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CBSE exam SA 2 class 9th solutions of important maths questions of last year’s question papers.
Here CBSE exam SA 2 class 9th 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 9th 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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