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**Q.1 If the difference between 62% and 80% of a number is 198, then the difference between 92% and 56% of the number will be:**

1. 3564

##### 2. 360

##### 3. 396

##### 4. 1100

##### Correct answer is **Option 3 i.e. 396**

##### Let the number be x, then

##### According to the question

##### x × (80/100) – (62/100) = 198

##### ⇒ 18x/100 = 198

##### ⇒ x = 198 × (100/18)

##### ⇒ x = 1100

##### ⇒ 1100 × (92/100) – 1100 × (56/100)

##### ⇒ 11 (92 – 56)

##### ⇒ 11 × 36

##### ⇒ 396

**Short Trick:**

##### ⇒ 80% – 62% = 18%

##### ⇒ 92% – 56% = 36%

##### ⇒ 18% = 198

##### ⇒ 36% = 198 × 2 = **396**

**Q.2 Two racers run at a speed of 100 m/min and 120 m/min, respectively. If the second racer takes 10 minutes less than the first to complete the run, then how long is the race?**

##### 1. 2 km

2. 4 km

3. 6 km

4. 1 km

##### The correct answer is__ option 3__: **6 Km**

##### Speed ratio of two racers = 100 : 120 = 5x : 6x

##### As we know,

##### Time is inversely proportional to speed.

##### Time ratio of two racers = 6x : 5x

##### ⇒ 6x – 5x = 10 min

##### ⇒ x = 10 min

##### Time taken by first racer to complete race = 6 × 10 = 60 min

##### Distance = Speed × Time

##### ∴ Distance = 60 × 100 = **6 km**

**Q.3 In the figure, what is the value of cot R?**

##### 1. 17/18

2. 8/15

3. 15/8

4. 15/17

##### Correct answer is **option 3 i.e. 15/8**

##### As we know,

##### PR^{2} = PQ^{2} + QR^{2}

##### ⇒ 17^{2} = 8^{2} + QR^{2}

##### ⇒ QR = 289 – 64

##### ⇒ QR = √225 = 15

##### ⇒ Cot R = QR/PQ = **15/8**

**Q.4 If 2 sin θ – 8 cos**^{2} θ + 5 = 0, 0° < θ < 90°, then what is the value of (tan 2θ + cosec 2θ)?

^{2}θ + 5 = 0, 0° < θ < 90°, then what is the value of (tan 2θ + cosec 2θ)?

##### 1. \frac { 5\sqrt { 3 } }{ 3 }

2. 2√3

3. 3√3

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

##### Correct answer is Option i.e. \frac { 5\sqrt { 3 } }{ 3 }

**Detailed Method:**

##### 2 sin θ – 8 cos^{2} θ + 5 = 0

##### ⇒ 2 sin θ – 8 (1 – sin^{2} θ) + 5 = 0

##### ⇒ 2 sin θ – 8 + 8 sin^{2} θ + 5 = 0

##### ⇒ 8 sin^{2} θ + 2 sin θ – 3 = 0

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

##### ⇒ 2 sin θ (4 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,

##### tan 2θ + cosec 2θ

##### ⇒ tan 60 + cosec 60

##### ⇒ √3 + 2/√3

##### ⇒ 5/√3

##### ⇒ 5√3/3

**Short Trick:**

##### 2 sin θ – 8 cos^{2} θ + 5 = 0

##### Put θ = 30 and satisfied, then

##### ⇒ tan 60 + cosec 60

##### ⇒ √3 + 2/√3

##### ⇒ 5/√3

##### ⇒ \frac { 5\sqrt { 3 } }{ 3 }

**Q.5 A dealer sold 6 sewing machines for Rs. 63,000 with a profit of 5%. For how much should be sell 8 machines if he intends to earn 15% profit?**

##### 1. Rs. 88,200

2. Rs. 92,400

3. Rs. 69,300

4. Rs. 92,000

##### Correct answer is option 4 i.e.Rs 92,000

##### SP of 6 sewing machines = Rs. 63,000

##### SP of 1 sewing machines = 63,000/6 = Rs. 10,500

##### CP of 1 sewing machine = 10,500 × (100/105) = 10,000

##### New SP of 1 sewing machine = 10,000 × (115/100) = 11,500

##### ∴ New SP of 8 sewing machines = 11,500 × 8 = 92,000

**Q.6 Two members are in the ratio 5 : 7. If the first number is 20, then the second number will be:**

1. 18

2. 22

3. 8

4. 28

##### Correct answer is option 4 i.e. 28

##### Ratio of two numbers = 5x : 7x

##### ⇒ 5x = 20

##### ⇒ x = 20/5

##### ⇒ x = 4

##### ∴ Second number = 7 × 4 = 28

**Q.7 A can finish a work in 20 days and B can finish the same work in 25 days. They began together, but B left the work after 5 days. How many more days will A take to finish the remaining work?**

##### 1. 11

2. 16

3. 8

4. 21

##### Correct answer is __option 1 i.e. 11__

##### A can complete the whole work in = 20 days

##### A’s 1 day’s work = 1/20

##### B can complete the whole work in = 25 days

##### B’s 1 day’s work = 1/25

##### Let A complete the whole work in x days.

##### According to the question

##### ⇒ 5/20 + 5/25 + x/20 = 1

##### ⇒ 1/4 + 1/5 + x/20 = 1

##### ⇒ (5 + 4 + x)/20 = 1

##### ⇒ 9 + x = 20

##### ⇒ x = 20 – 9

##### ⇒ x = 11 days

**Short Trick :**

##### Total work = 100

##### Work done by A and B in 5 days = (5 + 4) × 5 = 45

##### Remaining work = 100 – 45 = 55

##### ∴ Remaining work, A complete in = 55/5 = **11 days**

**Q.8 The number of students enrolled in different faculties in a school is as follows:**

## Science | ## Arts | ## Commerce | ## Vocational | ||||

## Boys | ## Girls | ## Boys | ## Girls | ## Boys | ## Girls | ## Boys | ## Girls |

## 35 | ## 18 | ## 25 | ## 47 | ## 45 | ## 40 | ## 10 | ## 30 |

##### The percentage of students studying in Science and Vocational Subjects is:

##### 1. 25%

2. 37.2%

3. 50%

4. 93%

##### Correct answer is option 2 i.e. **37.2%**

##### Total number of students in Science and Vocational = 35 + 18 + 10 + 40 = 93

##### Total number of students in all four subjects = 35 + 18 + 25 + 47 + 45 + 40 + 10 + 30 = 250

##### ∴ Percentage of students of Science and Vocational subjects = (93/250) × 100 = **37.2%**

**Q.9 What should replace * in the number 94*2357, so that number is divisible by 11?**

1. 7

2. 3

3. 1

4. 8

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

##### 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).

##### The number 94*2357 is divisible by 11, then

##### ⇒ 9 – 4 + * – 2 + 3 – 5 + 7 = 0

##### ⇒ 19 + * – 11 = 0

##### ⇒ (*) = – 8

##### ⇒ or * = 11 – 8 = 3

**Q. In the given figure, AP bisects ∠BAC. If AB = 4 cm, AC = 6 cm and BP = 3 cm, then the length of CP is:**

##### 1. 5 cm

2. 7 cm

##### 3. 3 cm

4. 4.5 cm

##### Correct answer is __Option 4. __**4.5**

__Option 4.__

##### If AP bisects ∠BAC, then

##### AB/AC = BP/PC

##### ⇒ 4/6 = 3/PC

##### ⇒ PC = 18/4

##### ⇒ PC = **4.5**

**Q.11 In ΔABC, if AB = AC and ∠BAC = 40°, then the measure of ∠B is: **

1. 50°

2. 60°

3. 40°

4. 70°

##### correct answer is **option 4 i.e. ****70°**

##### Given, ∠A = 40°

##### In ΔABC,

##### If AB = AC, then

##### ∠B = ∠C

##### ∠A + ∠B + ∠C = 180°

##### ⇒ 40° + ∠B + ∠B = 180°

##### ⇒ 2∠B = 180° – 40 = 140°

##### ⇒ ∠B = 140°/2 = **70°**

**Q.12 (a + b – c + d)**^{2} – (a – b + c – d)^{2} = ?

^{2}– (a – b + c – d)

^{2}= ?

##### 1. 4a (b + d – c)

2. 2a (b + c – d)

3. 4a (b – d + c)

4. 2a (a + b – c)

##### Correct answer is **Option 1 i.e. ****4a (b + d – c)**

##### (a + b – c + d)^{2} – (a – b + c – d)^{2}

##### Put a = 1, b = 2, c = 3 and d = 4, then

##### ⇒ (1 + 2 – 3 + 4)^{2} – (1 – 2 + 3 – 4)^{2}

##### ⇒ 16 – 4

##### ⇒ 12

##### From option 2.

**4a (b + d – c)**

##### ⇒ 4 × 1 (2 + 4 – 3)

##### ⇒ 4 × 3

##### ⇒ 12 (satisfied)

**Q.13 The simple interest on a certain sum at the end of three years at 5% p.a. is Rs. 1,200. The compound interest on the same sum for the same period at the same rate is (interest compounded yearly).**

##### 1. Rs. 1,261

2. Rs. 820

3. Rs. 1,260

4. Rs. 1,800

##### Correct Answer is **option 1 i.e. Rs 1,261**

**Tree Method:**

##### CI for three years = 1200 + 60 + 1 = 1261

**Ratio method:**

##### Rate = 5%

##### SI for 3 years = 1200

##### SI for 1 year = 1200/3 = 400

##### 5% of 400 = 20

##### ⇒ 5% of 20 = 1

##### CI for 3 years = 3 : 3 : 1 = 3 × 400 + 3 × 20 + 1 × 1 = 1200 + 60 + 1 = 1261

**Detailed Method:**

##### R = 5%, t = 3 years SI = 1200

##### As we know,

##### SI = Prt/100

##### ⇒ 1200 = (P × 5 × 3)/100

##### ⇒ P = 8000

##### Again, As we know,

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

##### ⇒ A = 8000 (1 + 5/100)^{3}

##### ⇒ A = 8000 × (21/20) × (21/20) × (21/20)

##### ⇒ A = 9261

##### ⇒ CI = 9261 – 8000

##### ⇒ **CI = 1261**

**Q.14 In a circle with radius 5 cm, a chord is at a distance of 3 cm from the centre. The length of the chord is:**

1. 8 cm

2. 3 cm

3. 7 cm

4. 4 cm

##### Correct answer is **Option 1 i.e. 8 cm**

##### Radius of circle OA = 5 cm

##### Let the length of the chord be AB

##### OP = 3 cm

##### As we know,

##### OA^{2} = OP^{2} + AP^{2}

##### ⇒ 5^{2} = 3^{2} + AP^{2}

##### ⇒ AP^{2} = 25 – 9 = 16

##### ⇒ AP = √16 = 4 cm

##### ⇒ AB = 2AP = 2 × 4 = **8 cm**

**Q.15 The value of \frac { 36÷42of6×7+24×6÷18+3÷(2−6)−(4+3×2)÷8 }{ 21÷3of7 } is:**

##### 1. 7

2. \frac { 1 }{ 7 }

##### 3. 7\frac { 1 }{ 2 }

##### 4. 8\frac { 1 }{ 2 }

#### Correct answer is **Option 1 i.e. 7**

#### =\frac{36÷42of6×7+24×6÷18+3÷(2−6)−(4+3×2)÷8}{21÷3of7}

#### =\frac { 36÷42×6×7+24×\frac { 6 }{ 18 } +\frac { 3 }{ -4 } −(4+6)÷8 }{ 21÷3×7 }

#### =\frac { \frac { 36 }{ 42\times 6 } ×7+8-\frac { 3 }{ 4 } −10÷8 }{ 21÷21 }

#### =\frac { 1+8-\frac { 3 }{ 4 } -\frac { 10 }{ 8 } }{ 1 }

#### =1+8–\frac { 3 }{ 4 } -\frac { 5 }{ 4 }

#### =\frac { (4+32–3–5) }{ 7 }

#### =7

**Q.16 In the given figure, AP and BP are tangents to a circle with centre O. If ∠APB = 62° then the measure of ∠AQB is:**

1. 28°

2. 118°

3. 31°

4. 59°

##### Correct answer is **Option 3 i.e. **7 cm

##### If AP and BP are tangents to a circle with centre O, then

##### ∠OAP = ∠OBP = 90°

##### In quadrilateral APBO

##### ∠OAP + ∠OBP + ∠APB + ∠BOA = 360°

##### ⇒ 90° + 90° + 62° + ∠BOA = 360°

##### ⇒ ∠BOA = 360° – 242° = 118°

##### As we know,

##### ∠AQB = ∠BOA/2

##### ⇒ ∠AQB = 118°/2 = 59°

**Short Trick:**

##### ∴ ∠AQB = 90° – 62°/2 = 90° – 31° = **59°**

**Q.17 **From the following table, how many patients were in the age group 40 – 60?

## Age (years) | ## Less than 10 | ## Less than 20 | ## Less than 30 | ## Less than 40 | ## Less than 50 | ## Less than 60 | ## Less than 70 |

## No. of patients | ## 11 | ## 15 | ## 22 | ## 29 | ## 35 | ## 45 | ## 50 |

1. 16

2. 6

3. 45

4. 29

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

##### The patient in the age group 40 – 60 are = 45 – 29 = **16**

**Q18. ****The table below shows income (in rupees) for a particular month, together with their sources in respect of 5 employees A, B, C, D and E**

## Employee | ## A | ## B | ## C | ## D | ## E |

## Salary | ## 52,000 | ## 48,500 | ## 42,000 | ## 31,000 | ## 25,000 |

## Overtime | ## 0 | ## 0 | ## 1,500 | ## 2,500 | ## 3,200 |

## Arrears | ## 5,500 | ## 4.500 | ## 4,000 | ## 3.000 | ## 1,5000 |

## Bonus | ## 3,500 | ## 3,000 | ## 2,500 | ## 2,000 | ## 2,000 |

## Miscellaneous income | ## 5,000 | ## 3,000 | ## 2,000 | ## 1,500 | ## 0 |

## Total | ## 66,000 | ## 59,000 | ## 52,000 | ## 40,000 | ## 31,700 |

How many employees have their salary more than four times their other incomes?

How many employees have their salary more than four times their other incomes?

1. 4

2. 3

3. 2

4. 1

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

##### Salary of B = 48,500

##### Total income of B = 59,000

##### Other income of B excluding salary = 59,000 – 48,500 = 10,500

##### Four times of other income of B = 4 × 10,500 = 42,000

##### Salary of C = 42,000

##### Total income of C = 52,000

##### Other income of C excluding salary = 52,000 – 42,000 = 10,000

##### Four times of other income of C = 4 × 10,000 = 40,000

##### Now, we can say only two** employees B and C** whose salary more than four times their other incomes.

**Q.19 The value of -\frac { 5 }{ 2 } +\frac { 3 }{ 2 } \div 6\times \frac { 1 }{ 2 } is equal to:**

##### 1. – 19/8

2. – 9/8

3. – 1/12

4. – 1/3

##### Correct answer is **Option 1 i.e. -19/8**

##### -\frac { 5 }{ 2 } +\frac { 3 }{ 2 } \div 6\times \frac { 1 }{ 2 }

##### = (– 5/2) + 3/2 × 1/6 × 1/2

##### = (– 5/2) + 1/8

##### = (– 20 + 1)/8

##### = (– 19)/8

**Q.20 The average of five consecutive even numbers is M. If the next five even numbers are also included, the average of ten numbers will be:**

1. M + 5

2. M + 10

3. 10

4. 11

##### Correct answer is **Option 1 i.e. M+5**

##### Let the five consecutive even numbers are 2, 4, 6, 8, 10.

##### Average of consecutive even numbers = (2 + 4 + 6 + 8 + 10)/5 = 30/5 = 6

**Hint: **As we know, Average of any AP is the median so, the average = 6

##### So, M = 6

##### If the next five even numbers are also included, then the numbers are

##### 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

##### Average of 10 consecutive numbers = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20)/10 = 110/10 = 11

**Hint: **As we know, Average is the median so, the average = 11

##### We can write 11 = 6 + 5 or M + 5

**Q.21 The value of 27a**^{3} – 2√2b^{3} is equal to:

^{3}– 2√2b

^{3}is equal to:

##### 1. (3a – √2b)(9a^{2} – 2b^{2} – 3√2 ab)

2. (3a – √2b)(9a^{2} – 2b^{2} + 6√2 ab)

3. (3a – √2b)(9a^{2} + 2b^{2} + 3√2 ab)

4. (3a – √2b)(9a^{2} + 2b^{2} + 6√2 ab)

##### Correct answer is **Option 3 i.e. (3a – √2b)(9a**^{2} + 2b^{2} + 3√2 ab)

^{2}+ 2b

^{2}+ 3√2 ab)

##### As we know,

##### a^{3} – b^{3} = (a – b) (a^{2} + b^{2} + ab)

##### ⇒ 27a^{3} – 2√2 b^{3}

##### ⇒ (3a – √2 b) (9a^{2} + 2b^{2} + 3√2 ab)

**Q.22 The curved surface area of a hemisphere with radius 7 cm is: (Take π = 22/7)**

##### 1. 616 cm^{2}

2. 308 cm^{2}

3. 462 cm^{2}

4. 385 cm^{2}

##### Correct answer is **Option 2 i.e. ****308 cm**^{2}

^{2}

##### Radius of hemisphere r = 7 cm

##### As we know

##### ∴ Curved surface area of hemisphere = 2πr^{2} = 2 × (22/7) × 7× 7 = **308 cm**^{2}

^{2}

**Q.23 A person marked his goods at a price that would give him 40% profit. But he declared a sale and allowed 20% discount on the marked price. What is the profit percentage of the person in the whole transaction?**

##### 1. 12%

##### 2. 20%

##### 3. 32%

##### 4. 30%

##### Corrrect answer is **Option 2 i.e. **12%

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

##### MP of the article = 100 × (140/100) = Rs. 140

##### SP of the article = 140 × (80/100) = Rs. 112

##### Profit = 112 – 100 = 12

##### ∴ Profit percentage = 12/100 × 100 = 12%

**Q.24 **As per data in the table, what is the percentage of students who got 20 or more marks?

## Scores | ## 0 – 5 | ## 5 – 10 | ## 10 – 15 | ## 15 – 20 | ## 20 – 25 | ## 25 – 30 | ## 30 – 35 | ## 35 – 40 |

## No. of students | ## 13 | ## 15 | ## 18 | ## 12 | ## 14 | ## 19 | ## 6 | ## 3 |