CDS Elementary Mathematics 5 Feb 2017 Official Paper (Previous Year Paper)

Calculation:

$\sqrt {\frac{{0.064{\rm{;}} × {\rm{;}}6.25}}{{0.081{\rm{;}} × {\rm{;}}4.84}}} {\rm{;}}$

$\sqrt {\frac{{64{\rm{;}} × {\rm{;}}625}}{{81{\rm{;}} × {\rm{;}}484}}} {\rm{;}}$

⇒ (8 × 25)/(9 × 22) = 100/99

Putting x = -4 in every option:

1. x² – 7x + 44 = 16 + 28 + 44 = 88

2. x² + 7x – 44 = 16 – 28 – 44 = -56

3. x² – 7x – 44 = 16 + 28 – 44 = 0

4. x² + 7x + 44 = 16 – 28 + 44 = 32

Since x = -4 satisfies the 3^rd equation;

∴ (x + 4) is a factor of (x² – 7x – 44)

Given:

2x2 + 6x + k = 0 and k < 0,

Concept used:

If α and β are the roots of quadratic equation ax2 + bx + c = 0

Sum of root (α + β) = -b/a

Product of root (αβ) = c/a

Calculation:

2x² + 6x + k = 0

Sum of root

⇒ (​​α + β) = (-6/2) = -3     ----- (1)

⇒ Product of roots

⇒  αβ = (k/2) ​​      ----- (2)

Now the required value, 

⇒ (α/β) + (β/a) = (α² + β²)/αβ     

Using the identity (a² + b²) = (a + b)² – 2ab 

⇒  $\left( {\frac{{\rm{\alpha }}}{{\rm{\beta }}} + \frac{{\rm{\beta }}}{{\rm{\alpha }}}} \right) = {\rm{;}}\left[ {\frac{{{{\rm{\alpha }}^2}{\rm{;}} + {\rm{;}}{{\rm{\beta }}^2}}}{{{\rm{\alpha \beta }}}}} \right]$

⇒ ${\rm{;}}\left[ {\frac{{{{\left( {{\rm{\alpha ;}} + {\rm{;\beta }}} \right)}^2}{\rm{;}} - {\rm{;}}2{\rm{\alpha \beta }}}}{{{\rm{\alpha \beta }}}}} \right] = {\rm{;}}\left[ {\frac{{{{\left( {{\rm{\alpha ;}} + {\rm{;\beta }}} \right)}^2}}}{{{\rm{\alpha \beta }}}} - {\rm{;}}2} \right]$

Using equation (1) & (2)

$\begin{array}{l} ⇒ {\rm{;}}\left[ {\frac{{{{\left( { - 3} \right)}^2}}}{{\frac{{\rm{k}}}{2}}} - {\rm{;}}2} \right]\ = {\rm{;}}\left[ {\frac{{18}}{{\rm{k}}} - {\rm{;}}2} \right] \end{array}$

Since, k < 0

⇒ 18/k < 0

⇒ 18/k - 2  < - 2

∴ The maximum value of this expression will be -2.

Since HCF (b, c) = 1;

Suppose b = 2, c = 3 and d = 8 then a = b × c = 2 × 3 = 6;

Statement 1:

HCF(c, bd) = HCF (3, 16) = 1

And HCF(c, d) = HCF (3, 8) = 1

Hence HCF(c, bd) = HCF(c, d) is true.

Statement 1:

LCM(a, d) = LCM(6, 8) = 24

And LCM(c, bd) = LCM(3, 16) = 48

Hence LCM(a, d) = LCM(c, bd) is not true.

∴ Statement 1 only is correct.

Calculation:

$\frac{{{{\left( {0.35} \right)}^2} + {\rm{;}}0.70{\rm{;}} + {\rm{;}}1}}{{2.25}} + 0.19{\rm{;}}$

$\begin{array}{l} \Rightarrow \frac{{0.1225{\rm{;}} + {\rm{;}}0.70{\rm{;}} + {\rm{;}}1 + {\rm{;}}0.4275}}{{2.25}}\ \Rightarrow \frac{{2.25}}{{2.25}} = 1 \end{array}$

∴ The required value is 1

Shortcut Trick

We know that 2¹⁰  = 1024 which is greater than 1000 i.e. 10³

⇒ 2¹⁰  > 10³

⇒ (2¹⁰)⁴ > (10³)⁴

⇒ 2⁴⁰ > 10¹²

∴ Number of digits in 240 = 13

<br>

Alternate Method

Let N = 2⁴⁰

Taking ‘log’ both the sides;

⇒ log₁₀N = 40(log₁₀2)

⇒ log₁₀N = 40 × 0.301 = 12.04

Taking ‘anti-log’ both sides;

⇒ N = (10)^12.04

∴ Number of digits in 2⁴⁰ = 13

(a² – 5a + 3)x² + (3a – 1)x + 2 = 0;

Let the roots of this equation are ‘y’ and ‘2y’;

∴ Sum of roots = 3y = [-(3a – 1)/(a² – 5a + 3)]      ---- (1)

And Product of roots = 2y² = 2/(a² – 5a + 3)        ---- (2)

∴ (Product) / (Sum) = 2y²/3y = 2/-(3a – 1)

⇒ y = [-3/(3a – 1)]

From equation 1;

3y = [-(3a – 1)/(a² – 5a + 3)]

Putting the value of y;

⇒ 3 × [-3/(3a – 1)] = [-(3a – 1)/(a² – 5a + 3)]

⇒ 9a² – 45a + 27 = 9a² – 6a + 1

⇒ 45a – 6a = 26

⇒ 39a = 26

⇒ a = 2/3

Given:

(4444)4444 is divided by 9

Calculation:

$Rem(\frac{4444^{4444}}{9}) = Rem(\frac{7^{4444}}{9})$

When 7¹ is divided by 9 it leaves the remainder of 7.

When 7² is divided by 9 it leaves the remainder of 4.

When 7³ is divided by 9 it leaves the remainder of 1.

Hence, after 7, 4, and 1 the pattern repeats itself.

Now, when 4444 is divided by 3 it leaves the remainder of 1.

⇒ $Rem(\frac{7^{4444}}{9}) = Rem(\frac{(7)^{3(1481)+1}}{9})$

⇒ $Rem(\frac{(7\times7)^{3(1481)}}{9})$ = $\frac{7}{9}$

∴ remainder = 7

${\rm{x}} = \frac{{\sqrt {{\rm{a}} + {\rm{b;}}} -{\rm{;}}\sqrt {{\rm{a}} - {\rm{b}}} }}{{\sqrt {{\rm{a}} + {\rm{b}}} {\rm{;}} + {\rm{;}}\sqrt {{\rm{a}} - {\rm{b;}}} }}$

Applying componendo and dividendo;

$\Rightarrow \frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}} = \frac{{2\sqrt {{\rm{a}} + {\rm{b;}}} }}{{{\rm{;}}-2\sqrt {{\rm{a}} - {\rm{b;}}} }}$

Squaring both sides;

$\Rightarrow \frac{{{{\rm{x}}^2} + {\rm{;}}1{\rm{;}} + {\rm{;}}2{\rm{x}}}}{{{{\rm{x}}^2} + {\rm{;}}1 - {\rm{;}}2{\rm{x}}}} = \frac{{{\rm{a}} + {\rm{b}}}}{{{\rm{;a}} - {\rm{b}}}}$

Applying componendo and dividend again;

$\Rightarrow \frac{{2({{\rm{x}}^2} + {\rm{;}}1)}}{{4{\rm{x}}}} = \frac{{\rm{a}}}{{\rm{b}}}$

⇒ bx² + b = 2ax

⇒ bx² – 2ax + b = 0

$\begin{array}{l} \frac{{{{\left( {443{\rm{;}} + {\rm{;}}547} \right)}^2} + {\rm{;}}{{\left( {443{\rm{;}}--{\rm{;}}547} \right)}^2}{\rm{;}}}}{{443{\rm{;}} \times {\rm{;}}443{\rm{;}} + {\rm{;}}547{\rm{;}} \times {\rm{;}}547}}\ \Rightarrow \frac{{2{{\left( {443} \right)}^2} + {\rm{;}}2{{\left( {547} \right)}^2}{\rm{;}}}}{{{{\left( {443} \right)}^2}{\rm{;}} + {\rm{;}}{{\left( {547} \right)}^2}}} \end{array}$

⇒ 2

Formula used:

(a^m)ⁿ = a^mn

Shortcut Trick

Put t = 2 

⇒ x = 2 ¹/(2 – 1) = 2

⇒ y = 2 ²/(2 – 1) = 4

From option 3, xy = yx

⇒ 2⁴ = 4²

⇒ 16 = 16 which is correct. 

Except for this, all other options will not satisfy.

∴ Correct option is yx =  xy 

Alternate Method

We have,

x = t 1/(t – 1)       ----(1)

Also, y = t t/(t – 1)

Rightarrow

y = t ^{[1 + 1/(t – 1)]}

⇒ y = t × t ^{1/(t – 1)}

⇒ y = t × x        [From equation (1)]

⇒ t = y/x      ---- (2)

Again, y = t ^{t/(t – 1)}

⇒ y = [t ^{1/(t – 1)}]^t

⇒ y = x^t      ---- (3)       [From equation (1)]

From equations (2) and (3);

⇒ y = x^(y/x)

⇒ y^x =  xy 

∴ Correct option is yx =  xy

$\left( {\frac{{3{{\rm{A}}^2} + 4{\rm{B}}}}{{3{\rm{A}} - 4{{\rm{B}}^2}}}} \right)$

Since A : B = 3 : 4, suppose A = 3x and B = 4x;

Putting these values in the expression;

$\Rightarrow {\rm{;}}\left( {\frac{{27{{\rm{x}}^2} + {\rm{;}}16{\rm{x}}}}{{9{\rm{x}} - 64{{\rm{x}}^2}}}} \right)$

$\Rightarrow {\rm{;}}\left( {\frac{{27{\rm{x;}} + {\rm{;}}16}}{{9 - 64{\rm{x}}}}} \right)$

Now we can see that the value of the expression can’t be determined because the expression is not homogeneous.

Concept:

A – B = elements present in set A not in B.

A ∩ B = Elements common in set A & set B.

A ∪ B = Elements present either in set A or set B.

∅ denotes null set i.e. set with no elements

Calculations:

A = {x : x is a multiple of 7}

⇒ A =  {7, 14, 21, ………..}

B = {x : x is a multiple of 5}

⇒ {5, 10, 15, 20, ………..}

C = {x : x is a multiple of 35}

⇒ {35, 70, 105, ……..}

Checking all the options:

Option:1 (A – B) ∪ C

= {7, 14, 21, 28, 42   …..} ∪ {35, 70, 105,……..}

(A – B) ∪ C =  {7, 14, 21, 28, 35, …}

Hence it is not a null set.

Option:2  (A – B) – C

⇒ {7, 14, 21, 28, …..} – {35, 70, 105,……..}

⇒ A – B) – C = {7, 14, 21, 28, 42, 49….}

Hence it is not a null set.

Option:3  (A ∩ B) ∩ C

⇒ {35, 70, 105, 140……} ∩ {35, 70, 105, 140……..}

⇒ {35, 70, 105, 140……..}

Hence it is not a null set.

Option:4 (A ∩ B) – C

⇒ {35, 70, 105, 140……} – {35, 70, 105, 140……..}

⇒ (A ∩ B) – C = ∅

∴ (A ∩ B) – C  is the null set.

Formula used:

( a + b ) = a³ + b³ + 3ab( a + b )

Calculation:

x = 2 + 2^2/3 + 2^1/3

⇒ (x – 2) = (2^2/3 + 2^1/3)         -----(1)

Taking cube on both sides

⇒ x³ – 8 – 6x² + 12x = (22/3)³ + (21/3)³ + 3 × 2^2/3 × 2^1/3(2^2/3 + 2^1/3)

⇒ x3 – 8 – 6x2 + 12x = 22 + 2 + 3 × 2⁽2/3 + 1/3)(22/3 + 21/3)

⇒ x3 – 8 – 6x2 + 12x = 6 + 6(22/3 + 21/3)

⇒ x3 – 14 – 6x2 + 12x = 6 (x – 2)

From equation (1),

⇒ x³ – 14 – 6x² + 12x = 6 (22/3 + 21/3)

⇒ x³ – 14 – 6x² + 12x – 6x + 12 = 0

⇒ x³ – 6x² + 6x = 2

∴ The value of x3 – 6x2 + 6x is 2.

Shortcut TrickWe can put x = 25 & y = 1 

⇒ $\sqrt {\frac{{\rm{x}}}{{\rm{y}}}} = \frac{{24}}{5} + \sqrt {\frac{{\rm{y}}}{{\rm{x}}}}$

⇒ $\sqrt {\frac{{\rm{25}}}{{\rm{1}}}} = \frac{{24}}{5} + \sqrt {\frac{{\rm{1}}}{{\rm{25}}}}$

⇒  5    =   $\frac{24}{5}$ + $\frac{1}{5}$

⇒ 5 = 5 

So, x = 25,, y = 1 satisfies

⇒ xy = 25 × 1 = 25

∴ The correct option is 25

Alternate Method

$\sqrt {\frac{{\rm{x}}}{{\rm{y}}}} = \frac{{24}}{5} + \sqrt {\frac{{\rm{y}}}{{\rm{x}}}}$

$⇒ \sqrt {\frac{{\rm{x}}}{{\rm{y}}}} - \sqrt {\frac{{\rm{y}}}{{\rm{x}}}} = \frac{{24}}{5}$

$⇒ {\rm{;}}\frac{{{\rm{x}} - {\rm{y}}}}{{\sqrt {{\rm{xy}}} }} = \frac{{24}}{5}$

Taking square both sides;

$⇒ {\rm{;}}\frac{{{{\rm{x}}^2} + {{\rm{y}}^2} - 2{\rm{xy}}}}{{{\rm{xy}}}} = \frac{{576}}{{25}}$

$⇒ {\rm{;}}\frac{{{{\left( {{\rm{x;}} + {\rm{;y}}} \right)}^2} - {\rm{;}}4{\rm{xy}}}}{{{\rm{xy}}}} = \frac{{576}}{{25}}$

Putting the value of (x + y) = 26

$⇒ {\rm{;}}\frac{{676 - 4{\rm{xy}}}}{{{\rm{xy}}}} = {\rm{;}}\frac{{576}}{{25}}$

⇒ 16900 – 100xy = 576xy

⇒ 676xy = 1690

⇒ xy = 25

∴ Correct option is 25

Formula used:

1. m log n = log n^m
2. log (m/n) = log m - log n
3. 3. log1010  = 1
4. log mn = log m + log n

 

Calculation:

x log₁₀(10/3) + log₁₀3 = log₁₀(2 + 3^x) + x

⇒ log₁₀(10/3)^x + log₁₀3 = log10(2 + 3x) + log₁₀10^x

⇒ log₁₀ [(10/3)^x × 3] = log₁₀ [(2 + 3^x)10^x]

⇒ (3^-x) × 3 = (2 + 3^x)

⇒ 3/3^x = (2 + 3^x)

Suppose 3^x = p;

⇒ 3/p = (2 + p)

⇒ p² + 2p – 3 = 0

⇒ (p + 3) (p – 1) = 0

⇒ p = -3, 1

⇒ 3^x = 1

⇒ 3^x = 3⁰

∴  Correct answer is x = 0

Since α and β are the roots of the equation x² + px + q = 0;

∴ α + β = -p and αβ = q

Now

α² + β² = (α + β)² – 2αβ

Putting the values;

⇒ α² + β² = p² – 2q

Option 1 is correct.

a³ = 335 + b³ and a = 5 + b;

∴ a³ – b³ = 335 and a - b = 5

We know that;

(a³ – b³) = (a – b) (a² + b² + ab)

⇒ (a³ – b³) = (a – b) ((a – b)² + 3ab)

Putting the values;

⇒ 335 = 5 × (25 + 3ab)

⇒ 67 = (25 + 3ab)

⇒ 3ab = 42

⇒ ab = 14

And we know that (a – b) = 5

∴ a = 7 and b = 2

∴ a + b = 9

9^x 3^y = 2187

⇒ 3^{(2x + y)} = 3⁷

∴ (2x + y) = 7     ---- (1)

2^3x 2^2y – 4^xy = 0

⇒ 2^3x 2^2y = 4^xy

⇒ 2^{(3x + 2y)} = 2^2xy

∴ (3x + 2y) = 2xy      ---- (2)

Now eliminating the options:

Since (2x + y) = 7 hence (x + y) can’t be 1, 3 or 7.

Suppose (x + y) = 5 where x = 2 and y = 3;

Putting in both the equations;

(2x + y) = 4 + 3 = 7

And (3x + 2y) = 2xy

⇒ 6 + 6 = 2 × 2 × 3

⇒ 12 = 12

Hence the values x = 2 and y = 3 satisfy the equations.

∴ (x + y) = 5

Since the pair of linear equations kx + 3y + 1 = 0 and 2x + y + 3 = 0 intersect each other;

∴ (a₁/a₂) ≠ (b₁/b₂)

⇒ k/2 ≠ 3/1

⇒ k ≠ 6

Prime numbers less than 100 = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97}

∴ Number of prime numbers which are less than 100 = 25

Let the original weight of the diamond was ‘10x’ gm;

Since the diamond broke into four pieces with their weights in the ratio of 1 ∶ 2 ∶ 3 ∶ 4;

∴ Weights of four pieces = x, 2x, 3x & 4x;

Since the cost of diamond varies directly as the square of its weight;

∴ Cost of diamond pieces = x², 4x², 9x² and 16x²

And Cost of original diamond = (10x)² = 100x²

Since loss in total value of the diamond was Rs. 70000 on cutting;

∴ 100x² – x² – 4x² – 9x² – 16x² = 70000

⇒ 70x² = 70000

⇒ x² = 1000

∴ Cost of original diamond = 100x² = Rs. 100000

Given:

Speed of A= 5/3 m/s

A gives a start of 4 m to B

Time is taken by A = Time taken by B - 12 sec

Concept used:

Speed = Distance/Time 

Calculation:

Time taken by A to cover 100 m = [100/(5/3)] = 60 sec

Since, A beats B by 12 seconds,  therefore

B must have run for (60 + 12) = 72 sec.

A gives a start of 4 m to B, so

B runs only (100 – 4) = 96 m

∴ Speed of B = (96/72) = (4/3) m/s

Given:

3 women would as much work as 2 men;

Formula used:

M1 × D2 × H2 = M2 × D2 × H₂

Explanation:

Let number of days required be d days

Efficiency of each men and each women be M and W;

Ratio of efficiencies of Man and Woman = 3 ∶ 2

So, 2M = 3W

(15M × 21 × 8) = (21W × 6 × d)

Putting 2M = 3W;

⇒ (15M × 21 × 8) = (14M × 6 × d)

⇒ d = 30

∴ Number of days required = 30

Let x is the required number;

∴ (27 – x) / (35 – x) = 2/3

⇒ 81 – 3x = 70 – 2x

⇒ x = 11

Since mean of 5 numbers is 15;

∴ Sum of 5 number = 5 × 15 = 75

Mean of 6 numbers becomes 17

∴ Sum of 6 number = 6 × 17 = 102

∴ The included 6^th number = 102 - 75 = 27

Shortcut Trick Mean of marks of three groups will be the same as the mean of marks of the entire class.

⇒ $\frac{80+50 +x}{3}$ = 60

⇒  $\frac{130+x}{3}$ = 60

⇒ 130 + x = 180

⇒ x = 180 - 130 = 50

∴ The correct value is 50

 Alternate MethodFormula Used:

Mean of observations = Sum of observations / Number of observations 

Calculation:

Since the mean marks obtained by 300 students is  60;

⇒ Total marks of 300 students = 300 × 60 = 18000

Mean of top 100 students is  80;

⇒ Total marks of top 100 students = 100 × 80 = 8000

Mean of last 100 students is 50;

⇒ Total marks of last 100 students = 100 × 50 = 5000

⇒ Sum of marks of remaining 100 students

⇒ 18000 - 5000 - 8000 = 5000

⇒ Mean of remaining 100 students = 5000/100 = 50

∴ The correct value is 50

 

Mean = (∑X_iF_i) / (∑F_i)

⇒ Mean = (170 + 840 + 1600 + 70f + 1710) / (17 + 28 + 32 + f + 19)

⇒ Mean = (4320 + 70f) / (96 + f)

Since mean is given 50;

∴ 50 = (4320 + 70f) / (96 + f)

⇒ 5f + 480 = 432 + 7f

⇒ 2f = 48

⇒ f = 24

Calculation:

Four slices with angles 150°, 90°, 60° and 60°

When 60° is deleted, the remaining are 150°, 90°, 60° 

Total angle = 150° + 90° + 60° = 300° 

Hence, 300° will be considered 100% in new pie-chart

Ratio of angles without 60° = 150 ∶ 90 ∶ 60 = 5 ∶ 3 ∶ 2

⇒ $\frac{90^\circ}{300^\circ} \times 300^\circ$ = 30%

$\frac{60^\circ}{300^\circ} \times 300^\circ$ = 20%

$\frac{150^\circ}{300^\circ} \times 300^\circ$ = 50%  Largest

Hence, 50% of 360° = 180° 

∴ The largest slice has an angle 180°

Mean = 270 and Median = 220;

We know the formula∶

3 × Median = Mode + 2 × Mean

∴ Mode = 3 × Median - 2 × Mean

⇒ Mode = 3 × 220 - 2 × 270

⇒ Mode = 660 - 540 = 120

Since a, b, c, d, e, f, g be consecutive even numbers;

Suppose a = 2, b = 4, c = 6, d = 8, e = 10, f = 12 and g = 14.

Since j, k, l, m, n be consecutive odd numbers;

Suppose j = 1, k = 3, l = 5, m = 7 and n = 9.

∴ Average of all the numbers = (a + b + c + d + e + f + g + j + k + l + m + n)/12

Putting the values of every term as supposed above;

∴ Average of all the numbers = (56 + 25)/12 = 81/12 = 27/4

On checking with the options, any option will not be satisfied with this data.

∴ None of the above

From the given table∶

Number of pencils of quality A = 50/1 = 50

Number of pencils of quality B = x/1.5

Number of pencils of quality C = 20/2 = 10

Since average price per pencil is Rs. 1.25;

∴ 1.25 = (50 + x + 20)/(50 + x/1.5 + 10)

⇒ 75 + 5x/6 = 70 + x

⇒ x/6 = 5

⇒ x = 30

f₁ = 29 - 18 = 11

and f₂ = 45 - 29 = 16

We know that;

π radians = 180 degree

Since radian = R and degree = D

⇒ π (in radian) = 180 (in degree)

⇒ 1^o = π/180 

According to the question, 

D is the number of degrees multiply by both side

⇒ D = π D/180

R is the number of radian

R = π D/180

⇒ π D = 180R

Given:

9 tan²θ + 4 cot²θ

Concept Used:

Minimum value of a tan2θ + b cot2θ = 2√(ab)

Calculation:

Minimum value of 9 tan2θ + 4 cot2θ = 2√(ab)

⇒ 2√(9 × 4)

⇒ 2√36 

⇒ 12

∴ T​he minimum value of 9 tan2θ + 4 cot2θ is 12.

Formula Used:

tan2θ = (2 tanθ) / (1 - tan2θ)

sin²θ + cos2θ = 1

cos2θ = cos²θ - sin²θ 

Calculation:

We know that tan2θ = (2 tanθ) / (1 - tan²θ)

⇒ (x sinθ) = (y cosθ) = (z tan2θ)

Suppose (x sinθ) = (y cosθ) = (z tan2θ) = k

⇒ x = k/sinθ, y = k/cosθ, z = k/tan2θ

Putting the values in 4z² (x² + y²);

⇒ 4z²(x² + y²) = $\frac{4k^2}{tan^2\ 2\theta}(\frac{k^2}{sin^2\theta} + \frac{k^2}{cos^2\theta})$

⇒ 4z2(x2 + y2) = $\frac{4k^4\ cos^22\theta}{sin^22\theta}(\frac{1}{sin^2\theta\ cos^2\theta})$

$\begin{array}{l} ⇒ {\left( {\frac{{2{{\rm{k}}^2}({\rm{co}}{{\rm{s}}^2}{\rm{θ ;}} - {\rm{;si}}{{\rm{n}}^2}{\rm{θ }}}}{{\left( {2{\rm{sinθ cosθ }}} \right)\left( {{\rm{sinθ cosθ }}} \right)}}} \right)^2}\ ⇒ {\left( {\frac{{{{\rm{k}}^2}({\rm{co}}{{\rm{s}}^2}{\rm{θ ;}} - {\rm{;si}}{{\rm{n}}^2}{\rm{θ }}}}{{({\rm{si}}{{\rm{n}}^2}{\rm{θ }})({\rm{co}}{{\rm{s}}^2}{\rm{θ }})}}} \right)^2} \end{array}$

$\begin{array}{l} ⇒ {\left( {\frac{{{{\rm{k}}^2}}}{{({\rm{si}}{{\rm{n}}^2}{\rm{θ }})}} - \frac{{{{\rm{k}}^2}}}{{({\rm{co}}{{\rm{s}}^2}{\rm{θ }})}}} \right)^2}\ ⇒ {\left( {{{\rm{x}}^2} - {{\rm{y}}^2}} \right)^2} \end{array}$

∴ Correct option is (x2 - y2)2

Since cosθ₁ + cosθ₂ + cosθ₃ = 3

It is possible only when all three terms are equal to 1. (As maximum value of cosθ is 1)

∴ cosθ₁ = cosθ₂ = cosθ₃ = 1

∴ θ₁ = θ₂ = θ₃ = 0°

Putting the values in sinθ₁ + sinθ₂ + sinθ₃;

sinθ₁ + sinθ₂ + sinθ₃ = 0 + 0 + 0 = 0

Using hit and trial method∶

1. θ = 0° ∶ cosθ + tanθ = 1 + 0 = 1

2. θ = 30° ∶ cosθ + tanθ = √3/2 + 1/√3

3. θ = 45° ∶ cosθ + tanθ = 1/√2 + 1

4. θ = 60° ∶ cosθ + tanθ = 1/2 + √3

Here option 1 satisfies the given equation cosθ + tanθ = 1;

∴ θ = 0° is the correct answer.

Calculation:

${\rm{sinx;}}\sqrt {\frac{1}{{1 + {\rm{cosx}}}} + \frac{1}{{1 - {\rm{cosx}}}}}$

Put x = 90°;

$\begin{array}{l} \Rightarrow {\rm{sin}}90^\circ \sqrt {\frac{1}{{1 + {\rm{cos}}90}} + \frac{1}{{1 - {\rm{cos}}90}}} \ \Rightarrow 1\sqrt {1 + 1} \end{array}$

⇒ √2

$\begin{array}{l} \frac{{{\rm{co}}{{\rm{s}}^4}{\rm{A}} - {\rm{;si}}{{\rm{n}}^4}{\rm{A}}}}{{{\rm{co}}{{\rm{s}}^2}{\rm{A}} - {\rm{;si}}{{\rm{n}}^2}{\rm{A}}}}\ \Rightarrow \frac{{({\rm{co}}{{\rm{s}}^2}{\rm{A}} - {\rm{;si}}{{\rm{n}}^2}{\rm{A}}){\rm{;}}({\rm{co}}{{\rm{s}}^2}{\rm{A;}} + {\rm{;si}}{{\rm{n}}^2}{\rm{A}})}}{{\left( {{\rm{co}}{{\rm{s}}^2}{\rm{A}} - {\rm{;si}}{{\rm{n}}^2}{\rm{A}}} \right)}} \end{array}$

$\Rightarrow ({\rm{co}}{{\rm{s}}^2}{\rm{A;}} + {\rm{;si}}{{\rm{n}}^2}{\rm{A}}) = 1$

Given 7sin²x + 3cos²x = 4

⇒ 3sin²x + 3cos²x + 4sin²x = 4

⇒ 3(sin²x + cos²x) + 4sin²x = 4

⇒ 3 + 4sin²x = 4

⇒ 4sin²x = 1

⇒ sin²x = ¼

⇒ sinx = ½

∴ x = 30°

∴ tanx = 1/√3

In the given figure;

BC = 300 m and suppose DC = H m

In triangle ABC;

tan60 = BC/AC

⇒ √3 = 300/AC

⇒ AC = 300/√3 m

In triangle ADC;

tan45 = DC/AC

⇒ 1 = DC/AC

⇒ DC = AC = 300/√3

∴ H = 300/√3 = 100√3 m

∴ Height of the lower plane from the ground = 100√3 m

Given x = a cosθ + b sinθ and y = a sinθ - b cosθ;

(x² + y²) = (a cosθ + b sinθ)² + (a sinθ - b cosθ)²

⇒ a²cos²θ + b²sin²θ + 2sinθ cosθ + a²cos²θ + b²sin²θ - 2sinθ cosθ

⇒ a²(cos²θ + sin²θ) + b²(cos²θ + sin²θ)

⇒ a² + b²

Let the height of the tree = h m;

In the figure;

From triangle ADB;

tan45 = AB/DB

⇒ 1 = 90/DB

⇒ DB = 90 m

∴ CE = DB = 90 m

From triangle CAE;

tan30 = AE/CE

⇒ 1/√3 = AE/90

⇒ AE = 90/√3 = 30√3 m

∴ h = 90 - AE = (90 - 30√3) m

∴ Height of the tree = (90 - 30√3) m

Following condition is must in any triangle∶

(Sum of 2 sides) > Third Side

In option 4∶ (2, 3, 6)

(2 + 3) < 6

That means it does not fulfil the necessary condition.

∴ Triplets (2, 3, 6) does not represent the sides of a triangle.

Given:

Perimeter of rectangle = 10 cm

Area of rectangle = 4 cm²

Formula used:

Perimeter of rectangle = 2(l + b)

Area of rectangle = l × b

Where, l = length and b = breadth

Shortcut Trick

2(Length + breadth) = 10 

⇒ (Length + breadth) = 5    

And, Length × breadth = 4

Two positive numbers  whose sum & product are 5 & 4 will be 1, 4

or 4, 1 only.

Hence, from the option, 

∴ The required length is 4 cm.

<br>

Alternate Method

Let the length and breadth of the rectangle be l and b respectively

According to the question

Perimeter of rectangle = 2(l + b)

⇒ 2(l + b) = 10

⇒ (l + b) = 5                  ------(1)

And, lb = 4                      

⇒ l(5 − l) = 4          [From equation (1)]

⇒ 5l − l² = 4

⇒ l² − 5(l) + 4 = 0

⇒ l² − 4(l) − 1(l) + 4 = 0

⇒ l(l − 4) − 1(l − 4) = 0

⇒ (l − 4) (l − 1) = 0

⇒ l = 4 or 1

∴ The required length is 4 cm.​

Given:

The ratio of angle of a triangle = 2 : 4 : 3

Calculation:

Let the angles of a triangle be 2x, 4x & 3x respectively

According to the question

⇒ 2x + 4x + 3x = 180

⇒ 9x = 180

⇒ x = 20

∴ Smallest angle = 2 × 20 = 40°

Given:

The radius of cylindrical pipe = 1 cm

The length of the pipe = 14 m

Formula used:

The surface area of the cylinder = 2πr

Calculation:

According to the question

The radius of cylindrical pipe = 1 cm

The length or height of the cylindrical pipe

= 14 m = 1400 cm        (∵ 1 m = 100 cm)

By using the above formula, 

Surface area of cylinder = [2 × (22/7) × 1 × 1400] cm

⇒ (2 × 22 × 200) square cm

⇒ 8800 square cm

∴ The surface area of the pipe is 8800 square cm.

Given:

The ratio of area of two circular fields = 16 : 49

The radius of the bigger field = 14 m

Formula used:

Area of circle = πr²

Calculation:

Let the radius of the smaller and bigger fields are r₁ and r₂

Radius of the bigger field is given 14 m.

⇒ r₂ = 14 m

And, Ratio of areas = 16 ∶ 49

⇒ (πr₁²) ∶ (πr₂²) = 16 ∶ 49

⇒ (r₁²) ∶ (14²) = 16 ∶ 49

⇒ r₁² = (196 × 16)/49

⇒ r₁² = 64

⇒ r₁ = 8 m

∴ The radius of the smaller field is 8 m.

Concept used:

Midpoint theorem: The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.

Calculation:

According to the figure

We can see that P, Q, R, S are the mid-point of the sides AB, BC, CD, DA respectively.

⇒ PQ and SR will be parallel to AC and PS and QR are parallel to BD.

⇒ PQ = SR = ½(AC) and PS = QR = ½(BD)   [By Midpoint theorem]

Since in a rectangle, both the diagonals are always equal i.e. AC = BD.

⇒ PQ = SR = PS = QR

But the angles of the quadrilateral PQRS may not be 90°(As it is inside a rectangle whose adjacent side are not equal).

∴ The quadrilateral PQRS is a Rhombus, but not be a square.

Let the value of each instalment is Rs. z and the principals for 1^st and 2^nd year are P1 & P2 respectively;

For First year∶

z = P₁ (1 + 10/100)

⇒ P₁ = 10z/11

For Second year∶

z = P₂ (1 + 10/100)²

⇒ P₂ = 100z/121

Given that principal is Rs. 8400;

∴ P₁ + P₂ = 8400;

⇒ 10z/11 + 100z/121 = 8400

⇒ 210z/121 = 8400

⇒ z = 4840

∴ Value of each instalment = Rs. 4840

Let the present age of Leela = x years;

∴ Age of Leela at the time of marriage = (x – 6) years

∴ x = 5/4 (x – 6)

⇒ x/4 = 15/2

⇒ x = 30

∴ Present age of Leela = 30 years

∴ Present age of Leela’s son = 30/10 = 3 years

Let total units of work = 60 units (LCM of 12 and 30)

∴ Efficiency of (A + B) = 60/12 = 5

And Efficiency of B = 60/30 = 2

∴ Efficiency of A = 5 – 2 = 3

∴ Time taken by A alone to finish the work = 60/3 = 20 days

Solution∶

5^{(1 + x)} + 5^{(1 – x)} = 26

Using hit and trial method∶

1. -1, 1

Putting x = -1∶ 5^{(1 - 1)} + 5^{(1 + 1)} = 1 + 25 = 26

Putting x = 1∶ 5^{(1 + 1)} + 5^{(1 - 1)} = 25 + 1 = 26

∴ Both the values of x satisfies the given equation, hence option 1 is correct.

Shortcut Trick 

5 men ---10 days --- 1 work

⇒ 5 men --- 1 day --- 1/10 work

12 women ---15 days ---1 work

⇒ 12 women ---1 day ---1/15 work

⇒ 6 women ---1 day ---1/30 work

So,

⇒ 5men + 6 women in 1 day

⇒ (1/10) + (1/30) = (2/15) part of work

⇒ Total work gets completed in = $\frac{15}{2}$ = 7.5 days

∴ To complete the work, no. of days required is 7.5 days.

Alternate Method 

Formula used:

M₁D₁H₁ / W₁ = M₂D₂H₂ / W₂

M = No. of worker ,D = days ,H = hours ,W = Total work

Calculation:

Let efficiencies of a man and a woman are

M and W respectively -

⇒ 5M × 10 = 12W × 15

⇒ 5M = 18W

Let the number of days required to complete

the work by 5 men and 6 women be 'd'

⇒ (5M + 6W) × d = 12W × 15 

⇒ (18W + 6W) × d = 12W × 15  (5M = 18W)

⇒ 24 d = 180

⇒ d = 7.5

∴ To complete the work, no. of days required is 7.5 days.

Given: 

We know that distance between the two cities is 1080 km & trains meet mid-way, therefore, they meet at 540 km.

Formula used: 

Distance = speed × time

Shortcut Trick

Let the time is 6 a.m

Train from Delhi & Mumbai takes 12 h & 9 hr respectively to cover

Speed of the train from Delhi = 540/12 = 45 Km/hr

Speed of train from Mumbai= 540/9 = 60 Km/hr

Difference in speed = 60 - 45 = 15 Km/hr

∴ The correct option is 6 a.m.

Alternate Method

let the speed of the train travelling from Delhi to Mumbai (passenger train ) be s kmph & time taken is t hours.

Since Distance = speed × time 

⇒ 540 = s × t 

⇒ s = 540/t  t           ------ (1) 

Since express train travels 15 kmph faster

Speed of express train = s + 15

As the express train departs 3 hours late

Time is taken by express train = t - 3

⇒ 540 = (s + 15)(t - 3)

⇒ 540 = [(540/t)  + 15](t - 3)         [From equation (1)]

⇒ 540 = ( $\frac{540+15t}{t}$ )(t - 3)

⇒ 540 = 15( $\frac{36+t}{t}$)(t - 3)

⇒ 36 t =  t2 +33t -108

⇒  t2 - 3t  - 108 = 0

⇒  t² -12t + 9t -108 = 0

⇒ t( t - 12 ) + 9(t - 12) = 0

⇒ (t - 12) = 0 , (t + 9) = 0

⇒ t = 12 & t = - 9 (rejected)

⇒ Trains meet at 6pm + 12 hours = 6 am

∴ The correct option is 6 am.

Given: 

Total students = 49  & Ratio of girls : boys = 4 ∶ 3

Concept:

If quantity Q is  divided into m : n then 

First part = Q × $\frac{m}{m+n}$,

Second part = Q × $\frac{n}{m+n}$

Calculation:

According to the above concept

No. of girls = 49 × $\frac{4}{4+3}$ = 28

No. of boys = 49 × $\frac{3}{4+3}$ = 21

When 4 girls leave the class

Number of girls = 28 - 4 = 24

Ratio of girls and boys = 24 ∶ 21 = 8 ∶ 7

∴ The correct answer is 8:7

We know the formula∶

(a³ + b³) = (a + b) (a² + b² – ab)

That can be written as∶

(a³ + b³) = (a + b) ((a + b)² – 3ab)

Putting the given values;

(a³ + b³) = 5 × (25 – 18) = 35

Given :

Cost price of Mobile = Rs. 4875

Discount = 25%

Formula used :

Marked price = $\frac{Selling \ price}{(100 - Discount)} × 100$         ---- (1)

Calculations :

Using equation (1), we get

⇒ $\frac{4875}{(100 - 25)} × 100$

⇒ $\frac{4875}{75} × 100$

⇒ 65 × 100

⇒ 6500

∴ The original selling price of the mobile is Rs 6500.

Formula used: 

Distance = speed × time

To convert km/h to m/s multiply by $\frac{5}{18}$

Concept used:

If two objects are travelling with speed u & v

When they move in the same direction then

Relative speed = u - v 

When they move in the opposite direction 

Relative speed = u + v 

Calculation:

The relative speed of a train and the man:

⇒ 30 – 3 = 27 km/hr

⇒  27 × (5/18) = 15/2 m/s

Time is taken by train to cross the man

⇒ $\frac{225}{15/2 }$ = 30 seconds

∴ The correct option is 30 seconds.

Shortcut Trick

<br>

Alternate Method

Converting the fractions into decimals;

7/9 = 0.77

11/14 = 0.78

3/4 = 0.75

10/13 = 0.76

⇒ 0.78 > 0.77 > 0.76 > 0.75

⇒ 11/14 > 7/9 > 10/13 > 3/4

∴ The correct answer is (11/14).

Given: 

Rate = 4%, Time = 2 years

CI - SI = 10  

Formula used

(SI - CI) for 2 years = P(R/100)2

P = Principal, R = Rate

Calculation

Using the above formula,

10 = P(4/100)²

⇒ 10 = P/25² = P/625

⇒ P = 6250

∴ Sum of money = Rs. 6250

Formula used:

p % = p/100

(a - b)² = a² - 2ab + b²

Calculation:

a% of a + b% of b = 2% of ab

Using the above formula

⇒ a²/100 + b²/100 = 2ab/100

⇒ a² + b² = 2 ab

⇒ a² + b² - 2ab = 0

⇒ (a - b)² = 0

⇒ a = b

∴ b is 100 % of a.

Alternate Methoda% of a + b% of b = 2% of ab

It can be written as;

0.01a² + 0.01b² = 0.02 × ab

⇒ a² + b² = 2ab

⇒ a² + b² - 2ab = 0

⇒ (a – b)² = 0

⇒ a = b

∴ b is 100% of a.

Formula used:

If a is y % of b then

a = (y/100) × b

Calculation:

Let the total population be 900 

⇒ Total male population = (5/9) of 900 = 500

Also female population = 900 - 500 = 400

Married male = 30% of 500 = 150

Number of married females = Number of married males = 150

The percentage of unmarried females in the total population is

⇒ $\frac{400-150}{900}$ × 100

⇒ $\frac{250}{900}$× 100 = $27\frac{7}{9}$%

∴ The correct option is $27\frac{7}{9}$%

Calculation:

Let Sunil get x number of Rs. 7 sweet and y number of Rs. 10 sweet;

From the given condition;

7x + 10y = 200

The integer values of x and y that can satisfy this equation are;

Either x = 10 and y = 13 or x = 20 and y = 6

Since we want a maximum value of (x + y),

⇒ x = 20 and y = 6 will sufficient.

⇒ Maximum number of sweets = 20 + 6 = 26

∴ The correct option is 26.

Given:

x3 + 8, x2 + 5x + 6 and x3 + 4x2 + 4x

Concept used:

a³ + b³ = (a + b)(a² + b² - ab)

(a + b)2 = a² + 2ab + b²

The LCM is the product of all primes that are common to all numbers.

Calculation:

x³ + 8 = (x + 2)(x² + 4 – 2x)

x² + 5x + 6 = (x + 3)(x + 2)

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

∴ LCM is x (x + 2)² (x + 3) (x² – 2x + 4)

We know that;

Product of two numbers = (HCF of 2 numbers) × (LCM of 2 numbers)

⇒ pq = 1 × (LCM of p and q)

⇒ (LCM of p and q) = 1/pq = (pq)^-1 

Calculation:

$\sqrt[3]{{4\frac{{12}}{{125}}}}$

$\Rightarrow \sqrt[3]{{\frac{{512}}{{125}}}}$

$\Rightarrow \frac{8}{5} = {\rm{;}}1\frac{3}{5}$

∴ The required value is $1\frac{3}{5}$.

Formula Used:

Distance = speed × time

1km/hr = $\frac{5}{18}$ m/s , 

Calculation:

Distance between policeman and thief = 100 m;

Since the speed of the thief is 8 km/hr and,

That of the policeman is 10 km/hr

⇒ Relative speed = 10 – 8 = 2 km/hr = 2 × (5/18) = 5/9 m/s;

Time is taken by police to overtake the thief

⇒ 100/(5/9) = 180 sec

Hence, the thief will be caught after 180 sec

Therefore, the distance travelled by the thief before overtaken

= Speed of thief × time taken by a thief

=  8 × (5/18) × 180  = 400 m

∴ The correct option is 400m

Aman gets the roots (4, 3);

∴ The equation will be x² – 7x + 12 = 0

And Alok gets the roots (3, 2);

∴ The equation will be x² – 5x + 6 = 0

Now as given, Aman made a mistake in writing down the constant term and Alok made a mistake in writing down the coefficient of x;

∴ After looking at the 2 equations we can say that;

Correct constant term = 6 and correct coefficient of x = -7

∴ The correct equation will be x² – 7x + 6 = 0

∴ The correct roots of the equation are 6, 1.