Quadratic Equation Questions And Answers Quiz 19

By Shiva
May 06, 2017
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1 . Directions (Q. 1 - 5) : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
(1) if x > y
(2) if x ≥ y
(3) if x < y
(4) if x ≤ y
(5) if x = y or no relation can be established between x and y.

$Q. $
I. $ x^ 2 $ - 2x - 15 = 0
II. $y^ 2$ + 5y + 6 = 0

A.    $ x > y$
B.    x ≥ y
C.    $ x < y$
D.    $ x ≤ y$
2 . I. $ x^ 2 $ - x - 12 = 0
II. $y^ 2$ -3y + 2 = 0

A.    $ x > y$
B.    $ x \geq y $
C.    $ x < y$
D.    x = y or no relation can be established between ‘x’ and ‘y’.
3 . I.$ x - \sqrt{169} = 0$
II.$ y^2 - \sqrt{169} = 0$

A.    $ x > y$
B.    x ≥ y
C.    $ x < y$
D.    $ x < y$
4 . I. $x^2 - 32 = 112$
II. y - $\sqrt{256}$ = 0

A.    $ x > y$
B.    $ x \geq y $
C.    $ x < y$
D.    $ x \leq y$
5 . I. $ x^ 2$ - 25 = 0
II. $y^ 2$ - 9y + 20 = 0

A.    $ x > y$
B.    $ x \geq y $
C.    $ x < y$
D.    x = y or no relation can be established between ‘x’ and ‘y’.
6 . Directions (Q. 6 - 10): In the following questions, three equations numbered I, II and III are given. You have to solve all the equations either together or separately, or two together and one separately or by any other method and give answer
(1) if x = y > z
(2) if x < y = z
(3) if x < y > z
(4) if x = y = z or if none of the above relationship can be established.
(5) if x ≤ y < z

$Q.$
I. 3x + 5y = 69
II. 9x + 4y = 108
III. x + z = 12

A.    x = y > z
B.    x < y = z
C.    x < y > z
D.    x = y = z or if none of the above relationship can be established.
7 . I . $y = \sqrt{9^{3 \times {1\over3}} \times {9^{3 \times {1\over3}}}} $ = $\sqrt{9 \times 9}$ = 9
II. 2x + 5z = 54
III. 6x + 4z = 74

A.    x = y > z
B.    x < y = z
C.    x < y > z
D.    x ≤ y < z
8 . I. 2x + 3y + 4z = 66
II. 2x + y + 3z = 42
III. 3x + 2y + 4z = 63

A.    x = y > z
B.    x < y = z
C.    x < y > z
D.    x ≤ y < z
9 . I. $(x + z)^ 3$ = 1728 = $ 12^ 3$
II. 2x + 3y = 35
III. x - z = 2

A.    x = y > z
B.    x < y = z
C.    x < y > z
D.    x ≤ y < z
10 . I. 4x + 5y = 37
II. x + z = 8
III. 7x + 3y = 36

A.    x = y > z
B.    x < y = z
C.    x < y > z
D.    x ≤ y < z
Answers & Solutions
1 .    
Answer : Option B
Explanation :
I. $ x^ 2 $ - 2x - 15 = 0
or,$x^ 2 $ - 5x + 3x - 15 = 0
or, x(x - 5) + 3(x - 5) = 0
or,(x - 5) (x + 3) = 0
x = 5, -3

II. $y^ 2$ + 5y + 6 = 0
or, $y^ 2$ + 3y + 2y + 6 = 0
or, y(y + 3) + 2(y + 3) = 0
or,(y + 3)(y + 2) = 0
y = -3, -2

x ≥ y
2 .    
Answer : Option D
Explanation :
I. $ x^ 2 $ - x - 12 = 0
or, $x^ 2$ - 4x + 3x - 12 = 0
or, x(x - 4) + 3(x - 4) = 0
or, (x - 4) (x + 3) = 0
x = 4, -3

II. $y^ 2$ -3y + 2 = 0
or, $y^ 2$ - 2y - y + 2 = 0
or, y(y - 2) - 1 (y - 2) = 0
or, (y - 2)(y - 1) = 0

y = 2, l

Hence, no relation can be established
3 .    
Answer : Option B
Explanation :
I.$ x - \sqrt{169} = 0$
or, x = $\sqrt{169}$
x = 13

II.$ y^2 - \sqrt{169} = 0$
or, $y^2$ = $\sqrt{169}$
y = ± 13

Hence, x ≥ y
4 .    
Answer : Option C
Explanation :
I. $x^2 - 32 = 112$
or, $x^2 = 112 + 32 = 144$
or, x = $\sqrt{144}$
x = ± 12

II. y - $\sqrt{256}$ = 0
or, y = $\sqrt{256}$
y = 16

Hence, x < y
5 .    
Answer : Option D
Explanation :
I. $ x^ 2$ - 25 = 0
or, $ x^ 2$ = 25
or, $ x = \sqrt{25} $
x = ± 5

II. $y^ 2$ - 9y + 20 = 0
or, $ y^ 2$ - 5y - 4y + 20 = 0
or, y(y - 5) - 4(y - 5) = 0
or, (y - 5) (y - 4) = 0
y = 5, 4

Hence, no relation can be established.
6 .    
Answer : Option C
Explanation :
3x + 5y = 69 ... (i)
9x + 4y = 108 ... (ii)
x + z = 12 ... (iii)
Now, from (i) and (ii), we have
3x + 5y = 69 ... (i) × 4
9x + 4y = 108 ... (ii) × 5
---------------------
12x + 20y = 276
45x + 20y = 540
----------------------
- 33x = - 264

On subtracting, we get
or, 33x = 264
x = $264\over 33$ = 8
Putting the value of x in equation (i), we get
3 × 8 + 5y = 69
or, 5y = 69 - 24 = 45
y = $45\over 5$ = 9
Again, putting the value of x in equation (iii), we get
x + z = 12
or, z = 12 - 8 = 4

Hence, x < y > z
7 .    
Answer : Option C
Explanation :
I . $y = \sqrt{9^{3 \times {1\over3}} \times {9^{3 \times {1\over3}}}} $ = $\sqrt{9 \times 9}$ = 9 ...... (i)
II. 2x + 5z = 54 .. (ii)
III. 6x + 4z = 74
or, 3x + 2z = 37 ... (iii)
From equation (ii) × 2 - (iii) × 5, we get
4x + 10z = 108
15x + 10z = 185
-$\,\,\,\,\,\,\,\,$ - $\,\,\,\,\,\,\,\,\,\,$ -
-------------------- - 11x = - 77
or, 11x = 77
x = 7

Putting the value of x in equation (ii), we get
2 × 7 + 5z = 54
or, 5z = 40
z = 8

Hence, x < y > z
8 .    
Answer : Option B
Explanation :
I. 2x + 3y + 4z = 66 ... (i)
II. 2x + y + 3z = 42 ... (ii)
III.3x + 2y + 4z = 63 ... (iii)
From (iii) and (i),
x - y = - 3 ...(iv)

From equation (i) × 3 - equation (ii) × 4
6x + 9y + 12z = 198
8x + 4y + 12z = 168
- $\,\,\,$ -$\,\,\,\,\,\,\,$ -$\,\,\,\,\,\,\,\,\,\,\,\,$ -
----------------------
2x + 5y = 30 ... (v)

Solving equation (iv) and (v), we get
x = 5, y = 8

Now, on putting the value of x and y in equation (i),
10 + 24 + 4z = 66
or, 4z = 32
z = $32\over 4$ = 8

Hence, x < y = z
9 .    
Answer : Option A
Explanation :
I. $(x + z)^ 3$ = 1728 = $ 12^ 3$
or, x + z = 12 ...(i)
II. 2x + 3y = 35 ... (ii)
III. x - z = 2 ...(iii)
Now, equation (i) and (ii),
x = 7, z = 5

Putting the value x in question (ii) we have,
2 × 7 + 3y = 35
or, 3y = 35 - 14 = 21
or, y = $21\over 3$ = 7

Hence, x = y > z
10 .    
Answer : Option B
Explanation :
4x + 5y = 37 ... (i)
x + z = 8 ... (ii)
7x + 3y = 36 ... (iii)
From equation (i) and (iii),
4x + 5y = 37 ... (i) × 3
7x + 3y = 36 ... (ii) × 5
--------------------
or, 12x + 15y = 111
35x + 15y = 180
- $\,\,\,\,\,$ -$\,\,\,\,\,\,\,\,\,\,\,\,\,$ -
---------------------
-23x = - 69

x = 3
Putting the value of x in equation (i)
4 × 3 + 5y = 37
or, y = $25\over 5$ = 5
Now, putting the value of x in equation (ii)
z = 5.

Hence, x < y = z
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