CDS I - Mathematics Previous Year Question Paper 2018 - CDS

200 men can complete the quarter of work in 50 days

If they continue the work, time taken to complete the remaining work = 3 × 50 = 150 days
[∵ 3/4 th of the work is remaining]

But they have to complete in 100 days to finish in time
Time ∝ 1/Number of employees

Let the number of extra employees required to complete the work in time be x
∴ 150/100 = (200 + x)/200
300 = 200 + x
∴ x = 100

Speed of the train = 60 km/hr = 60 × 5/18 = 50/3 m/s

Distance travelled by the head of the train after it crosses the pole till the whole train crosses the pole = Length of the train
Distance = Speed × Time
50/3 × 30 = Length of the train
∴ Length of the train = 500 m

Let A’s share be Rs. x
∴ B’s share = x - 20 and C’s share = x + 20
Total amount = x + x - 20 + x + 20
120 = 3x
∴ x = 40
B’s share = x - 20 = Rs. 20

In inverse variation M × N = constant for all values of M and N

MN = 15 × -4 = -60
-6 × A = -60
∴ A = 10
2 × B = -60
∴ B = -30
C × 60 = -60
∴ C = -1

The extra amount he gains is due the difference in the interest rates.
Difference in the interest rate = 5.5 - 5 = 0.5%
Interest he gained = 5000 × 0.5/100 × 2 = Rs. 50 [∵ Interest = Principal × Rate/100 × Time]

Average age of father and son after 10 years = 25 + 10 = 35
Son’s age after 10 years = Son’s age after 7 years + 3
⇒ 17 + 3 = 20 years
Let the father’s age after 10 years be ‘x’
(x + 20) /2 = 35
x + 20 = 70
∴ x = 50 years

5 tractors can plough 5 hectares in 5 days
5 tractors can plough 50 hectares in 50 days
1 tractor can plough 10 hectares in 50 days

∴ Number of tractors needed to plough 100 hectares = 100/ (Number of hectares 1 tractor can plough in 50 days)
⇒ 100/10 = 10
∴ 10 tractors can plough 100 hectares in 50 days

Let the initial amount invested be Rs x and the amount is compounded at 25% rate

Amount at the end of n years = Principal × (1 + Rate/100)ⁿ
10000 = x × (1 + 25/100)³
10000 = x × 1.25³
10 × 10 × 10 × 10 = x × 1.25 × 1.25 × 1.25
8 × 8 × 8 × 10 = x [∵ 10/1.25 = 8]
x = Rs. 5120

459 = 17 × 27
Second option is terminating decimal, ∴ denominator has a prime factors of 2 or 5 only

We can evaluate the first decimal and other two decimals are obtained by dividing it by 10 and 100 respectively

Let x = 0.459459459⋯
1000x = 459.459459⋯
1000x = 459 + 0.459459⋯
1000x = 459 + x
∴ 999x = 459
x = 459/999 = (17 × 27) / (37 × 27) = 17/37

∴ 0.459459459⋯ is the decimal with 37 in the denominator when represented as rational number

The decrease in the income is due to the reduction in rate of interest.

Decrease in amount = Principal × Reduction in rate of interest/100 × Time
64 = Principal × 0.25/100 × 1
∴ Principal = Rs. 25600

If 0 < m < 1, log₁₀ m is negative so that is the least (All other terms will be positive)

For numbers less than 1, the term with lowest power is the largest
-1 < 1 < 2
∴ m² < m < m^-1
Final order = Log₁₀m < m² < m < m^-1

Let the amount received by the widow = Rs x
∴ Amount received by daughter = 2 × x = 2x
Amount received by son = 3 × 2x = 6x

Total amount = Amount received by widow + 5 × Amount received by sons + 4 × Amount received by daughter

39000 = x + 5 × 6x + 4 × 2x
39000 = x + 30x + 8x
39000 = 39x
∴ x = 1000
Widow receives Rs. 1000

Co-prime numbers have no prime factors in common so when expressed as fraction there will be no simplification

Let the numbers be A, B and C respectively
A × B = 286 and B × C = 770
A × B/ (B × C) = 286/770
A/C = 13/35
∴ A = 13 and C = 35
B = 286/A = 22
Sum of the numbers = 13 + 35 + 22 = 70

Let the two digits of the age of the women be y and x (yx)
Age of the women = 10y + x
Age of the men = 10x + y

Difference in their age = 10x + y - 10y - x
⇒ 9x - 9y

Sum of their ages = 10x + y + 10y + x = 11x + 11y
Difference = 1/11 × Sum
9x - 9y = 1/11 × (11x + 11y)
9x - 9y = x + y
8x = 10y

Since x and y can only be natural numbers from 1 to 9, the only possibility is x = 5 and y = 4
Difference in their age = 9x - 9y = 45 - 36 = 9 years

For crossing the relative displacement is same and equal to the sum lengths of the trains in both cases

∴ Time ∝ 1/Relative speed

Let the speed of the goods train be x and passenger train be kx, where k is required ratio

Relative speed when moving in opposite direction = kx - (-x) = (k + 1)x [Speeds are added]

Relative speed when moving in same direction = kx - x = (k - 1)x

(Time taken when moving in same direction) / (Time taken when moving in opposite direction) = (Relative Speed in opposite direction) / (Relative Speed in same direction)

3 = (k + 1)x / [(k - 1)x]
3 = (k + 1) / (k - 1)
3k - 3 = k + 1
2k = 4
k = 2
∴ The required ratio = 2 : 1

Any number multiplied with 5 ends with 5 if it is odd
Product of all odd prime number is an odd number where one of the factor is 5
∴ The unit digit of the product is 5

Let the amount of pure copper added be ‘x’ kg

Amount of copper in 20 kg of alloy A = 2/ (2 + 3) × 20 = 8 kg

Amount of copper in 28 kg of alloy B = 3/ (3 + 4) × 28 = 12 kg

Total amount of copper in alloy C = 8 + 12 + x = 20 + x

Total amount of alloy C produced = 20 + 28 + x = 48 + x
20 + x = 6/(6 + 7) × (48 + x)
20 + x = 6/13 × (48 + x)
260 + 13x = 288 + 6x
7x = 28
∴ x = 4 kg

ax² + bx + c = x (ax + b) + c
Remainder when divided by x = c
∴ c = 3

Let P(x) be a polynomial when divided by (x - a), then the remainder is P(a)

Remainder when divided by (x - 1) = P(1)
6 = a (1) + b(1) + c
6 = a + b + 3
∴ a + b = 3

The terms are in A.P. where the common difference is 1

For an AP, average is same as the middle most term

Middle most term = 5^th term = 55
9^th term = 5^th term + 4
⇒ 55 + 4 = 59
∴ The largest integer is 59

Let the average age of 5 students be x
Sum of ages of 15 students + Sum ages of 5 students = Sum of ages of 20 students

Sum = Average × Number of observations
15 × 19 + 5 × x = 20 × 18.5
285 + 5x = 370
5x = 85
∴ x = 17 years

Total volume = Flow rate of R × Time taken to empty

Total volume = (Flow rate of P + Flow rate of Q - Flow rate of R) × Time taken to fill if all the taps and drains are open

To use equation 1, two quantities are required
S1 gives flow rate of R = 6 litres per minute
S3 gives time taken to empty = 15 hours 20 minutes

∴ Volume of tank = 6 × (15 × 60 + 20) = 5520 litres     [∵ 1 hour = 60 minutes]

To use equation 2,
S1: gives the flow rate of R
S2 gives the total time require to fill if all taps and drains are opened.

∴ Volume = (12 + 10 - 6) × (5 × 60 + 45)
⇒ 16 × 345 = 5520 litres
Let the volume of the tank be V litres
By S3: Flow rate of Pipe R = V/ (15 × 60 + 20) = V/920

Substituting in S2
V = (12 + 10 - V/920) × (5 × 60 + 45)
V/345 = 22 - V/920
V/345 + V/920 = 22
168V/57960 + 63V/57960 = 22
231V/57960 = 22
∴ V = 5520 litres
∴ Any two statements are required to find the capacity of the tank

Let the distance between the two places be D km
Time taken = Distance/Speed

Speed in upstream = Speed of boat in still water - Speed of the stream = x - y

Speed in downstream = Speed of the boat in still water + Speed of the stream = x + y

Total time taken = Time taken upstream + Time taken downstream

z = D/ (x - y) + D/ (x + y)
z = [D(x + y) + D(x - y)] / [(x - y) (x + y)]
z = 2Dx / (x² - y²) [∵ (a - b) (a + b) = (a² - b²)]
∴ D = z(x² - y²)/2x

Let the distance between Delhi and Agra be D km

Time taken for round journey = D/60 + D/y       [∵ Time = Distance/Speed]
⇒ D(60 + y) / (60y) = 2D/Average speed
⇒ 2 × 60y = Average speed × (60 + y)
120y = 48 × (60 + y)
120y = 2880 + 48y
72y = 2880
∴ y = 40 km/h

Let the original cost price be Rs ‘x’
Profit = 32% of cost price
∴ Selling price = Cost price + Profit = 1.32x
New cost price = (100 + 20)/100 × x = 1.2x
New profit = Selling price - cost price
⇒ 1.32x - 1.2x = 0.12x
∴ Profit% = 0.12x/1.2x × 100 = 10%

Let D’s contribution be x
∴ A’s contribution = 3 × x = 3x
E’s contribution = ½ × 3x = 1.5x
B’s contribution = 1/3 × 1.5x = 0.5x
C’s contribution = 2/3 × 3x = 2x
Total contribution = 3x + 0.5x + 2x + x + 1.5x = 8x
B, C and E contribution = 0.5x + 2x + 1.5x = 4x
Difference between (A + D + E) - (B + C) = 3x + x + 1.5x - 0.5x - 2x = 3x
Share ∝ Contribution
B + C + E contribution/13500 = 4/3
∴ B + C + E contribution = 4/3 × 13500 = Rs. 18000

5¹⁷ + 5¹⁸ + 5¹⁹ + 5²⁰ = 5¹⁷ × (1 + 5 + 25 + 125)
⇒ 5¹⁷ × 156 = 5¹⁷ × 12 × 13
∴ The given expression is divisible by 13

a + b = 2c
∴ a - c = c - b
a/(a - c) + c/(b - c) = a/(a - c) - c/(a - c) [∵ b - c = -(a - c)]
⇒ (a - c)/(a - c) = 1

x^a = (y^1/a)^a = y
(x^a)^b = y^b = (z^1/b)^b = z
[(x^a)^b]^c = z^c = (x^1/c)^c = x
∴ x^abc = x
∴ abc = 1

2b = a + c
∴ b - c = a - b
x^{b - c} y^{c - a} z^{a - b} = x^{b - c} y^{c - a} z^{b - c}
⇒ (xz)^{b - c} y^{c - a}
⇒ (y²)^{b - c} y^{c - a}
⇒ (y)^{2b - c - a} = y⁰ = 1

If decimal expansion terminates, the denominator of the rational number is 2 or 5 or combinations for their powers

If the decimal expansion is non terminating but repeating, then it is a rational number with terms in prime factorization of the denominators to be numbers other than 2 or 5

If the decimal expansion is non terminating and non repeating, then it is an irrational number

Products of the roots of quadratic equation = c/a
In this case, product = 1 [∵ Roots are reciprocal of each other]
1 = r/p
∴ p = r

65x - 33y = 97     ----(1)
33x - 65y = 1     ----(2)
Equation 1 + 2
98x - 98y = 98
∴ x - y = 1 or x = y + 1
Substituting in equation 1
65y + 65 - 33y = 97
32y = 32
∴ y = 1 and x = 1 + 1 = 2
∴ xy = 2 × 1 = 2

(ab + xy)/bx = a/x + y/b
⇒ (x/a)^-1 + (b/y)^-1
⇒ (1 - c/z)^-1 + (1 - z/c)^-1
⇒ z/(z - c) + c/(c - z)
⇒ (z - c) / (z - c) = 1

(a² - 1)/a = 5
a - 1/a = 5

Cubing on both the sides
a³ - 1/a³ - 3 × a × 1/a × (a - 1/a) = 125 [∵ (a - b)³ = a³ - b³ - 3ab (a - b)]
a³ - 1/a³ - 3 × 5 = 125
∴ a³ - 1/a³ = (a⁶ - 1)/a³ = 125 + 15 = 140

(y + z - x)³ + (z + x - y)³ + (x + y -z)³ = (x + y + z - 2x)³ + (y + z + x - 2y)³ + (x + y + z - 2z)³
⇒ (-2x)³ + (-2y)³ + (-2z)³ [∵ x + y + z = 0]
⇒ -8 × (x³ + y³ + z³)
If x + y + z = 0, then x³ + y³ + z³ = 3xyz
∴ The given expression = -8 × 3xyz = -24xyz

If (x - a) is a factor of polynomial P(x), then P(a) = 0
∴ (-3)³ + 3 × (-3)² + 4 × (-3) + k = 0
-12 + k = 0
k = 12

We know that 30² = 900
31² = 961
32² = 1024
∴ 1024 is the smallest 4 digit perfect square

Sum of the coefficients of the polynomial is zero (3 + 4 - 7 = 0)
∴ 1 is the root of the polynomial

If their HCF = 21, then the numbers are form 21a and 21b respectively, where a and b are coprimes

LCM × HCF = Product of the numbers
3003 × 21 = 21a × 21b
∴ ab = 143 = 13 × 11 or 143 × 1

Since both the numbers are greater than 21
a = 13 and b = 11

So the numbers are 21 × 13 = 273 and 21 × 11 = 231
Sum of the numbers = 273 + 231 = 504

Sum of roots = -b/a
∴ α + β = -b/a
Product of roots = c/a
∴ αβ = c/a
1 + α + β + αβ = 1 - b/a + c/a = (a - b + c)/a
1 + α + β (α + 1) = (a - b + c)/a
(α + 1) (β + 1) = (a - b + c)/a

P(a) is the remainder obtained when the polynomial is divided by (x - a)
∴ 3(-1)³ + k(-1)² + 5(-1) - 6 = -7
k - 14 = -7
∴ k = 7

Degree is the highest power of the given polynomial

If f(x) and g(x) are added the maximum of them will be the degree unless some term cancels.

If p = q, and the coefficients are negative to each other, the first term may get cancelled. Then the degree is less than the maximum.

∴ Degree can either be equal or less than the max(p, q)

(√5 - √3) × (√5 + √3) = 5 - 3 = 2 [∵ (a - b) (a + b) = (a² - b²)]

(√5 - √3)/2 = 1/(√5 + √3)

∴ [(√5 - √3)/(√5 + √3)] - [(√5 + √3)/(√5 - √3)] = (√5 - √3)²/2 - (√5 + √3)²/2

⇒ [(√5 - √3)² - (√5 + √3)²]/2

⇒ [2√5 × -2√3]/2 [∵ (a - b) (a + b) = (a² - b²)]

⇒ -2√15

Multiplying numerator and denominator by x^a in first term, x^b in the second term and x^c in third term

The given term = x^a/(x^a + x^b + x^c) + x^b/(x^a + x^b + x^c) + x^c/(x^a + x^b + x^c)
⇒ (x^a + x^b + x^c) / (x^a + x^b + x^c) = 1

Let the number be x
x² + x = 20
x² + x - 20 = 0
x² + 5x - 4x - 20 = 0
x(x + 5) - 4(x + 5) = 0
(x + 5) (x - 4) = 0
∴ x = -5 or 4

Let the initial price of wheat and consumptions be x and y respectively

Let the new consumption be y’

Price of the wheat × Consumption = Budget
∴ xy = (100 + 25)/100 × x × y’
xy = 1.25xy’
∴ y’ = y/1.25 = 0.8y
Reduction in consumption = (y - 0.8y)/y × 100 = 20%

Number of students who appeared for examination = (1 - 1/25) × 2000 = 1920
Number students who passed = 11/20 × 1920 = 1056

10x = 0.9 + x
9x = 0.9
∴ x = 0.1

B/E = B/C × C/D × D/E
⇒ ¾ × 2/3 × ¾ = 3/8
∴ B ∶ E = 3 ∶ 8

10 women complete work in 12 days
∴ 5 women complete work in 12 × 2 = 24 days
5 men can complete the same work in 8 days
∴ Men are 3 times efficient as women
Three women can be replaced by a man
6 women and 3 men = 6/3 + 3 = 5 men
5 men can complete the work in 8 days
∴ Work is completed in 8 days

Circle is defined as a locus of all the points at the same distance from the centre.

The distance of all the midpoints of the radii will be at a distance of 8 cm from the centre
∴ It is also a circle but with radius 8 cm

Curved surface area of cone = π r l
17600 = 22/7 × 70 × l [∵ 1 m² = 10000 cm² and radius = diameter/2]
∴ l = 17600/220 = 80 cm
h = √(l² - r²)
⇒ √(80² - 70²)
⇒ √1500 = 10√15 cm 

Statement 1:
The orthocenter of an obtuse angled triangle lies outside the triangle

∴ It is false

Statement 2:
Centroid is the intersection point of all medians and all medians are necessarily lying inside the triangle, centroid lies inside the triangle

∴ Statement 2 is true

Statement 3:
Orthocenter is the intersection of altitudes. Since in a right triangle both the sides (except hypotenuse) all altitudes, their point of intersection lies on the triangle.

∴ Statement 3 is true

Statement 4:
If statement 2 is true, statement 4 is false
∴ Statements 2 and 3 are correct

Let the intersection point be O and the point which is equidistant from the two lines be B
In triangle AOB and COB
AB = CB [∵ The point is equidistant]
OB is common

∠BAO = ∠BCO = 90° [Distance is measured along perpendicular]
∴ AOB ≅ COB
∠AOB = ∠COB

∴ The locus of all the points at the equidistant from the lines is the angular bisector of the lines

But since there are two pairs of vertically opposite angles made, there will two such angle bisector
∴ Locus is a pair of straight lines

For congruency, all sides and all angles need to be equal to the corresponding sides and angles

When fixing the angle, the ratio of sides get fixed but while fixing the sides the angle gets fixed.

Statement 1 is false, because only the ratio of sides is fixed and the sides can be expanded and shrunk without changing angle. For example, all equilateral triangles have same angle, but different sides

Statement 2 is true, since fixing sides fix angles. (Cosine formula for triangle)

[SSS congruency]
Statement 3 is true:

With two sides and an included angle, third side is fixed [Cosine formula]

When all sides are fixed, then angles are also fixed.

[SAS congruency]
Statement 4 is true:

When two angles are equal, third angle is also equal [Property sum of angles of triangle]

So the ratios of the sides are equal

Since one of the sides is equal and the ratios are equal, all sides are equal
ASA congruency

Statement 2:
Sum of interior angles of a polygon = (n - 2) × 180 = (2n - 4) × 90

∴ Statement 2 is false

Statement 1:
Sum of interior angles of the polygon = n × 90° [∵ All angles are right angles]
(2n - 4) × 90 = n × 90
2n - 4 - n
∴ n = 4.
Statement 1 is true

Let the original length of the side of the square be x cm

Area = side²
If the length increases by 8cm, area increased by 120 cm²
(x + 8)² - x² = 120
x² + 16x + 64 - x² = 120 [∵ (a + b)² = a² + 2ab + b²]
16x = 56
∴ x = 3.5 cm

10 = 2 × 5, i.e. Both 2 and 5 have to be available to make it a multiple of 10

Since number of multiples of 2 is greater than the number of multiples of 5 in the interval, Number of multiples of 10 = Number of multiples of 5

Number of multiples of 5 = 5, 10, 15, 20, 25

But 25 = 5 × 5
∴ Power of 5 in the prime factorization of the given term = 6
∴ Largest power of 10 that divides the given product = 6 

Area to be painted in each room = 2 (lh + bh)
⇒ 2 × (6 × 2.5 + 4 × 2.5) = 50 m²

Total area to be painted = 5 × 50 = 250 m² [∵ 5 rooms are to be painted]

Amount of paint required in litres = 250/20 = 12.5 litres [∵ 1 L paints 20 m²]

If one can contains 1 L, 13 cans have to be purchased to get 12.5 litres [∵ Fraction is not possible]

Number of tiles in a row = Width/Side of square
⇒ 4.5/0.5 = 9 tiles [∵ 50cm = 0.5 m]

Number of tiles in a column = Length/Side of the square
⇒ 10/0.5 = 20 tiles
Total number of tiles required = 9 × 20
Number of packets required = 9 × 20/20 = 9 packets
Cost of nine packets = 9 × 100 = Rs. 900

When cube of maximum volume is cut from sphere, the diagonal of the cube = Diameter of the sphere
∴ √3 a = 2r
a/r = 2/√3
Volume of cube/Volume of sphere = a³/ (4/3 π r³)
⇒ 3/(4π) × (a/r)³
⇒ 3/(4π) × (2/√3)³
⇒ 2/(√3π)
∴ The required ratio = 2 ∶ √3π

Given 2πr/h = 3/1
∴ r : h = 3 : 2π
l = √(r² + h²)
⇒ r × √[1 + (h/r)²]
⇒ r × √[1 + (2π/3)²] = r/3 × √(4π² + 9)
Curved surface area of cone = πrl
⇒ πr × r/3 × √(4π² + 9)
⇒ [πr² √(4π² + 9)]/3

Circumference of the circle = Perimeter of the square
2πr = 4s
∴ s = πr/2
⇒ 22/7 × 98/2 = 154 cm

Sum of two sides of a triangle > third side
In triangle ABD,
AB + BD > AD       ---- 1
In triangle ACD
AC + CD > AD        ---- 2
Adding the inequalities 1 and 2
AB + BD + CD + AC > 2AD
AB + BC + AC > 2 AD
a + b + c > 2p
Similarly for other medians
a + b + c > 2q
a + b + c > 2r
Adding all the inequalities
3(a + b + c) > 2(p + q + r)

Let the centroid be O
In triangle AOB
AO + BO > AB
Centroid divides the median in the ratio 2 ∶ 1
∴ 2p/3 + 2q/3 > c
Similarly for other medians
2q/3 + 2r/3 > a
2p/3 + 2r/3 > b
Adding the inequalities
4/3(p + q + r) > (a + b + c)
∴ 4(p + q + r) > 3(a + b + c)

Side of the square = Diameter of the circle
∴ Radius of the circle, r = 2/√π × ½ = 1/√π
Area of the circle = π r²
⇒ π × 1/√π × 1/√π = 1 square units

Both the diagonals have same length,
∴ Product of length of diagonals = Square of the length of diagonal
Length of diagonal = √2 × length of side
Length of diagonal² = 2 × Length of side²
50 = 2 × s²
25 = s²
∴ s = 5 units

TSA of cylinder = 2πr(h + r)
352 = 2 × 22/7 × r (10 + r)
56 = r (10 + r)
4 × 14 = r × (10 + r)
∴ r = 4 cm
Diameter = 2 × radius = 8 cm

Diagonal of the square = √2 × side of the square
∴ Side of the square = 6√2/√2 = 6 cm
4 × Side of the square = 3 × Side of the triangle
4 × 6 = 3 × Side of the triangle
∴ Side of the equilateral triangle = 8 cm
Area of the equilateral triangle = √3/4 × side²
⇒ √3/4 × 8² = 16√3 cm²

Diagonal of the square = Diameter of the circle
√2 a = 2r
∴ r = a/√2

Area of the required region = Area of the circle - Area of the square
⇒ πr² - a²
⇒ π (a/√2)² - a² = (π - 2)a²/2 square units

Let O be the centre of the circle

In quadrilateral AOBX
∠A + ∠O + ∠B + ∠X = 360° [Angle sum property of quadrilateral]

90° + ∠O + 90° + 50° = 360° [∵ Tangent is perpendicular to the radius]
∴ ∠AOB = 130°
∠ACB = ½ × 130 = 65° [∵ Angle subtended by a chord at any point on the circle is half the angle subtended at centre]

Construct a perpendicular from A to the line CF, Let it meet BE at O and CF at P

Similarly a perpendicular from D is drawn to the line CF, let it meet BE and CF at R and S

In triangle ABO and ACP
∠A is common and ∠O = ∠P = 90°
ABO ∼ ACP
∴ AB/AC = AO/AP
In triangle DRE and DSF
∠D is common and ∠R = ∠S = 90°
DRE ∼ DSF
∴ DE/DF = DR/DS
Parallel lines are equally spaced
∴ AO = DR and AP = DS
∴ AB/AC = DE/DF
So statement 1 is correct
AB/ (AC - AB) = DE/ (DF - DE)
AB/BC = DE/EF
∴ AB × EF = BC × DE
So statement 2 is also correct

In triangle ADE and ABC
AD/AB = AE/EC = ½
∠A is common
∴ ADE ∼ ABC
Area(ADE) /Area (ABC) = (AD/AB)²
∴ Area(ADE) = (1/2)² × Area(ABC)
⇒ ¼ × √3/4 × l² = √3/16 × l² [∵ Area of equilateral triangle = √3/4 × l²]
Area of shared region = Area(ABC) - Area(ADC)
⇒ √3/4 × l^{2 -} √3/16 × l²
⇒ 3√3/16 × l²

∠OPT = 90° [∵ Radius is perpendicular to tangent]
∠OPQ = 90° - ∠QPT = 90° - α
∠OQP = 90° - α [∵ OQ = OP]
In triangle OQP
∠O + ∠Q + ∠P = 180°
∠O + 90 - α + 90 - α = 180
∴ ∠O = 2α

Construct OE perpendicular to AB
EB = ½ × AB = 5 cm [∵ Perpendicular from the centre bisects the chord]
In triangle OEB
OB² = OE² + EB²
13² = OE² + 5²
∴ OE = 12 cm
In triangle OEP
OP² = OE² + EP²
EP = EB - PB = 5 - 3 = 2 cm
OP² = 12² + 2² = 148
∴ OP = 2√37 cm

Angle = Percentage of the sector/100 × 360°
⇒ 16.1/100 × 360 = 57.96° ≈ 58°

Let the numbers be a and be respectively
(a + b)/2 = 10
∴ a + b = 20      ----1
√(ab) = 8
∴ ab = 64
(a - b) = √[(a + b)² - 4ab]
⇒ √(20² - 4 × 64) = √(400 - 256) = 12
∴ a - b = 12      ----2
Adding equations 1 and 2
2a = 32
∴ a = 16 and b = 20 - 16 = 4
∴ The numbers are 16 and 4

Sum of numbers = Average of the numbers × Number of observations

Sum of 11 observations = Sum of first 6 numbers + Sum of last 6 number - 6^th number [∵ 6^th observation is added twice]
11 × 11 = 10.5 × 6 + 11.5 × 6 - 6^th number
121 = 63 + 69 - 6^th number
∴ 6^th number = 132 - 121 = 11

sin² θ = 1 - cos² θ
∴ sin⁴θ - cos⁴ θ = (1 - cos²θ)² - cos⁴ θ
⇒ 1 - 2cos² θ + cos⁴ θ - cos⁴ θ [∵ (a - b)² = a² - 2ab + b²]
⇒ 1 - 2cos² θ

cot 1° cot 23° cot 45° cot 67° cot 89° = cot 1° cot 23° cot 45° cot (90 - 23°) cot (90 - 1)°
⇒ cot 1° cot 23° cot 45° tan 23° tan 1° [∵ cot (90 - A) = tan A]
⇒ (cot 1° tan 1°) × (cot 23° tan 23°) × cot 45°
⇒ 1 × 1 × 1 [∵ cot A tan A = 1 and cot 45° = 1]
⇒ 1

Hour hand completes one full revolution in 12 hours
∴ Angle described in 1 hour = 360°/12 = 30°
Angle described in 60 minutes = 30° [∵ 1 hour = 60 minutes]
∴ Angle described in 10 minutes = 30/60 × 10 = 5°

Statement 1:
Sec² θ - 1 = tan² θ and cosec² θ - 1 = cot² θ

∴ (sec² θ - 1) (1 - cosec² θ) = tan² θ (-cot² θ) = -1 [∵ tan θ × cot θ = 1]

∴ Statement 1 is wrong

Statement 2:
sin θ/(1 + cos θ) + (1 + cos θ)/sinθ = [sin² θ + (1 + cos θ)²]/[(1 +cos θ) sinθ]

⇒ [sin² θ + 1 +2cos θ + cos² θ]/[(1 +cos θ) sinθ] [∵ (a + b)² = a² + 2ab + b²]
⇒ [1 + 1 + 2cos θ]/[(1 + cos θ) sinθ] [∵ sin² θ + cos² θ = 1]
⇒ [2 + 2cos θ]/[(1 + cos θ) sinθ]
⇒ 2 [1 + cosθ]/[(1 + cos θ) sinθ]
⇒ 2/sinθ = 2 cosecθ [∵ 1/sin θ = cosec θ]
Statement 2 is correct

EO = l/2 [∵ O is the centre of the square]
∠AFD = 60°

AF = DF due to symmetry and also ∠AFD = 60°
∴ AFD is an equilateral triangle of side l
FE = Altitude of an equilateral triangle = √3l/2
In triangle EFO
FO² = EF² - EO² [∵ Pythagoras theorem]
∴ h² = (√3l/2)² - (l/2)² = l²/2

∠ACB = α and ∠BCD = β initially
In triangle BCD, cot β = BC/BD
∴ BC = BD cot β = h cot β
In triangle ABC, tan α = AB/BC
∴ AB = BC tan α
H - h = h cotβ × tan α      ----1
∴ H = h (cotβ tan α + 1)      ----2

Now height changes to h’ and the angles get interchanged
∴ H = h’ (cot α tanβ + 1)      ----3

Equating the RHS of equations 2 and 3

h(cotβ tan α + 1) = h’(cot α tan β + 1)

h (1/tanβ × tanα + 1) = h’ (1/tan α × tan β + 1) [∵ tan A = 1/cot A]

h (tan α + tan β)/tan β = h’ (tan α + tan β)/tan α
∴ h’ = h/tan β × tan α

⇒ h cot β × tan α = H - h [From equation 1]
∴ h’ = H - h

Increase in height = h’ - h = H - h - h = H - 2h units

Shortcut:

Since the distance between the building does not change [BC]

If the angle interchanges corresponding height portion of the tallest building should be interchanged

Length of AB in original case = Length of BD in new case

H - h = h’
∴ h’ - h = H - 2h units

Secx cosecx = 2
1/cosx × 1/sinx = 2
∴ sinx cosx = ½
(sinx - cosx)² = sin²x + cos²x - 2sinx cosx
⇒ 1 - 2 × ½ = 0 [∵ sin²x + cos²x = 1]
∴ sinx = cos x
∴ sinx/cosx = 1
tanx = 1, cotx = 1/tanx = 1
∴ tanⁿx + cotⁿx = 1ⁿ + 1ⁿ = 2

cos x + cos² x = 1
cos x = 1 – cos² x
cos x = sin² x [∵ 1 – cos² A = sin² A]
Substituting cos x = sin² x in the main equation
sin² x + sin⁴ x = 1

(sin A + cos A)² = sin² A + cos² A + 2 sin A cos A [∵ (a + b)² = a² + b² + 2ab]

p² = 1 + 2 sin A cosA [∵ sin² A + cos² A = 1]
∴ sin A cos A = (p² – 1) /2
(sin A + cos A)³ = sin³ A + cos³ A + 3 sin A cos A (sin A + cos A)
[∵ (a + b)³ = a³ + b³ + 3ab (a + b)]
p³ = q + 3 × (p² – 1)/2 × p
p³ = q + 3p³/2 – 3p/2
(0 = q + p³/2 -3p/2) × 2
0 = 2q + p³ – 3p
∴ p³ – 3p + 2q = 0

x = (sec² θ – tan θ) / (sec² θ + tan θ)
Using componendo and dividend rule

(x + 1) / (1 – x) = [(sec² θ – tan θ) + (sec² θ + tan θ)] / [(sec² θ + tan θ) - (sec² θ – tan θ)]
⇒ sec²θ/tan θ = (1/cos² θ) / (sin θ/ cos θ)
⇒ 1/(sin θ cos θ)
⇒ 2/sin 2θ [∵ sin 2θ = 2 sin θ cos θ]
∴ (x + 1) / (1 – x) = 2/sin 2θ
Taking reciprocal on both the sides
(1 – x) / (x + 1) = sin 2θ/2
-1 ≤ Sin 2θ ≤ 1
∴ -1/2 ≤ (1 – x) / (x + 1) ≤ ½
-(x + 1) /2 ≤ (1 – x) ≤ (x + 1) /2
First inequality
-x/2 – ½ ≤ 1 – x
x/2 ≤ 3/2
∴ x ≤ 3
Second inequality
(1 – x) ≤ x/2 + ½
½ ≤ 3x/2
∴ x ≥ 1/3
The final condition is 1/3 ≤ x ≤ 3

½ × AC × BD = ½ × AB × BC [Equating area]
AC × BD = AB × BC
4 BD × BD = AB × BC [∵ AC = 4 BD]
Dividing by BC² both the sides
4 × BD/BC × BD/BD = AB/BC

In triangle BDC, sin C = BD/BC and In triangle ABC, tan C = AB/BC
4sin²C = tan C
4sin²C = sinC/cos C
4sinC cos C = 1
2sin C cos C = ½
Sin2C = ½ [∵ sin 2A = 2sinA cos A]
∴ 2C = 30° and C = 15°
tan C = tan 15°
To find tan 15°
Tan 2θ = 2tanθ/ (1 – tan² θ)
Let tan 15° be x
Tan 30° = 2x/(1 – x²)
1/√3 = 2x/(1 – x²)
1 – x² = 2√3x
x² + 2√3x – 1 = 0
Using quadratic formula and ignoring the negative term
x = (-2√3 + 4)/2 = 2 – √3

Area of the triangle ABC = ½ × Base × perpendicular = ½ × Hypotenuse × Height drawn on it

½ × AC × BC = ½ × AB × p
AC × BC = AB × p
ab = AB × p
Squaring on both sides
a²b² = AB² × p²
AB² = AC² + BC² [∵ Pythagoras theorem]
⇒ a² + b²
∴ a²b² = (a² + b²) × p²

In a cone h = √(l² – r²) = √(13² – 5²) = 12cm
Volume of the cone = 1/3 × π r² h
⇒ 1/3 × π × 5² × 12
⇒ 100π cm³

The greatest possible area is cut off,

∴ Both the circles touch each other and extend to the two ends of the diameter of the main circle

Sum of diameters of the circle = Diameter of the larger circle

2 × diameter of the small circle = Diameter of the larger circle

The radius of the small circle = ½ × Radius of the larger circle

Area ∝ Radius²

Area of small circle/Area of large circle = (Radius of small circle/Radius of a large circle)²

?/A = (½)²

∴ Area of small circle = A/4

Remaining area = Area of the large circle – 2 × Area of the small circle

⇒ A – 2 × A/4 = A/2

TSA = πr (l + r) whereas CSA = πrl
TSA/CSA = [πr (l + r)] / (πrl)
⇒ (l + r)/l
⇒ 1 + r/l = 1 + 1/3 = 4/3 [∵ r ∶ l = 1 ∶ 3]
∴ Required ratio = 4 ∶ 3

For same vertical angle,

Radius of small cone/Radius of large cone = Height of small cone/Height of large cone Slant height of small cone/Slant height of the large cone = k

Volume of small cone∶ Volume of frustum = 64 ∶ 61

∴ Volume small cone∶ Volume large cone = 64 ∶ (64 + 61) = 64 ∶ 125

Let the radii, height and slant heights of the small and large cones be r, h, l and R, H, L

Volume of small cone/Volume of large cone = 64/125

(1/3 π r² h) / (1/3 π R² H) = 64/125
(r/R)² × (h/H) = 64/125
k² × k = 64/125
k³ = 64/125
∴ k = 4/5

Curved surface area of small cone/Curved surface area of large cone = (π r l) / (π R L)
⇒ (r/R) × (l/L)
⇒ k × k = k² = 16/25

Curved surface area of small cone/Curved surface area of frustum = 16/ (25 – 16) = 16/9
∴ Required ratio 16 ∶ 9

Numerically surface area and volumes are equal
4πR² = 4/3 π R³
1 = R/3
∴ R = 3 cm

In triangle OQS amd PRS
∠OQS = ∠PRS =90° [∵ Radius is perpendicular to tangent]
∠QSO = ∠PSR [∵ Vertically opposite angles are equal]
∴ OQS ∼ PRS
OQ/PR = OS/PS
4.5/3.5 = OS/PS
9/7 = OS/PS
∴ PS = 7OS/9
OP = OS + PS
10 = OS + 7OS/9
10 = 16OS/9
∴ OS = 90/16 = 5.625 cm
And PS = 10 – 5.625 = 4.375 cm
In triangle OQS, OS² = OQ² + QS² [∵ Pythagoras theorem]
5.625² = 4.5² + OS²
∴ OS = 3.375 cm
In triangle PRS, PS² = PR² + SP² [∵ Pythagoras theorem]
4.375² = 3.5² + SP²
∴ SP = 2.625
∴ OP = OS + SP
⇒ 3.375 + 2.625 = 6 cm

When the cone of maximum volume cut out from cube of edge 2a, height of the cone = Edge of the cube = 2a
Radius of the cone = ½ × Edge of the cube =1/2 × 2a = a
Volume of the cone = 1/3 × π r² h
⇒ 1/3 × π × a² × 2a = 2πa³/3

Area of (m + 1)^th region = π (r_m+1² – r_m²)
⇒ πr_m² [(r_m+1/r_m)² – 1]
⇒ πr_m² × (p_m² – 1)
Since area of all the region is constant
p_m² – 1 ∝ 1/r_m²
So as m increases r_m increases, ∴ p_m decreases with the m

Area to be carpeted = ½ × (Area of the square floor – Area of the equilateral triangle – 4 × Area of book shelves)
⇒ ½ × (side² – √3/4 × side² – 4 × l × b)
⇒ ½ × (10² – √3/4 × 2² – 4 × 4 × 1)
⇒ ½ × (100 – √3 – 16)
√3 ≈ 1.732
∴ Area to be carpeted = 41.134 m²
Cost = 41.134 × 100 ≈ Rs 4113

Let the length of CB be x
∴ AC = AB – CB = 2 – x
AC² = AB × C
(2 – x)² = 2x
4 – 4x + x² = 2x
x² – 6x + 4 = 0
Using quadratic formula to solve and ignoring the root greater than 2
x = (6 – 2√5)/2 = 3 – √5 cm