Mathematics MCQ | Class 09 | Chapter 14
STATISTICS

Features

Action Plan

MCQ |Chapter 14 | Statistics | Class IX

Question: 1

The collection of information, collected for a purpose is called:

a) Mean

b) Median

c) Mode

d) Data

Answer: d

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       Mathematics MCQ | Class 09 | Chapter 15

PROBABILITY

Features

Action Plan

MCQ |Chapter 15 | Probability | Class IX

Question 1

1) The probability of
each event, when a coin is tossed for 1000 times with frequencies: Head: 455
& Tail: 545 is:

a)
0.455 & 0.545

b)
0.5 & 0.5

c)
0.45 & 0.55

d)
455 & 545

Answer: a

Question 2

Marks
obtained by a student in a test is shown in the table below.

Test no.	1	2	3	4	5
Marks	81	87	76	70	90

What is the probability that the student has scored more than
80?
a) 3/5
b) 4/5
c) 2/5
d) 1/2

Answer a

Question 3

When a die is thrown, what is
the probability of getting even number

a) 1/6

b) 1/3

c) 1/2

d) 2/3

Answer c

Question 4

The sum of all probabilities equal to:

a)
4

b)
1

c)
3

d)
2

Answer: b

Question 5

3) The probability of each
event lies between:

a)
1 & 2

b)
1 & 10

c)
0 & 1

d)
0 & 5

Answer: c

Question 6

Probability
of impossible event is 

a)
1

b)
0

c) Less
than 0

d)
Greater than 1

Answer
b

Question 7

If P(E) = 0.44, then P(not
E) will be:

a)
0.44

b)
0.55

c)
0.50

d)
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

Question 8

Probability
of certain (sure) event is

a) 1

b) 0

c)
Greater than 1

d)
Less than 0

Answer
a

Question 9

 If P(E) = 0.38, then
probability of event E, not occurring is:

a)
0.62

b)
0.38

c)
0.48

d)
1

Answer: a

Explanation: P(not E) = 1 – P(E) = 1-0.38 = 0.62

Question 10

Two coins are tossed simultaneously.
The probability of getting atmost one head is

a) 1/4

b) 3/4

c) 1/2

d) 1/4

Answer b

Question 11

The probability of drawing
an ace card from a deck of cards is:

a)
1/52

b)
1/26

c)
4/13

d)
1/13

Answer: d

Explanation: There are 4 aces in a deck of card.

Hence, the probability of taking one ace out of 52 cards = 4/52
= 1/13

Question 12

A coin is tossed 1000 times,
if the probability of getting a tail is 3/8, how many times head is obtained.

a) 525

b) 375

c) 625

d) 725

Answer b

Question 13

If the probability of an
event to happen is 0.3 and the probability of the event not happening is:

a)
0.7

b)
0.6

c)
0.5

d)
None of the above

Answer: a

Explanation: Probability of an event not happening = 1 – P(E)

P(not E) = 1 – 0.3 = 0.7

Question 14

A coin is tossed 1000 times,
if the probability of getting a tail is 3/8, how many times head is obtained.

a) 525

b) 375

c) 625

d) 725

Answer c

Question 15

 A dice is thrown. The
probability of getting 1 and 5 is:

a)
1/6

b)
2/3

c)
1/3

d)
1/2

Answer: c

Explanation: The probability of getting 1 and 5 = 2/6 = 1/3

Question 16

In a football match, Ronaldo
makes 4 goals from 10 panalty kicks. The probability of converting a penalty
kick into a goal by Ronaldo, is

a) 1/4

b) 1/6

c) 1/3

d) 2/5

Answer d

Question 17

A batsman hits boundaries
for 6 times out of 30 balls. Find the probability that he did not hit the boundaries.

a)
1/5

b)
2/5

c)
3/5

d)
4/5

Answer: d

Explanation: No. of boundaries = 6

No. of balls = 30

No. of balls without boundaries = 30 – 6 =24

Probability of no boundary = 24/30 = 4/5

Question 18

From a deck of 52 shuffled playing
cards, a card is drawn What is the probability of drawing a king or queen

a) 1/13

b) 2/13

c) 3/13

d 1/52

Answer b

Question 19

Three coins were tossed
200 times. The number of times 2 heads came up is 72. Then the probability of 2
heads coming up is:

a)
1/25

b)
2/25

c)
7/25

d)
9/25

Answer: d

Explanation: Probability = 72/200 = 9/25

Question 20

From a deck of 52 shuffled playing
cards, a card is drawn What is the probability of drawing a red face card

a) 1/13

b) 2/13

c) 3/26

d 1/26

Answer c

Question 21

What is the probability of getting an odd number less than 4, if a die is
thrown?

a) 1/6

b) 1/2

c) 1/3

d) 0

Answer: c

Explanation:
Sample space, S = {1, 2, 3, 4, 5, 6}

Favorable
outcomes = {1, 3}

Therefore, the
probability of getting an odd number less than 4 = 2/6 = 1/3.

Question 22

From a deck of 52 shuffled playing
cards, a card is drawn What is the probability of drawing a black king card

a) 1/13

b) 2/13

c) 3/26

d 1/26

Answer d

Question 23

What is the probability of impossible events?

a) 1

b) 0

c) More than 1

d) Less than 1

Answer: b

Explanation: The
probability of an impossible event is always 0.

Question 24

 Performing an event once is called

a) Sample

b) Trial

c) Error

d) None of the above

Answer: b

Explanation:
Performing an event once is called a trial.

Question 25

In a counting from 1 to 10,
what is the probability of finding a prime number

a) 1/2

b)  3/10

c) 2/5

d) 1/10

Answer c

Question 26

 A card is drawn from a well-shuffled deck of 52 cards. What is the probability
of getting a king of the red suits?

a) 3/36

b) 1/26

c) 3/26

d) 1/16

Answer: b

Explanation: In
a pack of 52 cards, there are a total of 4 king cards, out of which 2 are red
and 2 are black.

Therefore, in a
red suit, there are 2 king cards. 

Hence, the
probability of getting a king of red suits = 2/52 = 1/26.

Question 27

In a counting from 1 to 10,
what is the probability of finding a composite number

a) 1/2

b)  3/10

c) 2/5

d) 1/10

Answer a

Question 28

 Find the probability of a selected number is a multiple of 4 from the numbers
1, 2, 3, 4, 5, …15.

a) 1/5

b) 1/3

c) 4/12

d) 2/15

Answer: a

Explanation: S =
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

Multiples of 4
from the sample space = {4, 8, 12}

Therefore, the
probability of the selected number is a multiple of 5 is 3/15 = 1/5.

Question 29

In the word MATHEMATICS, what
is the probability of finding a vowel

a) 1/11

b) 2/11

c) 3/11

d) 4/11

Answer d

Question 30 

What is the probability of drawing a queen from the deck of 52 cards?

a) 1/26

b) 1/52

c) 1/13

d) 3/52

Answer: c

Explanation:
Total cards = 52

Number of queens
in a pack of 52 cards = 4

Hence, the
probability of drawing a queen from a deck of 52 cards = 4/52 = 1/13

Question 31

In the word MATHEMATICS, what
is the probability of finding a consonant

a) 5/11

b) 6/11

c) 4/11

d) 7/11

Answer d

Question 32

Which of the following cannot be the probability of an event?

a) 1

b) 0

c) 0.75

d) 1.3

Answer: d

Explanation: The
probability of an event always lies between 0 and 1.

Question 33

There are 4 green and 2 red balls in a basket. What is the probability of
getting the red balls? 

a) 1/2

b) 1/3

c) 1/5

d) 1/6

Answer: b

Explanation:
Total balls = 4 green + 2 red = 6 balls

No. of red balls
= 2.

Hence, the
probability of getting the red balls = 2/6 = 1/3

Question 34

Empirical probability is also known as 

a) Classic probability

b) Subjective probability

c) Experimental probability

d) None of the above

Answer: c

Explanation: Empirical
probability is also known as experimental probability.

Question 35

If two coins are tossed simultaneously, then what is the probability of getting
exactly two tails?

a) 1/4

b) 1/2

c) 1/3

d) None of the above

Answer: a

Explanation: If
two coins are tossed, then the sample space, S = {HH, HT, TH, TT}

Favorable
outcome (Getting exactly two tails) = {TT} 

Therefore, the
probability of getting exactly two heads = 1/4

 

      Mathematics MCQ | Class 09 | Chapter 10

CIRCLE

Features

Action Plan

MCQ |Chapter 10 | Circle | Class IX

Question 1

The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called
a) square
b) triangle
c) rectangle
d) circle

Answer d

Question 2

The centre of the circle lies in______ of the circle.
a) Interior
b) Exterior
c) Circumference
d) None of the above
Answer: a

Question 3

A circle divides the plane into __________ parts.
a) two
b) three
c) four
d) five
Answer b

Question 4

The longest chord of the circle is:
a) Radius
b) Arc
c) Diameter
d) Segment
Answer: c

Question 5

From the diagram given below, the shaded area is __________

a) major segment
b) minor segment
c) major sector
d) minor segment
Answer b

Question 6

How many circles can pass through three points which are not collinear?
a) One
b) Two
c) Three
d) Four
Answer a

Question 7

Equal _____ of the congruent circles subtend equal angles at the centres.

a) Segments
b) Radii
c) Arcs
d) Chords
Answer: d
Explanation: See the figure below:

Let ΔAOB and ΔCOD are two triangles inside the circle.

OA = OC and OB = OD (radii of the circle)

AB = CD (Given)

So, ΔAOB ≅ ΔCOD (SSS congruency)

∴  ∠AOB = ∠COD ……….. By CPCT

Hence, this prove the statement.

Question 8

The region between an arc and the two radii, joining the centre to the end points of the arc is called __________
a) arc
b) chord
c) sector
d) area of circle
Answer c

Question 9

When does both segments and both sectors become equal?
a) When two arcs are equal
b) When radius = 2 x chord
c) When two arcs are unequal
d) When chord = radius
Answer a

Question 10

If chords AB and CD of congruent circles subtend equal angles at their centres, then:
a) AB = CD
b) AB > CD
c) AB < AD
d) None of the above
Answer: a
Explanation:
In triangles AOB and COD,
∠AOB = ∠COD (given)
OA = OC and OB = OD (radii of the circle)
So, ΔAOB ≅ ΔCOD. (SAS congruency)
∴ AB = CD (By CPCT)

Question 11
From the diagram given below, if O is the centre of the circle and OL is perpendicular to chord AB, then which of the following is true?

a) AL > LB
b) AL < LB
c) AL ≠ LB
d) AL = LB
Answer d

Question 12

If there are two separate circles drawn apart from each other, then the maximum number of common points they have:
a) 0
b) 1
c) 2
d) 3
Answer: a

Question 13

What is the value of AC if radius of circle is 5cm and OB = 3cm?

a) 5cm

b) 8cm
c) 4cm
d) 2cm

Question 14

The
angle subtended by the diameter of a semi-circle is:

a) 90^o
b) 45^o
c) 180^o
d) 60^o
Answer: c

Explanation: Angle in a semi circle is always right

Question 15

Find the value of AB if radius of circle is 10cm and PQ = 12cm and RS = 16cm?

a) 14cm
b) 16cm
c) 4cm
d) 2cm
Answer d

Question 16
If AB and CD are two chords of a circle intersecting at point E, as per the given figure. Then:

a) ∠BEQ > ∠CEQ
b) ∠BEQ = ∠CEQ
c) ∠BEQ < ∠CEQ
d) None of the above
Answer: b

Explanation:

OM = ON …………….. (Equal chords are always equidistant from the centre)

OE = OE  …………….. (Common)

∠OME = ∠ONE …… (Each = 90^o)

So, ΔOEM ≅ ΔOEN ……… (by RHS congruence rule)

Hence, ∠MEO = ∠NEO …….. (by CPCT)

∴ ∠BEQ = ∠CEQ

Question 17

Find the radius of circle if AB = 7cm, RS = 6cm and PQ = 8cm.

a) 5cm

b) 8cm
c) 4cm
d) 20cm
Answer a

Explanation:

Let r be the radius of circle.
Since perpendicular from centre to a chord bisects the chord,
AR = RS/2 = 3cm and BP = PQ/2 = 4cm
In ΔBOP, ∠OBP = 90°
By Pythagoras theorem, OP² = OB² + BP²
⇒ r² = x² + (4)²
⇒ r² = x² + 16  ——————- (i)
Similarly, In ΔAOR, ∠OAR = 90°
By Pythagoras theorem, OR² = OA² + AR²
⇒ r² = (7 – x)² + (3)²
⇒ r² = (7 – x)² + 9  ———————— (ii)
From equation i and ii, x² + 16 = (7 – x)² + 9
⇒ x² + 16 = 7² – 2 x 7 x x + x² + 9
⇒ 2 x 7 x x = 49 – 16 + 9
⇒ x = 3
Substituting value of x in equation i, r² = 3² + 16
⇒ r = 5cm.

Question 18
If a line intersects two concentric circles with centre O at A, B, C and D, then:
a) AB = CD
b) AB > CD
c) AB < CD
d) None of the above
Answer: a
Explanation: See the figure below:

From the above fig., OM ⊥ AD.

Therefore, AM = MD …….. (1)

Also, since OM ⊥ BC, OM bisects BC.

Therefore, BM = MC ……… (2)

From equation (1) and equation (2)

AM – BM = MD – MC

∴ AB = CD

Question 19
Two circles of radius 13cm and 17cm intersect at A and B. What is the distance between the centres of the circles if AB = 24cm?

a) 15cm
b) 25cm
c) 22cm
d) 14cm
Answer d

Explanation:
Since line joining the centres of the circles bisects the common chord, AC = BC = 12cm.
Also, In ΔAOC, ∠OCA = 90°
By Pythagoras theorem, OA² = OC² + AC²
⇒ 15² = OC² + 12²
⇒ OC² = 225 – 144
⇒ OC = 9cm
Similarly, In ΔAO’C, ∠O’CA = 90°
By Pythagoras theorem, O’A² = O’C² + AC²
⇒ 13² = O’C² + 12²
⇒ OC² = 169 – 144
⇒ OC = 5cm
Now, OO’ = OC + O’C
⇒ OO’ = 9 + 5 = 14cm.

Question 20

In the below figure, the value of ∠ADC is:

a) 60°
b) 30°
c) 45°
d) 55°
Answer: c

Explanation: ∠AOC = ∠AOB + ∠BOC

So, ∠AOC = 60° + 30°

∴ ∠AOC = 90°

An angle subtended by an arc at the centre of the circle is twice the angle subtended by that arc at any point on the rest part of the circle.

So, ∠ADC = 1/2∠AOC

= 1/2 × 90° = 45°

Question 21
In the given figure, find angle OPR.

a) 20°
b) 15°
c) 12°
d) 10°
Answer: d

Explanation: Major arc PR subtend Reflex ∠POR at the centre of the circle and ∠PQR in the remaining part of the circle

So, Reflex ∠POR = 2 × ∠PQR,

We know the values of angle PQR as 100°

So, Reflex ∠POR = 2 × 100° = 200°

∴ ∠POR = 360° – 200° = 160° 

Now, in ΔOPR,

OP and OR are the radii of the circle

So, OP = OR

Also, ∠OPR = ∠ORP = x

By angle sum property of triangle, we know:

∠POR + ∠OPR + ∠ORP = 180°

160° + x + x  = 180°

2x = 180° – 160°

2x = 20° ⇒ x = 10°

Question 22

In the given figure, ∠AOB = 90º and ∠ABC = 30º, then ∠CAO is equal to:

a) 30º
b) 45º
c) 60º
d) 90º
Answer: c
Explanation:

Given that ∠AOB = 90º and ∠ABC = 30º
OA = OB (Radii of the circle)
Let ∠OAB = ∠OBA = x
In the triangle OAB,
∠OAB + ∠OBA + ∠AOB = 180° (By using the angle sum property of triangle)
⇒ x + x + 90° = 180°
⇒ 2x = 180° – 90°
⇒ x = 90°/ 2 = 45°
Therefore, ∠OAB = 45° and ∠OBA = 45°
By using the theorem, “ the angles subtended by arcs at the centre of the circle double the angle subtended at the remaining part of the circle”, we can write
∠AOB = 2∠ACB
This can also be written as,
∠ACB = ½ ∠AOB = (½) × 90° = 45°
Now, apply the angle sum property of triangle on the triangle ABC,
∠ACB + ∠BAC + ∠CBA = 180°
45° + 30° + ∠BAC = 180°
75° + ∠BAC = 180° ⇒ ∠BAC = 180° – 75°
⇒ ∠BAC = 105°
∠CAO + ∠OAB = 105°
∠CAO + 45° = 105°
⇒ ∠CAO = 105° – 45° = 60°
Question 23

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to:

a) 30º
b) 40º
c) 50º
d) 80º
Answer: c

Explanation:

Given that ABCD is a cyclic quadrilateral and ∠ADC = 140º.

We know that the sum of opposite angles of a cyclic quadrilateral is 180°.

Hence, ∠ADC + ∠ABC = 180° 

140° + ∠ABC = 180°

∠ABC = 180° – 140° = 40° 

Now  ∠ACB = 90° . . . . . . .  Angles in the semi circle is right

By using the angle sum property of triangle in the triangle, ABC, 

∠CAB + ∠ABC + ∠ACB = 180°

∠CAB + 40° + 90° = 180° 

∠CAB + 130°   = 180° 

∠CAB = 180° – 130°

∠CAB = 50° 

Therefore, ∠CAB or ∠BAC =50°.

Question 24

 In the given figure, if ∠OAB = 40º, then ∠ACB is equal to

a) 40º
b) 50º
c) 60º
d) 70º
Answer: b
Explanation:

Given that ∠OAB = 40º,

In  the triangle OAB, 

OA = OB (radii)

 ∠OAB = ∠OBA  = 40°  …… (∴ angles opposite to equal sides are equal)

Now, by using the angle sum property of triangle, we can write

∠AOB + ∠OBA + ∠BAO = 180° 

∠AOB + 40° + 40° = 180° 

∠AOB = 180 – 80° = 100° 

Since, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle

∠AOB = 2 ∠ACB 

∠ACB = ½ ∠AOB

Hence, ∠ACB = 100°/2 = 50°.

Question 25

In the given figure, if ∠ABC = 20º, then ∠AOC is equal to:

a) 10º
b) 20º
c) 40º
d) 60º
Answer: c
Explanation:
Given that, ∠ABC = 20º.
since the angle subtended by an arc at the centre of the circle is double the angle subtended at the remaining part.
∠AOC = 2∠ABC
∠AOC = 2 × 20° = 40°.

 Mathematics MCQ Class XI Chapter-11

Conic Section

MCQ BASED ON CIRCLE CLASS XI

Question 1

Find the equation of circle with centre at origin and radius
5 units.
a) x² + y² = 25
b) x² + y² = 5
c) x² = 25
d) y² = 25

Answer a

Question 2

The point (6, 2) lie ___________ the circle x² + y² – 2x -4y – 36 = 0.
a) inside circle
b) outside circle
c) on the circle
d) either inside or outside

Answer c

Question 3

Find the equation of circle with centre at (2, 5) and radius
5 units.
a) x² + y²   + 4x – 10y + 4 = 0
b) x² + y²  – 4x – 10y +
4 = 0
c) x² + y²  + 4x + 10y +
4 = 0
d) x² + y²  + 4x – 10y –
4 = 0

Answer b

Question 4

Find the centre of the circle with equation x² + y² – 4x – 10y + 4 =
0.
a) (-2, 5)
b) (-2, -5)
c) (2, -5)
d) (2, 5)

Answer: d
Explanation: Comparing the equation with general
form x² + y² + 2gx + 2fy + c =
0, we get
2g = – 4 ⇒ g = – 2
2f = -10 ⇒ f = – 5
c = 4
Centre is at (-g, -f) i.e. (2, 5).

Question 5

Find the radius of the circle with equation x² + y² – 4x – 10y + 4 =
0.
a) 25 units
b) 20 units
c) 5 units
d) 10 units

Answer: c

Question 6

The centre of
the circle 4x² + 4y² – 8x + 12y – 25 = 0 is

a) (-2, 3)

b) (1, -3/2)

c) (-4, 6)

d) (4, -6)

Answer: (b)
(1, -3/2)

Explanation:

Given circle
equation: 4x² + 4y² – 8x + 12y – 25 = 0

⇒ x² + y² –
(8x/4) + (12y/4) – (25/4) = 0

⇒ x² +y² -2x
+3y -(25/4) = 0 …(1)

As we know that the
general equation of a circle is x² + y² + 2gx + 2fy + c = 0,
and the centre of the circle = (-g, -f)

Hence, by comparing
equation (1) and the general equation,

2g = -2,  ⇒ g = -1

2f = 3,    ⇒ f = 3/2

Now, substitute the
values in the centre of the circle (-g, -f), we get,

Centre = (1, -3/2).

Question 7

If a circle pass through (2, 0) and (0, 4) and centre at
x-axis then find the radius of the circle.
a) 25 units
b) 20 units
c) 5 units
d) 10 units

Answer: c
Explanation: Equation of circle with centre at
x-axis (a, 0) and radius r units is
(x – a)² + (y)² = r²
⇒ (2 – a)² + (0)² = r²
And (0 – a)² + (4)² = r²
⇒ (a – 2)² = a² + 4² ⇒ (- 2)(2a
– 2) =16 ⇒ a – 1 = – 4 ⇒ a = – 3

So, r² = (2 + 3)² = 5 ²  ⇒ r = 5 units.

Question 8
The center of the circle 4x² + 4y² – 8x + 12y –
25 = 0 is?
(a) (2,-3)
(b) (-2,3)
(c) (-4,6)
(d) (4,-6)

Answer a

Question 9

If a circle pass through (4, 0) and (0, 2) and centre at
y-axis then find the radius of the circle.
a) 25 units
b) 20 units
c) 5 units
d) 10 units

Answer: c

Question 10

The point (1, 4) lie ___________ the circle x² + y² – 2x – 4y + 2 = 0.
a) inside circle
b) outside circle
c) on the circle
d) either inside or outside

Answer b

Question 11
The radius of the circle 4x² + 4y² – 8x + 12y –
25 = 0 is?
(a) √57/4
(b) √77/4
(c) √77/2
(d) √87/4

Answer c

MCQ BASED ON PARABOLA CLASS XI

Question 1

Find the focus of parabola with equation y² = 100x.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)

Answer c

Question 2

Find the focus of parabola with equation y² = – 100x.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)

Answer d

Question 3

Find the focus of parabola with equation x² = 100y.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)

Answer a

Question 4

The focus of the
parabola y² = 8x is

a) (0,
2)

b) (2,
0)

c) (0,
-2)

d) (-2,
0)

Answer: (b)
(2, 0)

Explanation:

Given parabola
equation y² = 8x …(1)

Here, the
coefficient of x is positive and the standard form of parabola is y² =
4ax …(2)

Comparing (1) and
(2), we get

4a = 8

a = 8/4 = 2

We know that the
focus of parabolic equation y² = 4ax is (a, 0).

Therefore, the
focus of the parabola y² =8x is (2, 0).

Question 5

The length of
the latus rectum of x² = -9y is equal to

a)  3 units

b) – 3 units

c)  9/4 units

d)  9 units

Answer: (d) 9
units

Explanation:

Given parabola
equation: x² = – 9y …(1)

Since the
coefficient of y is negative, the parabola opens downwards.

The general
equation of parabola is x² = – 4ay…(2)

Comparing (1) and
(2), we get

-4a = -9

a = 9/4

We know that the
length of latus rectum = 4a = 4(9/4) = 9.

Therefore, the
length of the latus rectum of x² = -9y is equal to 9 units.

Question 6

The parametric
equation of the parabola y² = 4ax is

a) x = at; y = 2at

b) x = at²;
y = 2at

c) x = at²;
y² = at³

d) x = at²;
y = 4at

Answer: (b) x
= at²; y = 2at

Question 7

Find the equation of latus rectum of parabola y² = 100x.

a) x = 25
b) x = – 25
c) y = 25
d) y = – 25

Answer a

Question 8

Find the equation of latus rectum of parabola x² = – 100y.
a) x = 25
b) x = – 25
c) y = – 25
d) y = 25

Answer c

Question 9

Find the equation of directrix of parabola y²=-100x.
a) x = 25
b) x = – 25
c) y = – 25
d) y = 25

Answer a

Question 10
If a parabolic reflector is 20 cm in diameter
and 5 cm deep then the focus of parabolic reflector is
(a) (0 0)
(b) (0, 5)
(c) (5, 0)
(d) (5, 5)

Answer c

Question 11

The equation of parabola with vertex at origin the axis is
along x-axis and passing through the point (2, 3) is
(a) y² = 9x
(b) y² = 9x/2
(c) y² = 2x
(d) y² = 2x/9

Answer b

Question 12

Find the vertex of the parabola y² =
4ax.
a) (0, 4)
b) (0, 0)
c) (4, 0)
d) (0, -4)

Answer b

Question 13

Find the equation of axis of the parabola y² = 24x.
a) x = 0
b) x = 6
c) y = 6
d) y = 0

Answer d

Question 14

Find the equation of axis of the parabola x² = 24y.
a) x = 0
b) x = 6
c) y = 6
d) y = 0

Answer a

Question 15

Find the length of latus rectum of the parabola y² = 40x.

a) 4 units
b) 10 units
c) 40 units
d) 80 units

Answer c

Question 16

At what point of the parabola x² = 9y is the abscissa three times that of ordinate
(a) (1, 1)
(b) (3, 1)
(c) (-3, 1)
(d) (-3, -3)

Answer b

MCQ BASED ON ELLIPSE CLASS – XI

Question 1

An ellipse has ___________ vertices and ____________ foci.
a) two, one
b) one, one
c) one, two
d) two, two

Answer d

Question 2

Find the coordinates of foci of ellipse (x/25)² + (y/16)² = 1.
a) (±3, 0)
b) (±4, 0)
c) (0, ±3)
d) (0, ±4)

Answer a

Question 3
In an ellipse, the distance between its foci is
6 and its minor axis is 8 then its eccentricity is
(a) 4/5
(b) 1/√52
(c) 3/5
(d) 1/ 2

Answer c

Question 4
A rod of length 12 CM moves with its and always
touching the co-ordinate Axes. Then the equation of the locus of a point P on
the road which is 3 cm from the end in contact with the x-axis is
(a) x²/81 + y²/9 = 1
(b) x²/9 + y²/81 = 1
(c) x²/169 + y²/9 = 1
(d) x²/9 + y²/169 = 1

Answer a

Question 5

Find the coordinates of foci of ellipse (x/16)² + (y/25)² = 1.
a) (±3, 0)
b) (±4, 0)
c) (0, ±3)
d) (0, ±4)

Answer c

Question 6

What is major axis length for ellipse (x/25)² + (y/16)² = 1?
a) 5 units
b) 4 units
c) 8 units
d) 10 units

Answer d

Question 7

For the ellipse
3x² + 4y² = 12, the length of the latus rectum is:

a)  2/5
b)  3/5
c)  3
d) 4

Answer: (c) 3

Explanation:

Given ellipse
equation: 3x² + 4y² = 12

The given equation
can be written as (x²/4) + (y²/3) = 1…(1)

Now, compare the
given equation with the standard ellipse equation: (x²/a²)
+ (y²/b²) = 1, we get

a = 2 and b = √3

Therefore, a >
b.

If a>b, then the
length of latus rectum is 2b²/a

Substituting the
values in the formula, we get

Length of latus
rectum = [2(√3)²] /2 = 3

Question 8

What is minor axis length for ellipse (x/25)² + (y/16)² = 1?
a) 5 units
b) 4 units
c) 8 units
d) 10 units

Answer c

Question 9

What is equation of latus rectums of ellipse (x/25)² + (y/16)² = 1?

a) x = ±3
b) y = ±3
c) x = ±2
d) y = ±2

Answer a

Question 10

In an ellipse,
the distance between its foci is 6 and the minor axis is 8, then its
eccentricity is

a) 1/2

b)  1/5

c)   3/5

d)   4/5

Answer: (c)
3/5

Explanation:

Given that the
minor axis of ellipse is 8.(i. e) 2b = 8. So, b=4.

Also, the distance
between its foci is 6. (i. e) 2ae = 6

Therefore, ae = 6/2
= 3

We know that b² =
a²(1-e²)

b² =
a² – a²e²

b² =
a² – (ae)²

Now, substitute the
values to find the value of a.

(4)² =
a² -(3)²

16 = a² –
9

a² =
16+9 = 25.

So, a = 5.

The formula to
calculate the eccentricity of ellipse is e = √[1-(b²/a²)]

e = √[1-(4²/5²)]

e = √[(25-16)/25]

e = √(9/25) = 3/5.

Question 11

A man running a race course notes that the sum of the
distances from the two flag posts from him is always 10 meter and the distance
between the flag posts is 8 meter. The equation of posts traced by the man is
(a) x²/9 + y²/5 = 1
(b) x²/9 + y2 /25 = 1
(c) x²/5 + y²/9 = 1
(d) x²/25 + y²/9 = 1

Answer d

MCQ BASED ON HYPERBOLA CLASS XI

Question 1

A hyperbola has ___________ vertices and ____________ foci.

a) two, one

b) one, one

c) one, two

d) two, two

Answer d

Question 2

Find the coordinates of foci of hyperbola (x/9)²−(y/16)²=1.

a) (±5,0)

b) (±4,0)

c) (0,±5)

d) (0,±4)

Answer a

Question 3
The equation of a hyperbola with foci on the
x-axis is
(a) x²/a² + y²/b² = 1
(b) x²/a² – y²/b² = 1
(c) x² + y² = (a² + b²)
(d) x² – y² = (a² + b²)

Answer b

The eccentricity of hyperbola is
a. e =1
b. e > 1
c. e < 1
d. 0 < e < 1
Answer: (b) e > 1

Question 5

Find the coordinates of foci of hyperbola (y/16)²−(/x9)²=1.

a) (±5,0)

b) (±4,0)

c) (0,±5)

d) (0,±4)

Answer c

Question 6

What is transverse axis length for hyperbola (x/9)²−(y/16)²=1?

a) 5 units

b) 4 units

c) 8 units

d) 6 units

Answer d

Question 7

What is conjugate axis length for hyperbola (x/9)²−(y/16)²=1?

a) 5 units

b) 4 units

c) 8 units

d) 10 units

Answer c

Question 8

What is equation of latus rectums of hyperbola (x/9)²−(y/16)²=1?

a) x = ±5

b) y = ±5

c) x = ±2

d) y = ±2

Answer a 

Question 9

The length of
the transverse axis is the distance between the ____.

a)   Two vertices

b)   Two Foci

c)   Vertex and the
origin

d)   Focus and the
vertex

Answer: (a)
Two vertices