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 201920″ 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 1
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Question 1 of 10
1. Question
Category: Mathematics(x + 4) is a factor of which one of the following expressions?
Correct
C. x^{2} – 7x – 44
Putting x = 4 in every option:
1) x^{2} – 7x + 44 = 16 + 28 + 44 = 88
2) x^{2} + 7x – 44 = 16 – 28 – 44 = 56
3) x^{2} – 7x – 44 = 16 + 28 – 44 = 0
4) x^{2} + 7x + 44 = 16 – 28 + 44 = 32
Since x = 4 satisfies the 3^{rd} equation;
∴ (x + 4) is a factor of (x^{2} – 7x – 44)
Incorrect
C. x^{2} – 7x – 44
Putting x = 4 in every option:
1) x^{2} – 7x + 44 = 16 + 28 + 44 = 88
2) x^{2} + 7x – 44 = 16 – 28 – 44 = 56
3) x^{2} – 7x – 44 = 16 + 28 – 44 = 0
4) x^{2} + 7x + 44 = 16 – 28 + 44 = 32
Since x = 4 satisfies the 3^{rd} equation;
∴ (x + 4) is a factor of (x^{2} – 7x – 44)

Question 2 of 10
2. Question
Category: MathematicsIf α and β are the roots of the quadratic equation 2x^{2} + 6x + k = 0, where k < 0, then what is the maximum value of (α/β + β/α)?
Correct
B. 2
2x^{2} + 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
2x^{2} + 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 2^{40}? (Given that log_{10}2 = 0.301)
Correct
B. 13
Let N = 2^{40}
Taking ‘log’ both the sides;
⇒ log_{10}N = 40(log_{10}2)
⇒ log_{10}N = 40 × 0.301 = 12.04
Taking ‘antilog’ both the sides;
⇒ N = (10)^{12.04}
∴ Number of digits in 2^{40 }= 13
Incorrect
B. 13
Let N = 2^{40}
Taking ‘log’ both the sides;
⇒ log_{10}N = 40(log_{10}2)
⇒ log_{10}N = 40 × 0.301 = 12.04
Taking ‘antilog’ both the sides;
⇒ N = (10)^{12.04}
∴ Number of digits in 2^{40 }= 13

Question 5 of 10
5. Question
Category: MathematicsIf one root of (a^{2} – 5a + 3)x^{2} + (3a – 1)x + 2 = 0 is twice the other, then what is the value of ‘a’?
Correct
A. 2/3
(a^{2} – 5a + 3)x^{2} + (3a – 1)x + 2 = 0;
Let the roots of this equation are ‘y’ and ‘2y’;
∴ Sum of roots = 3y = [(3a – 1)/(a^{2} – 5a + 3)] — (1)
And Product of roots = 2y^{2} = 2/(a^{2} – 5a + 3) — (2)
∴ (Product) / (Sum) = 2y^{2}/3y = 2/(3a – 1)
⇒ y = [3/(3a – 1)]
From equation 1;
3y = [(3a – 1)/(a^{2} – 5a + 3)]
Putting the value of y;
⇒ 3 × [3/(3a – 1)] = [(3a – 1)/(a^{2} – 5a + 3)]
⇒ 9a^{2} – 45a + 27 = 9a^{2} – 6a + 1
⇒ 45a – 6a = 26
⇒ 39a = 26
⇒ a = 2/3
Incorrect
A. 2/3
(a^{2} – 5a + 3)x^{2} + (3a – 1)x + 2 = 0;
Let the roots of this equation are ‘y’ and ‘2y’;
∴ Sum of roots = 3y = [(3a – 1)/(a^{2} – 5a + 3)] — (1)
And Product of roots = 2y^{2} = 2/(a^{2} – 5a + 3) — (2)
∴ (Product) / (Sum) = 2y^{2}/3y = 2/(3a – 1)
⇒ y = [3/(3a – 1)]
From equation 1;
3y = [(3a – 1)/(a^{2} – 5a + 3)]
Putting the value of y;
⇒ 3 × [3/(3a – 1)] = [(3a – 1)/(a^{2} – 5a + 3)]
⇒ 9a^{2} – 45a + 27 = 9a^{2} – 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;
4^{1}/9 = Remainder will be 4;
4^{2}/9 = Remainder will be 7;
4^{3}/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 4^{3} 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;
4^{1}/9 = Remainder will be 4;
4^{2}/9 = Remainder will be 7;
4^{3}/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 4^{3} 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 + 2^{2/3 }+ 2^{1/3}, then what is the value of x^{3} – 6x^{2} + 6x?
Correct
B. 2
Given
x = 2 + 2^{2/3 }+ 2^{1/3}
⇒ (x – 2) = (2^{2/3 }+ 2^{1/3})
Taking cube both the sides;
⇒ x^{3} – 8 – 6x^{2} + 12x = 4 + 2 + 3 × 2^{2/3 }× 2^{1/3} (2^{2/3 }+ 2^{1/3})
⇒ x^{3} – 14 – 6x^{2} + 12x = 6 (x – 2)
⇒ x^{3} – 14 – 6x^{2} + 12x – 6x + 12 = 0
⇒ x^{3} – 6x^{2} + 6x = 2
Incorrect
B. 2
Given
x = 2 + 2^{2/3 }+ 2^{1/3}
⇒ (x – 2) = (2^{2/3 }+ 2^{1/3})
Taking cube both the sides;
⇒ x^{3} – 8 – 6x^{2} + 12x = 4 + 2 + 3 × 2^{2/3 }× 2^{1/3} (2^{2/3 }+ 2^{1/3})
⇒ x^{3} – 14 – 6x^{2} + 12x = 6 (x – 2)
⇒ x^{3} – 14 – 6x^{2} + 12x – 6x + 12 = 0
⇒ x^{3} – 6x^{2} + 6x = 2

Question 9 of 10
9. Question
Category: MathematicsWhat is the solution of the equation x log_{10}(10/3) + log_{10}3 = log_{10}(2 + 3^{x}) + x?
Correct
D. 0
x log_{10}(10/3) + log_{10}3 = log_{10}(2 + 3^{x}) + x
⇒ x log_{10}10 – x log_{10}3 + log_{10}3 = log_{10}(2 + 3^{x}) + x
⇒ x – x log_{10}3 + log_{10}3 = log_{10}(2 + 3^{x}) + x
⇒ log_{10}3^{x} + log_{10}3 = log_{10}(2 + 3^{x})
⇒ log_{10}(3^{x}) × 3 = log_{10}(2 + 3^{x})
⇒ (3^{x}) × 3 = (2 + 3^{x})
⇒ 3/3^{x }= (2 + 3^{x})
Suppose 3^{x} = p;
⇒ 3/p = (2 + p)
⇒ p^{2} + 2p – 3 = 0
⇒ (p + 3) (p – 1) = 0
⇒ p = 3, 1
∴ 3^{x} = 1
⇒ 3^{x} = 3^{0}
∴ x = 0
Incorrect
D. 0
x log_{10}(10/3) + log_{10}3 = log_{10}(2 + 3^{x}) + x
⇒ x log_{10}10 – x log_{10}3 + log_{10}3 = log_{10}(2 + 3^{x}) + x
⇒ x – x log_{10}3 + log_{10}3 = log_{10}(2 + 3^{x}) + x
⇒ log_{10}3^{x} + log_{10}3 = log_{10}(2 + 3^{x})
⇒ log_{10}(3^{x}) × 3 = log_{10}(2 + 3^{x})
⇒ (3^{x}) × 3 = (2 + 3^{x})
⇒ 3/3^{x }= (2 + 3^{x})
Suppose 3^{x} = p;
⇒ 3/p = (2 + p)
⇒ p^{2} + 2p – 3 = 0
⇒ (p + 3) (p – 1) = 0
⇒ p = 3, 1
∴ 3^{x} = 1
⇒ 3^{x} = 3^{0}
∴ x = 0

Question 10 of 10
10. Question
Category: MathematicsIf a^{3} = 335 + b^{3} and a = 5 + b, then what is the value of a + b (given that a > 0 and b > 0)?
Correct
B. 9
a^{3} = 335 + b^{3} and a = 5 + b;
∴ a^{3} – b^{3} = 335 and a – b = 5
We know that;
(a^{3} – b^{3}) = (a – b) (a^{2} + b^{2} + ab)
⇒ (a^{3} – b^{3}) = (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
a^{3} = 335 + b^{3} and a = 5 + b;
∴ a^{3} – b^{3} = 335 and a – b = 5
We know that;
(a^{3} – b^{3}) = (a – b) (a^{2} + b^{2} + ab)
⇒ (a^{3} – b^{3}) = (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
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