**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**. 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.

**You can also study NCERT Solutions**

**Exercise 4.3- Linear equations in two variables**

**Exercise 4.1 & 4.2- Linear equation in two variables**

**Exercise.9.3- Areas of parallelograms and triangles**

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

**Section A**

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

**Answer. The total surface area of the cube is = **

**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**

**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**

**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) = **

**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 (b) 32 (c) 16 (d) 8 **

**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 **

**Hence the answer is (c) 16 .**

**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**

**Therefore the radius of the circle is**

**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 **=

**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**

**Therefore the probability that a student get marks more than 74% marks is **

**Q15. Given , find the value of a and b.**

**Answer.We are given that**

**Taking L.H.S**

**Multiplying the denominator and numerator of the above expression by the conjugate of the denominator **

**Hence the value of a and b are 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**

**Therefore the probability that winner is girl =**

**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 **

**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 = **

**= 100π**

**Therefore the volume of solid so formed is 100π **

**How to calculate recoil velocity of the gun**

**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 |

**Hence mean salary of worker is Rs49000.**

CONTINUE…

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