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**Q.1 The value of \frac { 8÷[(8−3)÷{ (4÷4of8)+4−4×4÷8 }−2] }{ 8×8÷4−8÷8of2−7 } is :**

1. \frac { 2 }{ 17}

##### 2. \frac { 8 }{ 3}

##### 3. \frac { 17 }{ 8}

##### 4. \frac { 16 }{ 170}

#### Correct answer is **Option 2 i.e.****\frac { 8 }{ 3} **

#### \frac { 8÷[(8−3)÷{ (4÷4of8)+4−4×4÷8 }−2] }{ 8×8÷4−8÷8of2−7 }

#### = \frac { 8÷[5÷{ (4÷32)+4−4×\frac { 1 }{ 2 } }−2] }{ 8×2−8÷16−7 }

#### = \frac { 8\div \left[ 5\div \left\{ \frac { 1 }{ 8 } +4-2 \right\} -2 \right] }{ 16-\frac { 1 }{ 2 } -7 }

#### = \frac { 8÷\left[ 5÷\frac { 17 }{ 8 } -2 \right] }{ 9-\frac { 1 }{ 2 } }

#### = \frac { 8\div \left[ 5\times \frac { 8 }{ 17 } -2 \right] }{ 2-\frac { 1 }{ 2 } -7 }

#### = \frac { 8\div \left[ \frac { 40 }{ 17 } -2 \right] }{ \frac { 17 }{ 2 } }

#### = \frac { 8\div \frac { 6 }{ 17 } }{ \frac { 17 }{ 2 } }

#### = \frac { 8\times \frac { 17 }{ 6 } }{ \frac { 17 }{ 2 } }

#### = 8\times \frac { 17 }{ 6 } \times \frac { 2 }{ 17 }

#### = **\frac { 8 }{ 3 } **

**Q.2 A sum of Rs. 8,000 invested at 10% p.a. amounts to Rs. 9,261 in a certain time, interest compounded half–yearly. What will be the compound interest (in Rs) on the same sum for the same time at double the earlier rate of interest, when interest is compounded annually?**

##### 1. Rs. 2,520

##### 2. Rs. 2,560

##### 3. Rs. 2,500

4. Rs. 2,480

##### The correct answer is__ option 2 i.e __**Rs. 2,560**

##### P = 8000, A = 9261 and r = 10%

##### If interest compounded half–yearly, then = 10/2 = 5%

##### As we know,

##### A = P (1 + r/100)^{t}

##### ⇒ 9261 = 8000 (1 + 5/100)^{t}

##### ⇒ (1 + 1/20 )^{t} = 9261/8000

##### ⇒ (21/20)^{t} = (21/20)^{3}

##### ⇒ t = 3 years

##### As given, interest compounded half–yearly, then

##### Time = 3/2 years [1 year and 6 month]

##### Now,

##### P = 8000, r = 10 × 2 = 20% and t = 3/2

##### As we know,

##### A = P (1 + r/100)^{t}

##### A = 8000 (1 + 20/100) × (1 + 10/100)

##### ⇒ A = 8000 × 120/100 × 110/100

##### ⇒ A = 10560

##### CI = 10560 – 8000 = 2560

**Short Trick:**

##### As we know,

##### P : A for 1 year = ∛8000 : ∛9261 = 20 : 21

##### Percentage = 1/20 × 100 = 5% (satisfied)

##### Time = 1 year 6 month

##### Now, rate = 20%

##### Rate for 1 year 6 month = 20 + 10 + (20 × 10)/100 = 30 + 2 = 32%

##### ⇒ 32% of 8000

##### ⇒ 2560

**Q.3 If secθ + tanθ = p, 0°< θ < 90°, then \frac { { p }^{ 2 }+1 }{ { p }^{ 2 }-1 } is equal to :**

##### 1. sin θ

##### 2. 2 cosec θ

##### 3. cosec θ

##### 4. cos θ

##### Correct answer is **option 1 i.e. sin θ**

##### sec θ + tan θ = P —— (1)

##### As we know,

##### (sec^{2}θ – tan^{2} θ) = 1

##### ⇒ (sec θ + tan θ) (sec θ – tan θ) = 1

##### ⇒ sec θ – tan θ = 1/p —— (2)

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

##### ⇒ 2 sec θ = p + 1/p

##### ⇒ 2 sec θ = (p^{2} + 1)/P

##### ⇒ sec θ = (P^{2} + 1)/2p

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

##### ⇒ sec θ = H/B = (p^{2} + 1)/2p

##### ⇒ H = (p^{2} + 1) and B = 2p

##### As we know,

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

##### ⇒ (p^{2} + 1)^{2} = (2p)^{2} + P^{2}

##### ⇒ p^{2} = p^{4} + 1 + 2p^{2} – 4p^{2}

##### ⇒ p^{2} = p^{4} + 1 – 2p^{2}

##### ⇒ p^{2} = (p^{2} – 1)^{2}

##### ⇒ p = (p^{2} – 1)

##### As we know

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

**Q.4 In ∆ABC, ∠B = 72° and ∠C = 44°, Side BC is produced to D. The bisectors of ∠B and ∠ACD meet at E. what is the measure of ∠BEC?**

##### 1. 32°

2. 46°

3. 36°

4. 58°

##### Correct answer is **Option 1 i.e. 32°**

**Short trick:**

##### ∠B = 72° and ∠C = 44°

##### ∠A = 180° – 72° – 44° = 64°

##### ∠BEC = ∠B/2 = 64°/2 = 32°

**Detailed Method:**

##### In ΔABC,

##### ∠B = 72° and ∠C = 44°

##### ⇒ ∠A + ∠B + ∠C = 180°

##### ⇒ 72° + ∠B + 44° = 180°

##### ⇒ ∠B = 180° – 72° – 44° = 64°

##### ⇒ ∠ACD = 180° – ∠C = 180° – 44° = 136°

##### ⇒ ∠ACE = ∠ACD/2 = 136°/2 = 68°

##### ⇒ ∠BCE = ∠BCA + ∠ACE = 44° + 68° = 112°

##### ⇒ ∠EBC = ∠B/2 = 72°/2 = 36°

##### In ΔBCE,

##### ∠EBC + ∠BCE + ∠BEC = 180°

##### ⇒ ∠BEC = 180° – 36° – 112° = 32°

**Q.5 The given table represents the production of different types of motorcycles (in thousands) over a period of six years. Study the table carefully and answer the question that follows.**

##### During 2015, the production of which type of motorcycle was more than 25% of the total production of all types of motorcycles in 2017?

##### 1. A

2. C

3. B

4. D

##### Correct answer is option 4 i.e. D

##### Total production of motorcycles in 2017 = 80 + 96 + 100 + 104 = 380

##### ⇒ 25% of 380

##### ⇒ 380 × 1/4

##### ⇒ 95

##### In 2015, D = 102 is only more than 95

##### Now it clear in 2015 production of type D was more than 25% more than total production in 2017.

**Q.6 The average age of a number of persons in a group was calculated as 35 years, which was 2.5 years more than the correct average as there was an error in recording the age to two persons as 38.5 years and 40 years instead of 29 years and 22 years respectively. The number of persons in the group was:**

1. 15

##### 2. 13

##### 3. 12

##### 4. 11

##### Correct answer is option 4 i.e.** ****11**

##### Let the number of persons in the group be x, then

##### Wrong average of age of persons in the group = 35

##### Wrong sum of age of person in the group = 35

##### Correct average of age of persons in the group = 35 – 2.5 = 32.5

##### Correct sum of age of persons in the group = 32.5x

##### According to the question

##### 35x – 32.5x = 38.5 + 40 – 29 – 22

##### ⇒ 2.5x = 27.5

##### ⇒ x = 27.5/2.5

##### ⇒ x = 11

**Short Trick:**

##### Total number of persons in the group = (38.5 + 40 – 29 – 22)/2.5 = 11

**Q.7 A sum of Rs. x was divided between A, B, C and D in the ratio 1/3 : 1/5 : 1/6 : 1/9. If the difference between the shares of B and D is Rs. 832, then the value of x is:**

##### 1. Rs. 7,384

2. Rs. 7,488

3. Rs. 7,696

4. Rs. 7,592

##### Correct answer is __option 4 i.e. __**Rs. 7,592**

##### Ratio of sum of A, B, C and D = 1/3 : 1/5 : 1/6 : 1/9 = 90/3 : 90/5 : 90/6 : 90/9 = 30 : 18 : 15 : 10

##### According to the question

##### ⇒ 18 – 10 = 8 unit

##### ⇒ 8 unit = 832

##### ⇒ 1 unit = 104

##### ⇒ Total sum = 30 + 18 + 15 + 10 = 73 unit

##### ⇒ 73 unit = 104 × 73 = 7592

##### So, total sum is 7592.

**Q.8 The area of a field in the shape of a regular hexagon is 2400√3 m**^{2}. The cost of fencing the field at Rs. 16.80/metre is:

^{2}. The cost of fencing the field at Rs. 16.80/metre is:

##### 1. Rs. 4,032

2. Rs. 3,528

3. Rs. 4,536

4. Rs. 3,024

##### Correct answer is option 1 i.e. **Rs. 4,032**

##### As we know,

##### Perimeter of the regular hexagon = 6a

##### Area of regular hexagon = 6 × (√3/4) × a^{2}

##### 6 × (√3/4) × a^{2} = 2400√3

##### a^{2}= 2400√3 × 4/√3 × 1/6

##### ⇒ a^{2} = 1600

##### ⇒ a = √1600

##### ⇒ a = 40

##### Perimeter of regular hexagon = 6 × 40 = 240 m

##### Cost of fencing 1 m = Rs. 16.80

##### ∴ Cost of fencing 2400 m = 240 × 16.80 = Rs. 4032

**Q.9 If the selling price of an article is 8% more than the cost price and the discount offered is 10% on the marked price of the article, then what is the ratio of the cost price to the marked price?**

1. 8 : 9

2. 5 : 6

3. 4 : 5

4. 3 : 4

##### Correct answer is option 2 i.e ** ****5 : 6**

**Short Tric****k:**

##### Ratio of CP to MP = (100% – Discount) : (100 + Profit) = (100 – 10) : (100 + 8) = 90 : 108 = 5 : 6

**Detailed Method:**

##### Let the cost price be Rs. 100

##### SP of the article = 100 × (108/100) = Rs. 108

##### MP of the article = 108 × (100/90) = 120

##### ∴ Ratio of CP to SP = 100 : 120 = 5 : 6

**Q10. If 12x**^{2} – 21x + 1 = 0, then what is the value of 9x^{2} + (16x^{2})^{–1 }?

^{2}– 21x + 1 = 0, then what is the value of 9x

^{2}+ (16x

^{2})

^{–1 }?

##### 1. \frac { 417 }{ 16 }

2. \frac { 453 }{ 8 }

##### 3. \frac { 465 }{ 16 }

4. \frac { 429 }{ 8 }

##### Correct answer is __Option 1. i.e. __**\frac { 417 }{ 16 } **

__Option 1. i.e.__

##### 12x^{2} – 21x+1 = 0

##### Divide by 4x

##### ⇒ 3x – 21/4 + 1/4x = 0

##### ⇒ 3x + 1/4x = 21/4

##### Squaring on both sides

##### ⇒ 9x^{2} + 1/16x^{2} + 2 × 3x × (1/4x) = 441/16

##### ⇒ 9x^{2} + 1/16x^{2} = 441/16 – 3/2

##### ⇒ 9x^{2} + 1/16x^{2} = (441 – 24)/16

##### ⇒ 9x^{2} + 1/16x^{2} = 417/16

**Q.11 If 2x**^{2} + y^{2} +8z^{2}– 2√2xy + 4√2yz – 8zx = (Ax + y + Bz)^{2}, then the value of (A^{2 }+ B^{2}– AB)is:

^{2}+ y

^{2}+8z

^{2}– 2√2xy + 4√2yz – 8zx = (Ax + y + Bz)

^{2}, then the value of (A

^{2 }+ B

^{2}– AB)is:

##### 1. 16

##### 2. 6

##### 3. 14

##### 4. 18

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

##### As we know,

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

##### ⇒ 2x^{2} + y^{2} + 8z^{2}– 2√2xy + 4√2yz – 8zx = (Ax + y + Bz)^{2}

##### ⇒ (√2 x)^{2} + y^{2} + (2√2 z)^{2} – 2 × (√2 x) × y + 2 × y × 2√2z – 2 × 2√2 z × √2 x = (Ax + y + Bz)^{2}

##### ⇒ (√2x –y – 2√2 z)^{2} = (Ax + y + Bz)^{2}

##### Comparing on both sides

##### A = √2 and B = – 2√2

##### Now,

##### (A^{2} + B^{2} – AB)

##### ⇒ [(√2)^{2} + (–2√2)^{2} – √2 × (–2√2)]

##### ⇒ [2 + 8 + 4]

##### ⇒ 14

**Q.12 A sells an article to B at a loss of 20%, B sells it to C at a profit of 12.5% and C sells it to D at a loss of 8%. If D buys it for Rs. 248.40, then what is the difference between the loss incurred by A and C’?**

##### 1. Rs. 36.80

##### 2. Rs. 39.20

##### 3. Rs. 38.40

##### 4. Rs. 42.60

##### Correct answer is **Option 3 i.e. ****Rs. 38.40**

##### Let A buy an article be 100 unit,

##### A sell the article to B = 100 × (80/100) = 80 unit

##### B sell the article to C = 80 × (9/8) = 90 unit

##### C buy the article = 90 unit

##### C sell the article to D = 90 × (92/100) = 82.8 unit

##### D buy the article = 82.8 unit

##### Loss of A = 100 – 80 = 20 unit

##### Loss of C = 90 – 82.8 = 7.2 unit

##### Difference of loss A and C = 20 – 7.2 = 12.8 unit

##### ⇒ 82.8 unit = Rs. 248.40

##### ⇒ 1 unit = 248.40/82.8 = Rs. 3

##### ⇒ 12.8 unit = Rs. 3 × 12.8 = 38.40

**Q.13 Places A and B are 144 km apart. Two cars start simultaneously, one from A and the other from B. If they move in the same direction, they meet after 12 hours, but if they move towards each other they meet after 9/8 hours. The speed (in km/h) of the car moving at a faster speed is:**

##### 1. 72

##### 2. 60

##### 3. 70

##### 4. 64

##### Correct Answer is **option 3 i.e. ****Rs. 70**

##### Distance between A and B = 144

##### Let the speed of first and second car be x km/hr and y km/hr

##### If both cars going same direction, then relative speed = (x – y) km/hr

##### If both cars going opposite direction, then relative speed = (x + y) km/hr

##### As we know,

##### Speed = Distance/Time

##### According to the question

##### (x + y) = 144/(9/8)

##### ⇒ (x + y) = 128 –––(1)

##### ⇒ (x – y) = 144/12

##### ⇒ (x – y) = 12 –––(2)

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

##### ⇒ x = 70

##### So, speed of first car = 70 km/hr

**Q.14 Four men and 6 women can complete a certain piece of work in 5 days whereas three men and 4 women can complete it in 7 days. How many men should assist 25 women to complete 2\frac { 1 }{ 2 } times the same work in 5 days?**

1. 4

2. 8

3. 10

4. 5

##### Correct answer is **Option 4 i.e. 5**

##### Let efficiency of 1 man and 1 woman be x and y respectively.

##### Total work = (4x + 6y) × 5 —— (1)

##### Total work = (3x + 4y) × 7 —— (2)

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

##### (4x + 6y) × 5 = (3x + 4y) × 7

##### ⇒ 20x + 30y = 21x + 28y

##### ⇒ 21x – 20x = 30y – 28y

##### ⇒ x = 2y

##### ⇒ x : y = 2 : 1

##### Efficiency of 1 man = 2

##### Efficiency of 1 woman = 1

##### Total work = (4x + 6y) × 5 = (4 × 2 + 6 × 1) × 5 = 70

##### 2 times of total work = 70 × 5/2 = 175

##### Let the number of men be n, then

##### According to the question

##### (2n + 1 × 25) 5 = 175

##### ⇒ 2n + 25 = 175/5 = 35

##### ⇒ 2n = 35 – 25 = 10

##### ⇒ n = 10/2 = 5

##### So, the number of men are 5.

**Q.15 How many numbers are there from 200 to 800 which are neither divisible by 5 nor by 7?**

##### 1. 410

2. 411

##### 3. 407

##### 4. 413

##### Correct answer is **Option 2 i.e. 411**

##### Total number between 199 and 800 or (from 200 to 800) = 800 – 200 + 1 = 601

##### Total number from 200 to 800 which are divisible by 5 = 800/5 – 199/5 = 160 – 39 = 121

##### Total number from 200 to 800 which are divisible by 7 = 800/7 – 199/7 = 114 – 28 = 86

##### Total number from 200 to 800 which are divisible by 35 = 800/35 – 200/35 = 22 – 5 = 17

##### ∴ Number neither divisible by 5 nor 7 = 601 – 121 – 86 + 17 = 411

**Q.16 If 7 cos**^{2}θ + 3sin^{2}θ = 6, 0° < θ < 90°, then the value of \frac { { cot }^{ 2 }2θ+{ sec }^{ 2 }2θ }{ { tan }^{ 2 }2θ−{ sin }^{ 2 }2θ } Is :

^{2}θ + 3sin

^{2}θ = 6, 0° < θ < 90°, then the value of \frac { { cot }^{ 2 }2θ+{ sec }^{ 2 }2θ }{ { tan }^{ 2 }2θ−{ sin }^{ 2 }2θ } Is :

##### 1. \frac { 28 }{27 }

##### 2. \frac { 52 }{27 }

##### 3. \frac { 26 }{15 }

##### 4. \frac { 49 }{45 }

##### Correct answer is **Option 2 i.e. ****\frac { 52 }{27 } **

##### 7 cos^{2}θ + 3sin^{2}θ = 6,

##### ⇒ 4 cos^{2} θ + 3 cos^{2}θ + 3sin^{2}θ = 6

##### ⇒ 4 cos^{2 }θ + 3 = 6

##### ⇒ 4 cos^{2} θ = 6 – 3

##### ⇒ 4 cos^{2} θ = 3

##### ⇒ cos^{2} θ = 3/4

##### ⇒ cos θ = √(3/4) = √3/2

##### ⇒ cos θ = cos 30

##### ⇒ θ = 30

##### Now,

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

##### = **\frac { { cot }^{ 2 }60+{ sec }^{ 2 }60 }{ { tan }^{ 2 }60−{ sin }^{ 2 }60 } **

##### = \frac { { \left( \frac { 1 }{ \sqrt { 3 } } \right) }^{ 2 }+{ 2 }^{ 2 } }{ { \left( \sqrt { 3 } \right) }^{ 2 }-\quad { \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 } }

##### = \frac { \frac { 1 }{ 3 } +4 }{ 3-\frac { 3 }{ 4 } }

##### = \frac { \frac { 13 }{ 3 } }{ \frac { 9 }{ 4 } }

##### = \frac { 13 }{ 3 } \times \frac { 4 }{ 9 }

##### = \frac { 52 }{ 27 }

**Q.17 In ∆ABC, AC = 8.4 cm and BC = 14 cm, P is a point on AB such that CP = 11.2 cm and ∠ACP =∠B. what is the length (in cm) of BP?**

##### 1. 2.8

2. 4.12

3. 3.78

4. 3.6

##### Correct answer is **Option 3 i.e. 3.78**

##### From the following figure,

##### If ∠ACP = ∠B and ∠A common, then

##### ΔACP∼ΔABC

##### As we know,

##### AC/AB = PC/BC

##### ⇒ 8.4/AB = 11.2/14

##### ⇒ AB = (8.4 × 14)/11.2 = 10.5

##### Again,

##### AP/AC = PC/BC

##### ⇒ AP/8.4 = 11.2/14

##### ⇒ AP = (11.2 × 8.4)/14 = 6.72

##### As we know,

##### ∴ BP = AB – AP = 10.5 – 6.72 = 3.78

**Q18. The value of \frac { (cos9°+sin81°)(sec9°+cosec81°) }{ \left( { 2sin }^{ 2 }63°+1+{ 2sin }^{ 2 }27° \right) } is :**

##### 1. 2

##### 2. \frac { 4}{ 3 }

##### 3. 1

##### 4. \frac { 1}{ 2 }

##### Correct answer is **Option 2 i.e. ****\frac { 4}{ 3 } **

**\frac { (cos9°+sin81°)(sec9°+cosec81°) }{ \left( { 2sin }^{ 2 }63°+1+{ 2sin }^{ 2 }27° \right) } **

##### = \frac { [cos9°+cos(90°−81°)][sec9°+sec(90°−81°)] }{ { 2sin }^{ 2 }63°+1+2\left[ { sin }^{ 2 }\left( 90°-63° \right) \right] }

##### = \frac { [cos9°+cos9°][sec9°+sec9°] }{ { 2sin }^{ 2 }63°+1+2{ cos }^{ 2 }63° }

##### = \frac { 2cos9×2sec }{ 2\left( { sin }^{ 2 }63°+{ cos }^{ 2 }63° \right) +1 }

##### As we know,

##### ⇒ cosθ secθ = 1

##### ⇒ sin2θ + cos2 θ = 1

##### = \frac { 2cos9×2sec }{ 2\left( { sin }^{ 2 }63°+{ cos }^{ 2 }63° \right) +1 }

##### = \frac { 4 }{ 3 }

**Q.19 Two circles of radii 7 cm and 5 cm intersect each other at P and Q, and the distance between their centres is 10 cm. The length (in cm) of the common chord PQ is:**

##### 1. \frac { 2\sqrt { 74 } }{ 5 }

2. \frac { 4\sqrt { 66 } }{ 5 }

3. \frac { 3\sqrt { 66 } }{ 5 }

4. \frac { 3\sqrt { 74 } }{ 5 }

##### Correct answer is **Option 2 i.e. **** \frac { 4\sqrt { 66 } }{ 5 } **

##### From the following figure,

##### AP = 7 cm, PB = 5 cm, and AB = 10 cm

##### Let AE = x cm, so EB = (10 – x) cm

##### The length of the common chord PQ

##### In right angled ΔAPE,

##### AP^{2} = PE^{2} + AE^{2}

##### PE^{2} = 7^{2} – x^{2} –––(1)

##### In right angled ΔPBE,

##### PB^{2} = PE^{2} + BE^{2}

##### PE^{2} = 5^{2} – (10 – x)^{2}

##### PE^{2} = 25 – 100 – x^{2} + 20x —— (2)

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

##### 7^{2} – x^{2} = 25 – 100 – x^{2} + 20x

##### ⇒ 20x = 100 + 49 – 25

##### ⇒ x = 124/20

##### ⇒ x = 31/5

##### From equation (1)

##### PE^{2} = 49 – (31/5)^{2} = 49 – 961/25

##### ⇒ PE^{2} = (1225 – 961)/25 = 264/25

##### ⇒ PE = √264/25

##### ⇒ PE = 2√66/5

##### As we know,

##### PQ = 2PE

##### ⇒ PQ = 2 × 2√66/5

##### ⇒ PQ = 4√66/5

**Q.20 A, B and C donate 8%, 7% and 9%, of their salaries, respectively to a charitable trust. The salaries of A and B are same and the different between their donations is Rs. 259. The total donation of A and B is Rs. 1,185 more than that of C. the total donation of A and C is what percentage of the total salaries of A, B and C? (Correct to one decimal place)**

1. 6.4%

2. 7.1%

3. 6.2%

4. 5.8%

##### Correct answer is **Option 4 i.e. ****5.8%**

##### Let salaries of A and B be Rs. x and salary of C be Rs. y, then

##### 8x/100 – 7x/100 = 259

##### ⇒ x/100 = 259

##### ⇒ x = 25900

##### Salaries of A and B is Rs. 25900.

##### Donation of A and B = 25900 × (8 + 7)/100 = 25900 × 15/100 = 3885

##### Donation of C = 3885 – 1185 = 2700

##### y × (9/100) = 2700

##### ⇒ y = 30,000

##### Sum of salaries of A, B and C = 25900 + 25900 + 30000 = 81800

##### Donation of A = 259 × (8/100) = 2072

##### Total donation of A and C = 2072 + 2700 = 4772

##### Percentage of total donation of total salaries of A, B and C = (4772/81800) × 100 = 5.8%

**Short Trick:**

##### Let salaries of A, B be 100%

##### ⇒ 8% – 7% = 259

##### ⇒ 1% = 259

##### ⇒ 8% = 2072

##### ⇒ 15% = 3885

##### ⇒ 100% = 25900

##### Donation of C = 3885 – 1185 = 2700

##### ⇒ 9% = 2700

##### ⇒ 100% = 30,000

##### Total donation of A and C = 2072 + 2700 = 4772

##### Total salaries of A, B and C = 25900 + 25900 + 30000 = 81800

##### ∴ Required percentage = 4772/81800 × 100 = 5.8%

**Q.21 The given table represents the production of different types of motorcycles (in thousands) over a period of six years. Study the table carefully and answer the question that follows.**

##### By what percentage is the total production of type A motorcycles over six years, less than the production of all types of motorcycles in 2013 and 2016?

##### 1. 31.6

##### 2. 32.8

##### 3. 30.5

##### 4. 32.2

##### Correct answer is **Option 3 i.e. ****30.5**

##### Total production of type A motorcycles over six years = 95 + 84 + 85 + 89 + 80 + 98 = 531

##### Total production of all types of motorcycles in 2013 and 2016 = 95 + 98 + 104 + 103 + 89 + 88 + 92 + 95 = 764

##### ∴ Required percentage = [(764 – 531)/764] × 100 = 30.5%

**Q.22 If x + y + z = 3, and x**^{2} + y^{2} + z^{2} = 101, then what is the value of \sqrt { { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }−3xyz } ?

^{2}+ y

^{2}+ z

^{2}= 101, then what is the value of \sqrt { { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }−3xyz } ?