**SSC CGL 2019 EXAM PAPER : Held on 05-March-2020 Shift-2 (Quantative Aptitude)**

Table of Contents

**Q.1 ABCD is a cyclic quadrilateral in which sides AD and BC are produced to meet at P, and sides DC and AB meet at Q when produced. If ∠A = 60° and ∠ABC = 72°, then ∠DPC –∠BQC = ?**

1. 40°

##### 2. 24°

##### 3. 36°

##### 4. 30°

##### Correct answer is **Option 3 i.e. 36°**

##### ∠A = 60° and ∠ABC = 72°

##### In ΔAPB

##### ∠PAC + ∠ABP + ∠APB = 180°

##### ∠ABP = 180° – 60° – 72° = 48°

##### ∠ABP = ∠DPC = 48°

##### As we know, In cyclic quadrilateral, sum of opposite angles are 180°.

##### ∠ABC + ∠ADC = 180°

##### ∠ADC = 180° – 72° = 108°

##### In ΔADQ

##### ∠DAQ + ∠ADQ + ∠AQD = 180°

##### ∠AQD = 180° – 60° – 108° = 12°

##### Difference between ∠DPC and ∠BQC = 48° – 12° = 36°

**Q.2 The average of some numbers is 54.6. If 75% of the numbers are increased by 5.6 each, and the rest are decreased by 8.4 each, then what is the average of the numbers so obtained ?**

##### 1. 55.8

##### 2. 55.6

##### 3. 56.7

4. 56.3

##### The correct answer is__ option 3 i.e 56.7__

##### Let there are 4 numbers, each value of 54.6

##### Average of 4 numbers = 54.6

##### Sum of 4 numbers = 54.6 × 4 = 218.4

##### According to the question,

##### New average of 4 numbers = 3 (54.6 + 5.6) + 1 (54.6 – 8.4) = 3 × 60.2 + 46.2 = 180.6 + 46.2 = 226.8

##### New average of 4 numbers = 226.8/4 = 56.7

**Short Trick:**

##### Let there are 4 numbers.

##### New average of 4 numbers = 54.6 + (3 × 5.6 – 1 × 8.4)/4 = 54.6 + 8.4/4 = 54.6 + 2.1 = 56.7

**Q.3 To complete a certain task, X is 40% more efficient than Y, and Z is 40% less efficient than Y. Working together, they can complete the task in 21 days. Y and Z together worked for 35 days. The remaining work will be completed by X alone in:**

##### 1. 4 days

##### 2. 5 days

##### 3. 8 days

##### 4. 6 days

##### Correct answer is **option 2 i.e. 5 days**

##### Let efficiency of y be 5 unit

##### Efficiency of X = 5 × (7/5) = 7

##### Efficiency of Z = 5 × (3/5) = 3

##### Total work = (5 + 7 + 3) × 21 = 315

##### Work done by y and Z in 35 days = (5 + 3) × 35 = 280

##### Remaining work = 315 – 280 = 35

##### ∴Remaining work will be completed by X alone in = 35/7 = 5 days

**Q.4 If 11 sin**^{2} θ – cos^{2} θ + 4 sin θ – 4 = 0, 0° < θ < 90°, then what is the value of [latex]frac { cos2θ+cot2θ }{ sec2θ−tan2θ }[/latex]

^{2}θ – cos

^{2}θ + 4 sin θ – 4 = 0, 0° < θ < 90°, then what is the value of [latex]frac { cos2θ+cot2θ }{ sec2θ−tan2θ }[/latex]

##### 1. [latex]frac { 12+5sqrt { 3 } }{ 3 } [/latex]

2. [latex]frac { 10+5sqrt { 3 } }{ 3 } [/latex]

3. [latex]frac { 12+7sqrt { 3 } }{ 6 } [/latex]

4. [latex]frac { 10+7sqrt { 3 } }{ 6 } [/latex]

##### Correct answer is **Option 3 i.e. frac { 12+7sqrt { 3 } }{ 6 } **

##### 11 sin^{2} θ – cos^{2} θ + 4 sin θ – 4 = 0

##### Put θ = 30, and (satisfied)

##### Now,

**frac { cos2θ+cot2θ }{ sec2θ−tan2θ }**

**= ****frac { cos60+cot60 }{ sec60−tan60 }**

##### = frac { frac { 1 }{ 2 } +frac { 1 }{ sqrt { 3 } } }{ 2-sqrt { 3 } }

##### = frac { 2+sqrt { 3 } }{ 2sqrt { 3 } (2-sqrt { 3 } ) } times frac { 2+sqrt { 3 } }{ 2+sqrt { 3 } }

##### = frac { 7+4sqrt { 3 } }{ 2sqrt { 3 } }

##### = frac { 12+7sqrt { 3 } }{ 6 }

**Q.5 A and B together borrowed a sum of Rs. 51,750 at an interest rate of 7% p.a. compound interest in such a way that to settle the loan, A paid as much amount after three years as paid by B after 4 years from the day of borrowing. The sum (in Rs.) borrowed by A was:**

##### 1. 26,750

2. 25,000

3. 24,860

4. 25,650

##### Correct answer is option 1 i.e. 25,650

**Short Trick:**

##### As we know, the difference between years 4 and 3 = 1 year, then

##### A : B = 107 : 100

##### 107 + 100 = 207 unit

##### 207 unit = 51750

##### 1 unit = 250

##### 107 unit = 26750

**Q.6 What x is added to each of 10, 16, 22 and 32, the numbers so obtained in this order are in proportion. What is the mean proportional between the numbers (x + 1) and (3x + 1)?**

1. 15

##### 2. 10

##### 3. 12

##### 4. 9

##### Correct answer is option 1 i.e. **15**

##### If, x is added to each of 10, 16, 22 and 32, the numbers so obtained in this order are in proportion, then

##### According to the question

##### (10 + x) : (16 + x) :: (22 + x) : (32 + x)

##### (10 + x)/(16 + x) = (22 + x)/(32 + x)

##### We can solve this using cross multiplication or Using Componendo or Dividendo but hare we can solve this using equal difference between 16 – 10 = 6 and 32 – 22 = 10. For makes calculation easy and saves time.

##### 6 × 5 = 30 and 10 × 3 = 30

##### Now,

##### 5 (10 + x) = 3 (22 + x)

##### 50 + 5x = 66 + 3x

##### 2x = 16

##### x = 8

##### Now,

##### Let the mean proportional between the numbers (x + 1) and (3x + 1) be A, then

##### (x + 1) : A :: A : (3x + 1)

##### (8 + 1)/A = A/(3 × 8 + 1)

##### A^{2} = 9 × (24 + 1)

##### A = √(9 × 25)

##### A = 3 × 5 = 15

**Q.7 Reema sold 48 articles for Rs. 2,160 and suffered a loss of 10%. How many articles should she sell for Rs. 2,016 to earn a profit of 12%?**

##### 1. 28

2. 36

3. 40

4. 32

##### Correct answer is __option 2 i.e.____ 36__

__36__

##### SP of 48 articles = Rs. 2160

##### CP of 48 articles = 2160 × (100/90) = Rs. 2400

##### CP of 1 article = 2400/48 = Rs. 50

##### New sp of 1 article = 50 × (112/100) = 56

##### In Rs. 56 we can sold = 1 article

##### In Rs. 1 we can sold = 1/56 article

##### In Rs. 2016 we can sold = 1/56 × 2016 = 36 articles

**Q.8 If ****[latex]{20 x }^{ 2 } – 30x + 1 = 0[/latex], then what is the value of ****[latex]{ 25x }^{ 2 }+frac { 1 }{ { 16x }^{ 2 } }[/latex] **

##### 1. [latex]58frac { 3 }{ 4 } [/latex]

2. [latex]53frac { 1 }{ 2 } [/latex]

3. [latex]58frac { 1 }{ 2 } [/latex]

4. [latex]53frac { 3 }{ 4 } [/latex]

##### Correct answer is option 4 i.e. **53frac { 3 }{ 4 } **

##### 20x^{2} – 30x + 1 = 0

##### Divided by 4x

##### 5x – 15/2 + 1/4x

##### 5x + 1/4x = 15/2

##### As we know,

##### (5x + 1/4x)^{2} = 25x^{2} + 1/16x^{2} + 2 × 5x × (1/4x)

##### (15/2)^{2} = 25x^{2} + 1/16x^{2} + 10/4

##### 25x^{2} + 1/16x^{2} = 225/4 – 10/4 = 215/4 = 53frac { 3 }{ 4 }

**Q.9 When a positive integer is divided by d, the remainder is 15. When ten times of the same number is divided by d, the remainder is 6. The least possible value of d is:**

1. 9

2. 12

3. 18

4. 16

##### Correct answer is option 4 i.e ** ****16**

##### As given, Remainder is 15 so we can say value of d is greater than 15. Only 16 and 18 is greater than 15.

##### Taking, d = 16 from the options

##### Ten times of 15 = 150

##### When we divide 150 by 16 we get remainder 6 (condition satisfied). So the value of d = 16.

**Q10. In ΔABC, ∠B = 90°. If points D and E are on side BC such that BD = DE = EC, then which of the following is true?**

##### 1. 5 AE^{2} = 2 AC^{2} + 3 AD^{2}

2. 8 AE^{2} = 3 AC^{2} + 5 AD^{2}

##### 3. 5 AE^{2} = 3 AC^{2} + 2AD^{2}

4. 8 AE^{2} = 5 AC^{2} + 3 AD^{2}

##### Correct answer is __Option 2. i.e. __**8 AE**^{2} = 3 AC^{2} + 5 AD^{2}

__Option 2. i.e.__

^{2}= 3 AC

^{2}+ 5 AD

^{2}

**Short trick:**

##### In this type of question let the numerical values for savings time, calculation, and pressure.

##### As we know, the triplet 3, 4, and 5.

##### Let AB = 4 and BC = 3 and AC = 5 and BD = DE = EC = 1 cm and BE = 2 cm

##### Using Pythagoras theorem, AD = √17 and AE = √20

##### Now, check the options one by one and satisfied RHS and LHS.

##### From option 3.

##### 8 AE^{2} = 3 AC^{2} + 5 AD^{2}

##### 8 × 20 = 3 × 25 + 5 × 17

##### 160 = 160 (satisfied)

**Detailed Method:**

##### In ΔABC

##### AC^{2} = AB^{2} + BC^{2}

##### AC^{2} = AB^{2} + (3BD)^{2}

##### AC^{2} = AB^{2} + 9BD^{2} —(1)

##### In ΔABE

##### AE^{2} = AB^{2} + BE^{2}

##### AE^{2} = AB^{2} + (2BD)^{2}

##### AE^{2} = AB^{2} + 4BD^{2} —(2)

##### In ΔABD

##### AD^{2} = AB^{2} + BD^{2}

##### AD^{2} = AB^{2} + BC^{2} —(3)

##### Add equation after multiplying by 3 in equation (1) and by 5 in equation (3)

##### 3AC^{2} + 5AD^{2} = 8AB^{2} + 32 BD^{2} —(4)

##### Multiply by 8 in equation (2)

##### 8AE^{2} = 8AB^{2} + 32BD^{2} —(5)

##### From equation (4) and equation (5)

##### 8AE^{2} = 3AC^{2} + 5AD^{2}

**Q.11 The Value of [latex]frac { 3 }{ 5 } times 1frac { 7 }{ 8 } div 1frac { 1 }{ 3 } offrac { 3 }{ 16 } -left( 3frac { 1 }{ 5 } div 4frac { 1 }{ 2 } of5frac { 1 }{ 3 } right) times 2frac { 1 }{ 2 } +frac { 1 }{ 2 } +frac { 1 }{ 8 } div frac { 1 }{ 4 } [/latex] is :**

##### 1. [latex]4frac { 1 }{ 8 }%[/latex]

##### 2. [latex]5frac { 1 }{ 6 }%[/latex]

##### 3. [latex]5frac { 5 }{ 6 }%[/latex]

##### 4. [latex]4frac { 1 }{ 3 }%[/latex]

##### Correct answer is **Option 1 i.e. ****16frac { 2 }{ 3 }%**

** frac { 3 }{ 5 } times 1frac { 7 }{ 8 } div 1frac { 1 }{ 3 } offrac { 3 }{ 16 } -left( 3frac { 1 }{ 5 } div 4frac { 1 }{ 2 } of5frac { 1 }{ 3 } right) times 2frac { 1 }{ 2 } +frac { 1 }{ 2 } +frac { 1 }{ 8 } div frac { 1 }{ 4 } **

##### = frac { 3 }{ 5 } times frac { 15 }{ 8 } div frac { 4 }{ 3 } offrac { 3 }{ 16 } -left( frac { 16 }{ 5 } div frac { 9 }{ 2 } offrac { 16 }{ 3 } right) times frac { 5 }{ 2 } +frac { 1 }{ 2 } +frac { 1 }{ 2 }

##### = frac { 3 }{ 5 } times frac { 15 }{ 8 } div frac { 1 }{ 4 } -left( frac { 16 }{ 5 } div 24 right) times frac { 5 }{ 2 } +1

##### = frac { 3 }{ 5 } times frac { 15 }{ 8 } times 4-left( frac { 16 }{ 5 } div frac { 1 }{ 24 } right) times frac { 5 }{ 2 } +1

##### = frac { 9 }{ 2 } -left frac { 2 }{ 15 } right times frac { 5 }{ 2 } +1

##### = frac { 9 }{ 2 } -left frac { 1 }{ 3 } right +1

##### = frac { 27-2+6 }{ 6 }

##### = frac { 31 }{ 6 }

##### = 5frac { 1 }{ 6 }

**Q.12 The given table represents the sale (in thousands) of cars by four companies A, B, C and D in six years. Study the table and answer the question that follows.**

##### The total number of cars sold by company C in 2018 exceeds the average number of cars sold by company A during 2014 to 2018 by:

##### 1. 12,000

##### 2. 16,000

##### 3. 15,000

##### 4. 14,000

##### Correct answer is **Option 4 i.e. ****16000**

##### Total number of cars sold by company A during 2014 to 2018 = 52 + 61 + 72 + 52 + 63 = 300

##### Average cars sold by A during 2014 to 2018 = 300/5 = 60

##### Cars sold by company C in 2018 = 76

##### The total number of cars sold by company C in 2018 exceeds the average number of cars sold by company A during 2014 to 2018 by = 76 – 60 = 16

##### The given table represents the sale (in thousands), so 16 = 16000

**Q.13 ****The given table represents the sale (in thousands) of cars by four companies A, B, C and D in six years. Study the table and answer the question that follows.**

##### The total number of cars sold by company B during 2015, 2017 and 2018 is what percentage less than the total number of cars sold by company C in 2013, 2015, 2017 and 2018?

##### 1. 40

##### 2. [latex]33frac { 1 }{ 3 } [/latex]

##### 3. [latex]16frac { 2 }{ 3 } [/latex]

##### 4. 50

##### Correct Answer is **option 2 i.e. **33frac { 1 }{ 3 }

##### To ease our calculation part, we can neglect 1000 part.

##### Total number of cars sold by company B during 2015, 2017 and 2018 = 60 + 53 + 67 = 180

##### Total numbers of cars sold by company C in 2013, 2015 and 2017 and 2018 = 65 + 66 + 63 + 76 = 270

##### Required percentage = [(270 – 180)/270] × 100 = 33frac { 1 }{ 3 }

**Q.14 ****The given table represents the sale (in thousands) of cars by four companies A, B, C and D in six years. Study the table and answer the question that follows.**

##### What is the ratio of the total number of cars sold by companies A, B and D in 2017 to the total number of cars sold by all four companies in 2018?

##### 1. 3 : 4

2. 6 : 13

3. 9 : 14

4. 18 : 23

##### Correct answer is **Option 3 i.e. ****9 : 14**

##### Total number of cars sold by companies A, B and D in 2017 = 52 + 53 + 75 = 180

##### Total number of cars sold by all four companies in 2018 = 63 + 67 + 76 + 74 = 280

##### Required ratio = 180 : 280 = 9 : 14

**Q.15 ****The given table represents the sale (in thousands) of cars by four companies A, B, C and D in six years. Study the table and answer the question that follows.**

##### The total number of cars sold by companies A in 2017 and C in 2013 is what percentage of the total number of cars sold by all four companies in 2013 and 2016? (correct to one decimal place)

##### 1. 23.3

2. 24.2

##### 3. 25.6

##### 4. 23.8

##### Correct answer is **Option 1 i.e. 23.3%**

##### Total number of cars sold by companies A in 2017 and C in 2013 = 52 + 65 = 117

##### Total number of cars sold by all four companies in 2013 and 2016 = 45 + 63 + 65 + 67 + 72 + 58 + 70 + 63 = 503

##### Required percentage = [117/503] × 100 = 23.3%

**Q.16 In ΔPQR, PQ = 24 cm and ∠Q = 58°. S and T are the points on side PQ and PR, respectively, such that ∠STR = 122°.If PS = 14 cm and PT = 12 cm, then the length of RT is:**

##### 1. 15 cm

##### 2. 16 cm

##### 3. 16.4 cm

##### 4. 14.8 cm

##### Correct answer is **Option 2 i.e. 16 cm**

##### Given, PS = 14 and PT = 12 and PQ = 24

##### In ΔPQR and In ΔPST

##### ∠P = ∠P = common

##### ∠PQR = ∠PTS = 58°

##### ΔPQR ∼ ΔPTS

##### As we know,

##### PR/PS = PQ/PT

##### PR/14 = 24/12

##### PR = 28

##### RT = PR – PT = 28 – 12 = 16 cm

**Q.17 If 5 cos θ – 12 sin θ = 0, then what is the value of [latex]frac { 1+sinθ+cosθ }{ 1-sinθ+cosθ } [/latex] is ?**

#### 1. [latex]frac { 3 }{ 4 } [/latex]

#### 2. [latex]frac { 5 }{ 4 } [/latex]

#### 3. [latex]frac { 3 }{ 2 } [/latex]

#### 4. [latex]frac { 5 }{ 2 } [/latex]

##### Correct answer is **Option 3 i.e. **** frac { 3 }{ 2 } cm **

##### 5 cosθ – 12 sin θ = 0

##### ⇒5 cos θ = 12 sinθ

##### ⇒sinθ/cosθ = 5/12

##### ⇒tanθ = 5/12

##### As we know,

##### Hypotenuse = H, Perpendicular = P and Base = B

##### tan θ = P/B = 5/12

##### As we know,

##### H^{2} = P^{2} + B^{2}

##### ⇒H^{2} = 12^{2} + 5^{2} = 144 + 25 = 169

##### ⇒H = √169 = 13 cm

##### sinθ = P/H = 5/13 and cos θ = B/H = 12/13

##### Now,

##### (1 + sinθ + cos θ)/(1 – sinθ + cosθ)

##### ⇒(1 + 5/13 + 12/13)/(1 – 5/13 + 12/13)

##### ⇒(13 + 12 + 5)/(13 – 5 + 12)

##### ⇒30/20

##### ⇒3/2

**Q18. [latex]frac { { x }^{ 2 }{ left( x-4 right) }^{ 2 } }{ { left( x+4 right) }^{ 2 }-4x } div frac { { left( { x }^{ 2 }-4x right) }^{ 3 } }{ { left( x+4 right) }^{ 2 } } times frac { 64-{ x }^{ 3 } }{ 16-{ x }^{ 2 } }[/latex] is equal to**

##### 1. [latex]frac { x-4 }{ x+4 } [/latex]

##### 2. [latex]frac { x+4 }{ x(4-x) } [/latex]

##### 3. [latex]frac { x+4 }{ x(x-4) } [/latex]

##### 4. [latex]frac { x+4 }{ (x-4) } [/latex]

#### Correct answer is **Option 3 i.e. ****frac { x+4 }{ x(x-4) } **

**frac { { x }^{ 2 }{ left( x-4 right) }^{ 2 } }{ { left( x+4 right) }^{ 2 }-4x } div frac { { left( { x }^{ 2 }-4x right) }^{ 3 } }{ { left( x+4 right) }^{ 2 } } times frac { 64-{ x }^{ 3 } }{ 16-{ x }^{ 2 } }**

#### Put x = 2

#### = frac { { 2 }^{ 2 }{ left( 2-4 right) }^{ 2 } }{ { left( 2+4 right) }^{ 2 }-4times 2 } div frac { { left( 2^{ 2 }-4times 2 right) }^{ 3 } }{ { left( 2+4 right) }^{ 2 } } times frac { 64-{ 2 }^{ 3 } }{ 16-{ 2 }^{ 2 } }

#### = frac { 4×4 }{ 36−8 } div frac { −64 }{ { 36 } } times frac { 64-8 }{ 16-4 }

#### = frac { 16 }{ 28 } times frac { 36 }{ { -64 } } times frac { 56 }{ 12 }

#### = frac { -3 }{ 2 }

#### From option 3

#### frac { x+4 }{ x(x-4) }

#### = frac { 2+4 }{ 2(2-4) }

#### = frac { -6 }{ 4 }

#### = frac { -3 }{ 2 }

**Q.19 Diameter AB of a circle with centre O is produced to a point P such that PO = 16.8 cm. PQR is a secant that intersects the circle at Q and R such that PQ = 12 cm and PR = 19.2 cm. The length of AB (in cm) is:**

##### 1. 15.8

2. 14.4

3. 15.2

4. 14.2

##### Correct answer is **Option 2 i.e. 31°**

##### Given, PO = 16.8, PQ = 12 and PR = 19.2

##### From the following figure

##### Let the radius of the circle AO = OB = x, then

##### PA = (16.8 + x) and PB = (16.8 – x)

##### As we know,

##### PB × PA = PQ × PR

##### (16.8 – x) (16.8 + x) = 12 × 19.2

##### 16.8^{2} – x^{2} = 12 × 19.2

##### x^{2} = 16.8^{2} – 12 × 19.2 = 12^{2} (1.4^{2} – 1.6) = 144 × (1.96 – 1.60)

##### x = √(144 × 0.36)

##### x = 7.2

##### Hence, 2x = 14.4

##### ∴AB = 14.4

**For easy calculation**

##### Given,

##### PO = 16.8, PQ = 12 and PR = 19.2

##### Divided by 12, then

##### PO = 1.4, PQ = 1 and PR = 1.6

##### Let the radius of the circle AO = OB = x, then

##### PA = (1.4 + x) and PB = (1.4 – x)

##### As we know,

##### PB × PA = PQ × PR

##### (1.4 – x) (1.4 + x) = 1 × 1.6

##### 1.96 – x^{2} = 1.6

##### x^{2} = 0.36

##### x = √0.36

##### x = 0.6

##### 2x = 1.2

##### AB = 1.2

##### Multiply by 12 in 1.2 and get correct answer = 1.2 × 12 = 14.4

**Q.20 What is the value of [latex]frac { cosec(78°−θ)−sec(12°+θ)−tan(67°+θ)+cot(23°−θ) }{ tan13°tan37°tan45°tan53°tan77° } [/latex] ?**

1. 2

##### 2. 1

3. 0

##### 4. -1

##### Correct answer is **Option 3 i.e. 0**

##### [cosec (78 – θ) – sec (12 + θ) – tan (67 + θ) + cot (23 – θ)]/[tan13 cot 37 tan 45 tan53 tan77]

##### As we know,

##### cosec (90 – θ) = sec θ and tanθ cotθ = 1 and tan 45 = 1 and tan θ tan (90 – θ) = 1

##### ⇒ [cosec (78 – θ) – cosec {90 – (12 + θ)} – tan (67 + θ) + tan {90 – (23 – θ)}]/1

##### ⇒ [cosec (78 – θ) – cosec (78 – θ) – tan (67 + θ) + tan (67 + θ)]

##### ⇒ 0

**Q.21 A and B spend 60% and 75% of their incomes, respectively. If the savings of A are 20% more than that of B, then by what percentage is the income of A less than the income of B?**

##### 1. 25

##### 2. 15

##### 3. 20

##### 4. 10

##### Correct answer is **Option 1 i.e. 25**

##### Using option 1

##### let income of B be 40

##### Income of A = 30

##### Savings of A = 30 × (40/100) = 12

##### Savings of B = 40 × (25/100) = 10

##### Savings of A is more than B by = (12 – 10)/10 × 100 = 20% (satisfied)

**Detailed Method:**

##### Let the income of A and B be Rs. x and y respectively.

##### Savings of A = 40% of x

##### Savings of B = 25% of y

##### According to the question

##### (40% of x) = (25% of y) × (120/100)

##### x × 40% = y × 25% × 6/5

##### x : y = 3 : 4

##### Let income of A = 3 and income of B = 4

##### Income of A is less than that of B by = [(4 – 3)/4] × 100 = 25%

**Q.22 A boat can go 3.6 km upstream and 5.4 km downstream in 54 minutes, while it can go 5.4 km upstream and 3.6 km downstream in 58.5 minutes. The time (in minutes) taken by the boat in going 10 km downstream is:**

##### 1. 50

2. 48

3. 45

4. 54

##### Correct answer is **Option 1 i.e. 50**

**Short Trick:**

##### U = upstream and D = downstream

##### 3.6(U) + 5.4(D) = 54 and 5.4(U) + 3.6(D) = 58.5

##### Divide by 18

##### 2U + 3D = 30 and 3U + 2D = 32.5

##### Now, Multiply by 3 in (1) and by 2 in (2), and then subtract (2) from (1), then

##### 5D = 25 min

##### ∴10D = 50 min

**Detailed Method:**

##### Let the speed of boat be x km/hr and speed of current be y km/hr

##### Upstream speed = (x – y) km/hr

##### Downstream speed = (x + y) km/hr

##### According to the question

##### 3.6/(x – y) + 5.4/(x + y) = 54/60

##### 5.4/(x – y) + 3.6/(x + y) = 58.5/60

##### Let 1/(x – y) = a and 1/(x + y) = b, then

##### 3.6a + 5.4b = 54/60 —(1)

##### 5.4a + 3.6b = 58.5/60 —(2)

##### Add equation (1) and equation (2)

##### 9 (a + b) = 112.5/60

##### ⇒a + b = 12.5/60

##### ⇒a + b = 5/24 —(3)

##### Subtract equation (2) from equation (1)

##### 1.8 (b – a) = -4.5/60

##### ⇒b – a = -2.5/60

##### ⇒b – a = -1/24 —(4)

##### Add equation (3) and equation (4)

##### 2b = 4/24

##### ⇒b = 1/12

##### ⇒(x + y) = 12 km/hr

##### Downstream speed = 12 km/hr

##### ∴Time taken to cover 10 km distance with downstream = 10/12 hr or (10/12) × 60 = 50 min

**Q.23 If x**^{4} + x^{2}y^{2} + y^{4} = 273 and x^{2} – xy + y^{2} = 13, then the value of xy is:

^{4}+ x

^{2}y

^{2}+ y

^{4}= 273 and x

^{2}– xy + y

^{2}= 13, then the value of xy is: