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1 . |
Directions (Q. Nos. 1-5) : In the following questions two equations numbered I and II are
given. You have to solve both the equations and—
Give answer
(1) if x > y
(2) if x $\geq$ y
(3) if x < y
(4) if x $\leq$ y
(5) if x = y or the relationship cannot be established
$Q.$
I. $\sqrt{1225x} + \sqrt{4900} = 0$
II. $(81)^{1\over 4} y + (343)^{1\over 3} = 0$
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| A). $ x > y $
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| B). $ x < y $
| | C). $ x \geq y $
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| D). $ x \leq y$
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Answer & Explanation
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Answer : Option
A
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| Explanation : |
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I. $\sqrt{1225x} + \sqrt{4900} = 0$
35x + 70 = 0
x = - $70\over 35$ = -2
II. 3y + 7 = 0
y = - $7\over 3$
$x > y$
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2 . |
I. $18\over x^2$ + $6\over x^2$ - $12\over x^2$ = $8\over x^2$
II. $y^
3 $
+ 9.68 + 5.64 = 16.95 |
| A). $ x > y $
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| B). $ x \geq y $
| | C). $ x \leq y$
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| D). x = y or no relation can be established between ‘x’ and ‘y’. |
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Answer & Explanation
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Answer : Option
D
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| Explanation : |
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I. $18 + 6 x - 12\over x^2$ = $8\over x^2$
or , x = $1\over 3$ = 0.333
II. $y^
2$ = 16.95 - 9.68 - 5.64 = 1.63
y = ±1.277 |
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3 . |
I. $(2)^5 + (11)^3\over 6$ = $x^3$
II. $4y^3 = - (589 \div 4 ) + 5 y^3$ |
| A). $ x > y $
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| B). $ x \geq y $
| | C). $ x < y $
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| D). $ x \leq y$
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Answer & Explanation
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Answer : Option
A
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| Explanation : |
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I. $x^3 = $ $32 + 1331\over 6$ = $1363\over6$
II. $5y^3 - 4y^3 =$ $589\over 4$
or $y^3 =$ $589\over 4$
$x > y$
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4 . |
I. $12x^
2$
+ llx + 12 = 10x
2
+22x
II. $13y^
2$
- 18y + 3 = 9y
2
- 10y |
| A). $ x > y $
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| B). $ x \geq y $
| | C). $ x < y $
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| D). $ x \leq y$
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Answer & Explanation
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Answer : Option
B
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| Explanation : |
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I. $2x
^2 $
- llx + 12 = 0
or, x = 4, $3\over 2$
II. $ 4y^
2$
- 8y + 3 = 0
y = $3\over 2$ , $1\over 2$
$x \geq y$ |
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5 . |
I. $(x^{7\over 5} \div 9)$ = $169 \div y{3\over 5}$
II. $y^{1\over 4} \times y^{1\over 4} \times 7$ = $273 \div y^{1\over 2}$ |
| A). $ x > y $
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| B). $ x \geq y $
| | C). $ x < y $
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| D). $ x \leq y$
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Answer & Explanation
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Answer : Option
D
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| Explanation : |
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$(x^{7\over 5} \div 9)$ = $169 \div x{3\over 5}$
$(x^{7\over 5} \times x{3\over 5}$ $ = 169 x 9
$(x^{7+3\over 5} = 1521
$x^2$ = 1521
x = ± 39
II. $y ^{1\over 4} $ + $y^{1\over 4} $ + $y^{1\over 2} $ = $273\over 7$
or, y $1\over 4$ + $1\over 4$ + $1\over 2$ = 39
y = 39
x $\leq$ y
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6 . |
Directions (Q. 6 - 10): 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. x = $\sqrt[
4]
{2401}$
II.$ 2y^
2$
- 9y - 56 = 0 |
| A). $ x > y $
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| B). $ x < y $
| | C). $ x \geq y $
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| D). x = y or no relation can be established between ‘x’ and ‘y’. |
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Answer & Explanation
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Answer : Option
D
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| Explanation : |
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I. x = $\sqrt[
4]
{2401}$
$\Rightarrow$ x = 7
II.$ 2y ^
2$
- 16y + 7y - 56 = 0
2y(y - 8) + 7(y - 8) = 0
(2y + 7) (y - 8) = 0
y = 8 , - $7\over 2$
Hence, no relation exists between x and y. |
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7 . |
I. $5x^
2$
+ 3x - 14 = 0
II.$ 2y^
2$
- 9y + 10 = 0 |
| A). $ x > y $
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| B). $ x < y $
| | C). $ x \geq y $
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| D). $ x \leq y$
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Answer & Explanation
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Answer : Option
B
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| Explanation : |
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I. $ 5x^
2$
+ 10x - 7x - 14 = 0
or, 5x(x + 2) - 7(x + 2) = 0
or, (x + 2) (5x - 7) = 0
x = - 2, $7\over 5$
II. $2y ^
2$
- 4y - 5y + 10 = 0
or, 2y(y - 2) - 5(y - 2) = 0
or, (2y - 5)(y - 2) = 0
or, y = 2, $5\over 2$
$x < y$ |
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8 . |
I. $8x ^
2$
+ 31x + 21 = 0
II. $5y ^2$
+ 11y - 36 = 0 |
| A). $ x > y $
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| B). $ x < y $
| | C). $ x \leq y$
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| D). x = y or no relation can be established between ‘x’ and ‘y’. |
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Answer & Explanation
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Answer : Option
D
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| Explanation : |
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I. $8x^
2$
+ 24x + 7x + 21 = 0
or, 8x(x + 3) + 7(x + 3) = 0
or, (x + 3) (8x + 7) = 0
x = - 3, - $7\over 8$
II. $5y^
2$
+ 20y - 9y - 36 = 0
or, 5y(y + 4) - 9(y + 4) = 0
or, (y + 4) (5y - 9) = 0
y = -4, $9\over 5$
Hence, no relation exists between x and y. |
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9 . |
I. 3x - y = 12
II. y = $\sqrt{
1089}$ |
| A). $ x > y $
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| B). $ x < y $
| | C). $ x \geq y $
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| D). $ x \leq y$
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Answer & Explanation
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Answer : Option
B
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| Explanation : |
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I. y = $\sqrt{
1089}$ $\Rightarrow$ y = 33
II. x = $12 + y\over 3$ = $12 + 33 \over 3 $ = $45\over 3$ = 15
$x < y$ |
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10 . |
I. $15x^
2$
+ 68x + 77 = 0
II. $3y^
2$
+ 29y + 68 = 0 |
| A). $ x > y $
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| B). $ x < y $
| | C). $ x \geq y $
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| D). $ x \leq y$
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Answer & Explanation
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Answer : Option
A
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| Explanation : |
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I. $15x^
2$
+ 68x + 77 = 0
or, $15x^
2$
+ 35x + 33x + 77 = 0
or, 5x(3x + 7) + 11(3x + 7) = 0
or, (5x + 11) (3x + 7) = 0
x = -$7\over 3$ , -$11\over 5$
II. $3y^
2$
+ 29y + 68 = 0
or, $ 3y^
2$
+ 12y + 17y + 68 = 0
or, 3y(y + 4) + 17(y + 4) = 0
or, (y + 4) (3y + 17) = 0
y = -4, -$17\over 3$
$x > y$ |
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