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


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30 Questions MCQ Test CDS (Combined Defence Services) Mock Test Series 2024 - CDS II - Mathematics Previous Year Question Paper 2016

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CDS II - Mathematics Previous Year Question Paper 2016 - Question 1

If the roots of the equation lx2 + mx + m = 0 are in the ratio p ∶ q, then   is equal to

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 1

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 2

If √(3x2 - 7x - 30) - √(2x2 - 7x - 5) = x - 5 has α and β as its roots, then the value of αβ is

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 2

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

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CDS II - Mathematics Previous Year Question Paper 2016 - Question 3

If p/x + q/y = m and q/x + p/y = n, then what is x/y equal to?

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 3

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) 

CDS II - Mathematics Previous Year Question Paper 2016 - Question 4

If a2 - by - cz = 0, ax - b2 + cz = 0 and ax+ by - c2 = 0, then what is the value of x/(a + x) + y/(b + y) + z/(c + z) will be

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 4

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 5

If the equations x2 - px + q = 0 and x2 + qx - p = 0 have a common root, then which one of the following is correct?

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 5

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 6

If x = 21/3 + 2-1/3, then the value of 2x3 - 6x - 5 is equal to 

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 6

Let a = 21/3 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 × 21/3 × 2-1/3 × x [∵ x = a + b]
x3 = 5/2 + 3x
2x3 = 5 + 6x
∴ 2x3 - 6x - 5 = 0

CDS II - Mathematics Previous Year Question Paper 2016 - Question 7

The sum and difference of two expressions are 5x2 - x - 4 and x2 + 9x - 10 respectively. The HCF of the two expressions will be

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 7

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)

CDS II - Mathematics Previous Year Question Paper 2016 - Question 8

If (s - a) + (s - b) + (s - c) = s, then the value of [(s - a)2 + (s - b)2 + (s - c)2 + s2] / (a2 + b2 + c2) will be

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 8

(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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 9

If the polynomial x6 + px5 + qx4 - x2 - x - 3 is divisible by (x4 - 1), then the value of p2 + q2 is 

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 9

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 10

Let p and q be non-zero integers. Consider the polynomial A(x) = x2 + px + q. It is given that (x - m) and (x - km) are simple factors of A(x), where m is a non-zero integer and k is positive integer, k ≥ 2. Which one of the following is correct?

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 11

Let m be a non-zero integer and n be a positive integer. Let R be the remainder obtained on dividing the polynomial xn + mn by (x – m). Then

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 11

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 12

If 4x2y = 128 and 33x 32y – 9xy = 0, then the value of x + y can be equal to 

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 12

4x2y = 128
22x 2y = 128 [∵ 4 = 22]
2(2x + y) = 27 [∵ am × an = a(m + n) and 128 = 27]
Equating the powers
2x + y = 7       ---- (1)
33x 32y - 9xy = 0
33x 32y - 32xy = 0 [∵ 9 = 32]
33x + 2y = 32xy
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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 13

If the linear factors of ax2 - (a2 + 1) x + a are p and q, then p + q is equal to

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 13

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) 

CDS II - Mathematics Previous Year Question Paper 2016 - Question 14

If x = [√(a + 2b) + √(a - 2b)]/[√(a + 2b) - √(a - 2b)], then bx2 - ax + b is equal to (given that b ≠ 0)

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 14

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 15

If a3 = 117 + b3 and a = 3 + b, then the value of a + b is (given that a > 0 and b > 0)

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 15

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 16

If the sum of the roots of ax2 + bx + c = 0 is equal to the sum of the squares of their reciprocals, then which one of the following relation is correct?

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 16

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 17

Consider the following statements in respect of the expression Sn = n(n + 1)/2
where n is an integer
I. There are exactly two values of n for which Sn = 861
II. Sn = S-(n + 1) and hence for any integer m, we have two values of n for which Sn = m

Q. Which of the above are not perfect squares?

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 17

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 Sn = 861
∴ Statement I is true

STATEMENT II:
Sn = S-(n + 1) and hence for any integer m, we have two values of n for which Sn = 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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 18

Consider the following statements in respect of two different non-zero integers p and q:
I. For (p + q) to be less than (p - q), q must be negative
II. For (p + q) to be greater than (p - q), both p and q must be positive

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

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 18

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 19

If a/b = b/c = c/d, then which of the following is/are correct?
I. (b3 + c3 + d3) / (a3 + b3 + c3) = d/a
II. (a2 + b2 + c2) / (b2 + c2 + d2) = a/d

Select the correct answer using the code given below.

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 19

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 20

710 - 510 is divisible by

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 20

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 21

Let a two digit number be k times the sum of its digits. If the number formed by interchanging the digits is m times the sum of the digits, then the value of m is

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 21

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 

CDS II - Mathematics Previous Year Question Paper 2016 - Question 22

A man walking at 5 km/hr noticed that a 225 m long train coming in the opposite direction crossed him in 9 seconds. The speed of the train is

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 22

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 23

(sin 35°/cos 55°)2 - (cos 55°/sin 35°)2 + 2sin 30° is equal to 

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 23

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 24

A cyclist moves non-stop from A to B, a distance of 14 km, at a certain average speed. If his average speed reduces by 1 km per hour, he takes 20 minutes more to cover the same distance. The original average speed of the cyclist is

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 24

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
x- 7x + 6x - 42 = 0
x(x - 7) + 6(x - 7) = 0
(x + 6) (x - 7) = 0
x cannot be negative
∴ x = 7 km/hr

CDS II - Mathematics Previous Year Question Paper 2016 - Question 25

If a sum of money at a certain rate of simple interest per year doubles in 5 years and at a different rate of simple interest per year becomes three times in 12 years, then the difference in the two rates of simple interest per year is 

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 25

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%

CDS II - Mathematics Previous Year Question Paper 2016 - Question 26

If each interior angle of a regular polygon is 140°, then the number of vertices of the polygon is equal to

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 26

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 27

Let the triangles ABC and DEF be such that ∠ABC = ∠DEF, ∠ACB = ∠DFE and ∠BAC = ∠EDF. Let L be the midpoint of BC and M be the point of EF. Consider the following statements:

Statement I: Triangle ABL and DEM are similar
Statement II: Triangle ALC is congruent to triangle DMF even if AC ≠ DF

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

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 27

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.

CDS II - Mathematics Previous Year Question Paper 2016 - Question 28

The number of rounds that a wheel of diameter 7/11 meter will make in traversing 4 km will be

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 28

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 29

The base of an isosceles triangle is 300 units and each of its equal sides 170 units. Then area of the triangle is

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 29

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

CDS II - Mathematics Previous Year Question Paper 2016 - Question 30

Four equal discs are placed such that each one touches two others. If the area of empty space enclosed by them is 150/847 square centimeter, then the radius of each disc is equal to

Detailed Solution for CDS II - Mathematics Previous Year Question Paper 2016 - Question 30


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

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