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Class 9 Mathematics MCQs of all chapters provided here with answers are based on the NCERT curriculum and the latest CBSE syllabus. Objective questions are always helpful for quick revision before exams. Students can evaluate their preparedness by practising these questions. In section “A” of the exam question paper, the student has to solve 10 MCQs, each carrying 1 mark (total 10 marks of full marks).
Chapter 1: Number Systems
In this chapter, we learn different types of numbers (natural, whole, real, rational, irrational, integers, etc.), and operations like representations of numbers on the number line, identities of irrational numbers, rationalization, laws of radicals, etc.
MCQs on Number Systems
Ques. √25 is __________ number.
- A rational
- An irrational
- Neither rational nor irrational
- None of the above
Answer: a
Explanation: √25 = 5
Ques. √6 x √27 is equal to:
- 9√2
- 3√3
- 2√2
- 9√3
Answer: a
Explanation: √6 x √27 = √(6 x 27) = √(2 x 3 x 3 x 3 x 3) = (3 x 3)√2 = 9√2
Chapter 2: Polynomials
Here, we get to know about types of polynomials, degree of polynomials, the value of polynomials, remainder theorem, factor theorem, factorization, etc.
Also Check:
MCQs on Polynomials
Ques. The coefficient of x² in 3x³+2x²-x+1 is:
- 1
- 2
- 3
- -1
Answer: b
Explanation: The coefficient of x² in equation 3x³+2x²-x+1 is the multiple of x².
Ques. 3x²-7x+5 is a polynomial in:
- One Variable
- Two Variables
- Three Variables
- None of the above
Answer: a
Explanation: 3x²-7x+5 can be written as 3x²-7x+5x0. Hence, we can see that x is the only variable having powers as whole numbers: 2,1 and 0.
Chapter 3: Coordinate Geometry
Coordinate Geometry is about the Cartesian system which includes the cartesian plane, coordinate axes, quadrants, points in different quadrants. Here we learn to plot a graph, representation of points on the cartesian plane, etc.
MCQs on Coordinate Geometry
Ques. If the coordinates of a point are (0, -4), then it lies in:
- X-axis
- Y-axis
- At origin
- Between x-axis and y-axis
Answer: b
Explanation: Since x=0 and y=-4. Hence, the point will lie in the negative y-axis 4 units far from the origin.
Ques. If the coordinates of a point are (-3, -4), then it lies in:
- First quadrant
- Second quadrant
- Third quadrant
- Fourth quadrant
Answer: c
Explanation: Since, x = -3 and y = -4, then if we plot the point in a plane, it lies in the third quadrant.
Chapter 4: Linear Equations with Two Variables
This chapter and chapter-2 as Algebra are very important for the students as it will help them in solving linear equations with two variables.
Also Read:
MCQs on Linear Equations with Two Variables
Ques. The solution to equation x-2y = 4 is:
- (0,2)
- (2,0)
- (4,0)
- (1,1)
Answer: c
Explanation: Putting x=4 and y = 0, on the L.H.S. of the given equation, we get; 4-2(0) = 4 – 0 = 4
Which is equal to R.H.S.
Ques. Point (3, 4) lies on the graph of the equation 3y = kx+7. The value of k is:
- 4/3
- 5/3
- 3
- 7/3
Answer: b
Explanation: 3y = kx + 7
Here, x = 3 and y = 4
Hence, (3×4) = (kx3) + 7
12 = 3k+7
3k = 12–7
3k = 5
k = 5/3.
Chapter 5: Introduction to Euclid’s Geometry
Here, we learn Euclid’s Axioms and Postulates. Euclid's 5 Postulates.
Check Further:
MCQs on Introduction to Euclid’s Geometry
Ques. A point has _______ dimension.
- One
- Two
- Three
- Zero
Answer: d
Explanation: A point is always dimensionless.
Ques. Which of these statements does not satisfy Euclid’s axiom?
- Things that are equal to the same thing are equal.
- If equals are added to equals, the wholes are equal.
- If equals are subtracted from equals, the remainders are equal.
- The whole is less than the part.
Answer: d.
Chapter 6: Lines and Angles
This chapter is about different types of lines and angles and their properties.
Also Read:
MCQs on Lines and Angles
Ques. A reflex angle is:
- More than 90 degrees
- Equal to 90 degrees
- More than 180 degrees
- Equal to 180 degrees
Answer: c
Ques. The angles of a triangle are in the ratio 2: 4 : 3. The smallest angle of the triangle is
- 20°
- 40°
- 60°
- 80°
Answer: b
Explanation: We know that the sum of the interior angles of a triangle is 180°.
Given that, the angles of a triangle are in the ratio of 2:4:3,
Hence, 2x+4x+3x = 180°
9x = 180°
x= 20°
Therefore, 2x = 2(20) = 40°
4x = 4(20) = 80°
3x = 3(20) = 60°
Hence, the angles are 40°, 80°, and 60°.
Therefore, the smallest angle of a triangle is 40°.
Chapter 7: Triangles
In this chapter, we learn about different types of triangles and their properties.
Also Check:
MCQs on Triangles
Ques. In triangle ABC, if AB=BC and ∠B = 70°, ∠A will be:
- 70°
- 110°
- 55°
- 130°
Answer: c
Explanation: Given, AB = BC,
Hence, ∠A=∠C
And ∠B = 70°
By the angle-sum property of the triangle we know:
∠A+∠B+∠C = 180°
2∠A+∠B=180°
2∠A = 180-∠B = 180-70 = 110°
∠A = 55°
Ques. In a right triangle, the longest side is:
- Perpendicular
- Hypotenuse
- Base
- None of the above
Answer: b
Explanation: In triangle ABC, right-angled at B.
∠B = 90
By angle sum property, we know:
∠A + ∠B + ∠C = 180
Hence, ∠A + ∠C = 90
So, ∠B is the largest angle.
Therefore, the side (hypotenuse) opposite to the largest angle will be the longest one.
Chapter 8: Quadrilaterals
Here, we learn about different types of quadrilaterals and their properties.
Check Important Links for
MCQs on Quadrilaterals
Ques. A diagonal of a parallelogram divides it into two congruent:
- Square
- Parallelogram
- Triangles
- Rectangle
Answer: c
Ques. The three angles of a quadrilateral are 75°, 90°, and 75°. The fourth angle is
- 90°
- 95°
- 105°
- 120°
Answer: d
Explanation: We know that the sum of angles of a quadrilateral is 360º.
Let the unknown angle be x.
Therefore, 75°+90°+75°+x = 360°
x = 360° – 240° = 120°.
Chapter 9: Circles
Here, we learn about the properties and theorems of circles.
Read Also:
MCQs on Circles
Ques. The angle subtended by the diameter of a semicircle is:
- 90
- 45
- 180
- 60
Answer: c
Explanation: The semicircle is half of the circle; hence the diameter of the semicircle will be a straight line subtending 180 degrees.
Ques. If chords AB and CD of congruent circles subtend equal angles at their centres, then:
- AB = CD
- AB > CD
- AB < AD
- None of the above
Answer: a
Explanation: Take the reference of the figure from the above question.
In triangles AOB and COD,
∠AOB = ∠COD (given)
OA = OC and OB = OD (radii of the circle)
So, ΔAOB ≅ ΔCOD. (SAS congruence)
∴ AB = CD (By CPCT)
Chapter 10: Constructions
Here, we learn how to construct different angles, triangles, etc. simply with a ruler and a compass.
Also Check:
MCQs on Constructions
Ques. A triangle ABC with AB = 4 cm and ∠A= 60° and ∠B= 40° is constructed. Then what is the measurement of ∠C?
- 40°
- 60°
- 80°
- 100°
Answer: c
Explanation: By using the angle-sum property of a triangle,
∠A+∠B+∠C = 180°
60°+40°+∠C = 180°
∠C = 180°-100°= 80°.
Ques. The side lengths of 2 cm, 3 cm, and 4 cm can be the sides of
- Scalene triangle
- Isosceles triangle
- Equilateral triangle
- None of the above
Answer: a
Explanation: The side length of 2 cm, 3 cm, and 4 cm can be the sides of a scalene triangle, as all the sides have different measurements.
Chapter 11: Area of Parallelograms and Triangles
Here, we learn to calculate the areas of different parallelograms and triangles.
Also Read: Area of Parallelograms
MCQs on Area of Parallelograms and Triangles
Ques. If ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 10 cm, AE = 6 cm, and CF = 5 cm, then AD is equal to:
- 10cm
- 6cm
- 12cm
- 15cm
Answer: c
Explanation: Given,
AB = CD = 10 cm (Opposite sides of a parallelogram)
CF = 5 cm and AE = 6 cm
Now, Area of parallelogram = Base × Altitude
CD × AE = AD × CF
10 × 6 = AD × 5
AD = 60/5
AD = 12 cm
Ques. The area of a parallelogram with base “b” and height “h” is
- b×h square units
- b² square units
- h² square units
- b+h square units
Answer: a
Explanation: The area of a parallelogram with base “b” and height “h” is b×h (i.e.) A = base × height square units.
Chapter 12: Heron’s Formula
Here we learn about Heron’s formula for calculating the area of a triangle that has sides of different lengths.
MCQs on Heron’s Formula
Ques. If the perimeter of an equilateral triangle is 180 cm. Then its area will be:
- 900 cm²
- 900√3 cm²
- 300√3 cm²
- 600√3 cm²
Answer: b
Explanation: Given, Perimeter = 180 cm
3a = 180 (Equilateral triangle)
a = 60 cm
Semi-perimeter = 180/2 = 90cm
Now, as per Heron’s formula,
A=s(s-a)(s-b)(s-c)
Hence, if we put values here, we get:
A = 900√3
Ques. The area of an equilateral triangle having a side length equal to √3/4cm is:
- 2/27 sq.cm
- 2/15 sq.cm
- 3√3/64 sq.cm
- 3/14 sq.cm
Answer: c
Explanation: Here, a = b = c = √3/4
Find the semi-perimeter of the triangle and use Heron’s formula to find the answer.
Chapter 13: Surface Areas and Volumes
This chapter explains in detail the calculations and formulas concerning the surface areas and volumes of many 2D and 3D geometric figures.
Check Areas for
MCQs on Surface Areas and Volumes
Ques. The surface area of a cuboid-shaped box having length=80 cm, breadth=40cm and height=20cm is:
- 11200 sq.cm
- 13000 sq.cm
- 13400 sq.cm
- 12000 sq.cm
Answer: a
Explanation: surface area of the box = 2(lb + bh + hl)
S.A. = 2[(80 × 40) + (40 × 20) + (20 × 80)]
= 2[3200 + 800 + 1600]
= 2 × 5600 = 11200 sq.cm.
Ques. The total surface area of a cube is 96 cm². The volume of the cube is:
- 8 cm³
- 512 cm³
- 64 cm³
- 27 cm³
Answer: c
Explanation: We know that the TSA of the cone = 6a².
6a² = 96 cm²
a² = 96/6 = 16
a =4 cm
The volume of cone = a³ cubic units
V = 4³ = 64cm³.
Chapter 14: Statistics
Here, we learn about different kinds of data and graphical representations of data.
Also Check:
MCQs on Statistics
Ques. The value which appears very frequently in a data set is called:
- Mean
- Median
- Mode
- Central tendency
Answer: c
Ques. Find the range of the following data: 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20.
- 10
- 15
- 18
- 26
Answer: d
Explanation: Range = Maximum value–Minimum value
Range = 32-6 = 26.
Chapter 15: Probability
Here, we learn to calculate the probability of an event happening.
Also Read:
MCQs on Probability
Ques. The probability of drawing an ace card from a deck of cards is:
- 1/52
- 1/26
- 4/13
- 1/13
Answer: d
Explanation: There are 4 aces in a deck of cards.
Hence, the probability of taking one ace out of 52 cards = 4/52 = 1/13
Ques. f P(E) = 0.44; then P(not E) will be:
- 0.44
- 0.55
- 0.50
- 0.56
Answer: d
Explanation: We know;
P(E) + P(not E) = 1
0.44 + P(not E) = 1
P(not E) = 1–0.44 = 0.56.
Formulas to Remember
Geometry
| Geometric Figure | Area | Perimeter |
| Rectangle | A = l × w | P = 2 × (l + w) |
| Triangle | A = ( ½) b h | P = a + b + c |
| Trapezoid | A = (½) h (b1 + b2) | P = a + b + c + d |
| Parallelogram | A = b h | P =2 (a+b) |
| Circle | A = r2 | C = 2 r |
Algebra
| (a + b)2 = a2 + 2ab + b2 |
| (a - b)2 = a2 - 2ab + b2 |
| a2 - b2 = (a + b) (a - b) |
| (a + b)3 = a3 + b3 + 3ab(a + b) |
| (a - b)3 = a3 - b3 - 3ab(a - b) |
| (a + b + c)2 = a + b + c + 2ab + 2bc + 2ca |
| a3 + b3 = (a + b) (a - ab + b) |
| a3 - b3 = ( a - b) (a + ab + b) |
| a2 + b2 = ½ [(a + b) + (a - b)] |
Surface Area and Volumes
| Shape | Surface Area | Volume |
| Cuboid | 2(lb + bh +lh) l=length, b=breadth, h=height | lbh |
| Cube | 6a2 | a3 |
| Cylinder | 2\(\Pi\)r(h+r) r=radius of circular bases, h=height of cylinder | \(\Pi\)r2h |
| Cone | \(\Pi\)r(l+r) r=radius of base, l=slant height; also l2=h2+r2, where h is the height of the cone. | (\(\Pi\))\(\Pi\)r2h |
| Sphere | 4\(\Pi\)r2 | (4/3)\(\Pi\)r3 |
Heron’s Formula
| Area of triangle using three sides = s(s-a)(s-b)(s-c) |
| Semi-perimeter, s = (a+b+c)/2 |
Polynomial
| P(x) = anxn + an-1xn-1 - an-2xn-2 + …+ ax + a0 |
Statistics
| Measure of Central Tendency | |
| Mean | Sum of Observation/Total number of observation = ????X/n |
| Median | [(n+1)/2]th term [For odd number of observations] Mean of (n/2)th term and (n/2+1)th term [For even number of observations] |
| Mode | The value which is repeated for maximum time in a data set |
Probability
| Empirical Probability = Number of trials with expected outcome/Total number of trials. |
Sample Questions
Ques. Represent 9.3 on the number line. (2 marks)
Ans. Draw a line segment AB = 9.3 units and extend it to C such that BC = 1 unit.
Find the midpoint of AC and mark it as O.
Draw a semicircle taking O as center and AO as radius. Draw BD ⊥ AC.
Draw an arc taking B as center and BD as radius meeting AC produced at E such that BE=BD= 9.3 units.
Ques. Write the coefficients of x2 in each of the following (2 marks)
(i) 2 + x2 + x
(ii) 2 – x2 + x3
(iii) π/2x2 + x
(iv) √2x – 1
Ans.
(i) The given polynomial is 2 + x2 + x.
The coefficient of x2 is 1.
(ii) The given polynomial is 2 – x2 + x3.
The coefficient of x2 is -1.
(iii) The given polynomial is π2x2 + x.
The coefficient of x2 is π2
(iv) The given polynomial is √2x – 1.
The coefficient of x2 is 0.
Ques. In which quadrant or on which axis do each of the points (-2, 4),(3, -1), (-1, 0),(1, 2) and (-3, -5) lie? Verify your answer by locating them on the Cartesian plane. (3 marks)
Ans. The point (-2, 4) has a negative abscissa and a positive ordinate.
∴ (-2,4) lies in the 2nd quadrant.
The point (3, -1) has a positive abscissa and a negative ordinate.
∴ (3, -1) lies in the 4th quadrant.
The point (-1, 0) has a negative abscissa and zero ordinate.
∴ The point (-1, 0) lies on the negative x-axis.
Point (1, 2) is having the abscissa as well as the ordinate positive.
∴ Point (1,2) lies in the 1st quadrant.
The point (-3, -5) is having the abscissa as well as the ordinate negative.
∴ Point (-3, -5) lies in the 3rd quadrant.
These points are plotted in the Cartesian plane as shown in the following figure as A(-2, 4); B(3, -1); C(-l, 0); D(l, 2), and E (-3, -5).
Ques. Find the value of k, if x = 2, y = 1 ¡s a solution of the equation 2x + 3y = k. (2 marks)
Ans. We have 2x + 3y = k
putting x = 2 and y = 1 in 2x+3y = k,we get
2(2) + 3(1) ⇒ k = 4 + 3 – k ⇒ 7 = k
Thus, the required value of k is 7.
Ques. If a point C lies between two points A and B such that AC = BC, then prove that AC = ½ AB, explain by drawing the figure. (2 marks)
Ans. We have,
AC = BC [Given]
∴ AC + AC = BC + AC
[If equals are added to equals, then wholes are equal]
or 2AC = AB [? AC + BC = AB]
or AC = ½ AB
Ques. In the figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c. (3 marks)
Ans. Since XOY is a straight line.
∴ b+a+∠POY= 180°
But ∠POY = 90° [Given]
∴ b + a = 180° – 90° = 90° …(i)
Also a : b = 2 : 3 ⇒ b = 3a/2 …(ii)
Now from (i) and (ii), we get
3a/2 + A = 90°
⇒ 5a/2 = 90°
⇒ a = 90° /5 × 2 =36°
From (ii), we get
b = 3/2 x 36° = 54°
Since XY and MN intersect at O,
∴ c = [a + ∠POY] [Vertically opposite angles]
or c = 36° + 90° = 126°
Thus, the required measure of c = 126°.
Ques. AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB. (2 marks)
Ans. In ?BOC and ?AOD, we have
∠BOC = ∠AOD
BC = AD [Given]
∠BOC = ∠AOD [Vertically opposite angles]
∴ ?OBC ≅ ?OAD [By AAS congruence]
⇒ OB = OA [By C.P.C.T.]
I.e O is the midpoint of AB.
Thus, CD bisects AB.
Ques. ABCD is a quadrilateral in which P, Q, R, and S are mid-points of the sides AB, BC, CD, and DA (see figure). AC is a diagonal. Show that (3 marks)
(i) SR || AC and SR = ½ AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Ans. 
(i) In ?ACD, we have
∴ S is the midpoint of AD and R is the midpoint of CD.
SR =½ AC and SR || AC …(1)
[By mid-point theorem]
(ii) In ?ABC, P is the AB and Q is the midpoint of BC.
PQ = ½ AC and PQ || AC …(2)
[By mid-point theorem]
From (1) and (2), we get
PQ = ½ AC = SR and PQ || AC || SR
⇒ PQ = SR and PQ || SR
(iii) In a quadrilateral PQRS,
PQ = SR and PQ || SR [Proved]
∴ PQRS is a parallelogram.
Ques. If a line intersects two concentric circles (circles with the same centre) with centre 0 at A, B, C, and D, prove that AB = CD (see figure). (2 marks)
Ans. Given: Two circles with the common center O.
A line D intersects the outer circle at A and D and the inner circle at B and C.
To Prove : AB = CD.
Construction:
Draw OM ⊥ l.
Proof: For the outer circle,
OM ⊥ l [By construction]
∴ AM = MD …(i)
[Perpendicular from the center to the chord that bisects the chord]
Ques. Construct a ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 35 cm. (3 marks)
Ans. 
Steps of Construction:
Step I: Draw BX
Step II: Along BX, cut off a line segment BC = 8 cm.
Step III: At B, construct ∠CBY = 45°
Step IV: FromBX, cut off BD = 3.5 cm (= AB – AC)
Step V: Join DC.
Step VI: Draw PQ, perpendicular bisector of DC, which intersects BYat A.
Step VII: Join AC.
Thus, ?ABC is the required triangle
Ques. In figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD. (2 marks)
Ans. We have, AE ⊥ DC and AB = 16 cm
? AB = CD [Opposite sides of parallelogram]
∴ CD = 16 cm
Now, area of parallelogram ABCD = CD x AE
= (16 x 8) cm² = 128 cm² [? AE = 8 cm]
Since, CF ⊥ AD
∴ Area of parallelogram ABCD = AD x CF
⇒ AD x CF = 128 cm
⇒ AD x 10 cm = 128 cm2 [? CF= 10 cm]
⇒ AD = 128/10 cm = 12.8 cm
Thus, the required length of AD is 12.8 cm.
Ques. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m, and 120 m (see figure). The advertisements yield an earning of ?5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay? (2 marks)
Ans. Let the sides of the triangular will be
a = 122m, b = 12cm, c = 22m
Semi-perimeter, s = a+b+c/2
(122+120+22/4)m = 264/2 m = 132m
The area of the triangular side wall
Rent for 1 year (i.e. 12 months) per m2 = Rs. 5000
∴ Rent for 3 months per m2 = Rs. 5000 x 3/12
= Rent for 3 months for 1320 m² = Rs. 5000 x 3/12 x 1320 = Rs. 16,50,000.
Ques. The floor of a rectangular hall has a perimeter of 250 m. If the cost of painting the four walls at the rate of ?10 per m² is ?15000, find the height of the hall. [Hint: Area of the four walls = Lateral surface area]. (2 marks)
Ans. A rectangular hall means a cuboid.
Let the length and breadth of the hall be l and b, respectively.
∴ Perimeter of the floor = 2(l + b)
⇒ 2(l + b) = 250 m
? Area of four walls = Lateral surface area = 2(1 + b) x h, where h is the height of the hall = 250 h m²
Cost of painting the four walls
= Rs. (10 x 250 h) = Rs. 2500h
⇒ 2500 h = 15000 ⇒ h = 15000/2500 = 6
Thus, the required height of the hall = 6 m.
Ques. Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows: (2 marks)
Prepare a frequency distribution table for the data given above.
Ans. The required frequency distribution table is
Ques. In a particular section of class IX, 40 students were asked about the month of their birth and the following graph was prepared for the data obtained. (2 marks)
Find the probability that a student in the class was born in August.
Ans. From the graph, we have
Total number of students born in various months = 40
Number of students born in August = 6
∴ Probability of a Class-IX student who was born in August = 6/40 + 3/20.
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