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CDS I - Mathematics Previous Year Question Paper 2019 - 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 2019

CDS I - Mathematics Previous Year Question Paper 2019 for CDS 2024 is part of CDS (Combined Defence Services) Mock Test Series 2024 preparation. The CDS I - Mathematics Previous Year Question Paper 2019 questions and answers have been prepared according to the CDS exam syllabus.The CDS I - Mathematics Previous Year Question Paper 2019 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 2019 below.
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CDS I - Mathematics Previous Year Question Paper 2019 - Question 1

Direction: Consider the following for the next three (03) items :
A cube is inscribed in a sphere. A right circular cylinder is within the cube touching all the vertical faces. A right circular once is inside the cylinder. Their heights are same and the diameter of the cone is equal to that of the cylinder.

Q. What is the ratio of the volume of the sphere to that of cone?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 1


The top view of the given assembly will look like the figure above
Outermost is the sphere. Inside that there is a cube and within that there is a cone and cylinder with same radius.
Here side of cube = a
Diameter of Sphere = body diagnol = √3 a
Radius of sphere = √3 a/2 = r1
Height of Cylinder = Height of cone = side of cube = a = h
Radius of cylinder = Radius of cone = side of cube/2 = a/2 = r2 (as shown in the figure)
Volume of sphere/volume of cone = 

CDS I - Mathematics Previous Year Question Paper 2019 - Question 2

Direction: Consider the following for the next three (03) items :
A cube is inscribed in a sphere. A right circular cylinder is within the cube touching all the vertical faces. A right circular once is inside the cylinder. Their heights are same and the diameter of the cone is equal to that of the cylinder.​
Q. What is the ratio of the volume of the cube to that of the cylinder ?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 2


The top view of the given assembly will look like the figure above
Outermost is the sphere. Inside that there is a cube and within that there is a cone and cylinder with same radius.
Here side of cube = a Diameter of Sphere = body diagnol = √3 a
Radius of sphere = √3 a/2 = r1
Height of Cylinder = Height of cone = side of cube = a = h
Radius of cylinder = Radius of cone = side of cube/2 = a/2 = r2 (as shown in the figure)

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CDS I - Mathematics Previous Year Question Paper 2019 - Question 3

Direction: Consider the following for the next three (03) items :
A cube is inscribed in a sphere. A right circular cylinder is within the cube touching all the vertical faces. A right circular once is inside the cylinder. Their heights are same and the diameter of the cone is equal to that of the cylinder.
1) The surface area of the sphere is √5
Times the curved surface area of the cone.
2) The surface area of the cube is equal to the curved surface area of the cylinder.
Q. Which of the above statements is/are correct?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 3


The top view of the given assembly will look like the figure above
Outermost is the sphere. Inside that there is a cube and within that there is a cone and cylinder with same radius.
Here side of cube = a
Diameter of Sphere = body diagnol = √3 a
Radius of sphere = √3 a/2 = r1
Height of Cylinder = Height of cone = side of cube = a = h
Radius of cylinder = Radius of cone = side of cube/2 = a/2 = r2 (as shown in the figure)

CDS I - Mathematics Previous Year Question Paper 2019 - Question 4

Direction: Consider the following for the next three (03) items:
ABCD is a quadrilateral with AB = 9 cm, BC = 40 cm, CD = 28 cm, DA = 15 cm and angle ABC is a right – angel 
Q. What is the area of triangle ADC?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 4


CDS I - Mathematics Previous Year Question Paper 2019 - Question 5

Direction: Consider the following for the next three (03) items:
ABCD is a quadrilateral with AB = 9 cm, BC = 40 cm, CD = 28 cm, DA = 15 cm and angle ABC is a right – angel
Q. What is the area of quadrilateral ABCD?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 5


Area of quadrilateral ABCD = area of triangle ADC + area of triangle ABC
= 126 + ½ * 9 * 40 = 306 cm2

CDS I - Mathematics Previous Year Question Paper 2019 - Question 6

Direction: Consider the following for the next three (03) items:
ABCD is a quadrilateral with AB = 9 cm, BC = 40 cm, CD = 28 cm, DA = 15 cm and angle ABC is a right – angel
Q. What is the difference between perimeter of triangle ABC and perimeter of triangle ADC?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 6


Perimeter of triangle ABC – Perimeter of triangle ADC = (9+40+41)-(15+28+41) = 6cm

CDS I - Mathematics Previous Year Question Paper 2019 - Question 7

Direction: Consider the following for the next two (02) items:
An equilateral triangle ABC is inscribed in a circle of radius 20√3
Q. What is the length of the side of the triangle?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 7


Radius of circumcircle of an equilateral triangle = side / √3  
R = a/√3  
a = R√3  = 20√3  * √3  = 60cm

CDS I - Mathematics Previous Year Question Paper 2019 - Question 8

Direction: Consider the following for the next two (02) items:
An equilateral triangle ABC is inscribed in a circle of radius 20√3
Q. The centroid of the triangle ABC is at a distance d from the vertex A. What is d equal to?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 8


For equilateral triangle circumcenter and centroid are the same points
So distance from vertex = radius of circumcircle = 20√3

CDS I - Mathematics Previous Year Question Paper 2019 - Question 9

Direction: Consider the following for the next two (02) items:
The sum of length, breadth and height of a cuboid is 22 cm and the length of its diagonal is 14 cm.
Q. What is the surface area of the cuboid?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 9

Let lengths, breadth and height of cuboid be l, b and h respectively
According to question l+b+h = 22cm……(i)
and √(l2+b2+h2) = 14cm …..(ii)
Surface area of cuboid = 2(lb+bh+lh)
Squaring eq (i) gives
l2+b2+h2 + 2(lb+bh+lh) = 484
Substituting l2+b2+h2 from eq (i)
2(lb+bh+lh) = 484-196 = 288 cm2

CDS I - Mathematics Previous Year Question Paper 2019 - Question 10

Direction: Consider the following for the next two (02) items:
The sum of length, breadth and height of a cuboid is 22 cm and the length of its diagonal is 14 cm.
Q. If S is sum of the cubes of the dimensions of the cuboid and V is its volume, then what is (S-3V) equal to?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 10

Let lengths, breadth and height of cuboid be l, b and h respectively
According to question
l+b+h = 22cm……(i)
and √(l2+b2+h2) = 14cm …..(ii)
S = l3+b3+h3 and  V = lbh
S-3V = l3+b3+h3 - 3 lbh = (l+b+h)( l2+b2+h2-[lb+bh+lh])…(iii)
As we know
Squaring eq (i) gives
l2+b2+h2 + 2(lb+bh+lh) = 484
Substituting l2+b2+h2 from eq (i)
2(lb+bh+lh) = 484-196 = 288 cm2
lb+bh+lh = 144 cm2
Putting this in eq (iii) we get
22(196-144) = 22*52 = 1144cm2

CDS I - Mathematics Previous Year Question Paper 2019 - Question 11

A race has three parts. The speed and time required to complete the individual parts for a runner is displayed on the following chart:​

Q. What is the average speed of this runner?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 11

Average speed = Total Distance / Total time 
 
= (45+64+75)/23 = 184/23 
= 8 kmph

CDS I - Mathematics Previous Year Question Paper 2019 - Question 12

If   then which one of the following statements is correct?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 12

a/(b+c) = b/(c+a) = c/(a+b)
Taking reciprocal and adding 1 to each ratio we get;
(b+c)/a + 1 = b/(c+a) + 1 = c/(a+b) + 1
Or (a+b+c)/a = (a+b+c)/b = (a+b+c)/c
So this can only be equal when a=b=c or a+b+c = 0
When a=b=c we get a/(b+c) = ½
When a+b+c = 0 we get b+c = -a
So a/(b+c) = -1
So the ratios are ½ or -1

CDS I - Mathematics Previous Year Question Paper 2019 - Question 13

The number 3521 is divided by 8. What is the remainder?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 13

3521/8 
As we know 32=9 will leave remainder = 1 when divided by 8
So 3521/8 = [(32)260 * 3]/8 = 1*3/8 = 3/8 Thus remainder is 3

CDS I - Mathematics Previous Year Question Paper 2019 - Question 14

A prime number contains the digit X at unit’s place. How many such digits of X are possible?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 14

For prime no units place cannot be occupied by even number except for 2 Thus no of digits occupying unit digit of prime numbers = 6 (1,2,3,5,7,9) Example 2,3,5,7,11,19 in itself are prime numbers

CDS I - Mathematics Previous Year Question Paper 2019 - Question 15

If an article is sold at a gain of 6% instead of a loss of 6% the seller gets Rs. 6 more. What is the cost price of the article?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 15

Let CP be Rs x
Then
1.06x – 0.94x = 6
So x = Rs 50

CDS I - Mathematics Previous Year Question Paper 2019 - Question 16

A field can be reaped by 12 men or 18 women in 14 days. In how many days can 8 men and 16 women reap it?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 16

12 men or 18 women can complete in 14 days
8 men and 16 women can complete in how many days
12men = 18 women (Comparing efficiencies)
1men = 18/12 = 1.5 women
8 men and 16 women = 12women + 16 women = 28 women
18 women completes in 14 days
1 woman completes in 14*18 days
28 women completes in (14*18)/28 days = 9 days

CDS I - Mathematics Previous Year Question Paper 2019 - Question 17

If 3x = 4y = 12z, then z is equal to 

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 17

3x = 4y = 12z
Taking log of all 3 we get
xln3 = yln4 = zln12 = k
z = k/ln12 = k / ln(3*4) = k/ln3 + ln4 = k / (k/x +k/y) = xy / (x+y)

CDS I - Mathematics Previous Year Question Paper 2019 - Question 18

If (4a +7b) (4c-7d) = (4a -7b) (4c +7d, then which one of the following is correct?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 18

(4a+7b)(4c-7d) = (4a-7b)(4c+7d)
(4a+7b)/(4a-7b) = (4c+7d)/(4c-7d)
Using componendo and dividendo
(4a+7b)+(4a-7b) / (4a+7b)-(4a-7b) = (4c+7d)+(4c-7d) / (4c+7d)-(4c-7d)
Or 8a/14b = 8c/14d
Or a/b = c/d

CDS I - Mathematics Previous Year Question Paper 2019 - Question 19

GiGiven that the polynomial (x2 + ax + b) leaves that same remainder when by (x – 1) or (x + 1) What are the values of a and b respectively?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 19

Since x2 + ax + b when divided by x-1 or x+1 leaves the same remainder
So on putting x=1 and x=-1 we get the same value
1+a+b = 1-a+b
2a=0
a=0
here b can take any value as it will always get cancelled out

CDS I - Mathematics Previous Year Question Paper 2019 - Question 20

Tushar takes 6 hours to complete a piece of work, while Amar completes the same work in 10 hours. If both of them work together, then what is the time required to complete the work?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 20

Let them take x hours working together
1/x = 1/10 + 1/6 = 8/30
X= 30/8 hours = 15/4 hours = 3hours 45 minutes

CDS I - Mathematics Previous Year Question Paper 2019 - Question 21

What is the value of 

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 21


2 + √t = t
Or t-2 = √t
Squaring both sides
t = t2 – 4t + 4
or t2 – 5t + 4 = 0
Or t = 4,1 Now t cannot be equal to 1 as it is clear that it is always greater than 2
So t = 4

CDS I - Mathematics Previous Year Question Paper 2019 - Question 22

In an examination, 52% candidates failed in English and 42% failed in Mathematics. If 17% failed in both the subjects, then what percent passed in both the subjects?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 22


No of students failed in English only = 52 – 17 = 35
No of students failed in maths only = 42 – 17 = 25
Total no of failed students in either of the subjects = 35+17+25 = 77
No of passed student in both subjects = 100 – 77 = 23

CDS I - Mathematics Previous Year Question Paper 2019 - Question 23

A man who recently died left a sum of Rs. 3,90,000 to be divided among his wife, five sons and four daughters. He directed that each son should receive 3 times as much as each daughter receives and that each daughter receives and that each daughter should receive twice as much as their mother receives. What was the wife’s share?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 23

Let his wife get a share of Rs x
Each of the 4 daughters get = Rs 2x
Each of the 5 sons get = Rs 6x
So x + 4*2x + 5*6x = 390000
So 39x = 390000
X= 10000 = wife’s share

CDS I - Mathematics Previous Year Question Paper 2019 - Question 24

What is the least number of complete years in which a sum of money put out at 40% annual compound interest will be more than trebled?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 24

A = P(1 + 1/R100)^t
3P < P(1 + 40/100)^t
3 < (1.4)^t
When t = 3 ; 1.4^3 = 2.744
And when t = 4; 1.4^4 = 3.8416
T=4 is the answer

CDS I - Mathematics Previous Year Question Paper 2019 - Question 25

A person divided a sum of Rs. 17,200 into three parts and invested at 5%. 6% and 9^ per annum simple interest. At the end of two years, he got the same interest on each part of money. What is the money invested at 9%?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 25

Let sum invested @ 5% be P1, @ 6% be P2 then @ 9% = 17200-(P1+P2)
So according to question
P1*5*2/100 = P2*6*2/100 or P1 = (6/5) P2
Also  P2*6*2/100 = [17200-(P1+P2)]*9*2/100
Or  2 P2 = [17200 – (11/5)P2] * 3
Or (2 + 33/5)P2 = 17200 * 3
P2 = 17200 * 3 * 5 / 43 = 6000
So P1 = 6/5 P2 = 7200
So Sum invested @ 9% = 17200 –(6000+7200) = Rs 4000

CDS I - Mathematics Previous Year Question Paper 2019 - Question 26

The corners of a square of side ‘a’ are cut away so as to form a regular octagon. What is the side of the octagon?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 26


Let side of hexagon be x
AE2 + AL2 = LE2 
Since we are forming a regular octagon so AE = AL = FB = BG and so on
So AE = SB = x/√2
AE + EF + FB = side of square = a (Given)
So x/√2 + x + x/√2 = a
X = a/(√2+1) = a(√2 - 1)

CDS I - Mathematics Previous Year Question Paper 2019 - Question 27

Three consecutive integers form the lengths of a right-angled triangle. How many sets of such three consecutive integers is/are possible?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 27

let n-1, n, n+1 be 3 consecutive integers
So
(n+1)2 = n2 + (n-1)2
(n+1)2- (n-1)2 = n2
4n = n2
So n = 0 or n = 4
n can’t be 0 as n-1 will be negative then
So 3,4 and 5 is the only triplet formed

CDS I - Mathematics Previous Year Question Paper 2019 - Question 28

Two circles are drawn with the same centre. The circumference of the smaller circle is 44 cm and that of the bigger circle is double the smaller one. What is the area between these two circles?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 28



 

CDS I - Mathematics Previous Year Question Paper 2019 - Question 29

A rectangular red carpet of size 6 ft × 12 ft has a dark red border 6 inches wide. What is the area of the dark red border?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 29

Initially carpet is 6×12 = 72 sq feet
Since red border is 6 inches wide from all 4 side
So area without border = 5 × 11 = 55 sq feet
Area of border = total – area without border = 72 – 55 = 17 sq feet

CDS I - Mathematics Previous Year Question Paper 2019 - Question 30

The perimeter of a right-angled triangle is k times the shortest side. If the ratio of the other side to hypotenuse is 4 : 6, then what is the value of k?

Detailed Solution for CDS I - Mathematics Previous Year Question Paper 2019 - Question 30

Let other side and hypotenuse be 4x and 5x respectively
Shortest side2 + (4x)2 = (5x)2
Shortest side = 3x
According to question
K*3x = 12x
So k = 4

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