# SSC CGL Held on 8th August 2017 Shift 3

### SSC CGL Held on 8th August 2017 Shift 3

**Staff Selection Commission – Combined Graduate Level Examination**, often referred to as **SSC CGL** is an examination conducted to recruit staff to various posts in ministries, departments and organisations of the Government of India. It is conducted by the Staff Selection Commission for selecting staff for various Group B and Group C posts. The Staff Selection Commission was established in **1975**.

Candidates applying for the various posts need to have a bachelor’s degree from a recognised university at the time of applying. Age of the candidate must be between 18 and 32 years (depending upon post applied).

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**Q.1)** *Which one is the largest among the fractions ^{5}⁄_{113} , ^{7}⁄_{120} , ^{13}⁄_{145} and ^{17}⁄_{160} ?*

a) ^{5}⁄_{113}

b) ^{7}⁄_{120}

c) ^{13}⁄_{145}

d) ^{17}⁄_{160}

## Answer

Correct Answer:** (d)** ^{17}⁄_{160}

## Solution

*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.)*

**Q.2)** *Nirmit can do ^{2}⁄_{3} of a job in 18 days. Kashish is twice as efficient as Nirmit. In how many days Kashish will complete the job ?*

a) ^{29}⁄_{4}

b) ^{27}⁄_{2}

c) ^{31}⁄_{2}

d) ^{13}⁄_{2}

## Answer

Correct Answer:** (b)** ^{27}⁄_{2}

## Solution

*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.*

**Q.3)** *40 men took a dip in a pool 30 m long and 25 m broad. If the average water displaced by a man is 5 m ^{3}, then what will be the rise (in cm) in level of the pool ?*

a) 25

b) 26.66

c) 27.33

d) 28

## Answer

Correct Answer:** (b)** 26.66

## Solution

*As given, one man can displaced volume of water = 5 m ^{3}*

*Therefore, 40 men can displaced volume of water = 5 × 40 = 200 m ^{3} *

*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.)*

**Q.4)** *An article is listed at Rs 2375. A man purchases it at two successive discounts of 50% and 25% and spends Rs 165 on repairing of article. If he sells the article at a profit of 62.5%, then what is the selling price (in Rs) of the article ?*

a) 1467.6

b) 1492.6

c) 1715.39

d) 1467.6

## Answer

Correct Answer:** (c)** 1715.39

## Solution

*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.)*

**Q.5)** *The length, breadth and height of a cuboid are in the ratio 19 : 11 : 13. If length is 30 cm more than height, then what is the volume (in cm ^{3}) of this cuboid ?*

a) 81510

b) 89665

c) 195300

d) 339625

## Answer

Correct Answer:** (d)** 339625

## Solution

*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 cm^{3} (Ans.) *

**Q.6)** *Average age of a team having 12 players is 23 years. If the age of the coach is also included, then the average age increases by 2 years. What is the age (in years) of the coach ?*

a) 41

b) 47

c) 49

d) 51

## Answer

Correct Answer:** (c)** 49

## Solution

*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.)*

**Q.7)** *An item is sold at two successive gains of 30% and 20%. If the final selling price is Rs 31200, then what is the cost price (in Rs) ?*

a) 15000

b) 20000

c) 22250

d) 24000

## Answer

Correct Answer:** (b)** 20000

## Solution

*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.)*

**Q.8)** *The number of trees in a town is 17640. If the numbers of trees increases annually at the rate of 5%, then how many trees were there 2 years ago ?*

a) 14000

b) 15000

c) 16000

d) 19450

## Answer

Correct Answer:** (c)** 16000

## Solution

*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.*

**Q.9)** *Aman and Kapil starts from Delhi and Gwalior respectively towards each other at same time. They meet at Mathura and then take 196 minutes and 225 minutes respectively to reach Gwalior and Delhi. If speed of Aman is 30 km/hr, then what is the speed (in km/hr) of Kapil ?*

a) 28

b) 30

c) ^{225}⁄_{7}

d) ^{392}⁄_{15}

## Answer

Correct Answer:** (a)** 28

## Solution

*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.)*

**Q.10)** *The simple interest on a sum of money for 10 years is Rs 3130. If the principal becomes 5 times after 5 years, then what will be the total interest (in Rs) obtained after 10 years ?*

a) 6260

b) 7825

c) 9390

d) 15650

## Answer

Correct Answer:** (c)** 9390

## Solution

**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.)*

**Q.11)** *If ^{(11 − 13x)}⁄_{x} + ^{(11 − 13y)}⁄_{y} + ^{(11 − 13z)}⁄_{z} = 5 , then what is the value of ^{1}⁄_{x} + ^{1}⁄_{y} + ^{1}⁄_{z} ?*

a) 1

b) ^{13}⁄_{11}

c) ^{13}⁄

d) 4

## Answer

Correct Answer:** (d)** 4

## Solution

* 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.)*

**Q.12)** *If 2x + ^{9}⁄_{x} = 9, then what is the minimum value of x^{2} + ^{1}⁄_{x2} ?*

a) ^{95}⁄_{36}

b) ^{97}⁄_{36}

c) ^{86}⁄_{25}

d) ^{623}⁄_{27}

## Answer

Correct Answer:** (b)** ^{97}⁄_{36}

## Solution

*Given, *

*2x + ^{9}⁄_{x} = 9*

*⇒ 2x ^{2} + 9 = 9x
*

*⇒ 2x ^{2} − 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 *

*x ^{2} + ^{1}⁄_{x2} *

*= 3 ^{2} + ^{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 *

*x ^{2} + ^{1}⁄_{x2} *

*= ( ^{3}⁄_{2 })^{2} + ( ^{2}⁄_{3 })^{2} *

*= ^{9}⁄_{4} + ^{4}⁄_{9} *

*= ^{97}⁄_{36} (Ans.) *

**Q.13)** *If ^{(5x − y)}⁄_{(5x + y)} = ^{3}⁄_{7}, then what is the value of ^{(4x2 + y2 − 4xy)}⁄_{(9x2 + 16y2 + 24xy)} ?*

a) 0

b) ^{3}⁄_{7}

c) ^{18}⁄_{49}

d) ^{1}⁄_{6}

## Answer

Correct Answer:** (a)** 0

## Solution

*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,*

*4x ^{2} + y^{2} − 4xy*

*= 4x ^{2} + (2x)^{2} − 4.x.(2x)*

*= 4x ^{2} + 4x^{2} − 8x^{2} *

*= 0*

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

**Q.14)** *If (x + y) ^{2} = xy + 1 and x^{3} − y^{3} = 1, then what is the value of x − y ?*

a) 1

b) 0

c) − 1

d) 2

## Answer

Correct Answer:** (a)** 1

## Solution

*Given,*

*(x + y) ^{2} = xy + 1*

*⇒ x ^{2} + 2xy + y^{2} = xy + 1*

*⇒ x ^{2} + xy + y^{2} = 1*

*⇒ (x − y)(x ^{2} + xy + y^{2}) = 1 × (x − y) [multiplying by (x − y) both sides]*

*⇒ x ^{3} − y^{3} = x − y*

*⇒ 1 = x − y [∵ given, x ^{3} − y^{3} = 1]*

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

**Q.15)** *In the given figure, PQRS is a square and SRT is an equilateral triangle. What is the value (in degrees) of ∠SOR ? *

a) 45

b) 55

c) 60

d) 75

## Answer

Correct Answer:** (d)** 75

## Solution

* *

* 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.)*

**Q.16)** *ABCD is a parallelogram in which AB = 7 cm, BC = 9 cm and AC = 8 cm. What is the length (in cm) of other diagonal ?*

a) 14

b) 14√2

c) 7

d) 7√2

## Answer

Correct Answer:** (a)** 14

## Solution

*We can easily prove that,*

** d _{1}^{2} + d_{2}^{2} = 2 (a^{2} + b^{2}) **

*where, *

**d _{1}** and

**d**are two

_{2}**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,*

**d _{1}** = 8 cm,

**d**= ?,

_{2}**a**= AB = 7 cm and

**b**= BC = 9 cm

*Therefore, according to the above formula, we have*

** 8 ^{2} + d_{1}^{2} = 2 ( 7^{2} + 9^{2} ) **

* ⇒ d _{1}^{2} = 196*

** ⇒ d _{1 }= **√196

**= 14 cm (Ans.)**

**Q.17)** *How many diagonals are there in octagon ?*

a) 12

b) 14

c) 20

d) 24

## Answer

Correct Answer:** (c)** 20

## Solution

**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.)*

**Q.18)** *A square is inscribed in a quarter circle in such a way that two of its adjacent vertices on the radius are equidistant from the centre and other two vertices lie on the circumference. If the side of square is √(5/2) cm, then what is the radius (in cm) of the circle ?*

a) 2

b) 2.5

c) 5

d) 10

## Answer

Correct Answer:** (b)** 2.5

## Solution

*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*

*SP ^{2} = OS^{2} + OP^{2} [applying Pythagoras theorem]*

*⇒ √( ^{5}⁄_{2})^{2} = x^{2} + x^{2} *

*∴ x ^{2} = ^{5}⁄_{4} *

*Now, diagonal of square, SQ = side √2 = √( ^{5}⁄_{2}) × √2 = √5 cm.*

*Again, in right-angled ∆SOQ, we have*

*OQ ^{2} = OS^{2} + SQ^{2} [applying Pythagoras theorem] *

*⇒ OQ ^{2} = x^{2} + (√5)^{2} *

*⇒ OQ ^{2} = ^{5}⁄_{4} + (√5)^{2} *

*⇒ OQ = ^{5}⁄_{2} cm. = 2.5 cm*

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

**Q.19)** *If sec ^{2}θ + tan^{2}θ = ^{5}⁄_{3}, then what is the value of tan 2θ ?*

a) 2√3

b) √3

c) ^{1}⁄_{√3}

d) cannot be determined

## Answer

Correct Answer:** (b)** √3

## Solution

*Given,*

*sec ^{2}θ + tan^{2}θ = ^{5}⁄_{3} *

*⇒ 1 + tan ^{2}θ + tan^{2}θ = ^{5}⁄_{3} [since, sec^{2}θ = 1 + tan^{2}θ]
*

*⇒ 3 + 6 tan ^{2}θ = 5*

*⇒ 6 tan ^{2}θ = 2*

*⇒ tan θ = ^{1}⁄_{√3} ⇒ θ = 30°*

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

**Q.20)** *A tower is broken at a point P above the ground. The top of the tower makes an angle 60° with the ground at Q. From another point R on the opposite side of Q angle of elevation of point P is 30°. If QR = 180 m, then what is the total height (in metres) of the tower ?*

a) 90

b) 45√3

c) 45(√3 + 1)

d) 45(√3 + 2)

## Answer

Correct Answer:** (d)** 45(√3 + 2)

## Solution

*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.)
*

**Q.21)** *If sin θ + sin 5θ = sin 3θ and 0 < θ < ^{π}⁄_{2}, then what is the value of θ (in degrees) ?*

a) 30°

b) 45°

c) 60°

d) 75°

## Answer

Correct Answer:** (a)** 30°

## Solution

*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.)*

**Q.22)** *The table below represent the cost, revenue and tax rate for XYZ Ltd. for a period of 8 years. Cost and revenue are given in Rs. ‘000 crores.
Profit for any year = revenue − cost
Profit after tax for any year = profit of that year − tax of that year
Tax on any year = tax rate of that year × profit of the year *

How much tax (in Rs ‘000 crores) was paid by XYZ limited in Y7 ?

a) 90

b) 99

c) 118.8

d) 126

## Answer

Correct Answer:** (c)** 118.8

## Solution

*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.)*

**Q.23)** *The table below represent the cost, revenue and tax rate for XYZ Ltd. for a period of 8 years. Cost and revenue are given in Rs. ‘000 crores.
Profit for any year = revenue − cost
Profit after tax for any year = profit of that year − tax of that year
Tax on any year = tax rate of that year × profit of the year *

Which of the following is correct about profit after tax for year Y2, Y6 and Y8 ?

a) Y8 > Y6 = Y2

b) Y6 > Y2 > Y8

c) Y8 > Y6 > Y2

d) Y6 = Y8 > Y2

## Answer

Correct Answer:** (c)** Y8 > Y6 > Y2

## Solution

**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.)*

* *

**Q.24)** *The table below represent the cost, revenue and tax rate for XYZ Ltd. for a period of 8 years. Cost and revenue are given in Rs. ‘000 crores.
Profit for any year = revenue − cost
Profit after tax for any year = profit of that year − tax of that year
Tax on any year = tax rate of that year × profit of the year *

How many distinct values of yearly profit are there ?

a) 3

b) 4

c) 5

d) 6

## Answer

Correct Answer:** (c)** 5

## Solution

**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.*

**Q.25)**

Profit for any year = revenue − cost

Profit after tax for any year = profit of that year − tax of that year

Tax on any year = tax rate of that year × profit of the year

What is the total sum (in ‘000 crores) of profit after tax for Y1 to Y8 ?

a) 1763.6

b) 1803.2

c) 1820.2

d) 1872.4

## Answer

Correct Answer:** (c)** 1820.2

## Solution

**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.)*

** If you still have any query then feel free to contact or comment below. I will definitely try to solve your issue. Thank you.**

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