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

Taking the square roots;

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 3^rd equation;
∴ (x + 4) is a factor of (x² – 7x – 44)

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)

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.

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

(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

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.

Applying componendo and dividendo;
Squaring both sides;

Applying componendo and dividend again;

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 = x^t      ---- (2)
From equation 1 and 2;
⇒ y = x^(y/x)
∴ x^y = y^x

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. 

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.

Given
x = 2 + 2^2/3 + 2^1/3
⇒ (x – 2) = (2^2/3 + 2^1/3)

Taking cube both the sides;
⇒ x³ – 8 – 6x² + 12x = 4 + 2 + 3 × 2^2/3 × 2^1/3 (2^2/3 + 2^1/3)
⇒ x³ – 14 – 6x² + 12x = 6 (x – 2)
⇒ x³ – 14 – 6x² + 12x – 6x + 12 = 0
⇒ x³ – 6x² + 6x = 2

Taking square both sides;

Putting the value of (x + y) = 26

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

x log₁₀(10/3) + log₁₀3 = log₁₀(2 + 3^x) + x
⇒ x log₁₀10 – x log₁₀3 + log₁₀3 = log₁₀(2 + 3^x) + x
⇒ x – x log₁₀3 + log₁₀3 = log₁₀(2 + 3^x) + x
⇒ log₁₀3^-x + log₁₀3 = log₁₀(2 + 3^x)
⇒ log₁₀(3^-x) × 3 = log₁₀(2 + 3^x)
⇒ (3^-x) × 3 = (2 + 3^x)
⇒ 3/3^x = (2 + 3^x)
Suppose 3^x = p;
⇒ 3/p = (2 + p)
⇒ p² + 2p – 3 = 0
⇒ (p + 3) (p – 1) = 0
⇒ p = -3, 1
∴ 3^x = 1
⇒ 3^x = 3⁰
∴ x = 0

Since α and β are the roots of the equation x² + px + q = 0;
∴ α + β = -p and αβ = q
Now
α² + β² = (α + β)² – 2αβ
Putting the values;
⇒ α² + β² = p² – 2q
Option 1 is correct.

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

9^x 3^y = 2187
⇒ 3^{(2x + y)} = 3⁷
∴ (2x + y) = 7     ---- (1)
2^3x 2^2y – 4^xy = 0
⇒ 2^3x 2^2y = 4^xy
⇒ 2^{(3x + 2y)} = 2^2xy
∴ (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

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

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

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)² = 100x²

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

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

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.

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

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 6^th number = 102 - 75 = 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

Mean = (∑X_iF_i) / (∑F_i)
⇒ 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

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°

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