CDS Elementary Mathematics 7 Feb 2021 Official Paper (Previous Year Paper)

Calculation:

⇒ For country A = (38 - 35)/35 × 100 = 8.5%

⇒ For country B = (47 - 45)/45 × 100 = 4.4%

⇒ For country C = (93 - 88)/88 × 100 = 5.6%

⇒ For country D = (79 - 75)/75 × 100 = 5.3%

⇒ For country E = (60.9 - 58)/58 × 100 = 5%

∴ From above, in A, C, D and E countries production of cars increased by more than or equal to 5% in 2016 over 2015.

Calculation:

⇒ 50% marks of total marks = 80 × 50/100 = 40

⇒ Total number of students who have obtained less than or equal to 40 marks

= 5 + 8 + 10 + 13 = 36

⇒ Total students = 90 

⇒ Required percentage = 36/90 × 100 = 40%

∴ 40% is the correct option.

Shortcut Trick  

 -1, 2, 2, 3, 6, 8, 9

The middle number is 3

Alternate Method 

Concept: 

Median:

Case 1: If the number of observations (n) is even

${\rm{Median;}} = {\rm{;}}\frac{{{\rm{value;of;}}{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{;observation;}} + {\rm{;;value;of;}}{{\left( {\frac{{\rm{n}}}{2}{\rm{;}} + 1} \right)}^{{\rm{th}}}}{\rm{;observation}}}}{2}$

Case 2: If the number of observations (n) is odd

${\rm{Median;}} = {\rm{value;of;}}{\left( {\frac{{{\rm{n}} + 1}}{2}} \right)^{{\rm{th}}}}{\rm{;observation}}$

Calculation:

Data arranged in ascending order = -1, 2, 2, 3, 6, 8, 9.

∴ Median = The number in the middle = 3

Important Points

If the number of observations is odd, the number in the middle of the list is the median.

Given:

First ten composite number = 4, 6, 8, 9, 10, 12, 14, 15, 16, 18.

Formula:

Arithmetic mean = total sum of sample/ number of sample

Calculation:

Arithmetic mean of first ten composite numbers

⇒ (4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18)/10

⇒ 112/10 = 11.2

∴ Arithmetic mean of the first ten composite numbers is 11.2.

Important Points

Composite numbers can be defined as the whole numbers that have more than two factors.

Calculation:

⇒ Total new marks of five students = (21 + 5) + (27 + 5) + (19 + 5) + (26 + 5) + (32 + 5) = 150

∴ Required revised average marks = 150/5 = 30

Given:

Mean of m observations = p

Mean of n observations = q

Mean of (m + n) observations = c 

Concept:

Mean: The mean is defined as the sum of the observations

divided by the total number of observations. 

Calculation:

According to the question,

⇒ Total sum of m observations = pm

⇒ Total sum of n observations = qn

⇒ Total sum of m and n observations = pm + qn

⇒ c = (pm + qn)/(m + n)

As, p ≤ q,

∴ From above, p ≤ c ≤ q is option is correct.

Calculation:

⇒ Total production of type1 product = 160 + 200 + 135 + 240 + 180 + 210 = 1125

⇒ Total production of type2 product = 80 + 150 + 35 + 95 + 110 + 150 = 620

⇒ Total production of type3 product = 70 + 85 + 44 + 120 + 85 + 100 = 504

⇒ Total production of type4 product = 90 + 160 + 95 + 80 + 95 + 92 = 612

⇒ Total production of type5 product = 75 + 100 + 85 + 120 + 115 + 110 = 605

∴ From above, type3 product is least produced over 2000 - 2005.

Calculation:

⇒ Total tablets produced in 2000 = 160 + 80 + 70 + 90 + 75 = 475

⇒ Total tablets produced in 2001 = 200 + 150 + 85 + 160 + 100 = 695

⇒ Total tablets produced in 2002 = 135 + 35 + 44 + 95 + 85 = 394

⇒ Total tablets produced in 2003 = 240 + 95 + 120 + 80 + 120 = 655

⇒ Total tablets produced in 2004 = 180 + 110 + 85 + 95 + 115 = 585

⇒ Total tablets produced in 2005 = 210 + 150 + 100 + 92 + 110 = 662

For (2003, 2005), difference = 662 - 655 = 7

For (2001, 2005), difference = 695 - 662 = 33

For (2003, 2004), difference = 655 - 585 = 70

For (2000, 2002), difference = 475 - 394 = 81

∴ For (2003, 2005), the difference in total number of tablets produced between them is minimum.

Shortcut TrickHere it is asked about the ratio and we know that in ratio we reduce all the values to their standard form.

⇒ Decrease in production in 2004 compared to 2001 = 20

⇒ Decrease in production in 2000 compared to 2001 = 40

⇒ Ratio = 20 : 40 = 1 : 2 

∴ The correct answer is 1 : 2.

Calculation:

⇒ Total tablets produced in 2000 = 160 + 80 + 70 + 90 + 75 = 475

⇒ Total tablets produced in 2001 = 200 + 150 + 85 + 160 + 100 = 695

⇒ Total tablets produced in 2004 = 180 + 110 + 85 + 95 + 115 = 585

⇒ Percentage drop in production in 2004 compared to 2001 = (695 - 585)/695 × 100 = 110/695 × 100

⇒ Percentage drop in production in 2000 compared to 2001 = (695 - 475)/695 × 100 = 220/695 × 100

∴ Required ratio = (110/695 × 100) : (220/695 × 100) = 1 : 2

Calculation:

For 2000, 160 = 70 + 90

For 2001, 200 < 85 + 160

For 2002, 135 < 44 + 95

For 2003, 240 > 120 + 80

For 2004, 180 = 85 + 95

For 2005, 210 > 100 + 92

∴ From options, in 2003, the production of Type 1 is more than the sum of the production of Type 3 and Type 4.

Given:

Ratio of sides of triangle is x : (x - 1) : (x - 18)

Formula:

Pythagoras theorem,

(diagonal)² = (side1)² + (side2)²

Calculation:

Let sides of right angled triangle be ax, a(x - 1), a(x - 18) respectively.

(ax)² = [a(x - 1)]² + [a(x - 18)]²

⇒ x² = x² - 2x + 1 + x² - 36x + 324

⇒ x² - 38x + 325 = 0

⇒ x = 25 or x = -13

Sides of right angled triangle = 25, 24 and 7

∴ Perimeter of right angled triangle = 25 + 24 + 7 = 56 units

Shortcut TrickPerimeter = x + x - 1 + x - 18 = 3x - 19

Now, we can see, answer (i.e. perimeter) should be divisible by 3 if

we add 19.

From option, 56 + 19 = 75

Which is divisible by 3.

∴ The perimeter of the triangle is 56.

Calculation:

AM = MN = NB

Δ AMP and Δ ABC are similar triangle as ∠AMP = ∠ABC = 90⁰ and MP ∥ BC and AM : AB = 1 : 3

⇒ MP = BC/3 = 12/3 = 4 cm

Similarly, ΔANQ and ΔABC are similar triangle as ∠ANQ = ∠ABC = 90⁰ and NQ ∥ BC and AN : AB = 2 : 3

⇒ NQ = 2BC/3 = 2 × 12/3 = 8 cm

∴ (PM + QN) = 4 + 8 = 12 cm

Calculation:

ACBD is a cyclic quadrilateral.

∠DAC = ∠ACB = ∠CBD = ∠BDA = 90^º all are angles in semicircle.

⇒ AO = DP = BP = CP = radius of a circle.

ΔAPD and ΔDPB are an isosceles triangles.

⇒ ∠PAD = ∠PDA = 60º

⇒ ∠APD = 180º - 2 × 60º = 60º

⇒ ∠APD + ∠BPD = 180º......(Linear angles)

∴ ∠BPD = 180º - 60º = 120º

Given:

ΔABC is an equilateral 

Concept used:

The angles subtended by an arc on the circumference on the same side of the arc are equal.

All the angles of the equilateral triangle are equal and the measure of each angle is 60°.

Calculation:

ΔABC is an equilateral 

∠CAB = 60° 

The angles subtended by an arc on the circumference on the same side of the arc are equal

∠CAB = ∠CDB = 60°

Calculation:

Let AM = AN = a

CM = 4 - a and BN = 6 - a

CO = 4 - a and BO = 6 - a

Then,

⇒ CO + BO = 8

⇒ 4 - a + 6 - a = 8

⇒ a = 1

Radius of three circles = (a = 1), (4 - a = 3) and (6 - a = 5)

∴ Sum of radii of three circles = 1 + 3 + 5 = 9cm

​Formula:

Tangent-secant theorem,

PT² = PA × PB

Calculation:

 

We know  that,

PT2 = PA × PB

⇒ 12² = 9 × PB

⇒ 9 × PB = 144

⇒ PB = 16 cm

∴ AB = 16 - 9 = 7cm

Shortcut Trick We know (5, 12, 13) is a Pythagorean Triplet.

Here, Hypotenuse = 13

Also, 5 + 12 +13 = 30 cm which is the perimeter of the triangle.

So, Area = (1/2) × 5 × 12 = 30 cm²

∴ The area of the triangle = 30 cm².

Calculation:

⇒ hypotenuse + base + height = 30

⇒ base + height = 17

Using pythagoras theorem,

⇒ hypotenuse² = base² + height²

⇒ base² + height² = 169

Solving,

⇒ Base × height = 60

∴ Area of right angled triangle = 1/2 × base × height = 1/2 × 60 = 30 cm²

Calculation:

Using pythagoras theorem,

⇒ AB = √(36 + 64) = 10cm

⇒ Area of Δ ABC = 1/2 × 6 × 8 = 24 cm²

⇒ Area of Δ ABC = 1/2 × p × AB

⇒ 1/2 × p × AB = 24

⇒ p = 4.8

∴ CP = p = 4.8cm

Calculation:

ABCD is a trapezium.

Draw OF || BA || CD, intersecting AD in F.

FO || AB (from figure)

⇒ DO/OB = DF/FA [basic proportionality theorem]

FO || DC (from figure)

⇒ CO/OA = DF/FA [basic proportionality theorem]

⇒ DO/OB = CO/OA

So, diagonals of a trapezium divide each other proportionally.

ΔAOB and ΔDOC are similar triangle

⇒ A(ΔAOB) : A(ΔDOC) =  AB² : DC²

∴ A(ΔAOB) : A(ΔDOC) = 9 : 4

Calculation:

Let the circle touches the sides AB, BC, CD and DA of the quadrilateral ABCD at points N, O, P and M respectively.

Since the lengths of the tangents drawn from an external point are equal. So

⇒ DP = DM (tangents on circle from point D)

⇒ CP = CO (tangents on circle from point C)

⇒ BN = BO (tangents on circle from point B)

⇒ AN = AM (tangents on circle from point A)

Adding,

⇒ DP + CP + BN + AN = DM + CO + BO + AM

⇒ CD + AB = AD + BC

∴ AD = 12 + 9 - 8 = 13cm

Given:

Formula:

Curved surface area of cone = πrl

Calculation:

Let length of cloth be a m.

Area of cloth = a × 3 = 3a m²

Diameter of tent = 12 m

Radius of tent = 12/2 = 6 m

Curved surface area of cloth = 22/7 × 6 × 7 = 132 m²

Then,

⇒ 3a = 132

⇒ a = 44 m

∴ Length of cloth is 44 m.

Given:

Diameter of sphere = 6cm

Radius of sphere = 3cm

Radius of  vessel = 6cm

Formula:

Volume of sphere = 4/3 × πr³

Volume of cylinder = πr²h

Calculation:

⇒ Volume of sphere = 4/3 × π × 3³ = 36π cm³

⇒ Volume of vessel = π × 6² × h = 36πh cm³

⇒ 36π = 36πh

⇒ h = 1cm

∴ 1cm water level will be raised.

Given:

Radius of circle = 21cm

Formula:

Let l = length of arc, A = area of arc and θ = sector angle and r = radius of circle

⇒ l = θ/360 × 2πr

⇒ A = θ/360 × πr²

Calculation:

⇒ 55 = θ/360 × 2 × 22/7 × 21

⇒ θ = 150º

⇒ Area of sector = 150/360 × 22/7 × 21²

∴ Area of sector is 577.5 cm²

Given:

Radius of wire = 70 cm

Formula:

⇒ Circumference of circle = 2πr

⇒ Perimeter of rhombus = 4 × side

Calculation:

⇒ Circumference of wire = 2 × 22/7 × 70 = 440 cm

⇒ Circumference of wire = perimeter of rhombus

∴ Side of rhombus = 440/4 = 110cm

Given:

The perimeter of semicircular park = 360 m

Formula:

Circumference of semicircular park = πr + 2r

Area of semicircular park = πr²/2

Calculation:

360 = r(22/7 + 2)

⇒ r = 70 m

⇒ Area of park = 22/7 × 1/2 × 70² = 7700 m²

∴ Area of park is 7700 m².

Given:

Concept:

The diagonals of a Trapezium divide each other proportionally.

Calculation:

⇒ AP : PC = BP : PD

⇒ 4/(4x - 4) = (2x - 1) /(2x + 4)

⇒ 8x + 16 = 8x² - 4x - 8x + 4

⇒ 8x² - 20x - 12 = 0

⇒ 4(2x² – 5x – 3) = 0 

⇒ 2x² – 6x + x – 3 = 0 

⇒ 2x(x – 3) + 1(x – 3) = 0 

⇒ (2x + 1) (x – 3) = 0 

⇒ x = -1/2, 3

∴ The value of x is 3.

Concept used:

Ratio of sides, perimeter, median, circumradius, inradius and square root of area of two similar triangle will be equal always.

Calculation:

ΔABC ~ ΔDEF

⇒ AB/DE = BC/EF = AC/DF = (AB + BC + AC)/(DE + EF + DF)

⇒ AB + BC + AC = 40

⇒ DE + EF + DF = 30

⇒ AB/DE = BC/EF = AC/DF = 40/30 = 4/3

Then,

Let BC and EF be 4a and 3a respectively.

Let AC and DF be 4b and 3b respectively.

∴ (BC + CA) : (EF + FD) = (4a + 4b) : (3a + 3b) = 4 : 3

Given:

The vertical angle of an isosceles triangle is equal.

The ratio of the area of both triangles = 4.84 ∶ 5.29

Concept:

Properties of similar triangles:

Angle – Angle similarity

Side – Angle – Side similarity

Side – Side – Side similarity

If two triangles are similar, the ratio of areas is equal to the

ratio of squares of corresponding sides.

So, $\rm\frac{\Delta_1}{\Delta_2}=\frac{\left(S_1\right)^2}{\left(S_2\right)^2}$

Calculation:

Let Δ ABC and Δ DEF are isosceles triangle.

⇒ AB/BC = 1

⇒ DE/EF = 1

⇒ AB/AC = DE/EF

⇒ AB/DE = AC/EF

⇒ ∠ A = ∠ C

ΔABC and ΔDEF are similar triangles.

⇒ A(ΔABC) : A(ΔDEF) = (height of ΔABC)² : (height of ΔDEF)²

⇒ 4.84 : 5.29 = (height of ΔABC)² : (height of ΔDEF)²

∴ Ratio of their corresponding heights = 22 : 23

Given:

Δ ABC is right-aa ngle triangle and  AD ⊥ BC

BD = 8 cm, DC = 12.5 cm

Concept:

In right ∆ ABC, ∠ A = 90˚ and AD is perpendicular to BC then

 AD2 = BD × CD

Calculation:

Using the above concept

AD² = BD × CD

⇒ AD² = 8 × 12.5

∴ AD = 10cm

Given:

Let r = radius of sphere and l = side of cube

Formula:

⇒ Surface area of cube = 6l²

⇒ Surface area of sphere = 4πr²

Calculation:

⇒ 6l² = 4πr²

⇒ l/r = √(4π/6)

Then,

⇒ x² : y² = [l³]² : [(4/3 × π × r³)]²

⇒ x² : y² = (3/4π)² × (l/r)⁶

⇒ x² : y² = (3/4π)² × (4π/6)^{(6 × 1/2 = 3)}

∴ x² : y² = π  : 6

Given:

⇒ Volume of cone = volume of hemisphere

⇒ Radius of cone = radius of hemisphere

Formula:

⇒ Volume of cone = 1/3 × πr²h

⇒ Volume of hemisphere = 2/3 × πr³

Calculation:

⇒ 1/3 × πr²h = 2/3 × πr³

∴ h : r = 2 : 1

Given:

Radius of sphere = 60/2 = 30mm = 3cm

Length of wire = 144 cm.

Formula:

Volume of sphere = 4/3 × πr³

Volume of cylinder = πr²L

Calculation:

The volume of solid sphere

= 4/3 × π × 3³ = 36π cm³

⇒ 36π = π × r² × 144

⇒ r = 0.5cm

⇒ Radius of wire = 0.5cm

∴ Diameter of wire = 2 × 0.5 = 1cm

Given:

r : h = 2 : 3

Volume of cylinder = 1617 cm³

Formula:

Let radius & height of the cylinder are r & h respectively. 

1. Volume of cylinder = πr²h

2. Curved surface area of cylinder = 2πrh

Shortcut Trick

Let the radius be 2x and height be 3x

Volume of cylinder = π(2x)².3x

⇒ 1617 = 22/7 × 4x² × 3x

⇒ x³ = 343/8

⇒ x = 7/2

So, r = 2x= 7 cm

h = 3x = 21/2 cm

Now, Curved surface area 

⇒ 2 × 22/7 × 7 × 21/2

⇒ 463 cm2

∴ The correct answer is option (3)

Alternate Method

Calculation:

We know that,  

Volume of cylinder = πr2h

⇒ 1617 = π × r² × h

⇒ 1617 = (22/7)  × r2 × h

⇒ 1617 × (7/22) =  r2 × h

⇒ r²h = 514.5     ----(1)

But according to question,

r/h = 2/3

⇒ h = 3r/2        ----(2)

Put this value in equation (1)

r2 × (3r/2)  = 514.5

⇒ r³ = 343

⇒ r = 7 cm.

Hence, 

h = 3 × 7/2 = 21/2

⇒ h = 10.5 cm.

Hence, curved surface area 

A = 2πrh

⇒ A = 2 × (22/7) × 7 × (21/2)

⇒ A = 463 cm².

∴ The curved surface area of cylinder 462 cm².

Given:

⇒ Height of cylindrical pipe = 14cm

⇒ Volume of metal used = 99 cm

Formula:

Let r = radius of cylinder and h = height of cylinder

⇒ Volume of cylinder = πr²h

⇒ Curved surface area of cylinder = 2πrh

Calculation:

⇒ 2πRh - 2πrh = 44

⇒ R - r = 1/2       ----(1)

Then,

⇒ πR²h - πr²h = 99

⇒ πh(R2 - r2) = 99

⇒ (22/7) × 14 ×(R2 - r2) = 99

⇒ (R² - r²) = 9/4

⇒ (R + r)(R - r) = 9/4

⇒ (R + r) × (1/2) = 9/4

⇒ (R + r) = 9/2 = 4.5

∴ (R + r) = 4.5

Given:

⇒ Edge of cube = 24cm

⇒ Edge of small cube = 12cm

⇒ Edge of small cube = 16cm

Let edge of other third cube be m cm.

Formula:

Let a = edge of cube

⇒ Surface area of cube = 6a²

⇒ Volume of cube = a³

Calculation:

⇒ Volume of large cube = 24³ = 13824 cm³

⇒ 13824 = 12³ + 16³ + m³

⇒ m = 20cm

∴ Surface area of third cube = 6 × 20² = 2400 cm²

Given:

Internal radius of vessel = 5cm

Height of vessel = 24cm

Formula:

Let Inner radius = r, Outer radius = R and h = height and l = slant = height

Volume of cone = 1/3 × πr²h

Volume of cylinder = πr²h

Calculation:

Volume of conical vessel = (1/3 × π × 5² × 24) cm³

Volume of cylindrical vessel = (π × 10² × height) cm³

Then,

(1/3 × π × 5² × 24) = (π × 10² × height)

Height of cylindrical vessel = 2 cm

∴ water rises by 2cm.

Given:

⇒ Side of cube = 22cm

⇒ Height of cone = 21cm

Formula:

⇒ Volume of cube = l³

⇒ Volume of cone = 1/3 × πr²h

Calculation:

⇒ Volume of cone formed = 22³ = 10648 cm³

⇒ 10648 = 1/3 × π × r² × 21

∴ Radius of cone = 22cm

Given:

Height of cone = 24

The curved surface area of cone = 550 cm²

Concept:

Pythagoras theorem:

It states that in a right-angled triangle, the square of the hypotenuse is

equal to the sum of the squares of the other two sides

The curved surface area of cone = πrl

Where,

r = radius of the cone, h = Height and  l = slant height

Calculation:

Let, r, l, and h are radius, height and slant height of cone respectively.

According to question,

Curved surface area of cone = 550 cm2

⇒ 550 = πrl      

⇒ rl × 3.14 = 550       (∵ π  = 3.14)

⇒ rl  ≈ 175 

⇒ r = 175/l   ----(1)

Using pythagoras theorem,

⇒ l² = h² + r²

⇒ l² - r² = 576     (∵ h = 24 cm)

⇒ l2 - (175/l)2 = 576      [From equation (1)]

⇒ l⁴ - 576 l² - 30625 = 0

⇒ l4 - 625 l² + 49 l2 - 625 × 49 = 0

⇒ l²(l² - 625) + 49(l² - 625) = 0

⇒ (l2 - 625)(l2 + 49) = 0

⇒ l² = 625 and l² = -49 (not possible)

⇒ l² = 625

⇒ l = 25 cm.

Hence,

r = 175/l

⇒ r = 7 cm

∴ Required ratio = r : l = 7 : 25

Shortcut Trick Using Pythagoras triplet 7, 24 and 25.

⇒ Height = 24 cm

​⇒ Slant height = 25 cm and radius = 7 cm  [using triplet]

⇒ Radius : Slant height = 7 : 25

∴ The correct answer is option (3).

Shortcut Trick 

2πR₁ = 44 (rolling along length)

⇒ R₁ = 7 cm

2πR₂ = 22 (rolling along width)

⇒ R₂ = 7/2 cm

⇒ x : y = 7² × 22 : (7/2)² × 44 = 2 : 1

∴ x : y = 2 : 1

Alternate Method 

Given:

Length of paper = 44cm

Width of paper = 22cm

Formula:

Volume of cylinder = πr²h

Calculation:

Case: 1

When x = volume of cylinder then

Height (h₁) = 22 cm and

Circumference = 44 = 2π(r₁)

r₁ = 7 cm

Case: 2

When y = volume of cylinder then

Height (h₂) = 44cm

Circumference = 22 = 2π(r₂)

r₂ = 7/2

We know that,

Volume of cylinder = πr2h

⇒ x = π × 7² × 22

⇒ y = π × (7/2)² × 44 =  π × 72 × 22

∴ x : y = 2 : 1

Shortcut Trick 

Mass of the shell = (4/3) × (22/7) × (6³ - 5³) × 3

⇒ Mass of the shell = (88/7) × 91 = 1144 g

∴ Mass of shell = 1144/3 × 3 = 1144 g

Alternate Method 

Given:

⇒ Density of shell = 3 g/cm³

⇒ Inner radius = 5cm

⇒ Outer radius = 6cm

Formula:

⇒ Volume of sphere = 4/3 × πr³

Calculation:

⇒ Volume of shell = 4/3 × 22/7 × (6³ - 5³) = 1144/3 cm³

∴ Mass of shell = 1144/3 × 3 = 1144 g

Concept:

For the interval θ ∈ [0, π/2]:

If θ₁ > θ_2, Then

sin θ₁ > sin θ₂    

cos θ1 < cos θ2   

Complementary angle formula:

sin(π/2 - θ) = cos θ 

cos(π/2 - θ) = sin θ 

Calculation:

1. sin 1° < cos 57°

⇒ sin 1° = sin (90 - 89) = cos 89°

⇒ cos 89° < cos 57°

This statement is true.

2. cos 60° > sin 57°

⇒ sin 57° = sin (90 - 57) = cos 33°

⇒ cos 60° > cos 33°

This statement is false.

∴ From above, option 1 is correct.

Given:

p = sec θ - tan θ and

q = cosec θ + cot θ

Formulae Used:

• sec θ = 1/cos θ
• tan θ = sin θ/cos θ
• cot θ = cos θ/sin θ
• sin²θ + cos²θ = 1

 

Calculation:

Let the required value be x.

⇒ x = p + q(p - 1)

⇒ x = p + qp - q     ---(1)

⇒ p = sec θ - tan θ

⇒ p = $\frac{1-sin\ θ}{cos\ θ}$       ---(2)

⇒ q = cosec θ + cot θ

⇒ q = $[\frac{1+cos\ θ}{sin\ θ}]$    ----(3)

Put the value of p, q in equation (1)

⇒ x = $\frac{1-sin\ θ}{cos\ θ}$ + [$\frac{1-sin\ θ}{cos\ θ}$] × [$\frac{1+cos\ θ}{sin\ θ}$] - $[\frac{1+cos\ θ}{sin\ θ}]$

⇒ x = $\frac{1-sin\ θ}{cos\ θ}$ - $[\frac{1+cos\ θ}{sin\ θ}]$ + [$\frac{1-sin\ θ}{cos\ θ}$] × [$\frac{1+cos\ θ}{sin\ θ}$] 

⇒ x = $\frac{sin\ θ-sin^2\ θ-cos\ θ-cos^2\ θ}{sin\ θ \ cos\ θ}$ + $\frac{1-sin^\ θ+cos\ θ-sin\ θ \ cos\ θ}{sin\ θ \ cos\ θ}$

⇒ x = $\frac{sin\ θ-1-cos\ θ+1-sin^\ θ+cos\ θ-sin\ θ \ cos\ θ}{sin\ θ \ cos\ θ}$

⇒ x = $\frac{-sin\ θ \ cos\ θ}{sin\ θ \ cos\ θ}$

∴ x = -1 

Shortcut Trick Put θ = 45° 

⇒ p = sec 45° - tan 45° = √2 - 1

​⇒ q = cosec 45° + cot 45° = √2 + 1

⇒ p + q(p - 1) = p - q + pq

⇒ (√2 - 1) - (√2 + 1) + (√2 - 1)(√2 + 1)

​⇒ -2 + 1

​⇒ -1 

∴ The correct answer is option (1).

Identity used:

cosec²θ – cot²θ = 1

Calculation:

⇒ cosec θ - cot θ = m      ---- (1)

⇒ (cosec θ - cot θ)(cosec θ + cot θ)/(cosec θ + cot θ) = m

⇒ (cosec²θ - cot²θ)/(cosec θ + cot θ) = m

⇒ 1/(cosec θ + cot θ) = m

⇒ cosec θ + cot θ = 1/m      ---- (2)

By adding (1) and (2), we'll get

2cosecθ = m + 1/m

∴ cosec θ = m/2 + 1/2m

Calculation:

⇒ tan A = BC/AC = a/b

⇒ tan B = AC/BC = b/a

⇒ tan A + tan B = a/b + b/a = (a² + b²)/ab

Using pythagoras theorem,

⇒ c² = a² + b²

∴ tan A + tan B = c²/ab

Formula Used:

sin²A + cos²A = 1

cos (A - B) = cos A cos B + sin A sin B

Calculation:

We have,

cos α + cos β = 2

Squaring,

⇒ cos²α + 2cosα .cosβ + cos²β = 4 ......(1)

⇒ sin α + sin β = 0

Squaring,

⇒ sin²α + 2sinα.sinβ + sin²β = 0......(2)

Adding (1) & (2),

2cos α . cos β + 2 sin α . sin β 

2(cos α . cos β + sin α . sin β)

By using the above formula 

⇒ 2cos(α - β) = 2

⇒ cos(α - β) = 1

⇒ α - β = cos^-1(1) = 0

⇒ α = β

⇒ ? = cos 2α - cos 2β

∴ ? = 0

Shortcut Trick 

Put θ = 60^°

2 + (1/2) = 5/2 (LHS = RHS)

Sin²θ = (√3/2)² = 3/4

∴ The correct answer is 3/4.

Alternate Method 

Formula used:

• cos θ = 1/sec θ
• sin2 θ + cos2 θ = 1

 

Calculation:

We have,

⇒ sec θ + cos θ = 5/2

⇒ 1/cos θ  + cos θ = 5/2

Above equation satisfy for θ = 60°, 

Hence,

sin²θ = sin²60° = (√3/2)² 

sin2θ = 3/4

Alternate Method

We have,

⇒ sec θ + cos θ = 5/2

⇒ 1/cos θ  + cos θ = 5/2

⇒ (1 + cos² θ)/cos θ = 5/2

⇒ 2 + 2cos² θ = 5cosθ

⇒ 2cos²θ - 5cosθ + 2 = 0

⇒ 2 cos2θ - 4cos θ - cos θ  + 2 = 0

⇒ 2cos θ(cos θ - 2) -1(cos θ - 2) = 0

⇒ (2cos θ - 1)(cos θ - 2) = 0

⇒ cosθ = 1/2

⇒ cos² θ = 1/4

We know that,

⇒ sin² θ + cos² θ = 1

∴  sin² θ = 1 - 1/4 = 3/4

Shortcut Trick 

Put θ = 45

⇒ (1 + 1 - √2)(1 + 1 + √2)

⇒ (2² - √2²) = (4 - 2) = 2

∴ (1 + cot θ - cosec θ)(1 + tan θ + sec θ) = 2

Alternate Method 

Formula used:

• sin θ/cos θ = tan θ
• cos θ/sin θ = cot θ
• sin2θ + cos2θ = 1
• cosec θ = 1/sin θ
• sec θ = 1/cos θ
• (a - b)² = a² - 2ab + b²

 

Calculation:

Let the required value be x.

⇒ x = (1 + cot θ - cosec θ)(1 + tan θ + sec θ)

⇒ x = (1 + cosθ/sinθ - 1/sinθ)(1 + sinθ/cosθ + 1/cosθ)

⇒ x = (sinθ + cosθ - 1)(cosθ + sinθ + 1)/(sinθ.cosθ)

⇒ x = [(sin θ + cosθ)² - 1]/(sinθ.cosθ)

⇒ x = [sin²θ + cos²θ + 2sinθ.cosθ - 1]/(sinθ.cosθ)

∴ (1 + cot θ - cosec θ)(1 + tan θ + sec θ) = 2

Formula used:

• (a - b)² = a² - 2ab + b²
• (a + b)2 = a2 + 2ab + b2
• sec2θ - tan2θ = 1

 

Calculation:

Given that,

6 + 8 tan θ = sec θ 

Multiply with 6 on both sides,

36 + 48 tan θ = 6sec θ       ----(1)

Also, we have

8 - 6 tan θ = k sec θ

Multiply with 8 on both sides,

64 - 48 tan θ = 8k sec θ      -----(2)

Adding equation (1) & (2), we get

36 + 48 tan θ + 64 - 48 tan θ  = 6sec θ + 8k sec θ

⇒ 100 = sec θ(6 + 8k) 

⇒ sec θ = 100/(6 + 8k)      -----(3)

Since, 6 + 8 tan θ = sec θ 

⇒ 6 + 8 tan θ = 100/(6 + 8k)  

⇒ 8 tan θ = 100/(6 + 8k)  - 6

⇒ 8 tan θ = (100 - 36 - 48k)/(6 + 8k)

⇒ 8 tan θ = 8(8 - 6k)/(6 + 8k)

⇒ tan θ = (8 - 6k)/(6 + 8k)       ----(4)

We know that, 

sec²θ - tan²θ = 1

Hence, from equations  (3) and (4)

[100/(6 + 8k)]² -  [(8 - 6k)/(6 + 8k)]² = 1 

⇒ [100²  - (8 - 6k)²]/(6 + 8k)² = 1

⇒ 10000 - 64 - 36k² + 96k = 36 + 64k² + 96k

⇒ 10000 - 100 = 100k²

⇒ 9900 = 100k²

⇒ k² = 99

∴ The value of k² is 99.

Calculation:

⇒ Let length of pole be AB metre.

⇒ ∠ ACD and ∠ BCD = ∠m

BD = DA, as D is the midpoint of AB

Δ ADC and Δ BDC are congruent triangles.

⇒ ∠ ADC = ∠ BDC

⇒ ∠ ADC + ∠ BDC = 180 ......(Linear pair angles)

⇒ ∠ ADC = ∠ BDC = 90

In Δ BDC,

⇒∠ BCD = m = 180 - 90 - 30 = 60

⇒ sin 60 = BD/x

⇒ BD = √3x/2

⇒ AB = 2BD = √3x

∴ Length of pole is √3x metre.

Calculation:

The values of tangent can be found using the given table.

Let AB is a height of pole and ECBD is a rectangular plot.

⇒ cot 60 = BD/AB

⇒ 1/√3 = BD/AB

⇒ cot 45 = BC/AB

⇒ BC = AB

As, ECBD is a rectangle and BE is a diagonal.

BD = EC and DE = BC

Using pythagoras theorem,

⇒ BE² = BD² + DE²

⇒ BE² = BD² + BC²

Then,

⇒ cot θ = BE/AB

⇒ cot θ = √(BD² + BC²)/AB

⇒ cot θ = √(AB²/3 + AB²)/AB

∴ cot θ = 2/√3

Formulae Used:

log₁₀(m)ⁿ = n × log10 (m)

log10(ab) = log10(a) + log10(b)

log10(a/b) = log10(a) – log10(b)

Calculation:

log10 31.25 = log₁₀ (3125/100)

⇒ log₁₀ (3125) – log₁₀ (100)

⇒ log10 (5)⁵ – log10 (10)²

⇒ 5[log10 (5)] – 2[log10 (10)]

⇒ 5[log10 (10/2)] – 2

⇒ 5[log10 (10) – log10 (2)] – 2

⇒ 5 – 5[log10 (2)] – 2

⇒ 3 – 5[log10 (2)]

∴ log10 31.25 = 3 – 5 log10 2

Shortcut Trick 

√(15 - 4√14) = √(8 + 7 - 4√14)

⇒ √{(2√2)² + (√7)² - 2 × 2√2 × √7}

⇒ √(2√2 - √7)² = (2√2 - √7)

∴ The square root of 915 – 4√14) is (2√2 – √7)

Alternate Method 

Let 15 – 4√14 = a

⇒ a = 8 + 7 – (2 × 2 × √14)

⇒ a = 8 + 7 – [2 × √(2² × 14)]

⇒ a = (√8)² + (√7)² – (2 × √56)

⇒ a = (√8)2 + (√7)2 – (2 × √8 × √7)

⇒ a = (√8)2 – (2 × √7 × √8) + (√7)2 

⇒ a = (√8 – √7)²

Taking square root on both the sides, we get:

√a = √8 – √7 = 2√2 – √7 

∴ The square root of 915 – 4√14) is (2√2 – √7)

Concept:

Natural numbers: Positive integers which start from 1 and ends at

infinity, such as: 1,2,3,4,5,6,7,8,9,10....∞ 

Calculation:

Let the two alternate numbers be x and (x + 2)

Reciprocals  of two number is

$\frac{1}{x}+\frac{1}{x+2}=\frac{7}{24}$

⇒ $\frac{x+2+x}{x(x+2)}=\frac{7}{24}$

⇒ 24 × (2x + 2) = 7 × (x2 + 2x)

⇒ 48x + 48 = 7x2 + 14x

⇒ 48x – 14x = 7x2 – 48

⇒ 34x = 7x2 – 48

⇒ 7x2 – 34x – 48 = 0

⇒ (x – 6) [x + (8/7)] = 0

⇒ x = 6, -(8/7)

We are given natural numbers only. So, no negative number will be allowed for x.

Hence, the next natural number = 6 + 2 = 8

∴ The sum of the numbers = 6 + 8 = 14

Given equation is 52n - 1

On substituting the value of n = 1,

⇒ 5⁽2×1) - 1 = 25 – 1 = 24

We have:

24 = 2 × 2 × 2 × 3

The factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Hence, total number of factors = 8

∴  52n - 1 is always divisible by a minimum of eight natural numbers

Formulae Used:

log₁₀ x^m = m × log₁₀ x 

log₁₀ (ab) = log10 a + log10 b 

log10 (a/b) = log10 a – log10 b

Calculation:

5x - 3 = 8

5x - 3 = 2³

Taking log₁₀ on both sides,

(x – 3)log₁₀ 5 = 3log₁₀ 2

⇒ x – 3 =  3log10 2/log10 5

⇒ x = (3log10 2/log10 5) + 3

⇒ x = 3[(log10 2/log10 5) + 1]

⇒ x = 3[(log10 2 + log10 5)/log10 5]

⇒ x = 3{[log10 (2 × 5)]/log10 5}

⇒ x = 3(1/log10 5)

⇒ x = 3/log10 5

⇒ x = 3/[log10 (10/2)]

⇒ x = 3/(log10 10 – log₁₀ 2)

∴ x = 3/(1 – log10 2)

Given:

3sin2 θ + 4cos2 θ

Concept used:

sin2 θ + cos2 θ = 1

Calculation:

⇒ 3sin2 θ + 3cos2 θ + cos² θ 

⇒ 3 × (sin2 θ + cos2 θ) + cos2 θ

⇒ (3 × 1) + cos2 θ = 3 + cos2 θ

The least value that cos² θ can take is 0

So, the required least value = 3 + 0 = 3

∴ The least value of 3sin2 θ + 4cos2 θ = 3

Important Points

It must be remembered here that:

Although cos θ can take any value from -1 to +1, but the value of cos² θ can never be negative.

Hence, the minimum value of the cos² θ will be zero.

Concept:

2sin θ cos θ = sin 2θ 

-1 ≤ sin θ ≤ 1

Calculation:

We have,

sin θ cos θ = k

⇒ $\frac{1}{2}$ (2sin θ cos θ) = k

⇒ sin 2θ = 2k       ---(1)

We know that,

-1 ≤ sin 2θ ≤ 1

But, $0 \le \theta \le \frac{\pi }{2}$

0 ≤ sin 2θ ≤ 1

⇒ 0 ≤ 2k ≤ 1

⇒ 0 ≤ k ≤ 0.5    

∴  0 ≤ k ≤ 0.5

Additional Information

We have:

p = sin2 θ + cos4 θ, for 0 ≤ θ ≤ (π/2)

The values of sin θ and cos θ vary according to the following table:

For θ = 0° ⇒ p = 0 + 1 = 1

For θ = 30° = (π/6) radians ⇒ p = (1/2)² + (√3/2)⁴ = (1/4) + (9/16) = 13/16 = 0.8125

For θ = 45° = (π/4) radians ⇒ p = (1/√2)2 + (1/√2)4 = (1/2) + (1/4) = 3/4 = 0.75

For θ = 60° = (π/3) radians ⇒ p = (√3/2)2 + (1/2)4 = (3/4) + (1/16) = 13/16 = 0.8125

For θ = 90° = (π/2) radians ⇒ p = 1 + 0 = 1

Hence the range of values for p is:

0.75 ≤ p ≤ 1 

So, the value of p can neither be less than (3/4) nor greater than 1.

∴ Neither the statement 1 nor statement 2 is correct

Important Points

Remember about the conversion from degrees to radians:

180° = π radians

So, 1° = (π/180) radians

Formula used:

(a + b)² = a² + 2ab + b²

Calculation:

Let 

y = 2 – 2 sin x – sin2 x

⇒ y = 3 - (sin2 x + 2 sin x + 1)

We know that,

(a + b)2 = a2 + 2ab + b2

⇒ y = 3 - (sin x + 1)²    ----(1)

We know that,

-1 ≤ sin θ ≤ 1

Also, we have 0 ≤ x ≤ (π/2). Hence,

sin x ∈ [0, 1]

For maxima, sin x = 0

From equation (1), 

(y)_max = 3 - (1)² = 2

For minima, sin x = 1

⇒ (y)_min = 3 - (1 + 1)² = -1

Hence, required ratio is

y_max/y_min = 2/-1 = -2

∴ ratio of the greatest to the smallest value is -2.

The given equation is:

x2 + y2 - 2xy sin2 θ = 0

⇒ x2 + y2 = 2xy sin2 θ      ----(i)

We can use the following table for the values of sin θ for the various values of θ:

From the table, it is clear that:

If we substitute any value of θ other than 90°, we don not get any real solution for x and y.

If θ = 90°, ⇒ x2 + y2 = 2xy

For, x = y, ⇒ x2 + x2 = 2x² 

∴ The condition is satisfied for x = y, for real values of x and y.

Given:

 The successive discounts of 20%, 10%, and 5% respectively

Formula used:

 Total discount = [x + y – (x y) ÷ 100] %

Calculation

First, we take 20% and 10%

Total discount = [20 + 10 – (20 × 10) ÷ 100] %

Total discount = 30 – 2 = 28 %

Now we take 28% and 5 %

Total discount = [28 + 5 – (28 × 5) ÷ 100] %

Total discount 33 – 1.40 = 31.6%

Hence, overall discount = 31.6%

Given:

The simple Interest (SI) = Rs 500, Time = 5 years, and Rate = 5%

Formula used:

We know the formula,

SI = (P × R × T) ÷ 100

Where P = principle, T = time, and R = Rate

Calculation

We know that formula,

SI = (P × R × T) ÷ 100

Putting the value,

500 = (5 × 5 × P) ÷ 100

P = (500 × 100) ÷ 25

P = 20 × 100 = 2000

Hence, Principle = Rs 2000

Shortcut TrickIn case of Compounded annually questions,

The difference between compound interest and simple interest is given as:

Principal amount = Difference × (100/Rate)²

⇒ Principal amount = 72 × (100/12)²

⇒ Principal amount = (72 × 25 × 25)/9 = 5000

∴ The correct answer is 5000 Rs.

Alternate Method

Formula used:

We know that formula, A = P(1 + r/100)²

CI = A – P and SI = PRT/ 100

Calculation       

Let principle amount = Rs x

We know that, A = P(1 + r/ 100)^t

A = x (1 + 12/100)²

A = x (1.12)²

A = 1.2544 x

CI = A – P

CI = 1.2544x – x = 0.2544x ……..(1)

SI = PRT/100

SI = x × 12 × 2 / 100

SI = 0.24x …….(2)

According to question,

CI – SI = 72

0.2544x – 0.24x = 72

X = 720000/144 = 5000

Hence, the value principle = Rs 5000

Given:

 The distance travels by trains 600 km and 900 km

Taking time 5 hours and 10 hours

Formula used:

 Average speed of the trains = total distance ÷ total time

Calculation

Total distance = 600 + 900 = 1500 km

Total time = 10 + 5 = 15 hours

Now average of the trains = 1500 ÷ 15 = 100 km/hours

Hence, average of the trains = 100 km/hours

Given:

 Walking at 4/5 of his usual speed, a man is 12 minutes late for his office

Calculation

Let the usual time be x minute

Walking at 4/5^th of his usual speed, he will take time 5/4 of the usual time

Now, 5/4^th of the usual time = usual time + 12 minute

5/4 x = x + 12                     

5/4x – x = 12

1/4 x = 12

X = 48 minute

Hence, he usually time take = 48 minute

Shortcut Trick

 

⇒ 1 unit = 12 min. [given]

⇒ usual time = 4 unit = 4 × 12

⇒ usual time = 48 min.

∴ The correct answer is option (1).

Given:

 The cost price of 100 mangoes is equal to the selling price of 80 mangoes

Calculation

Let selling price = cost price

According to question,

100 × CP = 80 × SP

CP ÷ SP = 80/100

CP ÷ SP = 4/5

Profit percentage = (5 – 4) ÷ 4 × 100

Profit percentage = 1 ÷ 4 × 100

Profit percentage = 25%

Hence, profit percentage = 25%

Shortcut TrickAfter selling 80 mangoes the seller recovers the cost price,

Hence, Profit = 100 - 80 = 20

Cost price will be 80 because that is total number of mangoes seller sold

Profit % = (Profit × 100)/Cost

Profit % = (20 × 100)/80 = 25

∴ The correct answer is 25%.

Shortcut Trick 

X : Y = 3 : 4

X : Z = 5 : 4

⇒ X : Y : Z = 15 : 20 : 12

⇒ % Less = (8/20) × 100 = 40%

∴ Z is 40% cheaper than Y

Alternate Method 

Given:

 X sells his goods 25% cheaper than Y and 25% dearer than Z

Calculation

Let price to y goods = x

Then price of y goods = x – 25 ÷ 100 = x – x/4 = 3x/4 ….. (1)

Let price of x sells = y + 25y/100 = y + y/4 = 5y/4 ….(2)

From the equation 1 and 2

⇒ 3x/4 = 5y/4

⇒ 3x= 5y

⇒ Y = 0.6x

Then y = 60% of x

Percentage of Z’s good cheaper than y = 100 – 60 = 40%

∴ Z is 40% cheaper than Y

Alternate Method

Let the price of y be 100

As X is 25% cheaper, rate of x = 100 - 25% of 100 = 75

Ax X is 25% dearer than Z

Price of X = Z + 25% of Z

⇒ 75 = (5/4) × Z

⇒ Z = 60

Z is 100 - 60 = 40 cheaper than Y

⇒ Z is (40/100) × 100 = 40% cheaper than Y

∴ Z is 40% cheaper than Y

Shortcut Trick

                          before             after

   liquid                 1                     1

   water                 3                     4

⇒ Quantity of added water = (80/4) × 1 = 20 litres

∴ The correct answer is 20 litres.

Alternate Method 

Given:

 In a mixture of 80 liters of a liquid and water, 25% of the mixture is the liquid

Calculation

Liquid amount = 80 × 25 ÷ 100

Liquid amount = 20 liters

Let x be the amount of water be added

According to question,

⇒ 20 = 20% of (80 + x)

⇒ 20 = 20 ÷ 100 (80 + x)

⇒ 100 = 80 + x

⇒ X = 20 liters

∴ The correct answer is 20 litres.

Shortcut Trick 

(20 × 20) / 20 = ( 16 × D₂) / 16

⇒ D₂ = 20 days

∴ The correct answer is 20 days.

Alternate Method 

Given:

 if 20 persons can clean 20 floors in 20 days

Formula used:

We know that, M₁ D₁ / W₁ = M₂ D₂ / W₂

Calculation

⇒ D₂ = 20 × 20 × 16 / 20 × 16

⇒ D₂ = 20 days

∴ The correct answer is 20 days.

Given:

the work done by (x - 1) men in (x + 1) days : work done by (x + 2) men in (x - 1) days = 9 : 10

Calculation       

According to question,

(x - 1)(x + 1)/(x + 2) (x - 1) = 9/10

⇒ (x + 1)/(x + 2) = 9/ 10

⇒ 10x + 10 = 9x +18

⇒ x = 8

∴ The value of x is 8.

Given:

The expressions 9x² + 8kx + 16 = 0

Calculation       

For real and equal roots D = 0

We know that, b² – 4ac = 0

Put the value, b = 8k, a = 9 and c = 16

b² – 4ac = 0

(8k)² – 4 × 16 × 9 = 0

64k² – 576 = 0

K² = 576/64 = 9

K = -3, +3

Hence, the value of k = -3, +3

Given:

Given that x + 1/ x = 8

Calculation       

The expression x + 1/ x = 8

Squaring both sides

x² + 1/x² = 64 – 2 = 62

∴ The value of x² + 1/x² is 62.

Given:

The equation 4x² - 2kx + 3k = 0 has equal roots

Calculation       

The equation 4x² - 2kx + 3k = 0

Here, a = 4, b = -2k, and c = 3k

D = b² – 4ac = 0

Put the value,

(-2k)² – 4 × 4 × 3K = 0

4k² – 48k = 0

4k (k – 12) = 0

K = 0

And (k - 12) = 0

K = 12

Hence, the value of K = 0, 12

Shortcut Trick

Sum of roots = -b / a 

-(-6) / p = 6

⇒ p = 1

Product of roots = c / a

⇒ q / p = 6

⇒ q = 6

⇒ p + q = 7

Therefore the correct answer is 7.

Alternate Method 

Given:

 The expression px² – 6x + q

Sum of root = 6 and product of root = 6

Formula used:

Sum of root = - b/a and product of root = c/a

Calculation

The expression px² – 6x + q

Here, a = p, b = -6 and c = q

Sum of roots = - b/a = - (- 6)/p

6 = 6/p

⇒ p = 1

Now, product of roots = c/a = q/p

6 = q/1 (put the value of p)

Then q = 6

Hence, (p + q) = 6 + 1 = 7   

Therefore the correct answer is 7.

Calculation:

Let, 

P(x) = 4x3 + 12x2 - x - 3

⇒ P(x) = 4x²(x + 3) - 1(x + 3)

⇒ P(x) = (4x² - 1)(x + 3)

We know that, (a² - b²) = (a + b)(a - b)

⇒ P(x) = (2x - 1)(2x + 1)(x + 3)

We can see that,  (2x - 1) & (2x + 1) are factor of polynomial. ​

∴ 4x3 + 12x2 - x - 3 is divisible by both (2x + 1) and (2x - 1)