CDS Elementary Mathematics 4 Feb 2018 Official Paper (Previous Year Paper)

200 men can complete the quarter of work in 50 days

If they continue the work, time taken to complete the remaining work = 3 × 50 = 150 days

[∵ ¾ th of the work is remaining]

But they have to complete in 100 days to finish in time

Time ∝ 1/Number of employees

Let the number of extra employees required to complete the work in time be x

∴ 150/100 = (200 + x)/200

300 = 200 + x

∴ x = 100

Shortcut Trick

Length of the train = Speed of train in m/s × Time taken to cross a stationary pole.

Convert speed: 60 km/h = 60 × (5 / 18) = 50 / 3 m/s.

Length = (50 / 3) × 30 = 50 × 10 = 500 meters.

∴ The correct answer is 500 m.

Alternate Method

Given:

Speed of the train = 60 km/h

Time to cross the pole = 30 seconds

Formula Used:

Distance = Speed × Time

To convert km/h to m/s, multiply by (5 / 18)

Pole

Train Length (L) = ?

Speed = 60 km/h

Time to cross = 30 seconds

Calculations:

⇒ Speed in m/s = 60 × (5 / 18)

⇒ Speed = 300 / 18 = 50 / 3 m/s

⇒ Distance covered = Speed × Time

⇒ Length of train = (50 / 3) × 30

⇒ Length of train = 50 × 10 = 500 m

∴ The correct answer is 500 m.

Additional Information

Crossing a Point Object

When a train crosses a stationary object with negligible width like a pole, standing man, or tree, the total distance covered is equal to the length of the train itself.

Crossing an Extended Object

When a train crosses a bridge, platform, or another train, the total distance covered is the sum of their individual lengths (Distance = L₁ + L₂).

Unit Conversion Trick

To rapidly convert a speed from km/h to m/s, simply multiply by 5/18. To convert backward from m/s to km/h, multiply by 18/5.

Calculation:

Let A’s share be Rs. x

⇒ B’s share = x - 20 and C’s share = x + 20

Total amount = x + x - 20 + x + 20

⇒ 120 = 3x

⇒ x = 40

∴ B’s share = x - 20 = Rs. 20

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

MN = 15 × -4 = -60

-6 × A = -60

∴ A = 10

2 × B = -60

∴ B = -30

C × 60 = -60

∴ C = -1

Given :

Principal = Rs 5000

Rate₁ = 5%

Rate₂ = 5.5%

Time = 2 years

Formula used :

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

Calculations :

Interest when rate is 5%, using equation (1)

⇒ Simple interest = $\frac{5000 \times 5 \times 2}{100}$

⇒ Rs 500

Interest when rate is 5.5%, using equation (1)

⇒ Simple interest = $\frac{5000 \times 5.5 \times2}{100}$

⇒ Rs 550

The required amount gained

⇒ 550 - 500

⇒ Rs 50

∴ The amount gained by him this transaction is Rs 50.

Age of son after 7 years = 17 years

⇒ Present age of son = 17 - 7 = 10 years

Average of the present ages of father and son = 25

⇒ Sum of the present ages of father and son = 25 × 2 = 50

⇒ Present age of father = 50 - 10 = 40 years

After 10 years age of father = 40 + 10 = 50 years

∴ The age of the father after 10 years will be 50 years.

5 tractors can plough 5 hectares in 5 days

5 tractors can plough 50 hectares in 50 days

1 tractor can plough 10 hectares in 50 days

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

⇒ 100/10 = 10

∴ 10 tractors can plough 100 hectares in 50 days

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

Amount at the end of n years = Principal × (1 + Rate/100)ⁿ

10000 = x × (1 + 25/100)³

10000 = x × 1.25³

10 × 10 × 10 × 10 = x × 1.25 × 1.25 × 1.25

8 × 8 × 8 × 10 = x [∵ 10/1.25 = 8]

x = Rs. 5120

459 = 17 × 27

Second option is terminating decimal, ∴ denominator has a prime factors of 2 or 5 only

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

Let x = 0.459459459⋯

1000x = 459.459459⋯

1000x = 459 + 0.459459⋯

1000x = 459 + x

∴ 999x = 459

x = 459/999 = (17 × 27) / (37 × 27) = 17/37

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

Given :

Reduction in rate of interest = 0.25%

Decrease in amount = Rs 64

Formula used :

Decrease in amount = $\frac{Principal \ \times \ rate \ reduction \ in \ rate \ of \ interest \ \times \ time}{100}$     ---(1)

Calculations :

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

Using equation (1), we get

⇒ 64 = $\frac{Principal \ \times \ .25 \ \times \ 1}{100}$

⇒ $Principal = \frac{6400}{0.25}$

⇒  Principal = Rs. 25600

∴ Original annual income is Rs 25600.

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

For numbers less than 1, the term with lowest power is the largest

-1 < 1 < 2

∴ m² < m < m^-1

Final order = Log₁₀m < m² < m < m^-1

Let the amount received by the widow = Rs x

∴ Amount received by daughter = 2 × x = 2x

Amount received by son = 3 × 2x = 6x

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

39000 = x + 5 × 6x + 4 × 2x

39000 = x + 30x + 8x

39000 = 39x

∴ x = 1000

Widow receives Rs. 1000

Given:

Three numbers which are co-prime to each other, are such that the product of the first two 286 and that of the last two is 770.

Concept:

(LCM × HCF) = Product of the numbers

Calculation:

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

Let the numbers be A, B and C respectively

A × B = 286 and B × C = 770

A × B/ (B × C) = 286/770

A/C = 13/35

∴ A = 13 and C = 35

B = 286/A = 22

Sum of the numbers = 13 + 35 + 22 = 70

∴ The sum of three terms is 70.

Let the two digits of the age of the women be y and x (yx)

Age of the women = 10y + x

Age of the men = 10x + y

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

⇒ 9x - 9y

Sum of their ages = 10x + y + 10y + x = 11x + 11y

Difference = 1/11 × Sum

9x - 9y = 1/11 × (11x + 11y)

9x - 9y = x + y

8x = 10y

Since x and y can only be natural numbers from 1 to 9, the only possibility is x = 5 and y = 4

Difference in their age = 9x - 9y = 45 - 36 = 9 years

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

∴ Time ∝ 1/Relative speed

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

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

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

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

3 = (k + 1)x / [(k - 1)x]

3 = (k + 1) / (k - 1)

3k - 3 = k + 1

2k = 4

k = 2

∴ The required ratio = 2 : 1

Any number multiplied with 5 ends with 5 if it is odd

Product of all odd prime number is an odd number where one of the factor is 5

∴ The unit digit of the product is 5

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

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

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

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

Total amount of alloy C produced = 20 + 28 + x = 48 + x

20 + x = 6/(6 + 7) × (48 + x)

20 + x = 6/13 × (48 + x)

260 + 13x = 288 + 6x

7x = 28

∴ x = 4 kg

ax² + bx + c = x (ax + b) + c

Remainder when divided by x = c

∴ c = 3

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

Remainder when divided by (x - 1) = P(1)

6 = a (1) + b(1) + c

6 = a + b + 3

∴ a + b = 3

Shortcut Trick

Concept:

Consecutive numbers:

• They usually have a difference of 1 between every two numbers.
• If the terms are in  AP (i.e. common difference same), and the total number of terms is odd then the average is the same as the middle most term.

Calculation:

Using the above concept

Average = 5th term = 55

Largest integer = 9th term = 55 + 4 = 59

∴ The largest integer is 59

Alternate Method

Let the integers  are

a - 4, a - 3, a - 2, a - 1, a, a + 1, a + 2, a + 3, a + 4

Sum of 9 integers = 9a

Average = {Sum of Observations} ÷ {Total number of Observations}

Average of 9 integers = 9a/9 = 55

⇒ a = 5

⇒ Largest integer = a + 4 = 55 = 4 = 59

∴ The largest integer is 59

Formula used:

Average = Sum of observation / Number of observations

Calculation

We know that,

Average = Sum of observation / Number of observations

Sum = Average × Number of observations

Therefore, 

Total age of 15 students =(15 × 19) = 285 years

Total age of (15 + 5) i.e. 20 students = (20 × 18.5) = 370 years

Therefore, the total age of 5 new students = 370 - 285 =  85 years

Now, average of 5 new students = 85/5 = 17 year.

∴ The average age of the 5 newly admitted students is 17 years.

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

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

To use equation 1, two quantities are required

S1 gives flow rate of R = 6 litres per minute

S3 gives time taken to empty = 15 hours 20 minutes

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

To use equation 2,

S1: gives the flow rate of R

S2 gives the total time require to fill if all taps and drains are opened.

∴ Volume = (12 + 10 - 6) × (5 × 60 + 45)

⇒ 16 × 345 = 5520 litres

Let the volume of the tank be V litres

By S3: Flow rate of Pipe R = V/ (15 × 60 + 20) = V/920

Substituting in S2

V = (12 + 10 - V/920) × (5 × 60 + 45)

V/345 = 22 - V/920

V/345 + V/920 = 22

168V/57960 + 63V/57960 = 22

231V/57960 = 22

∴ V = 5520 litres

∴ Any two statements are required to find the capacity of the tank

Let the distance between the two places be D km

Time taken = Distance/Speed

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

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

Total time taken = Time taken upstream + Time taken downstream

z = D/ (x - y) + D/ (x + y)

z = [D(x + y) + D(x - y)] / [(x - y) (x + y)]

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

∴ D = z(x² - y²)/2x

Given:

Average speed = 48kmph

Concept used:

When equal distance is travelled with 2 different speed x and y,

then the average speed = 2xy/(x + y)

Here, x = 60 kmph and y is other speed

2 × 60 × y/(60 + y) = 48

⇒ y = 40 kmph

Detailed Solution:

Let the distance between Delhi and Agra be D km

Time taken for round journey = D/60 + D/y            

               [∵ Time = Distance/Speed]

⇒ D(60 + y) / (60y) = 2D/Average speed

⇒ 2 × 60y = Average speed × (60 + y)

120y = 48 × (60 + y)

120y = 2880 + 48y

72y = 2880

∴ y = 40 km/h

Shortcut Trick let initial cost price = 100

Initial selling price  = 132

New Cost price = 100 + 20 % of 100 = 120

New Selling price  = old selling price = 132

∴ Profit % = $\frac{132-120 }{120}$× 100 = 10%

∴ Correct option is 10%.

 Alternate MethodCalculation:

Let the original cost price be Rs ‘x’

Profit = 32% of cost price

∴ Selling price = Cost price + Profit

= x + (32/100) % of x = 1.32x

New cost price = (100 + 20)/100 × x = 1.2x

New profit = Selling price - cost price

⇒ 1.32x - 1.2x = 0.12x

∴ Profit % = (0.12x/1.2x) × 100 = 10%

∴ The correct option is 10%.

Let D’s contribution be x

∴ A’s contribution = 3 × x = 3x

E’s contribution = ½ × 3x = 1.5x

B’s contribution = 1/3 × 1.5x = 0.5x

C’s contribution = 2/3 × 3x = 2x

Total contribution = 3x + 0.5x + 2x + x + 1.5x = 8x

B, C and E contribution = 0.5x + 2x + 1.5x = 4x

Difference between (A + D + E) - (B + C) = 3x + x + 1.5x - 0.5x - 2x = 3x

Share ∝ Contribution

B + C + E contribution/13500 = 4/3

∴ B + C + E contribution = 4/3 × 13500 = Rs. 18000

5¹⁷ + 5¹⁸ + 5¹⁹ + 5²⁰ = 5¹⁷ × (1 + 5 + 25 + 125)

⇒ 5¹⁷ × 156 = 5¹⁷ × 12 × 13

∴ The given expression is divisible by 13

a + b = 2c

∴ a - c = c - b

a/(a - c) + c/(b - c) = a/(a - c) - c/(a - c) [∵ b - c = -(a - c)]

⇒ (a - c)/(a - c) = 1

x^a = (y^1/a)^a = y

(x^a)^b = y^b = (z^1/b)^b = z

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

∴ x^abc = x

∴ abc = 1

2b = a + c

∴ b - c = a - b

x^{b - c} y^{c - a} z^{a - b} = x^{b - c} y^{c - a} z^{b - c}

⇒ (xz)^{b - c} y^{c - a}

⇒ (y²)^{b - c} y^{c - a}

⇒ (y)^{2b - c - a} = y⁰ = 1

Concept:

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

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

Now,

A non-terminating, non-repeating decimal is a decimal number that continues endlessly, with no group of digits repeating endlessly.

Decimals of this type cannot be represented as fractions, and as a result, are irrational numbers.

Example:

π is a non-terminating, non-repeating decimal. π = 3.141 592 653 5...

e is a non-terminating, non-repeating decimal. e = 2.718 281

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

Products of the roots of quadratic equation = c/a

In this case, product = 1 [∵ Roots are reciprocal of each other]

1 = r/p

∴ p = r

65x - 33y = 97     ----(1)

33x - 65y = 1     ----(2)

Equation 1 + 2

98x - 98y = 98

∴ x - y = 1 or x = y + 1

Substituting in equation 1

65y + 65 - 33y = 97

32y = 32

∴ y = 1 and x = 1 + 1 = 2

∴ xy = 2 × 1 = 2

(ab + xy)/bx = a/x + y/b

⇒ (x/a)^-1 + (b/y)^-1

⇒ (1 - c/z)^-1 + (1 - z/c)^-1

⇒ z/(z - c) + c/(c - z)

⇒ (z - c) / (z - c) = 1

(a² - 1)/a = 5

a - 1/a = 5

Cubing on both the sides

a³ - 1/a³ - 3 × a × 1/a × (a - 1/a) = 125 [∵ (a - b)³ = a³ - b³ - 3ab (a - b)]

a³ - 1/a³ - 3 × 5 = 125

∴ a³ - 1/a³ = (a⁶ - 1)/a³ = 125 + 15 = 140

(y + z - x)³ + (z + x - y)³ + (x + y -z)³ = (x + y + z - 2x)³ + (y + z + x - 2y)³ + (x + y + z - 2z)³

⇒ (-2x)³ + (-2y)³ + (-2z)³ [∵ x + y + z = 0]

⇒ -8 × (x³ + y³ + z³)

If x + y + z = 0, then x³ + y³ + z³ = 3xyz

∴ The given expression = -8 × 3xyz = -24xyz

If (x - a) is a factor of polynomial P(x), then P(a) = 0

∴ (-3)³ + 3 × (-3)² + 4 × (-3) + k = 0

-12 + k = 0

k = 12

We know that 30² = 900

31² = 961

32² = 1024

∴ 1024 is the smallest 4 digit perfect square

Given:

The polynomial 3x3 + 4x2 - 7 = 0

Calculation:

Option(1), x = 0

P(x) = 3x3 + 4x2 - 7 = 0

⇒ P(0) = 3(0)³ + 4(0)² - 7 

⇒ P(0) = - 7

This is not the zero of the polynomial 3x3 + 4x2 - 7 = 0

Option(2), x = 1

P(x) = 3x3 + 4x2 - 7 = 0

⇒ P(1) = 3(1)3 + 4(1)2 - 7 

⇒ P(1) = 7 - 7

⇒ P(1) = 0

∴ x = 1 the zero of the polynomial 3x3 + 4x2 - 7 = 0.

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

LCM × HCF = Product of the numbers

3003 × 21 = 21a × 21b

∴ ab = 143 = 13 × 11 or 143 × 1

Since both the numbers are greater than 21

a = 13 and b = 11

So the numbers are 21 × 13 = 273 and 21 × 11 = 231

Sum of the numbers = 273 + 231 = 504

Sum of roots = -b/a

∴ α + β = -b/a

Product of roots = c/a

∴ αβ = c/a

1 + α + β + αβ = 1 - b/a + c/a = (a - b + c)/a

1 + α + β (α + 1) = (a - b + c)/a

(α + 1) (β + 1) = (a - b + c)/a

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

∴ 3(-1)³ + k(-1)² + 5(-1) - 6 = -7

k - 14 = -7

∴ k = 7

Degree is the highest power of the given polynomial

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

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

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

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

(√5 - √3)/2 = 1/(√5 + √3)

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

⇒ [(√5 - √3)² - (√5 + √3)²]/2

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

⇒ -2√15

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

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

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

Let the number be x

x² + x = 20

x² + x - 20 = 0

x² + 5x - 4x - 20 = 0

x(x + 5) - 4(x + 5) = 0

(x + 5) (x - 4) = 0

∴ x = -5 or 4

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

Let the new consumption be y’

Price of the wheat × Consumption = Budget

∴ xy = (100 + 25)/100 × x × y’

xy = 1.25xy’

∴ y’ = y/1.25 = 0.8y

Reduction in consumption = (y - 0.8y)/y × 100 = 20%

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

Number students who passed = 11/20 × 1920 = 1056

Let x = 0. 9̅

Multiply both sides by 10 

⇒ 10x = 9.9̅  = 9.999999999999..... ⇒ 10x = 9 + 0.999999999999999

⇒ 10x = 9 + 0.9̅  ⇒ 10x = 9 + x         (∵ x = 0. 9̅)

⇒ 9x = 9 ⇒ x = 1 

Hence, required difference = 1 - 0.9 = 0.1

B/E = B/C × C/D × D/E

⇒ ¾ × 2/3 × ¾ = 3/8

∴ B ∶ E = 3 ∶ 8

10 women complete work in 12 days

∴ 5 women complete work in 12 × 2 = 24 days

5 men can complete the same work in 8 days

∴ Men are 3 times efficient as women

Three women can be replaced by a man

6 women and 3 men = 6/3 + 3 = 5 men

5 men can complete the work in 8 days

∴ Work is completed in 8 days

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

The distance of all the midpoints of the radii will be at a distance of 8 cm from the centre

∴ It is also a circle but with radius 8 cm

Curved surface area of cone = π r l

17600 = 22/7 × 70 × l [∵ 1 m² = 10000 cm² and radius = diameter/2]

∴ l = 17600/220 = 80 cm

h = √(l² - r²)

⇒ √(80² - 70²)

⇒ √1500 = 10√15 cm

Concept:

Orthocenter: 

• Point of intersection of altitudes that are drawn perpendicular from the vertex to the opposite sides of a triangle.
• The orthocenter lies inside the triangle for an acute angle triangle and outside the triangle for an obtuse angle triangle.
• For a right-angled triangle, it lies on the vertex of the right angle.

Centroid:

• The centroid of a triangle is a point of concurrency of the medians of a triangle.
• The centroid of a triangle is always within a triangle.
• Medians are the line segments that are drawn from the vertex to the mid-point of the opposite side of the vertex.

Explanation:

Statement 1:

The orthocenter of an obtuse-angled triangle lies outside the triangle

Hence, statement 1 is incorrect.

Statement 2:

Centroid is the intersection point of all medians and all medians are

necessarily lying inside the triangle, centroid lies inside the triangle

Hence, statement 2 is correct.

Statement 3:

The orthocenter is the intersection of altitudes. Since in a right

triangle both the sides (except hypotenuse) are altitudes, their point

of intersection lies on the triangle.

Hence, statement 3 is correct.

Statement 4:

We know that the centroid of a triangle is always within a triangle.

Hence, statement 4 is incorrect.

∴ Statements 2 and 3 are correct.

Let the intersection point be O and the point which is equidistant from the two lines be B

In triangle AOB and COB

AB = CB [∵ The point is equidistant]

OB is common

∠BAO = ∠BCO = 90° [Distance is measured along perpendicular]

∴ AOB ≅ COB

∠AOB = ∠COB

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

But since there are two pairs of vertically opposite angles made, there will two such angle bisector

∴ Locus is a pair of straight lines

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

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

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

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

[SSS congruency]

Statement 3 is true:

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

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

[SAS congruency]

Statement 4 is true:

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

So the ratios of the sides are equal

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

ASA congruency

Statement 2:

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

∴ Statement 2 is false

Statement 1:

Sum of interior angles of the polygon = n × 90° [∵ All angles are right angles]

(2n - 4) × 90 = n × 90

2n - 4 - n

∴ n = 4.

Statement 1 is true

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

Area = side²

If the length increases by 8cm, area increased by 120 cm²

(x + 8)² - x² = 120

x² + 16x + 64 - x² = 120 [∵ (a + b)² = a² + 2ab + b²]

16x = 56

∴ x = 3.5 cm

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

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

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

But 25 = 5 × 5

∴ Power of 5 in the prime factorization of the given term = 6

∴ Largest power of 10 that divides the given product = 6

Area to be painted in each room = 2 (lh + bh)

⇒ 2 × (6 × 2.5 + 4 × 2.5) = 50 m²

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

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

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

Number of tiles in a row = Width/Side of square

⇒ 4.5/0.5 = 9 tiles [∵ 50cm = 0.5 m]

Number of tiles in a column = Length/Side of the square

⇒ 10/0.5 = 20 tiles

Total number of tiles required = 9 × 20

Number of packets required = 9 × 20/20 = 9 packets

Cost of nine packets = 9 × 100 = Rs. 900

When cube of maximum volume is cut from sphere, the diagonal of the cube = Diameter of the sphere

∴ √3 a = 2r

a/r = 2/√3

Volume of cube/Volume of sphere = a³/ (4/3 π r³)

⇒ 3/(4π) × (a/r)³

⇒ 3/(4π) × (2/√3)³

⇒ 2/(√3π)

∴ The required ratio = 2 ∶ √3π

Given 2πr/h = 3/1

∴ r : h = 3 : 2π

l = √(r² + h²)

⇒ r × √[1 + (h/r)²]

⇒ r × √[1 + (2π/3)²] = r/3 × √(4π² + 9)

Curved surface area of cone = πrl

⇒ πr × r/3 × √(4π² + 9)

⇒ [πr² √(4π² + 9)]/3

Circumference of the circle = Perimeter of the square

2πr = 4s

∴ s = πr/2

⇒ 22/7 × 98/2 = 154 cm

Sum of two sides of a triangle > third side

In triangle ABD,

AB + BD > AD       ---- 1

In triangle ACD

AC + CD > AD        ---- 2

Adding the inequalities 1 and 2

AB + BD + CD + AC > 2AD

AB + BC + AC > 2 AD

a + b + c > 2p

Similarly for other medians

a + b + c > 2q

a + b + c > 2r

Adding all the inequalities

3(a + b + c) > 2(p + q + r)

Let the centroid be O

In triangle AOB

AO + BO > AB

Centroid divides the median in the ratio 2 ∶ 1

∴ 2p/3 + 2q/3 > c

Similarly for other medians

2q/3 + 2r/3 > a

2p/3 + 2r/3 > b

Adding the inequalities

4/3(p + q + r) > (a + b + c)

∴ 4(p + q + r) > 3(a + b + c)

Given:

Side of the square = 2/√π 

Concept:

Area of circle = πr²

Calculation:

Diameter of the largest circular disc cut from a square of side 2/√π = Side of the Square

⇒ 2/√π units

So, Area of the Circular Disc = πr²

⇒ π × {2/(2√π)}²

⇒ π × (1/√π)2

⇒ π × (1/π)

⇒ 1 square units.

∴ The area of the largest circular disc cut from a square of side 2/√π units is 1 square units.

Both the diagonals have same length,

∴ Product of length of diagonals = Square of the length of diagonal

Length of diagonal = √2 × length of side

Length of diagonal² = 2 × Length of side²

50 = 2 × s²

25 = s²

∴ s = 5 units

Formula used: 

Total surface area of cylinder = 2πr(h + r)

Calculation:

Given that,

The total surface area of the cylinder = 352 square cm.

⇒ 352 = 2 × 22/7 × r (10 + r)

⇒ 56 = r (10 + r)

⇒ 4 × 14 = r × (10 + r)

⇒  r = 4 cm

∴ Diameter = 2 × radius = 8 cm

Concept:

Perimeter = Sum of sides

Diagonal of square = (side)√2

Area of equilateral triangle = √3/4 × side2

Calculation:

Diagonal of the square = √2 × side of the square

⇒  Side of the square = 6√2/√2 = 6 cm

According to the question

The perimeter of a triangle = perimeter o square

4 × Side of the square = 3 × Side of the triangle

4 × 6 = 3 × Side of the triangle

Side of the equilateral triangle = 8 cm

Area of the equilateral triangle = √3/4 × side²

⇒ √3/4 × 8² = 16√3 cm²

∴ Area of the equilateral triangle 16√3 cm²

Diagonal of the square = Diameter of the circle

√2 a = 2r

∴ r = a/√2

Area of the required region = Area of the circle - Area of the square

⇒ πr² - a²

⇒ π (a/√2)² - a² = (π - 2)a²/2 square units

Let O be the centre of the circle

In quadrilateral AOBX

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

90° + ∠O + 90° + 50° = 360° [∵ Tangent is perpendicular to the radius]

∴ ∠AOB = 130°

∠ACB = ½ × 130 = 65° [∵ Angle subtended by a chord at any point on the circle is half the angle subtended at centre]

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

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

In triangle ABO and ACP

∠A is common and ∠O = ∠P = 90°

ABO ∼ ACP

∴ AB/AC = AO/AP

In triangle DRE and DSF

∠D is common and ∠R = ∠S = 90°

DRE ∼ DSF

∴ DE/DF = DR/DS

Parallel lines are equally spaced

∴ AO = DR and AP = DS

∴ AB/AC = DE/DF

So statement 1 is correct

AB/ (AC - AB) = DE/ (DF - DE)

AB/BC = DE/EF

∴ AB × EF = BC × DE

So statement 2 is also correct

In triangle ADE and ABC

AD/AB = AE/EC = ½

∠A is common

∴ ADE ∼ ABC

Area(ADE) /Area (ABC) = (AD/AB)²

∴ Area(ADE) = (1/2)² × Area(ABC)

⇒ ¼ × √3/4 × l² = √3/16 × l² [∵ Area of equilateral triangle = √3/4 × l²]

Area of shared region = Area(ABC) - Area(ADC)

⇒ √3/4 × l^{2 -} √3/16 × l²

⇒ 3√3/16 × l²

∠OPT = 90° [∵ Radius is perpendicular to tangent]

∠OPQ = 90° - ∠QPT = 90° - α

∠OQP = 90° - α [∵ OQ = OP]

In triangle OQP

∠O + ∠Q + ∠P = 180°

∠O + 90 - α + 90 - α = 180

∴ ∠O = 2α

Let the point of intersection of AB and CD is P 

Construct OE perpendicular to AB

EB = ½ × AB = 5 cm [∵ Perpendicular from the centre bisects the chord]

In triangle OEB

OB² = OE² + EB²

13² = OE² + 5²

∴ OE = 12 cm

In triangle OEP

OP² = OE² + EP²

EP = EB - PB = 5 - 3 = 2 cm

OP² = 12² + 2² = 148

∴ OP = 2√37 cm

Angle = Percentage of the sector/100 × 360°

⇒ 16.1/100 × 360 = 57.96° ≈ 58°

Let the numbers be a and b respectively

(a + b)/2 = 10

∴ a + b = 20      ----1

√(ab) = 8

∴ ab = 64

(a - b) = √[(a + b)² - 4ab]

⇒ √(20² - 4 × 64) = √(400 - 256) = 12

∴ a - b = 12      ----2

Adding equations 1 and 2

2a = 32

∴ a = 16 and b = 20 - 16 = 4

∴ The numbers are 16 and 4

Formula Used:

Mean or Average:

Sum of observations / Total number of observations

Calculation:

Using the above formula

Sum of 11 observations = Sum of first 6 numbers + Sum of last 6 number - 6^th number [∵ 6^th observation is added twice]

11 × 11 = 10.5 × 6 + 11.5 × 6 - 6^th number

121 = 63 + 69 - 6^th number

⇒ 6^th number = 132 - 121 = 11

∴ 11 is the correct option.

sin² θ = 1 - cos² θ

∴ sin⁴θ - cos⁴ θ = (1 - cos²θ)² - cos⁴ θ

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

⇒ 1 - 2cos² θ

cot 1° cot 23° cot 45° cot 67° cot 89° = cot 1° cot 23° cot 45° cot (90 - 23°) cot (90 - 1)°

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

⇒ (cot 1° tan 1°) × (cot 23° tan 23°) × cot 45°

⇒ 1 × 1 × 1 [∵ cot A tan A = 1 and cot 45° = 1]

⇒ 1

Hour hand completes one full revolution in 12 hours

∴ Angle described in 1 hour = 360°/12 = 30°

Angle described in 60 minutes = 30° [∵ 1 hour = 60 minutes]

∴ Angle described in 10 minutes = 30/60 × 10 = 5°

Statement 1:

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

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

∴ Statement 1 is wrong

Statement 2:

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

⇒ [sin² θ + 1 +2cos θ + cos² θ]/[(1 +cos θ) sinθ] [∵ (a + b)² = a² + 2ab + b²]

⇒ [1 + 1 + 2cos θ]/[(1 + cos θ) sinθ] [∵ sin² θ + cos² θ = 1]

⇒ [2 + 2cos θ]/[(1 + cos θ) sinθ]

⇒ 2 [1 + cosθ]/[(1 + cos θ) sinθ]

⇒ 2/sinθ = 2 cosecθ [∵ 1/sin θ = cosec θ]

Statement 2 is correct

EO = l/2 [∵ O is the centre of the square]

∠AFD = 60°

AF = DF due to symmetry and also ∠AFD = 60°

∴ AFD is an equilateral triangle of side l

FE = Altitude of an equilateral triangle = √3l/2

In triangle EFO

FO² = EF² - EO² [∵ Pythagoras theorem]

∴ h² = (√3l/2)² - (l/2)² = l²/2

∠ACB = α and ∠BCD = β initially

In triangle BCD, cot β = BC/BD

∴ BC = BD cot β = h cot β

In triangle ABC, tan α = AB/BC

∴ AB = BC tan α

H - h = h cotβ × tan α      ----1

∴ H = h (cotβ tan α + 1)      ----2

Now height changes to h’ and the angles get interchanged

∴ H = h’ (cot α tanβ + 1)      ----3

Equating the RHS of equations 2 and 3

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

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

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

∴ h’ = h/tan β × tan α

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

∴ h’ = H - h

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

Shortcut:

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

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

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

H - h = h’

∴ h’ - h = H - 2h units

Secx cosecx = 2

1/cosx × 1/sinx = 2

∴ sinx cosx = ½

(sinx - cosx)² = sin²x + cos²x - 2sinx cosx

⇒ 1 - 2 × ½ = 0 [∵ sin²x + cos²x = 1]

∴ sinx = cos x

∴ sinx/cosx = 1

tanx = 1, cotx = 1/tanx = 1

∴ tanⁿx + cotⁿx = 1ⁿ + 1ⁿ = 2

cos x + cos² x = 1

cos x = 1 – cos² x

cos x = sin² x [∵ 1 – cos² A = sin² A]

Substituting cos x = sin² x in the main equation

sin² x + sin⁴ x = 1

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

p² = 1 + 2 sin A cosA [∵ sin² A + cos² A = 1]

∴ sin A cos A = (p² – 1) /2

(sin A + cos A)³ = sin³ A + cos³ A + 3 sin A cos A (sin A + cos A)

[∵ (a + b)³ = a³ + b³ + 3ab (a + b)]

p³ = q + 3 × (p² – 1)/2 × p

p³ = q + 3p³/2 – 3p/2

(0 = q + p³/2 -3p/2) × 2

0 = 2q + p³ – 3p

∴ p³ – 3p + 2q = 0

x = (sec² θ – tan θ) / (sec² θ + tan θ)

Using componendo and dividend rule

(x + 1) / (1 – x) = [(sec² θ – tan θ) + (sec² θ + tan θ)] / [(sec² θ + tan θ) - (sec² θ – tan θ)]

⇒ sec²θ/tan θ = (1/cos² θ) / (sin θ/ cos θ)

⇒ 1/(sin θ cos θ)

⇒ 2/sin 2θ [∵ sin 2θ = 2 sin θ cos θ]

∴ (x + 1) / (1 – x) = 2/sin 2θ

Taking reciprocal on both the sides

(1 – x) / (x + 1) = sin 2θ/2

-1 ≤ Sin 2θ ≤ 1

∴ -1/2 ≤ (1 – x) / (x + 1) ≤ ½

-(x + 1) /2 ≤ (1 – x) ≤ (x + 1) /2

First inequality

-x/2 – ½ ≤ 1 – x

x/2 ≤ 3/2

∴ x ≤ 3

Second inequality

(1 – x) ≤ x/2 + ½

½ ≤ 3x/2

∴ x ≥ 1/3

The final condition is 1/3 ≤ x ≤ 3

½ × AC × BD = ½ × AB × BC [Equating area]

AC × BD = AB × BC

4 BD × BD = AB × BC [∵ AC = 4 BD]

Dividing by BC² both the sides

4 × BD/BC × BD/BD = AB/BC

In triangle BDC, sin C = BD/BC and In triangle ABC, tan C = AB/BC

4sin²C = tan C

4sin²C = sinC/cos C

4sinC cos C = 1

2sin C cos C = ½

Sin2C = ½ [∵ sin 2A = 2sinA cos A]

∴ 2C = 30° and C = 15°

tan C = tan 15°

To find tan 15°

Tan 2θ = 2tanθ/ (1 – tan² θ)

Let tan 15° be x

Tan 30° = 2x/(1 – x²)

1/√3 = 2x/(1 – x²)

1 – x² = 2√3x

x² + 2√3x – 1 = 0

Using quadratic formula and ignoring the negative term

x = (-2√3 + 4)/2 = 2 – √3

Given:

AB =c , BC = a, CA = b and CD = p

Concept used:

Area of triangle = (1/2) Base × height

Pythagoras theorem:

Hypotenuse² = Base² + Perpendcular

Calculation:

Using the above formula

Area of triangle ABC = $\frac{1}{2}$ × AB × CD = $\frac{1}{2}$cp

Again,

Area of triangle ABC = $\frac{1}{2}$× BC × AC = $\frac{1}{2}$ ab

⇒  $\frac{1}{2}$ cp = $\frac{1}{2}$ ab

⇒ cp = ab                    ------(1)

Now, In right angled triangle ABC,

AB2 = AC2 + BC2

⇒ c2 = b2 + a2

⇒ (ab/p)2 = b2 + a2       [From equation (1)]

∴  a2b2 = p2(a2 + b2)

In a cone h = √(l² – r²) = √(13² – 5²) = 12cm

Volume of the cone = 1/3 × π r² h

⇒ 1/3 × π × 5² × 12

⇒ 100π cm³

The greatest possible area is cut off,

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

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

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

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

Area ∝ Radius²

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

?/A = (½)²

∴ Area of small circle = A/4

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

⇒ A – 2 × A/4 = A/2

Given:

ratio of radius of base and slant height (r : l)= 1 : 3

Formula used:

Total Surface area(TSA) = πr (l + r) 

Curved surface area(CSA) = πrl

Calculation:

TSA/CSA = [πr (l + r)] / (πrl)

⇒ (l + r)/l

⇒ 1 + r/l = 1 + 1/3 = 4/3 [∵ r ∶ l = 1 ∶ 3]

∴ Required ratio = 4 ∶ 3

For same vertical angle,

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

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

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

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

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

(1/3 π r² h) / (1/3 π R² H) = 64/125

(r/R)² × (h/H) = 64/125

k² × k = 64/125

k³ = 64/125

∴ k = 4/5

Curved surface area of small cone/Curved surface area of large cone = (π r l) / (π R L)

⇒ (r/R) × (l/L)

⇒ k × k = k² = 16/25

Curved surface area of small cone/Curved surface area of frustum = 16/ (25 – 16) = 16/9

∴ Required ratio 16 ∶ 9

Numerically surface area and volumes are equal

4πR² = 4/3 π R³

 1 = R/3

∴ R = 3 cm

In triangle OQS amd PRS

∠OQS = ∠PRS =90° [∵ Radius is perpendicular to tangent]

∠QSO = ∠PSR [∵ Vertically opposite angles are equal]

∴ OQS ∼ PRS

OQ/PR = OS/PS

4.5/3.5 = OS/PS

9/7 = OS/PS

∴ PS = 7OS/9

OP = OS + PS

10 = OS + 7OS/9

10 = 16OS/9

∴ OS = 90/16 = 5.625 cm

And PS = 10 – 5.625 = 4.375 cm

In triangle OQS, OS² = OQ² + QS² [∵ Pythagoras theorem]

5.625² = 4.5² + OS²

∴ OS = 3.375 cm

In triangle PRS, PS² = PR² + SP² [∵ Pythagoras theorem]

4.375² = 3.5² + SP²

∴ SP = 2.625

∴ OP = OS + SP

⇒ 3.375 + 2.625 = 6 cm

Shortcut TrickTransverse Common tangent = √{d² - (r₁ + r₂)²} 

⇒ {√10² - (4.5 + 3.5)²} 

⇒ √(10² - 8² )

⇒ √36

⇒ 6 cm

∴ The length of the transverse common tangent is 6 cm.

When the cone of maximum volume cut out from cube of edge 2a, height of the cone = Edge of the cube = 2a

Radius of the cone = ½ × Edge of the cube =1/2 × 2a = a

Volume of the cone = 1/3 × π r² h

⇒ 1/3 × π × a² × 2a = 2πa³/3

Area of (m + 1)^th region = π (r_m+1² – r_m²)

⇒ πr_m² [(r_m+1/r_m)² – 1]

⇒ πr_m² × (p_m² – 1)

Since area of all the region is constant

p_m² – 1 ∝ 1/r_m²

So as m increases r_m increases, ∴ p_m decreases with the m

Given:

Side of square floor = 10 m

Side of  equilateral triangular table = 2 m

Dimension of book = 4 m × 1 m × 9 m

Formula used:

Area of square = (Side)²

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

Area rectangle = Length × Width

Calculation:

Area to be carpeted

= ½ × (Area of the square floor – Area of the equilateral triangle – 4 × Area of bookshelves)

⇒ A = ½ × (side² – √3/4 × side² – 4 × l × b)

⇒ A = ½ × (10² – √3/4 × 2² – 4 × 4 × 1)

⇒ A = ½ × (100 – √3 – 16)

⇒ A = 41.134 m²     (∵ √3 ≈ 1.732)

Hence,

Cost = 41.134 × 100 ≈ Rs 4113

∴ Cost = 41.134 × 100 =  Rs 4113.