MCQ in Mathematics
For IX, X, XI and XII Classes
Multiple Choice Questions (MCQ) in Mathematics for 9th, 10th, 11th and 12th standard and case study based questions with answers. Chapter-wise important MCQ in mathematics.
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Multiple Choice Questions (MCQ) in Mathematics for 9th, 10th, 11th and 12th standard and case study based questions with answers. Chapter-wise important MCQ in mathematics.
Mathematics
Q 1) In an Arithmetic Progression, if a = 28, d = – 4, n = 7, then an is:
a) 4
b) 5
c) 3
d) 7
Q 21) The sum of all two digit odd numbers is
a) 2575
b) 2475
c) 2524
d) 2425
Q 28) If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?
(a) 30
(b) 33
(c) 37
(d) 38
Q 31) If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are
(a) 2, 4, 6
(b) 1, 5, 3
(c) 2, 8, 4
(d) 2, 3, 4
Q 32) If the sum of the first m terms of an AP is n and the sum of its n terms is m, then the sum of its (m + n) terms will be
(a) m + n
(b) -(m + n)
(c) m – n
(d) 0
Q 42) If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?
(a) 30
(b) 33
(c) 37
Q 1) Which one of the following is not a quadratic equation?
(a) (x + 2)22 = 2(x + 3)
(b) x22 + 3x = (–1) (1 – 3x)22
Ans: c
(a) 2x² + x – 1 = 0
(b) 2x² – x – 1 = 0
(c) 2x² + x + 1 = 0
(d) 2x² – x + 1 = 0
(a) 4
(b) 3
(c) 3.5
(d) -3
Ans: c
Q 13) The sum of the roots of the quadratic equation 3x2 – 9x + 5 = 0 is
(a) 3
(b) 6
(c) – 3
(d) 2
Q 21) Equation of (x + 1)2 – x2 = 0 has number of real roots equal to:
Q 29) The equation 2x² + kx + 3 = 0 has two equal roots, then the value of k is
(a) ± √6
(b) ± 4
(c) ± 3√2
(d) ± 2√6
Q 31) A quadratic equation ax2 + bx + c = 0 has no real roots, if
(a) b2 – 4ac > 0
(b) b2 – 4ac = 0
(c) b2 – 4ac < 0
(d) b2 – ac < 0
Q 1) The probability of event equal to zero is called;
(a) Certain event
(b) Sure Event
(c) Impossible event
(d) Independent event
Q 2) What is the probability of a sure event?
a) 0
b) 1
c) 2
d) 3
Q 3) The probability that cannot exist among the following:
(a) 2 / 3
(b) -1.5
(c) 15%
(d) 0.7
Explanation:
Q 21) The probability that a non leap year selected at random will contain 53 Sundays is
(a) 1/7
(b) 2/7
(c) 3/7
(d) 5/7
Ans: a
Explanation:
Non-leap year = 365 days
365 days = 52 weeks + 1 day
For 52 weeks, number of Sundays = 52
1 remaining day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Total possible outcomes = 7
The number of favorable outcomes = 1 Thus, the probability of getting 53 Sundays = 1/7
Q 33)
A man is known to speak truth 3 out of 4 times. He throws a die and a number other than six comes up. Find the probability that he reports it is a six.
a) 3/ 4
b) 1/ 4
c) 1/ 2
d) 1
Ans: b
Explanation:
Probability of telling a truth = 3/4
Q 36) An integer is chosen at random from 1 to 100, Find the probability that it is a prime number
a) 4/ 25
b) 8/ 25
c) 1/ 4
d) 2/ 5
Q 37) A jar contains 24 marbles, some are green and some others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/ 3. Find the number of blue marbles.
a) 12
b) 10
c) 8
d) 16
Q 38) Three coins are tossed together. Find the probability of getting at most 2 tails.
a) 7/ 8
b) 3/ 4
c) 1/ 8
d) 5/ 8
Ans: a
Solution Hint
When 3 coins are tossed then possible outcomes are as follows
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTH}
P(Getting at most 2 tail) = P(≤ two tails) = P(2 tail) + P(1 tail) + P(no tail)
(a) 1/45
(b) 2/15
c) 4/45
(d) 1/9
Q 44) Two fair coins are tossed. What is the probability of getting at the most one head?
(a) 3 ⁄ 4
(b) 1 ⁄ 4
(c) 1 ⁄ 2
(d) 3 / 8
Ans: a
Solution Hint
Possible outcomes are (HH), (HT), (TH), (TT)
Favorable outcomes(at the most one head) are (HT), (TH), (TT) So probability of getting at the most one head =3/4
Solution Hint
Number of Possible outcomes are 26
Favorable outcomes are M, A, T, H, E, I, C, S probability = 8/26 = 4/13
MATHEMATICS
Q 1) What is the formula for the circumference of a circle?
a) C = 2πd
b) C = 2πr
c) C = 2πa
d) C = 2πs
(a) r2
(b) 1/2r2
(c) 2r2
(d) √2r2
Ans: a
(a) 256 cm2
(b) 128 cm2
(c) 642 cm2
(d) 64 cm2
Ans: b
Solution Hint
Radius of circle = 8 cm
Diameter of circle = 16 cm = diagonal of the square
Let “a” be the triangle side, and the hypotenuse is 16 cm
Using Pythagoras theorem, we can write
162 = a2 + a2 ⇒ 256 = 2a2 ⇒ a2 = 256/2 ⇒ a2 = 128 = Area of a Square
Q 21) Find the radius of the wheel if the wheel rotates 100 times to cover 500 m.
a) 0.07 m
b) 0.47 cm
c) 0.79 m
d) 0.57 cm
(b) 462 cm2
(c) 386 cm2
(d) 512 cm2
Ans: b
b) 7 cm
c) 21 cm
d) 10.5 cm
Ans: a
Solution Hint Find the radius by using the relation :
Q 35) The difference between circumference and diameter of a ring is 10 cm. Find the radius of the ring.
a) 3.07 cm
b) 0.37 cm
c) 2.33 cm
d) 4.57 cm
Ans: c
Solution Hint Circumference – Diameter = 10 cm (∵ Diameter = 2 × radius)
2πr – 2r = 10 cm ⇒ 2r(π – 1) = 10 cm
Q 36) The difference between the circumference and radius of a circle is 37 cm then area of the circle is
a) 111 cm2
b) 184 cm2
c) 154 cm2
d) 259 cm2
Solution Hint
a)
20 %
b)
18 %
c)
19 %
d)
11 %
Ans:
c
Solution Hint
Take diameter = 2r and radius = r
Find area by taking radius r
Find new area by taking radius = (2r + 20% of 2r)
Find the difference of two areas
Required
Percentage =
Ans: b
a)
42 cm2
b)
48 cm2
c)
38.5 cm2
d)
77 cm2
Ans: c
a)
54 cm2
b)
42 cm2
c)
48 cm2
d)
36 cm2
Ans: b
a)
16.3 cm2
b)
17.3 cm2
c)
18.8 cm2
d)
18.3 cm2
Ans: d
b)
231 cm2
c)
308 cm2
d)
316 cm2
Ans: c
a) 54 cm2
b)
42 cm2
c)
56 cm2
d)
48 cm2
Ans: b
b)
1416 cm2
c)
1308 cm2
d)
1386 cm2
Ans: d
b)
154 cm2
c)
308 cm2
d)
77 cm2
Ans: a
a)
36 cm2
b)
18 cm2
c)
12 cm2
d)
9 cm2
Ans: d
a)
256 cm2
b)
128 cm2
c)
64 cm2
d)
77 cm2
Ans: b
Q 52) In a circle of diameter 42cm ,if an arc subtends an angle of 60 ̊ at the centre where 𝜋 = 22/7,then the length of the arc is
(a) 22/7 cm
(b) 11cm
(c) 22 cm
(d) 44 cm
Ans: c
Q 53) If the circumference of a circle increases from 2𝜋 to 4𝜋 then its area _____ the original area
(a) Half
(b) Double
(c) Three times
(d) Four times
Ans: d
Q 54) If the difference between the circumference and the radius of a circle is 37cm , 𝜋 = 22/7, the circumference (in cm) of the circle is
(a) 154
(b) 44
(c) 14
(d) 7
Ans: b
Q 55) The perimeter of a semicircular protractor whose radius is ‘r’ is
(a) 𝜋 + 2r
(b) 𝜋 + r
(c) 𝜋 r
(d) 𝜋 r + 2r
Ans: d
Q 56) Area
of a sector of a circle is 1/6 to the area of circle. Find the degree measure
of its minor arc.
(a)
90 ̊
(b)
60 ̊
(c)
45 ̊
(d)
30 ̊
Ans: b
Q 57) The
number of revolutions made by a circular wheel of radius 0.7m in rolling a
distance of 176m is
(a)
22
(b)
24
(c)
75
(d)
40
Ans: d
Q 58) In
the figure given below, ABCD is a square of side 14 cm with E, F, G and H as
the mid points of sides AB, BC, CD and DA respectively. The area of the shaded
portion is
(b)
49 cm2
(c)
98 cm2
(d)
49π/2 cm2
Ans:
c
Solution Hint
Shaded area = Area of semicircle + (Area of half square – Area
of two quadrants)
=
Area of semicircle + (Area of half square – Area of semicircle)
= Area of half square = x 14 x14 = 98cm2
(b)
(π/6 – √3/4) cm2
(c)
4(π/6 – √3/4) cm2
(d)
8(π/6 – √3/4) cm2
Ans: d
Solution Hint
Let O be the center of
the circle. OA = OB = AB = 1cm.
So ∆OAB is an equilateral triangle and ∴ ∠AOB = 60°
Required Area= 8x Area of
one segment with r = 1cm, θ = 60°
(a)
50√2 cm
(b)
100/π cm
(c)
50√2/π cm
(d)
100√2/π cm
Ans: c
a) Right-angled triangles b) Any type of triangles
c) Acute angled triangles d) Obtuse angled triangles
Q 21) Evaluate (cosec θ – cot θ) (cosec θ + cot θ).
a) 0 b) 1 c) 2 d) 3
Q 22) Evaluate cosec θ sec θ.
a) cos θ + tan θ b) cos θ – tan θ
c) tan θ – cot θ d) cot θ + tan θ
Q 23) (Sin 30° + cos 60°) – (sin 60° + cos 30°) is equal to:
(a) 0
(b) 1 + 2√3
(c) 1-√3
(d) 1+√3
Q 25) Evaluate sec2 A + (1 + tan A) (1 – tan A).
Q 28) The value of (Cos θ + sin θ)2 + (Cos θ – sin θ)2 is
a) -2
b) 0
c) 1
d) 2
Q 31) sin (90° – A) and cos A are:
(a) Different
(b) Same
(c) Not related
(d) None of the above
Q 42) The value of (tan 1° tan 2° tan 3° … tan 89°) is
(a) 0
(b) 1
(c) 2
(d) ½
Ans: b
Explanation:
tan 1° tan 2° tan 3°…tan 89°
= [tan 1° tan 2°…tan 44°] tan 45° [tan (90° – 44°) tan (90° – 43°)…tan (90° – 1°)]
= [tan 1° tan 2°…tan 44°] [cot 44° cot 43°…cot 1°] × [tan 45°]
= [(tan 1°× cot 1°) (tan 2°× cot 2°)…(tan 44°× cot 44°)] × [tan 45°]
= 1 × 1 × 1 × 1 × …× 1 {since tan A × cot A = 1 and tan 45° = 1} = 1
a) 2
b) -1
c) 1
d) 0
Ans: c
a) -30°
b) 30°
c) 90°
d) 60°
Ans: a
a) 2
b) 0
c) 1
d) ∞
Ans: c
a) 0
b) 2
c) 1
d) 3
Ans: c
a) sin
2A = 2sinA
b) tan
A =
c)
sin(A + B) = sin A + sin B
d) (sin
A)2 = sin2 A
Ans: d
a)
b)
c)
d)
Ans: a
a) tan 60o
b) cos 30o
c) tan 30o
d) sin 30o
Ans: c
a) 24o, 18o
b) 18o, 24o
c) 24o, 28o
d) 20o, 24o
Ans: b
a) 24o, 18o
b) 18o, 26o
c) 28o, 24o
d) 24o, 30o
Ans: a
a) 0o
b) 30o
c) 60o
d) 90o
Ans: d
a) 210
b)
2
c) 25
d) 1
Ans: b
Solution
Hint
tanθ + 1/ tanθ = 2 ⇒ tan2 θ – 2tan θ + 1 = 0
Factorise
this equation and then solving it we get θ = 45o
Using this value
of θ and find the required value
a) 215
b)
1
c)
2
d) 0
Ans: c
Solution
Hint
sinθ + cosecθ = 2 = 1 + 1
sinθ = 1 and cosecθ = 1
sin15 θ + cosec 15 θ = (sinθ) 15 + (cosec θ) 15 = (1)15 + (1)15 = 1 + 1 = 2
a) 45o
b) 30o
c) 60o
d) 90o
Ans: a
Solution
Hint
Dividing
both side by cos2 θ we get
Sec2 θ + tan2 θ = 3 tan θ
1 + tan2 θ + tan2 θ = 3 tan θ
2 tan2 θ – 3 tanθ + 1 = 0
Factorise
this equation we get tan θ = 1 or 1/2
tanθ = 1 ⇒ θ = 45o
a) 1
b) – 1
c) 2
d) 3
Ans: a
Solution
Hint:
(sec θ – tan θ)(sec θ+ tan θ ) = sec2 θ – tan2 θ = 1
a) 3
b) 3/ 2
c) 2/
3
d) 0
Ans: d
Solution
Hint
sin θ +
cos θ = 1
Squaring
on both side and find the value of sin θ
cos θ
We get sin θ cos θ = 0 ⇒ 3 x sin θ cos θ = 0
Q 62) If sinθ = x and secθ = y , then tanθ is
(a)
xy
(b)
x/y
(c)
y/x
(d)
1/xy
Ans: a
Q 63) What
is the value of (tanθ cosecθ)2 – (sinθ secθ)2
(a)
-1
(b)
0
(c)
1
(d)
2
Ans: c
Q 64) Given
that sinθ = a/b ,then tanθ is equal to
Q 65) If
x = 2sin2θand y = 2cos2θ+ 1 then x + y is
(a)
3
(b)
2
(c)
1
(d)
1/2
Ans: a
(a)
-1
(b)
0
(c)
1
(d)
2
Ans: c
(a)
3/4
(b)
5/12
(c)
4/3
(d)
12/5
Ans: d
(a)
-1
(b)
0
(c)
1
(d)
√3/2
Ans:
b
Solution Hint
tan A = √3 = tan 60° so ∠A = 60°, Hence ∠C = 30°.
cos A cos C – sin A sin C = (1/2)x (√3/2) –
(√3/2)x (1/2) = 0
(a)
0
(b)
1/2
(c)
1
(d)
√3/2
Ans:
a
Solution Hint
1x + 1x + 2x = 180°, x = 45°.
∠A , ∠B and ∠C are 45°, 45° and
90°resp.
Using these values to find
the value of the given function.
(a)
0
(b)
1/3
(c)
2/3
(d)
3 ⁄ 4
Ans: a
(a)
0
(b)
2
(c)
20
(d)
220
Ans:
b
Solution Hint
tan α + cot α = 2 gives α = 45°. So tan α = cot α = 1
tan20α + cot 20α
= 120 + 120 = 1 + 1 = 2
(a)
-1, 1
(b)
0, 1
(c)1,
2
(d)
-1, -1
Ans:
c
Solution Hint
1+ sin2α = 3 sinα cos α
sin2α + cos2α + sin2α = 3 sinα
cos α
2 sin2α – 3sinα cos α + cos2α = 0
(2sinα – cos α)( sinα – cosα) =0
⸫ cotα
= 2 or cotα = 1
(a)
0o
(b)
90o
(c)
45o
(d)
30o
Ans: b
(a)
2
(b)
1/2
(c)
1/3
(d)
1/4
Ans: b
Q 1) Which of the following triangles have the same side lengths?
(a) Scalene
(b) Isosceles
(c) Equilateral
(d) None of these
a) True
b) False
Ans: a
Explanation The two figures shown are similar because they have same shape but are different in sizes. The second pentagon is smaller in size as compared to a pentagon, but the basic structure of both the figures is same i.e. both are pentagon.
Q 21) In two similar triangles ∆ABC and ∆DEF, AB = 15cm, DE = 5cm. If AL and DM are the altitudes of the triangles ABC, DEF respectively, then what will be the ratio of their altitudes?
Q 1)What will be the nature of the graph lines of the equations 5x – 2y = 9 and 15x – 6y = 1 ?
a) Parallel
b) Coincident
c) Intersecting
d) Perpendicular to each other
Ans: a
Q 19) A pair of linear equations a1x + b1y + c1 = 0; a2x + b2y
+ c2 = 0 is said to be inconsistent, if
Q 21) If the lines given by 2x + ky = 1 and 3x – 5y = 7 are parallel, then the value of k is
a) -10/3
b) 10/3
c) -13
d) -7
Ans a
Q1) How many points will the graph of x2 + 2x + 1 will cut the x-axis?
a) 3
b) 2
c) 1
d) 0
Ans: b
Q2) If the graph of a polynomial cuts the x-axis at 3 points, then the polynomial is a) Linear
b) Quadratic
c) Cubic
d) Biquadratic
Ans: c
Q3) What will be the nature of the zeros of a quadratic polynomial if it cuts the x-axis at two different points?
a) Real
b) Distinct
c) Real, Distinct
d) Complex
a) x2 + 2x + 5
Ans: a
Q5) Which of the following is not a polynomial ?
a) x2 + 5x + 10
b) x – 2 + 2x + 4
c) x12 + 10x
d) 5x + 4
Q6) If the zeros of a polynomial are 3 and -5, then they cut the x-axis at ____ and _____ points.
a) (8, 0) and (-4, 0)
b) (3, -3) and (-5, 5)
c) (-3, 0) and (5, 0)
d) (3, 0) and (-5, 0)
Ans: d
Q7) The sum and product of zeros of a quadratic polynomial are 10 and 5/2 respectively. What will be the quadratic polynomial?
a) 2x2 – 20x + 10
b) 2x2 – x + 5
c) 2x2 – 20x + 5
d) x2 – 20x + 5
Q9) If α and β are the zeros of 3x2 – 5x – 15, then the value of αβ is
a) -5
b) – 10
c) – 15
d) – 20
Ans: a
Q 10) What will be the value of other zero, if one zero of the quadratic polynomial is 5 and the sum of the zeros is 10?
a) 10
b) 5
c) -5
d) -10
Ans: b
Q 11) The value of a and b, if the zeros of x2 + (a + 5)x – (b – 4) are -5 and 9 will be
a) 47, -5
b) -5, 47
c) -9, 49
d) -4, 45
Q13) If α, β and γ are the zeros of 5x3 + 10x2 – x + 20, then the value of αβγ is
a) -1
b) 5
c) -10
d) – 4
Ans: d
Q14) If α, β and γ are the zeros of 2x3 – 6x2 + 5x + 2, then the value of α + β + γ is
a) 0
b) 1
c) 3
d) 2
Ans c
Q15) If the two zeros of the polynomial x3 – 9x2 -x + 9, are 1 and 9, then the third zero is
a) 9
b) 1
c) 2
d) -1
Ans: d
Q 16) If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, then
(a) a = -7, b = -1
(b) a = 5, b = -1
(c) a = 2, b = -6
(d) a =0, b = -6
Ans: d
Q 18) A quadratic polynomial, whose zeroes are -4 and -5, is
(a) x² – 9x + 20
(b) x² + 9x + 20
(c) x² – 9x – 20
(d) x² + 9x – 20