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

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 = 

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)

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)

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

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

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

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

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

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

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

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

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

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

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

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

3^x = 4^y = 12^z
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)

(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

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

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

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

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

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

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

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

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)

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

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

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