| Answers & Solutions | | 1 . | Answer : Option C | Explanation : | eqn (I) × 7 eqn (II) × 11 $\,\,\,$77x + 35y = 819 - 77x ± 143y = 1683 ------------------------------ - 108y = - 864 y = 8, x = 7
ie x < y | | | 2 . | Answer : Option C | Explanation : | I. $6x^ 2$ + 21x + 30x + 105 = 0 or, 3x(2x + 7) + 15(2x + 7) = 0 or, (3x + 15) (2x + 7) = 0 x = -5 , -$7\over 2$
II. $ 2y^ 2$ + 12y + 13y + 78 = 0 or, 2y(y + 6) + 13(y + 6) = 0 or, (2y + 13) (y + 6) = 0 y = -$13\over 2$ , -6
$x < y$ | | | 3 . | Answer : Option C | Explanation : | eqn (I) × 4 eqn (II) × 7
24x + 28y = 208 98x ± 28y = 245 - ---------------------- - 74x = - 37 x = $1\over 2$, y = 7
$x < y$ | | | 4 . | Answer : Option B | Explanation : | I. $x^ 2$ + 5x + 6x + 30 = 0 or, x(x + 5) + 6(x + 5) = 0 or, (x + 5) (x + 6) = 0 x = - 5, - 6
II. $y^ 2$ + 12y + 36 = 0 or, $(y + 6)^ 2$ = 0 or, y + 6 = 0 y = - 6
ie x $\geq$ y | | | 5 . | Answer : Option D | Explanation : | I. $2x^ 2$ + 2x - x - 1 = 0 or, 2x(x + 1) - 1(x + 1) = 0 or, (2x - 1) (x + 1) = 0 x = $1\over 2$ , -1
II. $2y^ 2$ - 2y - y + 1 = 0 or, 2y(y - 1) - 1(y - 1) = 0 or, (2y - 1)(y - 1) = 0 y = $1\over 2$, 1
i.e., $x \leq y$ | | | 6 . | Answer : Option A | Explanation : | 7x + 6y + 4z = 122 ... (i) 4x + 5y + 3z = 88 ... (ii) 9x + 2y + z = 78 ... (iii) From (i) and (ii) 5x - 2y = 14... (iv) From (ii) and (iii) 23x + y = 146 ... (v) From (iv) and (v), x = 6, y = 8 Putting the value of x and y in eqn (i), we get z = 8
:. x < y = z | | | 7 . | Answer : Option C | Explanation : | 7x + 6y = 110 ... (i) 4x + 3y = 59 ... (ii) x + z = 15 ... (iii) From eqn (i) and (ii), x = 8, y = 9 Put the value of x in eqn (iii). Then, z = 7
x < y > z | | | 8 . | Answer : Option D | Explanation : | x = $\sqrt{(6^2)^{1\over 2} \times (6^4)^{1\over 4}}$ $\sqrt{6\times6}$ = 6 ..(i) 2y + 3z = 33 ... (ii) 6y + 5z = 71 ... (iii) From eqn (ii) and (iii), y = 6 and z = 7
x = y < z | | | 9 . | Answer : Option D | Explanation : | 8x + 7y = 135 ... (i) 5x + 6y = 99 ... (ii) 9y + 8z = 121 ... (iii) From eqn (i) and (ii), x = 9, and y = 9 Putting the value of y in eqn (iii), z = 5
:. x = y > z | | | 10 . | Answer : Option D | Explanation : | $(x + y)^ 3$ = 1331 or, x + y = 11 ... (i) $(x + y)^ 2$ = 121 $(x - y)^ 2$ + 4xy = 121 x - y = 3... (ii) [value of xy from eqn (iii)] From eqn (i) and (ii), x = 7, y = 4 Put the value x and y in the eqn x - y + z = 0 7 - y + z = 0 3 + z = 0 z = -3
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