y + 20 $= 0 | | A). x > y | | B). x < y | | C). x $\geq$ y | | D). x $\leq$ y | | Answer & Explanation | Answer : Option A | | | | Explanation : | | | I . $3x^ 2$ + 9x + 4x + 12 = 0
or, 3x(x + 3) + 4(x + 3) = 0
or, (3x + 4) (x + 3) = 0
x = $-4\over 3$, -3
II.$ y^ 2$ + 5y + 4y + 20 = 0
or, y(y + 5) + 4(y + 5) = 0
or, (y + 4) (y + 5) = 0
y = - 4, - 5
$ x > y$ | | | |
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| 4 . | I. $8x^ 2 $ - 15x + 7 = 0
II. $2y^ 2 $ - 7y + 6 = 0 | | A). x > y | | B). x < y | | C). x $\geq$ y | | D). x $\leq$ y | | Answer & Explanation | Answer : Option B | | | | Explanation : | | | I. $8x^ 2$ - 8x - 7x + 7 = 0
or, 8x(x - 1) -7(x - 1) = 0
or, (8x - 7) (x - 1) = 0
x = $7\over 8$ , 1
II. $2y^ 2$ - 4y - 3y + 6 = 0
or, 2y(y - 2) -3(y - 2) = 0
or, (y - 2) (2y - 3) = 0
y = 2 , $3\over 2$
$x < y$ | | | |
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| 5 . | I. 7x - 3y = 13
II. 5x + 4y = 40 | | A). x > y | | B). x < y | | C). x $\geq$ y | | D). x $\leq$ y | | Answer & Explanation | Answer : Option B | | | | Explanation : | | | Eqn (I) × 4 + Eqn (II) × 3
28x - 12y = 52
15x + 12y = 120
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43x $\,\,\,\,\,\,\,\,\,\,\,$ = 172
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x = 4 & y = 5
$x < y $ | | | |
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| 6 . | I.$ 2x^ 2 $ - 11x + 15 = 0
II.$ 21y^2$ - 23y + 6 = 0 | | A). x > y | | B). x < y | | C). x $\geq$ y | | D). x = y or no relation can be established between ‘x’ and ‘y’. | | Answer & Explanation | Answer : Option A | | | | Explanation : | | | I. $ 2x^ 2 $ - 6x - 5x + 15 = 0
or,2x(x - 3) - 5(x - 3) = 0
or, (2x - 5) (x - 3) = 0
x = 3 , $5\over 2$
II. $21y^ 2$ - 14y - 9y + 6 = 0
or,7y(3y - 2) - 3 (3y - 2) = 0
or,(7y - 3)(3y - 2) = 0
y = $3\over 7$ , $2\over 3$
$x > y$ | | | |
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| 7 . | I. $5x^ 2 $ - 16x + 11= 0
II. $5y^ 2 $ - 3y - 2 = 0 | | A). x>y | | B). x $\geq$ y | | C). x< td> | | D). x $\leq$ y | | Answer & Explanation | Answer : Option B | | | | Explanation : | | | I. $ 5x^ 2 $ - 5x - 11x + 11 = 0
or,5x(x - 1) - 11(x - 1) = 0
or,(x - 1) (5x - 11) = 0
x = 1 , $11\over 5$
II. $5y^ 2$ - 5y + 2y - 2 = 0
or, 5y (y - 1) + 2(y - 1) = 0
or, (5y + 2)(y - 1) = 0
y = 1 , - $2\over 5$
x $\geq$ y | | | |
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| 8 . | I. $x^ 2 $ + 11x + 28 = 0
II. $2y^ 2 $ + 13y + 20 = 0 | | A). x > y | | B). x < y | | C). $ x \geq y $ | | D). $ x \leq y$ | | Answer & Explanation | Answer : Option D | | | | Explanation : | | | I. $x^ 2$ + 4x + 7x + 28 = 0
or, x(x + 4) +7(x + 7) = 0
or, (x + 4) (x + 7) = 0
x = - 4, - 7
II. $2y^ 2$ + 8y + 5y + 20 = 0
or, 2y(y + 4) + 5(y + 4) = 0
or, (y + 4) (2y + 5) = 0
y = -4, -$5\over 2$
x $\leq$ y | | | |
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| 9 . | I. $6x^ 2$ + 29x + 35 = 0
II. $3y^ 2 $ + 19y + 30 = 0 | | A). $ x > y $ | | B). $ x < y $ | | C). $ x \geq y $ | | D). $ x \leq y$ | | Answer & Explanation | Answer : Option A | | | | Explanation : | | | I. $6x ^ 2$ + 15x + 14x + 35 = 0
or, 3x(2x + 5) + 7(2x + 5) = 0
or,(3x + 7) (2x + 5) = 0
x = - $7\over 3$ , - $5\over 2$
II. $3y^ 2$ + 9y + 10y + 30 = 0
or, 3y(y + 3) +10(y + 3) = 0
or,(3y + 10) (y + 3) = 0
y = 3, -$10\over 3$
$ x > y $ | | | |
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| 10 . | I. 2x + 5y = 6
II. 5x + 11y = 9 | | A). $ x > y $ | | B). $ x < y $ | | C). $ x \geq y $ | | D). $ x \leq y$ | | Answer & Explanation | Answer : Option B | | | | Explanation : | | | eq (I) × 5 - eq (II) × 2
10x + 25y = 30
10x ± 22y = 18
- $\,\,\,\,\,\,$ - $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$-
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$\,\,\,\,\,\,\,\,\,\,\,$ 3y = 12
y = 4 and x = - 7
$x < y$
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