Quadratic Equations Test

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Quadratic Equations Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following values of x satisfy the quadratic equation 2x²= 3x?

A
0, 2/3

B
0, 3/2

C
0, -2/3

D
-3/2, 0

E
1, 0

Solution

2x²= 3x
2x²- 3x = 0
x(2x - 3) = 0
Therefore, x = 0 or 2x - 3 = 0
i.e. x = 0 or x = 3/2
Question 2
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The roots of the quadratic equation 5x²- 14x - 3 = 0 are

A
1/5, 3

B
5, 1/3

C
- 1/5, 3

D
1/5, - 3

E
- 1/5, - 3

Solution

5x²- 14x – 3 = 0
5x²- 15x + x - 3 = 0
5x(x - 3) + 1(x - 3) = 0
(5x + 1) (x - 3) = 0
Therefore, x = - 1/5 OR x = 3
Question 3
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Which of the following is one of the roots of the quadratic equation 12(1 - y²) = 7y?

A
4/3

B
-4/3

C
-3/4

D
-3

E
4

Solution

12(1 - y²) = 7y

12y²+ 7y - 12 = 0

12y²+ 16y - 9y - 12 = 0

4y(3y + 4) - 3(3y + 4) = 0

(3y + 4)(4y - 3) = 0

Therefore, y = -4/3 or y = 3/4
Question 4
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What are the roots of the quadratic equation px² – qx + r = 0, where p, q and r are real numbers?

A

B

C

D

Solution
Question 5
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Which of the following equations has -84 as its discriminant?

A
2x² + 2x + 11 = 0

B
2x² + 2x - 11 = 0

C
-4x² + 2x + 11 = 0

D
2x² + 2x - 22 = 0

Solution

For a quadratic equation ax² + bx + c = 0, discriminant (D) = b² - 4ac.
Consider the quadratic equation 2x² + 2x + 11 = 0.
Here, a = 2, b = 2 and c = 11
So, D = b² - 4ac = (2)² - 4 2 11 = -84
Question 6
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Solve for x:

A

B

C

D
None of these

Solution

Let (x + 1)/x = y.
So, the equation becomes 1/y + y = 34/15 or 15y² - 34y + 15 = 0.
On solving the equation, we get y = 3/5 or y = 5/3.
This gives x as - 5/2 or 3/2. Hence, option (3) is correct.
Question 7
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Find the roots of .

A
1, -3

B
-1, 3

C
1, 3

D
-1, -3

E
None of these

Solution

x² + 4x - 2 = 8x - 5

x² - 4x + 3 = 0

x =

= 2 1

x = 1, 3
Question 8
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Which of the following is the solution set of the quadratic equation ?

A
4, 3

B
–3, 4

C
–4, 3

D
, 1

E
–3, –4

Solution

(2y – 5)(y + 1) = (y – 1)(3y – 7)
2y²– 3y – 5 = 3y²– 10y + 7
y²– 7y + 12 = 0
y²– 3y – 4y + 12 = 0
(y – 3)(y – 4) = 0
y = 3, 4
Question 9
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Two submersible pumps working together can fill a reservoir with water in 10 hours. The larger pump can fill the reservoir in 16 hours less if the smaller pump were to work alone. How will you represent the above situation mathematically? Assume that x is the number of hours required by the smaller pump to fill the reservoir.

A
x² + 36x + 160 = 0

B
x² – 36x – 160 = 0

C
x² – 36x + 160 = 0

D
x² + 36x – 160 = 0

Solution

Let x be the number of hours it takes to fill the reservoir using the smaller pump only.
Then, x – 16 is the time in hours to fill the reservoir using the larger pump only.
The smaller pump fills of the reservoir in 1 hour.
The larger pump fills reservoir in 1 hour.
In 10 hours working together, those will fill the complete reservoir.
A.T.Q.
∴
10x + 10x – 160 = x² – 16x
x² – 16x – 20x + 160 = 0
x² – 36x + 160 = 0
Question 10
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The base of a right-angled triangle is 2 cm less than the perpendicular. The length of the hypotenuse is 10 cm. Which of the following equations represents this situation? Assume that x represents the length of the perpendicular.

A
2x² + 4x – 96 = 0

B
2x² – 4x – 96 = 0

C
2x² – 104 = 0

D
2x² – 4x + 4 = 0

Solution

Let the perpendicular (p) be x cm. According to the question, base (b) = (x – 2) cm, hypotenuse (h) = 10 cm.
Also, according to Pythagoras theorem,
h² = b² + p²

So, (10)² = (x – 2)² + x²
100 = x² – 2 (2) x + 2² + x²
2x² – 4x + 4 – 100 = 0
2x² – 4x – 96 = 0
Thus, 2x² – 4x – 96 = 0 is the required equation.
Question 11
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The area of a rectangle is 168 cm². The length of the rectangle is 2 cm more than its breadth. Which of the following quadratic equations represents this situation? Assume that x represents the breadth of the rectangle.

A
x² + 2x = –168

B
x² + 2x = 168

C
2x² + 4x = 168

D
x² – 2x = 168

Solution

Given: Area (A) of the rectangle = 168 cm²
Let breadth (b) of the rectangle be x cm.
So, according to the question, length (l) of the rectangle = (x + 2) cm
Area of a rectangle = Length × Breadth
168 = x ×(x + 2)
or x² + 2x = 168 is the required equation.
Question 12
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Joy buys some packets of colours for Rs. 300. If the cost of each packet had been decreased by Rs. 5, then he would have bought 5 more packets. How much does each packet cost?

A
Rs. 30

B
Rs. 25

C
Rs. 20

D
Rs. 15

E
Rs. 10

Solution

Let the cost of each packet be Rs. x.
Number of packets bought for Rs. 300 = y

So, y = … (1)

If the cost of each packet is decreased by Rs. 5, then
Number of packets = y + 5 =

Comparing both the equations,

x² - 5x - 300 = 0
x = -15, 20
Negative value is rejected; so x = 20.
Question 13
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B's age is the square of the age of A. After 5 years, B will be 3 times as old as A. What is the difference between their ages?

A
5 years

B
10 years

C
15 years

D
20 years

E
25 years

Solution

Let the age of A be x.
So, B's age = x²
After 5 years, x² + 5 = 3(x + 5)
x² + 5 – 3x – 15 = 0
x² – 3x – 10 = 0 x = -2, 5
x is +ve, so x = 5
So, A's age = 5 years
B's age = 25 years
Difference = (25 – 5) years = 20 years
Thus, answer option 4 is correct.
Question 14
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Each side of a square measures 4 cm more than each side of another square. If the sum of their areas is 400 sq. cm, then what is the area of the larger square?

A
144 cm²

B
169 cm²

C
196 cm²

D
256 cm²

E
225 cm²

Solution

Let each side of the smaller square measure x cm.
Then, length of each side of the larger square = (x + 4) cm
According to problem:
(x + 4)² + x² = 400 {Area of square = Side × Side}
x² + 8x + 16 + x² = 400
Or 2x² + 8x - 384 = 0
x² + 4x - 192 = 0
Or (x + 16)(x - 12) = 0
x = 12
So, length of each side of the larger square = (12 + 4) cm = 16 cm
Area = (16)² cm²= 256 cm²
Question 15
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A rectangular park is to be designed having breadth 4 m less than its length. Its area is to be 5 m² more than that of an isosceles triangle having an altitude of 14 m and a base equal to the breadth of the rectangular park. Which of the following quadratic equations represents this situation? Assume that 'x' represents the breadth of the rectangular park.

A
x² - 3x + 5 = 0

B
x² - 3x - 5 = 0

C
x² + 3x + 5 = 0

D
x² + 3x - 5 = 0

Solution

Let b be 'x' m.
= (x + 4) m
Area of the rectangular park
= x(x + 4) m²
= (x² + 4x) m²
Base of the isosceles triangle = x m
Area of the isosceles triangle =
According to question,
x² + 4x = 7x + 5
x² - 3x - 5 = 0