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

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

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

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

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]

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

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

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

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

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

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° 

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

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

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%

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

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

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

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

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

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

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

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

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

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

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

∴ It is secondary data

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

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

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

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

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

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