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**Q.1 ∆ABC, ∠A = 90°, M is the midpoint of BC and D is a point on BC such that AD⊥ BC. If AB = 7 cm and AC = 24 cm, then AD : AM is equal to:**

1. 32 : 43

##### 2. 24 : 25

##### 3. 336 : 625

##### 4. 168 : 275

##### Correct answer is **Option 3 i.e. ****336 : 625**

##### In right angled ∆ABC, 90° at A

##### BC^{2} = AB^{2} + AC^{2} = 7^{2} + 24^{2} = 49 + 576 = 625

##### ⇒ BC = √625 = 25

##### If, M is the midpoint of BC, then

##### AM = MC = BM = 25/2

##### As we know,

##### AD × BC = AB × AC

##### ⇒ AD × 25 = 7 × 24

##### ⇒ AD = (7 × 24)/25

##### ∴ Ratio of AD : AM = (7 × 24)/25 : 25/2 = **336 : 625**

**Q.2 On simplification,** ** \frac { { x }^{ 3 }-{ y }^{ 3 } }{ { x[(x+y) }^{ 2 }-3xy] } \div \frac { { y[(x−y) }^{ 2 }+3xy] }{ { x }^{ 3 }+{ y }^{ 3 } } \times \frac { { (x+y) }^{ 2 }-{ (x−y) }^{ 2 } }{ {x}^{2}-{y}^{2} } is equal to :**

##### 1. 4

2. 1

3. \frac {1}{2}

4. \frac {1}{4}

#### The correct answer is__ option 1 i.e __ **4**

#### As we know,

#### (x^{3} + y^{3}) = (x + y) (x^{2} + y^{2} – xy)

#### (x^{3} – y^{3}) = (x – y) (x^{2} + y^{2} + xy)

#### (x + y)^{2} – (x – y)^{2} = 4xy

#### [(x – y)^{2} + 3xy] = [x^{2} + y^{2} – 2xy + 3xy] = [x^{2} + y^{2} + xy]

#### [(x + y)^{2} – 3xy] = [x^{2} + y^{2} + 2xy – 3xy] = [x^{2} + y^{2} – xy]

#### Now,

#### \frac { { x }^{ 3 }-{ y }^{ 3 } }{ { x[(x+y) }^{ 2 }+3xy] } \div \frac { { y[(x−y) }^{ 2 }-3xy] }{ { x }^{ 3 }+{ y }^{ 3 } } \times \frac { { (x+y) }^{ 2 }-{ (x−y) }^{ 2 } }{ { x }^{ 2 }-{ y }^{ 2 } }

#### =\frac { { (x−y)(x }^{ 2 }+{ y }^{ 2 }+xy) }{ { x[(x }^{ 2 }+{ y }^{ 2 }+xy)] } \div \frac { { y[(x }^{ 2 }+{ y }^{ 2 }-xy)] }{ { (x+y)(x }^{ 2 }+{ y }^{ 2 }-xy) } \times \frac { 4xy }{ (x−y)(x+y) }

#### =\frac { { (x−y)(x }^{ 2 }+{ y }^{ 2 }+xy) }{ { x[(x }^{ 2 }+{ y }^{ 2 }+xy)] } \times \frac { { (x+y)(x }^{ 2 }+{ y }^{ 2 }-xy) }{ { y(x }^{ 2 }+{ y }^{ 2 }-xy) } \times \frac { 4xy }{ (x−y)(x+y) }

#### =**4**

**Q.3 If 5x+\frac { 1 }{ 3x }=4 then what is the value of { 9x }^{ 2 }+\frac { 1 }{ { 25x }^{ 2 } } ?**

##### 1. \frac {174 }{125 }

##### 2. \frac {119 }{25 }

##### 3. \frac {144 }{125 }

##### 4. \frac {114 }{25 }

##### Correct answer is **option 4 i.e. \frac {114 } {25 }**

##### 5x + 1/3x = 4

##### Multiply by 3/5

##### 3x + 1/5x = 12/5

##### Squaring on both sides

##### (3x + 1/5x)^{2} = (12/5)^{2}

##### ⇒ 9x^{2} + 1/25x^{2} + 2 × 3x × (1/5x) = 144/25

##### ⇒ 9x^{2} + 1/25x^{2} = 144/25 – 6/5

##### ⇒ 9x^{2} + 1/25x^{2} = (144 – 30)/25 = **114/25**

**Q.4 The given table represents the number of computers sold by four dealers A, B, C and D during the first six months of 2016. Study the table carefully and answer the question that follows.**

##### 1. 25.3

2. 21.2

3. 24.4

4. 17.5

##### Correct answer is **Option 2 i.e. 21.2**

##### Total computers sold by dealer A from February to June = 94 + 85 + 108 + 98 + 95 = 480

##### Total number of computers sold by all the dealers in June = 95 + 108 + 102 + 91 = 396

##### ∴ Required percentage [(480 – 396)/396] × 100 = 21.2

**Q.5 The average of twelve numbers is 45.5. The average of the first four numbers is 41.5 and that of the next five numbers is 48. The 10**^{th} number is 4 more than the 11^{th} number and 9 more than the 12^{th} number. What is the average of the 10^{th} and 12^{th} numbers?

^{th}number is 4 more than the 11

^{th}number and 9 more than the 12

^{th}number. What is the average of the 10

^{th}and 12

^{th}numbers?

##### 1. 47

2. 47.8

3. 46

4. 46.5

##### Correct answer is option 4 i.e. **46.5**

##### Average of 12 numbers in = 45.5

##### Sum of 12 numbers = 45.5 × 12 = 546

##### Average of first 4 numbers = 41.5

##### Sum of first 4 numbers = 41.5 × 4 = 166

##### Average of next 5 numbers = 48

##### Sum of next 5 numbers = 48 × 5 = 240

##### Sum of 10^{th}, 11^{th} and 12^{th} number = 546 – 166 – 240 = 140

##### Let the 10^{th} number be x

##### 11^{th} number = x – 4

##### 12^{th} number = x – 9

##### According to the question

##### x + x – 4 + x – 9 = 140

##### ⇒ 3x – 13 = 140

##### ⇒ 3x = 140 + 13

##### ⇒ 3x = 153

##### ∴ x = 153/3 = 51

##### Average of 10^{th} and 12^{th} number = (51 + 51 – 9)/2 = 93/2 = 46.5

**Q.6 The value of \frac { tan30°cosec60°+tan60°sec30° }{ { sin }^{ 2 }30°+{ 4cot }^{ 2 }45°-{ sec }^{ 2} 60°} is:**

1. \frac { 2 }{ 3 }

##### 2. \frac { 32 }{ 99 }

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

##### 4. \frac { 32 }{ 3 }

##### Correct answer is option 4 i.e.** ****\frac { 32 }{ 3 }**

**\frac { tan30°cosec60°+tan60°sec30° }{ { sin }^{ 2 }30°+{ 4cot }^{ 2 }45°-{ sec }^{ 2} 60°}**

##### =\frac { \frac { 1 }{ \sqrt { 3 } } \times \frac { 2 }{ \sqrt { 3 } } +\sqrt { 3 } \times \frac { 2 }{ \sqrt { 3 } } }{ \frac { 1 }{ { 2 }^{ 2 } } +4\times 1-{ 2 }^{ 2 } }

##### =\frac { \frac { 2 }{ 3 } +2 }{ \frac { 1 }{ 4 } +4-4 }

##### =\frac { 8 }{ 3 } \times 4

##### = \frac { 32 }{ 3 }

**Q.7 Sonu saves 15% of her income. If her income increases by 20% and she still saves the same amount as before, then what is the percentage increase in her expenditure? (correct to one decimal place)**

##### 1. 23.5

2. 22.8

3. 23.8

4. 24.2

##### Correct answer is __option 1 i.e. __**23.5**

##### Let the income of Sonu be Rs. 100

##### Savings of Sonu = 100 × (15/100) = Rs. 15

##### Expenditure of Sonu = 100 – 15 = Rs. 85

##### New income of Sonu = 100 × (120/100) = Rs. 120

##### If savings amount same, then new expenditure = 120 – 15 = 105

##### Her expenditure increased by = 105 – 85 = 20

##### Her expenditure increased by = 20/85 × 100 ≈ 23.5%

**Q.8 When 732 is divided by a positive integer x, the remainder is 12. How many values of x are there?**

##### 1. 19

2. 18

3. 20

4. 16

##### Correct answer is option 3 i.e. **20**

##### When 732 is divided by a positive integer x, the remainder is 12. So,

##### 732 – 12 = 720

##### So 720 will be completely divisible by x

##### 720 = 2^{4} × 3^{2} × 5^{1}

##### Total number of factors of 720 = (4 + 1) × (2 + 1) × (1 + 1) = 5 × 3 × 2 = 30

##### Number of factors 12 or less than 12 are (10) = 1, 2, 3, 4, 5, 6, 8, 9, 10, 12

##### So x cannot have these values because x is greater than 12

##### So, we can say that there are 30 – 10 = **20** possible values of x

**Q.9 The circumference of the base of a conical tent is 66 m. If the height of the tent is 36 m, what is the area (in m**^{2}) of the canvas used in making the tent? (Take π = 22/7)

^{2}) of the canvas used in making the tent? (Take π = 22/7)

1. 1237.5

2. 1171.5

3. 1254

4. 1155

##### Correct answer is option 1 i.e **1237.5**

##### Height of the conical tent h = 36 m

##### As we know

##### Curved surface area of conical tent = πrl

##### Circumference of the base of a conical tent = 2 πr

##### 2 πr = 66

##### ⇒ 2 × (22/7) × r = 66

##### ⇒ r = 66 × (7/22) × (1/2)

##### ⇒ r = 21/2

##### As we know,

##### l^{2} = r^{2} + h^{2}

##### ⇒ l^{2} = (21/2)^{2} + 36^{2}

##### ⇒ l^{2} = 441/4 + 1296

##### ⇒ l^{2} = (441 + 5184)/4 = 5625/4

##### ⇒ l = √5625/4

##### ⇒ l = 75/2

##### ∴ Curved surface area of conical tent = πrl = (22/7) × (21/2) × (75/2) = 1237.5 m^{2}

**Q10. If a + b + c = 11, ab + bc + ca = 3 and abc = -135, then what is the value of a**^{3} + b^{3} + c^{3}?

^{3}+ b

^{3}+ c

^{3}?

##### 1. 827

2. 929

##### 3. 823

4. 925

##### Correct answer is __Option 1. i.e. 827 __

__Option 1. i.e. 827__

##### a + b + c = 11, ab + bc + ca = 3 and abc = -135

##### 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} – 3 × (-135) = 11 [11^{2} – 3 × 3]

##### ⇒ a^{3} + b^{3} + c^{3} + 405 = 11 × [121 – 9] = 11 × 112

##### ⇒ a^{3} + b^{3} + c^{3} = 1232 – 405 = **827**

**Q.11 ∆ABC, AB = AC. A circle drawn through B touches AC at D and intersects AB at P. If D is the midpoint of AC and AP = 2.5 cm, then AB is equal to:**

1. 9 cm

2. 10 cm

3. 7.5 cm

4. 12.5 cm

##### correct answer is **option 2 i.e. 10cm**

##### AB = AC

##### AC = 2 AD

##### As we know,

##### AD^{2} = AP × AB

##### ⇒ AD^{2} = 2.5 × AC

##### ⇒ AD^{2} = 2.5 × 2AD

##### AD = 5 cm

##### ⇒ AC = 2 × 5 = 10 cm

##### ∴ AB = AC = 10 cm

**Q.12 If 5sin**^{2}θ + 14 cosθ = 13, 0° < θ < 90°, then what is the value of \frac { secθ+cotθ }{ cosecθ+tanθ } ?

^{2}θ + 14 cosθ = 13, 0° < θ < 90°, then what is the value of \frac { secθ+cotθ }{ cosecθ+tanθ } ?