CDS Elementary Mathematics 18 Nov 2018 Official Paper (Previous Year Paper)

We can write,

⇒ 16 = 2⁴

⇒ 36 = 2² × 3²

⇒ 45 = 3² × 5

⇒ 48 = 2⁴ × 3

⇒ LCM of the numbers = Product of highest powers of prime factors = 2⁴ × 3² × 5 = 720

Now, Highest 4-digit number = 9999

On dividing 9999 by 720, remainder = 639

∴ Required number = 9999 - 639 = 9360

Given:

x = y^a, y = zb and z = x^c 

Calculation:

x = ya

Substituting value of ‘y’,

⇒ x = (z^b)^a = z^ab

Substituting value of ‘z’,

⇒ x = (x^c)^ab

⇒ x = x^abc

Equating powers of ‘x’,

⇒ abc = 1

∴ Value of abc is 1

Given,

⇒ x = 2 + 2^2/3 + 2^1/3

⇒ x - 2 = 2^2/3 + 2^1/3       ----(1)

Taking cube on both sides,

⇒ (x - 2)³ = (2^2/3 + 2^1/3)³

∵ (a + b)³ = a³ + b³ + 3ab(a + b)

⇒ x³ - 8 - 6x² + 12x = 4 + 2 + 6(2^2/3 + 2^1/3)

Substituting from (1),

⇒ x³ - 8 - 6x² + 12x = 6 + 6(x - 2)

⇒ x³ - 6x² + 12x - 6x = 6 + 8 - 12

⇒ x³ - 6x² + 6x = 2

We can write, 33 = 3 × 11

A number is divisible by 11, if the difference between the sum of its odd digits and the sum of its even digits is either 0 or 11,

⇒ Required difference = (X + Y + X) - (X + X) = Y

∵ Y is a digit, hence, it cannot be 11,

⇒ Y = 0

Now, for a number to be divisible by 3, the sum of its digits should be divisible by 3,

⇒ Sum of digits = X + X + 0 + X + X = 4X

In order for the sum of digits to be divisible by 3, digit X should be a multiple of 3,

⇒ X is either 3, 6 or 9

⇒ Required number = 33033, 66066 or 99099

∴ 3 five-digit numbers are possible of the given form

For a number to be divisible by 3, the sum of its digits should be divisible by 3,

⇒ Sum of digits = X + Y + 2 + 3 + 5 = X + Y + 10

Also, given X + Y ≤ 5,

Thus, to satisfy the above two conditions, (X + Y) is either equal to 2 or 5,

Also, X cannot be equal to 0, otherwise the number will not be a 5-digit number

When (X + Y) = 2, possible pairs are (1, 1) and (2, 0)

When (X + Y) = 5, possible pairs are (1, 4), (2, 3), (3, 2), (4, 1) and (5, 0)

∴ Total number of p

ossible pairs of (X, Y) = 7

Note: (0, 2) and (0, 5) should not be counted because otherwise the number will not remain a five-digit number.

Given,

⇒ x² - 6x - 27 > 0

⇒ x² - 9x + 3x - 27 > 0

⇒ x(x - 9) + 3(x - 9) > 0

⇒ (x - 9)(x + 3) > 0

As we know, when ab > 0, then there are two cases:

⇒ Either a > 0 and b > 0

⇒ Or a < 0 and b < 0

Considering the first case,

⇒ (x - 9) > 0 and (x + 3) > 0

⇒ x > 9 and x > - 3

⇒ x > 9

Considering the second case,

⇒ (x - 9) < 0 and (x + 3) < 0

⇒ x < 9 and x < - 3

⇒ x < - 3

∴ x < - 3 or x > 9

Given:

The given number is 38808

Concept:

Let K be the composite number, then

K = xa × yb × zc

Number of total factors = (a + 1) × (b + 1) × (c + 1)

Calculation:

38808 = 2³ × 3² × 7² × 11

Total no. of divisors of 38808

⇒  (3 + 1)(2 + 1)(2 + 1)(1 + 1)

⇒  4 × 3 × 3 × 2 = 72

Excluding 1 and itself, required no. of divisors

⇒  72 - 2 = 70

∴ Required number of divisior is 70.

Let the other polynomial be f(x),

As we know,

Product of two numbers = Product of their LCM & HCF

⇒ (x² - 4x - 21) × f(x) = (x + 3)(x³ - 9x² - x + 105)

⇒ (x + 3)(x - 7) × f(x) = (x + 3)(x³ - 9x² - x + 105)

⇒ f(x) = (x³ - 9x² - x + 105)/(x - 7)

∴ f(x) = (x² - 2x - 15)

Given, (α + β) = - q/p and αβ = r/p

Option 1 ⇒ 1/(α + β) = 1/(- q/p) = - p/q

Option 2 ⇒ 1/α + 1/β = (α + β)/αβ = (- q/p)/(r/p) = - q/r

Option 3 ⇒ - 1/αβ = - 1/(r/p) = - p/r

Option 4 ⇒ αβ/(α + β) = (r/p)/(- q/p) = - r/q

Now,

∵ p < r

Option 1 > Option 4

∵ q < r

⇒ Option 3 > Option 1

∵ p < q

⇒ Option 3 > Option 2

∴ Option 3 ⇒ - 1/αβ is the greatest

Let A can alone complete the work in ‘x’ days.

⇒ A’s 1-day work = 1/x

∵ A and B together completed the work in 5 days.

⇒ B’s 1-day work = 1/5 - 1/x

Now, if A worked twice & B worked 1/3^rd in a day, time taken = 3 days

⇒ 2(1/x) + (1/3)(1/5 - 1/x) = 1/3

⇒ 2/x + 1/15 - 1/3x = 1/3

⇒ 5/3x = 4/15

⇒ x = 5/3 × 15/4 = 25/4 days

∴ A can alone complete the work in 25/4 days

∵ (a³ + b³) = (a + b)(a² - ab + b²)

Putting a = x², b = 1/x², we get,

⇒ (x⁶ + 1/x⁶) = (x² + 1/x²)(x⁴ - 1 + 1/x⁴)

⇒ k = (x⁴ - 1 + 1/x⁴)

Let the three consecutive natural numbers be (x - 1), x, (x + 1)

⇒ Sum of squares = (x - 1)² + x² + (x + 1)² = 110

⇒ x² - 2x + 1 + x² + x² + 2x + 1 = 110

⇒ 3x² = 110 - 2 = 108

⇒ x² = 108/3 = 36

⇒ x = √36 = 6 (neglecting negative values)

⇒ The three consecutive numbers are 5, 6 and 7

∴ Sum of their cubes = 5³ + 6³ + 7³ = 125 + 216 + 343 = 684

Given, HCF of p & q = 16

Let p = 16x and q = 16y

Product of p & q = 16x × 16y = 7168

⇒ xy = 28

⇒ xy = 2 × 2 × 7

∵ p > 60 and q > 60,

Neither x nor y will be equal to 2,

⇒ x and y can have only two possible values, either 4 or 7

Hence, either p = 64 and q = 112 or p = 112 and q = 64

∴ Sum of the numbers = 64 + 112 = 176

According to the logarithmic rules,

⇒ log_a x = (log_b x)/(log_b a)

⇒ log_a xy = log_a x + log_a y

⇒ log_a xⁿ = n log_a x

⇒ log₁₀ 10 = 1

Hence,

⇒ log₁₀₀ (0.72)

= log₁₀ (0.72)/ log₁₀ (100)

= log₁₀ (2³ × 3² × 10^{- 2})/log₁₀ (10²)

= [log₁₀ (2³) + log₁₀ (3²) + log₁₀ (10^{- 2})]/log₁₀ (10²)

= [3 log₁₀ (2) + 2 log₁₀ (3) - 2 log₁₀ (10)]/[2 log₁₀ (10)]

= [(3 × 0.3010) + (2 × 0.4771) - 2]/2

= (0.903 + 0.9542 - 2)/2

= - 0.1428/2

= - 0.0714

= - 1 + 0.9286

= $\bar 1.9286$            [∵ bar 1 = $\bar 1$ represents - 1]

Suppose a^x = b^y = c^z = p

⇒ a = p^1/x

⇒ b = p^1/y

⇒ c = p^1/z

Now,

⇒ abc = 1

⇒ p^1/x × p^1/y × p^1/z = 1

⇒ p^{(1/x + 1/y + 1/z)} = p⁰

Equating powers of ‘p’,

⇒ (1/x + 1/y + 1/z) = 0

For a quadratic equation of form (ax² + bx + c) = 0,

Sum of roots = - b/a

Product of roots = c/a

Given roots of equation as ‘α’ and ‘β’,

⇒ (α + β) = - b/a

⇒ αβ = c/a

Now,

⇒ 1/(aα + b) + 1/(aβ + b)

= (aβ + b + aα + b)/(a²αβ + αab + βab + b²)

= [a(α + β) + 2b]/[a²αβ + ab(α + β) + b²)

= [a(- b/a) + 2b]/[a²(c/a) + ab(- b/a) + b²]

= (- b + 2b)/(ac - b² + b²)

= b/ac

We can write,

⇒ XYZ = 100X + 10Y + Z

⇒ YZX = 100Y + 10Z + X

⇒ ZXY = 100Z + 10X + Y

Sum of the numbers

= 100X + 10Y + Z + 100Y + 10Z + X + 100Z + 10X + Y

= 111X + 111Y + 111Z

= 111(X + Y + Z)

Hence, the sum of the numbers is divisible by both 111 and (X + Y + Z)

∴ Statement 1 and 2, both are correct.

Given, 1/m + 1/n - 1/mn = 2/5

⇒ (n + m - 1)/mn = 2/5

⇒ 5n + 5m - 5 = 2mn

Putting m = 1,

⇒ 5n + 5 - 5 = 2n

⇒ n = 0 (Not valid as m & n are positive integers)

Putting m = 2,

⇒ 5n + 10 - 5 = 4n

⇒ n = - 5 (Not valid as m & n are positive integers)

Putting m = 3,

⇒ 5n + 15 - 5 = 6n

⇒ n = 10

Similarly if we take n = 3 we get m = 10

Putting m = 4,

⇒ 5n + 20 - 5 = 8n

⇒ 3n = 15

⇒ n = 5

Similarly if we take n = 4 we get m = 5

Hence, possible values of (m, n) ⇒ (3, 10), (10, 3), (4, 5), (5, 4)

∴ Number of possible values = 4

Considering option 1:

Suppose (a^{q - r} b^{r - p} c^{p - q}) = abc

Equating powers, we get,

⇒ q - r = 1

⇒ r - p = 1

⇒ p - q = 1

Adding the three equations, we get,

⇒ 0 = 3, which is not valid

Considering option 2:

Suppose (a^{q - r} b^{r - p} c^{p - q}) = xyz

Substituting for a, b & c, we get,

⇒ (xy^{p - 1})^{q - r} × (yz^{q - 1})^{r - p} × (zx^{r - 1})^{p - q} = xyz

⇒ (x)^{[q - r + (r - 1)(p - q)]} × (y)^{[(p - 1)(q - r) + r - p]} × (z)^{[(q - 1)(r - p) + p - q)]} = xyz

⇒ (x)^{[q - r + pr - qr - p + q]} × (y)^{[pq - pr - q + r + r - p]} × (z)^{[qr - pq - r + p + p - q)]} = xyz

⇒ (x)^{2q - p - r + pr - qr)} × (y)^{(2r - p - q + pq - pr)} × (z)^{(2p - q - r + qr - pq)} = xyz

Equating powers, we get,

⇒ 2q - p - r + pr - qr = 1

⇒ 2r - p - q + pq - pr = 1

⇒ 2p - q - r + qr - pq = 1

Adding the three equations, we get,

⇒ 0 = 3, which is also not valid

Considering option 3:

Suppose (a^{q - r} b^{r - p} c^{p - q}) = 0

This is not possible as raising a non-zero number to any power will never be equal to 0

∴ The correct answer is none of the above.

Alternate solution∶

a = xy^{p - 1}, b = yz^{q - 1}, c = zx^{r - 1},

⇒ a^{q - r} b^{r - p} c^{p - q} = (xy^{p - 1})^{(q - r)}.(yz^{q - 1})^{(r - p)}.(zx^{r - 1})^{(p - q)}

⇒ x^{(q - r + r - p + p - q)} y^{(pq - pr - q + r + qr - pq - r + p + pr - qr - p + q)}

⇒ x⁰.y⁰

⇒ 1

∴ The correct answer is none of the above.

Let the number of sides of the two polygons be ‘5x’ and ‘4x’

∵ Measure of interior angle of regular polygon with ‘n’ sides = (n - 2) × 180°/n

Given, difference between interior angles = 9°

⇒ [(5x - 2) × 180°/5x] - [(4x - 2) × 180°/4x] = 9°

⇒ [(5x - 2) × 36°/x] - [(4x - 2) × 45°/x] = 9°

⇒ x180° - 72° - x180° + 90° = x9°

⇒ 18° = x9°

⇒ x = 2

⇒ One of them is a decagon (10 sides) and other is an octagon (8 sides)

Now,

Sum of exterior angles of a regular polygon = 360°

⇒ Sum of exterior angles of the two polygons = 360° + 360° = 720°

∴ Statements 2 and 3 are correct.

For any quadratic equation of form (ax² + bx + c),

If a > 0, then the minimum value of equation = c - b²/4a

Given, equation ⇒ (2x² + 5x + 5),

Here, a = 2 > 0

Also, b = 5 and c = 5,

⇒ Minimum value of equation = 5 - (25/8) = 15/8

Concept:

If H is the Harmonic mean of numbers a and b and is given by

 ${\rm{H}} = \frac{{2{\rm{ab}}}}{{{\rm{a}} + {\rm{b}}}}$

Calculation:

Given that,

H is the harmonic mean between P and Q.

$⇒\ \frac{2}{H}=\frac{1}{p}+\frac{1}{Q}$

⇒ H/P + H/Q = 2

Additional Information AM, GM, Formulas:

If A is the arithmetic mean of numbers a and b and is given by

⇔ ${\rm{A}} = \frac{{{\rm{a;}} + {\rm{;b}}}}{2}$

If G is the geometric mean of the numbers a and b and is given by 

⇔ ${\rm{G}} = \sqrt {{\rm{ab}}}$

Relation between AM, GM and HM

1. G2 = AH 2. AM  ≥  GM  ≥  HM

Given numbers, 1, - 1, 2, - 2, 3, - 3, 4, - 4

Let the sum of all their possible products be ‘x’,

As we know, (a + b)² = a² + b² + 2ab

Also, (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

Hence, we can write,

⇒ Square of sum of ‘n’ numbers = Sum of their squares + 2(Sum of all possible products taken two at a time)

Now,

Square of sum of given numbers = (± 1 ± 2 ± 3 ± 4)² = 0

Sum of their squares = 1 + 1 + 4 + 4 + 9 + 9 + 16 + 16 = 60

⇒ 0 = 60 + 2x

⇒ x = - 60/2 = - 30

∴ Required sum = - 30

According to remainder theorem, when any polynomial f(x) is divided by (x - a), the remainder obtained is given as f(a),

Let f(x) = 3x³ - 2x²y - 13xy² + 10y³

Putting x = 2y,

⇒ f(2y) = 3(8y³) - 2y(4y²) - 13y²(2y) + 10y³ = 24y³ - 8y³ - 26y³ + 10y³ = 0

∴ Required remainder = 0

Given, ab + bc + ca = 0

⇒ (a² - bc) = a² + ab + ca = a(a + b + c)

⇒ (b² - ca) = b² + ab + bc = b(a + b + c)

⇒ (c² - ab) = c² + bc + ca = c(a + b + c)

Hence,

$\frac{{\left( {{b^2} - ca} \right)\left( {{c^2} - ab} \right) + \left( {{a^2} - bc} \right)\left( {{c^2} - ab} \right) + \left( {{a^2} - bc} \right)\left( {{b^2} - ca} \right)}}{{\left( {{a^2} - bc} \right)\left( {{b^2} - ca} \right)\left( {{c^2} - ab;} \right)}}$

$\Rightarrow \frac{1}{{\left( {{a^2} - bc} \right)}} + \frac{1}{{\left( {{b^2} - ca} \right)}} + \frac{1}{{\left( {{c^2} - ab;} \right)}}$

$\Rightarrow \frac{1}{{a\left( {a + b + c} \right)}} + \frac{1}{{b\left( {a + b + c} \right)}} + \frac{1}{{c\left( {a + b + c;} \right)}}$

$\Rightarrow \frac{{bc + ac + ab}}{{abc\left( {a + b + c} \right)}}$

⇒ 0

Given:

Rate = 5% p.a.

Compound interest for second year = Rs. 210

Assumption:

Let the principal amount be Rs. ‘P’

Formula:

Compound interest after ‘n’ years = Principal × [(1 + Rate/100)ⁿ – 1]

Calculation:

CI after 2 years – CI after 1 year = Rs. 210

⇒ [P × {(1 + 5/100)² – 1}] – [P × {(1 + 5/100) – 1}] = 210

⇒ P × (1.05² – 1 – 0.05) = 210

⇒ P × (1.1025 – 1 – 0.05) = 210

⇒ P = 210/0.0525

⇒ P = Rs. 4000

∴ Required principal amount = Rs. 4000

Short trick∶

Required percentage = 100% – (50 + 40 – 15)% = 25%

Detailed solution∶

Let the total number of students appearing in examination be ‘x’

No. of students that failed in English = 50% of x = 0.5x

No. of students that failed in Hindi = 40% of x = 0.4x

No. of students that failed in both = 15% of x = 0.15x

⇒ No. of students that either failed in English or Hindi or both = 0.5x + 0.4x – 0.15x = 0.75x

⇒ No. of students that passed both in English & Hindi = x – 0.75x = 0.25x

∴ Percentage of students that passed both in English & Hindi = (0.25x/x) × 100 = 25%

Given:

Length of train = 100 m

Length of platform = 100 m

Time to cross the platform = 10 sec

Formula used:

Speed = Distance/Time

To convert m/sec into km/hr, multiply the number by 18/5 

Calculation:

Distance covered by train in passing the platform

⇒  Length of train + Length of platform

⇒ 100 + 100 = 200 m

Using the above formula

⇒ Speed of train = 200/10 = 20 m/sec.

But 1 m/s = 18/5 km/hr

⇒  Speed of train = 20 × 18/5 = 72 kmph.

∴ The speed of the train is 72 kmph.

Topper’s approach∶

Average speed = $\frac{{20 + 10 + 30}}{{\frac{{20}}{{40}} + \frac{{10}}{{10}} + \frac{{30}}{{40}}}} = 26.67$ kmph

Detailed solution∶

∵ Average speed = Total distance covered/Total time taken

Time taken in covering first 20 km = 20/40 = 0.5 hrs.

Time taken in covering next 10 km = 10/10 = 1 hr.

Time taken in covering last 30 km = 30/40 = 0.75 hrs.

⇒ Total time taken = 0.5 + 1 + 0.75 = 2.25 hrs.

⇒ Total distance covered = 20 + 10 + 30 = 60 km

∴ Average speed of entire journey = 60/2.25 = 26.67 kmph.

Suppose the speed of B = speed of D = ‘x’ m/sec.

In the race of 1000 m, A beats B by 150 m,

⇒ Distance covered by B = 1000 – 150 = 850 m

∵ Time taken by both participants will be equal,

⇒ Time taken = 1000/(Speed of A) = 850/x

⇒ Speed of A = (20x/17) m/sec.

Similarly, in race of 3000 m, C beats D by 400 m,

⇒ Distance covered by D = 3000 – 400 = 2600 m

⇒ Time taken = 3000/(Speed of C) = 2600/x

⇒ Speed of C = (15x/13) m/sec.

Now, in the race of 6000 m,

∵ Speed of A > Speed of C

This means that A will win the race,

⇒ Time taken by A to complete the race = 6000/(20x/17) = (5100/x) sec.

⇒ Distance covered by C in the same time = (15x/13) × (5100/x) = 5884.62 m

∴ In race of 6000 m, A beats C by = 6000 – 5884.62 = 115.38 m

Given:

Sum of current ages of Father, Mother, Sonu, Savita & Sonia = 96 years

Concept used:

We will be using the fundamental concept of ratio.

Calculation:

Let Sonu’s current age be ‘x’ years

When Sonu was born, i.e., ‘x’ years ago,

Sum of ages of Father, Mother, Savita & Sonia = 66 years

⇒ 66 = 96 – 5x

⇒ x = 30/5 = 6 years

⇒ Sonu’s current age = 6 years

⇒ Father’s current age = 6 × 6 = 36 years

∴ Father’s age after 12 years = 36 + 12 = 48 years

Let ‘B’ takes ‘x’ days to finish the work alone

⇒ No. of days A will take to do the work alone = (x – 10)

Now,

Work efficiency of A = 3 × Work efficiency of B

⇒ 1/(x – 10) = 3 × 1/x

⇒ x = 3x – 30

⇒ 2x = 30

⇒ x = 30/2 = 15

∴ B will alone finish the work in 15 days

Total no. of children = 85

Given, total no. of girls = 70% of total no. of boys

⇒ 85 – Total no. of boys = 0.7 × Total no. of boys

⇒ Total no. of boys = 85/1.7 = 50

Now,

No. of children playing only table tennis = 40% of 85 = 34

No. of children playing badminton & table tennis both = 12

⇒ No. of children playing only badminton = 85 – 34 – 12 = 39

∵ No. of boys playing only badminton = 50% of No. of boys = 0.5 × 50 = 25

∴ No. of girls playing only badminton = 39 – 25 = 14

Given,

Price of X + Price of Y = Rs. 130      ----(1)

Price of X + Price of Y + 9% of Price of Y = Rs. 136.75      ----(2)

Subtracting eq. (1) from (2), we get,

⇒ 9/100 × Price of Y = 136.75 – 130 = 6.75

⇒ Price of Y = 100/9 × 6.75 = Rs. 75

∴ Price of article Y before sales tax = Rs. 75

Let the total worth of the property of Mr. Sharma be Rs. ‘x’

Initial property share of his wife = x/2

Property share of Ravi = Property share of Raj = (1/2) × (x/2) = x/4

After death of Ravi,

Property share of Raj = x/4 + (1/2) × (x/4) = x/4 + x/8 = 3x/8

After death of Raj,

Property share of their mother = x/2 + (1/2) × (3x/8) = x/2 + 3x/16 = 11x/16

But, she owns Rs. 88000 worth of the property,

⇒ 11x/16 = 88000

⇒ x = 16/11 × 88000 = Rs. 128000

∴ Total worth of property of Mr. Sharma = Rs. 128000

Let the cost of one bottle of lemon juice be Rs. ‘x’

⇒ Cost of one bottle of orange juice = Rs. 2x

X bought 4 bottles of lemon juice, while Y bought 1 bottle of orange juice,

Total cost of drink = 4(x) + 2x = Rs. 6x

Now, the cost was divided equally between X, Y & Z,

⇒ Share of each of them = 6x/3 = Rs. 2x

But, Z contributed Rs. 50 as his share,

⇒ 2x = 50

∴ Cost of 1 bottle of orange juice = Rs. 2x = Rs. 50

Let the ages of the mother and daughter ten years ago be ‘3x’ years and ‘x’ years respectively

Ratio of their current ages = (3x + 10)/(x + 10)

Ratio of their ages after ten years = (3x + 20)/(x + 20)

But, ratio of their ages after ten years = 13 ∶ 7

⇒ (3x + 20)/(x + 20) = 13/7

⇒ 21x + 140 = 13x + 260

⇒ 8x = 120

⇒ x = 120/8 = 15

⇒ Current age of mother = (3x + 10) = 45 + 10 = 55 years

⇒ Current age of daughter = (x + 10) = 15 + 10 = 25 years

Total no. of boys in class = 60

No. of boys who play chess = 45

∵ Each boy of class play at least one of the two games,

∴ No. of boys who play carrom only = 60 – 45 = 15

Let the equal amount borrowed be Rs. ‘P’

Time = 4 years

∵ Simple interest = (Principal × Rate × Time)/100

Simple interest at 5% = (P × 5 × 4)/100 = Rs. 0.2P

Simple interest at 4% = (P × 4 × 4)/100 = Rs. 0.16P

⇒ Total interest after 4 years = 0.2P + 0.16P = Rs. 0.36P

But, total interest = Rs. 405

⇒ 0.36P = 405

⇒ P = 405/0.36 = Rs. 1125

∴ Amount borrowed = Rs. 1125

Note: This question has some error in the official paper. The question asks for the total amount borrowed which should be equal to 2P. The answer is given according to the answer key.

12 men build a wall in 10 days working for 8 hours per day

⇒ Total no. of hours 12 men took = 10 × 8 = 80 hours

⇒ No. of hours 1 man will take = 12 × 80 = 960 hours

⇒ No. of hours 8 men will take = 960/8 = 120 hours

But, 8 men have only 8 days to finish the work,

∴ No. of hours per day 8 men must work = 120/8 = 15 hours

Total cost price of milk = 28 × 8.5 = Rs. 238

Gain of milk vendor = 12.5% of 238 = Rs. 29.75

∵ Selling price = Cost price + Profit

⇒ Total selling price of milk & water mixture = 238 + 29.75 = Rs. 267.75

Now, let the milk vendor added ‘x’ litres of water to the milk

∵ The milk vendor sold the mixture at the same rate,

⇒ Total selling price of milk & water mixture = (28 + x) × 8.5 = 267.76

⇒ 28 + x = 267.75/8.5

⇒ 28 + x = 31.5

⇒ x = 31.5 – 28 = 3.5 litres

∴ The milk vendor added 3.5 litres of water to the milk

As we know, the minute and hour hand of a correct clock meet after every 720/11 minutes

In the given case, the minute and hour hand are meeting after every 72 minutes

This means that the clock is running slow and losing time

Time the clock loses after every 72 minutes = 72 – 720/11 = 72/11 min.

Time the clock loses after every hour = 72/11 × 60/72 = 60/11 min.

∵ A day has 24 hours,

∴ Time the clock loses in a day = 24 × 60/11 = 1440/11 min. = $130\frac{{10}}{{11}}$ minutes

Error: In the official question paper, the question is incorrect. We have made changes in the given options and have provided the answer accordingly.

Let the owner meet the thief ‘x’ km from his house

Speed of thief = 40 kmph

⇒ Time taken by the thief to cover ‘x’ km = x/40 hrs.

Speed of owner = 60 kmph

⇒ Time taken by the owner to cover ‘x’ km = x/60 hrs.

But, the owner started half hour after the theft,

⇒ x/40 = x/60 + 1/2

⇒ 60x = 40x + 1200

⇒ 20x = 1200

⇒ x = 1200/20 = 60 km

⇒ Time elapsed after theft = x/40 = 60/40 = 1.5 hrs.

∴ The owner met the thief 60 km from his house and 1.5 hours after the theft

Given:

X and Y together can finish a job in 6 days.

X can alone do the same job in 12 days. 

Calculation:

(X + Y), 1  day work = 1/6

⇒ X, 1  day work = 1/12

⇒ Y, 1 day work = 1/6 – 1/12 = 1/12

∴ Y will alone do the same job in 12 days

Given:

12 persons can paint 10 identical rooms in 16 days.

Calculations:

To paint 10 rooms, 12 persons take = 16 days

To paint 1 room, 12 persons will take = 16/10 = 8/5 days

To paint 1 room, 1 person will take = 12 × 8/5 = 96/5 days

To paint 1 room, 8 persons will take = 1/8 × 96/5 = 12/5 days

To paint 20 rooms, 8 persons will take = 20 × 12/5 = 48 days

∴ 8 persons will take 48 days to paint 20 rooms.

According to the logarithmic rules,

⇒ log_a xy = log_a x + log_b x

⇒ log_a xⁿ = n log_a x

⇒ log₁₀ 10 = 1

Let x = (0.2)²⁵,

⇒ log₁₀ x = log₁₀ (0.2)²⁵

⇒ log₁₀ x = 25 log₁₀ (0.2)

⇒ log₁₀ x = 25 log₁₀ (2 × 10^{– 1})

⇒ log₁₀ x = 25 [log₁₀ 2 – log₁₀ 10]

⇒ log₁₀ x = 25 [0.30103 – 1]

⇒ log₁₀ x = 25 × – 0.69897

⇒ log₁₀ x = – 17.47425

Its characteristics is 17.

∴ No. of zeros after the decimal point in the value of (0.2)²⁵ = n = 17

Calculation:

Let the ages of the father & son be ‘x’ years and ‘y’ years

respecthe tively

So, ratio of the sum & difference between their ages

⇒  (x + y)/(x – y) = 11/3

⇒ 3x + 3y = 11x – 11y

⇒ 8x = 14y

⇒ x/y = 7/4             ----(1)

⇒ Ratio of their current ages = 7 ∶ 4

Clearly, Statement 1 is incorrect.

Now, a son will attain twice his present age after ‘y’ years.

The father and son's ages after y year are (x + y) & (y + y)

Hence, the ratio of father and son's present age

⇒ Ratio of their ages after ‘y’ years = (x + y)/(y + y)

⇒ (x/2y) + (y/2y)

⇒ (1/2)[x/y + 1]

From equation (1)

⇒(1/2)[7/4 + 1] = 11/8

⇒ (x + y)/(2y) = 11/8

Therefore, Statement 2 is correct

∴ Only statement 2 is correct

Concept:

Draw the constraints to find the feasible region:

• To draw the inequalities, first, draw the equation form of the inequalities.
• Now check the region which we have to choose depending on the sign of inequality.
• To check which region we need to choose put (0,0) in both the inequality. and check whether this inequality is satisfying or not.
• If it is satisfying the inequality then take the region containing (0,0) else the opposite side of (0,0).

Calculation:

Given: x + y ≥ 5

⇒ y ≥ – x + 5

• The given linear inequality covers the area (shaded region) above the straight line y = – x + 5 as shown below,

Similarly, x – y ≤ 3

⇒ – y ≤ – x + 3

⇒ y ≥ x – 3

• The given linear inequality covers the area (shaded region) above the straight line y = x – 3 as shown below,

• Hence, the solution of the two linear inequalities is the common area (shaded region) covered by the two inequalities as shown below,

∴ The solution to the given linear inequalities lies in the first and second quadrants.

Given, x² – y² = 0

⇒ x² = y²      ----(1)

Also, (x – a)² + y² = 1

Substituting from (1), we get,

⇒ (x – a)² + x² = 1

⇒ x² – 2ax + a² + x² = 1

⇒ 2x² – 2ax + (a² – 1) = 0      ----(2)

As we know, if a quadratic equation of form ax² + bx + c = 0 have only one positive solution, its Discriminant = b² – 4ac = 0

Hence, for the eq. (2),

⇒ Discriminant = (– 2a)² – 4(2)(a² – 1) = 0

⇒ 4a² – 8a² + 8 = 0

⇒ 4a² = 8

⇒ a² = 8/4 = 2

⇒ a = ±√2

∵ There is only one positive solution,

∴ a = √2

Given, α, β and γ are the zeros/roots of the polynomial f(x) = ax³ + bx² + cx + d,

As we know, a cubic polynomial can be written in terms of its roots as,

⇒ a(x – α)(x – β)(x – γ) = ax³ – a(α + β + γ)x² + a(αβ + βγ + γα)x – aαβγ

Comparing with the given equation, we get,

⇒ (α + β + γ) = – b/a

⇒ (αβ + βγ + γα) = c/a

⇒ αβγ = – d/a

Now, using the algebraic identity,

⇒ (α + β + γ)² = α² + β² + γ² + 2(αβ + βγ + γα)

Substituting values, we get,

⇒ (– b/a)² = (α² + β² + γ²) + 2(c/a)

⇒ b²/a² = (α² + β² + γ²) + 2c/a

⇒ (α² + β² + γ²) = b²/a² – 2c/a

⇒ (α² + β² + γ²) = (b² – 2ac)/a²

Formula used:

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) +

n(A ∩ B ∩ C)

Calculation:

Let A, B & C denotes Biology, Physics & Chemistry respectively.

Consider a total 100 % students.

We know that, 

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) +

n(A ∩ B ∩ C)

⇒ 100 - 2 = 55 + 62 + 60 - 25 - 30 - 28 + n (A ∩ B ∩ C) 

⇒ n (A ∩ B ∩ C)  = 98 - 94 

⇒  n (A ∩ B ∩ C) = 4%

∴ 4 % percentage of students passed in all three subjects.

Alternate Method 

Total passed student = 98%

⇒ 7 + x + 2 + x + 2 + x + 30 - x + 25 - x+ 28 - x + x = 98%

⇒ 94 + x = 98

⇒ x = 4 %

∴ 4 % percentage of students passed in all three subjects.

Calculation:

Total passed student = 98%

⇒ 7 + x + 2 + x + 2 + x + 30 - x + 25 - x+ 28 - x + x = 98%

⇒ 94 + x = 98

⇒ x = 4 %

Percent of student pass in exactly one subject

⇒  7 + x + 2 + x + 2 + x

⇒ 11 + 3x

⇒ 11 + 3 ×  4 = 23 %

∴ 23 % students passed in exactly one subject.

Formula used:

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) +

n(A ∩ B ∩ C)

Calculation:

Let A, B & C denotes Biology, Physics & Chemistry respectively.

Consider a total 100 % students.

We know that, 

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) +

n(A ∩ B ∩ C)

⇒ 100 - 2 = 55 + 62 + 60 - 25 - 30 - 28 + n (A ∩ B ∩ C) 

⇒ n (A ∩ B ∩ C)  = 98 - 94 

⇒  n (A ∩ B ∩ C) = 4%

Hence, a number of students passed at least two subject

⇒ $\frac{25-4 + 30-4 + 28 - 4 + 4}{100}\times 100$

⇒ $\frac{83 - 2\times 4}{100}\times 360$ = 270

∴ 270 students passed in at least two subjects.

Alternate Method 

Total passed student = 98%

⇒ 7 + x + 2 + x + 2 + x + 30 - x + 25 - x+ 28 - x + x = 98%

⇒ 94 + x = 98

⇒ x = 4 %

Hence, a number of students passed at least two subject

⇒ $\frac{25-x + 30-x + 28 - x + x}{100}\times 100$

⇒ $\frac{83 - 2x}{100}\times 360$

⇒ $\frac{83 - 2\times 4}{100}\times 360$ = 270

∴ 270 students passed in at least two subjects.

Formula used:

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) +

n(A ∩ B ∩ C)

Calculation:

Let A, B & C denotes Biology, Physics & Chemistry respectively.

Consider a total 100 % students.

We know that, 

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) +

n(A ∩ B ∩ C)

⇒ 100 - 2 = 55 + 62 + 60 - 25 - 30 - 28 + n (A ∩ B ∩ C) 

⇒ n (A ∩ B ∩ C)  = 98 - 94 

⇒  n (A ∩ B ∩ C) = 4% = passed in all three subject.

Hence,

Percentage of students passed in Physice & Biology 

⇒  (25 – 4)% = 21%

Percentage of students passed in Physics and Chemistry = 30%

Therefore, required ratio = 30 ∶ 21 = 10 ∶ 7

∴ 10 : 7 is the required ratio.

Alternate Method  

Total passed student = 98%

⇒ 7 + x + 2 + x + 2 + x + 30 - x + 25 - x+ 28 - x + x = 98%

⇒ 94 + x = 98

⇒ x = 4 %

Therefore, required ratio = 30 ∶ (25 - 4)  = 10 ∶ 7

∴ 10 : 7 is the required ratio.

On a nominal scale, the measured data is classified using labels and they do not have any numerical value.

On an ordinal scale, the measured data is classified using ranks and orders and the difference between the values is not important.

On an interval scale, both order and the exact difference between the measured values are considered.

On a ratio scale, an absolute zero is also present along with the order and difference between measured data.

∵ The ratings of a hotel are classified using ranks in terms of stars, which is a measure of its quality

∴ Ratings of a hotel are measured on an ordinal scale

Short trick∶

Using allegation method

Required ratio = 3 ∶ 2

Detailed solution∶

Let the number of students in section A be ‘x’ and that in section B be ‘y’

Total marks of section A = 65x

Total marks of section B = 70y

But, the total marks of both sections combined = 67(x + y)

⇒ 65x + 70y = 67x + 67y

⇒ 2x = 3y

⇒ x/y = 3/2

∴ Required ratio = 3 ∶ 2

As we know, the median is the magnitude of the middle term when all the terms are arranged in order of their magnitude (either ascending or descending)

Given, the median of 19 observations = 30

When two more observations are taken, observation of value 8 is added to one side of the median as 8 < 30, while observation with value 32 is added to the other side of the median as 30 < 32

Hence, the median will remain the same

∴ The median of 21 observations = 30

A frequency polygon is a type of graphical representation of data and is obtained by joining the mid-points of the histogram

∴ As the number of observations and the number of classes of data increase, the shape of the frequency polygon becomes increasingly smooth.

Pooled mean is simply known as the weighted mean

Hence, when the two means are pooled,

Pooled mean = x̅ = (n₁x̅₁ + n₂x̅₂)/(n₁ + n₂)

⇒ x̅ = n₁x̅₁/(n₁ + n₂) + n₂x̅₂/(n₁ + n₂)

∵ n₁ > n₂

The first term will be nearly equal to x̅₁ and the second term will be less than x̅₂

∴ x̅₁ < x̅ < x̅₂

Median is the magnitude of the middle term when all the terms in the series are arranged in order of their magnitudes (either ascending or descending). Hence, the median is a positional average, where the middlemost value of the series represents the whole series.

Due to this property, the median is not dependent on the magnitude of all the terms, i.e., if a term smaller than the median is replaced by another term that is also smaller than the median, then the median remains unchanged. So, the median is not influenced by the extreme values of the distribution. Hence, the median is the best-suited measure in case of an open-ended frequency distribution.

∴ Both Statement I and Statement II are true and Statement II is the correct explanation of Statement I

Formula used:

sin θ = √(1 – cos2 θ)

tan θ = sin θ/cos θ

Calculation:

Since, sin θ = √(1 – cos² θ)

⇒ sin θ = √(1 – 1/5) = √(4/5) = 2/√5

⇒ tan θ = sin θ/cos θ = (2/√5) / (1/√5)

⇒ tan θ = 2

⇒ 2 tan θ/(1 – tan² θ)

= (2 × 2) / (1 – 4)       [∴ tan θ = 2]

= 4/(– 3)

∴ $\frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$ = (– 4/3)

When the value of θ ranges as 0 < θ < 90°, the value of cos θ ranges as 0 > cos θ > 1

Similarly,

When the value of φ ranges as 0 < φ < 90°, the value of cos φ ranges as 0 > cos φ > 1

Now given,

⇒ cos θ < cos φ

⇒ θ > φ

In the figure given below,

Considering ∆PQR,

⇒ tan 45° = Perpendicular/Base = PQ/QR

⇒ 1 = (h + r)/QR

⇒ QR = (h + r)

∵ The flag is just visible from the point R, PR is a tangent to the hemispherical dome

⇒ QT = Radius of dome = r

⇒ ∠QTR = ∠QTP = 90°

Considering ∆QTR,

⇒ ∠TQR = ∠QRT = 45°

This means that ∆QTR is an isosceles triangle

⇒ RT = QT = r

Similarly, considering ∆QTP,

⇒ ∠TQP = ∠QPT = 45°

This means that ∆QTP is an isosceles triangle

⇒ PT = QT = r

⇒ PR = PT + RT = 2r

Now, applying Pythagoras theorem in ∆PQR,

⇒ PR² = PQ² + QR²

⇒ (2r)² = (h + r)² + (h + r)²

⇒ 4r² = 2(h + r)²

⇒ 2r² = (h + r)²

⇒ r√2 = h + r

⇒ h = r(√2 – 1)

⇒ h/r = (√2 – 1)

Given, sin (A + B) = √3/2

⇒ sin (A + B) = sin 60°

⇒ A + B = 60°      ----(1)

Also, cos B = √3/2

⇒ cos B = cos 30°

⇒ B = 30°

Substituting in (1),

⇒ A = 60° – 30° = 30°

∴ tan (2A – B) = tan (60° – 30°) = tan 30° = 1/√3

Considering statement 1,

⇒ [cos θ/(1 – sin θ)] + [cos θ/(1 + sin θ)] = 4

⇒ [cos θ (1 + sin θ) + cos θ (1 – sin θ)]/(1 – sin² θ) = 4

⇒ 2 cos θ/cos² θ = 4                     [∵ sin² θ + cos² θ = 1]

⇒ cos θ = 1/2

⇒ θ = 60° [∵ 0 < θ < 90°]

⇒ Statement 1 is correct

Considering statement 2,

⇒ 3 tan θ + cot θ = 5 cosec θ

⇒ 3 (sin θ/cos θ) + (cos θ/sin θ) = 5/sin θ

⇒ 3 sin² θ + cos² θ = 5 cos θ

⇒ 3 – 3 cos² θ + cos² θ = 5 cos θ [∵ sin² θ + cos² θ = 1]

⇒ 2 cos² θ + 5 cos θ – 3 = 0

⇒ (2 cos θ – 1)(cos θ + 3) = 0

⇒ cos θ = 1/2, – 3

⇒ θ = 60° [∵ 0 < θ < 90°]

⇒ Statement 2 is correct

∴ Both statements 1 and 2 are correct

Considering statement 1,

The value of cos² θ lies in the range [0, 1]

The value of 1 – cos² θ also lies in the range [0, 1]

⇒ 0 ≤ (1 – cos² θ) ≤ 1

⇒ 0 ≤ (p² + q²)/2pq ≤ 1

⇒ 0 ≤ (p² + q²)/2pq

⇒ 0 ≤ (p² + q²), is always true as square terms are always greater than zero

Similarly,

⇒ (p² + q²)/2pq ≤ 1

⇒ (p² + q²) ≤ 2pq

⇒ (p² + q²) – 2pq ≤ 0

⇒ (p – q)² ≤ 0

∵ A square term can never be less than zero,

⇒ (p – q)² = 0

⇒ p = q

⇒ Statement 1 is correct

Considering statement 2,

Given, tan² θ = 4pq/(p + q)² – 1

⇒ sec² θ = 4pq/(p + q)² [∵ sec² θ = 1 + tan² θ]

⇒ cos² θ = (p + q)²/4pq

The value of cos² θ lies in the range [0, 1]

⇒ 0 ≤ (p + q)²/4pq ≤ 1

⇒ 0 ≤ (p + q)²/4pq

⇒ 0 ≤ (p + q)², is always true as square terms are always greater than zero

Similarly,

⇒ (p + q)²/4pq ≤ 1

⇒ p² + q² + 2pq ≤ 4pq

⇒ p² + q² – 2pq ≤ 0

⇒ (p – q)² ≤ 0

∵ A square term can never be less than zero,

⇒ (p – q)² = 0

⇒ p = q

⇒ Statement 2 is correct

∴ Both statements 1 and 2 are correct

Considering statement 1

Using AM-GM inequality

$\frac{{cos\theta + sec\theta }}{2} \ge \sqrt {cos\theta .sec\theta }$

$\Rightarrow \frac{{cos\theta + sec\theta }}{2} \ge \sqrt 1$

⇒ cos θ + sec θ ≥ 2

Hence statement 1 is correct.

Considering statement 2

sec² θ + cosec² θ = 1 + tan² θ + 1 + cot² θ = 2 + tan² θ + cot² θ

Using AM-GM inequality

$\frac{{ta{n^2}\theta + co{t^2}\theta }}{2} \ge \sqrt {{{\tan }^2}\theta .{{\cot }^2}\theta }$

⇒ tan² θ + cot² θ ≥ 2

∴ sec² θ + cosec² θ = 2 + tan² θ + cot² θ ≥ 4

Hence, statement 2 is also correct.

Alternate solution∶

Considering statement 1,

Suppose cos θ + sec θ = 1.5

⇒ cos θ + 1/cos θ = 3/2

⇒ 2(cos² θ + 1) = 3 cos θ

⇒ 2 cos² θ – 3 cos θ + 3 = 0

For a quadratic equation of form ax² + bx + c, x = $x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

⇒ cos θ = [3 ± √(9 – 24)]/4

⇒ cos θ = (3 ± √– 15)/4, which is not possible as the value of cos θ is always a real number

⇒ Statement 1 is correct

Considering statement 2,

Suppose sec² θ + cosec² θ < 4

⇒ 1/cos² θ + 1/sin² θ < 4

⇒ sin² θ + cos² θ < 4 sin² θ cos² θ

⇒ 1 < sin² 2θ, which is not possible as the value of sin² x always lie between [0, 1]

⇒ Statement 2 is correct

∴ Both statements 1 and 2 are correct.

Given, sin² x + sin x = 1

⇒ sin x = 1 – sin² x

⇒ sin x = cos² x       ----(1)

As we know, (a + b)³ = a³ + 3a²b + 3ab² + b³

⇒ cos¹² x + 3 cos¹⁰ x + 3 cos⁸ x + cos⁶ x

= (cos⁴ x + cos² x)³ [Substituting from eq. (1)]

= (sin² x + cos² x)³                   [∵ sin² θ + cos² θ = 1]

= 1³

= 1

Given, (3sin θ + 5cos θ) = 4

Squaring both sides,

⇒ (3sin θ + 5cos θ)² = 4²

⇒ 9sin² θ + 25cos² θ + 30sin θ cos θ = 16 [∵ (a + b)² = a² + b² + 2ab]

⇒ 30sin θ cos θ = 16 – 9sin² θ – 25cos² θ      ----(1)

Now,

(3cos θ – 5sin θ)²

= 9cos² θ + 25sin² θ – 30sin θ cos θ

= 9cos² θ + 25sin² θ – 16 + 9sin² θ + 25cos² θ [Substituting from eq. (1)]

= 34(sin² θ + cos² θ) – 16

= 34 – 16 [∵ sin² θ + cos² θ = 1]

= 18

Given, cot θ (1 + sin θ) = 4m

⇒ (cos θ/sin θ) (1 + sin θ) = 4m     ----(1)

Also, cot θ (1 – sin θ) = 4n

⇒ (cos θ/sin θ) (1 – sin θ) = 4n     ----(2)

Multiplying (1) and (2), we get,

⇒ (cos² θ/sin² θ) (1 – sin² θ) = 16mn [∵ cos² θ = 1 – sin² θ]

⇒ cos⁴ θ/16sin² θ = mn      ----(3)

Now,

Using equation (1) & (2), we can write,

⇒ 16(m² + n²) = (cos² θ/sin² θ) (1 + sin² θ + 2 sin θ + 1 + sin² θ – 2 sin θ)

⇒ (m² + n²) = (1/8) (cos² θ/sin² θ) (1 + sin² θ)

⇒ (m² + n²)² = (1/64) (cos⁴ θ/sin⁴ θ) (1 + sin² θ)²

Comparing with equation (3),

⇒ (m² + n²)² ≠ mn

Also, (m² + n²)² ≠ m²n²

Again using equation (1) & (2), we can write,

⇒ 16(m² – n²) = (cos² θ/sin² θ) (1 + sin² θ + 2 sin θ – 1 – sin² θ + 2 sin θ)

⇒ (m² – n²) = (1/4) (cos² θ/sin² θ) (sin θ)

⇒ (m² – n²)² = (1/16) (cos⁴ θ/sin⁴ θ) (sin² θ)

⇒ (m² – n²)² = (cos⁴ θ/16sin² θ)

Comparing with equation (3),

⇒ (m² – n²)² = mn

Let the given triangle be ∆ABC, as shown below,

⇒ BC = base = (u² – v²)

⇒ AC = hypotenuse = (u² + v²)

According to Pythagoras theorem,

⇒ (AC)² = (AB)² + (BC)²

⇒ (u² + v²)² = (AB)² + (u² – v²)²

⇒ u⁴ + v⁴ + 2u²v² = (AB)² + u⁴ + v⁴ – 2u²v²

⇒ (AB)² = 4u²v²

⇒ AB = 2uv

Now,

Area of triangle = 1/2 × base × height

⇒ 1/2 × BC × AB = 2016

⇒ (u² – v²) × uv = 2016

⇒ uv(u + v)(u – v) = 32 × 7 × 9

We can write,

⇒ uv(u + v)(u – v) = 16 × 9√2 × 7√2

Thus, we get, u = 8√2 and v = √2,

∴ Perimeter of triangle = AB + BC + AC = 2uv + u² – v² + u² + v²

= 2uv + 2u² = 32 + 256 = 288 units

Let the radius of the circle be ‘r’ and the side of square be ‘a’ in the figure given below,

Area of equilateral triangle = A = (√3/4) × (side)² = (√3/4)l²

Semi-perimeter of triangle = s = 3l/2

∵ Radius of incircle of triangle = Area/semi-perimeter

⇒ r = (√3/4)l² / (3l/2)

⇒ r = l/2√3

Now,

Diagonal of square = Diameter of circle

⇒ (side of square) × √2 = 2 × radius of circle

⇒ a√2 = l√3

⇒ a = l√6

∴ Area of square = (side)² = a² = l²/6

Given, length of room = l = 6 m

Breadth of room = b = 4 m

Height of room = h = 2.5 m

⇒ Total area of walls of 1 room = 2lh + 2bh = 30 + 20 = 50 m²

⇒ Total area of walls in 5 identical rooms = 5(50) = 250 m²

Now, two rooms have one square window each of side 2.5 m,

⇒ Area of the two windows = 2 × (2.5)² = 12.5 m²

⇒ Total area to be painted = 250 – 12.5 = 237.5 m²

∵ Area of 20 m² can be painted in 1 can,

∴ Total no. of cans required = 237.5/20 = 11.875 ≅ 12

Let ABCD be the parallelogram T and WXYZ be the parallelogram S as shown below,

Considering statement 1,

As we know, the line segments joining the mid-points of the sides of a triangle divides the triangle into four equal triangles

∵ In ∆ABD, AW = WB & AZ = ZD

⇒ Area (∆AWZ) = 1/4 × Area (∆ABD)       ----(1)

Also, in ∆BCD, BX = XC & CY = YD

⇒ Area (∆CXY) = 1/4 × Area (∆BCD)      ----(2)

Adding (1) and (2), we get,

⇒ Area (∆AWZ) + Area (∆CXY) = 1/4 × [Area (∆ABD) + Area (∆BCD)]

⇒ Area (∆AWZ) + Area (∆CXY) = 1/4 × [Area (ABCD)]      ----(3)

Similarly,

⇒ Area (∆DYZ) = 1/4 × Area (∆ADC)      ----(4)

⇒ Area (∆BWX) = 1/4 × Area (∆ABC)      ----(5)

Adding (4) and (5), we get,

⇒ Area (∆DYZ) + Area (∆BWX) = 1/4 × [Area (ABCD)] ----(6)

Now, adding (3) and (6), we get,

⇒ Area (∆AWZ) + Area (∆CXY) + Area (∆DYZ) + Area (∆BWX) = 1/2 × [Area (ABCD)]

⇒ Area (ABCD) – Area (WXYZ) = 1/2 × [Area (ABCD)]

⇒ 1/2 × Area (ABCD) = Area (WXYZ)

⇒ Area (ABCD)/Area (WXYZ) = 2

⇒ (Area of parallelogram T)∶ (Area of parallelogram S) = 2∶ 1

⇒ Statement 1 is correct

Considering statement 2,

According to the mid-point theorem, the length of the line segment joining the mid-points of two sides of a triangle is half the length of the third side

In ∆ABD, WZ = 1/2 × BD

In ∆ABC, WX = 1/2 × AC

In ∆BCD, XY = 1/2 × BD

In ∆ADC, YZ = 1/2 × AC

Adding the four equations, we get,

⇒ WZ + WX + XY + YZ = AC + BD

⇒ Perimeter (WXYZ) = Sum of diagonals of ABCD

⇒ Perimeter of parallelogram S is equal to the sum of diagonals of parallelogram T

⇒ Statement 2 is not correct

∴ Only statement 1 is correct

Formula∶

For a triangle with sides of lengths a, b, and c units,

Semi-perimeter = s = (a + b + c)/2

Area of triangle = √[s(s – a)(s – b)(s – c)]

Calculation∶

Hence, for the given triangle,

⇒ s = (5 + 6 + 7)/2 = 18/2 = 9 cm

∴ Area of given triangle = √(9 × 4 × 3 × 2) = √36 × √3 × √2

= 6 × 1.732 × 1.414 = 14.69 ≅ 14.7 cm²

As we know, the area of circle = π × (Radius)²

Given, area of circular plot = 144π m²

⇒ Radius of circular plot = √144 = 12 m

Now,

Width of path = 5 m

⇒ Radius of circular plot including the path surrounding it = 12 + 5 = 17 cm

∴ Required area = π × (17)² = 289π m²

Given, the lateral surface area of cone = 462 cm²

Also, the slant height of cone = 35 cm

∵ Lateral surface area of cone = π × radius × slant height

⇒ 462 = (22/7) × radius × 35

∴ Radius of base of cone = 21/5 = 4.2 cm

Let the angle between the generator and the axis of the cone be ‘θ’, shown below,

⇒ Radius of cone = OB = r (let)

When the semi-circle is rolled to form a cone, slant height of cone = Radius of semi-circle

⇒ AB = R (let)

Also, the circumference of the base of cone = Circumference of semi-circle

⇒ 2πr = πR

⇒ r/R = 1/2

Now, considering ∆OAB,

⇒ sin θ = OB/AB = r/R

⇒ sin θ = 1/2

⇒ sin θ = sin 30°

∴ θ = 30°

Let the radius of the base of the cylinder be ‘r’ cm and its height be ‘h’ cm

Given,

Lateral surface area of cylinder = (Total surface area of cylinder) /2

⇒ 2πrh = (2πrh + 2πr²) /2

⇒ 2πrh – πrh = πr²

⇒ πrh = πr²

⇒ h = r

∵ Height = h = π cm

∴ Radius of the base of cylinder = π cm