Correct Answer: (d) ¹⁷⁄₁₆₀

We have,

 ⁵⁄₁₁₃ = 0.044….

 ⁷⁄₁₂₀ = 0.058….

¹³⁄₁₄₅ = 0.089….

¹⁷⁄₁₆₀ = 0.106….

Therefore, the required largest fraction = ¹⁷⁄₁₆₀   (Ans.)

Correct Answer: (b) ²⁷⁄₂

Let, total job be 1 part.

As given, Nirmit can do ²⁄₃ of a job in 18 days.

Therefore, Nirmit can complete the job in ( 18 × ³⁄₂ ) = 27 days.

Again, Kashish is 2 times efficient than Nirmit.

Hence, Kashish will take = ²⁷⁄₂ days to complete the whole job. (Ans.)

Note : Efficient means less time to do a job.

Correct Answer: (b) 26.66

As given, one man can displaced volume of water = 5 m³

Therefore, 40 men can displaced volume of water = 5 × 40 = 200 m³ 

Now, rise in level of the pool,

= ^{volume of water}⁄_{(length × breadth)}

= ^{200 m3}⁄_{(30 m × 25 m)}

= 0.2666 m (approx)

= 26.66 cm (approx)   (Ans.)

Correct Answer: (c) 1715.39

Net discount = [ 50 + 25 − ^{(50 × 25)}⁄₁₀₀ ] = 62.5 %

Note :

(1) If x and y are two successive discount then total discount percentage = [ x + y − ^{(x × y)}⁄₁₀₀ ]

(2) If x and y are two successive gain then total gain percentage = [ x + y + ^{(x × y)}⁄₁₀₀ ]

He purchases the article at,

= Rs. 2375 × ^{(100 − 62.5)}⁄₁₀₀

= Rs. 890.625

Repairing Cost = Rs. 165

Therefore, actual cost price (C.P.) of the article = Rs. (890.625 + 165) = Rs. 1055.625

His profit percentage = 62.5 %, ⇒ C.P. = 100 and S.P. = 100 + 62.5 = 162.5

Hence, required selling price (S.P.) of the article,

= Rs. [ (actual S.P.) × ( ^S.P.⁄_C.P. ) ]

= Rs. 1055.625 × ^{(100 + 62.5)}⁄₁₀₀ 

= Rs. 1715.39     (Ans.)

Correct Answer: (d) 339625

Let, length, breadth, and height of the cuboid is be 19x cm, 11x cm, and 13x cm respectively.

Now, according to the statement of the question, we have

19x = 13x +  30

 ⇒ x = 5

∴ Length = 19 × 5 = 95 cm

Breadth = 11 × 5 = 55 cm and

Height = 13 × 5 = 65 cm

We know, volume of a cuboid = length × breadth × height

= 95 cm × 55 cm × 65 cm

= 339625 cm³   (Ans.)

Correct Answer: (c) 49

Shortcut Approach :

∵ Average age is increased by 2 years among 13 members after joining the coach.

Therefore, Age of the coach,

= (13 × 2 + 23 ) years

= 49 years.   (Ans.)

Correct Answer: (b) 20000

Net gain = [ 30 + 20 + ^{(30 × 20)}⁄₁₀₀ ] = 56 %

Note :

(1) If x and y are two successive gain then total gain percentage = [ x + y + ^{(x × y)}⁄₁₀₀ ]

(2) If x and y are two successive discount then total discount percentage = [ x + y − ^{(x × y)}⁄₁₀₀ ]

In 56 % gain, cost price (C.P.) = 100 and selling price (S.P.) = 100 + 56 = 156

Hence, required C.P. of the item,

= Rs. [ (actual C.P.) × ( ^C.P.⁄_S.P. ) ]

= Rs. 31200 × ¹⁰⁰⁄₁₅₆

= Rs. 20,000     (Ans.)

Correct Answer: (c) 16000

Given,

The number of trees at present (P) = 17640

Rate of increase (annually) (r) = 5 %

Time (t) = 2 years

Therefore,

The number of trees before 2 years,

= P [ ¹⁰⁰⁄_{( 100 + r )} ] ^t

= 17640 [ ¹⁰⁰⁄_{( 100 + 5 )} ] ² 

= 17640 × ¹⁰⁰⁄₁₀₅ × ¹⁰⁰⁄₁₀₅  

= 16,000  (Ans.)

Note :

If the question is asked for “the number after ‘t’ years“, then the required  number will be = P [ ^{( 100 ± r )}⁄₁₀₀ ] ^t

but,

If the question is asked for “the number before ‘t’ years“, then the required  number will be = P [ ¹⁰⁰⁄_{( 100 ± r )} ] ^t

Where ‘+’ for rate of increase and ‘−’ for rate of decrease.

Correct Answer: (a) 28

Given,

Time taken by Aman to reach Gwalior from their meeting point (i.e. Mathura) = 196 minutes and

Time taken by Kapil to reach Delhi from Mathura = 225 minutes

Let, time taken by both of them to reach Mathura from their respective starting points = ‘t’ minutes.

∴ t = √(196 × 225)

⇒ t = 14 × 15 = 210 minutes.

Now,

Total time taken by Aman to reach Gwalior from Delhi = 196 + 210 = 406 minutes and

Total time taken by Kapil to reach Delhi from Gwalior = 225 + 210 = 435 minutes

As, speed of Aman = 30 km/hr = ³⁰⁄₆₀ km/minutes = 0.5 km/min

Distance between Gwalior to Delhi = Speed × Time = 0.5 × 406 = 203 km

∴ Speed of Kapil = ^Distance⁄_Time

= ^{203 km}⁄_{435 min}

= ^{( 203 × 60 ) km}⁄_{435 hr} = 28 km/hr   (Ans.)

Correct Answer: (c) 9390

Shortcut Approach :

Given,

Simple Interest (S.I.) for 10 years = Rs. 3130

We know that, S.I. is same at every year.

Therefore, S.I. at the end of the 5 year,

= Rs. ³¹³⁰⁄₁₀ × 5 = Rs. 1565 

If principal becomes 5 times after 5 years then S.I. for the remaining 5 years also becomes 5 times i.e. Rs. (5 × 1565) = Rs. 7825

Therefore, total interest obtained after 10 years,

= Rs.(7825 + 1565)

= Rs. 9390   (Ans.)

Correct Answer: (d) 4

 Given,

^{(11 − 13x)}⁄_x + ^{(11 − 13y)}⁄_y + ^{(11 − 13z)}⁄_z = 5

 ⇒ ¹¹⁄_x − 13 + ¹¹⁄_y − 13 + ¹¹⁄_z − 13 = 5

 ⇒ 11 (¹⁄_x + ¹⁄_y + ¹⁄_z) = 5 + 39

 ⇒ ¹⁄_x + ¹⁄_y + ¹⁄_z = 4   (Ans.)

Correct Answer: (b) ⁹⁷⁄₃₆

Given,

2x + ⁹⁄_x = 9

⇒ 2x² + 9 = 9x

⇒ 2x² − 9x + 9 = 0

⇒ (x − 3)(2x − 3) = 0

Now,

either, (x − 3) = 0 ⇒ x = 3

or, (2x − 3) = 0 ⇒  x = ³⁄₂ 

Now, if we take x = 3, then the required

x² + ¹⁄_x² 

= 3² + ¹⁄_3² 

= 9 + ¹⁄₉ 

= ⁸²⁄₉ (but there has no such option in the given question)

But, if we take x = ³⁄₂ , then the required

x² + ¹⁄_x² 

= ( ³⁄₂ )² + ( ²⁄₃ )² 

= ⁹⁄₄ + ⁴⁄₉ 

= ⁹⁷⁄₃₆    (Ans.)

Correct Answer: (a) 0

Given,

^{(5x − y)}⁄_{(5x + y)} = ³⁄₇

 or, 7 (5x − y) = 3 ( 5x + y)

or, 35x − 7y = 15x + 3y

or, 35x − 15x = 3y + 7y

or, 20x = 10y

or, 2x = y

Now, the value of numerator part in the required question means,

4x² + y² − 4xy

= 4x² + (2x)² − 4.x.(2x)

= 4x² + 4x² − 8x² 

= 0

Therefore, the required value of ^{(4x2 + y2 − 4xy)}⁄_{(9x² + 16y² + 24xy)} = 0 (Ans.)

Correct Answer: (a) 1

Given,

(x + y)² = xy + 1

⇒ x² + 2xy + y² = xy + 1

⇒ x² + xy + y² = 1

⇒ (x − y)(x² + xy + y²) = 1 × (x − y)    [multiplying by (x − y) both sides]

⇒ x³ − y³ = x − y

⇒ 1 = x − y    [∵ given, x³ − y³ = 1]

Therefore, the required value of (x − y) = 1 (Ans.)

Correct Answer: (d) 75

 As, we know that diagonal of a square bisects its angle.

Therefore, ∠RSQ = 45°

Again, as ∆SRT is an equilateral triangle.

Therefore, ∠SRT = 60°

Now, in ∆SOR,

∠SOR + ∠OSR + ∠SRO = 180°   [using angle sum property of triangle]

 ⇒ ∠SOR + 45° + 60° = 180°

 ∴ ∠SOR = 75° (Ans.)

Correct Answer: (a) 14

We can easily prove that,

 d₁² + d₂² = 2 (a² + b²) 

where,

d₁ and d₂ are two diagonals and ‘a‘ and ‘b‘ are two sides.

Note: The above formula is valid for all type quadrilateral like ‘Parallelogram‘; ‘Rhombus‘; ‘Rectangle‘; ‘Square‘ and ‘Kite‘ but not for irregular quadrilateral and ‘Trapezium‘.

Now given,

d₁ = 8 cm, d₂ = ?, a = AB = 7 cm and b = BC = 9 cm

Therefore, according to the above formula, we have

 8² + d₁² = 2 ( 7² + 9² ) 

 ⇒ d₁² = 196

 ⇒ d₁ = √196 = 14 cm  (Ans.)

Correct Answer: (c) 20

Number of diagonals = ^{n(n − 3)}⁄₂       where, n = number of sides

In an Octagon, n = 8

Therefore, the required number of diagonals in an Octagon,

= ^{8(8 − 3)}⁄₂  

= 20   (Ans.)

Correct Answer: (b) 2.5

Let, O be the center of the circle and SPQR is a square inscribed in it in such a way that points S and P are equidistant from center ‘O’ means OS = OP = x (say)

 ∠PSQ = 45°       [diagonal of square bisects the angle]

Again, ∆SOP is an isosceles right-angled triangle.

Therefore, ∠OSP = 45°

Now, ∠OSQ = ∠OSP + ∠PSQ = 45° + 45°

∠OSQ = 90°

Now, in ∆SOP, we have

SP² = OS² + OP²      [applying Pythagoras theorem]

⇒ √(⁵⁄₂)² = x² + x² 

∴ x² = ⁵⁄₄

Now, diagonal of square, SQ = side √2 = √(⁵⁄₂) × √2 = √5 cm.

Again, in right-angled ∆SOQ, we have

OQ² = OS² + SQ²      [applying Pythagoras theorem]

⇒ OQ² = x² + (√5)² 

⇒ OQ² = ⁵⁄₄ + (√5)² 

⇒ OQ = ⁵⁄₂ cm. = 2.5 cm

 ∴ The required radius of the circle = 2.5 cm.   (Ans.)

Correct Answer: (b) √3

Given,

sec²θ + tan²θ = ⁵⁄₃ 

⇒ 1 + tan²θ + tan²θ  = ⁵⁄₃   [since, sec²θ = 1 + tan²θ]

⇒ 3 + 6 tan²θ = 5

⇒ 6 tan²θ = 2

⇒ tan θ = ¹⁄_√3 ⇒ θ = 30°

Now, tan 2θ = tan (2 × 30)° = tan 60° = √3  (Ans.)

Correct Answer: (d) 45(√3 + 2)

Let, MQ’ be the height of the tower.

In right-angled ∆PMQ,

^PM⁄_MQ = tan 60° ⇒ √3

Therefore, PM = √3 MQ

Again, in right-angled ∆PMR,

^PM⁄_RM = tan 30° ⇒  ¹⁄_√3 

⇒ √3 PM = RM

⇒ √3.(√3 MQ) = RM

Therefore, RM = 3MQ

Now, Given

RQ = 180 m

⇒ RM + MQ = 180

⇒ 3MQ + MQ = 180

Therefore, MQ = 45 m

Therefore, PM = √3 MQ ⇒ √3 × 45 m

Again, in right-angled ∆PMQ,

^MQ⁄_PQ = cos 60° ⇒  ¹⁄₂

⇒ PQ = 2MQ ⇒ 2 × 45 = 90 m

We know that, PQ = PQ’ = 90 m

Therefore, the required height of the tower,

MQ’ = MP + PQ’

      = √3 × 45 m + 90 m

      = 45 (√3 + 2) m

MQ’ = 45 (√3 + 2) m   (Ans.)

Correct Answer: (a) 30°

Given,

sin θ + sin 5θ = sin 3θ

2 sin{ ^{( 5θ + θ )}⁄₂ } × cos{ ^{( 5θ − θ )}⁄₂ } = sin 3θ     [using the formula, sin C + sin D = 2 sin{ ^{( C + D )}⁄₂ } × cos{ ^{( C − D )}⁄₂ }]

⇒ 2 sin 3θ × cos 2θ = sin 3θ

⇒ sin 3θ (2 cos 2θ − 1) = 0

Now,

Either, sin 3θ = 0 ⇒ θ = 0°  [the value of θ is not acceptable because of the given condition i.e. 0 < θ < ^π⁄₂ ]

Or, (2 cos 2θ − 1) = 0

⇒ cos 2θ = ¹⁄₂ ⇒ cos 60°

⇒ 2θ = 60°

∴ θ = 30°  (Ans.)

Correct Answer: (c) 118.8

Tax (in Rs. ‘000 crores) was paid by XYZ Ltd. in Y7,

= tax rate for Y7 × (revenue − cost) for Y7

= 33 % × (1600 − 1240 )

= ³³⁄₁₀₀ × 360

= 118.8  (Ans.)

Correct Answer: (c) Y8 > Y6 > Y2

Calculation of Profit after Tax :

Tax (in Rs. ‘000 crores) for Y2,

= 22 % × (1100 − 850)

= ²²⁄₁₀₀ × 250

= 55

Therefore, Profit (in Rs. ‘000 crores) after tax for Y2,

= (1100 − 850) − (55)

= 195

Similarly,

Profit (in Rs. ‘000 crores) after tax for Y6,

= (1500 − 1200) − (90)

= 210

Profit (in Rs. ‘000 crores) after tax for Y8,

= (1850 − 1400) − (67.5)

= 382.5

Conclusion:

Y8 > Y6 > Y2    (Ans.)

Correct Answer: (c) 5

Calculation of Profit for any year :

Profit (in Rs. ‘000 crores) for Y1 = 800 − 600 = 200

Profit (in Rs. ‘000 crores) for Y2 = 1100 − 850 = 250

Profit (in Rs. ‘000 crores) for Y3 = 1200 − 900 = 300

Profit (in Rs. ‘000 crores) for Y4 = 1200 − 950 = 250

Profit (in Rs. ‘000 crores) for Y5 = 1350 − 1050 = 300

Profit (in Rs. ‘000 crores) for Y6 = 1500 − 1200 = 300

Profit (in Rs. ‘000 crores) for Y7 = 1600 − 1240 = 360

Profit (in Rs. ‘000 crores) for Y8 = 1850 − 1400 = 450

Now, we can see from the above data, there are 5 different values of profit on any year which are 200, 250, 300, 360 and 450.

So, there are 5 distinct values of profit.  (Ans.)

Note :

Distinct Number : The term distinct number is used to refer to a number in a set that is not equal to another number.

Correct Answer: (c) 1820.2

Calculation of Profit after Tax :

Tax (in Rs. ‘000 crores) for Y1,

= 20 % × (800 − 600)

= ²⁰⁄₁₀₀ × 200

= 40

Therefore, Profit (in Rs. ‘000 crores) after tax for Y1,

= (800 − 600) − (40)

= 160

Similarly,

Profit (in Rs. ‘000 crores) after tax for Y2,

= (1100 − 850) − (55)

= 195

Profit (in Rs. ‘000 crores) after tax for Y3,

= (1200 − 900) − (66)

= 234

Profit (in Rs. ‘000 crores) after tax for Y4,

= (1200 − 950) − (62.5)

= 187.5

Profit (in Rs. ‘000 crores) after tax for Y5,

= (1350 − 1050) − (90)

= 210

Profit (in Rs. ‘000 crores) after tax for Y6,

= (1500 − 1200) − (90)

= 210

Profit (in Rs. ‘000 crores) after tax for Y7,

= (1600 − 1240) − (118.8)

= 241.5

Profit (in Rs. ‘000 crores) after tax for Y8,

= (1850 − 1400) − (67.5)

= 382.5

Therefore, net profit (in Rs. ‘000 crores) after tax for Y1 to Y8,

= 160 + 195 + 234 + 187.5 + 210 + 210 + 241.5 + 382.5

= 1820.5   (Ans.)