1. The roots of quadratic equation 5x2 – 4x + 5 = 0 are
ANSWER=
C. Not real
2. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
ANSWER=
B. 8
Explanation:
Let the number be x
Then according question
3. Rohini had scored 10 more marks in her mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test?
ANSWER=
C. 15
Let her actual marks be x
Therefore, 9 (x + 10) = x2
⇒x2 – 9x – 90 = 0
⇒x2 – 15x + 6x – 90 = 0
⇒x(x – 15) + 6 (x – 15) = 0
⇒(x + 6) (x – 15) = 0
Therefore x = – 6 or x =15
Since x is the marks obtained, x ≠ – 6. Therefore, x = 15.
4. A takes 6 days less than B to finish a piece of work. If both A and B together can finish the work in 4 days, find the time taken by B to finish the work.
ANSWER=
A.12 days
5. In the quadratic equation 5x2 – 4x + 5 = 0, which one of them has the following roots?
ANSWER=
D. Not real
Explanation: Calculate b2 – 4ac in order to find out the nature.
So, b2 – 4ac will be
= 42 – 4 x 5 x 5
= 16 – 100
= -84 < 0
6. When a natural number is multiplied by 12, it becomes 160 times its reciprocal. Calculate the number.
ANSWER=
A. 8
Explanation: Assume the number as x
Then according to the given question,
x + 12 = 160/x
⇒ x2 + 12x – 160 = 0
⇒ x2 + 20x – 8x – 160 = 0
⇒ (x + 20) (x – 8) = 0
⇒ x = -20, 8
We only consider positive values because the number is natural.
7. 300 is the result of two consecutive integral multiples of 5. Find out the numbers.
ANSWER=
B. 15, 20
Explanation: Let 5n and 5(n + 1) be the consecutive integral multiples, with n being a positive integer.
In answer to the question:
5n × 5(n + 1) = 300
⇒ n2 + n – 12 = 0
⇒ (n – 3) (n + 4) = 0
⇒ n = 3 and n = – 4
n = – 4 will be eliminated because n is a positive natural number.
As a result, the numbers 15 and 20 are used.
8. Rohini could have gotten 10 additional marks out of a possible 30 on her math test, which would have been the square of her actual score 9 times. How many marks did she receive on the exam?
ANSWER=
C. 15
Explanation: Assume her actual marks will be x
As 9 (x + 10) = x2
⇒x2 – 9x – 90 = 0
⇒x2 – 15x + 6x – 90 = 0
⇒x(x – 15) + 6 (x – 15) = 0
⇒(x + 6) (x – 15) = 0
So, x = – 6 or x =15
As x is assumed to be as the marks obtained, x ≠ – 6. Hence, x will be = 15.
9. A right triangle's altitude is 7 cm less than its base. The other two sides of the triangle are equal to: If the hypotenuse is 13 cm, the other two sides are equal to:
ANSWER=
A. Base=12cm and Altitude=5cm
Explanation: Let base will be as x cm.
So, Altitude = (x – 7) cm
As we know that in a right triangle,
Base2 + Altitude2 = Hypotenuse2 (Pythagoras theorem)
∴ x2 + (x – 7)2 = 132
On solving the above equation, we will get-
⇒ x = 12 or x = – 5
As the side of the triangle cannot be negative.
So, base = 12cm and altitude = 12 - 7 = 5cm
10. If one of the roots of the equation 4x2-2x+k-4=0 is reciprocal to the other, then k will have the value as:
ANSWER=
D. 8
Explanation: α x 1/α = (k-4)/4
k-4 = 4
k = 8
11. For a quadratic equation, the maximum number of roots is equal to
ANSWER=
C. 2
Explanation: Because the degree of a quadratic equation is 2, the maximum number of roots for a quadratic equation is equal to 2.
12. In the following quadratic equation, 2x2 – √5x + 1 = 0 has
ANSWER=
B. No real roots
Explanation: Given equation 2x2 – √5x + 1 = 0
When compared to a quadratic equation in its usual form,
a = 2, b = -√5, c = 1
Now, b2 – 4ac = (-√5)2 – 4B.A.
= 5 – 8
= -3 < 0
Hence, the given equation will be having no real roots.
13. Which of the following is not a quadratic equation
ANSWER=
B. x² + x3 + 2 = 0
Explaination:Reason: Since it has degree 3.
14. The quadratic equation has degree
ANSWER=
C. 2
Explaination:Reason: A quadratic equation has degree 2.
15. The cubic equation has degree
ANSWER=
C. 3
Explaination:Reason: A cubic equation has degree 3.
16. A bi-quadratic equation has degree
ANSWER=
D. 4
Explaination:Reason: A bi-quadratic equation has degree 4.
17. The polynomial equation x (x + 1) + 8 = (x + 2) {x – 2) is
ANSWER=
A. linear equation
Explaination:Reason: We have x(x + 1) + 8 = (x + 2) (x – 2)
⇒ x² + x + 8 = x² – 4
⇒ x² + x + 8- x² + 4 = 0
⇒ x + 12 = 0, which is a linear equation.
18. The equation (x – 2)² + 1 = 2x – 3 is a
ANSWER=
B. quadratic equation
Explaination:Reason: We have (x – 2)² + 1 = 2x – 3
⇒ x² + 4 – 2 × x × 2 + 1 = 2x – 3
⇒ x² – 4x + 5 – 2x + 3 = 0
∴ x² – 6x + 8 = 0, which is a quadratic equation.
19. The quadratic equation whose roots are 1 and
ANSWER=
B. 2x² – x – 1 = 0
20. The quadratic equation whose one rational root is 3 + √2 is
ANSWER=
D. x² – 6x + 7 = 0
Explaination:Reason: ∵ one root is 3 + √2
∴ other root is 3 – √2
∴ Sum of roots = 3 + √2 + 3 – √2 = 6
Product of roots = (3 + √2)(3 – √2) = C.² – (√2)² = 9 – 2 = 7
∴ Required quadratic equation is x² – 6x + 7 = 0
21. The sum of the roots of the quadratic equation 3×2 – 9x + 5 = 0 is
ANSWER=
C. -3
Explaination:Reason: Here a = 3, b = -9, c = 5
∴ Sum of the roots =−ba=−(−9)3=3
22. Mohan and Sohan solve an equation. In solving Mohan commits a mistake in constant term and finds the roots 8 and 2. Sohan commits a mistake in the coefficient of x. The correct roots are
ANSWER=
A. 9,1
Explaination:Reason: Correct sum = 8 + 2 = 10 from Mohan
Correct product = -9 x -1 = 9 from Sohan
∴ x² – (10)x + 9 = 0
⇒ x² – 10x + 9 = 0
⇒ x² – 9x – x + 9
⇒ x(x – 9) – 1(x – 9) = 0
⇒ (x-9) (x-l) = 0 .
⇒ Correct roots are 9 and 1.
23. If the roots of px2 + qx + 2 = 0 are reciprocal of each other, then
ANSWER=
D. p = 2
Explaination:Reason: here α = 1β
∴ αβ = 1
⇒ 2p = 1
∴ p = 2
24. If one root of the quadratic equation 2x² + kx – 6 = 0 is 2, the value of k is
ANSWER=
B. -1
Explaination:Reason: Scice x = 2 is a root of the equation 2x² + kx -6 = 0
∴ 2B.² +kB. – 6 = 0
⇒ 8 + 2k – 6 = 0
⇒ 2k = -2
∴ k = -1
25. The roots of the equation 7x² + x – 1 = 0 are
ANSWER=
A. real and distinct
Explaination:Reason: Here a = 2, b = 1, c = -1
∴ D = b² – 4ac = A.² – 4 × 2 × (-1) = 1 + 8 = 9 > 0
∴ Roots of the given equation are real and distinct.
26. The equation 12x² + 4kx + 3 = 0 has real and equal roots, if
ANSWER=
A. k = ±3
Explaination:Reason: Here a = 12, b = 4k, c = 3
Since the given equation has real and equal roots
∴ b² – 4ac = 0
⇒ (4k)² – 4 × 12 × 3 = 0
⇒ 16k² – 144 = 0
⇒ k² = 9
⇒ k = ±3
27. If -5 is a root of the quadratic equation 2x² + px – 15 = 0, then
ANSWER=
C. p = 7
Explaination:Reason: Since – 5 is a root of the equation 2x² + px -15 = 0
∴ 2(-5)² + p (-5) – 15 = 0
⇒ 50 – 5p -15 = 0
⇒ 5p = 35
⇒ p = 7
28. If the roots of the equations ax² + 2bx + c = 0 and bx² – 2√ac x + b = 0 are simultaneously real, then
ANSWER=
B. b2 = ac
Explaination:Reason: Given equations have real roots, then
D1 ≥ 0 and D2 ≥ 0
(2b)² – 4ac > 0 and (-2√ac)² – 4B.b ≥ 0
4b² – 4ac ≥ 0 and 4ac – 4b2 > 0
b² ≥ ac and ac ≥ b²
⇒ b² = ac
29. The roots of the equation (b – c) x² + (c – a) x + (a – b) = 0 are equal, then
ANSWER=
D. 2b = a + c
Explaination:Reason: Since roots are equal
∴ D = 0 => b² – 4ac = 0
⇒ (c – a)² -4(b – c) (a – b) = 0
⇒ c² – b² – 2ac -4(ab -b² + bc) = 0 =>c + a-2b = 0 => c + a = 2b
⇒ c² + a² – 2ca – 4ab + 4b² + 4ac – 4bc = 0
⇒ c² + a² + 4b² + 2ca – 4ab – 4bc = 0
⇒ (c + a – 2b)² = 0
⇒ c + a – 2b = 0
⇒ c + a = 2b
30. A chess board contains 64 equal squares and the area of each square is 6.25 cm². A border round the board is 2 cm wide. The length of the side of the chess board is
ANSWER=
C. 24 cm
31. One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Their present ages are
ANSWER=
A. 7 years, 49 years
32. The sum of the squares of two consecutive natural numbers is 313. The numbers are
ANSWER=
A. 12, 13
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