CDS Elementary Mathematics 23 Oct 2016 Official Paper (Previous Year Paper)

${\rm{lo}}{{\rm{g}}_{10}}{[1 - {{ 1 - {\left( {1 - {\rm{;}}{{\rm{x}}^2}} \right)^{ - 1}}} ^{ - 1{\rm{;}}}}]^{ - \frac{1}{2}}} = 1$

$\Rightarrow {\rm{lo}}{{\rm{g}}_{10}}{[1 - {{ 1 - \frac{1}{{1 - {{\rm{x}}^2}}}{\rm{;}}} ^{ - 1{\rm{;}}}}]^{ - \frac{1}{2}}} = 1$

$\Rightarrow {\rm{lo}}{{\rm{g}}_{10}}{[1 - {{ \frac{{1 - {\rm{;}}{{\rm{x}}^2} - {\rm{;}}1}}{{1 - {{\rm{x}}^2}}}{\rm{;}}} ^{ - 1{\rm{;}}}}]^{ - \frac{1}{2}}} = 1$

$\Rightarrow {\rm{lo}}{{\rm{g}}_{10}}{[1 - {{ \frac{{ - {{\rm{x}}^2}}}{{1 - {{\rm{x}}^2}}}{\rm{;}}} ^{ - 1{\rm{;}}}}]^{ - \frac{1}{2}}} = 1$

$\Rightarrow {\rm{lo}}{{\rm{g}}_{10}}{[1 + {{ \frac{{{\rm{;}}1 - {{\rm{x}}^2}}}{{{{\rm{x}}^2}}}{\rm{;}}} ^{\rm{;}}}]^{ - \frac{1}{2}}} = 1$

$\Rightarrow {\rm{lo}}{{\rm{g}}_{10}}{[\frac{{{\rm{;}}{{\rm{x}}^2} + {\rm{;}}1 - {{\rm{x}}^2}}}{{{{\rm{x}}^2}}}{{\rm{;}}^{\rm{;}}}]^{ - \frac{1}{2}}} = 1$

$\Rightarrow {\rm{lo}}{{\rm{g}}_{10}}{[{\rm{;}}\frac{1}{{{{\rm{x}}^2}}}{{\rm{;}}^{\rm{;}}}]^{ - \frac{1}{2}}} = 1$

$\Rightarrow {\rm{lo}}{{\rm{g}}_{10}}{\rm{x}} = 1$

∴ x = 10

4x² – 16x + λ/4 = 0

⇒ 16x² – 64x + λ = 0

Sum of root = α + β = 64/16 = 4

Product of root = αβ = λ/16

Given 1 < α < 2 and 2 < β < 3

⇒ Value of αβ should be between 2 and 6

But we need to consider that α + β = 4

⇒ 3 < αβ < 4

Putting αβ = λ/16

⇒ 3 < λ/16 < 4

⇒ 48 < λ < 64

∴ λ can take values from 49 to 63 i.e. 15 no. of values.

$\frac{{{6^2} + {\rm{;}}{7^2} + {\rm{;}}{8^2} + {\rm{;}}{9^2} + {\rm{;}}{{10}^2}}}{{\sqrt {7 + {\rm{;}}4\sqrt 3 } - {\rm{;}}\sqrt {4{\rm{;}} + {\rm{;}}2\sqrt 3 } }}$

$\Rightarrow \frac{{36{\rm{;}} + {\rm{;}}49{\rm{;}} + {\rm{;}}64{\rm{;}} + {\rm{;}}81{\rm{;}} + {\rm{;}}100}}{{\sqrt {{{\left( {2{\rm{;}} + {\rm{;}}\sqrt 3 } \right)}^2}} {\rm{;}} - \sqrt {{{\left( {1{\rm{;}} + {\rm{;}}\sqrt 3 } \right)}^{2{\rm{;}}}}} }}$

$\Rightarrow \frac{{330}}{{2{\rm{;}} + {\rm{;}}\sqrt 3 {\rm{;}} - {\rm{;}}1 - {\rm{;}}\sqrt 3 }}$

$\therefore \frac{{{6^2} + ;{7^2} + ;{8^2} + ;{9^2} + ;{{10}^2}}}{{\sqrt {7; + ;4\sqrt 3 ;} - ;\sqrt {4; + ;2\sqrt 3 } }} = 330/1 = 330$

$\frac{1}{{x; + ;1}} + \frac{1}{{y; + ;1;}} + \frac{1}{{z; + ;1}}$

Multiplying x in first part, y in second part and z in third part.

$\Rightarrow \frac{x}{{{x^2}; + ;x}} + \frac{y}{{{y^2}; + ;y;}} + \frac{z}{{{z^2}; + ;z}}$

Putting the given values

$\Rightarrow \frac{x}{{y; + ;z; + ;x}} + \frac{y}{{z; + ;x; + ;y}} + \frac{z}{{x; + ;y; + ;z}}$

$\therefore \frac{{x; + ;y; + ;z}}{{y; + ;z; + ;x}} = 1$

5P9 + 3R7 + 2Q8 = 1114

We can see that, the sum of last digits i.e. (5 + 3 + 2) = 10 and 1 is added as carry

And 2 is added as carry in the sum of middle digits of the numbers (P + Q + R)

⇒ (P + Q + R + 2) will be 11 only.

⇒ P + Q + R = 9

For maximum value of Q, values of P and R should be minimum

If P = R = 0 then Q = 9

∴ Maximum value of Q = 9

We can observe that: The difference in divisor and reminder is same in every case.

(2 – 1) = (3 – 2) = (4 – 3) = (5 – 4) = (6 – 5) = 1

LCM of 2, 3, 4, 5 and 6 = 60

⇒ Required number = 60 – 1 = 59

Set A = [(60 – 1), (120 – 1), (180 – 1), ….] = [59, 119, 179, …..]

Between 0 and 100, only 59 is such integer.

∴ Only one integer between 0 and 100 belongs to the set A.

Let the number be x

⇒ 8x – x/8 = 2016

⇒ 64x – x = 2016 × 8

⇒ 63x = 16128

⇒ x = 256

∴ The number is 256.

x² + ax + b = 0

For real roots, necessary condition is:

a² – 4b ≥ 0

⇒ a² ≥ 4b

And (a, b) $\epsilon$ {1, 2, 3, 4}

Possible pairs of a & b according to the above conditions:

(2, 1), (3, 1), (4, 1), (3, 2), (4, 2), (4, 3), (4, 4)

∴ There are 7 possible pairs of a and b.

Suppose the numbers are ‘a’ and ‘b’ with a > b

a² + b² = 208 and a² = 18b

Solving the equations

⇒ b² + 18b = 208

⇒ b² + 18b – 208 = 0

⇒ (b + 26)(b – 8) = 0

⇒ b = 8, b = -26

If b = 8

a² = 18b = 18 × 8 = 144

⇒ a = 12

∴ Difference of the larger and the smaller number = 12 – 8 = 4

A = {7, 8, 9, 10, 11, 12} and B = {7, 10, 14, 15}

(A – B) = {8, 9, 11, 12}

(B – A) = {14, 15}

∴ Number of elements in (A – B) and (B – A) are 4 and 2 respectively.

Using hit and trial method:

1. 4.65 × 10 = 46.5

2. 4.65 × 20 = 93

3. 4.65 × 21 = 97.65

4. 4.65 × 25 = 116.25

∴ He will be able to save an exact number of rupees in 20 days.

Given:

Two men, A and B run a 4 km race on a course 0.25 km round.

Their speeds are in the ratio 5 : 4,

Formula used:

Distance = Speed × Time

Calculation:

Number of rounds to be completed = 4/0.25

⇒ 16

Since the ratio of speeds of A and B is 5 : 4

When A covers 5 rounds, B will complete 4 rounds and they will meet.

A will pass B after 5^th, 10^th & 15^th round.

∴ The winner passes the other thrice.

Calculations:

Observing all the options

Check option (1), we get

1. 15/1600 = 0.009375

Now, check option (2), we get

2. 23/8 = 2.875

Now, Check option (3), we get

3. 35/50 = 0.7

Now, Check option (4), we get

4. 17/6 = 2.8333333……

Here we can see that 17/6 has non-terminating and repeating decimal expression.

∴ The non-terminating and repeating decimal is 17/6.

Concept:

If α and β are the two zeros of the quadratic polynomial ax2 + bx + c, then:

ax2 + bx + c = (x - α)(x - β) = x2 - (α + β)x + αβ = 0

Sum of roots = α + β = −b/a 

Product of roots = αβ = c/a.

Calculation:

25x² – 15x + 2

⇒ 25x² – 10x – 5x + 2

⇒ 5x(5x – 2) – 1(5x – 2)

⇒ (5x – 1)(5x – 2)

⇒ x = 1/5 and 2/5

⇒ α and β will be equal to 1/5 and 2/5.

If zeros are (2α)^-1 and (2β)^-1

Then the zeros of that polynomial are 5/2 and 5/4

∴ Polynomial = (x – 5/2)(x – 5/4) = (x² – 15x/4 + 25/8) = (8x² – 30x + 25)

$\Rightarrow \frac{{{2^{100}}}}{{101}} = ;\frac{{{2^{\left( {101; - ;1} \right)}}}}{{101}}$

[If a^{(n – 1)} is divided by ‘n’ where ‘a’ and ‘n’ are co-prime numbers then remainder will be 1]

Hence here also ‘2’ and ‘101’ are co-prime numbers.

∴ Remainder = 1

Suppose the total number of workers = x

⇒ Number of women workers = x/3

⇒ Number of men workers = 2x/3

⇒ Number of married women = (x/3) × ½ = x/6

⇒ Number of married men = (2x/3) × 3/4 = x/2

∴ Ratio of married women to married men = (x/6) : (x/2) = 1 : 3

B alone can do the work in 30 hours and A can do 50% more work than B in same time;

⇒ Time taken by A to complete the work alone = 30/1.5 = 20 hours

Suppose total work = 60 units (LCM of 20 & 30)

 ⇒ Efficiency of A = 60/20 = 3

⇒ Efficiency of B = 60/30 = 2

⇒ Unit of work done by B in 12 hours = 12 × 2 = 24 units

⇒ Remaining work = 60 – 24 = 36 units

∴ Time taken by (A + B) to complete remaining work = 36/5 = 7.2 hours

Suppose the distance be ‘x’ km and speed be ‘n’ km/minutes

⇒ Increased speed = 1.2n

⇒ x/n – x/1.2n = 20

⇒ (0.2x/1.2n) = 20

⇒ x/n = 120 minutes

Here ‘x/n’ is the time taken to cover the distance with the original speed.

∴ Train will take 120 minutes to cover the same distance with the original speed.

Given Cost of 2.5 kg rice is Rs. 125

⇒ Cost of 1 kg rice = 125/2.5 = Rs. 50

⇒ Cost of 4 kg pulses = cost of 9 kg rice = 50 × 9 = Rs. 450

⇒ Cost of 1 kg pulses = 450/4 = Rs. 112.5

⇒ Cost of 1.5 kg tea = Cost of 14 kg pulses = 112.5 × 14 = Rs. 1575

⇒ Cost of 1 kg tea = 1575/1.5 = Rs. 1050

⇒ Cost of 5 kg nuts = Cost of 2 kg tea = 1050 × 2 = Rs. 2100

∴ Cost of 11 kg nuts = (11/5) × 2100 = Rs. 4620

Ratio in which the profit will be shared = Investment × Time for which investment was made

⇒ (1 × 12) : (2 × 11) : (3 × 10) : (4 × 9) : (5 × 8) : (6 × 7) : (7 × 6) : (8 × 5) : (9 × 4) : (10 × 3) : (11 × 2) : (12 × 1)

⇒ 12 : 22 : 30 : 36 : 40 : 42 : 42 : 40 : 36 : 30 : 22 : 12

Here we can see that shares of both F and G in the profit will be maximum.

x² + px + q = 0

If a and b are the roots of the equation

⇒ Sum of roots = (p + q) = -p     ----1

⇒ Product of roots = pq = q     ----2

On solving equation 1 and 2, we get

p = 1 and q = -2

∴ p = 1 only

1729 = 1728 + 1 = 12³ + 1³ and 1729 = 1000 + 729 = 10³ + 9³

∴ It can be written as the sum of cubes of two positive integers in two ways only.

Let the cost price of the item = Rs. x

⇒ Increased cost price = marked price = Rs. 1.2x

⇒ Overall selling price = 1.2x × 0.9 = Rs. 1.08x

Profit = Selling price - Cost price = 1.08x - x = Rs. 0.08x

∴ Percentage gain = 0.08x/x × 100 = 8%

HCF of p and q is 1

1. Both p and q may be prime

Suppose p = 2 and q = 3

⇒ HCF of p and q = 1

⇒ Statement 1 is correct

2. One number may be prime, and the other is composite.

Suppose p = 3 and q = 4

⇒ HCF of p and q = 1

⇒ Statement 2 is correct

3. Both the numbers may be composite

Suppose p = 9 and q = 10

⇒ HCF of p and q = 1

⇒ Statement 3 is correct.

∴ All three statements are correct.

$\sqrt {1 + \frac{1}{{{1^{2}}}} + \frac{1}{{{2^{2}}}}} + \sqrt {1 + \frac{1}{{{2^{2}}}} + \frac{1}{{{3^{2}}}}} + \ldots \ldots .. + \sqrt {1 + \frac{1}{{{{2007}^{2}}}} + \frac{1}{{{{2008}^{2}}}}}$

n^th term of this series can be written as

${T_n} = \sqrt {1 + \frac{1}{{{n^{2}}}} + \frac{1}{{{{\left( {n + 1} \right)}^{2}}}}}$ 

$\Rightarrow \sqrt {\frac{{{n^2}{{\left( {n + 1} \right)}^2} + {n^2} + {{\left( {n + 1} \right)}^2}}}{{{n^2}{{\left( {n + 1} \right)}^2}}}}$ 

$\Rightarrow \sqrt {\frac{{{n^2}{{\left( {n + 1} \right)}^2} + 2{n^2} + 2n + 1}}{{{n^2}{{\left( {n + 1} \right)}^2}}}}$ 

$\Rightarrow \sqrt {\frac{{{n^2}{{\left( {n + 1} \right)}^2} + 2n\left( {n + 1} \right) + 1}}{{{n^2}{{\left( {n + 1} \right)}^2}}}}$ 

$\Rightarrow \sqrt {\frac{{{{\left( {n\left( {n + 1} \right) + 1} \right)}^2}}}{{{n^2}{{\left( {n + 1} \right)}^2}}}}$ 

$\Rightarrow \frac{{n\left( {n + 1} \right) + 1}}{{\left[ {n\left( {n + 1} \right)} \right]}}$ 

$\Rightarrow 1 + \frac{1}{{\left[ {n\left( {n + 1} \right)} \right]}}$

$\Rightarrow 1 + \frac{{n + 1 - n}}{{\left[ {n\left( {n + 1} \right)} \right]}}$ 

$\Rightarrow 1 + \frac{1}{n} - \frac{1}{{n + 1}}$

Sum of the series = $1 + \frac{1}{1} - \frac{1}{2} + 1 + \frac{1}{2} - \frac{1}{3} + 1 + \frac{1}{3} - \frac{1}{4} \ldots \ldots \ldots \ldots \ldots .1 + \frac{1}{{2007}} - \frac{1}{{2008}}$

Sum of the series = $1 + \frac{1}{1} + 1 + 1 + 1 \ldots \ldots \ldots 1 - \frac{1}{{2008}}$

∴ Sum of the series = $2008 - \frac{1}{{2008}}$

Formula used:

Speed = Distance / Time

Speed of current = $\frac{1}{{2}}$(Downstream speed - Upstream speed)

Calculation:

Using the above formula,

Downstream speed = $\frac{20}{{2}}$ = 10 km/hr

Upstream speed = $\frac{4}{{2}}$ = 2 km/hr

⇒ Speed of current = $\frac{1}{{2}}$(Downstream speed - Upstream speed)

⇒ Speed of the current = $\frac{10 - 2}{{2}}$ = 4 km/hr

∴ The speed of the current is 4 km/hr

Since x and y are two odd integers, their squares will be odd always and sum of two odd numbers is always even.

⇒ Statement 1 is correct.

Suppose x = 1 and y = 3

⇒ x² + y² = 1 + 9 = 10

It is even but not divisible by 4.

∴ Only statement 1 is correct.

2^{x + 2} × 27^{x/ (x - 1)} = 9

⇒ 2^{x + 2} × 3^{3x/ (x - 1)} = 3²

⇒ 2^{x + 2} = 3^{[2 - {(3x/(x - 1))}]}

⇒ 2^{x + 2} = 3^{[-(x + 2)/(x - 1)]}

Taking log

⇒ (x + 2)log2 = -[(x + 2)/(x - 1)]log3

⇒ x - 1 = -(log3/log2)

⇒ x = 1 - (log3/log2)

And

(x + 2) = 0 ⇒ x = -2

∴ Roots of the equation = -2, 1 - (log3/log2)

2¹²² + 4⁶² + 8⁴² + 4⁶⁴ + 2¹³⁰

⇒ 2¹²² + 2¹²⁴ + 2¹²⁶ + 2¹²⁸ + 2¹³⁰

⇒ 2¹²² (1 + 4 + 16 + 64 + 256)

⇒ 2¹²² × 341

Since 341 is divisible by 11

∴ 2¹²² + 4⁶² + 8⁴² + 4⁶⁴ + 2¹³⁰ is divisible by 11 only

Let the maximum marks for the examination be ‘n’.

Candidate scoring x% marks in an examination fails by ‘a’ marks but another candidate who scores y% marks gets ‘b’ marks more than the minimum pass marks;

∴ (n × x)/100 + a = (n × y)/100 – b

⇒ (a + b) = n/100 × (y – x)

$\Rightarrow {\rm{n}} = {\rm{}}\frac{{100\left( {{\rm{a;}} + {\rm{;b}}} \right)}}{{{\rm{y;}} - {\rm{;x}}}}$

2p + 3q = 12

⇒ 2p = (12 - 3q)      ---- 1

4p² + 4pq - 3q² = 126

⇒ (2p)² + 2p × 2q - 3q² = 126       ---- 2

Putting the value of 2p from equation 1 in equation 2

⇒ (12 - 3q)² + 2q(12 - 3q) - 3q² = 126

⇒ 144 + 9q² - 72q + 24q - 6q² - 3q² = 126

⇒ 48q = 18

⇒ q = 3/8

Putting this into equation 1

⇒ 2p = 12 - 9/8

⇒ p = 87/16

∴ p + 2q = 87/16 + 6/8 = 99/16

Suppose x = 3³⁰

Taking log both the side

⇒ log₁₀x = 30 log₁₀3

⇒ log₁₀x = 30 × 0.4771

⇒ log₁₀x = 14.313

⇒ x = (10)^14.313

Here we can see that the number will have (14 + 1) = 15 digit.

∴ n = 15

Given:

7139

Calculation:

(7¹³⁹) = (7^{136 + 3}) = (7^{(4 × 34 + 3)}) = 7^{4 × 34} × 7³

We know that cyclicity of 7 is 4 only, so (7^{4 × 34}) will have the unit digit 1

⇒ Unit digit of (7¹³⁹) = Unit digit of (1 × 7³) = Unit digit of 343

∴ Unit digit of (7¹³⁹) is 3.

Since 4x + 3a = 0

⇒ x = -3a/4

Putting this value in the expression

$\frac{{{x^2}; + ;ax; + ;{a^2}}}{{{x^3}; - ;{a^3}}} - ;\frac{{{x^2} - ;ax; + ;{a^2}}}{{{x^3}; + ;{a^3}}}$

$\Rightarrow ;\frac{{\frac{{9{a^2}}}{{16}}; - ;\frac{{3{a^2}}}{4}; + ;{a^2}}}{{\frac{{ - 27{a^3}}}{{64}}; - ;{a^3}}} - \frac{{\frac{{9{a^2}}}{{16}}; + ;\frac{{3{a^2}}}{4}; + ;{a^2}}}{{\frac{{ - 27{a^3}}}{{64}}; + ;{a^3}}}$

$\Rightarrow ;\frac{{\frac{{13{a^2}}}{{16}}}}{{\frac{{ - 91{a^3}}}{{64}}}} - \frac{{\frac{{37{a^2}}}{{16}}}}{{\frac{{37{a^3}}}{{64}};}}$

$\Rightarrow ;\frac{4}{{ - 7a}} - \frac{4}{{a;}}$

$\therefore \frac{{{x^2}; + ;ax; + ;{a^2}}}{{{x^3}; - ;{a^3}}} - ;\frac{{{x^2} - ;ax; + ;{a^2}}}{{{x^3}; + ;{a^3}}} = ;\left( { - 32} \right)/7{\rm{a}}$

With the help of ratios only, it is not possible to determine who saves more.

Amount put by X for first 3 months = Rs. 7000

⇒ Amount put by X for 4^th to 6^th month = 7000 - (2/7) × 7000 = 7000 - 2000 = Rs. 5000

⇒ Amount put by X for 7^th to 12^th month = 5000 + (3/5) × 2000 = Rs. 6200

⇒ Amount put by X for whole year = (7000 × 3 + 5000 × 3 + 6200 × 6) = Rs. 73200

Amount put by Y for whole year = 6000 × 12 = Rs. 72000

⇒ Ratio in which the profit would be shared = 732 : 720 = 61 : 60

The profit at the end of the year is Rs. 726

∴ Share of X = (61/121) × 726 = Rs. 366

Suppose the capacity of tank = 6 units (LCM of 2 and 6)

⇒ Efficiency of X = 6/2 = 3 and Efficiency of Y = 6/6 = 1

Pipe X was opened at 10 am and pipe Y is opened at 11 am

Units of water filled by X in 1 hour (10 a.m. to 11 a.m.) = 3 units

⇒ Remaining units of water to be filled = 6 - 3 = 3 units

⇒ Time taken by (X + Y) to fill the remaining tank = ¾ hours = 45 minutes

∴ Tank will get filled at 11:45 a.m.

Let the length of train be ‘x’ km and that of another train be ‘x/2’ km and length of the platform be ‘y’

Train travelling at 48 km/hr crosses another train half its length and travelling in opposite direction in 12 sec

⇒ Relative speed = 48 + 42 = 90 km/hr

⇒ (x + x/2)/(48 + 42) = 12/3600

⇒ 3x/2 = 1080/3600

⇒ 3x = 2160/3600

⇒ x = 720/3600 = 0.2 km = 200 m

The train crosses railway platform in 45 s

⇒ (x + y)/48 = 45/3600

⇒ (0.2 + y) = 2160/3600

⇒ y = 0.6 - 0.2 = 0.4 = 400 m

∴ Length of the platform = 400 m.

Speed = Distance/time

⇒ Time ∝ 1/Speed

Ratio of speed = 2 : 3 : 4

Suppose the distance is 12x units (LCM of 2, 3 & 4)

∴ Ratio of time taken to cove same distance = (12x/2 : 12x/3 : 12x/4) = 6 : 4 : 3

Let the number of boys be x and number of girls be y

Mean age of combined group of boys and girls is 18 years

Total Number of boys and girls = x + y

⇒ Total combined age of Boys and Girls = 18(x + y)

⇒ Total age of boys = 20x

⇒ Total age of girls = 16y

⇒ 20x + 16y = 18(x + y)

⇒ 2x = 2y

⇒ x = y

∴ The percentage of boys in the group is 50%

When the square of side x is inscribed in the circle of radius ‘r’

By Pythagoras, x² + x² = 4r²

⇒ 2x² = 4r²

⇒ x = r√2

When the equilateral triangle of side y is inscribed in the circle of radius ‘r’

In triangle AOD

Sin 30° = OD/r

⇒ OD = r/2

⇒ (y/2)² + (r/2)² = r²

⇒ y²/4 + r²/4 = r²

⇒ 3r²/4 = y²/4

⇒ y = r√3

∴ x : y = r√2 : r√3 = $\sqrt {\frac{2}{3}}$

Suppose the radius and height of the cylinder is ‘r’ and ‘h’

Ratio of the curved surface area to the total surface area of a right circular cylinder is 1 : 2

⇒ (2πrh) : (2πrh + 2πr²) = 1 : 2

⇒ h : (h + r) = 1 : 2

⇒ h = r

Since total surface area is 616 cm²

⇒ (2πrh + 2πr²) = 616

Putting h = r

⇒ 2π(r² + r²) = 616

⇒ 4r² = 196

⇒ r = 7 cm

⇒ h = r = 7 cm

∴ Volume of the cylinder = πr²h = (22/7) × 7 × 7 × 7 = 1078 cm³

The volume of the copper cube = 1 m³

Let the length of side of square bar is ‘x’ m

⇒ Volume of the square bar 9 m long = x² × 9 = (9x²) m³

Since the copper cube is converted into the square bar, their volume must be same

⇒ 9x² = 1

⇒ x = 1/3

When an exact cube is cut off from the bar, it’s side will be x only

⇒ Volume of the cube = (1/3)³ = 1/27 m³

∴ Weight of the cube = (1/27) × 9000 = 1000/3 kg

Radius of the conical tent and height is 8.4 m and 3.5 m respectively

⇒ Volume of the conical tent = (π/3)r²h = 22/7 × 1/3 × 8.4 × 8.4 × 3.5 = 258.72 m³

Since the space required for the wheat in each bag is 1.96 m³

∴ Number of bags that can be emptied in the conical tent = 258.72/1.96 = 132

BC = 5 cm, AC = 12 cm and ∠BCD = 30°

Since it is a right-angle triangle, AB = 13 cm

Let CD = x cm

⇒ Area of triangle ABC = ½ × 5 × 12 = 30 cm²

⇒ Area of triangle BCD = ½ × BC × CD × Sin30 = ½ × 5 × CD × ½ = 5CD/4 cm²

⇒ Area of triangle ACD = ½ × AC × CD × Sin60 = ½ × 12 × CD × √3/2 = 3√3CD cm²

In the figure                                                             

Area of triangle ABC = Area of triangle BCD + Area of triangle ACD

⇒ 30 = 5CD/4 + 3√3CD

⇒ (5 + 12√3)CD = 120

∴ CD = $\frac{{120}}{{5; + ;12\sqrt 3 }}$ cm

Suppose r and R are the radius of the circles

Sum of the areas of the circles = 130π cm²

⇒ π(R² + r²) = 130π

⇒ (R² + r²) = 130     ---- (1)

Distance between their centres = 14 cm

⇒ R + r = 14      ---- (2)

From equation (1) and (2)

⇒ (14  -r)² + r² = 130

⇒ 196 + r² - 28r + r² = 130

⇒ 2r² - 28r + 66 = 0

⇒ r² - 14r + 33 = 0

⇒ r² - 11r - 3r + 33 = 0

⇒ (r - 11)(r - 3) = 0

⇒ r = 11 & 3

⇒ R = 11 and r = 3

∴ Difference in the radii of the circles = 11 – 3= 8 cm

In the figure, AB is the diameter of the circle and XY is the chord

Since the diameter intersects the chord perpendicularly, chord will be divided into 2 equal parts of length 1 unit.

⇒ PY = 1 unit and Radius OY = 3 units

⇒ OP² = 9 – 1

⇒ OP = √8 units

⇒ AP = AO + OP = 3 + √8 and BP = BO – OP = 3 – √8

∴ AP : BP = 3 + √8 : 3 – √8

Radius of the circle = 5/22 m

Circumference of the circle = 2πr = 2 × 22/7 × 5/22 = 10/7 m

∴ Number of rounds in traversing 7 km = (7000) / (10/7) = 4900

Let the height of the cylinder be H and radius be r m

⇒ Radius of the hemisphere = r

Given that height of the building is three times the radius of the base of the cylinder

⇒ (H + r) = 3r

⇒ H = 2r

Since the building contains $67\frac{1}{{21}};$m³ of air

⇒ Volume of the building = $67\frac{1}{{21}}$ m³ = 1408/21

⇒ (πr²H) + (2/3)πr³ = 1408/21

Putting H = 2r

⇒ (2πr³) + (2/3)πr³ = 1408/21

⇒ (8/3) × 22/7 ×r³ = 1408/21

⇒ r³ = 8

⇒ r = 2

∴ Height of the building = H + r = 4 + 2 = 6 m

Let the breadth and length of the room be ‘x’ and ‘2x’;

Height of the room is given 4 m;

Area of 4 walls = 2(l + b) × h

⇒ 120 = 2(3x) × 4

⇒ 3x = 15

⇒ x = 5 m

Breadth = 5 m and Length = 10 m

∴ Area of the floor = 50 m²

Ratio of radius of the base and height = 2 ∶ 3;

Let the radius and height be 2x and 3x respectively;

Given that volume is 1617 cm³;

⇒ πr²h = 1617

⇒ (22/7) × 4x² × 3x = 1617

⇒ x³ = 42.875

⇒ x = 3.5 cm

⇒ Radius = 7 cm and Height = 10.5 cm;

∴ Total surface area of the cylinder = (2πr² + 2πrh)

⇒ 2 × 22/7 × (7² + 7 × 10.5)

⇒ 770 cm²

∵ Diameter of the vessel = 14 cm

∴ Radius = 7 cm;

We need to find out the surface area of the spherical vessel;

Surface area of spherical vessel = 4πr²

⇒ 4 × 22/7 × 7²

⇒ 616 cm²

∵ The cost of painting the spherical vessel is Rs. 8008;

∴ Cost of painting per square centimetre = 8008/616 = Rs. 13

Concept:

The lengths of tangents drawn from an external point to a circle are equal 

Calculation:

Hence, the sum of the opposite side

⇒ AB + CD = (AE + EB) + (DG + GC)

Similarly, a sum of the two remaining

⇒ AD + BC = (AH + HD) + (BF + FC)

By using the above concept

AE = AH, BE = BF, DH = DG & CG = CF    

⇒ AD + BC = (AE + DG) + (BE + CG)

⇒ AD + BC = (AE + EB) + (DG + GC)

Hence, AB + CD = AD + BC

∴ The Sum of a pair of opposite sides equals the sum of another pair of opposite sides.

Let the side of square = x and the radius of circle = r;

∵ The circle and the square have the same perimeter;

∴ 4x = 2πr

⇒ x = πr/2

Area of circle = πr²

Area of square = x² = (πr/2)² = π²r²/4

⇒ Area of square = (π/4) × Area of circle

So, the Area of the circle is greater than the area of the square.

∴ It is evident that the Area of circle is larger.

Given:

Height of glass = 24 cm.

Diameters of its bottom and top circular ends

= 4 cm and 18 cm respectively

Formula used:

The volume of the frustum of a cone = (π/3) × H × (R2 + r2 + Rr)

Calculation:

Given that height is 24 cm and diameters of bottom and top circular ends are 4 cm and 18 cm respectively;

∴ H = 24 cm, R = 18/2 = 9 cm and r = 4/2 = 2 cm;

We know the formula;

Volume of the frustum of a cone = (π/3) × H × (R² + r² + Rr)

∵ It is given that the capacity of the glass as πx cm³ that means the volume of the frustum of the cone is πx cm³;

∴ πx = (π/3) × 24 × (81 + 4 + 18)

⇒ x = 8 × 103

⇒ x = 824

Let the length of side of first equilateral triangle = x;

As shown in the figure, when an equilateral triangle is drawn inside another equilateral triangle joining the mid-points of the sides, the sides of that triangle will be half of the side of the 1^st equilateral triangle;

In this way;

Side of 2^nd triangle = x/2;

Side of 3^rd triangle = x/4;

Side of 4^th triangle = x/8;

Side of 5^th triangle = x/16;

Side of 6^th triangle = x/32;

Side of 7^th triangle = x/64;

∴ Ratio of the area of 4^th to 7^th triangle = [√3/4 × (x/8)²] : [√3/4 × (x/64)²] = 64 : 1

Let the rainfall is x m;

Radius and height of the cylindrical vessel are 1 m and 3.5 m;

∴ Volume of the cylindrical vessel = π r²h = 22/7 × 1 × 3.5 = 11 m²

Area of the roof = 22 × 20 = 440 m²

∴ Volume of the cylindrical vessel = Area of the roof × x

⇒ 11 = 440 × x

⇒ x = 0.025 m = 2.5 cm

∴ The rainfall is 2.5 cm.

In the figure, R and r are the radius of the two cones and H and h are the heights;

∴ r/R = h/H

Volume of new cone : Volume of original cone = 1 : 64;

∴ 1/3(πr²h) : 1/3(πR²H) = 1 : 64

⇒ r²h : R²H = 1 : 64

Putting r/R = h/H;

⇒ h³ : H³ = 1 : 64

⇒ h : H = 1 : 4

∵ H = 60 cm (Given)

∴ h = 60/4 = 15 cm

∴ Height from the base at which the section is made = H – h = 60 – 15 = 45 cm

In the figure;

Two tangents OP and OQ inclined at 60° are drawn, they will make 90° angle with centre;

Given, Radius of the circle = PC = QC = 3 cm;

In ΔOCP;

tan 30° = PC/OP

⇒ 1/√3 = 3/OP

⇒ OP = 3√3 cm

Similarly OQ = 3√3 cm

∴ Length of each tangent = 3√3 cm

Formula used:

The volume of a sphere = 4/3 × πr³

Calculation:

Given that, 

Radius = 3 cm 

The volume of a sphere = 4/3 × πr³

⇒ 4π/3 × 3 × 3 × 3

⇒ 36π cm³

Sides of a triangle are √(a² + b²), √(c² + a²) and (b + c);

Taking a = b = c = 1;

Sides of a triangle will be √2, √2 and 2;

This will be a right-angled triangle with 2 as hypotenuse;

∴ Area of the triangle = ½ × √2 × √2 = 1

Checking the options by putting a = b = c = 1;

1. √(a² + b² + c²)/2 = √3/2
2. √(a²b² + b²c² + c²a²)/2 = √3/2
3. a(b + c)/2 = 1
4. √3(a²b² + b²c² + c²a²)/2 = 3/2

∴ Option 3 gives the area = 1;

∴ Area of the triangle equal to a(b + c)/2.

In the figure;

ABC is the largest triangle in the semi-circle (with diameter as one of its side);

∴ Area of the triangle = 1/2 × 2r × r = r² square units

Circumference of front wheel = 2π × x = 2πx;

∴ Distance covered by front wheel when it revolves n times = (n × 2πx) cm

Circumference of rear wheel = 2π × y = 2πy;

Suppose, real wheel revolves ‘m’ times;

∴ Distance covered by rear wheel when it revolves m times = (m × 2πy) cm

∵ the distance covered is same;

∴ (n × 2πx) = (m × 2πy)

⇒ m = nx/y

∴ Real wheel revolves ‘nx/y’ times.

We need to find out the total surface area of the tent;

Total surface area of tent = Curved surface area of cylinder + Curved surface area of cone

⇒ 1980 + πrl

⇒ 1980 + 22/7 × 63 × 65

⇒ 1980 + 12870

⇒ 14850 m²

∴ 14850 square metres of canvas was used.

Radius of cylinder = 63 m and height = 5 m;

Curved surface area of the cylinder = 2πrh

⇒ 2 × 22/7 × 63 × 5

⇒ 1980 m²

In the figure;

Radius of cylinder and cone will be equal to 126/2 = 63 m

Height of cylindrical part is 5 m and total height is 21 m;

∴ Height of conical part = 21 – 5 = 16 m;

∴ Slant height (l) = √(63² + 16²)

⇒ √(3969 + 256)

⇒ √4225

⇒ 65 m

Diameter of the wheel is 80 cm;

∴ Circumference of the wheel = 2 × 22/7 × 40 = 1760/7 cm

Speed of the car is 66 km/hr;

∴ Distance covered by the car in 10 minutes = 66 × 10/60 = 11 km

∴ Number of revolution = (11 × 100000)/(1760/7) = 4375

∴ 4375 complete revolution each wheel makes in 10 minutes.

Sides of the triangle are 12 cm, 13 cm and 5 cm;

∵ It is a Pythagoras triplet (13² = 12² + 5²);

∴ It will be a right angled triangle with hypotenuse 13;

∴ Area of the triangle = ½ × 12 × 5 = 30 cm²

Let the side of the square is ‘x’ and radius of circle is ‘y’;

∵ the perimeter of a circle is equal to that of a square;

∴ 4x = 2πr

⇒ x = 11r/7

Ratio of area of circle to that of square = πr² : x²

⇒ (22/7)r² : (121/49)r²

⇒ 14 : 11

In the figure, AB is the diameter and CD is parallel and half of AB;

∴ AB = 2r and CD = r;

Locus of the middle point of CD is a circle with radius EO;

In ΔEOC;

EO = √(r² – (r/2)²)

⇒ EO = √3r/2

sinθ + cosθ = √7/2

Squaring both sides;

⇒ sin²θ + cos²θ + 2sinθ.cosθ = 7/4

⇒ 1 + 2sinθ.cosθ = 7/4

⇒ sinθ.cosθ = 3/8

We can write;

sin²θ + cos²θ = (sinθ – cosθ)² + 2sinθ.cosθ

⇒ (sinθ – cosθ)² = 1 – 3/4 = 1/4

⇒ (sinθ – cosθ) = 1/2

sinx + sin²x = 1

⇒ sinx = 1 – sin²x

⇒ sinx = cos²x

Again,

sinx + sin²x = 1

Squaring both side;

⇒ sin⁴x + 2sin³x + sin²x = 1

⇒ cos⁸x + 2cos⁶x + cos⁴x = 1 {Putting cos²x = sinx}

In the figure,

The car’s position changes from B to D and it travelled ‘x’ m;

In ΔACD;

tan 60° = 100/AD

⇒ √3 = 100/AD

⇒ AD = 100/√3 m

In ΔABC;

tan 30° = 100/AB

⇒ 1/√3 = 100/AB

⇒ AB = 100√3 m

∴ x = AB – AD = 100√3 – 100/√3 = 200/√3 = 200√3/3 m

∴ Distance travelled by car during this time = 200√3/3 m

In the figure, AB is the tower and CD is the distance between two men on either side of tower;

In ΔABD

tan 60° = AB/AD

⇒ √3 = 75/AD

⇒ AD = 75/√3

tan 30° = AB/AC

⇒ 1/√3 = 75/AC

⇒ AC = 75√3

∴ CD = 75√3 + 75/√3 = 75√3 + 25√3 = 100√3

∴ Distance between two men on either side of tower = 100√3 m

cosec²68° + sec²56° – cot²34° – tan²22°

⇒ cosec²68° + sec²56° – cot²(90 – 56)° – tan²(90 – 68)°

⇒ cosec²68° + sec²56° – tan²56° – cot²68°

We know the formulas:

cosec²θ – cot²θ = 1 and sec²θ – tan²θ = 1

∴ cosec²68° + sec²56° – tan²56° – cot²68° = 1 + 1 = 2

2y cosθ = x sinθ

⇒ 2y = x tanθ      ---- (1)

And

2x secθ – y cosecθ = 3

From equation (1);

⇒ 2x secθ – (x/2) tanθ.cosecθ = 3

⇒ 2x secθ – (x/2) secθ = 3

⇒ (3x/2) secθ = 3

⇒ x = 2cosθ

From equation (1);

2y = 2cosθ.tanθ

⇒ y = sinθ

Putting x = 2cosθ and y = sinθ in x² + 4y²;

⇒ x² + 4y² = 4cos²θ + 4sin²θ

⇒ x² + 4y² = 4

sinθ + cosθ = (1 + √3)/2

Squaring both sides

⇒ sin²θ + cos²θ + 2sinθ.cosθ = (1 + 3 + 2√3)/4

⇒ 1 + 2sinθ.cosθ = 1 + √3/2

⇒ 2sinθ.cosθ = √3/2

⇒ sinθ.cosθ = √3/4

Now

tanθ + cotθ

⇒ (sinθ/cosθ + cosθ/sinθ)

⇒ (sin²θ + cos²θ)/sinθcosθ