Find below the important questions for CBSE Class 9 Mathematics Annual Exam 2019:
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Some sample questions from the set of Important 1 Mark Questions for CBSE Class 9 Mathematics Exam, are given below:
Q. If (1, −2) is a solution of the equation 2x – y = p, then find the value of p.
Sol.
2x − y = p
Putting x = 1, y = −2, in above equation, we get:
2(1) – (−2) = p
⟹ p = 4
Q. If the graph of equation 2x + ky = 10k, intersect x-axis at point (2, 0) then find value of k.
Sol.
Here, 2x + ky = 10k ...(i)
At point (2, 0), equation (i) becomes:
2(2) + k(0) = 10k
⟹ 4 = 10k
⟹ k = 10/4 = 0.4
Q. Find the radius of largest sphere that is carved out of the cube of side 8 cm.
Sol.
The largest sphere can be carved out from a cube, if we take diameter of the sphere equal to edge of the cube.
∴ Diameter of the sphere = 8 cm
Thus, radius of the sphere = 8/2 = 4 cm
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Some sample questions from the set of Important 2 Marks Questions for CBSE Class 9 Mathematics Exam, are given below:
Q. Find the total surface area of a cube, whose volume is 3√3a3 cubic units.
Sol.
Since volume of a cube = 3√3a3
⟹ (side)3 = 3√3a3
⟹ (side)3 = √3 × √3 × √3 × a3
⟹ (side)3 = (√3a)3
⟹ side= √3a
Now, total surface area of cube =6 (side)2
= 6(√3a)2
= 18a2sq. units
Q. If two opposite angles of a parallelogram are (63 − 3x)° and (4x − 7)°. Find all the angles of the parallelogram.
Sol.
In a parallelogram, the opposite angles are equal.
∴ (63 − 3x)° = (4x − 7)°
⟹ 4x + 3x = 63 +7
⟹ 7x = 70
⟹ x = 10
(63 − 3x)° = 33°
(4x − 7)° = 33°
Now, sum of all interior angles of a parallelogram = 360°
∴ Sum of the other two opposite angles = 360° − (33° + 33°) = 360° − 66° = 294°
∴ Each of the other two opposite angles = 294/2 = 147°
Hence the four angles of a parallelogram are 33°, 147°, 33°, 147°
Some sample questions from the set of Important 3 Marks Questions for CBSE Class 9 Mathematics Exam, are given below:
Q. Suman spins two spinners, one of which is labeled 1, 2 and 3 and the other are labeled A, B, C and D. Find the probability of:
(i) Stopping at 2 and C.
(ii) Stopping at 3 and either B or D
(iii) Stopping at any number and A
Sol.
Sample space = {A1, A2, A3, B1, B2, B3, C1, C2, C3, D1, D2, D3}
Total possible outcome = 12
Probability of stopping at 2 and C = 1/12
Probability of stopping at 3 and either B or D = 2/12 = 1/6
Probability of stopping at any number and A = 3/12 = 1/4
Q. Construct an angle of 135o at the initial point of a given ray and justify the construction.
Sol.
Steps of construction:
1. Draw a ray OA.
2. With its initial point O as centre and any radius, draw an arc BE, cutting OA and its produced part at C and E respectively.
3. With centre B and same radius (as in step 2), draw an arc, cutting the arc BE at C.
4. With C as centre and the same radius, draw an arc cutting the arc BE at D.
5. With C and D as centres, and any convenient radius (more than 1/2 CD), draw two arcs intersecting at P.
6. Join OP.
Then ∠AOP = ∠EOP = 90o
7. Draw OQ as the bisector of ∠EOP.
Then, ∠AOQ =135o
Justification :
By construction ∠AOP = 90o
Thus, ∠AOP = ∠EOP = 90o
Also, OQ is drawn as the bisector of ∠EOP
Therefore, ∠POQ =1/2∠EOP =1/2 × 90o = 45o
Thus, ∠AOQ = ∠AOP + ∠POQ = 90o + 45o = 135o
Hence justified.
Some sample questions from the set of Important 4 Marks Questions for CBSE Class 9 Mathematics Exam, are given below:
Q. Q. If a + b + c = 6 and ab + bc + ca = 11, find the value of a3 +b3 +c3 − 3abc.
Sol.
(a + b + c)2 = a2 + b2 +c2 +2(ab + bc + ca)
(6)2 = a2 + b2 +c2 + 2 × 11
a2 + b2 +c2 = 36 – 22 = 14
a3 +b3 +c3 − 3abc = ( a + b + c)[ a2 + b2 +c2 −(ab + bc + ca)]
= 6 × (14 − 11)
= 6 × 3 = 18
Q. The polynomials ax3 – 3x2 +4 and 2x3 – 5x +a when divided by (x – 2) leave the remainders pand q respectively. If p – 2q = 4, find the value of a.
Sol.
Let, f(x) = ax3 – 3x2 +4
And g(x) = 2x3 – 5x +a
When f(x) and g(x) are divided by (x – 2) the remainders are p and q respectively.
⟹ f(2) = p and g(2) = q
⟹ f(2) = a × 23 – 3 × 22 + 4
⟹ p = 8a – 12 + 4
⟹ p = 8a – 8 ....(i)
And g(2)= 2 × 23 – 5 × 2 + a
⟹ q = 16 – 10 + a
⟹ q = 6 + a ....(ii)
But p – 2q = 4 (Given)
⟹ 8a – 8 – 2(6 + a) = 4 (Using equations (i) and (ii))
⟹ 8a – 8 – 12 − 2a = 4
⟹ 6a – 20 = 4
⟹ 6a = 24
⟹ a = 24/6
⟹ a = 4
Q. Construct a ΔABC in which BC = 3.8 cm, ∠B = 45oand AB + AC = 6.8cm.
Sol.
Steps of Construction
1. Draw BC = 38 cm.
2. Draw a ray BX making an ∠CBX = 45°.
3. From BX, cut off line segment BD equal to AB + BC i.e., 6.8 cm.
4. Join CD.
5. Draw the perpendicular bisector of CD meeting BD at A.
6. Join CA to obtain the required
Justification:
Clearly, A lies on the perpendicular bisector of CD.
∴ AC = AD
Now, BD = 6.8 cm
⟹ BA + AD = 6.8 cm
⟹ AB + AC = 6.8 cm
Hence, is the required triangle.
(6)2 = a2 + b2 +c2 + 2 × 11
a2 + b2 +c2 = 36 – 22 = 14
a3 +b3 +c3 − 3abc =( a + b + c)[ a2 + b2 +c2−(ab + bc + ca)]
= 6 × (14 − 11)
= 6 × 3 = 18
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