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# Quant Quiz On Quadratic Equation Questions Day 5 Bag

• For which of the following equations the value of X is less than Y(X<Y)
I.5x + 2y = 31; 3x + 7y = 36
II.2x² -15x + 27 = 0; 5y² – 26y + 33 = 0
III.25/√x + 9/√x = 17√x; √y/3 + 5√y/6 = 3/√y
1.Only I
2.Only II
3.Only III
4.Both II and III
5.None

Explanation :
25/√x + 9/√x = 17√x
34 = 17x
x = 2
21√y/18 = 3/√y
y = 18/7 = 2.57
• For which of the following equations the value of X is less than or equal to Y (X≤ Y)
I.X2– 4X + 3= 0; Y2 – 8Y + 15 = 0
II. 3X2 – 19X + 28= 0; 4Y2 – 29Y + 45 = 0
III.x2 – (16)2 = (23)2 – 56; y1/3 – 55 + 376 = (18)2
1.Only I
2.Only II
3.Both I and III
4.Both II and III
5.All follow

Answer – 3.Both I and III
Explanation :
From I, (x-3)(x-1) = 0 X=1,3
(y-5)(y-3) = 0 Y = 5,3
From III, x2 – (16)2 = (23)2 – 56
x2 = 729
x = ± 27
y1/3 – 55 + 376 = (18)2
y = 33 = 27
• For which of the following equations the value of X is greater than or equal to Y (X ≥ Y)
I.X2 -3X – 4 = 0; 3Y2 – 10Y+8 = 0
II.√(x + 6) = √121 – √36; y2 + 112 = 473
III.5x2– 7x – 6 = 0; 5y2 + 23y + 12 = 0
1.Only I
2.Only II
3.Both I and III
4.Both II and III
5.None follow

Answer – 4.Both II and III
Explanation :
From II, √(x + 6) = √121 – √36
x + 6 = 25; x = 19
y2 + 112 = 473
y2 = 361; y = ± 19
From III, 5x2 – 7x – 6 = 0; x = -3/5, 2
5y2 + 23y + 12 = 0; y = -4, -3/5
• For which of the following equations the value of X is greater than Y(X>Y)
I.3X2+23X + 44 = 0; 3Y2 + 20Y +33 = 0
II.3X2+29 X +56 = 0; 2Y+ 15Y + 25 = 0
III.3X2– 16X + 21 = 0; 3Y2 – 28Y + 65 = 0
1.Only I
2.Only II
3.Both I and III
4.Both II and III
5.None follow

Explanation :
From I, X = -4, -11/3; Y = -3, -11/3
From II, X=-7, -8/3; Y = -5, -5/2
From III,X= 3,7/3; Y = 5,13/3
• For which of the following equations the value of X=Y or relationship cannot be established.
I.1/x + 1/(x-10) = 8/75; 132/y – 132/(y + 11) = 1
II.(3x – 2)/y = (3x + 6)/(y + 16); (x + 2)/(y + 4) = (x + 5)/(y + 10)
III. x² – 4x – 21 = 0; y² – 35y + 306 = 0
1.Only I
2.Only II
3.Both I and III
4.Both II and III
5.All follow

Explanation :
From I, x = 25, 3.75; y = -44, 33
• For which of the following equations the value of Y is greater than X(Y>X)
I.7x + 6y + 4z = 122; 4x + 5y + 3z = 88; 9x + 2y + z = 78
II.7x + 6y = 110; 4x + 3y = 59
III.2x + 5y = 23.5, 5x+ 2y = 22
1.Only I
2.Only II
3.Both I and II
4.Both II and III
5.All follow

Explanation :
From I, x = 6, y = 8
From II, x = 8; y = 9
From III, x = 3, y = 3.5
• For which of the following equations the value of X is less than or equal to Y (X≤ Y)
I.[48 / x4/7] – [12 / x4/7] = x10/7; y³ + 783 = 999
II.9/√x + 19/√x = √x; y– [(28)11/2 /√y] = 0
III.6X2 – 7X + 2 = 0; 12Y2 – 7Y+1 = 0
1.Only I
2.Only II
3.Both I and II
4.Both I and III
5.All follow

Explanation :
From I, x = ± 6; y = 6
• For which of the following equations the value of X is greater than or equal to Y (X ≥ Y)
I.3X2 +17X + 10=0; 10Y2 + 9Y+2 = 0
II.X2 +X – 56 = 0; Y2 – 17Y+72 = 0
III.X2– 12X + 35= 0; Y2 + Y – 30 = 0
1.Only I
2.Only II
3.Both I and II
4.Both I and III
5.None of these

Explanation :
Only III follows, From III, X = 7,5; Y = 5, -6
• For which of the following equations the value of X is greater than Y(X>Y)
I.2X+ 15X + 25= 0; 3Y2 + 29Y + 56 = 0
II.3X2 – 19X + 28= 0; 4Y2 – 29Y + 45 = 0
III.X2 – 13X + 42= 0; Y– 9Y + 20 = 0
1.Only I
2.Only II
3.Only III
4.Both II and III
5.None of these

Explanation :
From III, X=7,6; Y =5,4
• For which of the following equations the value of X=Y or relationship cannot be established.
I.x2 – 18x + 72= 0; 3y2 + 7y + 4 = 0
II.2x2 + x – 36 = 0; 2y2 – 13y + 20 = 0
III.4X2 – 19X + 12= 0; 3Y2 + 8Y + 4 = 0
1.Only I
2.Only II
3.Both I and II
4.Both II and III
5.None of these

Explanation :
From II, x = 4 , -9/2; y = 4, 5/2

• 5/√x + 7/√x = √x
4/√y + 6/√y = √y
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

A. X > Y
Explanation:

5/√x + 7/√x = √x
1/√x (5 + 7) = √x => x = 12
4/√x + 6/√x = √x => y = 10
• x² – 300 = 325
y – √144 = √169
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

D. X ≤ Y
Explanation:

x² – 300 = 325
x = ± 25
y – √144 = √169
y = 12 + 13 = 25
• x² – 115/2/√x = 0
y² – 135/2/√y = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

B. X < Y
Explanation:

x² – 115/2/√x = 0
x5/2 = 115/2
x = 11
y² – 135/2/√y = 0 => y = 13
• √(x + 20) = √256 – √121
y² + 576 = 697
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

E. X = Y or relation cannot be established
Explanation:

√(x + 20) = √256 – √121
x + 20 = 25 => x = 5
y² + 576 = 697
y = ± 11
• y² – x² = 32
y – x = 4
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

B. X < Y
Explanation:
y² – x² = 32
y – x = 4
(y + x) (y – x) = 32
(y + x) 4 = 32
y + x = 8
y – x = 4
y = 16 and x = 12
• x² + 12x + 32 = 0
y² + 19y + 90 = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

A. X > Y
Explanation:

x² + 12x + 32 = 0
x = – 4, – 8
y² + 19y + 90 = 0
y = -9, -10
• x² – 15x + 56 = 0
y² + 17y + 72 = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

A. X > Y
Explanation:

x² – 15x + 56 = 0
x = 8, 7
y² + 17y + 72 = 0
y = -8, -9
• x² – 23x + 132 = 0
y² + 13y + 42 = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

A. X > Y
Explanation:

x² – 23x + 132 = 0
x = 11, 12
y² + 13y + 42 = 0
y = -6, -7
• x² – 32x + 255 = 0
y² – 35y + 306 = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

D. X ≤ Y
Explanation:

x² – 32x + 255 = 0
x = 17, 15
y² – 35y + 306 = 0
y = 17, 18
• x² – 21x + 108 = 0
y² – 17y + 72 = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

C. X ≥ Y
Explanation:

x² – 21x + 108 = 0
x = 11, 9
y² – 17y + 72 = 0
y = 9, 8

Here I am giving you some examples of quadratic equations.

Ex 1: x2 + 3x – 10 = 0 and y2 – 7y + 12 = 0

By solving we get x = -5, 2 and y = 3, 4
Put these values as on a number line as:
-5(x)           2(x)     3(y)      4(y)
By this, it is clear that x is always less than y. So, we will say x < y.

Ex 2: x2 – 8x + 12 = 0 and y2 + 3y – 10 = 0

By solving we get x = 2, 6 and y = -5, 2
Put these values as on a number line as:
-5(y)             2(y,x)         6(x)
When x = 2, x = y=2 and x > y = -5, so x ≥ y.
When x = 6, x > y = 2 and x > y = -5.
By this, it is clear that either x is greater than y always or is equal to y at value 2. So, we will say x ≥ y.

Ex 3: x2 + 3x – 4 = 0 and y2 – y – 6 = 0

By solving we get x = -4, 1 and y = -2, 3
Put these values as on a number line as:
-4(x)             -2(y)         1(x)       3(y)
When there are values in between, one cannot find the relationship. So, we will say that no relationship exists between x and y.

Ex 4: x2 – 5x + 6 = 0 and y2 – 7y + 10 = 0

By solving we get x = 2, 3 and y = 2, 5
Put these values as on a number line as:
2(x,y)        3(x)            5(y)
When x = 2, x = y=2 and x < y = 5, so x ≤ y.
When x = 3, x > y = 2 and x < y = 5.
By these values, one cannot find the exact relationship. So, we will say that no relationship exists between x and y.