**SSC CGL 2019 EXAM PAPER : Held on 04-March-2020 Shift-2 (Quantative Aptitude)**

Table of Contents

**Q.1 One–third of goods are sold at a 15% profit, 25% of the goods are sold at a 20% profit and the rest at a 20% loss. If the total profit of Rs. 138.50 is earned on the whole transaction, then the value (in Rs.) of the goods is:**

1. Rs. 8,310

##### 2. Rs. 7,756

##### 3. Rs. 8,030

##### 4. Rs. 8,587

##### Correct answer is **Option 1 i.e. ****Rs. 8,310**

##### Let total number of articles be 120 and cost price of 120 articles be 120 unit

##### 1/3^{rd} of 120 = 40 and 1/4^{th} of 120 = 30 and Rest = 120 – 40 – 30 = 50

##### Total profit = 40 × (15/100) + 30 × (20/100) – 50 × (20/100) = 6 + 6 – 10 = 2 unit

##### 2 unit = 138.50

##### ⇒ 1 unit = 138.50/2

##### ⇒ 120 unit = (138.50/2) × 120 = **Rs. 8310**

**Q.2 If 7 sin**^{2} θ – cos^{2} θ + 2 sin θ = 2, 0° < θ < 90°, then the value of [latex]frac { sec2θ+cot2θ }{ cosec2θ+tan2θ }[/latex] is:

^{2}θ – cos

^{2}θ + 2 sin θ = 2, 0° < θ < 90°, then the value of [latex]frac { sec2θ+cot2θ }{ cosec2θ+tan2θ }[/latex] is:

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

##### 2. [latex]frac { 2 }{ 5 } (1+sqrt {3 }) [/latex]

##### 3. [latex]frac { 1 }{ 5 } (1+2sqrt { 3 }) [/latex]

4. 1

#### The correct answer is__ option 3 i.e __**frac { 1 }{ 5 } (1+2sqrt { 3 }) **

#### 7 sin^{2} θ – cos^{2} θ + 2 sin θ = 2

#### ⇒ 7 sin^{2} θ – (1 – sin^{2} θ) + 2 sin θ = 2

#### ⇒ 7 sin^{2} θ – 1 + sin^{2} θ + 2 sin θ = 2

#### ⇒ 8 sin^{2} θ + 2 sin θ – 3 = 0

#### ⇒ 8 sin^{2 }θ + 6 sin θ – 4 sin θ – 3 =0

#### ⇒ 2 sin θ (sin θ + 3) – 1 (4 sin θ – 3) = 0

#### ⇒ (2 sin θ – 1) (4 sin θ + 3) = 0

#### Taking,

#### 2 sin θ – 1 = 0

#### ⇒ sin θ = 1/2

#### ⇒ sin θ = sin 30

#### ⇒ θ = 30

#### Now,

#### frac { sec2θ+cot2θ }{ cosec2θ+tan2θ }

#### = frac { sec60+cot60 }{ cosec60+tan60}

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

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

#### = frac { 2sqrt { 3 } +1 }{ 5 }

**Short Trick:**

#### 7 sin^{2} θ – cos^{2} θ + 2 sin θ = 2

#### Put θ = 30 and satisfied, then

#### = frac { sec60+cot60 }{ cosec60+tan60}

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

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

#### = frac { 2sqrt { 3 } +1 }{ 5 }

**Q.3 Two chord AB and CD of a circle are produced to intersect each other at a point P outside the circle. If AB = 7 cm, BP = 4.2 cm and PD = 2.8 cm, then the length of CD is:**

##### 1. 15.8 cm

##### 2. 14 cm

##### 3. 14.6 cm

##### 4. 12 cm

##### Correct answer is **option 2 i.e. 14 cm**

##### Given,

##### PB = 4.2 cm and AB = 7 cm

##### AP = 4.2 + 7 = 11.2

##### PD = 2.8 cm and Let CD = x.

##### PC = 2.8 + x

##### As we know,

##### AP × PB = PC × PD

##### 11.2 × 4.2 = 2.8 × (x + 2.8)

##### ⇒ x + 2.8 = (11.2 × 4.2)/2.8

##### ⇒ x + 2.8 = 16.8

##### ⇒ x = 16.8 – 2.8

**⇒ x = 14 cm**

**Q.4 The given table represents the revenue (in Rs. Crores) of a company from the sale of four products A, B, C and D in 6 years. Study the table carefully and answer the question that follows.**

##### The number of years in which the revenue of the company from the sale of product D is more than the average revenue from the sale of product A over six years, is:

##### 1. 4

2. 1

3. 2

4. 3

##### Correct answer is **Option 3 i.e. 2**

##### Total revenue of sale from product A over six year = 98 + 94 + 80 + 95 + 110 + 115 = 592

##### Average revenue of sale from product A over six year = 592/6 = 98.66

##### There are 2 years 2013 and 2017 in which the revenue of the company from the sale of product D is more than the average revenue from the sale of product A over six years.

**Q.5 If P **= [latex]frac { { x }^{ 4 }-8x }{ { x }^{ 3 }-{ x }^{ 2 }-2x } [/latex],Q = [latex]frac { { x }^{ 2 }+2x+1 }{ { x }^{ 2 }-4x-5 } [/latex],and R = [latex]frac { { 2x }^{ 2 }+4x+8 }{ x-5 } [/latex] then (P × Q) ÷ R is equal to:

##### 1. 1/2

2. 4

3. 2

4. 1

#### Correct answer is option 1 i.e. **frac { 1 }{ 2 }**

#### P × Q ÷ R

#### frac { { x }^{ 4 }-8x }{ { x }^{ 3 }-{ x }^{ 2 }-2x } times frac { { x }^{ 2 }+2x+1 }{ { x }^{ 2 }-4x-5 } div frac { { 2x }^{ 2 }+4x+8 }{ x-5 }

#### = frac { x({ x }^{ 3 }-{ 2 }^{ 3 }) }{ x({ x }^{ 2 }−x−2) } times frac { { x }^{ 2 }+2x+1 }{ { x }^{ 2 }−4x−5 } times frac { (x−5) }{ 2({ x }^{ 2 }+2x+4) }

#### = frac { x(x−2)({ x }^{ 2 }+4+2x) }{ x({ x }^{ 2 }−2x+x−2) } times frac { { x }^{ 2 }+2×x×1+{ 1 }^{ 2 } }{ { x }^{ 2 }−5x+x−5 } times frac { (x−5) }{ 2({ x }^{ 2 }+2x+4) }

#### = frac { (x−2)({ x }^{ 2 }+4+2x) }{ x(x−2)+1(x−2) } times frac { { (x+1) }^{ 2 } }{ x(x−5)+1(x−5) } times frac { (x−5) }{ 2({ x }^{ 2 }+2x+4) }

#### = frac { (x−2) }{ (x−2)(x+1) } times frac { { (x+1) }^{ 2 } }{ (x−5)(x+1) } times frac { (x−5) }{ 2 }

#### = frac { 1 }{ 2 }

**Q.6 If the 6–digit numbers x35624 and 1257y4 are divisible by 11 and 72, respectively, then what is the value of (5x – 2y)?**

1. 13

##### 2. 14

##### 3. 10

##### 4. 12

##### Correct answer is option 2 i.e.** ****14**

##### Divisibility law of 11: If the difference of the alternating sum of digits of the number is a multiple of 11 (e.g. 2343 is divisible by 11 because 2 – 3 + 4 – 3 = 0, which is a multiple of 11).

##### Divisibility law of 8 ⇒ A number is divisible by 8 if the number formed by its last three digits is divisible by 8

##### Divisibility law of 9 ⇒ A number is divisible by 9 if the sum of its digits is divisible by 9.

##### The 6–digit numbers x35624 is divisible by 11, then

##### x – 3 + 5 – 6 + 2 – 4 = 0

##### ⇒ x + 7 – 13 = 0

##### ⇒ x = 6

##### The 6–digit number 1257y4 is divisible by 72, so we can say that the number is also divisible by 8 and 9, then

##### ⇒ 1 + 2 + 7 + y + 4

##### ⇒ 19 + y

##### If we put y = 8, then the number becomes 27 which is divisible by 9.

##### 5x – 2y

##### ⇒ 5 × 6 – 2 × 8

##### ⇒ 30 – 16

##### ⇒ 14

**Q.7 If a + b + c = 7 and ab + bc + ca = –6, then the value of a**^{3} + b^{3} + c^{3} – 3abc is:

^{3}+ b

^{3}+ c

^{3}– 3abc is:

##### 1. 469

2. 472

3. 463

4. 479

##### Correct answer is __option 1 i.e. __**469**

##### As we know,

##### a^{3} + b^{3} + c^{3} – 3abc = (a + b + c) [(a + b + c)^{2} – 3 (ab + bc + ca)]

##### ⇒ a^{3} + b^{3} + c^{3} – 3abc = 7 [7^{2} – 3 (–6)] = 7 [49 + 18] = 7 × 67 = **469**

**Q.8 The given table represents the revenue (in Rs. Crores) of a company from the sale of four products A, B, C and D in 6 years. Study the table carefully and answer the question that follows.**

##### The total revenue of the company from the sale of products B, C and D in 2014 is what percentage of the total revenue from the sale of products C and D in 6 years?

##### 1. 25

2. 28

3. 18

4. 20

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

##### Total revenue of the company from the sale of products B, C and D in 2014 = 92 + 96 + 92 = 280

##### Total revenue of the company from the sale of products C and D in 6 years = 82 + 98 + 96 + 88 + 93 + 103 + 74 + 102 + 92 + 93 + 97 + 102 = 1120

##### There are 2 years in which the revenue of the company from the sale of product D is more than the average revenue from the sale of product A over six years.

##### ∴ Required Percentage = (280/1120) × 100 = 25%

**Q.9 The expression 3 sec**^{2} θ tan^{2} θ + tan^{6} θ – sec^{6} θ is equal to:

^{2}θ tan

^{2}θ + tan

^{6}θ – sec

^{6}θ is equal to:

1. -2

2. 2

3. -1

4. 1

##### Correct answer is option 3 i.e –**1**

**Short Trick:**

##### 3 sec^{2} θ tan^{2} θ + tan^{6} θ – sec^{6} θ

##### ⇒ Put θ = 45

##### ⇒ 3 sec^{2} 45 tan^{2} 45 + tan^{6} 45 – sec^{6} 45

##### ⇒ 3 × (√2)^{2} × 1 + 1 – (√2)^{6}

##### ⇒ 3 × 2 + 1 – 8

##### ⇒ 7 – 8

##### ⇒ (–1)

**Detailed Solution:**

##### 3 sec^{2} θ tan^{2} θ + tan^{6} θ – sec^{6} θ

##### ⇒ 3 sec^{2} θ tan^{2} θ + (tan^{2} θ)^{3} – (sec^{2} θ)^{3}

##### ⇒ 3 sec^{2} θ tan^{2} θ + (tan^{2} θ – sec^{2} θ) (tan^{4} θ + sec^{4} θ + tan^{2} θ sec^{2} θ)

##### ⇒ 3 sec^{2} θ tan^{2} θ + (–1) [tan^{4} θ + sec^{4} θ – 2 tan^{2} θ sec^{2} θ + 3tan^{2} θ sec^{2} θ]

##### ⇒ 3 sec^{2} θ tan^{2} θ + (–1) [(tan^{2} θ – sec^{2} θ)^{2} + 3 tan^{2} θ sec^{2} θ]

##### ⇒ 3 sec^{2} θ tan^{2} θ + (–1) [(–1)^{2} + 3 sec^{2} θ tan^{2} θ]

##### ⇒ 3 sec^{2} θ tan^{2} θ + (–1) [1 + 3 sec^{2} θ tan^{2} θ]

##### ⇒ 3 sec^{2} θ tan^{2} θ – 1 – 3 sec^{2} θ tan^{2} θ

##### ⇒ (–1)

**Q10. PRT is a tangent to a circle with centre O, at the point R on it. Diameter SQ of the circle is produced to meet the tangent at P and QR is jointed. If ∠QRP = 28°, then the measure of ∠SPR is:**

##### 1. 62°

2. 29°

##### 3. 32°

4. 34°

##### Correct answer is __Option 4. i.e. __**34°**

__Option 4. i.e.__

##### ∠QRP = 28°

##### As we know,

##### ∠ORP = 90° [tangents at R]

##### ⇒ ∠ORQ = ∠ORQ – ∠QRP = 90° – 28° = 62°

##### ⇒ ∠OQR = ∠ORQ = 62° [OQ = OR = radius]

##### As we know,

##### ∠OQR = ∠ QRP + ∠QPR

##### ⇒ 62 = 28 + ∠QPR

##### ⇒ ∠QPR = 62 – 28 = 34

##### ⇒ ∠SPR = ∠QPR = 34

**Q.11 ****The income of A is 60% less than that of B, and the expenditure of A is equal to 60% of B’s expenditure. If A’s income is equal to 70% of B’s expenditure, then what is the ratio of the saving of A and B?**

##### 1. 3 : 8

##### 2. 4 : 7

##### 3. 2 : 15

##### 4. 5 : 9

##### Correct answer is **Option 3 i.e. ****2 : 15**

##### Let the income of B be Rs. 5x

##### Income of A = 5x × (2/5) = Rs. 2x

##### Let the expenditure of B be Rs. 5y

##### Expenditure of A = 5y × (3/5) = 3y

##### According to the question

##### 2x = 5y × (70/100)

##### ⇒ 2x = y × (7/2)

##### ⇒ 4x = 7y

##### ⇒ x : y = 7 : 4

##### Savings of A = 2x – 3y = 2 × 7 – 3 × 4 = 14 – 12 = 2

##### Savings of B = 5x – 5y = 5 (7 – 4) = 5 × 3 = 15

##### ∴ Ratio of savings A to B = 2 : 15

**Q.12 In ΔABC, D and E are the points on sides AC and BC, respectively such that DE || AB. F is a point on CE such that DF ∥ AE. If CE = 6 cm, and CF = 2.5 cm, then BC is equal to:**

##### 1. 12 cm

##### 2. 14 cm

##### 3. 14.4 cm

##### 4. 15.6 cm

##### Correct answer is **Option 3 i.e. ****14.4 cm**

##### In ΔCAE

##### CF/CE = CD/AC

##### CD/AC = 2.5/6 = 25/60

##### ⇒ CD/AC = 5/12 —— 1)

##### Let CF = 5x and CE = 12x

##### In ΔCAB,

##### CE/BC = CD/AC

##### CD/AC = 6/BC —— 2)

##### From equation (1) and equation (2)

##### 6/BC = 5/12

##### ⇒ BC = (12 × 6)/5

##### ⇒ BC = 72/5

##### ⇒ BC = 14.4

**Q.13 The difference in compound interest on a certain sum at 10% p.a. for one year, when the interest is compounded half-yearly and yearly, is Rs. 88.80. What is the simple interest on the same sum for [latex]1frac { 2 }{ 3 }[/latex] years at the same rate?**

##### 1. Rs. 5,980

##### 2. Rs. 5,916

##### 3. Rs. 5,986

##### 4. Rs. 5,920

##### Correct Answer is **option 4 i.e. ****Rs. 5,920**

##### Let total sum be 100%

##### Interest rate for 1 year = 10%

##### Interest rate for 6 month = 5%

##### Effective interest rate when interest compounded half–yearly = 5 + 5 + (5 × 5)/100 = 10.25%

##### Difference of rate = 10.25% – 10% = 0.25%

##### 0.25% = 88.80

##### ⇒ 1% = 88.80/0.25

##### ⇒ 100% = 35520

##### Now,

##### P = 35520, r = 10% and t =1frac { 2 }{ 3 } year = 5/3 year

##### As we know,

##### SI = Prt/100

##### ∴ SI = (35520 × 10 × 5)/(100 × 3) = **5920.**

**Q.14 A cylindrical vessel of radius 30 cm and height 42 cm is full of water. Its contents are emptied into a rectangular tub of length 75 cm and breadth 44 cm. The height (in cm) to which the water rises in the tub is: (Take π = 22/7)**

1. 30

2. 45

3. 36

4. 40

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

##### Radius of cylindrical vessel r = 30 cm

##### Height of cylindrical vessel h = 42 cm

##### Length of rectangular tub L = 75 cm

##### Breadth of rectangular tub B = 44 cm

##### Le the height of the rectangular tub be H cm

##### As we know,

##### Volume of cylindrical vessel = πr^{2}h

##### The volume of rectangular tub = L × B × H

##### According to the question

##### 75 × 44 × H = (22/7) × 30 × 30 × 42

##### ⇒ H = (22 × 30 × 30 × 6)/(75 × 44)

##### ⇒ H = **36**

**Q.15 The value of [latex]frac { { tan }^{ 2 }θ−{ sin }^{ 2 }θ }{ 2+tan ^{ 2 }{ θ } +cot ^{ 2 }{ θ } } [/latex] is :**

##### 1. cosec^{6} θ

2. cos^{4} θ

##### 3. sin^{6} θ

##### 4. sec^{4} θ

##### Correct answer is **Option 3 i.e. ****sin**^{6} θ

^{6}θ

**Short Trick: **

** Put θ = 45**

**frac { { tan }^{ 2 }θ−{ sin }^{ 2 }θ }{ 2+tan ^{ 2 }{ θ } +cot ^{ 2 }{ θ } } **

##### = frac { { tan }^{ 2 }45-{ sin }^{ 2 }45 }{ 2+{ tan }^{ 2 }45+{ cot }^{ 2 }45 }

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

##### = frac { frac { 1 }{ 2 } }{ 4 }

##### = frac { 1 }{ 8 }

**From option 3.**

##### sin^{6} θ

##### ⇒ sin^{6} 45

##### ⇒ (1/√2)^{2}

##### ⇒ 1/8

**Q.16 The given table represents the revenue (in Rs. Crores) of a company from the sale of four products A, B, C and D in 6 years. Study the table carefully and answer the question that follows.**

##### By what percentage is the total revenue of the company from the sale of products A, B and D in 2012 and 2013 more than the total revenue from the sale of product B in 2013 to 2016? (Correct to one decimal place)

##### 1. 45.4

2. 31.2

3. 43.6

4. 44.5

##### Correct answer is **Option 1 i.e. 45.4**

##### Total revenue from the sale of products A, B and D in 2012 and 2013 = 98 + 74 + 74 + 94 + 96 + 102 = 538

##### Total revenue from the sale of products B in 2013 to 2016 = 96 + 92 + 84 + 98 = 370

##### Required percentage = (538 – 370)/370 × 100 = 45.4%

**Q.17 The ratio of boys and girls in a group is 7 : 6. If 4 more boys join the group and 3 girls leave the group, then the ratio of boys to girls becomes 4 : 3. What is the total number of boys and girls initially in the group?**

##### 1. 91

2. 117

3. 104

4. 78

##### Correct answer is **Option 3 i.e. 104**

##### The ratio of boys and girls in a group is = 7x : 6x

##### According to the question

##### (7x + 4)/(6x – 3) = 4/3

##### ⇒ 3 (7x + 4) = 4 (6x – 3)

##### ⇒ 21x + 12 = 24x – 12

##### ⇒ 24x – 21x = 12 + 12

##### ⇒ 3x = 24

##### ⇒ x = 24/3 = 8

##### ∴ Number of boys and girls in the group were = 7x + 6x = 13x = 13 × 8 = **104**

**Q18. The time taken by a boat to travel 13 km downstream is the same as time taken by it to travel 7 km upstream. If the speed of the stream is 3 km/h, then how much time (in hours) will it take to travel a distance of 44.8 km in still water?**

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

##### 2. [latex]4frac { 13}{ 25 } [/latex]

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

##### 4. [latex]4frac { 12}{ 25 } [/latex]

##### Correct answer is **Option 4 i.e. ****4frac { 12}{ 25 } **

##### Speed of the stream = 3 km/hr

##### Let speed of the boat be x km/hr

##### As we know,

##### Time = Distance/Speed

##### According to the question

##### 13/(x + 3) = 7/(x – 3)

##### ⇒ 13 (x – 3) = 7 (x + 3)

##### ⇒ 13x – 39 = 7x + 21

##### ⇒ 13x – 7x = 21 + 39

##### ⇒ 6x = 60

##### ⇒ x = 60/6 = 10

##### So, speed of boat = 10 km/hr

##### ∴ Time taken to cover 44.8 km distance in still water = 44.8/10 = **4frac { 12}{ 25 } **

**Q.19 The given table represents the revenue (in Rs. Crores) of a company from the sale of four products A, B, C and D in 6 years. Study the table carefully and answer the question that follows.**

##### What is the ratio of the total revenue of the company in 2014 from the sale of all the four products to the total revenue from the sale of product C in 2014 to 2017?

##### 1. 14 : 23

2. 7 : 10

3. 7 : 9

4. 18 : 19

##### Correct answer is **Option 4 i.e. 18 : 19**

##### Total revenue of the company in 2014 from the sale of all the four products = 80 + 92 + 96 + 92 = 360

##### Total revenue from the sale of product C in 2014 to 2017 = 96 + 88 + 93 + 103 = 380

##### ∴ Required ratio = 360 : 380 = 18 : 19

**Q.20 The average weight of some students in a class was 58.4 kg. When 5 students having the average weight 62.8 kg joined the class, the average weight of all students in the class increased by 0.55 kg. The number of students initially in the class, were:**

1. 25

2. 40

3. 35

4. 30

##### Correct answer is **Option 3 i.e. 35**

##### Let the total students in the class be x, then

##### Average weight of x students in the class = 58.4

##### Sum of weight of x students in the class = 58.4x

##### Average weight of 5 students = 62.8 kg

##### Sum of weight of 5 students = 62.8 × 5 = 314

##### According to the question

##### 58.4x + 314 = (x + 5) (58.4 + 0.55)

##### ⇒ 58.4x + 314 = (x + 5) 58.95

##### ⇒ 58.4x + 314 = 58.95x + 294.75

##### ⇒ 58.95x – 58.4x = 314 – 294.75

##### ⇒ 0.55x = 19.25

##### ⇒ x = 19.25/0.55

##### ⇒ x = 35

**Short Trick:**

##### Let total number of students be x after joining 5 students, then

##### 0.55x = 5 (62.8 – 58.4)

##### ⇒ x = (5 × 4.4)/0.55

##### ⇒ x = 40

##### ∴ Number of students in the class were = 40 – 5 = 35

**Q.21 The value of [latex]frac { 5frac { 1 }{ 2 } div 3frac { 2 }{ 3 } offrac { 1 }{ 4 } +(5frac { 1 }{ 9 } -7frac { 7 }{ 8 } div 9frac { 9 }{ 20 } )times frac { 9 }{ 11 } }{ 5÷5offrac { 1 }{ 10 } −10×10÷20 }[/latex] is equal to:**

#### 1. [latex]1frac { 4 }{ 5 }[/latex]

#### 2. [latex]9frac { 1 }{ 2 }[/latex]

#### 3. [latex]3frac { 4 }{ 5 }[/latex]

#### 4. [latex]1frac { 9 }{ 10 }[/latex]

#### Correct answer is **Option 4 i.e. ****1frac { 9 }{ 10 }**

#### frac { 5frac { 1 }{ 2 } div 3frac { 2 }{ 3 } offrac { 1 }{ 4 } +(5frac { 1 }{ 9 } -7frac { 7 }{ 8 } div 9frac { 9 }{ 20 } )times frac { 9 }{ 11 } }{ 5÷5offrac { 1 }{ 10 } −10×10÷20 }

#### = frac { frac { 11 }{ 2 } div frac { 11 }{ 3 } times frac { 1 }{ 4 } +(frac { 46 }{ 9 } -frac { 63 }{ 8 } div frac { 189 }{ 20 } )times frac { 9 }{ 11 } }{ 5÷5×frac { 1 }{ 10 } -10times frac { 1 }{ 2 } }

#### = frac { frac { 11 }{ 2 } div frac { 11 }{ 12 } +(frac { 46 }{ 9 } -frac { 63 }{ 8 } times frac { 20 }{ 189 } )times frac { 9 }{ 11 } }{ 5÷frac { 1 }{ 2 } -5 }

#### = frac { frac { 11 }{ 2 } times frac { 12 }{ 11 } +(frac { 46 }{ 9 } -frac { 5 }{ 6 } )times frac { 9 }{ 11 } }{ 5times 2-5 }

#### = frac { 6+(frac { 92-15 }{ 18 } )times frac { 9 }{ 11 } }{ 10-5 }

#### = frac { 6+frac { 77 }{ 18 } times frac { 9 }{ 11 } }{ 5 }

#### = frac { 6+frac { 7 }{ 2 } }{ 5 }

#### = frac { frac { 19 }{ 2 } }{ 5 }

#### = frac { 19 }{ 10 }

#### = **1frac { 9 }{ 10 } **

**Q.22 In ΔPQR, ∠Q = 85° and ∠R = 65°. Points S and T are on the sides PQ and PR, respectively such that ∠STR = 95° and the ratio of the QR and ST is 9 : 5. If PQ = 21.6 cm, then the length of PT is:**

##### 1. 12 cm

2. 10.5 cm

3. 9.6 cm

4. 9 cm

##### Correct answer is **Option 1 i.e. ****12 cm**

##### Given, QR : ST = 9 : 5, PQ = 21.6 cm, ∠Q = 85°, ∠R = 65 and ∠STR = 95°

##### From the following figure

##### ΔPTS ∼ PQR

##### ⇒ QR/TS = PQ/PT

##### ⇒ 9/5 = 21.6/PT

##### ⇒ PT = (21.6 × 5)/9

##### ⇒ PT = 12 cm

**Q.23 A and B, working together, can complete a work in d days. Working alone, A takes (8 + d) days and B takes (18 + d) days to complete the same work. A works for 4 days. The remaining work will be completed by B alone, in:**

##### 1. 18 days

##### 2. 24 days

##### 3. 16 days

##### 4. 20 days

##### Corrrect answer is **Option 2 i.e. ****24 days**

##### According to the question

##### 1/(8 + d) + 1/(18 + d) = 1/d

##### ⇒ (18 + d + 8 + d)/(144 + 26d + d^{2}) = 1/d

##### ⇒ 144 + 26d + d^{2} = 26d + 2d^{2}

##### ⇒ d^{2} = 144

##### ⇒ d = √144

##### ⇒ d = 12 days

##### A and B together, can complete the whole work = 12 days

##### A alone can complete the whole work in = 12 + 8 = 20 days

##### B alone can complete the whole work in = 12 + 18 = 30 days.

##### Let the remaining work B complete in x days, then

##### According to the question

##### 5/20 + x/30 = 1

##### ⇒ x/30 = 1 – 1/4

##### ⇒ x/30 = 4/5

##### ⇒ x = 24 days

**Short Trick:**

##### D = √(8 × 18)

##### ⇒ D = 12

##### A and B together, can complete the whole work = 12 days

##### A alone can complete the whole work in = 12 + 8 = 20 days

##### B alone can complete the whole work in = 12 + 18 = 30 days.

##### Total work = 60

##### Work done by A in 4 days = 4 × 3 = 12

##### Remaining work = 60 – 12 = 48

##### ∴ Remaining work, B complete in = 48/2 = 24 days

**Q.24 The marked price of an article is Rs. 740. After two successive discounts of 15% and x%, it is sold for Rs. 566.10. What is the value of x?**

##### 1. 10

2. 12

3. 5

4. 20

##### Correct answer is **Option 1 i.e. ****10 **

##### MP of the article = Rs. 740

##### After getting 15% discount the first SP of the article = 740 × (85/100) = 629

##### Second SP of the article = Rs. 566.10

##### Second discount = 629 – 566.10 = 62.9

##### ∴ Second discount percentage = (62.9/629) × 100 = 10%

**Q.25 If 30x**^{2} – 15x + 1 = 0, then what is the value of 25x^{2} + (36x^{2})^{–1 }?

^{2}– 15x + 1 = 0, then what is the value of 25x

^{2}+ (36x

^{2})

^{–1 }?