**CBSE exam SA 2 class 9**^{th }**solutions of important maths questions of last year’s question papers.**

Here **CBSE exam SA 2 class 9 ^{th }solutions of important maths questions of last year’s question papers** are given here which has been asked in the

**last years CBSE exam**. These questions are taken from the

**last**6

**years’ question papers.**

**The important maths questions**of

**last**

**year’s question paper classs 9**help the students to get the idea of the type of

**question**asked in the maths paper of the CBSE board. All questions are solved here by a step by step method, so every student can understand the solutions of each question.

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**NCERT Solutions of Class 9 Science : Chapter 1 to Chapter 15**

**CBSE exam SA 2 class 9**^{th }**solutions of important maths questions of last year’s question papers.**

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## Section A

Q1. Find the total surface area of a cube of edge 10 cm.

Answer. The total surface area of the cube is = 6a²

side(a) = 10 cm

The total surface area of the cube is = 6 × 10² =6 ×100 = 600

Hence the total surface area of the cube is = 600 sq.cm

Q2. Find the mean of the first 5 natural numbers.

Answer. The first 5 natural numbers are 1,2,3,4 and 5

= ∑X/N = (1+2 +3 +4 +5)/5 = 15/5 =3

Q3.Write a linear equation representing a line which is parallel to the y-axis and is at a distance of 2 units on the left side of y-axis.

Answer.In left side of y-axis, the x coordinate is negative

The line is at a distance of 2 units left of y axis will pass through (–2,0)

Since the line is parallel to the y-axis then the equation of this line is x =–2

Q4.If the point (3,4) lies on the graph of linear equation 3y =ax + 7, find the value of a.

Answer.Since the point (3,4) lies on the graph of linear equation 3y =ax + 7 then the coordinates (3,4) must satisfy the equation

3(4) = a(3) + 7

3a = 12 – 7 = 5

a = 5/3

Q5.O is the center of a circle that passes through P,Q,R and S as shown in the figure .SR is produced to x . If ∠QRX =133°,find x.

Answer.

PQRS is a cyclic quadrilateral

∴ ∠QPS +∠QRS = 180°…….(i)(opposite angles of cyclic quadratic angles)

∠QRS + ∠QRX = 180°(linear pair)

∠QRS + 133° = 180°

∠QRS = 180° – 133° =47°

putting the value of ∠QRS in (i)

∠QPS + 47° = 180°

∠QPS = 180° – 47° = 133°

4x + 13 = 133°

4x = 120

x = 30

Q6.Write the edge of cube in term of its volume.

Answer. The volume of the cube(V) =a³

side of the cube (a) =

Q7. The equation of x-axis is

(a) x = 0 (b) y = 0 (c) x = y (d) x ≠5

Answer. Since the coordinates of y on x axis is 0,so the equation of x axis is y= 0

Q8. Two parallelograms on equal base and between the same parallel lines. The ratio of their areas is

(a)1:2 (b) 2 : 1 (c) 3 : 1 (d) 1 : 1

Answer. The area of parallelograms on the same base and between the same parallel lines are equal, so the ratio of their areas will be (d) 1: 1

**NCERT Solutions Class 10 Science from chapter 1 to 16**

Q9.The volume of cuboid formed by joining two cubes of side 2 cm is:

(a) 64 cm³ (b) 32cm³ (c) 16cm³ (d) 8cm³

Answer.

The length of so formed cuboid = 2 +2 = 4 cm

breadth = 2 cm, height = 2 cm

So,volume of the cuboid = l × b× h =4 × 2×2 =16 cm³

Hence the answer is (c) 16 cm³.

Q10.If the chord of a circle is equal to the radius of the circle then the angle subtended by it at the center is

(a)30° (b) 120° (c) 50° (d)60°

Answer. Joining both ends of the chord a triangle formed which will be an equilateral because the radius is equal to the length of the chord, therefore each angle will be of 60°. Hence the angle subtended at the center by it is equal to 60°.

**Section B**

Q11. The surface area of a sphere is 154 cm². Find its radius.

Answer. The surface area of a sphere = 4πr²

4πr² = 154

4 ×(22/7) ×r² = 154

r² = (154×7)/(22×4) =49/4

r = 7/2

Therefore the radius of the circle is 7/2 cm

Q12. In the figure, ABCD is parallelogram AE ⊥DC and CF⊥ AD. If AB =16 cm,AE =8 cm and CF = 10 cm . Find AD.

Answer. Area of ABCD = AD × CF = AD × 10

As base AB area of ABCD = AB× AE = 16 × 8 =128

10 AD = 128

AD = 12.8

Hence the length of AD = 12.8 cm

**CBSE 10 maths most important questions of science with solutions**

Q13.A cuboidal water tank is 8 m long ,4m wide and 7 m deep .How many litres of water can it hold .

Answer. The volume of cuboidal water tank is = l × b× h = 8×4 ×7=224

The volume of water tank = 224 m³

Since 1m³ = 1000 l

224 m³ = 224/1000 = 0.224 l

Therefore the cuboidal tank can hold 0.224 l water.

Q14.The percentage of marks obtained by a student in unit tests are given as follows.

Unit test | I | II | III | IV |

Percentage of marks | 68 | 75 | 85 | 58 |

On the basis of this data, find the probability that in a unit test student scored more than 74 % marks.

Answer. The student got marks in 4 test, therefore total possible outcomes are =4

The student got more than 74% marks in II and III tests since there are 2 tests in which he got more than 74% marks, so favorable outcome of this event = 2

Let the event of getting more than 74% marks→ E

P(E) = Favourable outcome/Total possible outcome

P(E) = 2/4 = 1/2

Therefore the probability that a student get marks more than 74% marks is 1/2

Q15. Given (7+√3)/(7-√3), find the value of a and b.

Answer.We are given that

(7+√3)/(7-√3) = a + b√3

Taking L.H.S

(7+√3)/(7-√3)

Multiplying the denominator and numerator of the above expression by the conjugate of the denominator (7+√3)

(7+√3) (7+√3)/(7-√3)(7+√3)

=(7²+ √3²+ 14√3)/(7²- √3²)

=(49 +3 +14√3)/(49-3)

=(52 +14√3)/46

(52 +14√3)/46 = a + b√3

a = 52/46 = 26/23, b = 14√3/46 = 7√3/23

Hence the value of a and b are 26/23 and 7√3/23 respectively.

Q16.Out of 35 students participating in a dance competition, 10 are boys. What is the probability that the winner is a girl?

Answer. Total number of students are =35, so total possible outcomes are =35

Number of boys are = 10

Number of girls = 35 – 10 = 25, so favourable outcomes are 25

Let E is the event of winner is girl

P(E) = Favourable outcome/Total possible outcome

P(E) = 25/35 = 5/7

Therefore the probability that winner is girl =5/7

Q17.Find the mean of the first 5 composite numbers.

Answer. First 5 composite numbers are 4,6,8,9 and 10

Let the mean of First 5 composite numbers is

= (4 +6 +8 +9 +10)/5 = 37/5 = 7.4

Therefore the mean of first 5 composite numbers is =7.4

Q18.In the given figure, ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD and BA is produced to F . If ∠ABC is 70° and ∠FAE =30°, Find ∠BCD.

Answer.ABCD is a cyclic quadrilateral ∠ABC and ∠ADC are its opposite angles

∠ABC + ∠ADC = 180°

70° + ∠ADC = 180°

∠ADC = 110°

∵AE∥CD and AD is transversal

∴ ∠DAE = ∠ADC ( corresponding angle)

∠DAE = 110°

∠FAE + ∠DAE + ∠DAB = 180°

30° + 110° +∠DAB = 180°

∠DAB = 180° –140°= 40°

∠DAB + ∠BCD = 180°

40° + ∠BCD = 180°

∠BCD = 40°

Section C

Q19. A right triangle with sides 5 cm, 12 cm, and 13 cm is revolved about the side 12 cm. Find the volume of the solid so formed.

Answer.

We are given a right triangle which is to be revolved about the side 12 cm ,the solid so formed will be a cone whose height(h) = 12 cm, radius(r) =5 cm and slant height(l) = 13cm

Volume of cone =(1/3) πr²h

= (1/3) ×π×5×5×12

= 100π

Therefore the volume of solid so formed is 100π cm³

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Q20.Find the mean salary of 70 workers of a factory from the following table.

Salary | No of workers |

30000 | 20 |

40000 | 15 |

50000 | 12 |

60000 | 9 |

70000 | 6 |

80000 | 5 |

90000 | 3 |

Answer.

Salary(x) | No of workers(f) | fx |

30000 | 20 | 600000 |

40000 | 15 | 600000 |

50000 | 12 | 600000 |

60000 | 9 | 540000 |

70000 | 6 | 420000 |

80000 | 5 | 400000 |

90000 | 3 | 270000 |

ΣF =70 | ΣFX=3430000 |

Mean = ∑FX/∑F

Mean = 3430000/70

Mean = 49000

Hence mean salary of worker is Rs49000.

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