CDS II - Mathematics Previous Year Question Paper 2016 - CDS MCQ

Let the roots be pt and qt respectively
Sum of roots = -b/a
(p + q)t = -m/l
Product of roots = c/a
pqt² = m/l

⇒ [(p + q) t] / [√(pqt²)] + √(m/l)
⇒ (-m/l) / [√(m/l)] + √(m/l)
∴ -√(m/l) + √(m/l) = 0

Let 3x² - 7x – 30 = t² and 2x² - 7x - 5 = u²
t² - u² = x² – 25          ---- (1)
t - u = x – 5        ---- (2) [∵ Given]
Divide equation 1 by 2
t + u = x + 5           --- (3) [∵ (a² – b²) / (a – b) = a + b]
Add equation 2 and 3
2t = 2x
∴ t = x
t² = x²
3x² - 7x – 30 = x²
2x² - 7x – 30 = 0
Product of roots = c/a = -30/2 = -15

p/x + q/y = m
∴ (py + qx) /xy = m
py + qx = mxy       ---- (1)
q/x + p/y = n
∴ qy + px = nxy      ---- (2)
Divide equation 1 by 2
(py + qx) / (qy + px) = m/n
Dividing the numerator and denominator by y
[p + q(x/y)] / [q + p(x/y)] = m/n
Let x/y = t
(p + qt)/(q + pt) = m/n
Cross multiplying
np + nqt = mq + mpt
np – mq = t(mp - nq)
∴ t = (np - mq) / (mp - nq) 

a² - by - cz = 0       ---- (1)
ax - b² + cz = 0      ---- (2)
ax + by - c² = 0      ---- (3)

Add equation 2 and 3 and subtract 1 from it
2ax + 2by + 2cz - a² - b² - c² = 0
a² + b² + c² - 2ax - 2by - 2cz = 0
Adding x² + y² + z² on both the sides
a² + b² + c² - 2ax - 2by - 2cz + x² + y² + z² = x² + y² + z²
(a² - 2ax + x²) + (b² - 2by + y²) + (c² - 2cz + z²) = x² + y² + z²
(a - x)² + (b - y)² + (c - z)² = x² + y² + z²
Solution of this equation
x = a/2, y = b/2, z = c/2
or a = 2x, b = 2y and c = 2z
x/(a + x) + y/(b + y) + z/(c + z) = x/(2x + x) + y/(2y + y) + z/(2z + z)
⇒ 1/3 + 1/3 + 1/3 = 1

Let the common root be t
Then t² - pt + q = 0        ---- (1)
t² + qt - p = 0       ---- (2)
Subtract equation 1 from 2
(q + p)t - (q + p) = 0
∴ t = 1
Substituting in equation 1
1 - p + q = 0
∴ p - q - 1 = 0

Let a = 2^1/3 and b = 2^-1/3
∴ a³ = 2 and b³ = ½
∴ x = a + b
Cubing on both sides

x³ = a³ + b³ + 3ab(a + b) [∵ (a + b)³ = a³ + b³ + 3ab(a + b)]
x³ = 2 + ½ + 3 × 2^1/3 × 2^-1/3 × x [∵ x = a + b]
x³ = 5/2 + 3x
2x³ = 5 + 6x
∴ 2x³ - 6x - 5 = 0

Let the two expressions be P and Q
P + Q = 5x² - x - 4
P - Q = x² + 9x - 10

Adding two equations 2P = 6x² + 8x -14
∴ P = 3x² + 4x - 7
⇒ 3x² + 7x - 3x - 7
⇒ x(3x + 7) - 1(3x + 7)
∴ (x - 1) (3x + 7)

Subtracting two equations
Q = 2x² - 5x + 3
⇒ 2x² - 3x - 2x + 3
⇒ x(2x - 3) - 1(2x - 3)
⇒ (x - 1) (2x - 3)
∴ HCF = (x - 1)

(s - a) + (s - b) + (s - c) = s

Rearranging the terms

2s = a + b + c

[(s - a)² + (s - b)² + (s - c)² + s²] / [(a² + b² + c²)]

Adding and subtracting (a² + b² + c²) in the numerator

[(s - a)² + (s - b)² + (s - c)² + s² - (a² + b² + c²) + (a² + b² + c²)] / (a² + b² + c²)

⇒ [(s - a)² + (s - b)² + (s - c)² + s² - (a² + b² + c²)] / (a² + b² + c²) + (a² + b² + c²) / (a² + b² + c²)

Rearranging the terms

⇒ {[(s - a)² - a²] + [(s - b)² - b²] + [(s - c)² - c²] + s²} / {(a² + b² + c²)} + 1

⇒ [(s² - 2as + s² - 2bs + s² - 2cs + s²)] / [(a² + b² + c²)] + 1

⇒ [4s² - 2s(a + b + c)] / [(a² + b² + c²)] + 1

⇒ [(4s² - 2s × 2s)] / [(a² + b² + c²)] + 1 [∵ 2s = a + b + c]

⇒ 0 + 1 = 1

x⁶ + px⁵ + qx⁴ - x² - x - 3 = x⁶ - x² + px⁵ - x + qx⁴ - 3
⇒ x² (x⁴ - 1) + x (px⁴ - 1) + 3 [(q/3)x⁴ - 1]
For it to be a multiple of x⁴ - 1
p = 1 and q = 3
Then the expression becomes
x² (x⁴ - 1) + x (x⁴ - 1) + 3 (x⁴ - 1)
⇒ (x⁴ - 1) (x² + x + 3)
∴ p² + q² = 1 + 9 = 10

If (x - m) and (x - km) are the factors of the polynomial, then m and km are roots of the polynomial

Sum of roots = -b/a
m + km = -p
m(k + 1) = -p       ---- (1)
Product of roots = c/a
m × km = q
km² = q       ---- (2)
Divide equation (2) by square of equation (1)
(k + 1)²/k = p²/q
∴ (k + 1)²q = kp²

xⁿ + mⁿ = xⁿ – mⁿ + 2mⁿ
For any positive integer ‘n’, xⁿ – yⁿ is divisible by x – y
The remainder when xⁿ + mⁿ is divided by (x – y), R = 2mⁿ
∴ R is an even number

4^x2^y = 128
2^2x 2^y = 128 [∵ 4 = 2²]
2^{(2x + y)} = 2⁷ [∵ a^m × aⁿ = a^{(m + n)} and 128 = 2⁷]
Equating the powers
2x + y = 7       ---- (1)
3^3x 3^2y - 9^xy = 0
3^3x 3^2y - 3^2xy = 0 [∵ 9 = 3²]
3^{3x + 2y} = 3^2xy
Equating powers
3x + 2y = 2xy
3x = 2xy - 2y
3x = 2y (x - 1)
∴ y = 3x/[2(x - 1)]
Substituting in equation 1
2x + 3x/[2(x - 1)] = 7
[(4x² - 4x + 3x)] / [2(x - 1)] = 7
∴ 4x² - x = 14x - 14
4x² - 15x + 14 = 0
4x² - 7x - 8x + 14 = 0
x(4x - 7) - 2(4x - 7) = 0
(x - 2) (4x - 7) = 0
∴ x = 2 or x = 7/4
Substituting in equation 1
y = 3 or y = 7/2
∴ x + y = 2 + 3 = 5 or 7/4 + 7/2 = 21/4
From the given option x + y = 5 is the right choice

ax² - (a² + 1)x + a = ax² - a²x - x + a
⇒ ax(x - a) - 1(x - a)
⇒ (x - a) (ax - 1)
∴ The linear factors are (x - a) and (ax - 1)
Sum of the factors = x - a + ax - 1
⇒ x - 1 + ax - a
⇒ (x - 1) + a (x - 1)
∴ (x - 1) (1 + a) 

Multiplying by [√(a + 2b) + √(a - 2b)] in the numerator and denominator
⇒ [√(a + 2b) + √(a - 2b)]² / [a + 2b - a + 2b] [∵ (u + v) (u - v) = u² - v²]
⇒ [a + 2b + a - 2b + 2 √(a² - 4b²)]/4b [∵ (u + v)² = u² + v² + 2uv]
⇒ [2a + 2√(a² - 4b²)]/4b
⇒ [a + √(a² - 4b²)]/2b
∴ 2bx = [a + √(a² - 4b²)]
2bx - a = √(a² - 4b²)
Squaring on both the sides
4b²x² + a² - 4abx = a² - 4b² [∵ (u + v)² = u² + v² + 2uv]
4b²x² - 4abx = - 4b²
Divide both sides by 4b
bx² - ax = -b
∴ bx² - ax + b = 0

a³ - b³ = 117 and a - b = 3
(a - b)³ = a³ - b³ - 3ab(a - b)
3³ = 117 - 3ab × (3)
27 = 117 - 9ab
9ab = 90
∴ ab = 10
(a + b)² = (a - b)² + 4ab
⇒ (a + b)² = 3² + 4 × 10 = 49
∴ (a + b)² = 49
a + b = 7

Let the roots be α and β
Sum of roots = -b/a
α + β = -b/a
Product of the roots = c/a
αβ = c/a
Sum of square of their reciprocal = 1/α² + 1/β²
⇒ (α² + β²)/(α²β²)
⇒ [(α + β)² - 2αβ]/(α²β²)
Sum of roots = Sum of square of their reciprocal
α + β = [(α + β)² - 2αβ]/(α²β²)
α²β² (α + β) = (α + β)² - 2αβ
(c/a)² × (-b/a) = (-b/a)² - 2c/a
-bc²/a³ = b²/a² - 2c/a
-bc²/a³ = (b² - 2ac)/a²
-bc² = [a(b² - 2ac)]
Rearranging the terms
∴ ab² + bc² = 2a²c

STATEMENT I:
n(n + 1)/2 = 861
n(n + 1) = 1722
n² + n - 1722 = 0
n² + 42n - 41n - 1722 = 0
n(n + 42) - 41(n + 42) = 0
(n - 41) (n + 42) = 0
∴ n = 41 or -42

There are two integer values of n for which S_n = 861
∴ Statement I is true

STATEMENT II:
S_n = S_{-(n + 1)} and hence for any integer m, we have two values of n for which S_n = m

It is false, because not for all integral values of m, there exists an integral value of n

For example, If m = 5
n(n + 1)/2 = 5
n² + n - 10 = 0
Finding roots using quadratic equation
n = (1 ± √41)/2, not an integer

p + q < p - q
q < - q
2q < 0
∴ q is negative
Statement I is true
p + q > p - q
2q > 0
∴ Only q must be positive
p can be either positive or negative

For example
Let p = -5 and q = 3
p + q = -2
p - q = -8
-2 > -8
∴ Statement II is false

Let a/b = b/c = c/d = k
a = bk
b = ck
∴ a = ck²
c = dk
b = dk² and a = dk³

Statement I:
(b³ + c³ + d³) / (a³ + b³ + c³) = (d³k⁶ + d³k³ + d³) / (d³k⁹ + d³k⁶ + d³k³)
⇒ [d³ × (k⁶ + k³ + 1)] / [d³k³ × (k⁶ + k³ + 1)]
⇒ 1/k³
⇒ d/(dk³) = d/a [∵a = dk³]
∴ Statement I is true

Statement II:
(a² + b² + c²) / (b² + c² + d²) = (d²k⁶ + d²k⁴ + d²k²) / (d²k⁴ + d²k² + d²)
⇒ [d²k² × (k⁴ + k² + 1)] / [d²× (k⁴ + k² + 1)]
⇒ k² = (dk²)/d = b/d
∴ Statement II is wrong

7¹⁰ - 5¹⁰ = (7⁵ + 5⁵) (7⁵ - 5⁵) [∵ (a² - b²) = (a + b) (a - b)]
7⁵ = 343 × 49 = 16807
5⁵ = 625 × 5 = 3125
7⁵ + 5⁵ = 16807 + 3125 = 19932
Sum of odd place digits = 2 + 9 + 1 = 12
Sum of even place digits = 9 + 3 = 12
Difference = 12 - 12 = 0
∴ The number is divisible by 11

Let the unit and ten’s digits be x and y respectively
The number is k times the sum of digits
10x + y = k(x + y)
(10 - k)x = (k - 1)y      ---- (1)
Reversed number is m times the sum of digits
10y + x = m(x + y)
(1 - m)x = (m - 10)y      ---- (2)
Divide equation 2 by 1
(10 - k) / (1 - m) = (k - 1) / (m - 10)
(10 - k) (m - 10) = (1 - m) (k - 1)
10m + 10k - 100 - km = -1 + m + k - km
10m - m + 10k - k = 100 - 1
9(m + k) = 99
m + k = 11
∴ m = 11 - k 

Relative displacement of the train = Length of the train
Relative speed = Relative distance/Time
⇒ 225/9 = 25 m/s = 25 × 18/5 = 90 km/hr
Relative speed = Speed of train + Speed of man
90 = Speed of train + 5
∴ Speed of train = 90 - 5 = 85 km/hr

sin θ = cos (90 - θ)
∴ sin 35° = cos (90 - 35) = cos 55°
∴ sin 35°/cos 55° = cos 55°/sin 35° = 1
(sin 35°/cos 55°)² - (cos 55°/sin 35°)² + 2 sin 30° = 1² - 1² + 2 × ½ = 1

Let the original average speed be x
∴ Reduced speed = x - 1

Difference in time = Time taken at average speed (x - 1) - Time taken at average speed x

20/60 = 14/(x - 1) - 14/x [∵ Time = Distance/Speed]
1/3 = 14 × [1/(x - 1) - 1/x]
1/3 = 14/[x(x - 1)]
x(x - 1) = 42
x² - x - 42 = 0
x² - 7x + 6x - 42 = 0
x(x - 7) + 6(x - 7) = 0
(x + 6) (x - 7) = 0
x cannot be negative
∴ x = 7 km/hr

Interest/Principal = Rate/100 × Time
In first case, Amount gets double
∴ Interest = Principal
1 = Rate/100 × 5
∴ Rate = 20%
In second case, amount gets tripled
∴ Interest = 2 × Principal
2 = Rate/100 × 12
∴ Rate = 50/3%
Difference = 20 - 50/3 = 10/3%

In a polygon, Exterior angle = 180° – Interior angle
⇒ Exterior angle = 180 – 140 = 40°

Sum of exterior angles = Number of sides × Value of each exterior angle
⇒ 360 = n × 40 [ ∵ Sum of exterior angles of any polygon is 360°]
∴ Number of sides, n = 360/40 = 9

Statement I:
In triangles ABC and DEF
∠ABC = ∠DEF and ∠ACB = ∠DFE [ ∵ Given]
⇒ ABC ∼ DEF [ ∵ AA ∼]
⇒ AB/DE = BC/EF
⇒ AB/DE = (BC/2)/(EF/2)
⇒ AB/DE = BL/EM [ ∵ L and M are midpoints of BC and EF respectively]
In triangle ABL and DEM
AB/DE = BL/DM [Proved]
∠ABC = ∠DEF [Given]
⇒ ABL ∼ DEM [ ∵ SAS ∼]
⇒ Statement I is true

Statement II:
For any two triangles to be congruent, their corresponding sides have to be equal
⇒ Statement II is false
∴ Only Statement I is true but statement II is false.

Distance traversed = Number of rotations × Perimeter of the wheel
⇒ 4000 = Number of rotations × πd [ ∵ 1 km = 1000m]
⇒ 4000 = Number of rotations × 22/7 × 7/11
⇒ 4000 = 2 × Number of rotations
∴ Number of rotations = 4000/2 = 2000

Sides of the triangle a = 170, b = 170 and c = 300
⇒ Semi-perimeter, s = (a + b + c) /2 = (170 + 170 + 300) /2 = 320 units
Using heron’s formula to calculate area
Area = √[s (s – a) (s – b) (s – c)]
⇒ √[320 × (320 – 170) × (320 – 170) × (320 – 300)
⇒ √(320 × 150 × 150 × 20)
⇒ √(6400 × 150 × 150)
∴ Area = 80 × 150 = 12000 square units

Let the radius be r
Area of empty space = Area of the square – 4 × Area of quadrant
⇒ Side of square = 2 × radius = 2r
⇒ 150/847 = Side² – 4 × ¼ × π r²
⇒ 150/847 = (2r)² – 22/7 r²
⇒ 150/847 = r² (4 – 22/7)
⇒ 150/847 = 6r²/7
⇒ r² = 25/121
∴ radius, r = 5/11 cm