CDS Exam  >  CDS Tests  >  CDS (Combined Defence Services) Previous Years Papers  >  CDS II - Mathematics Previous Year Question Paper 2017 - CDS MCQ

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


Test Description

30 Questions MCQ Test CDS (Combined Defence Services) Previous Years Papers - CDS II - Mathematics Previous Year Question Paper 2017

CDS II - Mathematics Previous Year Question Paper 2017 for CDS 2024 is part of CDS (Combined Defence Services) Previous Years Papers preparation. The CDS II - Mathematics Previous Year Question Paper 2017 questions and answers have been prepared according to the CDS exam syllabus.The CDS II - 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 II - Mathematics Previous Year Question Paper 2017 below.
Solutions of CDS II - Mathematics Previous Year Question Paper 2017 questions in English are available as part of our CDS (Combined Defence Services) Previous Years Papers for CDS & CDS II - Mathematics Previous Year Question Paper 2017 solutions in Hindi for CDS (Combined Defence Services) Previous Years Papers course. Download more important topics, notes, lectures and mock test series for CDS Exam by signing up for free. Attempt CDS II - Mathematics Previous Year Question Paper 2017 | 100 questions in 120 minutes | Mock test for CDS preparation | Free important questions MCQ to study CDS (Combined Defence Services) Previous Years Papers for CDS Exam | Download free PDF with solutions
CDS II - Mathematics Previous Year Question Paper 2017 - Question 1

A tent has been constructed which is in the form of a right circular cylinder surmounted by a right circular cone whose axis coincides with the axis of the cylinder. If the radius of the base is 50 m, the height of the cylinder is 10 m and the total height of the tent is 15 m, then what is the capacity of the tent in cubic meters?

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

Height of the cone = Total height – Height of the cylinder = 15 – 10 = 5 cm

Radius of the cone = Radius of the cylinder = 50 m

Total volume of tent = Volume of cylinder + Volume of cone
⇒ πr2H + 1/3 × πr2h
⇒ π502 × 10 + 1/3 × π502 × 5
⇒ 502 × π × (10 + 5/3)
⇒ 2500 × 35/3 × π
⇒ 87500π/3 cm3

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

Two rectangular sheets of sizes 2π × 4π and π × 5π are available. A hollow right circular cylinder can be formed by joining a pair of parallel sides of any sheets. What is the maximum possible volume of the cylinder that can be formed this way?

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

When a rectangle is formed into a cylinder, the length of the rectangle becomes the circumference of base and breadth becomes the height.

For first rectangle
2πr = 4π
∴ r = 2
h = 2π
Volume = π r2 h
⇒ π × 22 × 2π = 8π2
For second rectangle
2πr = 5π
∴ r = 5/2
h = π
Volume = π r2 h
⇒ π × (5/2)2 × π = 6.25π2
Maximum possible volume = 8π2

1 Crore+ students have signed up on EduRev. Have you? Download the App
CDS II - Mathematics Previous Year Question Paper 2017 - Question 3

In triangle ABC, the medians AD and BE intersect at G. A line DF is drawn parallel to BE such that F is on AC. If AC = 9 cm, then what is CF equal to?

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

In triangle CBE
DF is parallel to BE
∴ CD/BC = CF/CE [∵ BPT theorem]
AD and BE are medians of BC and AC respectively
∴ CD/BC = ½
⇒ CE = AC/2 = 9/2 = 4.5 cm
⇒ ½ = CF/4.5
∴ CF = 4.5/2 = 2.25 cm

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

In a triangle PQR, X is a point on PR and Y is a point on QR such that PR = 10 cm, RX = 4 cm, YR = 2 cm, QR = 5 cm. Which one of the following is correct?

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


XR = 4 cm
⇒ PX = PR – XR = 10 – 4 = 6 cm
⇒ YR = 2 cm
⇒ QY = QR – YR = 5 – 2 = 3 cm
⇒ XR/PR = 4/6 = 2/3
⇒ YR/QY = 2/3
∴ XR/PR = YR/PR
∴ XY is parallel to PQ [∵ Inverse BPT theorem]

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

Consider the following statements in respect of three straight lines A, B and C on a plane:
1. If A and C are parallel and B and C are parallel, then A and B are parallel
2. If A is perpendicular to C and B is perpendicular to C, then A and B are parallel
3. If the acute angle between A and C is equal to the acute angle between B and C; then A and B are parallel

Q. Which of the above statements are correct?

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

Since they are in a plane (Let us consider XY plane) all line is of form y = mx + c

Let the slopes of the lines A, B and C be a, b and c respectively

Lines are parallel if their slopes are equal and perpendicular if the product of their slopes is -1

Statement 1:
A and C are parallel
∴ a = c
B and C are parallel
∴ b = c
Comparing both the equations
⇒ a = b
∴ A and B are parallel too. Statement 1 is true

Statement II:
A and C are perpendicular
∴ ac = -1 or a = -1/c
B and C are perpendicular
∴ bc = -1 or b = -1/c
Comparing both the equations
a = b
∴ A is parallel to B
∴ Statement II is true

Statement III:
Statement III need not be true because the lines A and B can be on either side of line C forming same angle but are not parallel

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

The diagonals of a rhombus are of length 20 cm and 48 cm. What is the length of a side of the rhombus?

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


Let AC = 48 cm and BD = 20 cm
In a rhombus, diagonals bisect each other at right angles
∴ AO = 48/2 = 24 cm and BO = 20/2 = 10 cm

In triangle AOB
AB2 = AO2 + BO2 [∵ Pythagoras theorem]
⇒ 242 + 102
⇒ 676
∴ AB = 26 cm
∴ Side of the rhombus is 26 cm

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

An arc of a circle subtends an angle π at the centre. If the length of the arc is 22 cm, then what is the radius of the circle?
(Take π = 22/7)

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

Angle subtended by full circle at center = 2π
Angle subtended by the given arc = π
∴ The arc is a semi-circle
Arc length of a semi-circle = π r = 22
⇒ 22/7 × r = 22
∴ r = 7 cm

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

One fifth of the area of the triangle ABC is cut off by the line DE drawn parallel to BC such that D is on AB and E is one AC. If BC = 10 cm, then what is DE equal to?

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


In triangle ADE and ABC
∠ A is common

∠ ADE = ∠ ABE [∵ Corresponding angles are equal when DE is parallel to BC]
∴ ADE ∼ ABC
Area(ADE) /Area(ABC) = (DE/BC)2
⇒ 1/5 = (DE/10)2
∴ DE = 2√5 cm

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

There are 8 lines in a plane, no two of which are parallel. What is the maximum number of points at which they can intersect?

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

Each pair can have one intersection point
Number of pairs in 8 lines = 8C2 = 8! / (2! × 6!) = 28

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

A closed polygon has six sides and one of its angles is 30° greater than each of the other five equal angles. What is the value of the equal angles?

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

Let the equal angles be x and the larger angle be x + 30
Sum of angles = 5x + x + 30 = 6x + 30

Sum of angles of a six sides polygon = (n – 2) × 180 = (6 – 2) × 180 = 720
⇒ 6x + 30 = 720
⇒ 6x = 690
∴ x = 115° 

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

Consider the following statements:
(1) The point of intersection of the perpendicular bisectors of the sides of a triangle may lie outside the triangle.
(2) The point of intersection of the perpendicular drawn from the vertices to the opposite side of a triangle may lie on two sides.

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

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

Point of intersection of the perpendicular bisector is the circumcenter of the triangle.

It is possible for three chords to form a triangle but do not include the center when it is an obtuse angled triangle

∴ Statement 1 is possible

In a right triangle the perpendicular drawn from two vertices of the hypotenuse are the sides themselves. They meet on the vertex and lie on the side

∴ Statement 2 is possible
∴ Both 1 and 2 true

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

Directions: In a University, there are 1200 students studying four different subjects, Mathematics, Statistics, Physics and Chemistry. 20% of the total number of students are studying Mathematics, one-fourth of the total number of students are studying Physics, 320 students are studying Statistics. Three-fifth of the total number of students studying Chemistry are girls. 150 boys are studying Mathematics. 60% of the students studying Physics are boys. 250 girls are studying Statistics.

Q. What is the total number of boys studying Statistics and Physics?

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

Number of students studying Physics = ¼ × Total number of students

⇒ ¼ × 1200 = 300 students

Number of boys studying in physics = 60% of number of students studying Physics

⇒ 60/100 × 300 = 180 boys

Number of boys studying Statistics = Total number of students studying Statistics – Number of girls studying Statistics

⇒ 320 – 250 = 70 boys

∴ Total number of boys studying Statistics and Physics = 180 + 70 = 250 boys

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

Directions: In a University, there are 1200 students studying four different subjects, Mathematics, Statistics, Physics and Chemistry. 20% of the total number of students are studying Mathematics, one-fourth of the total number of students are studying Physics, 320 students are studying Statistics. Three-fifth of the total number of students studying Chemistry are girls. 150 boys are studying Mathematics. 60% of the students studying Physics are boys. 250 girls are studying Statistics.

Q. The number of girls studying Statistics is what percent (approximate) of the total number of students studying Chemistry?

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

Total number of students studying chemistry = 1200 – Number of students studying Physics – Number of students studying Mathematics – Number of students studying Statistics

⇒ 1200 – ¼ × 1200 – 20/100 × 1200 – 320
⇒ 1200 – 300 – 240 – 320 = 340 students

Required percentage = Number of girls studying Statistics/Number students studying Chemistry
⇒ 250/340 × 100 ≈ 73.5%

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

Directions: In a University, there are 1200 students studying four different subjects, Mathematics, Statistics, Physics and Chemistry. 20% of the total number of students are studying Mathematics, one-fourth of the total number of students are studying Physics, 320 students are studying Statistics. Three-fifth of the total number of students studying Chemistry are girls. 150 boys are studying Mathematics. 60% of the students studying Physics are boys. 250 girls are studying Statistics.

Q. In which subjects is the difference between the number of boys and girls is equal?

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

Mathematics:
Total number of students = 240 [From previous question]
Number of boys = 150
Number of girls = 240 – 150 = 90
Difference between number of boys and girls = 150 – 90 = 60

Chemistry:
Total number of students = 340
Number of girls = 3/5 × Total = 3/5 × 340 = 204
Number of boys = 340 – 204 = 136
Difference = 204 – 136 = 68 students

Statistics:
Total number of students = 320
Number of girls = 250
Number of boys = 320 – 250 = 70
Difference = 250 – 70 = 180

Physics:
Total number of students = 300
Number of boys = 60/100 × 300 = 180
Number of girls = 300 – 180 = 120
Difference = 180 – 120 = 60
∴ Difference is same (60) for Mathematics and Physics

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

Directions: In a University, there are 1200 students studying four different subjects, Mathematics, Statistics, Physics and Chemistry. 20% of the total number of students are studying Mathematics, one-fourth of the total number of students are studying Physics, 320 students are studying Statistics. Three-fifth of the total number of students studying Chemistry are girls. 150 boys are studying Mathematics. 60% of the students studying Physics are boys. 250 girls are studying Statistics.

Q. What is the difference between number of boys studying in Mathematics and number of girls studying in Physics?

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

Physics:
Total number of students = 300
Number of boys = 60/100 × 300 = 180
Number of girls = 300 – 180 = 120

∴ Difference between number of boys studying Mathematics and number of girls studying Physics = 150 – 120 = 30

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

Directions: In a University, there are 1200 students studying four different subjects, Mathematics, Statistics, Physics and Chemistry. 20% of the total number of students are studying Mathematics, one-fourth of the total number of students are studying Physics, 320 students are studying Statistics. Three-fifth of the total number of students studying Chemistry are girls. 150 boys are studying Mathematics. 60% of the students studying Physics are boys. 250 girls are studying Statistics.

Q. What is the ratio between total number of boys to the total number of girls? 

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

Total number of boys = 150 + 180 + 70 + 136 = 536
Total number of girls = 90 + 250 + 120 + 204 = 664
Required ratio = 536 : 664 = (67 × 8) : (83 × 8) = 67 : 83

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

Frequency density of a class is computed by the ratio

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

Frequency density is the ratio of the class frequency to the class width of the particular class.

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

A small company pays each of its 5 category ‘C’ workers Rs. 20,000, each of its 3 category ‘B’ workers Rs. 25,000 and a category ‘A’ worker Rs. 65,000. The number of workers earning less than the mean salary is

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

Total amount of money spent as salary = Salary for category A worker + Salary for category B workers + Salary for category C workers
⇒ 65000 + 3 × 25000 + 5 × 20000 = 240000

Average salary = Total salary/Total number of workers
⇒ 240000/9 = 26666. 67

All B and C category workers get salary less than mean salary

∴ The number of workers earning less than the mean salary = 5 + 3 = 8

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

A man travelled 12 km at a speed of 4 km/h and further 10 km at a speed of 5 km/hr. What was his average speed?

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

Total time taken = Time taken at a speed of 4 km/h + Time taken at a speed of 5 km/h

⇒ 12/4 + 10/5 = 5 hours [∵ Time = Distance/Speed]

Average speed = Total distance/Total time
⇒ (12 + 10) /5 = 22/5 = 4.4 km/h

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

The pie diagrams on the monthly expenditure of two families A and B are drawn with radii of two circles taken in the ratio 16 : 9 to compare their expenditures.

Q. Which one of the following is the appropriate data used for the above mentioned pie diagram?

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

In pie chart, Data ∝ Area
Area = πr2
∴ Income ∝ Radius2
Income of A/Income of B = (16/9)2 = 256/81

In option (3)
Income of A/Income of B = 25600/8100 = 256/81
∴ The appropriate data is Rs. 25,600 and Rs. 8,100

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

Consider the following statements:

Statement I:  The value of a random variable having the highest frequency is mode

Statement II: Mode is unique

Q. Which one of the following is correct in respect to the above statements? 

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

Mode is defined as the data with maximum frequency. In case two or more data have same frequencies, then both are mode

∴ Mode is not unique
For example, If the data set is 13, 14, 14, 15, 15 and 16
Both 14 and 15 are modes
∴ Statement I is true but Statement II is false

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

Which one of the following is not correct?

The proportion of various items in a pie diagram is proportional to the 

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

Data in a pie chart ∝ Area
Area = θ/360 × πr2
∴ Area ∝ θ
∴ Data ∝ θ, where θ is the angle of slice
Length of the curved surface = θ/360 × 2πr
∴ Length of curved surface of arc ∝ θ
∴ Data ∝ Length of the curved surface
Perimeter of slices = θ/360 × 2πr + 2r
∴ Perimeter of slices is not proportional to the data

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

The geometric mean of x and y is 6 and the geometric mean of x, y and z is also 6. Then the value of z is

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

Geometric mean of n numbers = (Product of numbers)1/n
⇒ Geometric mean of x and y = (xy)1/2
⇒ 6 = (xy)1/2
⇒ Squaring on both the sides
⇒ 36 = xy
Geometric mean of x, y and z = (xyz)1/3
⇒ 6 = (xyz)1/3
Cubing on both the sides
⇒ 216 = xyz
⇒ 216 = 36z
∴ z = 216/36 = 6

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

The total number of live births in a specific locality during different months of a specific year was obtained from the office of the Birth registrar. This set of data may be called

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

The data required was not collected directly from source but from others

∴ It is secondary data

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

The heights (in cm) of 5 students are 150, 165, 161, 144 and 155. What are the values of mean and median (in cm) respectively?

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

On arranging the heights in ascending order
144, 150 155, 161 and 165
Median = Middle most term = 155
Sum of the heights = 144 + 150 + 155 + 161 + 165 = 775
Mean = Sum/Number of observations = 775/5 = 155
∴ Mean and median are 155 and 155 respectively

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

The average height of 22 students of a class is 140 cm and the average height of 28 students of another class is 152 cm. What is the average height of students of both classes?

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

The difference in the averages of the first class and second class = 152 – 140 = 12

This difference from 28 students will be now distributed to all students

Increase in average = 28 × Difference/Number of students

⇒ 28 × 12/ (28 + 22) = 6.72
∴ New average = 140 + 6.72 = 146.72 cm

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

To maintain 8 cows for 60 days, a milkman spends Rs. 6400. To maintain 5 cows for n days, he spends Rs. 4800. What is the value of n?

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

Amount spent for maintaining 8 cows for 60 days = Rs. 6400

⇒ Amount spent for maintaining 8 cows for 1 day = 6400/60 = Rs. 1600/15

⇒ Amount spent for maintaining 1 cow for 1 day = (1600/15) /8 = Rs. 200/15

⇒ Amount spent for maintaining 5 cows for 1 day = 200/15 × 5 = 200/3

⇒ Amount spent for maintaining 5 cows for n days = 200n/3 = Rs. 4800

⇒ 4800 = 200n/3
∴ n = 4800 × 3/200 = 72 days

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

A student secure 40% of the marks to pass an examination. He gets only 45 marks and fails by 5 marks. The maximum marks are

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

Total marks required for passing = 45 + 5 = 50 marks
50 marks is 40% of the maximum marks
⇒ 50 = 40/100 × Maximum marks
∴ Maximum mark = 50 × 100/40 = 125 marks

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

What is the value of u in the system of equations 3 (2u + v) = 7uv, 3 (u + 3v) = 11uv?

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

3 (2u + v) = 7uv
On dividing by uv
⇒ 6/v + 3/u = 7       ---- 1
3 (u + 3v) = 11uv
On dividing by uv
⇒ 3/v + 9/u = 11     ---- 2
2 × Equation 2 – Equation 1
⇒ 6/v + 18/u – 6/v – 3/u = 22 – 7
⇒ 15/u = 15
∴ u = 15/15 = 1

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

Five years ago, Ram was three times as old as Shyam. Four years from now, Ram will be only twice as old as Shyam. What is the present age of Ram?

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

Shortcut:
Let Ram’s age be ‘x’
His age five years ago = 3 × (Shyam’s age five years ago)
∴ (x – 5) should be a multiple of 3
Only option that satisfies this condition is x = 32

Detailed solution:
Let the age of Ram’s and Shyam’s 5 years ago from present be x and y respectively
Then, Ram’s age was thrice as Shyam’s
⇒ x = 3y

Four years after from present, their age will x + 9 and y + 9

Now Ram’s age is twice the Shyam’s age
⇒ x + 9 = 2 × (y + 9)
⇒ 3y + 9 = 2y + 18 [∵ x = 3y]
∴ y = 9 and x = 3 × 9 = 27
∴ Ram’s present age = 27 + 5 = 32 years

View more questions
56 docs|18 tests
Information about CDS II - Mathematics Previous Year Question Paper 2017 Page
In this test you can find the Exam questions for CDS II - Mathematics Previous Year Question Paper 2017 solved & explained in the simplest way possible. Besides giving Questions and answers for CDS II - Mathematics Previous Year Question Paper 2017 , EduRev gives you an ample number of Online tests for practice

Top Courses for CDS

Download as PDF

Top Courses for CDS