CDS Elementary Mathematics 19 Nov 2017 Official Paper (Previous Year Paper)

Height of the cone = Total height – Height of the cylinder = 15 – 10 = 5 cm

Radius of the cone = Radius of the cylinder = 50 m

Total volume of tent = Volume of cylinder + Volume of cone

⇒ πr²H + 1/3 × πr²h

⇒ π50² × 10 + 1/3 × π50² × 5

⇒ 50² × π × (10 + 5/3)

⇒ 2500 × 35/3 × π

⇒ 87500π/3 cm³

When a rectangle is formed into a cylinder, the length of the rectangle becomes the circumference of base and breadth becomes the height.

For first rectangle

2πr = 4π

∴ r = 2

h = 2π

Volume = π r² h

⇒ π × 2² × 2π = 8π²

For second rectangle

2πr = 5π

∴ r = 5/2

h = π

Volume = π r² h

⇒ π × (5/2)² × π = 6.25π²

Maximum possible volume = 8π²

Concept:

Median: A line segment that joins a vertex to the mid-point of the

the side that is opposite to that vertex

Basic proportionality theorem(BPT): If a line is drawn parallel to

one side of a triangle intersecting the other two sides in distinct

points, then the other two sides are divided in the same ratio. 

Calculation:

Given that,

AD and BE are medians of BC and AC respectively.

⇒ BC = 2DC          ------(1)

Also, we have AC = 9 cm

⇒ AE = EC = 9/2 = 4.5

In triangle CBE,  DF ∥ BE

Using the Basic proportionality theorem

⇒ $\frac{DC}{BC} = \frac{FC}{EC}$

⇒  $\frac{DC}{2DC} = \frac{FC}{4.5}$

⇒ FC = 4.5/2 = 2.25 cm

∴ CF = 2.25 cm.

Given:

PR = 10 cm, RX = 4 cm, YR = 2 cm, QR = 5 cm

Concept:

Basic proportionality theorem(BPT): If a line is drawn parallel to

one side of a triangle intersecting the other two sides in distinct

points, then the other two sides are divided in the same ratio. 

Calculation:

We have XR = 4 cm

⇒ PX = PR – XR = 10 – 4

⇒ PX = 6 cm

⇒ YR = 2 cm

⇒ QY = QR – YR = 5 – 2

⇒ QY  = 3 cm

⇒ $\frac{XR}{PR} = \frac{4}{6} = \frac{2}{3}$        ----(1)

⇒ $\frac{YR}{QY} = \frac{2}{3}$                -----(2)

From equation (1) & (2)

$\frac{XR}{PR} = \frac{YR}{QY}$

Using Basic proportionality theorem (BPT)

⇒  XY ∥ PQ

∴ XY is parallel to PQ.

Since they are in a plane (Let us consider XY plane) all line is of form y = mx + c

Let the slopes of the lines A, B and C be a, b and c respectively

Lines are parallel if their slopes are equal and perpendicular if the product of their slopes is -1

Statement 1:

A and C are parallel

∴ a = c

B and C are parallel

∴ b = c

Comparing both the equations

⇒ a = b

∴ A and B are parallel too. Statement 1 is true

Statement II:

A and C are perpendicular

∴ ac = -1 or a = -1/c

B and C are perpendicular

∴ bc = -1 or b = -1/c

Comparing both the equations

a = b

∴ A is parallel to B

∴ Statement II is true

Statement III:

Statement III need not be true because the lines A and B can be on either side of line C forming same angle but are not parallel

Let AC = 48 cm and BD = 20 cm

In a rhombus, diagonals bisect each other at right angles

∴ AO = 48/2 = 24 cm and BO = 20/2 = 10 cm

In triangle AOB

AB² = AO² + BO² [∵ Pythagoras theorem]

⇒ 24² + 10²

⇒ 676

∴ AB = 26 cm

∴ Side of the rhombus is 26 cm

Angle subtended by full circle at center = 2π

Angle subtended by the given arc = π

∴ The arc is a semi-circle

Arc length of a semi-circle = π r = 22

⇒ 22/7 × r = 22

∴ r = 7 cm

Concept:

Similar triangle:

• Two triangles are similar if their corresponding angles are equal and their corresponding sides are within the same ratio (or proportion).
• The ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

Calculation:

In triangle ADE and ABC

∠ A is common

We  know that corresponding angles are equal

⇒ ∠ ADE = ∠ ABE 

Therefore ADE ∼ ABC

Using the property of similarity,

$\frac{Ar(Δ ADE)}{Ar(Δ ABC)} = (\frac{DE}{BC})^2$

According to the question 

Area of ΔABC = 5 Area of ΔADE and BC = 10 cm

⇒ $\frac{1}{5} = (\frac{DE}{10})^2$

⇒  DE² = 20 cm

∴ DE = 2√5

Each pair can have one intersection point

Number of pairs in 8 lines = ⁸C₂ = 8! / (2! × 6!) = 28

Given:

Number of sides in polygon = 6

One of its angles = (30°+ each of the other five equal angles)

Formula used:

Sum of angles of a polygon = (n – 2) × 180° 

Calculation:

Let the equal angles are x and the larger angle be x + 30

Sum of angles = 5x + x + 30 = 6x + 30

We know that,

Sum of angles of polygon = (n – 2) × 180° 

⇒ 6x + 30 = (6 – 2) × 180°      [∵ n =  6]

⇒ 6x + 30 = 720° 

⇒ 6x = 690° 

∴ x = 115°

Point of intersection of the perpendicular bisector is the circumcenter of the triangle.

It is possible for three chords to form a triangle but do not include the center when it is an obtuse angled triangle

∴ Statement 1 is possible

In a right triangle the perpendicular drawn from two vertices of the hypotenuse are the sides themselves. They meet on the vertex and lie on the side

∴ Statement 2 is possible

∴ Both 1 and 2 true

Number of students studying Physics = ¼ × Total number of students

⇒ ¼ × 1200 = 300 students

Number of boys studying in physics = 60% of number of students studying Physics

⇒ 60/100 × 300 = 180 boys

Number of boys studying Statistics = Total number of students studying Statistics – Number of girls studying Statistics

⇒ 320 – 250 = 70 boys

∴ Total number of boys studying Statistics and Physics = 180 + 70 = 250 boys

Total number of students studying chemistry = 1200 – Number of students studying Physics – Number of students studying Mathematics – Number of students studying Statistics

⇒ 1200 – ¼ × 1200 – 20/100 × 1200 – 320

⇒ 1200 – 300 – 240 – 320 = 340 students

Required percentage = Number of girls studying Statistics/Number students studying Chemistry

⇒ 250/340 × 100 ≈ 73.5%

Mathematics:

Total number of students = 240 [From previous question]

Number of boys = 150

Number of girls = 240 – 150 = 90

Difference between number of boys and girls = 150 – 90 = 60

Chemistry:

Total number of students = 340

Number of girls = 3/5 × Total = 3/5 × 340 = 204

Number of boys = 340 – 204 = 136

Difference = 204 – 136 = 68 students

Statistics:

Total number of students = 320

Number of girls = 250

Number of boys = 320 – 250 = 70

Difference = 250 – 70 = 180

Physics:

Total number of students = 300

Number of boys = 60/100 × 300 = 180

Number of girls = 300 – 180 = 120

Difference = 180 – 120 = 60

∴ Difference is same (60) for Mathematics and Physics

Physics:

Total number of students = 300

Number of boys = 60/100 × 300 = 180

Number of girls = 300 – 180 = 120

∴ Difference between number of boys studying Mathematics and number of girls studying Physics = 150 – 120 = 30

Total number of boys = 150 + 180 + 70 + 136 = 536

Total number of girls = 90 + 250 + 120 + 204 = 664

Required ratio = 536 : 664 = (67 × 8) : (83 × 8) = 67 : 83

Frequency density is the ratio of the class frequency to the class width of the particular class.

Total amount of money spent as salary = Salary for category A worker + Salary for category B workers + Salary for category C workers

⇒ 65000 + 3 × 25000 + 5 × 20000 = 240000

Average salary = Total salary/Total number of workers

⇒ 240000/9 = 26666. 67

All B and C category workers get salary less than mean salary

∴ The number of workers earning less than the mean salary = 5 + 3 = 8

Formula used:

Time = Distance/Speed

Average speed = Total distance / Total time

Calculation:

Total distance = 12 + 10 = 22 km

Since, time = distance/speed

Total time = 12/4 + 10/5 = 5 hours

By using the above formula 

∴ Average speed = 22/5 = 4.4 km/hr

Additional Information

1. Average Speed = 2ab/(a + b)

• Applicable when one travels at speed a for half the distance and speed b for the other half of the distance.
• In this case, the average speed is the harmonic mean of the two speeds. On similar lines, you can modify this formula for one-third distance.

2. Average Speed = (a + b)/2

• Applicable when one travels at speed a for half the time and speed b for another half of the time.
• In this case, the

In pie chart, Data ∝ Area

Area = πr²

∴ Income ∝ Radius²

Income of A/Income of B = (16/9)² = 256/81

In option (3)

Income of A/Income of B = 25600/8100 = 256/81

∴ The appropriate data is Rs. 25,600 and Rs. 8,100

Mode is defined as the data with maximum frequency. In case two or more data have same frequencies, then both are mode

∴ Mode is not unique

For example, If the data set is 13, 14, 14, 15, 15 and 16

Both 14 and 15 are modes

∴ Statement I is true but Statement II is false

Data in a pie chart ∝ Area

Area = θ/360 × πr²

∴ Area ∝ θ

∴ Data ∝ θ, where θ is the angle of slice

Length of the curved surface = θ/360 × 2πr

∴ Length of curved surface of arc ∝ θ

∴ Data ∝ Length of the curved surface

Perimeter of slices = θ/360 × 2πr + 2r

∴ Perimeter of slices is not proportional to the data

Geometric mean of n numbers = (Product of numbers)^1/n

⇒ Geometric mean of x and y = (xy)^1/2

⇒ 6 = (xy)^1/2

⇒ Squaring on both the sides

⇒ 36 = xy

Geometric mean of x, y and z = (xyz)^1/3

⇒ 6 = (xyz)^1/3

Cubing on both the sides

⇒ 216 = xyz

⇒ 216 = 36z

∴ z = 216/36 = 6

The data required was not collected directly from source but from others

∴ It is secondary data

Given:

5 students are 150, 165, 161, 144, and 155.

Formula used:

Mean = Sum of term/Number of term

Median = (n + 1)/2 th term, If n is odd

Calculation:

On arranging the heights in ascending order

144, 150 155, 161 and 165

Median = Middlemost term = 155

Sum of the heights = 144 + 150 + 155 + 161 + 165 = 775

Mean = Sum/Number of observations = 775/5 = 155

∴ Mean and median are 155 and 155 respectively

The difference in the averages of the first class and second class = 152 – 140 = 12

This difference from 28 students will be now distributed to all students

Increase in average = 28 × Difference/Number of students

⇒ 28 × 12/ (28 + 22) = 6.72

∴ New average = 140 + 6.72 = 146.72 cm

Amount spent for maintaining 8 cows for 60 days = Rs. 6400

⇒ Amount spent for maintaining 8 cows for 1 day = 6400/60 = Rs. 1600/15

⇒ Amount spent for maintaining 1 cow for 1 day = (1600/15) /8 = Rs. 200/15

⇒ Amount spent for maintaining 5 cows for 1 day = 200/15 × 5 = 200/3

⇒ Amount spent for maintaining 5 cows for n days = 200n/3 = Rs. 4800

⇒ 4800 = 200n/3

∴ n = 4800 × 3/200 = 72 days

Given :

Passing marks = 40%

Marks obtained by student = 45

Calculations :

Total marks required for passing = 45 + 5 = 50 marks

50 marks is 40% of the maximum marks

According to the question, we have

⇒ 50 = 40/100 × Maximum marks

⇒ Maximum marks = 50 × 100/40 = 125 marks

∴ The maximum marks are 125 marks.

3 (2u + v) = 7uv

On dividing by uv

⇒ 6/v + 3/u = 7       ---- 1

3 (u + 3v) = 11uv

On dividing by uv

⇒ 3/v + 9/u = 11     ---- 2

2 × Equation 2 – Equation 1

⇒ 6/v + 18/u – 6/v – 3/u = 22 – 7

⇒ 15/u = 15

∴ u = 15/15 = 1

Shortcut:

Let Ram’s age be ‘x’

His age five years ago = 3 × (Shyam’s age five years ago)

∴ (x – 5) should be a multiple of 3

Only option that satisfies this condition is x = 32

Detailed solution:

Let the age of Ram’s and Shyam’s 5 years ago from present be x and y respectively

Then, Ram’s age was thrice as Shyam’s

⇒ x = 3y

Four years after from present, their age will x + 9 and y + 9

Now Ram’s age is twice the Shyam’s age

⇒ x + 9 = 2 × (y + 9)

⇒ 3y + 9 = 2y + 18 [∵ x = 3y]

∴ y = 9 and x = 3 × 9 = 27

∴ Ram’s present age = 27 + 5 = 32 years

Let the cost price of chair and stool be Rs. x and Rs. y per unit

Cost of 4 chairs and 9 stools = Rs. 1340

⇒ 4x + 9y = 1340       ---- 1

Profit = Profit%/100 × CP

⇒ Profit he makes = 10/100 × 4x + 20/100 × 9y

⇒ 188 = 0.4x + 1.8y

⇒ 10 × 188 = [0.4x + 1.8y] × 10

⇒ 1880 = 4x + 18y

⇒ 1880 = 4x + 9y + 9y

⇒ 1880 = 1340 + 9y

⇒ 9y = 540

y = 60, Substituting in equation 1

⇒ 4x + 540 = 1340

∴ x = Rs. 200

∴ Amount he spends for chairs = 4x = Rs. 800

Solution for the given equation x + 5 = 5 is x = 0

∴ Set of the equations = {0}

Third option cannot be used because it doesn’t represent a set

ϕ means there is no such solution and cannot be used

ab + bc + ca = 0

⇒ -bc = a (b + c)

Similarly -ca = b (c + a) and -ab = c (a + b)

⇒ a²/(a² – bc) + b²/(b² – ca) + c²/(c² – ab) = a²/[a² + a (b + c)] + b²/[b² + b (c + a)] + c²/[c² + c (a + b)]

⇒ a²/[a (a + b + c)] + b²/[b (a + b + c)] + c²/[c (a + b + c)]

⇒ a/(a + b + c) + b/(a + b + c) + c/(a + b + c)

⇒ (a + b + c)/(a + b + c) = 1

Given :

35% students failed in Hindi

45% failed in English

20% students failed in both the subjects

Formula used :

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)        ------ (1)

Calculations :

Let the total number of students be 100.

Percentage of students failed = n(A ∪ B)

Percentage of students who failed in Hindi = n(A)

Percentage of students who failed in English = n(B)

Percentage of students who failed in both = n(A ∩ B)

Using equation (1), we get

⇒ Percentage of students failed = 45 + 35 - 20

⇒ 60%

Students passed in both subjects = 100 - 60 = 40%

∴ The percentage of students who passed in both the subjects is 40%.

Let (x – y), (y – z) and (z – x) be a, b and c respectively

⇒ a + b + c = x – y + y – z + z – x = 0

If a + b + c = 0, then a³ + b³ + c³ = 3abc

⇒ [(x – y) (y – z) (z – x)]/[(x –y)³ + (y – z)³ + (z – x)³] = [(x – y) (y – z) (z – x)]/3[(x – y) (y – z) (z – x)] = 1/3

Let $\sqrt {1 + \sqrt {1 + \sqrt {1 + \cdots } } } {\rm{;}} = {\rm{;x}}$

Squaring on both the sides

$\Rightarrow ;1 + \sqrt {1 + \sqrt {1 + \sqrt {1 + \cdots } } } {\rm{;}} = {\rm{;}}{{\rm{x}}^2}$

⇒ 1 + x = x²

⇒ x² – x – 1 = 0

Using quadratic equation and neglecting negative root

⇒ x = (1 + √5)/2

∴ 1 < x < 2

log₁₀ 8000 + log₁₀ 600 = log₁₀ (8 × 1000) + log₁₀ (6 × 100)

⇒ log₁₀ 8 + log₁₀ 1000 + log₁₀ 6 + log₁₀ 100 [∵ log(mn) = log m + log n]

⇒ 0.9031 + 3 + 0.7782 + 2 = 6.6813

After 24 days, let ‘x’ men leave

⇒ 16 days of work for 30 men is remaining

⇒ Remaining men needs 40 days to complete the work

⇒ Number of days taken ∝ 1/Number of men

⇒ Number of days taken × Number of men = constant

⇒ 16 × 30 = 40 × (30 – x)

⇒ 16 × 30/40 = 30 – x

⇒ 12 = 30 – x

⇒ x = 18

∴ 18 men leave after 24 days.

6 sheep are equal to 4 goats

⇒ 3 sheep are equivalent to 4/2 = 2 goats

⇒ 2 goats and 3 sheep are equivalent to = 2 + 2 = 4 goats

∴ 4 goats can graze the field in 50 days [Given]

The tank will get filled in 12 hours instead of 10 hours

2 hours of the extra water filled by the tap is emptied by the outlet in 5 hours

∴ 10 hours of the water by tap (Full tub) will be emptied in 10/2 × 5 = 25 hours

Let the distant between his school and house be ‘x’

Total time taken = Time taken from school to house + Time taken from house to school

50/60 h = x/12 + x/8 [∵ Speed = Distance/time]

50/60 = (2x + 3x) /24

50/60 = 5x/24

∴ x = 4 kms

∴ Total distance walked = 2x = 8 kms

Given:

78 divided into 3 parts which are proportional to 1, $\frac{1}{{3}}$ and $\frac{1}{{6}}$

Shortcut Trick1 + $\frac{1}{{3}}$ + $\frac{1}{{6}}$ = 78

⇒ 9/6 = 78

⇒ 1 = 78 × (6/9) = 52

⇒ $\frac{1}{{3}}$ = 52/3

Alternate Method

Lcm of 1, 3 & 6 is 6

Multiply the 

⇒  [1 ∶ $\frac{1}{{3}}$ ∶ $\frac{1}{{6}}$] × 6 = 6 : 2 : 1

Let the parts are 6x, 2x and x respectively

⇒ 6x + 2x + x = 78

⇒ 9x = 78

⇒ x = $\frac{78}{{9}}$

Hence, middle part = 2x = (2 × $\frac{78}{{9}}$) 

⇒ $\frac{156}{{9}}$ = $\frac{52}{{3}}$

∴ The required value is $\frac{52}{{3}}$.

Let the student studying in first and second standard be 2x and 3x respectively

Students studying in third standard = 5/4 × 3x = 3.75x

Total number of students studying = 2x + 3x + 3.75x

⇒ 350 = 8.75x

∴ x = 40

∴ Number of students in first and third standards = 5.75x = 5.75 × 40 = 230

Formula used:

For 2 year

CI - SI = P× [(R/100)²]

For 3 year

CI - SI = $P[(\frac{R}{100})^2× (\frac{300+R}{100})]$

Here also, P = principal amount, R = rate of interest

Calculation:

Given that, CI - SI = 15, R = 5%, n = 2

⇒ 15 = P(5/100)²

⇒ P = 15 × 400

∴ P = 6000 Rs.

Let the initial price be Rs x and consumption be y

Expenditure = xy [∵ Price(Income) = Expenditure/consumption]

12% increase in prices

⇒ New price = (100 + 12)/100 × x = 112x/100 = 28x/25

New consumption = Expenditure/New price

⇒ New consumption = (xy)/(28x/25)

⇒ New consumption = 25y/28

∴ Percentage decrease = (y – 25y/28)/y × 100

∴ Percentage decrease $= {\rm{;}}300/28{\rm{;}} = {\rm{;}}10\frac{5}{7}\%$

Speed = Distance/Time

Downstream Speed = 18/4 = 4.5 km/h

Upstream Speed = 18/10 = 1.8 km/h

∴ Statement 1 is true

Speed of the man in still water = ½ × (Speed in downstream + Speed in upstream)

⇒ ½ × (4.5 + 1.8) = 3.15 km/h

∴ Statement 2 is true

Speed of the stream = ½ × (Speed in downstream – Speed in upstream)

⇒ ½ × (4.5 – 1.8) = 1.35 km/h

∴ Statement 3 is true

∴ Statement 1, 2 and 3 are correct

5² + 12² = 13²

⇒ The given triangle is a right angle triangle

There are two acute angles but both of them will give the same results

Let us consider the angle between sides measuring 12cm and 13 cm

⇒ cosθ + sinθ = adjacent/hypotenuse + opposite/hypotenuse

⇒ 12/13 + 5/13 = 17/13

∴ cosθ + sinθ = 17/13

In the given interval, both sinθ and cosecθ are positive

For two positive numbers

A.M. ≥ GM

(sinθ + cosecθ)/2 ≥ √(sinθ cosecθ)

(sinθ + cosecθ)/2 ≥ 1 [∵ sinθ = 1/cosecθ]

∴ sinθ + cosecθ ≥ 2

But for it to be 2, sinθ = cosecθ = 1 or θ = π/2

But θ is less than π/2

∴ sinθ + cosecθ > 2

Concept:

• (a – b)2 = (a + b)2 – 4ab
• sin2θ + cos2θ = 1

Calculation:

Given: sinθ = (m2 – n2)/(m2 + n2) and 0 < θ < π/2

Here, we have to find the value of cos θ 

sinθ = (m² – n²)/(m² + n²)-----------(Given)

By squaring on both the sides we get

Sin²θ = (m² – n²)²/(m² + n²)² [∵ (a – b)² = (a + b)² – 4ab]

Sin²θ = [(m² + n²)² – 4m²n²]/(m² + n²)² [∵ (a – b)² = (a + b)² – 4ab]

Sin²θ = 1 – 4m²n²/(m² + n²)²

4m²n²/(m² + n²)² = 1 – sin²θ

4m²n²/(m² + n²)² = cos²θ [∵ sin²θ + cos²θ = 1]

By taking the square roots on both the sides we get

∴ 2mn/(m² + n²) = cosθ

Concept:

Theorem:  The angle subtended by an arc of a circle at its centre is

twice the angle it subtends anywhere on the circle's circumference.

Circumradius: The radius of the circle circumscribed around a triangle. 

The angles opposite to equal sides of a triangle are equal.

Calculation:

Given that, ∠A = 30 °, then using above theorem

∠O = 60° 

In triangle OBC, 

OB = OC = radius of the circle = 10 cm

Since, angles opposite to equal sides of a triangle are equal.

⇒  ∠OBC = ∠OCB = 60° 

⇒ OBC is an equilateral triangle

⇒ OB  OC =  BC = 10 cm

∴ Length of side BC is 10 cm.

Sin45° = cos45° = 1/√2

sin30° = cos60° = 1/2

sec45° = cosec45° = √2

tan45° = cot45° = 1

∴ A = (1/√2 – 1/2) / (1/√2 + 1/2) [(Numerator and denominator) × 2]

⇒ A = (√2 – 1) / (√2 + 1)

⇒ B = (√2 – 1) / (√2 + 1)

∴ A = B

Let the common root be ‘a’ and the other roots of the equation are ‘b’ and ‘c’ respectively.

Sum of roots = -b/a and Product of the roots = c/a

For the first quadratic equation

a + b = 5      ---- 1                                                                                                                                     

ab = α      ---- 2

For the second quadratic equation

a + c = 7      ---- 3

ac = 2α      ---- 4

On dividing equation 4 by equation 2

⇒ c/b = 2

⇒ c = 2b

On subtracting equation 3 and equation 1

⇒ c – b = 2

⇒ 2b – b = 2

∴ b = 2

Substituting in equation 1

a + 2 = 5

∴ a = 3

α = ab = 3 × 2 = 6

Number divisible by 2, 3, 4 and 5 should be divisible by their LCM

⇒ LCM of 2, 3, 4 and 5 = 60

Number of multiples of 60 in 1 to 1000 = Quotient of 1000/60 = 16

[1 × 60, 2 × 60, ⋯, 16 × 60 = 960]

∴ Total 16 numbers from 1 to 1000 are divisible by 2, 3, 4 and 5.

Detailed solution∶

Getting the first root by trial and error method∶

Since the sum of the coefficients is not zero, 1 is not a root

Let x = -2

⇒ -8 + 16 + 22 – 30 = 0

∴ -2 is or (x + 2) is a factor

Dividing x³ + 4x² – 11x – 30 by (x + 2)

Quotient = x² + 2x – 15

⇒ x² + 5x – 3x – 15 = 0

⇒ x (x + 5) – 3 (x + 5) = 0

⇒ (x – 3) (x + 5) = 0

Other factors are (x – 3) and (x + 5)

∴ (x + 2), (x – 3) and (x + 5) are the roots of the equation

Alternative method∶

Since all the given options have only 3, 5 and 2 in different signs, this shortcut can be used

Product of the roots = -d/a = 30, positive

Sum of the roots = -b/a = -4

For product to be positive, either all terms are positive or two terms are negative

But since sum is negative, two terms has to be negative

Trying all thee combination of two roots to be negative

2 – 5 – 3 = -6

3 – 5 – 2 = -4

Second combination matches the condition –b/a = 4

∴ The roots are 3, -5 and -2

∴ The factors are (x – 3), (x + 5) and (x + 2)

x = 111­ - - - - 1 (20 digits)

⇒ x = 1[10¹⁹ + 10¹⁸ + 10¹⁷ + 10¹⁶ + _ _ _ _ _ + 10² + 10¹ + 1]

Let [10⁹ + 10⁸ + 10⁷ + 10⁶ + _ _ _ _ _ + 10² + 10¹ + 1] = a

 ⇒ x = 1 × 10¹⁰ [10⁹ + 10⁸ + 10⁷ + 10⁶ + _ _ _ _ _ + 10² + 10¹ + 1] + a

⇒ x = 10¹⁰ a + a

Similarly, y = 3a and z = 2a

(x – y²)/z = [(10¹⁰ + 1) a – 9a²]/2a

⇒ a × [(10¹⁰ + 1) – 9a] /2a

⇒ [10¹⁰ + 1 – 9a] /2

⇒ [1 + 1] /2 = 1[∵ 10¹⁰ – 9a = 1]

∴ (x – y²)/z = 1

Let the roots be 3t and 2t respectively

Product of the roots = c/a

3t × 2t = 5/12

t² = 5/72

∴ t = ± √5/(6√2)

Sum of roots = -b/a

3t + 2t = -m/12

-60t = m

∴ m = -60 × ± √5/(6√2) = ± 5√10

Only positive value of m is required

∴ m = 5√10

Concept:

Degree is the highest power of the polynomial

Highest power of the product of polynomials = Product of the highest power of the two polynomials

Calculation:

⇒ for ex: a × x⁴ × b × x³ = ab x⁷

∴ Degree = 7

Concept used:

Remainder theorem: 

If a polynomial p(x) is divided by (x−a), then the remainder is a

constant given by p(a).

Calculation:

Let p(x) = 5x3 + 5x2 – 6x + 9 

Since, (x + 3) divide p(x), then, remainder will be p(-3).

⇒ p(-3) = 5 × (-3)3 + 5 × (-3)2 – 6 × (-3) + 9

⇒ p(-3) = -63

LCM × HCF = Product of the numbers

∴ LCM = Product of the numbers/HCF

⇒ [(x + y + z) p³] /p²

∴ Their LCM is (x + y + z) p

Concept:

$a\bar{b} = \frac{ab - a}{90}$

Calculation:

By using the above concept

$8\bar{3} = \frac{83 - 8}{90} = \frac{75}{90}$

$6\bar{2} = \frac{62 - 6}{90} = \frac{56}{90}$

Distance = 75/90 - 56/90 = (75 - 56)/90

∴ Distance = 19/90

Given:

The successive discount are 20%, 12.5% and 5%.

Formula:

Successive discount = X + Y - (XY)/100

Where,

X = First discount

Y = Second discount

Calculation:

Successive discount = X + Y - (XY)/100

⇒ 20 + 12.5 - 20 × 12.5/100

⇒ 32.5 - 2.5 = 30

Now, again

Successive discount = X + Y - (XY)/100

X = 30%, Y = 5%

⇒ 30 + 5 - (30 × 5)/100

⇒ 35 - 150/100

⇒ 35 - 1.5 = 33.5%

∴ The single equivalent discount is 33.5%.

Alternate Method

Let the Marked price be Rs 100 [Since only the percentage is asked, this assumption is not wrong]

⇒ Price after 20% discount = 100 – 20 = Rs. 80

⇒ Price after 12.5% discount = (100 – 12.5)/100 × 80 = Rs. 70

⇒ Price after 5% discount = (100 – 5)/100 × 70 = Rs. 66.5

Net discount = Rs. 100 – Rs. 66.5 = Rs. 33.5

∴ Discount % = 33.5/100 × 100 = 33.5%

Let the original number of mangoes he had be ‘x’

Number of mangoes which were not rotten = (100 – 5)/100 × x = 0.95x

Number of mangoes unsold out of them = (100 – 75)/100 × 0.95x = 0.25 × 0.95x

⇒ 95 = 0.25 × 0.95x

⇒ 100 = 0.25x

∴ x = 400

∴ He had 400 mangoes originally

Given:

The train crosses a km-stone in 12 seconds

Speed of train = 60 Km/hr

Concept:

Speed = Distance/Time

1 Km/hr = 5/18 m/s

Kilo-stones: Kilo-stones are road stones marking distances in

kilometers. The word is similar to the milestone but refers to

kilometers instead of miles.

Calculation:

Let the length of the train is L.

Speed of train = 60 km/hr 

Since a train crosses a km-stone in 12 seconds

⇒ 60 × (5/18) = L/12

⇒ L = 200 meters.

Now, time to cross 91 km of stone

⇒ t = $\frac{200 + 91 \times 1000}{60\times \frac{5}{18}}$

⇒ t = 5412 sec

∴  t = 1 h 30 min 12 sec.

Alternate MethodGiven that,

In 1 hr. train travels, 60 km 

60 km is traveled in 60 min

1 km is traveled in 1 min

Therefore,

90 km is traveled in 90 minutes i.e. 1 hour 30 minutes

The remaining 1 km in 12 seconds

∴ The total time taken is 1 hr 30 min 12 sec.

Confusion PointsKilo-stones are considered as the pole in the case of a train. So we

will neglect the length of the Kilo-stone.

Speed of A = 6 km/h = 6 × (5/18) = 5/3 m/s [∵ 1 m/s = km/h × 5/18]

Time taken by A to complete 100 m race = Distance/Speed

⇒ 100/ (5/3) = 60 seconds

Time taken by B to complete = 60 + 9 = 69 seconds

Distance covered by B = 100 – 8 = 92 m

Speed of B = 92/69 m/s

⇒ (92/69) × 18/5 = 4.8 km/h

∴ Speed of B is 4.8 km/h

8x³ – y³ = (2x)³ – (y)³

⇒ (2x – y) × [(2x)² + (2x) (y) + (y)²] [∵ (a³ – b³) = (a – b) (a² + ab + b²)]

⇒ (2x – y) × [4x² + 2xy + y²]

∴ (8x³ – y³) / (4x² + 2xy + y²) = (2x – y)

Let a/b = k

(x + 2) is the factor of x² + ax + b

⇒ -2 is the root of x² + ax + b

⇒ (-2)² + a × (-2) + b = 0

⇒ 2a – b = 4         ---- 1

⇒ -2 is the root of x² + bx + a

⇒ (-2)² + b × (-2) + a = 0

⇒ 2b – a = 4        ---- 2

Equating equations 1 and 2

⇒ 2a – b = 2b – a

⇒ a = b

∴ a ∶ b = 1

If a and b are equal, both the polynomials are same and should have common roots.

P(a) is the remainder obtained when the polynomial is divided by (x – a)

If the above equation is divisible by (ax – b) then it must be a factor of the equation

⇒ (ax – b) = 0

⇒ a (x – b/a) = 0

∴ The remainder = f(b/a)

(1) 2222

2222 = 2 × 1111

Since 1111 is not an even number, there will no ‘2’ in its prime factorisation

There is only one 2 in the prime factorization of 2222

∴ It is not a square number

(2) 11664

11664 = 8 × 1458

⇒ 16 × 729 = 4² × 27²

It is a square number

(3) 343343

343343 = 343 × 1001 = 7³ × 7 × 143

⇒ 7⁴ × 13 × 11

It is not a square number

(4) 220347

220347 = 9 × 24483

⇒ 3² × 3 × 8161

8161 is not divisible by 3

The power of 3 in its prime factorization = 3

∴ It is not a perfect square

∴ 1, 3 and 4 are not perfect square numbers

x³ – 2x² + 4x – 8 = x²(x – 2) + 4 (x – 2) = (x² + 4) (x – 2)

⇒ (x³ + 2x² + 4x + 8) = x²(x + 2) + 4 (x + 2) = (x² + 4) (x + 2)

∴ Required product = (x + 2) × (x – 2) × (x² + 4) (x – 2) × (x² + 4) (x + 2)

⇒ [(x² + 4) × (x + 2) (x – 2)]²

⇒ [(x² + 4) × (x² - 4)]² [∵ (a + b) (a – b) = (a² – b²)]

⇒ (x⁴ – 16)² [∵ (a + b) (a – b) = (a² – b²)]

x (x + 2) (x + 3) (x + 5) – 72

x (x + 5) × (x + 2) (x + 3) – 72

⇒ (x² + 5x) × (x² + 5x + 6) – 72

Let x² + 5x = t

⇒ t × (t + 6) – 72

⇒ t² + 6t – 72

⇒ t² + 12t – 6t – 72

⇒ t (t + 12) – 6(t + 12)

⇒ (t – 6) (t + 12)

⇒ (x² + 5x – 6) (x² + 5x + 12)

⇒ (x² + 6x – x – 6) (x² + 5x + 12)

⇒ [x (x + 6) – 1(x + 6)] (x² + 5x + 12)

∴ (x – 1) (x + 6) (x² + 5x + 12) are the factors

(x² + x – 2) = x² + 2x – x – 2

⇒ x (x + 2) – 1 (x + 2)

⇒ (x – 1) (x + 2)

f(x) = (x – 1) (x² + 3x + a)

∴ (x + 2) is a factor of (x² + 3x + a)

-2 is the zero of (x² + 3x + a)

⇒ 4 – 6 + a = 0

∴ a = 2

g(x) = (x + 2) (x² + 2x + b)

(x – 1) is a factor of (x² + 2x + b)

⇒ 1 is a zero of (x² + 2x + b)

⇒ 1 + 2 + b = 0

∴ b = -3

Given:

a, b, c and d are non-zero integers

a and c are coprime

Calculation:

ab divides cd

⇒ cd = k × ab

Since c and a are co primes, d should contain the prime factors of a

⇒ d = ma

∴ a is a factor of d

a (b – c) x² + b (c – a) x + c (a – b) = 0

divide the equation by abc

(1/c – 1/b) x² + (1/a – 1/c) x + (1/b – 1/a) = 0

For equal roots

B² – 4AC = 0

⇒ (1/a – 1/c)² – 4(1/c – 1/b) (1/b – 1/a) = 0

⇒ 1/a² + 1/c² - 2/ (ac) – 4 × [-1/b² + 1/b (1/a + 1/c) - 1/ (ac)] = 0

⇒ 1/a² + 1/c² – 2/ (ac) + 4/ (ac) + 4/b² - 4/b × (1/a + 1/c) = 0

⇒ 1/a² + 1/c² + 2/ (ac) + 4/b² – 4/b (1/a + 1/c) = 0

⇒ (1/a + 1/c)² + (2/b)² – 2 × 2/b × (1/a + 1/c) = 0 [∵ a² + b² + 2ab = (a + b)²]

⇒ (1/a + 1/c – 2/b)² = 0                      [∵ a² + b² - 2ab = (a - b)²]

∴ 1/a + 1/c - 2/b = 0

∴ 1/a + 1/c = 2/b

(a – x²) / (bx) – (b – x) /c = (c – x) /b – (b – x²) / (cx)

⇒ (a – x²) / (bx) + (b – x²) / (cx) = (c – x) /b + (b – x) /c

Taking LCM of the denominator and combining them into a single factor

[c × (a – x²) + b × (b – x²)] / (bcx) = [c × (c – x) + b × (b – x)] / (bc)

⇒ [ac - cx² + b² – bx²] /x = [c² – cx + b² – bx]

⇒ b² + ac – x² (b + c) = x × [(b² + c²) – x (b + c)]

⇒ b² + ac – x² (b + c) = (b² + c²) x – x² (b + c)

⇒ b² + ac = (b² + c²) x

∴ x = (b² + ac) / (b² + c²)

Nature of the root is determined by the determinant of the equation

B² – 4AC = 7² – 4 × (1) × [-14 (k² – 7/8)]

⇒ 49 + 56 × (k² – 7/8)

⇒ 49 + 56k² – 49 = 56k²

Determinant has to be a perfect square number for the equation to have a possibility to have integer roots

56k² is not a perfect square as 56 is not a perfect square

∴ No integer roots exist

500 = 13 × 38 + 6                    [∵ 13 × 38 = 494]

⇒ First number divisible by 13 = 13 × 39 = 507

1000 = 76 × 13 + 12 [∵ 76 × 13 = 988]

⇒ Last number divisible by 13 in the given interval = 13 × 76

⇒ Number of multiples = 76 – 39 + 1 = 38     [1 is added ∵ 507 is also included]

∴ There are 38 numbers between 500 and 1000 that are divisible by 13

Trick:

Angle between hour hand and minute hand is given by

[11/2 × minutes - 30 × hours]

[11/2 × 36 - 30 × 4] = 78°

Detailed solution:

Angle swept by hour hand in 1 minute = ½°

 Angle swept by minute hand in 1 minute = 6°

Lag between hour hand and minute hand = 11/2°

Angle between the hour hand and the minute hand at 4:36 pm = Lag for 36 minutes - Angle traversed by the hour hand until 4 pm = 198 - 120 = 78°

78° = 78°/180 × π radians [∵ π radians = 180°]

≈ 0.433π

⇒ 0.4π < 0.433π < 0.6π

∴ 2π/5 < θ < 3π/5

sec²θ + cosec²θ = 1 + tan²θ + 1 + cot²θ [∵ sec²A = tan²A + 1, cosec²A = cot²A + 1]

⇒ tan²θ + cot²θ + 2

⇒ tan²θ + cot²θ + 2 tanθ × cotθ [∵ tanθ × cot θ = 1]

⇒ (tanθ + cotθ)² = α²

∴ tanθ + cotθ = α

α ≥ 2 [∵ AM - GM inequality]

∴ Statement 1 is true

Statement 2:

1 + cosθ = 2 cos²(θ/2) and 1 - cosθ = 2 sin²(θ/2)

∴ (1 + cosθ)/(1 - cosθ) = [2 cos²(θ/2)]/[2 sin²(θ/2)]

⇒ cot²(θ/2) = x² [∵ cosA/sinA = cotA]

∴ x = cot(θ/2)

cot(θ/2) continuously decrease in the interval 0° - 45°

Its value at 0° is not defined whereas at 45°

⇒ cot(45/2) = cot 22.5° = √2 + 1

∴ x > 2 in the interval

∴ Statement 2 is true

Statement 3:

cosθ/(1 - tanθ) + sinθ/(1 - cotθ) = cosθ/(1 - sinθ/cosθ) + sinθ/(1 - cosθ/sinθ) [∵ cosA/sinA = cotA and sinA/cosA = tanA]

⇒ cos²θ/(cosθ - sinθ) + sin²θ/(sinθ - cosθ)

⇒ (cos²θ - sin²θ)/(sinθ - cosθ)

⇒ cosθ + sinθ [∵ (a² - b²)/(a - b) = a + b]

⇒ -√(a² + b²) ≤ a sinA + b cosA ≤ √(a² + b²)

∴ cosθ + sinθ ≤ √2

∴ Statement 3 is wrong

Only two statements are true

In triangle ACA’, ∠A’AC = 45° [Given]

∴ tan 45° = A’C/AC

⇒ 1 = x/d

∴ d = x

∴ h + x > d

∴ Statement I is true

⇒ ∠A’AC + ∠AA’C = ∠BA’B’ + ∠A’BB’ = 90° [∵ Angle sum property]

⇒ ∠A’AC = ∠AA’C = 45°

⇒ 2∠A’AC - ∠BA’B’ = ∠A’BB’ > 2∠A’AC - ∠A’AC [∵ ∠BA’B’ < ∠A’AC]

∴ ∠A’BB’ > ∠A’AC

∠A’BB’ > 45°

⇒ Angle of depression > 45°

∴ Statement II is false

In triangle TXY

∠X = 90° - α and ∠Y = 90° - β

Let the height be TM

⇒ TM/MX = tan(90 - α)

⇒ TM/MX = cotα = 1/tanα = 1/2

∴ MX = 2TM

Similarly

TM/MY = tan(90 - β)

⇒ TM/MY = cot(90 - β) = 1/tan β = 1/0.5

∴ MY = 0.5TM

XY = MX + MY

⇒ 200 = 2TM + 0.5TM

⇒ 200 = 2.5TM

∴ TM = 80 m

∴ Distance between the banks = 80 m

1^c = 1/π × 180° < 90°

When x and y are less than 90°

⇒ sin x > sin y, when x > y

⇒ 1^c > 1°

∴ sin1^c > sin1°

∴ sin1°/sin1^c < 1

Let the diameters of the circles be a and b cm respectively

⇒ a/b = 12/5

⇒ a²/b² = 144/25

Area ∝ Diameter²

⇒ Sum of area/Area of circle of diameter b = (a² + b²)/b² = (144 + 25)/25 = 169/25

⇒ Sum of area of the triangles of the circles is same as the circle with diameter 65 cm

∴ 169/25 = (65)²/b²

∴ b = 25 cm and a = 12/5 × 25 = 60 cm

Radii = 1/2 × Diameter

∴ Radii of the circle = 60/2 and 25/2 = 30 cm and 12.5 cm

Area of tape used for each edge = l × b = 4 × 0.5 = 2 cm²

Number of edges to be joined = 12

∴ Tape used = 12 × 2 = 24 cm²

Let their radii be a and b respectively (a < b)

Area ∝ radius²

Since areas are in ratio 1 : 9

a : b = 1 : 3

In triangle AOQ,

sin∠QAO = OQ/AO

∴ AO = a/sin∠QAO

In triangle APR,

Sin∠PAR = sin∠QAO = PR/AP

∴ AP = b/sin∠QAO

⇒ AP - AO = Sum of the radii

⇒ (b - a)/sin∠QAO = (b + a)

⇒ (3a - a)/(3a + a) = sin∠QAO

1/2 = sin∠QAO

∴ ∠QAO = 30°

∴ ∠BAC = 2 × 30 = 60° [∵ Line from the centre bisects the angle between the two tangents from a common point].

Perimeter of the completed figure = 3 × side of square + 2 × equal side of triangle

Perimeter of the original square = 4 × side of square

7/6 × 4 × side of square = 3 × side of square + 2 × equal side of triangle

5/3 × side of square = 2 × equal side of triangle

Let the side of the square be a and triangle be b

∴ Equal side of the triangle b = 5a/6

Height of the isosceles triangle drawn to non - equal side bisects the base

∴ h² = √[equal side² - (non - equal side/2)²] = √[b² - (a/2)²]

Area of the isosceles triangle = ½ × base × height

⇒ ½ × a × 2a/3 = 1/3 × a²

⇒ 1/3 × Area of the square

∴ Area of triangle : Area of the square = 1 : 3

GIVEN:

Sides of a triangle are 51 cm, 37 cm and 20 cm respectively.

CONCEPT:

Area of triangle.

FORMULA USED:

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

Where S is the semi perimeter

Semi - perimeter of the triangle is = (a + b + c)/2

CALCULATION:

a, b and c are sides of triangle

Semi - perimeter of the triangle is S = (a + b + c)/2

⇒ (51 + 37 + 20)/2 = 108/2 = 54 cm

Using heron’s formula

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

⇒ √[54 × (54 - 51) × (54 - 37) × (54 - 20)]

⇒ √(54 × 3 × 17 × 34)

⇒ √(9 × 6 × 3 × 17 × 2 × 17) = 3 × 6 × 17 = 306 cm²

Calculation:

In triangle PQR,

∠Q = 90° [∵ Tangent is perpendicular to the radius]

⇒ PQ = QR = r

Therefore, Δ PQR is an isosceles right triangle

∠QPR = 45°

Area of the region outside the circle = Area of triangle PQR - Area of sector

⇒ ½ × b × h - θ/360 × πr^{2 -}

⇒ ½ × r × r - 45/360 × πr²

∴ Required area is r²/2 - πr²/8

Formula used:

Area of triangle = $\frac{1}{2}$ Base × height

Calculation:

We know that

Area of triangle = $\frac{1}{2}$ Base × height

Area of triangle ABC = $\frac{1}{2}$ × AB × AC = $\frac{1}{2}$bc

Area of triangle ABC = $\frac{1}{2}$ × BC × AD = $\frac{1}{2}$ap

⇒ $\frac{1}{2}$bc = $\frac{1}{2}$ap

∴  p = bc/a

Let O be the centre of the circle and EF be a perpendicular drawn to the lines from O

Let the radius of the circle be r cm

Perpendicular drawn from the center bisects the chord

∴ AF = FB = AB/2 = 2 cm

DE = EC = CD/2 = 5 cm

In triangle AFO

⇒ OF = √(AO² - AF²) [∵ Pythagoras theorem]

⇒ OF = √(r² - 4)

In triangle OEC

⇒ OE = √(CO² - CE²)

⇒ OE = √(r² - 25)

⇒ EF = OE + OF

⇒ 3 = √(r² - 4) + √(r² - 25)

⇒ 3 - √(r² - 25) = √(r² - 4)

Squaring on both the sides

⇒ 9 + r² - 25 - 6√(r² - 25) = r² - 4

⇒ - 12 = - 6√(r² - 25)

⇒ 2 = √(r² - 25)

⇒ 4 = r² - 25

⇒ r² = 29

∴ r = √29 cm

since the gap is less than the radius thus both the chord will lie same side of the centre.

In triangle APB and PDC

∠APB = ∠CPD [∵ Vertically opposite angle]

∠ABP = ∠PCD [∵ Angle subtended by chord AD at any point on the circle is same]

∴ APB ∼ PDC

∴ area(APB)/area(PDC) = (AB/CD)²

⇒ 24/area(PDC) = (8/5)²

∴ Area of PDC = 24 × 25/64 = 9.375 cm²

Concept:

1. Pythagoras theorem: In a right-angled triangle

Hypotenuse² = Base² + Perpendicular²

2. All angles of an equilateral triangle are equal i.e. 60°.

3. In an equilateral triangle, altitude bisects the side

Calculation:

We know that, 

AB = BC = AC

Since, in an equilateral triangle, altitude bisects the side

⇒ AB = 2AD

In Δ BAD, using Pythagoras theorem

⇒ AB² = BD² + AD²

⇒ (2AD)² = BD² + AD²

⇒ 4AD2 = BD2 + AD2

∴  BD² = 3AD²

Since the distance between the centers is 13 cm, they touch each other.

CD is the common tangent

CA = AB and AD = AB [∵ Tangents drawn from a common point are same]

∴ CD = CA + AD = 2AB

⇒ ∠PAC = ∠BAP

⇒ ∠DAQ = ∠BAQ [∵ Line drawn from the center divides the angle between the tangents]

⇒ ∠CAP + ∠PAB + ∠BAQ + ∠DAQ = 180° [Angle is straight line is 180°]

⇒ 2 × (∠PAB + ∠BAQ) = 180°

∴ ∠PAB + ∠BAQ = 90°

⇒ tan∠PAB = PB/AB

⇒ tan∠BAQ = tan(90 - ∠PAB) = cot∠PAB = QB/AB

⇒ tan∠PAB × cot∠PAB = 1

⇒ PB/AB × QB/AB = 1

⇒ 9/AB × 4/AB = 1

⇒ AB² = 36

∴ AB = 6 cm

∴ CD = 2AB = 12 cm