**Q.1 If the number 1005×4 is completely divisible by 8, then the smallest integer in place of x will be:**

1. 0

2. 1

3. 4

4. 2

##### the correct answer is **option 1 i.e. 0**

##### Divisibility rule of 8 = A number divisible by 8 if its last three digits are divisible by 8

##### If the number 1005×4 is divisible by 8, then its last digit 5×4 also divisible by 8

##### 504 is divisible by 8 so the number 100504 is also divisible by 8.

##### Hence, **0** will be the correct answer.

**Q.2 The given table shows the number (in thousands) of cars of five different models A, B, C, D and E produced during years 2012-2017. Study the table and answer the question that follow.**

** If 2013 and 2014 are put together, which type of cars constitutes exactly 25% of the total number of cars produced in those 2 years?**

###### 1. E

2. B

3. D

4. C

##### The correct answer is__ option 3__: __D__.

__D__

Total production of cars in 2013 and 2014 = 130 + 170 = 300

Hence, 25% of 300 = 300 × 25/100 = 75

Production of car D in 2013 and 2014 are = 40 + 35 = 75

**Q.3 If x = 4 cos A + 5 sin A and y = 4 sin A – 5 cos A, then the value of x**^{2} + y^{2} is:

^{2}+ y

^{2}is:

1. 25

2. 41

3. 0

4. 16

#### Correct answer is **option 2 i.e. 41**

##### Given, x = 4 cos A + 5 sin A and y = 5 sin A – 5 cos A

##### Hence, x^{2} + y^{2} = (4 cos A + 5 sin A)^{2} + (4 sin A – 5 cos A)^{2}

##### ⇒ 16 cos^{2} A + 25 sin^{2} A + 40 cos A sinA + 16 sin^{2} A + 25 cos^{2}A – 40 sin A cos A

##### ⇒ 16 (sin^{2} A + cos^{2} A) + 25 (sin^{2} A + cos^{2} A)

##### ⇒ 16 + 25

##### ⇒ 41

**Q.4 The radius of a circular garden is 42 m. The distance (in m) covered by running 8 rounds around it, is: (Take π = 22/7)**

1. 2112

2. 1124

3. 3248

4. 4262

##### Correct answer is **option 1 i.e. 2112**

##### Radius of the circular garden r = 42 m

##### Circumference of circular garden = 2 πr = 2 × (22/7) × 42 = 264 m

##### 1 round cover the distance = circumference of the circular garden = 264 m

##### 8 round cover the distance = 8 × 264 = 2112 m

**Short trick:**

##### Answer divisible by 11 and 8 also. Only option is divisible by 11 and 8 is 2112.

**Q.5 ****The given table shows the number (in thousands) of cars of five different models A, B, C, D and E produced during years 2012-2017. Study the table and answer the question that follow.**

**The percentage decrease in the production of which type of car in 2017, with reference to 2016, was the maximum?**

1. D

2. E

3. A

4. C

##### Correct answer is option 3 i.e. A

Decreasing percentage of type car D in 2017 with reference to 2016 = [(38 – 22)/38] × 100 = 16/38 × 100 = 42.1%

Decreasing percentage of type car C in 2017 with reference to 2016 = (44 – 40)/44 × 100 = 4/44 × 100 = 9.09%

Decreasing percentage of type car A in 2017 with reference to 2016 = (36 – 12)/36 × 100 = 24/36 × 100 = 66.67%

Decreasing percentage of type car E in 2017 with reference to 2016 = (50 – 28)/50 × 100 = 22/50 × 100 = 44%

The percentage decrease in the production of type A car in 2017, with reference to 2016, was the maximum

**Q.6 If x, y, z are three integers such that x + y = 8, y + z = 13 and z + x = 17, then the value of x**^{2}/yz is:

^{2}/yz is:

1. 7/5

2. 18/11

3. 1

4. 0

##### Correct answer is option 2 i.e. 18/11

##### x + y = 8 —(1)

##### y + z = 13 —(2)

##### z + x = 17 —(3)

##### from equation (1)

##### y = 8 – x

##### Put y = 8 – x in equation (2)

##### 8 – x + z = 13

##### z – x = 13 – 8

##### z – x = 5 —(4)

##### From equation (3) and equation (4)

##### z = 11 and x = 6

##### Put x = 6 in equation (2)

##### y + 11 = 13

##### y = 2

##### Now,

##### x^{2}/yz

##### ⇒ (6 × 6)/(2 × 11)

##### ⇒ **18/11**

**Q.7 In ΔABC, MN∥BC, the area of quadrilateral MBCN = 130 cm**^{2}. If AN : NC = 4 : 5, then the area of ΔMAN is:

^{2}. If AN : NC = 4 : 5, then the area of ΔMAN is:

1. 40 cm^{2}

2. 45 cm^{2}

3. 32 cm^{2}

4. 65 cm^{2}

##### Correct answer is option 3 i.e. 32 cm^{2}

##### Given,

##### the area of quadrilateral MBCN = 130 cm^{2}.

##### AN : NC = 4 : 5

##### Hence, AC = 4 + 5 = 9

##### In ΔABC, If MN∥BC, then

##### Area of ΔAMN/area of ΔABC = (AN/AC)^{2}

##### Area of ΔAMN/area of ΔABC = (4/9)^{2} = 16/81

##### Area of ΔAMN = 16 unit and area of ΔABC = 81 unit

##### Now, Area of quadrilateral MBCN = area of ΔABC – area of ΔAMN

##### ⇒ 81 – 16 unit = 130 cm^{2}

##### ⇒ 65 unit = 130 cm^{2}

##### ⇒ 1 unit = 130/65 = 2 cm^{2}

##### ∴ Area of ΔAMN = 16 × 2 = 32 cm^{2}

**Q.8 A, B and C can individually complete a piece of work in 24 days, 15 days and 12 days, respectively. B and C started the work and worked for 3 days and left. The number of days required by A alone to complete the remaining work is:**

1. 11

2. 15 1/2

3. 18

4. 13 1/5

#### Correct answer is option 4 i.e. 13 1/5

##### Let, total work = LCM of (24, 15 and 12) = 120

##### Hence, efficiency of A = 5

##### Efficiency of B = 8

##### Efficiency of C = 10

##### Work done by B and C in 3 days = (8 + 10) × 3 = 54

##### Remaining work = 120 – 54 = 66

##### ∴ Remaining work done by A in = 66/5 = 13 1/5 days.

**Q.9 If x – y = 4 and xy = 45, then the value of x**^{3} – y^{3} is:

^{3}– y

^{3}is:

1. 822

2. 604

3. 151

4. 82

##### Correct answer is option 2 i.e. 604

**Short Trick:**

##### x – y = 4 and xy = 45

##### Put x = 9 and y = 5

##### x^{3} – y^{3}

##### ⇒ 9^{3} – 5^{3}

##### ⇒ 729 – 125

##### ⇒ 604

**Detailed solution:**

##### x – y = 4 and xy = 45

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

##### x^{3} – y^{3} = 4 [4^{2} + 3 × 45] [**Hint: **Answer will be divisible by 4, only 1 option 604 divisible by 4.]

##### x^{3} – y^{3} = 4 [16 + 135]

##### x^{3} – y^{3} = 4 × 151 = **604**

**Q.10 A person sells an article at 10% below its cost price. Had he sold it for Rs. 332 more, he would have made a profit of 20%. What is the original selling price (in Rs.) of the article?**

1. 996

2. 1,328

3. 1,028

4. 896

##### Correct answer is **Option 1. 996**

##### Let the Cost price of the article be Rs. 100x, then according to the question

##### 100x × (120/100) = 100x × (90/100) + 332

##### ⇒ 120x – 90x = 332

##### ⇒ 30x = 332

##### ⇒ x = 332/30

##### Selling price of the article = 90x = 90 × (332/30) = **996 **

**Q.11 If the base radius of 2 cylinders are in the ratio 3 : 4 and their heights are in the ratio 4 : 9, then the ratio of their volumes is:**

1. 2 : 1

2. 1 : 4

3. 1 : 2

4. 4 : 1

##### correct answer is **option 2 i.e. 1:4**

**As we** know,

##### Volume of cylinder = πr^{2}h

##### Ratio of radius of 2 cylinder, r_{1} : r_{2} = 3 : 4

##### Ratio of heights of 2 cylinder, h_{1} : h_{2} = 4 : 9

##### Ratio of volume of two cylinder, V1 : V2 = π(r_{1})^{2 }h_{1} : π(r_{2})^{2} h_{2} = 3^{2} × 4 : 4^{2} × 9 =

**1 : 4**

**Q.12 A train crosses a pole in 12 sec, and a bridge of length 170 m in 36 sec. Then the speed of the train is:**

1. 10.8 km/h

2. 30.75 km/h

3. 25.5 km/h

4.32.45 km/h

##### Correct answer is **Option 3 i.e. 25.5 km/hr**

**Short Trick:**

##### If train cross its length in 12 seconds and 170 m bridge in (36 – 12 = 24) seconds.

##### Speed of train = [170/24] × [18/5] = 25.5 km/hr

**Detailed method:**

##### Let the length of the train be x m.

##### As we know,

##### Speed = Distance/time

##### Speed = x/12 —(1)

##### Speed = (x + 170)/36 —(2)

##### Frome equation (1) and equation (2)

##### x/12 = (x + 170)/36

##### ⇒ 3x = x + 170

##### ⇒ 2x = 170

##### ⇒ x = 170/2

##### ⇒ x = 85 m

##### From equation (1)

##### Speed = 85/12 × (18/5) km/hr

##### ∴ Speed = 25.5 km/hr

**Q.13 If ‘+’ means ‘-‘, ‘-‘ means ‘+’, ‘×’ means ‘÷’ and ‘÷’ means ‘×’, then the value of ****is:**** **

##### 1. -15/19

2. 5/3

3. -5/3

4. 15/19

Correct Answer is **option 1 i.e. -15/19**

**Q.14 A shopkeeper marks the price of the article in such a way that after allowing 28% discount, he wants a gain of 12%. If the marked price is Rs. 224, then the cost price of the article is:**

1. Rs. 168

2. Rs. 144

3. Rs. 120

4. Rs. 196

##### Correct answer is **Option 2 i.e. Rs 144**

**Short Trick:**

##### Ratio of MP to CP = (100% + Profit%) : (100% – Discount%) = (100 + 12) : (100 – 28) = 112 : 72

##### 112 unit = 224

##### ⇒ 1 unit = 224/112 = 2

##### Now, 72 unit = 2 × 72 = 144

##### ∴ Cost price of the article is Rs. 144.

**Detailed Solution:**

##### MP of the article = Rs. 224

##### SP of the article = Rs. 224 × (72/100)

##### ∴ CP of the article = 224 × (72/100) × (100/112) = **Rs. 144**

**Q.15 Rs. 4,300 becomes Rs. 4,644 in 2 years at simple interest. Find the principle amount that will become Rs. 10,104 in 5 years at the same rate of interest**

1. Rs. 7,200

2. Rs. 8420

3. Rs. 9,260

4. Rs. 5,710

##### Correct answer is **Option 2 i.e. Rs 8420**

##### P = 4300, A = 4644 and t = 2 years

##### SI = 4644 – 4300 = 344

##### As we know,

##### SI = Prt/100

##### ⇒ 344 = (4300 × r × 2)/100

##### ⇒ r = 344/86

##### ⇒ r = 4%

##### Again,

##### A = 10,104, r = 4% and t = 5 years

##### As we know

##### A = P + Prt/100

##### ⇒ 10104 = P + (P × 4 × 5)/100

##### ⇒ 10104 = P + P/5

##### ⇒ 6P/5 = 10104

##### ⇒ P = 10104 × (5/6)

##### ∴ P = 8420

**Q.16 Out of 6 numbers, the sum of the first 5 numbers is 7 times the 6**^{th} number. If their average is 136, then the 6^{th} number is:

^{th}number. If their average is 136, then the 6

^{th}number is:

1.116

2. 102

3. 96

4. 84

##### Correct answer is **Option 2 i.e. 102**

##### Let the sum of 6^{th} number be x

##### The sum of first 5 numbers = 7x

##### According to the question

##### (x + 7x)/6 = 136

##### ⇒ 8x = 136 × 6

##### ⇒ x = (136 × 6)/8

##### ⇒ x = 102

##### ∴ The 6^{th} number is 102.

**Q.17 If A + B = 45°, then the value of 2(1 + tan A) (1 + tan B) is: **

1. 4

2. 0

3. 2

4. 1

##### Correct answer is **Option 1 i.e. 4**

##### 2(1 + tan A) (1 + tan B)

##### If A + B = 45°

##### Put A = 45° and B = 0°

##### 2 (1 + tan 45°) (1 + tan 0°)

##### ⇒ 2 (1 + 1) (1 + 0)

##### ⇒ 2 × 2 × 1

##### ⇒ 4

**Detailed solution:**

##### Taking tan both sides:

tan(A + B) = tan 45°

(tan A + tan B)/(1 – tan A tan B) = 1

tan A + tan B = 1 – tan A tan B

tan A + tan B + tan A tan B = 1

adding “1” on both sides

1+ tan A + tan B + tan A tan B = 1 + 1

(1 + tan A) + tan B(1 + tan A) = 2

(1 + tan A)(1 + tan B) = 2

##### ∴ 2 × (1 + tan A)(1 + tan B) = 2 × 2 = **4**

**Q18. If the length of a rectangle is increased by 40%, and the breadth is decreased by 20%, then the area of the rectangle increases by x%. Then the value of x is:**

1. 16

2. 20

3. 8

4. 12

##### Correct answer is **Option 4 i.e. 12**

**Short trick:**

##### Area increased by = 40 – 20 – [(40 × 20)/100] = 20 – 8 = 12%

**Detailed method:**

##### Let the length and breadth of the rectangle be 10 unit each.

##### Area of the rectangle = lb = 10 × 10 = 100 sq. unit

##### New increased length = 10 × (40/100) = 14 unit

##### New decreased length = 10 × (80/100) = 8 unit

##### New area of the rectangle = 14 × 8 = 112

##### Area increased by = 112 – 100 = 12

##### ∴ Area increased by (in %) = (12/100) × 100 = 12%

**Q.19 The area of ΔABC is 44 cm**^{2}. If D is the midpoint of BC and E is the midpoint of AB, then the area (in cm^{2}) of ΔBDE is:

^{2}. If D is the midpoint of BC and E is the midpoint of AB, then the area (in cm

^{2}) of ΔBDE is:

1. 5.5

2. 44

3. 11

4. 22

##### Correct answer is **Option 3 i.e. 11**

##### If D is the midpoint of BC and E is the midpoint of AB, then

##### DE ∥ AC

##### Let BC = 2 unit and BD = 1 unit

##### As we know,

##### area of ΔBDE/area of ΔBCA = (BD/BC)^{2}

##### ⇒ area of ΔBDE/44 = (1/2)^{2}

##### ∴ area of ΔBDE = (1/4) × 44 = 11 cm^{2}

**Q.20 The given table shows the number (in thousands) of cars of five different models A, B, C, D and E produced during years 2012-2017. Study the table and answer the question that follow.**

##### In the year 2015, which type of car constitutes exactly 20% of the total number of cars produced that year?

1. E

2. A

3. B

4. D

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

##### Total production of total car in 2015 = 100

##### Hence, 20% of 100 = 20

##### In the year 2015, E type of car constitutes exactly 20% of the total number of cars produced that year.

**Q.21 If A lies in the first quadrant and 6 tan A = 5, then the value of **

1. 16

2. 1

3. 4

4. -2

##### Correct answer is **Option 2 i.e. 1**

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

##### Given, 6 tan A = 5

##### ⇒ tan A = 5/6

##### ⇒ tan A = P/B = 5/6

##### ⇒ P = 5 and B = 6

##### Using Pythagoras theorem,

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

##### ⇒ H^{2} = 5^{2} + 6^{2}

##### ⇒ H^{2} = 25 + 36

##### ⇒ H^{2} = 61

##### ⇒ H = √61

##### sin A = P/H = 5/√61 and cos A = B/H = 6/√61

##### 8sinA−4cosAcosA+2sinA8sinA−4cosAcosA+2sinA

##### ⇒8×561√−4×661√661√+2×561√⇒8×561−4×661661+2×561

##### ⇒ (40 – 24)/(6 + 10)

##### ⇒ 16/16

##### ⇒ 1

**Short Trick:**

##### 6 tan A = 5

##### tan A = 5/6

##### sinA/cosA = 5/6

##### Let sin A = 5 and cos A = 6

##### ⇒8sinA−4cosAcosA+2sinA⇒8sinA−4cosAcosA+2sinA

##### ⇒8×5−4×66+2×5⇒8×5−4×66+2×5

##### ⇒ (40 – 24)/(6 + 10)

##### ⇒ 16/16

##### ⇒ 1

**Q.22 The ratio of the number of boys to the number of girls in a school of 640 students, is 5 : 3. If 30 more girls are admitted in the school, then how many more boys should be admitted so that the ratio of boys to that of the girls, becomes 14 : 9.**

1. 30

2. 25

3. 15

4. 20

##### Correct answer is **Option 4 i.e. 20**

##### Total students = 640

##### Ratio of boys to girls = 5 : 3

##### Number of boys = [5/8] × 640 = 400

##### Number of girls = 640 – 400 = 240

##### Let x boys admitted in the school.

##### According to the question

##### (400 + x)/(240 + 30) = 14/9

##### ⇒ 400 + x = [14/9] × 270

##### ⇒ x = 420 – 400

##### ⇒ x = 20

##### ∴ **20** students admitted in the school.

**Q.23 A, B and C are three points on a circle such that the angles subtended by the chord AB and AC at the centre O are 110° and 130°, respectively. The value of ∠BAC is:**

1. 70°

2. 75°

3. 60°

4. 65°

##### Corrrect answer is **Option 3 i.e. 60°**

##### from the following figure.

##### ∠AOB = 110° and ∠AOC = 130°

##### As we know,

##### ∠AOB + ∠AOC + ∠BOC = 360°

##### ∠BOC = 360° – 110° – 130° = 120°

##### ∠BAC = ∠BOC/2

##### ∴ ∠BAC = 120°/2 = **60°**

**Q.24 The given table shows the number (in thousands) of cars of five different models A, B, C, D and E produced during years 2012-2017. Study the table and answer the question that follow.**

##### The percentage increase in the total car in 2016 over 2012, is:

1. 50%

2. 45%

3. 33.33%

4. 62.33%

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

##### Total production of car in 2012 = 120

##### Total production of car in 2016 = 180

##### ∴ The percentage increase of total car in 2012 over 2016 = [(180 – 120)/120] × 100 = **50% **

**Q.25 If x**^{2a} = y^{2b} = z^{2c} ≠ 0 and x^{2} = yz, then the value of

1. 3

2. 3ac

^{2a}= y

^{2b}= z

^{2c}≠ 0 and x

^{2}= yz, then the value of