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CDS I - Mathematics Previous Year Question Paper 2016 - 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 2016

CDS I - Mathematics Previous Year Question Paper 2016 for CDS 2024 is part of CDS (Combined Defence Services) Mock Test Series 2024 preparation. The CDS I - Mathematics Previous Year Question Paper 2016 questions and answers have been prepared according to the CDS exam syllabus.The CDS I - Mathematics Previous Year Question Paper 2016 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 2016 below.
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CDS I - Mathematics Previous Year Question Paper 2016 - Question 1

What is the solution of 

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 1


CDS I - Mathematics Previous Year Question Paper 2016 - Question 2

If λ is an integer and α, β are the roots of 4x2 – 16x + λ/4 = 0 such that 1 < α < 2 and 2 < β < 3, then how many values can λ take?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 2

4x2 – 16x + λ/4 = 0
⇒ 16x2 – 64x + λ = 0
Sum of root = α + β = 64/16 = 4
Product of root = αβ = λ/16
Given 1 < α < 2 and 2 < β < 3
⇒ Value of αβ should be between 2 and 6
But we need to consider that α + β = 4
⇒ 3 < αβ < 4
Putting αβ = λ/16
⇒ 3 < λ/16 < 4
⇒ 48 < λ < 64
∴ λ can take values from 49 to 63 i.e. 15 no. of values.

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CDS I - Mathematics Previous Year Question Paper 2016 - Question 3

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 3


CDS I - Mathematics Previous Year Question Paper 2016 - Question 4

If x2 = y + z, y2 = z + x and z2 = x + y, then what is the value of 

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 4


Multiplying x in first part, y in second part and z in third part.

Putting the given values

CDS I - Mathematics Previous Year Question Paper 2016 - Question 5

What would be the maximum value of Q in the equation 5P9 + 3R7 + 2Q8 = 1114?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 5

5P9 + 3R7 + 2Q8 = 1114

We can see that, the sum of last digits i.e. (5 + 3 + 2) = 10 and 1 is added as carry

And 2 is added as carry in the sum of middle digits of the numbers (P + Q + R)

⇒ (P + Q + R + 2) will be 11 only.
⇒ P + Q + R = 9

For maximum value of Q, values of P and R should be minimum
If P = R = 0 then Q = 9
∴ Maximum value of Q = 9

CDS I - Mathematics Previous Year Question Paper 2016 - Question 6

A is a set of positive integers such that when divided by 2, 3, 4, 5 and 6 leaves the reminder 1, 2, 3, 4 and 5 respectively. How many integers between 0 and 100 belong to the set A?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 6

We can observe that: The difference in divisor and reminder is same in every case.
(2 – 1) = (3 – 2) = (4 – 3) = (5 – 4) = (6 – 5) = 1
LCM of 2, 3, 4, 5 and 6 = 60
⇒ Required number = 60 – 1 = 59
Set A = [(60 – 1), (120 – 1), (180 – 1), ….] = [59, 119, 179, …..]
Between 0 and 100, only 59 is such integer.
∴ Only one integer between 0 and 100 belongs to the set A.

CDS I - Mathematics Previous Year Question Paper 2016 - Question 7

In an examination, a student was asked to divide a certain number by 8. By mistake he multiplied it by 8 and got the answer 2016 more than the correct answer. What was the number?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 7

Let the number be x
⇒ 8x – x/8 = 2016
⇒ 64x – x = 2016 × 8
⇒ 63x = 16128
⇒ x = 256
∴ The number is 256.

CDS I - Mathematics Previous Year Question Paper 2016 - Question 8

In the quadratic equation x+ ax + b = 0, a and b can take any value from the set {1, 2, 3, 4}. How many pairs of values of a and b are possible in order that the quadratic equation has real roots?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 8

x+ ax + b = 0
For real roots, necessary condition is:
a2 – 4b ≥ 0
⇒ a2 ≥ 4b
And (a, b) ϵϵ {1, 2, 3, 4}
Possible pairs of a & b according to the above conditions:
(2, 1), (3, 1), (4, 1), (3, 2), (4, 2), (4, 3), (4, 4)
∴ There are 7 possible pairs of a and b.

CDS I - Mathematics Previous Year Question Paper 2016 - Question 9

The sum of squares of two positive integers is 208. If the square of the larger number is 18 times the smaller number, then what is the difference of the larger and smaller numbers.

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 9

Suppose the numbers are ‘a’ and ‘b’ with a > b
a2 + b2 = 208 and a2 = 18b
Solving the equations
⇒ b2 + 18b = 208
⇒ b2 + 18b – 208 = 0
⇒ (b + 26)(b – 8) = 0
⇒ b = 8, b = -26
If b = 8
a2 = 18b = 18 × 8 = 144
⇒ a = 12
∴ Difference of the larger and the smaller number = 12 – 8 = 4

CDS I - Mathematics Previous Year Question Paper 2016 - Question 10

Let A = {7, 8, 9, 10, 11, 12} and B = {7, 10, 14, 15}. What is the number of elements in (A – B) and (B – A) respectively?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 10

A = {7, 8, 9, 10, 11, 12} and B = {7, 10, 14, 15}
(A – B) = {8, 9, 11, 12}
(B – A) = {14, 15}
∴ Number of elements in (A – B) and (B – A) are 4 and 2 respectively.

CDS I - Mathematics Previous Year Question Paper 2016 - Question 11

A boy saves Rs. 4.65 daily. What is the least number of days in which he will be able to save an exact number of rupees?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 11

Using hit and trial method:
1) 4.65 × 10 = 46.5
2) 4.65 × 20 = 93
3) 4.65 × 21 = 97.65
4) 4.65 × 25 = 116.25
∴ He will be able to save an exact number of rupees in 20 days.

CDS I - Mathematics Previous Year Question Paper 2016 - Question 12

Two men, A and B run a 4 km race on a course 0.25 km round. If their speeds are in the ratio 5 : 4, how often does the winner pass the other? 

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 12

Number of rounds to be completed = 4/0.25 = 16
Since the ratio of speeds of A and B is 5 : 4
⇒ When A covers 5 rounds, B will complete 4 rounds and they will meet.
⇒ A will pass B after 5th, 10th & 15th round.
∴ The winner passes the other thrice.

CDS I - Mathematics Previous Year Question Paper 2016 - Question 13

Which one of the following rational numbers has non-terminating and repeating decimal expression?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 13

Observing all the options
1. 15/1600 = 0.009375
2. 23/8 = 2.875
3. 35/50 = 0.7
4. 17/6 = 2.8333333……
∴ Here we can see that 17/6 has non-terminating and repeating decimal expression.

CDS I - Mathematics Previous Year Question Paper 2016 - Question 14

If α and β are the two zeros of the polynomial 25x2 – 15x + 2, then what is a quadratic polynomial whose zeros are (2α)-1 and (2β)-1 ?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 14

25x2 – 15x + 2
⇒ 25x2 – 10x – 5x + 2
⇒ 5x(5x – 2) – 1(5x – 2)
⇒ (5x – 1)(5x – 2)
⇒ x = 1/5 and 2/5
⇒ α and β will be equal to 1/5 and 2/5.
If zeros are (2α)-1 and (2β)-1
Then the zeros of that polynomial are 5/2 and 5/4
∴ Polynomial = (x – 5/2)(x – 5/4) = (x2 – 15x/4 + 25/8) = (8x2 – 30x + 25)

CDS I - Mathematics Previous Year Question Paper 2016 - Question 15

What is the remainder when 2100 is divided by 101?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 15


[If a(n – 1) is divided by ‘n’ where ‘a’ and ‘n’ are co-prime numbers then remainder will be 1]
Hence here also ‘2’ and ‘101’ are co-prime numbers.
∴ Remainder = 1

CDS I - Mathematics Previous Year Question Paper 2016 - Question 16

In an office, one-third of the workers are women, half of the women are married, and one-third of the married women have children. If three-fourth of the men are married and one-third of the married men have children, then what is the ratio of married women to married men?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 16

Suppose total number of workers = x
⇒ Number of women workers = x/3
⇒ Number of men workers = 2x/3
⇒ Number of married women = (x/3) × ½ = x/6
⇒ Number of married men = (2x/3) × 3/4 = x/2
∴ Ratio of married women to married men = (x/6) : (x/2) = 1 : 3

CDS I - Mathematics Previous Year Question Paper 2016 - Question 17

A can do 50% more work than B in same time. B alone can do a piece of work in 30 hours. B starts working and had already worked for 12 hours when A joins him. How many hours should B and A work together to complete the remaining work? 

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 17

B alone can do the work in 30 hours and A can do 50% more work than B in same time;
⇒ Time taken by A to complete the work alone = 30/1.5 = 20 hours
Suppose total work = 60 units (LCM of 20 & 30)

⇒ Efficiency of A = 60/20 = 3
⇒ Efficiency of B = 60/30 = 2
⇒ Unit of work done by B in 12 hours = 12 × 2 = 24 units
⇒ Remaining work = 60 – 24 = 36 units
∴ Time taken by (A + B) to complete remaining work = 36/5 = 7.2 hours

CDS I - Mathematics Previous Year Question Paper 2016 - Question 18

When the speed of a train is increased by 20%, it takes 20 minutes less to cover the same distance. What is the time taken to cover the same distance with the original speed?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 18

Suppose the distance be ‘x’ km and speed be ‘n’ km/minutes
⇒ Increased speed = 1.2n
⇒ x/n – x/1.2n = 20
⇒ (0.2x/1.2n) = 20
⇒ x/n = 120 minutes

Here ‘x/n’ is the time taken to cover the distance with the original speed.

∴ Train will take 120 minutes to cover the same distance with the original speed.

CDS I - Mathematics Previous Year Question Paper 2016 - Question 19

The cost of 2.5 kg rice is Rs. 125. The cost of 9 kg rice is equal to that of 4 kg pulses. The cost of 14 kg pulses is equal to that of 1.5 kg tea. The cost of 2 kg tea is equal to that of 5 kg nuts. What is the cost of 11 kg nuts?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 19

Given Cost of 2.5 kg rice is Rs. 125
⇒ Cost of 1 kg rice = 125/2.5 = Rs. 50
⇒ Cost of 4 kg pulses = cost of 9 kg rice = 50 × 9 = Rs. 450
⇒ Cost of 1 kg pulses = 450/4 = Rs. 112.5
⇒ Cost of 1.5 kg tea = Cost of 14 kg pulses = 112.5 × 14 = Rs. 1575
⇒ Cost of 1 kg tea = 1575/1.5 = Rs. 1050
⇒ Cost of 5 kg nuts = Cost of 2 kg tea = 1050 × 2 = Rs. 2100
∴ Cost of 11 kg nuts = (11/5) × 2100 = Rs. 4620

CDS I - Mathematics Previous Year Question Paper 2016 - Question 20

There are 12 friends A, B, C, D, E, F, G, H, I, J, K and L who invested money in some business in the ratio of 1 : 2 : 3 : 4 : 5 : 6 : 7 : 8 : 9 : 10 : 11 : 12 and the duration for which they invested the money is in the ratio of 12 : 11 : 10 : 9 : 8 : 7 : 6 : 5 : 4 : 3 : 2 : 1 respectively. Who will get the maximum benefit at the end of the year?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 20

Ratio in which the profit will be shared = Investment × Time for which investment was made

⇒ (1 × 12) : (2 × 11) : (3 × 10) : (4 × 9) : (5 × 8) : (6 × 7) : (7 × 6) : (8 × 5) : (9 × 4) : (10 × 3) : (11 × 2) : (12 × 1)

⇒ 12 : 22 : 30 : 36 : 40 : 42 : 42 : 40 : 36 : 30 : 22 : 12

Here we can see that shares of both F and G in the profit will be maximum.

CDS I - Mathematics Previous Year Question Paper 2016 - Question 21

If p and q are the roots of x2 + px + q = 0, then which of the following is correct?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 21

x2 + px + q = 0
If a and b are the roots of the equation
⇒ Sum of roots = (p + q) = -p     ----1
⇒ Product of roots = pq = q     ----2
On solving equation 1 and 2, we get
p = 1 and q = -2
∴ p = 1 only

CDS I - Mathematics Previous Year Question Paper 2016 - Question 22

Which of the following is correct in respect to number 1729?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 22

1729 = 1728 + 1 = 123 + 13 and 1729 = 1000 + 729 = 103 + 93

∴ It can be written as the sum of cubes of two positive integers in two ways only.

CDS I - Mathematics Previous Year Question Paper 2016 - Question 23

A shopkeeper increases the cost price of an item by 20% and offers a discount of 10% on this marked price. What is his percentage gain?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 23

Let the cost price of the item = Rs. x
⇒ Increased cost price = marked price = Rs. 1.2x
⇒ Overall selling price = 1.2x × 0.9 = Rs. 1.08x
Profit = Selling price - Cost price = 1.08x - x = Rs. 0.08x
∴ Percentage gain = 0.08x/x × 100 = 8%

CDS I - Mathematics Previous Year Question Paper 2016 - Question 24

There are two numbers p and q such that their HCF is 1. Which of the following statements are correct?
1. Both p and q may be prime.
2. One number may be prime and the other composite.
3. Both the numbers may be composite.

Select the correct answer using the code given below:

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 24

HCF of p and q is 1
1. Both p and q may be prime
Suppose p = 2 and q = 3
⇒ HCF of p and q = 1
⇒ Statement 1 is correct

2. One number may be prime, and the other is composite.
Suppose p = 3 and q = 4
⇒ HCF of p and q = 1
⇒ Statement 2 is correct

3. Both the numbers may be composite
Suppose p = 9 and q = 10
⇒ HCF of p and q = 1
⇒ Statement 3 is correct.
∴ All three statements are correct.

CDS I - Mathematics Previous Year Question Paper 2016 - Question 25

What is 

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 25


nth term of this series can be written as


Sum of the series 
Sum of the series 
∴ Sum of the series 

CDS I - Mathematics Previous Year Question Paper 2016 - Question 26

A person can row downstream 20 km in 2 hours and upstream 4 km in 2 hours. What is the speed of the current?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 26

Downstream speed = 20/2 = 10 km/hr
Upstream speed = 4/2 = 2 km/hr
⇒ Speed of current = (Downstream speed - Upstream speed)/2
∴ Speed of the current = (10 - 2)/2 = 4 km/hr

CDS I - Mathematics Previous Year Question Paper 2016 - Question 27

Consider the following statements in respect of positive odd integers x and y:
1. x2 + y2 is even integer.
2. x2 + y2 is divisible by 4.

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

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 27

Since x and y are two odd integers, their squares will be odd always and sum of two odd numbers is always even.

⇒ Statement 1 is correct.
Suppose x = 1 and y = 3
⇒ x2 + y= 1 + 9 = 10
It is even but not divisible by 4.
∴ Only statement 1 is correct.

CDS I - Mathematics Previous Year Question Paper 2016 - Question 28

What are the roots of the equation 2x + 2 × 27x/x - 1 = 9?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 28

2x + 2 × 27x/ (x - 1) = 9
⇒ 2x + 2 × 33x/ (x - 1) = 32
⇒ 2x + 2 = 3[2 - {(3x/x - 1)}]
⇒ 2x+2 = 3[(x + 2) / (x - 1)]

Taking log
⇒ (x + 2)log2 = [(x + 2)/(x - 1)]log3
⇒ x - 1 = (log3/log2)
⇒ x = 1 + (log3/log2)
⇒ x = 1 - (log2/log3)

And
(x + 2) = 0 ⇒ x = -2
∴ Roots of the equation = -2, 1 - (log2/log3)

CDS I - Mathematics Previous Year Question Paper 2016 - Question 29

2122 + 462 + 842 + 464 + 2130 is divisible by which one of the following integers?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 29

2122 + 462 + 842 + 464 + 2130
⇒ 2122 + 2124 + 2126 + 2128 + 2130
⇒ 2122 (1 + 4 + 16 + 64 + 256)
⇒ 2122 × 341
Since 341 is divisible by 11
∴ 2122 + 462 + 842 + 464 + 2130 is divisible by 11 only

CDS I - Mathematics Previous Year Question Paper 2016 - Question 30

A candidate scoring x% marks in an examination fails by ‘a’ marks, while another candidate who scores y% marks gets ‘b’ marks more than the minimum pass marks. What is the maximum mark for the examination?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2016 - Question 30

Let the maximum marks for the examination be ‘n’.

Candidate scoring x% marks in an examination fails by ‘a’ marks but another candidate who scores y% marks gets ‘b’ marks more than the minimum pass marks;

∴ (n × x)/100 + a = (n × y)/100 – b
⇒ (a + b) = n/100 × (y – x)

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