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CDS I - Mathematics Previous Year Question Paper 2017 - CDS MCQ


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30 Questions MCQ Test CDS (Combined Defence Services) Mock Test Series 2024 - CDS I - Mathematics Previous Year Question Paper 2017

CDS I - Mathematics Previous Year Question Paper 2017 for CDS 2024 is part of CDS (Combined Defence Services) Mock Test Series 2024 preparation. The CDS I - Mathematics Previous Year Question Paper 2017 questions and answers have been prepared according to the CDS exam syllabus.The CDS I - Mathematics Previous Year Question Paper 2017 MCQs are made for CDS 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for CDS I - Mathematics Previous Year Question Paper 2017 below.
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CDS I - Mathematics Previous Year Question Paper 2017 - Question 1

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 1


Taking the square roots;

CDS I - Mathematics Previous Year Question Paper 2017 - Question 2

(x + 4) is a factor of which one of the following expressions?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 2

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)

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CDS I - Mathematics Previous Year Question Paper 2017 - Question 3

If 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)?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 3

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,



∴ Maximum value of this expression will be -2. (As the first term will be negative always)

CDS I - Mathematics Previous Year Question Paper 2017 - Question 4

Consider 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).

Q. Which of the above statements is/are correct?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 4

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.

CDS I - Mathematics Previous Year Question Paper 2017 - Question 5

What is the square root of 

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 5

CDS I - Mathematics Previous Year Question Paper 2017 - Question 6

What is the number of digits in 240? (Given that log102 = 0.301)

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 6

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

CDS I - Mathematics Previous Year Question Paper 2017 - Question 7

If one root of (a2 – 5a + 3)x2 + (3a – 1)x + 2 = 0 is twice the other, then what is the value of ‘a’?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 7

(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

CDS I - Mathematics Previous Year Question Paper 2017 - Question 8

What is the remainder when the number (4444)4444 is divided by 9?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 8

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.

CDS I - Mathematics Previous Year Question Paper 2017 - Question 9

 then what is bx2 – 2ax + b equal to (b ≠ 0)?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 9


Applying componendo and dividendo;
Squaring both sides;

Applying componendo and dividend again;

CDS I - Mathematics Previous Year Question Paper 2017 - Question 10

What is the value of 

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 10

CDS I - Mathematics Previous Year Question Paper 2017 - Question 11

If x = t 1/(t – 1) and y = t t/(t – 1), t > 0, t ≠ 1 then what is the relation of x and y?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 11

t/(t – 1) = 1 + 1/(t – 1)
∴ y = t t/(t – 1) = t [1 + 1/(t – 1)] = t × t 1/(t – 1)
⇒ y = t × x
∴ t = y/x      ---- (1)
Now
y = t t/(t – 1) = [t 1/(t – 1)]t
⇒ y = xt      ---- (2)
From equation 1 and 2;
⇒ y = x(y/x)
∴ xy = yx

CDS I - Mathematics Previous Year Question Paper 2017 - Question 12

If A : B = 3 : 4, then what is the value of the expression 

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 12


Since A : B = 3 : 4, suppose A = 3x and B = 4x;
Putting these values in the expression;

Now we can see that the value of the expression can’t be determined because the expression is not homogeneous. 

CDS I - Mathematics Previous Year Question Paper 2017 - Question 13

If 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}

Q. Then which of the following is null set?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 13

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.

CDS I - Mathematics Previous Year Question Paper 2017 - Question 14

If x = 2 + 22/3 + 21/3, then what is the value of x3 – 6x2 + 6x?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 14

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

CDS I - Mathematics Previous Year Question Paper 2017 - Question 15

and x + y = 26, then what is the value of xy?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 15


Taking square both sides;

Putting the value of (x + y) = 26

⇒ 16900 – 100xy = 576xy
⇒ 676xy = 1690
⇒ xy = 25

CDS I - Mathematics Previous Year Question Paper 2017 - Question 16

What is the solution of the equation x log10(10/3) + log103 = log10(2 + 3x) + x?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 16

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)
⇒ 3/3= (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

CDS I - Mathematics Previous Year Question Paper 2017 - Question 17

If α and β are the roots of the equation x2 + px + q = 0, then what is α2 + β2 equal to?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 17

Since α and β are the roots of the equation x2 + px + q = 0;
∴ α + β = -p and αβ = q
Now
α2 + β= (α + β)2 – 2αβ
Putting the values;
⇒ α2 + β= p2 – 2q
Option 1 is correct.

CDS I - Mathematics Previous Year Question Paper 2017 - Question 18

If a3 = 335 + b3 and a = 5 + b, then what is the value of a + b (given that a > 0 and b > 0)?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 18

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

CDS I - Mathematics Previous Year Question Paper 2017 - Question 19

If 9x 3y = 2187 and 23x 22y – 4xy = 0, then what can be the value of (x + y)? 

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 19

9x 3y = 2187
⇒ 3(2x + y) = 37
∴ (2x + y) = 7     ---- (1)
23x 22y – 4xy = 0
⇒ 23x 22y = 4xy
⇒ 2(3x + 2y) = 22xy
∴ (3x + 2y) = 2xy      ---- (2)
Now eliminating the options:
Since (2x + y) = 7 hence (x + y) can’t be 1, 3 or 7.
Suppose (x + y) = 5 where x = 2 and y = 3;
Putting in both the equations;
(2x + y) = 4 + 3 = 7
And (3x + 2y) = 2xy
⇒ 6 + 6 = 2 × 2 × 3
⇒ 12 = 12
Hence the values x = 2 and y = 3 satisfy the equations.
∴ (x + y) = 5

CDS I - Mathematics Previous Year Question Paper 2017 - Question 20

The pair of linear equations kx + 3y + 1 = 0 and 2x + y + 3 = 0 intersect each other, if

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 20

Since the pair of linear equations kx + 3y + 1 = 0 and 2x + y + 3 = 0 intersect each other;
∴ (a1/a2) ≠ (b1/b2)
⇒ k/2 ≠ 3/1
⇒ k ≠ 6

CDS I - Mathematics Previous Year Question Paper 2017 - Question 21

The number of prime numbers which are less than 100 is

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 21

Prime numbers less than 100 = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97}
∴ Number of prime numbers which are less than 100 = 25

CDS I - Mathematics Previous Year Question Paper 2017 - Question 22

The cost of diamond varies directly as the square of its weight. A diamond broke into four pieces with their weights in the ratio of 1 ∶ 2 ∶ 3 ∶ 4. If the loss in total value of the diamond was Rs. 70000, what was the price of the original diamond? 

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 22

Let the original weight of the diamond was ‘10x’ gm;

Since the diamond broke into four pieces with their weights in the ratio of 1 ∶ 2 ∶ 3 ∶ 4;

∴ Weights of four pieces = x, 2x, 3x & 4x;

Since the cost of diamond varies directly as the square of its weight;

∴ Cost of diamond pieces = x2, 4x2, 9x2 and 16x2

And Cost of original diamond = (10x)= 100x2

Since loss in total value of the diamond was Rs. 70000 on cutting;
∴ 100x– x2 – 4x2 – 9x2 – 16x2 = 70000
⇒ 70x2 = 70000
⇒ x2 = 10000
⇒ x = 100
∴ Cost of original diamond = 100x2 = Rs. 100000

CDS I - Mathematics Previous Year Question Paper 2017 - Question 23

In a 100 m race, A runs at a speed of 5/3 m/s. If A gives a start of 4 m to B and still beats him by 12 seconds, what is the speed of B?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 23

Time taken by A to cover 100 m = 100/(5/3) = 60 sec

Since A beats B by 12 seconds, B must have run for (60 + 12) = 72 sec.

And since A gives a start of 4 m to B, so B runs only (100 – 4) = 96 m;

∴ Speed of B = 96/72 = 8/6 = 4/3 m/s

CDS I - Mathematics Previous Year Question Paper 2017 - Question 24

If 15 men take 21 days of 8 hours each to do a piece of work, then what is the number of days of 6 hours each that 21 women would take, if 3 women would as much work as 2 men? 

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 24

Since 3 women would as much work as 2 men;

Ratio of efficiencies of Man and Woman = 3 ∶ 2
Or 2M = 3W

15 men take 21 days of 8 hours each to do a piece of work, let suppose 21 women take x days of 6 hours each to do that work;
∴ (15M × 21 × 8) = (21W × 6 × x)
Putting 2M = 3W;
⇒ (15M × 21 × 8) = (14M × 6 × x)
⇒ x = 30
∴ 21 women will take 30 days of 6 hours each to do the work.

CDS I - Mathematics Previous Year Question Paper 2017 - Question 25

What number must be subtracted from both the numerator and denominator of the fraction 27/35 so that it becomes 2/3? 

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 25

Let x is the required number;
∴ (27 – x) / (35 – x) = 2/3
⇒ 81 – 3x = 70 – 2x
⇒ x = 11

CDS I - Mathematics Previous Year Question Paper 2017 - Question 26

The mean of 5 numbers is 15. If one more number is included, the mean of 6 numbers becomes 17. What is the included number?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 26

Since mean of 5 numbers is 15;
∴ Sum of 5 number = 5 × 15 = 75
Mean of 6 numbers becomes 17
∴ Sum of 6 number = 6 × 17 = 102
∴ The included 6th number = 102 - 75 = 27

CDS I - Mathematics Previous Year Question Paper 2017 - Question 27

The mean marks obtained by 300 students in a subject are 60. The mean of top 100 students was found to be 80 and the mean of last 100 students was found to be 50. The mean marks of the remaining 100 students are

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 27

Since the mean marks obtained by 300 students in a subject are 60;

∴ Sum of marks of 300 students = 300 × 60 = 18000

Mean of top 100 students was found to be 80;

∴ Sum of marks of top 100 students = 100 × 80 = 8000

Mean of last 100 students was found to be 50;

∴ Sum of marks of last 100 students = 100 × 50 = 5000

∴ Sum of marks of remaining 100 students = 18000 - 5000 - 8000 = 5000

∴ Mean of remaining 100 students = 5000/100 = 50

CDS I - Mathematics Previous Year Question Paper 2017 - Question 28

Consider the following distribution

Q. If the mean of the above distribution is 50 then what is the value of f?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 28


Mean = (∑XiFi) / (∑Fi)
⇒ Mean = (170 + 840 + 1600 + 70f + 1710) / (17 + 28 + 32 + f + 19)
⇒ Mean = (4320 + 70f) / (96 + f)
Since mean is given 50;
∴ 50 = (4320 + 70f) / (96 + f)
⇒ 5f + 480 = 432 + 7f
⇒ 2f = 48
⇒ f = 24

CDS I - Mathematics Previous Year Question Paper 2017 - Question 29

In a pie diagram, there are four slices with angles 150°, 90°, 60° and 60°. A new pie diagram is formed by deleting one of the slices having angle 60° in the given pie diagram. In the new pie diagram

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 29

Ratio of angles without 60° = 150 ∶ 90 ∶ 60 = 5 ∶ 3 ∶ 2

When the slice having angle 60° is deleted from the pie diagram, it would be divided in the same ratio among other three slices;

150° slice will increased to [150° + (5/10) × 60] = 180°

90° slice will increased to [90° + (3/10) × 60] = 108°

60° slice will increased to [60° + (2/10) × 60] = 72°

∴ The largest slice has angle 180°

CDS I - Mathematics Previous Year Question Paper 2017 - Question 30

In an asymmetrical distribution, if the mean and median of the distribution are 270 and 220 respectively, then the mode of the data is

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2017 - Question 30

Mean = 270 and Median = 220;
We know the formula∶
3 × Median = Mode + 2 × Mean
∴ Mode = 3 × Median - 2 × Mean
⇒ Mode = 3 × 220 - 2 × 270
⇒ Mode = 660 - 540 = 120

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