Mathematics Practice Questions for CDS. Mathematics Language Quiz for CDS. Practice Set for CDS Exam. Welcome to the www.pridedefence.co online Mathematics Practice section. If you are preparing for NDA/CDS and AFCAT & Others Defence Exams, you will come across a section on Elementary Mathematics Section. Here we are providing you with Mathematics quiz on “Mathematics Practice Set for CDS Exam 2019-20″ for your daily practice.
This “Mathematics Practice Questions for CDS Exam” is also important for other banking exams such as NDA/CDS and AFCAT & Others Defence Exams and other competitive exams.
Mathematics Practice Questions Set for NDA/CDS & Defence Exams |Set- 2|
Quiz-summary
0 of 10 questions completed
Questions:
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
Information
Help:- Share this quiz to your Friends & FB Groups.
All the Best !!!
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 10 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Average score |
|
Your score |
|
Categories
- Mathematics 0%
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- Answered
- Review
-
Question 1 of 10
1. Question
Category: Mathematics(x + 4) is a factor of which one of the following expressions?
Correct
C. x2 – 7x – 44
Putting x = -4 in every option:
1) x2 – 7x + 44 = 16 + 28 + 44 = 88
2) x2 + 7x – 44 = 16 – 28 – 44 = -56
3) x2 – 7x – 44 = 16 + 28 – 44 = 0
4) x2 + 7x + 44 = 16 – 28 + 44 = 32
Since x = -4 satisfies the 3rd equation;
∴ (x + 4) is a factor of (x2 – 7x – 44)
Incorrect
C. x2 – 7x – 44
Putting x = -4 in every option:
1) x2 – 7x + 44 = 16 + 28 + 44 = 88
2) x2 + 7x – 44 = 16 – 28 – 44 = -56
3) x2 – 7x – 44 = 16 + 28 – 44 = 0
4) x2 + 7x + 44 = 16 – 28 + 44 = 32
Since x = -4 satisfies the 3rd equation;
∴ (x + 4) is a factor of (x2 – 7x – 44)
-
Question 2 of 10
2. Question
Category: MathematicsIf α and β are the roots of the quadratic equation 2x2 + 6x + k = 0, where k < 0, then what is the maximum value of (α/β + β/α)?
Correct
B. -2
2x2 + 6x + k = 0
Since α and β are the roots of this equation;
∴ Sum of roots = α + β = -6/2 = -3
And Multiplication of roots = αβ = k/2
Now,
Since k < 1;
∴ Maximum value of this expression will be -2. (As the first term will be negative always)
Incorrect
B. -2
2x2 + 6x + k = 0
Since α and β are the roots of this equation;
∴ Sum of roots = α + β = -6/2 = -3
And Multiplication of roots = αβ = k/2
Now,
Since k < 1;
∴ Maximum value of this expression will be -2. (As the first term will be negative always)
-
Question 3 of 10
3. Question
Category: MathematicsConsider the following sentences:
1. If a = bc with HCF(b, c) = 1, then HCF(c, bd) = HCF(c, d).
2. If a = bc with HCF(b, c) = 1, then LCM(a, d) = LCM(c, bd).
Which of the above statements is/are correct?
Correct
A. 1 only
Since HCF (b, c) = 1;
Suppose b = 2, c = 3 and d = 8 then a = b × c = 2 × 3 = 6;
Statement 1:
HCF(c, bd) = HCF (3, 16) = 1
And HCF(c, d) = HCF (3, 8) = 1
Hence HCF(c, bd) = HCF(c, d) is true.
Statement 1:
LCM(a, d) = LCM(6, 8) = 24
And LCM(c, bd) = LCM(3, 16) = 48
Hence LCM(a, d) = LCM(c, bd) is not true.
∴ Statement 1 only is correct.
Incorrect
A. 1 only
Since HCF (b, c) = 1;
Suppose b = 2, c = 3 and d = 8 then a = b × c = 2 × 3 = 6;
Statement 1:
HCF(c, bd) = HCF (3, 16) = 1
And HCF(c, d) = HCF (3, 8) = 1
Hence HCF(c, bd) = HCF(c, d) is true.
Statement 1:
LCM(a, d) = LCM(6, 8) = 24
And LCM(c, bd) = LCM(3, 16) = 48
Hence LCM(a, d) = LCM(c, bd) is not true.
∴ Statement 1 only is correct.
-
Question 4 of 10
4. Question
Category: MathematicsWhat is the number of digits in 240? (Given that log102 = 0.301)
Correct
B. 13
Let N = 240
Taking ‘log’ both the sides;
⇒ log10N = 40(log102)
⇒ log10N = 40 × 0.301 = 12.04
Taking ‘anti-log’ both the sides;
⇒ N = (10)12.04
∴ Number of digits in 240 = 13
Incorrect
B. 13
Let N = 240
Taking ‘log’ both the sides;
⇒ log10N = 40(log102)
⇒ log10N = 40 × 0.301 = 12.04
Taking ‘anti-log’ both the sides;
⇒ N = (10)12.04
∴ Number of digits in 240 = 13
-
Question 5 of 10
5. Question
Category: MathematicsIf one root of (a2 – 5a + 3)x2 + (3a – 1)x + 2 = 0 is twice the other, then what is the value of ‘a’?
Correct
A. 2/3
(a2 – 5a + 3)x2 + (3a – 1)x + 2 = 0;
Let the roots of this equation are ‘y’ and ‘2y’;
∴ Sum of roots = 3y = [-(3a – 1)/(a2 – 5a + 3)] —- (1)
And Product of roots = 2y2 = 2/(a2 – 5a + 3) —- (2)
∴ (Product) / (Sum) = 2y2/3y = 2/-(3a – 1)
⇒ y = [-3/(3a – 1)]
From equation 1;
3y = [-(3a – 1)/(a2 – 5a + 3)]
Putting the value of y;
⇒ 3 × [-3/(3a – 1)] = [-(3a – 1)/(a2 – 5a + 3)]
⇒ 9a2 – 45a + 27 = 9a2 – 6a + 1
⇒ 45a – 6a = 26
⇒ 39a = 26
⇒ a = 2/3
Incorrect
A. 2/3
(a2 – 5a + 3)x2 + (3a – 1)x + 2 = 0;
Let the roots of this equation are ‘y’ and ‘2y’;
∴ Sum of roots = 3y = [-(3a – 1)/(a2 – 5a + 3)] —- (1)
And Product of roots = 2y2 = 2/(a2 – 5a + 3) —- (2)
∴ (Product) / (Sum) = 2y2/3y = 2/-(3a – 1)
⇒ y = [-3/(3a – 1)]
From equation 1;
3y = [-(3a – 1)/(a2 – 5a + 3)]
Putting the value of y;
⇒ 3 × [-3/(3a – 1)] = [-(3a – 1)/(a2 – 5a + 3)]
⇒ 9a2 – 45a + 27 = 9a2 – 6a + 1
⇒ 45a – 6a = 26
⇒ 39a = 26
⇒ a = 2/3
-
Question 6 of 10
6. Question
Category: MathematicsWhat is the remainder when the number (4444)4444 is divided by 9?
Correct
A. 4
We know that;
41/9 = Remainder will be 4;
42/9 = Remainder will be 7;
43/9 = Remainder will be 1;
Now (4444)4444 = [(4444)3]1481 × (4444)1
When dividing this from 9 it can be written as: [(4)3]1481 × (4)1]
Since Remainder when 43 is divided by 9 is 1 hence dividing [(4)3]1481 × (4)1] by 9, Remainder will be 4 only.
Incorrect
A. 4
We know that;
41/9 = Remainder will be 4;
42/9 = Remainder will be 7;
43/9 = Remainder will be 1;
Now (4444)4444 = [(4444)3]1481 × (4444)1
When dividing this from 9 it can be written as: [(4)3]1481 × (4)1]
Since Remainder when 43 is divided by 9 is 1 hence dividing [(4)3]1481 × (4)1] by 9, Remainder will be 4 only.
-
Question 7 of 10
7. Question
Category: MathematicsIf A = {x : x is a multiple of 7},
B = {x : x is a multiple of 5} and
C = {x : x is a multiple of 35}
Then which of the following is null set?
Correct
D. (A ∩ B) – C
A = {x : x is a multiple of 7} = {7, 14, 21, 28, 35………..}
B = {x : x is a multiple of 5} = {5, 10, 15, 20, 25………..}
C = {x : x is a multiple of 35} = {35, 70, 105, 140……..}
Checking all the options:
(A – B) ∪ C = {7, 14, 21, 28, 42 …..} ∪ {35, 70, 105, 140……..} = {7, 14, 21, 28, 35, 42, 49…}
Hence it is not a null set.
(A – B) – C = {7, 14, 21, 28, 42 …..} – {35, 70, 105, 140……..} = {7, 14, 21, 28, 42, 49….}
Hence it is not a null set.
(A ∩ B) ∩ C = {35, 70, 105, 140……} ∩ {35, 70, 105, 140……..} = {35, 70, 105, 140……..}
Hence it is not a null set.
(A ∩ B) – C = {35, 70, 105, 140……} – {35, 70, 105, 140……..} = ∅
Hence it is a null set.
Incorrect
D. (A ∩ B) – C
A = {x : x is a multiple of 7} = {7, 14, 21, 28, 35………..}
B = {x : x is a multiple of 5} = {5, 10, 15, 20, 25………..}
C = {x : x is a multiple of 35} = {35, 70, 105, 140……..}
Checking all the options:
(A – B) ∪ C = {7, 14, 21, 28, 42 …..} ∪ {35, 70, 105, 140……..} = {7, 14, 21, 28, 35, 42, 49…}
Hence it is not a null set.
(A – B) – C = {7, 14, 21, 28, 42 …..} – {35, 70, 105, 140……..} = {7, 14, 21, 28, 42, 49….}
Hence it is not a null set.
(A ∩ B) ∩ C = {35, 70, 105, 140……} ∩ {35, 70, 105, 140……..} = {35, 70, 105, 140……..}
Hence it is not a null set.
(A ∩ B) – C = {35, 70, 105, 140……} – {35, 70, 105, 140……..} = ∅
Hence it is a null set.
-
Question 8 of 10
8. Question
Category: MathematicsIf x = 2 + 22/3 + 21/3, then what is the value of x3 – 6x2 + 6x?
Correct
B. 2
Given
x = 2 + 22/3 + 21/3
⇒ (x – 2) = (22/3 + 21/3)
Taking cube both the sides;
⇒ x3 – 8 – 6x2 + 12x = 4 + 2 + 3 × 22/3 × 21/3 (22/3 + 21/3)
⇒ x3 – 14 – 6x2 + 12x = 6 (x – 2)
⇒ x3 – 14 – 6x2 + 12x – 6x + 12 = 0
⇒ x3 – 6x2 + 6x = 2
Incorrect
B. 2
Given
x = 2 + 22/3 + 21/3
⇒ (x – 2) = (22/3 + 21/3)
Taking cube both the sides;
⇒ x3 – 8 – 6x2 + 12x = 4 + 2 + 3 × 22/3 × 21/3 (22/3 + 21/3)
⇒ x3 – 14 – 6x2 + 12x = 6 (x – 2)
⇒ x3 – 14 – 6x2 + 12x – 6x + 12 = 0
⇒ x3 – 6x2 + 6x = 2
-
Question 9 of 10
9. Question
Category: MathematicsWhat is the solution of the equation x log10(10/3) + log103 = log10(2 + 3x) + x?
Correct
D. 0
x log10(10/3) + log103 = log10(2 + 3x) + x
⇒ x log1010 – x log103 + log103 = log10(2 + 3x) + x
⇒ x – x log103 + log103 = log10(2 + 3x) + x
⇒ log103-x + log103 = log10(2 + 3x)
⇒ log10(3-x) × 3 = log10(2 + 3x)
⇒ (3-x) × 3 = (2 + 3x)
⇒ 3/3x = (2 + 3x)
Suppose 3x = p;
⇒ 3/p = (2 + p)
⇒ p2 + 2p – 3 = 0
⇒ (p + 3) (p – 1) = 0
⇒ p = -3, 1
∴ 3x = 1
⇒ 3x = 30
∴ x = 0
Incorrect
D. 0
x log10(10/3) + log103 = log10(2 + 3x) + x
⇒ x log1010 – x log103 + log103 = log10(2 + 3x) + x
⇒ x – x log103 + log103 = log10(2 + 3x) + x
⇒ log103-x + log103 = log10(2 + 3x)
⇒ log10(3-x) × 3 = log10(2 + 3x)
⇒ (3-x) × 3 = (2 + 3x)
⇒ 3/3x = (2 + 3x)
Suppose 3x = p;
⇒ 3/p = (2 + p)
⇒ p2 + 2p – 3 = 0
⇒ (p + 3) (p – 1) = 0
⇒ p = -3, 1
∴ 3x = 1
⇒ 3x = 30
∴ x = 0
-
Question 10 of 10
10. Question
Category: MathematicsIf a3 = 335 + b3 and a = 5 + b, then what is the value of a + b (given that a > 0 and b > 0)?
Correct
B. 9
a3 = 335 + b3 and a = 5 + b;
∴ a3 – b3 = 335 and a – b = 5
We know that;
(a3 – b3) = (a – b) (a2 + b2 + ab)
⇒ (a3 – b3) = (a – b) ((a – b)2 + 3ab)
Putting the values;
⇒ 335 = 5 × (25 + 3ab)
⇒ 67 = (25 + 3ab)
⇒ 3ab = 42
⇒ ab = 14
And we know that (a – b) = 5
∴ a = 7 and b = 2
∴ a + b = 9
Incorrect
B. 9
a3 = 335 + b3 and a = 5 + b;
∴ a3 – b3 = 335 and a – b = 5
We know that;
(a3 – b3) = (a – b) (a2 + b2 + ab)
⇒ (a3 – b3) = (a – b) ((a – b)2 + 3ab)
Putting the values;
⇒ 335 = 5 × (25 + 3ab)
⇒ 67 = (25 + 3ab)
⇒ 3ab = 42
⇒ ab = 14
And we know that (a – b) = 5
∴ a = 7 and b = 2
∴ a + b = 9
Note:
- Click view Questions button to view Explanation
- Ask your doubt in comment section we’ ll clear your doubts in caring way.
- If you find any mistake, please tell us in the comment section.