CDS Elementary Mathematics 8 Sep 2019 Official Paper (Previous Year Paper)

(X + Y) = 100 – 20 – 15 – 10 = 55     ---- (1)

(2X + Y) = 100 – 4 – 8 – 4 = 84     ---- (2)

From (1) and (2):

X = 29 and Y = 26

Mean = ∑x_if_i/∑f_i = 3760/100 = 37.6

Mode $= l; + ;\frac{{{f_1} - {f_0}}}{{\left( {{f_1} - {f_0}} \right); + ;\left( {{f_1} - {f_2}} \right)}} \times h$

h = Class difference = 10

l = Lower limit of model class = 40

Now,

Mode $= 40; + ;\frac{{58 - 4}}{{\left( {58 - 4} \right); + ;\left( {58 - 26} \right)}} \times 10$

$= 40; + ;\frac{{54}}{{54; + ;32}} \times 10; = ;40; + ;\frac{{54}}{{86}} \times 10$

= 40 + 6.28 = 46.28 = 46 (Approx)

Number of scooters of company Y sold by showroom E = (12% of 6400) – (8% of 3000) = 768 – 240 = 528

Number of scooters of both companies sold by showroom C = 15% of 6400 = 960

Required percent = $\frac{{528}}{{960}} \times 100; = ;55\%$

Number of scooters of both the companies sold by showroom B = 21% of 6400 = 1344

Number of scooters of company X sold by showroom A = 24% of 3000 = 720

Difference = 1344 – 720 = 624

Required percent = $\frac{{624}}{{720}} \times 100; = ;86\frac{2}{3}$

Number of scooters of company Y sold by the showrooms A, C, and E = (scooters of company X and Y sold by A, C, and E) - (scooters of company x sold by A, C, and E)

Number of scooters of company Y sold by the showrooms A, C and E = [(19 + 15 + 12)% of 6400] – [(24 + 20 + 8)% of 3000] = 2944 – 1560 = 1384

Required average = $\frac{{1384}}{3} = 461\frac{1}{3}$

Number of scooters of company X and Y sold by showroom A = 19% of 6400 = 1216

Number of scooters of company X sold by showrooms B and E together = (18 + 8)% of 3000 = 780

Required difference = 1216 – 780 = 436

Total days in a year = 365

Average number of pedestrians killed pedestrians killed per day in the year 2017 = 20475/365 = 56.095 ≈ 56

Pedestrians’ fatalities in 2014 = 12330

Pedestrians’ fatalities in 2017 = 20475

Difference = 20475 – 12330 = 8145

Required percent = (8145/12330) × 100 = 66.05% ≈ 66%

Total days in a year = 365

Average number of bikers killed daily in road accidents in the year 2017 = 48746/365 = 133.55 ≈ 134

Total days in a year = 365

Average number of cyclists killed daily in road accidents in 2017 = 3559/365 = 9.75 ≈ 10

Formula used:

1. a^m aⁿ = a^m+n
2. (a + b)³ = a³ + b³ + 3ab(a + b)
3. a³ + b³ = (a + b)(a² - ab + b²)

Calculation:

Suppose a = x² and b = y²

We have , a√a + b√b = 32

⇒ x3 + y3 = 32         ---- (1)

Also, a√b + b√a = 31.

⇒ √a√b(√a + √b) = 31

⇒ xy(x + y) = 31      ---- (2)

Using the identity (2)

⇒ (x + y)³ = x³ + y³ + 3xy(x + y)

Using equation (1) & (2)

⇒ (x + y)³ = 32 + 3 × 31 = 125 = 5³

⇒ (x + y) = 5

Again using identity (3)

⇒ x³ + y³ = (x + y)(x² + y² – xy) = 32

⇒  5(x² + y²) – 31 = 32

⇒ (x² + y²) = 63/5

⇒ (a + b) = 63/5

⇒ 5(a + b)/7 = (5/7) × (63/5) = 9

∴  5(a + b)/7 is equal to 9.

Formula used:

(a + b)² = a² +2ab + b²

Calculation:

Given that,

$x = \frac{{1; + ;\sqrt 3 }}{2}$

${x^2} = {\left( {\frac{{1; + ;\sqrt 3 }}{2}} \right)^2}$

${x^2} = \frac{1}{4}{\left( {1; + ;\sqrt 3 } \right)^2} = \frac{1}{4}\left( {1; + ;3; + ;2\sqrt 3 } \right) = \frac{{2; + ;\sqrt 3 }}{2}$

$y = {x^3} = {x^2} \times x = \frac{{2; + ;\sqrt 3 }}{2} \times \frac{{1; + ;\sqrt 3 }}{2} = \frac{{5; + ;3\sqrt 3 }}{4}$

${y^2} = {\left( {\frac{{5; + ;3\sqrt 3 }}{4}} \right)^2} = \frac{{25; + ;27; + ;30\sqrt 3 }}{{16}} = \frac{{52; + ;30\sqrt 3 }}{{16}}$

$8{y^2} = \frac{{8\left( {52; + ;30\sqrt 3 } \right)}}{{16}} = \left( {26; + ;15\sqrt 3 } \right)$

$20y = \frac{{20\left( {5; + ;3\sqrt 3 } \right)}}{4} = 25; + ;15\sqrt 3$

⇒ 8y² – 20y – 1 = (26 + 15√3) – (25 + 15√3) – 1

⇒ 8y2 – 20y – 1 = 0

Given:

HCF of two numbers = 12

Concept used:

HCF = the greatest number which divides each of the two or more numbers

Calculation:

Let the two numbers be ‘12a’ and ‘12b’ respectively where ‘a’ and ‘b’ are co-prime integers.

LCM of both the numbers = 12ab which must be divisible by 12.

Only 80 is not divisible by 12.

So, 80 can never be the HCF of both the numbers.

∴ The required answer is 80.

1. Unit digit = 17¹⁴⁷ = 7¹⁴⁷ = 7² = 49 = 9 [Since, cyclicity of 7 is ‘4’.]
2. Let the two odd numbers be ‘x’ and ‘x + 2’ respectively.

Now,

(x + 2)² – x² = x² + 4 + 4x – x² = 4(1 + x).

Hence, it will be divisible by 8 as ‘x’ is an integer.

3. Let the two consecutive odd numbers be ‘x’ and ‘x + 2’ respectively.

Now,

x(x + 2) + 1 = x² + 2x + 1 = (x + 1)².

Hence, it will be a perfect square.

Given :

Rate of interest = 20% per annum

First interest compounded half - yearly

Second interest compounded yearly

Time = 2 years

Formula used :

Amount = $P(1+\frac{R}{100})^t$        ----- (1)

Where, P = principal

R = rate and t = time

Calculations :

Let the amount invested in each scheme be 100x.

When interest in compounded half - yearly then

t = 2t and R = R/2, using equation (1)

Total amount from the first scheme

⇒ $100{\rm{x}} \times {\left( {1; + ;\frac{{20}}{{2 \times 100}}} \right)^{2 \times 2}}; = ;146.41{\rm{x}}$

Total amount from the second scheme 

⇒ $100{\rm{x}} \times {\left( {1; + ;\frac{{20}}{{100}}} \right)^2}; = ;144{\rm{x}}$

The required difference between two schemes

⇒ 146.41x – 144x = 482

⇒ 2.41x = 482

⇒ x = 200

⇒ 100x = Rs 20000

∴ Principle amount in each scheme is Rs. 20000.

Put k = 0 in the equation:

x³ – 7x + 6 = (x – 2)(x – 1)(x + 3)

Put k = 1 in the equation:

x³ + x² – 7x + 6 = x(x² + x – 7) + 6

Put k = 2 in the equation:

x³ + 2x² – 7x + 6 = x(x² + 2x – 7) + 6

Put k = 3 in the equation:

x³ + 3x² – 7x + 6 = x(x² + 3x – 7) + 6

Hence, k = 0

Time at which X completes a round = 252 seconds

Time at which Y completes a round = 308 seconds

Time at which Z completes a round = 198 seconds

Time after which they meet again at the starting point after the start = LCM of 252, 308, 198 = 2772 seconds = 46 minutes 12 seconds

Given:

Numbers are 1/3, 5/6, 2/9, 4/27

Concept used:

LCM of fractions = LCM of numerator/HCF of denominator 

Calculation:

LCM of 1/3, 5/6, 2/9, 4/27

⇒ LCM of (1, 5, 2, 4)/HCF of (3, 6, 9, 27)

⇒ 20/3

∴ Required answer is 20/3

x² + 5x + 6 = 0

⇒ x² + 3x + 2x + 6 = 0

⇒ x(x + 3) + 2(x + 3) = 0

⇒ (x + 3)(x + 2) = 0

⇒ x = -2 and -3

Put x = -2 in x² + kx + 1 = 0

⇒ (-2)² + k × (-2) + 1 = 0

⇒ 2k = 5

⇒ k = 5/2

Put x = -3 in x² + kx + 1 = 0

⇒ (-3)² + k × (-3) + 1 = 0

⇒ 3k = 10

⇒ k = 10/3

Given :

A lent Rs. 25000 to B

Rate = 7%

Time = 4 years

Formula used :

Simple interest = $\frac{PRT}{100}$        ----- (1)

P = principal, R = rate and T = time

Calculations :

Let the amount given to C be ‘x’.

According to the question, we have

Interest = Interest when  A lent Rs 25000 to B + 'x' lent to C

Using equation (1), we get

⇒ $\frac{(25000 × 7 × 4)}{100} + \frac{(x × 7 × 4)}{100} = 11200$

⇒ 0.28x = 11200 – 7000

⇒ x = Rs.15000

∴ Amount of money A lend to C is Rs 15000.

Let the selling price of both computers be ‘91x’ each.

Cost price of first computer = 91x × (100/130) = 70x

Cost price of second computer = 91x × (100/70) = 130x

Total cost price = 70x + 130x = 200x

Total selling price = 91x × 2 = 182x

Loss amount = 200x – 182x = 18x

Loss percent = (18x/200x) × 100 = 9%

Shortcut Trick

Saving 1 unit = 600 Rs

∴ Monthly income of A = 600 × 4 = 2400 Rs.

Alternate Method

Formula used: 

Income = Expenditure + Savings

Calculation:

Let, the monthly income of A and B be ‘4x’ and ’3x’ respectively.

Expenditure of A = (4x – 600)

Expenditure of B = (3x – 600)

According to the question,

⇒ $\frac{4x-600}{3x-600}$ = $\frac{3}{2}$

⇒ 8x – 1200 = 9x – 1800

⇒ x = 600

Monthly income of A = 4x = Rs. 2400

∴ Rs. 2400 is the correct option.

Given :

Ratio of train fare and bus fare = 3 : 4

The train fare increases by 20% and bus fare increases by 30%

Calculations :

Let the original train fare and the bus fare be ‘3x’ and ‘4x’ respectively.

Increase train fare = 120% of 3x

⇒ 3.6x

Increase bus fare = 130% of 4x

⇒ 5.2x

Required ratio = 3.6x: 5.2x

⇒ 9 : 13

∴ The ratio between revised train fare and revised bus fare is $\frac{9}{13}$.

Given:

Quotient = 182

Concept used:

Number = Divisor × quotient + remainder

Calculation:

According to the question,

Remainder = 182 – 175

⇒ Remainder = 7

Now,

N = 182 × 17 + 7

⇒ N = 3094 + 7

⇒ N = 3101

∴ The value of N is 3101

Given:

A stock of food grains is enough for 240 men for 48 days.

Formula used:

Work = Efficiency × Time

Calculation:

Since the stock of food grains is enough for 240 men for 48 days.

Total stock = 240 × 48

Suppose the same stock for 160 men will last for ‘N’ days.

Now,

⇒ 240 × 48 = 160 × N

⇒ N = 72 days

∴ For 72 days the same stock will last for 160 men.

Let the diameter and height of the cylinder are ‘2x’ each.

Volume of cylinder $= V = {\rm{\pi ;}}{\left( {\frac{{2x}}{2}} \right)^2}.2x = 2{\rm{\pi }}{x^3}$

Diameter of sphere of maximum possible volume = Diameter of cylinder = 2x

Volume of sphere $= \frac{4}{3};{\rm{\pi }}{\left( {\frac{{2x}}{2}} \right)^3} = \frac{4}{3}{\rm{\pi ;}}{x^3} = \frac{2}{3}{\rm{}} \times {\rm{}}2{\rm{\pi }}{x^3} = \frac{{2V}}{3}$

Volume of water remains in the vessel = Volume of the cylinder – Volume of the sphere

= V – 2V/3 = V/3

Now,

$\frac{{PQ}}{{QR}}; = ;\frac{{LM}}{{MN}}$

$\Rightarrow ;\frac{3}{9} = \frac{{LM}}{{10.5}}$

$\Rightarrow LM = \frac{{3{\rm{;}} \times {\rm{;}}10.5}}{9}$

⇒ LM = 3.5 cm

Area of circle = π × 4² = 16π cm²

Let the angle formed by arc be θ.

Angle of sector = 25.6 = $\frac{{\rm{\theta }}}{{360^\circ }} \times {\rm{\pi }} \times {\rm{;}}{4^2} = \frac{{16{\rm{\pi \theta }}}}{{360^\circ }}$

$\Rightarrow {\rm{}}\theta = \frac{{25.6{\rm{;}} \times {\rm{;}}360}}{{16{\rm{;}} \times {\rm{;\pi }}}} = {\frac{{576}}{{\rm{\pi }}}^0}$

⇒ θ = 3.2 Radian

Concept:

Orthocenter:

Point of intersection of altitudes that are drawn perpendicular from the vertex

to the opposite sides of a triangle. In a triangle, it is that point where all the

three altitudes of a triangle intersect.

• The orthocenter lies inside the triangle for an acute angle triangle.
• The orthocenter lies outside the triangle for an obtuse angle triangle.
• The orthocenter lies on the vertex of the right angle of the right triangle.

Explanation:

Using the above concept, we can conclude that, for a right-angled triangle

It's orthocentre lies on the triangle.

Suppose BC = a, AC = b and AB = c

We know that,

AX divides BC is the same ratio equivalent to the ratio of sides AB and AC respectively.

$BX; = ;BC{\rm{;}} \times {\rm{;}}\frac{{AB}}{{AB; + ;AC}}; = ;AB{\rm{;}} \times {\rm{;}}\frac{a}{{c; + ;b}}$

Since, we know that,

Sum of any two sides of a triangle is always greater than the third largest side.

c + b > a

Hence,

BX < AB

AB > BX

log₁₀(cos θ) + log₁₀(sin θ) + log₁₀(tan θ) + log₁₀(cot θ) + log₁₀(sec θ) + log₁₀(cosesc θ)

⇒ log₁₀(cos θ × sin θ × tan θ × cot θ × sec θ × cosesc θ)

$\Rightarrow {\rm{lo}}{{\rm{g}}_{10}}\left( {\cos {\rm{\theta }} \times {\rm{}}\sin {\rm{\theta }} \times {\rm{}}\tan {\rm{\theta ;}} \times {\rm{;}}\frac{1}{{{\rm{tan;\theta }}}}{\rm{}} \times {\rm{}}\frac{1}{{{\rm{cos;\theta }}}}{\rm{}} \times {\rm{}}\frac{1}{{{\rm{sin\theta }}}}} \right)$

⇒ log₁₀1 = 0

cos²x + cosx = 1

cosx = 1 – cos²x

cosx = sin²x

sin¹²x + 3sin¹⁰x + 3sin⁸x + sin⁶x

⇒ (sin⁴x)³ + 3 × sin⁴x × sin²x × (sin⁴x + sin²x) + (sin²x)³

⇒ (sin⁴x + sin²x)³

⇒ [sin²x(sin²x + 1)]³

⇒ [cosx(cosx + 1)]³

⇒ (cos²x + cosx)³

⇒ 1³ = 1

sinθ = 3/5 = Perpendicular/Hypotenuse

Base = $\sqrt {{5^2} - {3^2}} = 4$

cotθ = Base/Perpendicular = 4/3

x = cotθ = 4/3

Now,

1 + 3x + 9x² + 27x³ + 81x⁴ + 243x⁵

$\Rightarrow {\rm{}}1{\rm{}} + {\rm{}}3{\rm{}} \times {\rm{}}\frac{4}{3}{\rm{}} + {\rm{}}9{\rm{}} \times {\rm{}}{\left( {\frac{4}{3}} \right)^2}{\rm{;}} + {\rm{;}}27{\rm{}} \times {\rm{}}{\left( {\frac{4}{3}} \right)^3}{\rm{}} + {\rm{}}81{\rm{}} \times {\rm{}}{\left( {\frac{4}{3}} \right)^4}{\rm{}} + {\rm{}}243{\rm{}} \times {\rm{}}{\left( {\frac{4}{3}} \right)^5}$

⇒ 1 + 4 + 16 + 64 + 256 + 1024 = 1365

tanθ = h/x

⇒ tan(90° – θ) = 2h/y

cotθ = 2h/y

tanθ = y/2h

Now,

⇒ h/x = y/2h

⇒ 2h² = xy

$\frac{{\sin 19^\circ }}{{\cos 71^\circ }} + \frac{{\cos 73^\circ }}{{\sin 17^\circ }}$

$\Rightarrow {\rm{}}\frac{{\sin \left( {90^\circ - 71^\circ } \right)}}{{\cos 71^\circ }} + \frac{{\cos (90^\circ - 17^\circ )}}{{\sin 17^\circ }}$

$\Rightarrow {\rm{}}\frac{{cos71^\circ }}{{\cos 71^\circ }} + \frac{{\sin 17^\circ }}{{\sin 17^\circ }}$

= 1 + 1 = 2

Suppose AB = a, BC = b and AC = c

Since, ABPC is parallelogram, which means; AB = PC = a

ABCQ is parallelogram, which means; AB = CQ = a

Similarly, AQ = AR = BC = b

And, BR = BP = AC = c

Perimeter of ABC = a + b + c = 22

Perimeter of PQR = 2a + 2b + 2c = 2(a + b + c) = 2 × 22 = 44 cm

In a trapezium-

We know that: S₁ = S₃

S₁ = S₃ = 3 cm²

S₂ + S₃ = 8 cm²

S₂ = 5 cm²

We also know that:

${S_1} = \sqrt {{S_2}{S_4}}$

$\Rightarrow {\rm{}}3 = \sqrt {5{\rm{}} \times {\rm{}}{S_4}}$

⇒ 9 = 5S­₄

⇒ S₄ = 9/5 = 1.8 cm²

Let point P on the circle which means triangle will be in the semi-circle and we know that triangle in a semi-circle is always a right-angle triangle.

A = x, BP = y and AB = 2 × 6.5 = 13 cm

AP² + BP² = AB²

⇒ x² + y² = 13²

Options (2), (3) and (4) will satisfy the above relation.

So, option (1) will be our correct answer.

Since ΔABC and ΔPQR are similar.

Perimeter of ΔABC: Perimeter of ΔPQR = One side of ΔABC: Corresponding side of ΔPQR

⇒ 75 : 50 = x : 20

⇒ x = (75 × 20)/50

⇒ x = 30 cm

Concept:

When the side of an equilateral triangle is 'a'.

Its area, $A=\frac{\sqrt 3}{4}a^2$ 

And area of the normal triangle =  $\frac{1}{2}\times Base \times Height$

Since QRS is an equilateral triangle of side 10 cm. That means PQRS is a rhombus and diagonal of a rhombus are always perpendicular to each other.

Area of ΔQRS = $\frac{{\sqrt 3 }}{4}{a^2} = \frac{1}{2}{\rm{;}} \times {\rm{;}}QS{\rm{;}} \times {\rm{;}}OR$

$\frac{{\sqrt 3 }}{4}{\rm{}} \times {\rm{}}{10^2} = \frac{1}{2}{\rm{}} \times {\rm{}}10{\rm{}} \times {\rm{}}OR$

⇒ OR = 5√3 cm

⇒ PR = 2OR = 10√3 cm

Area of the first face of cuboid = x

Length of the first side = √x

Length of the second side = √y

Length of the third side = √z

Volume of cuboid = V = √x × √y × √z

⇒ V = √(xyz)

⇒ V² = xyz

Length of altitude of an equilateral triangle =$l = \frac{{\sqrt 3 }}{2}a$

⇒ a = 2l/√3

Area of equilateral triangle $= \frac{{\sqrt 3 }}{4}{a^2} = \frac{{\sqrt 3 }}{4}{\rm{}} \times {\rm{}}{\left( {\frac{{2l}}{{\sqrt 3 }}} \right)^2} = \frac{{\sqrt 3 {l^2}}}{3}$

Formula used:

The perimeter of the circle = 2πr

The perimeter of the sector = $2r+\frac{θ}{360^{\circ}}$

Where r is the radius of the circle and

θ is the subtended angle at the centre of the circle

Calculation:

Using the above formula, 

Length of arc = $2{\rm{π }} \times {\rm{}}20{\rm{}} \times {\rm{}}\frac{{150}}{{360}} = \frac{{50{\rm{π }}}}{3}$ cm

Let the length of the circle = R

The perimeter of circle = Length of arc

⇒  50π/3 = 2πR

⇒ R = 25/3 cm

∴ The radius of the circle will be 25/3 cm.

Since P, Q and R are mid-points of sides YZ, XY and XZ respectively.

QR = YZ/2

⇒ YZ = 2QR

Area of ΔXYZ : Area of ΔPQR = YZ² : QR²

⇒ 128 : Area of ΔPQR = (2QR)² : QR² = 4 : 1

⇒ Area of ΔPQR = 128/4 = 32 cm²

Similarly,

BC = QR/2

⇒ QR = 2BC

Area of ΔPQR : Area of ΔABC = QR² : BC²

⇒ 32 : Area of ΔABC = (2BC)² : BC² = 4 : 1

⇒ Area of ΔABC = 32/4 = 8 cm²

Given:

AC : AD = 4 : 9 , p ll q

Calculation: 

 Shortcut TrickΔBAC ∼ ΔEAD (By AA similarity)

⇒ Area of ΔBAC: Area of ΔEAD = AC2 : AD2= 16 : 81

 Alternate MethodConcept:

If two triangles are similar, the ratio of the area of triangles is equal to their corresponding ratios.

Calculation:

We have,p II q  and

∴ ∠ CBE = ∠ BED & ∠ BCD = ∠ CDE (Alternate angles)

By AA Similarity ΔBAC ∼ ΔEAD 

∴ $\frac{\Delta BAC}{\Delta EAD}=\frac{BA^{2}}{EA^{2}}=\frac{AC^{2}}{AD^{2}}=\frac{BC^{2}}{ED^{2}}$

⇒ $\frac{\Delta BAC}{\Delta EAD}=\frac{AC^{2}}{AD^{2}}$

⇒  $\frac{\Delta BAC}{\Delta EAD}=\frac{4^{2}}{9^{2}}$ = $\frac{16}{81}$

∴ 16 : 81 is the correct option.

Since an equilateral triangle and a square are constructed using metallic wires of equal length which means that the perimeter of the triangle and the square is same.

Let the side of the triangle and the square is ‘a’ and ‘b’ respectively.

Now,

3a = 4b

⇒ a/b = 4/3

Suppose a = 4x and b = 3x

Area of equilateral triangle = $\frac{{\sqrt 3 }}{4}{a^2} = 4\sqrt 3 {x^2}$

Area of square = b² = (3x)² = 9x²

∴ Required ratio = (4√3 x²) / (9x²) = 4√3 : 9

Since all the four sides of a parallelogram are of equal length, then it is a rhombus.

Area of rhombus = 1/2 × First diagonal × Second diagonal

Sum of the length of diagonals = 12 cm

First diagonal = 12 × (1/3) = 4 cm

Second diagonal = 12 × (2/3) = 8 cm

Area of parallelogram = Area of rhombus = (1/2) × 4 × 8 = 16 cm²

AB = 5 cm, BC = 10 cm

AC = $\sqrt {{5^2} + {{10}^2}} = 5\sqrt 5$ cm

BM is perpendicular to AC.

Area of ∆ABC = 1/2 × AB × BC = 1/2 × AC × BM

⇒ 5 × 10 = 5√5 × BM

⇒ BM = 2√5 cm

Let the height of the cylinders are ‘2h’ and ‘3h’ respectively.

Let the radii of the cylinders are ‘x’ and ‘y’ respectively.

Now,

Volume of the first cylinder = Volume of the second cylinder

πx² × 2h = πy² × 3h

⇒ x²/y² = 3/2

⇒ x : y = √3 : √2

Let the length and breadth of the rectangle is ‘x’ and ‘y’ respectively.

Original area of rectangle = xy

Increased length of rectangle = 120% of x = 1.2x

Increased breadth of rectangle = 110% of y = 1.1y

Increased area of rectangle = 1.2x × 1.1y = 1.32xy

Increment in the area = 1.32xy – xy = 0.32xy

Percent increment = (0.32xy/xy) × 100 = 32%

Hypotenuse is given as 10 cm.

Since it is a right-angle triangle, its maximum area will be possible only if the triangle is isosceles.

Other 2 sides when hypotenuse is 10 cm = √(10²/2) = 5√2 cm

Hence, area = 1/2 × 5√2 × 5√2 = 25 cm²

From the figure,

Diagonal of circle = diameter = 2r

Side of the square = 2r/√2 = r√2

Now,

Area of circle = πr²

Area of square = side² = (r√2)² = 2r²

Hence,

Ratio of area of circle to that of square = πr² : 2r² = 22 : 14 = 11 : 7

Given:

Diameter of 20 cm then, radius = 10 cm

The curved surface area is 1000 cm².

Formula used:

The surface area of the cylinder = 2πrh

Volume of cylinder = πr²h

Where r and h are the radius and height of the cylinder

Calculation:

Using the above formula

⇒ 2πrh = 1000

⇒ 2 × π × 10 × h = 1000

⇒ h = 50/π

Now,

Volume of cylinder = πr²h

⇒ π × 10 × 10 × 50/π = 5000 cm³

∴ Volume of the cylinder is 5000 cm3

Length of arc = 33 cm

Radius of arc = 14 cm

Then,

Angle subtended = l/r × 180/π

⇒ 33/14 × 180° × 7/22

⇒ 135°

In the semicircle:

a² + a²/4 = r²

⇒ 5a² = 4r²

⇒ a = 2r/√5

In the circle with radius r:

Side of square = r√2

∴ Required ratio = (2r/√5)² / (r√2)²

⇒ 2 : 5

On dividing (x⁴ - x² + 7x + 5) by (x + 2):

Quotient = x³ - 2x² + 3x + 1

Comparing the coefficients with ax³ + bx² + cx + d:

⇒ a = 1, b = -2, c = 3, d = 1

Sides: a = 30 cm, b = 28 cm and c = 16 cm

s = (a + b + c)/2 = (30 + 28 + 16)/2 = 74/2 = 37

Area of the triangle = $\sqrt {\left[ {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \right]}$

⇒ $\sqrt {\left[ {37\left( {37 - 30} \right)\left( {37 - 28} \right)\left( {37 - 16} \right)} \right]}$

⇒ $\sqrt {\left[ {37 \times 7 \times 9 \times 21} \right]}$

To get the value of this expression, we need the log values of 37, 21, 9, 7.

Suppose X = (0.001995)^1/8

Taking log:

⇒ log₁₀X = 1/8 log₁₀0.001995

⇒ log₁₀X = 1/8 log₁₀ (1995 × 10^-6)

⇒ log₁₀X = 1/8 [log₁₀ 1995 + log₁₀ 10^-6]

⇒ log₁₀X = 1/8 [3.3 - 6]

⇒ log₁₀X = -2.7/8 = -0.3375

⇒ X = $\frac{1}{{{{10}^{0.3375}}}}$

(x - a) (x - b) (x - c)

Expanding the expression:

⇒ (x² - ax - bx + ab) (x - c)

⇒ (x³ - ax² - bx² + abx - cx² + acx + bcx - abc)

⇒ (x³ - ax² - bx² + abx - cx² + acx + bcx - abc)

⇒ x³ - (a + b + c)x² + (bc + ca + ab)x - abc

Side of equilateral triangle = 7 cm

A = √3/4 × 7² = √3/4 × 49  $=\sqrt3:\times \frac{7}{2} \times \frac{7}{2}$ = 3.5 × 3.5 × √3

Now,

log₁₀A⁴ = 2 log₁₀A²

⇒ 2 log₁₀ (3.5 × 3.5 × √3)²

⇒ 2 log₁₀ [(3.5 × 3.5)² × 3]

⇒ 2 log₁₀ [(3.5)³ × 3.5 × 3]

⇒ 2 [log₁₀ 3.5³ + log₁₀10.5]

⇒ 2 [3log₁₀ 3.5 + log₁₀1050/100]

⇒ 2 [3log₁₀ 35/10 + log₁₀1050/100]

⇒ 2 [3(log₁₀ 35 - log₁₀ 10) + (log₁₀1050 - log₁₀ 100)]

Putting log₁₀1050 = 3.0212 and log₁₀35 = 1.5441

⇒ 2 [3 × (1.5441 - 1) + (3.0212 - 2)]

⇒ 2 [1.6323 + 1.0212]

⇒ 5.3070

For hollow sphere: r₁ = 6/2 = 3 cm and r₂ = 4/2 = 2 cm

Volume of the hollow sphere = 4/3π (r₁³ - r₂³)

⇒ 4π/3 × (27 - 8) = 76π/3

Since it is melted into a cone of base diameter 8 cm:

Volume of the cone = 76π/3

⇒ π/3 × 4² × h = 76π/3

⇒ h = 76/16 = 4.75 cm

For solid metallic cylinder: height = 10 cm and radius = 6 cm

Volume of the cylinder = π × 36 × 10 = 360π

Flat surface area = 2π × 36 = 72π

This cylinder is melted to make 2 cones with ratio of volumes 1 : 2

Hence,

Volume of smaller cone = 360 π/3 = 120π

Volume of larger cone = (360 π/3) × 2 = 240π

Suppose radiuses of smaller and larger cones are r and R, respectively;

Since the height of both cones is same as 10 cm;

Now,

π/3 × r² × 10 = 120 π

⇒ r = 6 cm

And

π/3 × R² × 10 = 240 π

⇒ R = 6√2 cm

Now,

Flat surface area of 2 cones = πr² + πR² = π × 36 + π × 36 × 2 = 108π

Hence,

Percentage increase = [108π - 72π]/72π = 50%

We know that:

(112, 15, 113) is a Pythagoras triplet with one number 15:

⇒ 113² = 112² + 15²

Hence,

Maximum perimeter = 112 + 113 + 15 = 240 cm

Given:

Length of rod = 24 feet

Formula used:

The surface area of each face of a cube with side a = a²

Calculation:

Since a skeleton cube has 12 sides:

Length of each side of the skeleton cube

= 24/12 = 2 feet

Hence, the area of one of the faces of the largest cube

= 2² = 4 square feet

Given:

EB = 25 cm and FC = 16 cm

Since CF and CP are tangents to the circle from the same point C, CP = CF = 16 cm

Similarly, BP = BE = 25 cm

BC = 16 + 25 = 41 cm

CQ is perpendicular to BA:

QB = 25 - 16 = 9 cm

In ΔBCQ:

CQ = √(41² - 9²) = 40 cm

Hence, diameter of the circle = 40 cm

Given:

Radii of three copper spheres = 3 cm, 4 cm, and 5 cm.

Formula used:

The volume of sphere = 4/3(πr³)

Calculation:

Sum of volumes of 3 copper spheres = Volume of new sphere.

⇒ 4/3(πr3) = 4π/3 × (27 + 64 + 125)

⇒ 4/3(πr3) = 4π/3 × 216

⇒ r3 = 6³

⇒ The radius of the larger sphere would be 6 cm.

Given:

Volume of a hemisphere is 155232 cm³

Formula Used:

Volume of Hemisphere = 2π/3 × r3

Calculation:

2π/3 × r³ = 155232

⇒ π × r³ = 232848

⇒ 22/7 × r³ = 232848

⇒ r³ = 74088

⇒ r = 42 cm

We can say that the bucket is in the form of frustum:

r = 3 cm and R = 6 cm, h = 7 cm

Now,

Volume of the frustum = π/3(r² + R² + rR)h

⇒ π/3(3² + 6² + 18) × 7

⇒ 22/21 × (3² + 6² + 18) × 7

⇒ 22 × (9 + 36 + 18)

⇒ 22/3 × 63

⇒ 462 cm³

Given:

Height  = 8 cm

The radius of its base = 6 cm

Formula used:

Slant height l = $\sqrt {{r^2} + {h^2}}$ 

Total surface area of cone = πr2 + πrl

Calculation:

Height = 8 cm and radius = 6 cm

Slant height l = $\sqrt {{8^2} + {6^2}}$ = 10 cm

Total surface area of cone = πr² + πrl

⇒ π (36 + 60)

∴  96π cm²

Given:

Six cubes, each with 12 cm edge are joined end to end.

Formula used:

Surface area of cuboid = 2 × (lb + bh + hl)

Calculation:

Length of cuboid = 12 × 6 = 72 cm

Breadth of cuboid = 12 cm

Height = 12 cm

Now, surface area of cuboid = 2 × (lb + bh + hl)

⇒ 2 × (72 × 12 + 12 × 12 + 12 × 72) cm²

⇒ 2 × (864 + 144 + 864) cm²

⇒ 2 × 1872 = 3744 cm²

∴ The surface area of resulting cuboid is 3744 cm²

x + 1/x = 26/5

⇒ x + 1/x = 5 + 1/5

Hence, x = 5 or 1/5

Hence, two such numbers are possible.

Statement 1: If p is relatively prime to each of q and r, then p is relatively prime to the product qr.

Suppose p = 8, q = 15 and r = 7

Here, p is relatively prime to each of q and r.

Now, product qr = 105

Here, p i.e. 8 is relatively prime to qr i.e. 105 also.

Hence, statement 1 is correct.

Statement 2: If p divides the product qr and if p divides q, then p must divide r.

Suppose p = 4, q = 8 and r = 7

Here, qr = 56

Here, p divides the product qr and p divides q also.

But p does not divide r here.

Hence, statement 2 is not correct.

Calculation:

Suppose the present ages of Radha and Rani are X and Y, respectively;

According to the question:

(X - 5) = 3(Y - 5)

⇒ 3Y - X = 10

⇒ X = 3Y - 10      -----(1)

And

(X - 1) = 2(Y - 1)

⇒ 2Y - X = 1

From equation (1)

⇒ 2Y - (3Y - 10) = 1

⇒ -Y + 10 = 1

⇒ Y = 9

From equation (1)

⇒ X = 3 × 9 - 10 = 17

∴ The age difference between Radha and Rani = 17 - 9 = 8

Suppose he buys x apples and y oranges:

5x + 7y = 500

Possible non-zero values of (x, y) pair:

(2, 70), (9, 65), (16, 60), (23, 55), (30, 50), (37, 45), (44, 40), (51, 35), (58, 30), (65, 25), (72, 20), (79, 15), (86, 10), (93, 5)

Hence, total 14 ways are possible.

Alternate Method

5x + 7y = 500

for x and y to be integers y should be divisible by 5.

Let y = 5a

So 5x + 35a = 500

x + 7a = 100

Minimum value of x = 1

So 7a = 99, a = 14( to be integer)

Given:

y ∝ 1/√x

Or

y = k/√x

Given x = 36 when y = 36;

⇒ 36 = k/6

⇒ k = 216

Now when y = 54:

54 = 216/√x

⇒ √x = 4

⇒ x = 16

Concept used:

(a + b)² = a² + 2ab + b²

Calculations:

16 + 6√7

⇒ 7 + 9 + 6√7

= (√7)² + 3² + 2 × 3 × √7

⇒ (3 + √7)²

∴ The square root is (3 + √7)