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MCQ Questions for Class 11 Accountancy Chapter 8 Bills of Exchange with Answers

Bills of Exchange Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below Online Education NCERT MCQ Questions for Class 11 Accountancy Chapter 8 Bills of Exchange with Answers Pdf free download. MCQ Questions for Class 11 Accountancy with Answers were prepared based on the latest exam pattern. We have provided Bills of Exchange Class 11 Accountancy MCQs Questions with Answers to help students understand the concept very well.

Class 11 Accountancy Chapter 8 Bills of Exchange MCQ With Answers

Accountancy Class 11 Chapter 8 MCQs On Bills of Exchange

Bill Of Exchange MCQ Chapter 8 Question 1.
On dishonor of a discounted bill whom does the bank look for payment
(a) Drawer
(b) Drawee
(c) Endorser
(d) Payee

Answer

Answer: (a) Drawer

Bills Of Exchange MCQ Chapter 8 Question 2.
The act for signing by the drawer on the book of the instruments for the purpose of transfer
(a) Acceptance of bill
(b) Cheque
(c) Endorsement
(d) Bill

Answer

Answer: (c) Endorsement

Bills Of Exchange Class 11 MCQ Chapter 8 Question 3.
Which balance is shown by a B/R Book
(a) Credit
(b) None
(c) Both
(d) Debit

Answer

Answer: (d) Debit

MCQ On Bills Of Exchange Chapter 8 Question 4.
On whom the trade bill drawn
(a) Seller
(b) Creditor
(c) Debtor
(d) Owner

Answer

Answer: (c) Debtor

MCQ On Bills Of Exchange Class 11 Chapter 8 Question 5.
The party who is entitled to receive the cash of a bill receivable is called
(a) Drawer
(b) Drawee
(c) Capitalist
(d) Bank

Answer

Answer: (a) Drawer

MCQ On Bill Of Exchange Chapter 8 Question 6.
Discounting Charges =
(a) Amount of Bill Discounted × Rate × Unexpired Period
(b) Amount of Bill Discounted × Rate / Unexpired Period
(c) Amount of Bill Discounted × Rate + Unexpired Period
(d) Amount of Bill Discounted + Rate × Unexpired Period

Answer

Answer: (d) Amount of Bill Discounted + Rate × Unexpired Period

A Bill Of Exchange Includes MCQ Chapter 8 Question 7.
According to Negotiable Instrument Act, 1881, which of the following refers to “an instrument in writing (not being a bank note or a currency note) containing unconditional undertaking, signed by the maker to pay on demand or at a fixed or determinable future time a certain sum of money only to or to the order of a certain person, or to the bearer of the instrument”?
(a) Promissory note
(b) Bearer debentures
(c) Cheque
(d) Bill of exchange

Answer

Answer: (d) Bill of exchange

Bill Of Exchange Class 11 MCQ Chapter 8 Question 8.
If the due date is public holiday what will be the due date of the bill
(a) Preceding day
(b) Following day
(c) The same day
(d) After two days

Answer

Answer: (a) Preceding day

Bills Of Exchange MCQs Class 11 Chapter 8 Question 9.
Fee paid in cash to Notary Public is charged by
(a) Holder of bill of exchange
(b) Drawee
(c) Drawer
(d) None

Answer

Answer: (a) Holder of bill of exchange

Question 10.
Kamal draws a bill on Sahil for Rs.3000. Kamal endorsed it to Rohan. Rohan endorsed it to Rakesh. The payee of the bill will be
(a) Kamal
(b) Rakesh
(c) Sahil
(d) Rohan

Answer

Answer: (b) Rakesh

Question 11.
Noting charges are paid by the ___ but these are recordable from the ____
(a) Drawer, Drawer
(b) Drawer, Drawee
(c) Drawee, Drawer
(d) Drawee, Drawee

Answer

Answer: (c) Drawee, Drawer

Question 12.
Person to whom the bill is endorsed called ____
(a) Endorsement
(b) Endorser
(c) Endorsee
(d) None

Answer

Answer: (b) Endorser

Question 13.
What are the parties to a bill of exchange
(a) Drawee, Drawer, Payee
(b) Drawee, Drawer, Debtor
(c) Payer, Drawee, Drawer
(d) Drawee, Drawer, Creditor

Answer

Answer: (a) Drawee, Drawer, Payee

Question 14.
Which bill is not allowed 3 days of grace
(a) Bill at the time of due date
(b) Bill at sight
(c) Bill after due date
(d) Bill before due date

Answer

Answer: (b) Bill at sight

Question 15.
The purpose of accommodation bill is :
(a) To finance actual purchase
(b) When both parties are in need of funds
(c) To facilitate trade transmission
(d) None

Answer

Answer: (b) When both parties are in need of funds

Question 16.
It is a period of time after which a bill becomes payable?
(a) Grace days
(b) Maturity
(c) Usance
(d) Tenor

Answer

Answer: (d) Tenor

Question 17.
If the drawer is in need of money and cannot wait till due date and receive the money form bank is called?
(a) Endorsement of bill
(b) Discounting of bill
(c) Retirement of bill
(d) Dishonor of bill

Answer

Answer: (b) Discounting of bill

Question 18.
In case if endorsement of bill the endorser debits?
(a) Creditor account
(b) Cash account
(c) Bill receivable account
(d) Bill payable account

Answer

Answer: (a) Creditor account

Question 19.
Which of the following is not true?
(a) There is no difference in appearance between trade and accommodation bill
(b) A bill of exchange must be accepted
(c) Drawee is maker of a bill
(d) Accommodation bill is for an imaginary transaction

Answer

Answer: (c) Drawee is maker of a bill

Question 20.
When an acceptor refuses to pay the amount of bill to the holder of bill on its maturity is called?
(a) Honored bill
(b) Dishonored bill
(c) Retired bill
(d) Endorsed bill

Answer

Answer: (b) Dishonored bill

Question 21.
A bill of exchange is an?
(a) A promise
(b) Unconditional order
(c) A request
(d) A order

Answer

Answer: (b) Unconditional order

Question 22.
A person who writes out the order to pay is called?
(a) Drawer
(b) Acceptor
(c) Payee
(d) Drawee

Answer

Answer: (a) Drawer

Question 23.
When a drawer discounts the bill, he gets?
(a) More than face value
(b) Less than face value
(c) Equal to face value
(d) None of above

Answer

Answer: (b) Less than face value

Question 24.
In case of term bill extra three days are given to acceptor are called?
(a) Days of bill payable
(b) Days if bill receivable
(c) Days of grace
(d) Days of tenor

Answer

Answer: (c) Days of grace

Question 25.
How many parties are involved in case of discounting of bill?
(a) Two
(b) Three
(c) Four
(d) Unlimited

Answer

Answer: (a) Two

Question 26.
Bill of Exchange has parties :
(a) 6
(b) 2
(c) 3
(d) 4

Answer

Answer: (c) 3
Explanation:
A bill of exchange consist of three parties namely:
(i) Drawer
(ii) Drawee
(iii) Payee.

Question 27.
The party which is ordered to pay the amount of bill of exchange is called :
(a) Drawee
(b) Payee
(c) Drawer
(d) None of these

Answer

Answer: (a) Drawee
Explanation:
The party upon whom the bill is drawn is called the drawee. He is the person to whom the bill is addressed and who is ordered to pay. He becomes an acceptor when he indicates his willingness to pay the bill.

Question 28.
The party which is entitled to receive the payment of bill of exchange is known as :
(a) Drawer
(b) Payee
(c) Drawee
(d) None of these

Answer

Answer: (b) Payee
Explanation:
A person to whom money is paid or is to be paid, especially the person to whom a cheque is made payable. A payee is a party in an exchange who receives payment. The payee is paid by cash, check, or another transfer medium by a payer.

The drawee is the party that pays the sum specified by the bill of exchange. The drawer is the party that obliges the drawee to pay the payee. The drawer and the payee are the same entity unless the drawer transfers the bill of exchange to a third-party payee.
So, payee is the correct option.

Question 29.
Due date of a bill of exchange drawn on 30th January, 2011 for one month will be :
(a) 5 Mar
(b) 3 Mar
(c) 29 Feb
(d) 4 Mar

Answer

Answer: (b) 3 Mar
Explanation:
Due date of a bill is only after the given period (in this case one month ) plus three days of grace . So the bill will be paid only after 1 month and 3 days i.e. on 3rd march.

Question 30.
The promissory note should be signed by
(a) Payee
(b) Drawee
(c) Drawer
(d) Promiser

Answer

Answer: (d) Promiser
Explanation:
Promissory Notes: A negotiable instrument is a document in writing. It is signed by a certain person who promises to pay another person a fixed sum of money on a fixed date.

Question 31.
On dishonor of a discounted bill who does the bank look for payment?
(a) Drawer
(b) Payee
(c) Endorser
(d) None

Answer

Answer: (a) Drawer
Explanation:
Drawer (the person who had received B/R) because he had discounted the bill from the bank and now he’s liable for it’s dishonour. But later he can claim this amount from drawee.

Question 32.
While calculating the due date of the bill, how many days are added to the period of the bill :
(a) 4 days
(b) 3 days
(c) 5 days
(d) Neither of these

Answer

Answer: (b) 3 days
Explanation:
3 days of grace are added to the period of bill while calculating the due date of the bill.

Question 33.
Encashing the bill before the date of its maturity is called :
(a) Dishonour of Bill
(b) Retirement of Bill
(c) Discounting of Bill
(d) Endorsement of Bill

Answer

Answer: (c) Discounting of Bill
Explanation:
When we encash a bill before it’s maturity, it’s generally discounted with bank, bank charges some discounting charges and thus the process is known as discounting of bill.

Question 34.
A bill of exchange renewed generally at the request of
(a) Drawee
(b) Bank
(c) Drawer
(d) None

Answer

Answer: (a) Drawee
Explanation:
When drawer want their money and drawee is not in the position to pay his money, then he wants some time to pay his money and so he requests to make a new bill to drawer.

Question 35.
A bill of exchange can not be
(a) Endorsed
(b) Accepted
(c) Refused
(d) Crossed

Answer

Answer: (c) Refused
Explanation:
A bill of exchange is a document used in transactions that orders the payer to pay a certain amount of money to the payee. It is a guarantee of payment on demand or on a specified date, and it cannot be refused or cancelled, like a check.

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Class 11 Accountancy With Answers MCQ:

MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers

Complex Numbers and Quadratic Equations Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Complex Numbers and Quadratic Equations Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 5 MCQ With Answers

Maths Class 11 Chapter 5 MCQs On Complex Numbers and Quadratic Equations

Complex Numbers Class 11 MCQ Question 1.
The value of √(-16) is
(a) -4i
(b) 4i
(c) -2i
(d) 2i

Answer

Answer: (b) 4i
Hint:
Given, √(-16) = √(16) × √(-1)
= 4i {since i = √(-1) }

Complex Numbers Class 11 MCQ Questions And Answers Question 2.
The value of √(-144) is
(a) 12i
(b) -12i
(c) ±12i
(d) None of these

Answer

Answer: (a) 12i
Hint:
Given, √(-144) = √{(-1) × 144}
= √(-1) × √(144)
= i × 12 {Since √(-1) = i}
= 12i
So, √(-144) = 12i

MCQ On Complex Numbers Class 11 Question 3:
The value of √(-25) + 3√(-4) + 2√(-9) is
(a) 13i
(b) -13i
(c) 17i
(d) -17i

Answer

Answer: (c) 17i
Hint:
Given, √(-25) + 3√(-4) + 2√(-9)
= √{(-1) × (25)} + 3√{(-1) × 4} + 2√{(-1) × 9}
= √(-1) × √(25) + 3{√(-1) × √4} + 2{√(-1) × √9}
= 5i + 3×2i + 2×3i {since √(-1) = i}
= 5i + 6i + 6i
= 17i
So, √(-25) + 3√(-4) + 2√(-9) = 17i

Class 11 Maths Chapter 5 MCQ With Answers Question 4.
if z lies on |z| = 1, then 2/z lies on
(a) a circle
(b) an ellipse
(c) a straight line
(d) a parabola

Answer

Answer: (a) a circle
Hint:
Let w = 2/z
Now, |w| = |2/z|
=> |w| = 2/|z|
=> |w| = 2
This shows that w lies on a circle with center at the origin and radius 2 units.

Class 11 Maths Chapter 5 MCQ Question 5.
If ω is an imaginary cube root of unity, then (1 + ω – ω²)⁷ equals
(a) 128 ω
(b) -128 ω
(c) 128 ω²
(d) -128 ω²

Answer

Answer: (d) -128 ω²
Hint:
Given ω is an imaginary cube root of unity.
So 1 + ω + ω² = 0 and ω³ = 1
Now, (1 + ω – ω²)⁷ = (-ω² – ω²)⁷
⇒ (1 + ω – ω²)⁷ = (-2ω²)⁷
⇒ (1 + ω – ω²)⁷ = -128 ω¹⁴
⇒ (1 + ω – ω²)⁷ = -128 ω¹² × ω²
⇒ (1 + ω – ω²)⁷ = -128 (ω³)⁴ ω²
⇒ (1 + ω – ω²)⁷ = -128 ω²

MCQ Of Complex Numbers Class 11 Question 6.
The least value of n for which {(1 + i)/(1 – i)}ⁿ is real, is
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (b) 2
Hint:
Given, {(1 + i)/(1 – i)}ⁿ
= [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}]ⁿ
= [{(1 + i)²}/{(1 – i²)}]ⁿ
= [(1 + i² + 2i)/{1 – (-1)}]ⁿ
= [(1 – 1 + 2i)/{1 + 1}]ⁿ
= [2i/2]ⁿ
= iⁿ
Now, in is real when n = 2 {since i2 = -1 }
So, the least value of n is 2

MCQ Of Chapter 5 Maths Class 11 Question 7.
Let z be a complex number such that |z| = 4 and arg(z) = 5π/6, then z =
(a) -2√3 + 2i
(b) 2√3 + 2i
(c) 2√3 – 2i
(d) -√3 + i

Answer

Answer: (a) -2√3 + 2i
Hint:
Let z = r(cos θ + i × sin θ)
Then r = 4 and θ = 5π/6
So, z = 4(cos 5π/6 + i × sin 5π/6)
⇒ z = 4(-√3/2 + i/2)
⇒ z = -2√3 + 2i

MCQ On Complex Numbers Class 11 Pdf Question 8:
The value of i^-999 is
(a) 1
(b) -1
(c) i
(d) -i

Answer

Answer: (c) i
Hint:
Given, i^-999
= 1/i⁹⁹⁹
= 1/(i⁹⁹⁶ × i³)
= 1/{(i⁴)²⁴⁹ × i³}
= 1/{1²⁴⁹ × i³} {since i⁴ = 1}
= 1/i³
= i⁴/i³ {since i⁴ = 1}
= i
So, i^-999 = i

MCQ Questions On Complex Numbers Class 11 Question 9.
Let z₁ and z₂ be two roots of the equation z² + az + b = 0, z being complex. Further assume that the origin, z₁ and z₁ form an equilateral triangle. Then
(a) a² = b
(b) a² = 2b
(c) a² = 3b
(d) a² = 4b

Answer

Answer: (c) a² = 3b
Hint:
Given, z₁ and z₂ be two roots of the equation z² + az + b = 0
Now, z₁ + z₂ = -a and z₁ × z₂ = b
Since z₁ and z₂ and z₃ from an equilateral triangle.
⇒ z₁² + z₂² + z₃² = z₁ × z₂ + z₂ × z₃ + z₁ × z₃
⇒ z₁²+ z₂² = z₁ × z₂ {since z₃ = 0}
⇒ (z₁ + z₂)² – 2z₁ × z₂ = z₁ × z₂
⇒ (z₁ + z₂)² = 2z₁ × z₂ + z₁ × z₂
⇒ (z₁ + z₂)² = 3z₁ × z₂
⇒ (-a)² = 3b
⇒ a² = 3b

Complex Numbers MCQs With Solution Question 10:
The complex numbers sin x + i cos 2x are conjugate to each other for
(a) x = nπ
(b) x = 0
(c) x = (n + 1/2) π
(d) no value of x

Answer

Answer: (d) no value of x
Hint:
Given complex number = sin x + i cos 2x
Conjugate of this number = sin x – i cos 2x
Now, sin x + i cos 2x = sin x – i cos 2x
⇒ sin x = cos x and sin 2x = cos 2x {comparing real and imaginary part}
⇒ tan x = 1 and tan 2x = 1
Now both of them are not possible for the same value of x.
So, there exist no value of x

Class 11 Complex Numbers MCQ Question 11.
The curve represented by Im(z²) = k, where k is a non-zero real number, is
(a) a pair of striaght line
(b) an ellipse
(c) a parabola
(d) a hyperbola

Answer

Answer: (d) a hyperbola
Hint:
Let z = x + iy
Now, z² = (x + iy)²
⇒ z² = x² – y² + 2xy
Given, Im(z²) = k
⇒ 2xy = k
⇒ xy = k/2 which is a hyperbola.

Class 11 Maths Complex Numbers MCQ Question 12.
The value of x and y if (3y – 2) + i(7 – 2x) = 0
(a) x = 7/2, y = 2/3
(b) x = 2/7, y = 2/3
(c) x = 7/2, y = 3/2
(d) x = 2/7, y = 3/2

Answer

Answer: (a) x = 7/2, y = 2/3
Hint:
Given, (3y – 2) + i(7 – 2x) = 0
Compare real and imaginary part, we get
3y – 2 = 0
⇒ y = 2/3
and 7 – 2x = 0
⇒ x = 7/2
So, the value of x = 7/2 and y = 2/3

Complex Numbers MCQs With Solution Pdf Question 13.
Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is imaginary
(a) θ = nπ ± π/2 where n is an integer
(b) θ = nπ ± π/3 where n is an integer
(c) θ = nπ ± π/4 where n is an integer
(d) None of these

Answer

Answer: (b) θ = nπ ± π/3 where n is an integer
Hint:
Given,
(3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 + 2i × sin θ)×(1 – 2i × sin θ)}/(1 – 4i² × sin² θ)
(3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 – 4sin² θ) + 8i × sin θ}/(1 + 4sin² θ) …………. 1
Now, equation 1 is imaginary if
3 – 4sin² θ = 0
⇒ 4sin² θ = 3
⇒ sin² θ = 3/4
⇒ sin θ = ±√3/2
⇒ θ = nπ ± π/3 where n is an integer

Class 11 Maths Chapter 5 MCQ Questions Question 14.
If {(1 + i)/(1 – i)}ⁿ = 1 then the least value of n is
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4
Hint:
Given, {(1 + i)/(1 – i)}ⁿ = 1
⇒ [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}]ⁿ = 1
⇒ [{(1 + i)²}/{(1 – i²)}]ⁿ = 1
⇒ [(1 + i² + 2i)/{1 – (-1)}]ⁿ = 1
⇒ [(1 – 1 + 2i)/{1 + 1}]ⁿ = 1
⇒ [2i/2]ⁿ = 1
⇒ iⁿ = 1
Now, iⁿ is 1 when n = 4
So, the least value of n is 4

Class 11 Maths MCQ Chapter 5 Question 15.
If arg (z) < 0, then arg (-z) – arg (z) =
(a) π
(b) -π
(c) -π/2
(d) π/2

Answer

Answer: (a) π
Hint:
Given, arg (z) < 0
Now, arg (-z) – arg (z) = arg(-z/z)
⇒ arg (-z) – arg (z) = arg(-1)
⇒ arg (-z) – arg (z) = π {since sin π + i cos π = -1, So arg(-1) = π}

Question 16.
if x + 1/x = 1 find the value of x²⁰⁰⁰ + 1/x²⁰⁰⁰ is
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (c) -1
Hint:
Given x + 1/x = 1
⇒ (x² + 1) = x
⇒ x² – x + 1 = 0
⇒ x = {-(-1) ± √(1² – 4 × 1 × 1)}/(2 × 1)
⇒ x = {1 ± √(1 – 4)}/2
⇒ x = {1 ± √(-3)}/2
⇒ x = {1 ± √(-1)×√3}/2
⇒ x = {1 ± i√3}/2 {since i = √(-1)}
⇒ x = -w, -w²
Now, put x = -w, we get
x²⁰⁰⁰ + 1/x²⁰⁰⁰ = (-w)²⁰⁰⁰ + 1/(-w)²⁰⁰⁰
= w²⁰⁰⁰ + 1/w²⁰⁰⁰
= w²⁰⁰⁰ + 1/w²⁰⁰⁰
= {(w³)⁶⁶⁶ × w²} + 1/{(w³)⁶⁶⁶ × w²}
= w² + 1/w² {since w³ = 1}
= w² + w³ /w²
= w² + w
= -1 {since 1 + w + w² = 0}
So, x²⁰⁰⁰ + 1/x²⁰⁰⁰ = -1

Question 17.
The value of √(-144) is
(a) 12i
(b) -12i
(c) ±12i
(d) None of these

Answer

Answer: (a) 12i
Hint:
Given, √(-144) = √{(-1)×144}
= √(-1) × √(144)
= i × 12 {Since √(-1) = i}
= 12i
So, √(-144) = 12i

Question 18.
If the cube roots of unity are 1, ω, ω², then the roots of the equation (x – 1)³ + 8 = 0 are
(a) -1, -1 + 2ω, – 1 – 2ω²
(b) – 1, -1, – 1
(c) – 1, 1 – 2ω, 1 – 2ω²
(d) – 1, 1 + 2ω, 1 + 2ω²

Answer

Answer: (c) – 1, 1 – 2ω, 1 – 2ω²
Hint:
Note that since 1, ω, and ω² are the cube roots of unity (the three cube roots of 1), they are the three solutions to x³ = 1 (note: ω and ω² are the two complex solutions to this)
If we let u = x – 1, then the equation becomes
u³ + 8 = (u + 2)(u² – 2u + 4) = 0.
So, the solutions occur when u = -2 (giving -2 = x – 1 ⇒ x = -1), or when:
u² – 2u + 4 = 0,
which has roots, by the Quadratic Formula, to be u = 1 ± i√3
So, x – 1 = 1 ± i√3
⇒ x = 2 ± i√3
Now, x³ = 1 when x³ – 1 = (x – 1)(x² + x + 1) = 0, giving x = 1 and
x² + x + 1 = 0
⇒ x = (-1 ± i√3)/2
If we let ω = (-1 – i√3)/2 and ω₂ = (-1 + i√3)/2
then 1 – 2ω and 1 – 2ω² yield the two complex solutions to (x – 1)³ + 8 = 0
So, the roots of (x – 1)³ + 8 are -1, 1 – 2ω, and 1 – 2ω²

Factor complex polynomials calculator is generally used to solve the complex equation so that it is easier to work with simpler terms.

Question 19.
(1 – w + w²)×(1 – w² + w⁴)×(1 – w⁴ + w⁸) × …………… to 2n factors is equal to
(a) 2ⁿ
(b) 2²ⁿ
(c) 2³ⁿ
(d) 2⁴ⁿ

Answer

Answer: (b) 2²ⁿ
Hint:
Given, (1 – w + w²)×(1 – w² + w⁴)×(1 – w⁴ + w⁸) × …………… to 2n factors
= (1 – w + w²)×(1 – w² + w )×(1 – w + w²) × …………… to 2n factors
{Since w⁴ = w, w⁸ = w²}
= (-2w) × (-2w²) × (-2w) × (-2w²)× …………… to 2n factors
= (2² w³)×(2² w³)×(2² w³) …………… to 2n factors
= (2²)ⁿ {since w³ = 1}
= 2²ⁿ

Question 20.
The modulus of 5 + 4i is
(a) 41
(b) -41
(c) √41
(d) -√41

Answer

Answer: (c) √41
Hint:
Let Z = 5 + 4i
Now modulus of Z is calculated as
|Z| = √(5² + 4²)
⇒ |Z| = √(25 + 16)
⇒ |Z| = √41
So, the modulus of 5 + 4i is √41

We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Complex Numbers and Quadratic Equations MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

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MCQ Questions for Class 11 Business Studies Chapter 2 Forms of Business Organisation with Answers

Forms of Business Organisation Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below Online Education NCERT MCQ Questions for Class 11 Business Studies Chapter 2 Forms of Business Organisation with Answers Pdf free download. MCQ Questions for Class 11 Business Studies with Answers were prepared based on the latest exam pattern. We have provided Forms of Business Organisation Class 11 Business Studies MCQs Questions with Answers to help students understand the concept very well.

Class 11 Business Studies Chapter 2 Forms of Business Organisation MCQ With Answers

Business Studies Class 11 Chapter 2 MCQs On Forms of Business Organisation

Forms Of Business Organisation Class 11 MCQ Question 1.
At least 10 adults, no maximum limit in case of ________
(a) Cooperative Society
(b) Joint Hindu Family
(c) Partnership
(d) Company

Answer

Answer: (a) Cooperative Society

MCQ Questions For Class 11 Business Studies Chapter 2 Question 2.
What is the limit of members in case of a Private Company?
(a) 2
(b) 7
(c) 10
(d) 50

Answer

Answer: (d) 50

Forms Of Business Organisation MCQ Question 3.
The life of sole proprietorship business is ___________
(a) Unstable
(b) Stable
(c) Very short life
(d) Long life

Answer

Answer: (a) Unstable

Class 11 Business Studies Chapter 2 MCQ Question 4.
Provision of residential accommodation to the members at reasonable rates is the objective of
(a) Consumers cooperative
(b) Credit cooperative
(c) Housing cooperative
(d) Producers cooperative

Answer

Answer: (c) Housing cooperative

Business Studies Class 11 Chapter 2 MCQ Question 5.
The capital of a company is divided into number of parts each one of which are called
(a) Share
(b) Dividend
(c) Profit
(d) Interest

Answer

Answer: (a) Share

Chapter 2 Business Studies Class 11 MCQ Question 6.
The maximum number of partners allowed in the banking business are
(a) Two
(b) Twenty
(c) No limit
(d) Ten

Answer

Answer: (d) Ten

Bst Class 11 Chapter 2 MCQ Question 7.
In a cooperative society the principle followed is
(a) One share one vote
(b) One man one vote
(c) No vote
(d) Multiple votes

Answer

Answer: (a) One man one vote

Ch 2 Bst Class 11 MCQ Question 8.
The structure in which there is separation of ownership and management as per law is called
(a) Company
(b) All business organisations
(c) Partnership
(d) Sole proprietorship

Answer

Answer: (a) Company

Ch 2 Business Studies Class 11 MCQ Question 9.
Co-operatives play an important role in:
(a) aerospace.
(b) agriculture.
(c) manufacturing.
(d) all of the above.

Answer

Answer: (b) agriculture.

MCQ On Forms Of Business Organisation With Answers Pdf Question 10.
Which of the following is a characteristic of a co-operative?
(a) profits are not subject to income tax.
(b) one vote per share.
(c) dividends are paid on a per share basis.
(d) all of the above.

Answer

Answer: (a) profits are not subject to income tax.

MCQ Of Forms Of Business Organisation Question 11.
A__________ provides for the greatest degree of continuity:
(a)general partnership.
(b) joint venture.
(c) corporation.
(d) sole proprietorship.

Answer

Answer: (c) corporation.

Business Studies Class 11 MCQ Chapter 2 Question 12.
Which of the following is probably the most important reason for incorporating?
(a)limited liability of shareholders.
(b) more money for investment.
(c) increased flexibility.
(d)shared management.

Answer

Answer: (a) limited liability of shareholders.

MCQ On Business Organisation With Answers Pdf Question 13.
A_____________partner is an owner who has unlimited liability and is active in managing the firm:
(a) senior partner.
(b) general partner.
(c) silent partner.
(d) limited partner.

Answer

Answer: (b) general partner.

Class 11 Bst Chapter 2 MCQs Question 14.
A ___________ is a business with two or more owners:
(a) corporation.
(b) conglomerate.
(c) partnership.
(d) public corporation.

Answer

Answer: (c) partnership.

Business Organisation MCQ Pdf Question 15.
Which of the following is an advantage of a sole proprietorship?
(a) ease of starting a business.
(b) being your own boss.
(c) pride of ownership.
(d) all of the above.

Answer

Answer: (d) all of the above.

Question 16.
The form of business organization that has the largest sales volume is the:
(a) partnership.
(b) corporation.
(c) cooperative.
(d) multinational.

Answer

Answer: (b) corporation.

Question 17.
In case of ___________ Registration is compulsory
(a) Sole Proprietorship
(b) Partnership
(c) None of these
(d) Company

Answer

Answer: (d) Company

Question 18.
Hindu Succession Act was passed in _____________
(a) 1952
(b) 1960
(c) 1932
(d) 1956

Answer

Answer: (d) 1956

Question 19.
A partner whose association with the firm is unknown to the general public is called
(a) Active partner
(b) Sleeping partner
(c) Nominal partner
(d) Secret partner

Answer

Answer: (d) Secret partner

Question 20.
The Head of the joint Hindu family business is called
(a) Manager
(b) Proprietor
(c) Karta
(d) Director

Answer

Answer: (c) Karta

Question 21.
Profits do not have to be shared. This statement refers to
(a) Company
(b) Sole proprietorship
(c) Joint Hindu family business
(d) Partnership

Answer

Answer: (b) Sole proprietorship

Question 22.
The board of directors of a joint stock company is elected by
(a) General public
(b) Government bodies
(c) Shareholders
(d) Employees

Answer

Answer: (c) Shareholders

Question 23.
The karta in Joint Hindu family business has
(a) No liability for debts
(b) Unlimited liability
(c) Joint liability
(d) Limited liability

Answer

Answer: (b) Unlimited liability

Question 24.
The most effective form of business organization for raising capital is the:
(a) joint venture.
(b) partnership.
(c) corporation.
(d) proprietorship.

Answer

Answer: (c) corporation.

Question 25.
The major advantage of a franchise is:
(a) training and management assistance.
(b) personal ownership.
(c) nationally recognized name.
(d) all of the above.

Answer

Answer: (d) all of the above.

Question 26.
Which of the following is an example of a non profit organization?
(a) Royal Bank.
(b) YMCA.
(c) BCE.
(d) Air Canada.

Answer

Answer: (b) YMCA.

Question 27.
Which of the following is an example of a public corporation?
(a) Imperial Oil.
(b) Hospital for Sick Children.
(c) Mouvement Caisse Desjardin.
(d) YMCA.

Answer

Answer: (a) Imperial Oil.

Question 28.
A partner who is not actually involved in the partnership but lends his name for public relations purposes is a:
(a) silent partner.
(b) general partner.
(c) nominal partner.
(d) dominant partner.

Answer

Answer: (c) nominal partner.

Question 29.
The main disadvantage of a general partnership is:
(a) the unlimited liability of the partners.
(b) disagreement amongst partners.
(c) shared management.
(d) difficulty of termination.

Answer

Answer: (a) the unlimited liability of the partners.

Question 30.
The simplest form of business ownership is a:
(a) proprietorship.
(b) partnership.
(c) corporation.
(d) cooperative.

Answer

Answer: (a) proprietorship.

One Mark Questions

1. Write any one type of forms of business organisations.

Answer

Answer: Partnership firm

2. Give the meaning of sole trading concern.

Answer

Answer: Sole trading concern refers to a form of business organisation which is owned, managed and controlled by an individual who is the recipient of ail profit bearer of all risks.

3. Give any one example to sole trading concern.

Answer

Answer: Home Healthcare

4. Who is Karta?

Answer

Answer: The head of the Hindu Joint Family also called the Karta

5. State the minimum & maximum members in partnership.

Answer

Answer: Minimum Number of members is 2 and maximum is 10 in case, of banking and 20 in case of business.

6. Write any one effect of non-registration of a partnership

Answer

Answer: No suit by a partner against other partners or firm.

7. Write any one type of partnership firm.

Answer

Answer: Limited Partnerships

8. Give the meaning of co-operative society.

Answer

Answer: The co-operative society is a voluntary association of persons, who join together with the motive of welfare.

9. State the minimum and maximum members required for the formation of a co-operative society.

Answer

Answer: Minimum of 5 and maximum is unlimited.

10. Name any one type of co-operative society.

Answer

Answer: Consumer co-operative society

11. State the liability of a sole trader.

Answer

Answer: Unlimited liability

12. State the liability of Karta.

Answer

Answer: Unlimited liability

13. Which act governs the partnership firms in India?

Answer

Answer: Indian Partnership Act 1932

14. Which act governs the co-operative societies in India?

Answer

Answer: Co-operative Societies Act, 1912

15. State the liability of members of co-operative societies.

Answer

Answer: Limited liability

16. State the voting principles in co-operative societies.

Answer

Answer: One man one vote

17. State the main objectives of co-operative societies.

Answer

Answer: Enhanced cooperation

18. State the liability of co-parceners.

Answer

Answer: The liability o f coparceners is always limited in nature.

19. Who is minor partner?

Answer

Answer: A partner one who have not attained the age of 18 years is called as minor partner.

20. Is registration of partnership compulsory?

Answer

Answer: As per Indian partnership Act 1932, it is not compulsory for registration of partnership.

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MCQ Questions for Class 12 Accountancy Chapter 7 Issue and Redemption of Debentures with Answers

Issue and Redemption of Debentures Class 12 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 12 Accountancy Chapter 7 Issue and Redemption of Debentures with Answers Pdf free download. MCQ Questions for Class 12 Accountancy with Answers were prepared based on the latest exam pattern. We have provided Issue and Redemption of Debentures Class 12 Accountancy MCQs Questions with Answers to help students understand the concept very well.

Class 12 Accountancy Chapter 7 Issue and Redemption of Debentures MCQ With Answers

Accountancy Class 12 Chapter 7 MCQs On Issue and Redemption of Debentures

MCQ On Issue Of Debentures Class 12 Question 1.
When all debentures are redeemed, balance in the Debenture Redemption Fund Account is transferred to:
(a) Capital Reserve
(b) General Reserve
(c) Profit & Loss Appropriation A/c
(d) None of these

Answer

Answer: (a) Capital Reserve

Issue And Redemption Of Debentures MCQ Class 12 Question 2.
According to SEBI guidelines, a Company will have to create debenture redemption reserve equivalent to the amount of the following percentage of debenture issued:
(a) 50%
(b) 25%
(c) 70%
(d) 100%

Answer

Answer: (b) 25%

Discount On Issue Of Debentures Is A MCQ Class 12 Question 3.
The balance of ‘Sinking Fund Account’ after the redemption of debentures is transferred to :
(a) Profit & Loss Account
(b) Profit & Loss Appropriation Account
(c) General Reserve Account
(d) Sinking Fund Account

Answer

Answer: (c) General Reserve Account

The Redemption Means MCQ Class 12 Question 4.
Profit on cancellation of own debentures is transferred to:
(a) Profit & Loss Account
(b) Profit & Loss Appropriation Account
(c) General Reserve Account
(d) Capital Reserve Account

Answer

Answer: (d) Capital Reserve Account

Question 5.
If debenture of ₹ 1,00,000 were issued for discount of ₹ 10,000, which are redeemable after four years. Then amount of discount to be written off from P. & L. Account each year is :
(a) ₹ 3,000
(b) ₹ 4,000
(c) ₹ 2,500
(d) ₹ 5,000

Answer

Answer: (c) ₹ 2,500

Question 6.
Debentures can be redeemed out of:
(a) Profit
(b) Capital
(c) Provision
(d) All of the above

Answer

Answer: (d) All of the above

Question 7.
Premium on redemption of debentures is a :
(a) Personal A/c
(b) Real A/c
(c) Nominal A/c
(d) Suspense A/c

Answer

Answer: (c) Nominal A/c

Question 8.
Premium on redemption of debentures is generally provided at the time of ……………
(a) Issue of debentures
(b) Redemption of debentures
(c) Writing off
(d) After 10 years

Answer

Answer: (b) Redemption of debentures

Question 9.
Debentures cannot be redeemed at:
(a) Par
(b) Premium
(c) Discount
(d) More than 10% premium

Answer

Answer: (c) Discount

Question 10.
If debentures purchased in open market are not immediately cancelled, they are treated as :
(a) Current Assets
(b) Current Liabilities
(c) Investment
(d) Capital

Answer

Answer: (c) Investment

Question 11.
Sources of finance of the redemption of debentures are:
(a) Redemption out of profits
(b) Redemption out of capital
(c) The proceeds from fresh issue of shares/debentures
(d) All the above

Answer

Answer: (d) All the above

Question 12.
A company issued 1,000, 12% debentures of ₹ 100 each at 10% premium. 12% stand for:
(a) Rate of dividend
(b) Rate of Tax
(c) Rate of interest
(d) None of these

Answer

Answer: (c) Rate of interest

Question 13.
A company should transfer to Debenture Redemption Reserve A/c at least what percent of the amount of debentures issued before the commencement of redemption of debentures-
(a) 50%
(b) 25%
(c) 15%
(d) 100%

Answer

Answer: (b) 25%

Question 14.
If redemption of debentures is made by conversion method, the amount to be transferred to ‘Debenture Redemption Reserve Account’ will be equal to…….percent of converted amounted.
(a) 40
(b) 50
(c) 60
(d) Not required

Answer

Answer: (d) Not required

Question 15.
Loss on Issue of Debenture Account is shown:
(a) On Assets side of Balance Sheet
(b) On Liabilities side of Balance Sheet
(c) On Credit side of P & L Account
(d) None of these

Answer

Answer: (a) On Assets side of Balance Sheet

Question 16.
Profit on sale of Sinking Fund Investment is transferred to:
(a) Profit & Loss Account
(b) General Reserve
(c) Sinking Fund Account
(d) Capital Reserve

Answer

Answer: (c) Sinking Fund Account

Question 17.
Premium on Redemption of Debentures A/c is:
(a) Asset
(b) Expenses
(c) Liability
(d) Revenue

Answer

Answer: (c) Liability

Question 18.
Interest on sinking fund investment is credited to :
(a) Profit & Loss A/c
(b) Sinking Fund A/c
(c) General Reserve A/c
(d) Sinking Fund Investment A/c

Answer

Answer: (b) Sinking Fund A/c

Question 19.
Sinking fund investment is:
(a) An Income
(b) An exause
(c) An Asset
(d) A Liability

Answer

Answer: (c) An Asset

Question 20.
When debentures are issued at par and are redeemable at a premium, the loss on such an issue is debited to :
(a) Profit & Loss A/c
(b) Debenture Application and Allotment A/c
(c) Loss on Issue of Debentures A/c
(d) Premium on Redemption A/c

Answer

Answer: (c) Loss on Issue of Debentures A/c

Question 21.
Own debentures are those debentures of the company which ?
(a) The company allots to its own promotors
(b) The company allots to its directors
(c) The company purchases from the markets and hold them as investments
(d) None of these

Answer

Answer: (c) The company purchases from the markets and hold them as investments

Question 22.
When debentures are redeemed out of profits, an equivalent amount is transferred to :
(a) General Reserve
(b) Debenture Redemption Reserve
(c) Capital Reserve
(d) Profit & Loss A/c

Answer

Answer: (b) Debenture Redemption Reserve

Question 23.
Profit on sale of debentures redemption fund investment in the first instance in credited to :
(a) Debenture Redemption Fund A/c
(b) Profit & Loss Appropriation A/c
(c) General Reserve A/c
(d) Sinking Fund A/c

Answer

Answer: (a) Debenture Redemption Fund A/c

Question 24.
When debentures are issued at a discount and are redeemable at a premium, which of the following accounts is debited at the time of issue ?
(a) Debentures A/c
(b) Premium on Redemption of Debentures A/c
(c) Loss on Issue of Debentures A/c
(d) Profit & Loss A/c

Answer

Answer: (c) Loss on Issue of Debentures A/c

Question 25.
According to SEBI guidelines what percentage of the amount of debentures must be transferred to Debenture Redemption Reserve, before the commencement of redumption of debentures, in case of convertible debentures ?
(a) 25%
(b) 50%
(c) 100%
(d) zero

Answer

Answer: (d) zero

Question 26.
Profit on redemption of debentures in transferred to which account ?
(a) Capital Reserve Account
(b) Sinking Gund Account
(c) General Reserve Account
(d) Profit & Loss Account

Answer

Answer: (a) Capital Reserve Account

Question 27.
Profit on cancellation of own debentures is :
(a) Revenue Profit
(b) Capital Profit
(c) Operating Profit
(d) Trading Profit

Answer

Answer: (b) Capital Profit

Question 28.
6,000 debentures of ₹ 10 each where discharged by issuing equity shares of ₹ 10 each at 20% premium. The number of shares issued will be :
(a) 50,000
(b) 60,000
(c) 5,000
(d) 6,000

Answer

Answer: (c) 5,000

Question 29.
Every company required to create DRR shall on or before the 30th April of each year, deposit or invest, a sum which shall not be less than…………of the amount of its debentures maturing (to be redeemed) during the year ending on 31st March of the next year.)
(a) 10%
(b) 15%
(c) 25%
(d) 50%

Answer

Answer: (b) 15%

Question 30.
A Sinking Fund is a part of:
(a) Fixed Liabilities
(b) Current Liabilities
(c) Reserves and Surplus
(d) Fixed Assets

Answer

Answer: (c) Reserves and Surplus

Question 31.
A company issued 1000, 12% debentures of ₹ 100 each at 10% premium. 12% stand for :
(a) Rate of Dividend
(b) Rate of Tax
(c) Rate of Interest
(d) None of these

Answer

Answer: (c) Rate of Interest

Question 32.
BST Ltd. want to redeem its 900, 10% debentures at 105% by converting them into shares of ₹ 10 each at ₹ 9 each. The number of shares to be issued will be :
(a) 9,000 Shares
(b) 10,500 Shares
(c) 10,000 Shares
(d) 8,500 Shares

Answer

Answer: (b) 10,500 Shares

Question 33.
‘Premium on Redemption of Debentures A/c is in the nature of:
(a) Personal A/c
(b) Real A/c
(c) Nominal A/c
(d) None of these

Answer

Answer: (c) Nominal A/c

Question 34.
Deep Ltd. issue 10,00,000, 7 % debentures of 100 Rs. each at a discount of 4%, redeemable after 5 years at a premium of 6%. Loss issue of debentures is :
(a) ₹ 10,00,000
(b) ₹ 6,00,000
(c) ₹ 16,00,000
(d) ₹ 4,00,000

Answer

Answer: (a) ₹ 10,00,000

Question 35.
Debenture is a :
(a) Loan certificate
(b) Cash certificate
(c) Credit certificate
(d) None of these

Answer

Answer: (a) Loan certificate

Question 36.
Debenture holders are the :
(a) Customers of the Company
(b) Owners of the Company
(c) Creditors of the Company
(d) None of these

Answer

Answer: (c) Creditors of the Company

Question 37.
In case of issue of debentures as a collateral security for loan from the bank which account will be debited :
(a) Bank Account
(b) Bank Loan Account
(c) Debentures Account
(d) Debentures Suspense Account

Answer

Answer: (d) Debentures Suspense Account

Question 38.
If debentures of ₹ 4,50,000 are issued for the consideration of net assets of ₹ 5,00,000 balance ₹ 50,000 will be credited to:
(a) Profit & Loss A/c
(b) Goodwill A/c
(c) General Reserve A/c
(d) Capital Reserve A/c

Answer

Answer: (d) Capital Reserve A/c

Question 39.
Debentures which are transferred by mere delivery are called:
(a) Registered Debentures
(b) First Debentures
(c) Bearer Debentures
(d) None of these

Answer

Answer: (c) Bearer Debentures

Question 40.
In the Balance Sheet of a Company, Debentures are shown under the head :
(a) Unsecured Loans
(b) Long-term Loans
(c) Current Liabilities
(d) Reserve and Surplus

Answer

Answer: (b) Long-term Loans

Question 41.
Discount on issue of Debentures is in the nature of:
(a) Revenue Loss
(b) Capital Loss
(c) Deferred Revenue Expenditure
(d) None of there

Answer

Answer: (b) Capital Loss

Question 42.
On liquidation of a company, principal amount of debentures is returned:
(a) First of all
(b) Last of all
(c) Before Equity Capital
(d) None of those

Answer

Answer: (a) First of all

Question 43.
Interest payable on debentures is :
(a) An appropriation of profits of the company
(b) A charge against profits of the company
(c) Transfer to Sinking Fund
(d) None of the above

Answer

Answer: (b) A charge against profits of the company

Question 44.
Debentures cannot be redeemed at:
(a) Premium
(b) Discount
(c) Par
(d) None of these

Answer

Answer: (b) Discount

Question 45.
Debentures represent:
(a) Director’s Share in a Company
(b) Investments by Equity Shareholders
(c) Long-term Debt of the Business
(d) None of these

Answer

Answer: (c) Long-term Debt of the Business

Question 46.
Debentures carries interest at: .
(a) 12% p.a.
(b) Fixed Rate
(c) 20% p.a.
(d) 6% p.a.

Answer

Answer: (b) Fixed Rate

Question 47.
Generally debentures are :
(a) Secured
(b) Unsecured
(c) Partly Secured
(d) None of these

Answer

Answer: (a) Secured

Question 48.
Premium on issue of debentures is a for Company :
(a) Revenue Receipt
(b) Profit
(c) Capital Receipt
(d) All of these

Answer

Answer: (c) Capital Receipt

Question 49.
For a company discount on issue of debentures is :
(a) Capital Loss
(b) Revenue Loss
(c) General Loss
(d) None of these

Answer

Answer: (a) Capital Loss

Question 50.
‘Premium on issue of debentures’ is shown in the Balance Sheet on:
(a) Assets side
(b) Liabilities side
(c) None of these
(d) All of these

Answer

Answer: (b) Liabilities side

Question 51.
The balance of discount on debentures is shown in the Balance Sheet on:
(a) Assets side
(b) Liabilities side
(c) None of these
(d) All of these

Answer

Answer: (a) Assets side

Question 52.
Rate of interest on debentures is :
(a) 12% p.a.
(b) 20% p.a.
(c) Fixed Rate
(d) 15% p.a.

Answer

Answer: (c) Fixed Rate

Question 53.
Discount on issue of Debentures should be written off:
(a) Out of Securities Premium Account
(b) Out of Capital Profits
(c) Out of Statements of Profit and Loss
(d) In the above order over the period of debentures

Answer

Answer: (d) In the above order over the period of debentures

Question 54.
F Ltd. purchased machinery for a book value of ₹ 4,00,000. The consideration was paid by issue of 10% Debentures of ₹ 100 each at a discount of 20%. The Debenture Account will be credited by :
(a) ₹ 4,00,000
(b) ₹ 5,00,000
(c) ₹ 3,20,000
(d) ₹ 4,80,000

Answer

Answer: (b) ₹ 5,00,000

Question 55.
Debenture holder gets:
(a) Di vidend
(b) Profit
(c) Interest
(d) Interest at fixed rate

Answer

Answer: (c) Interest

Question 56.
Loss on issue of debentures is generally written off in :
(a) 5 years
(b) 10 years
(c) 15 years
(d) Over the period of redemption

Answer

Answer: (d) Over the period of redemption

Question 57.
When debentures are issued as collateral security, which entry has to be passed ?
(a) Debenture Suspense A/c Dr.
To Debentures
(b) No entry has to be made
(c) (a) or (b)
(d) None of these

Answer

Answer: (c) (a) or (b)

Question 58.
Debenture holder receives:
(a) Dividend
(b) Interest
(c) Both Dividend and Interest
(d) Bonus

Answer

Answer: (b) Interest

Question 59.
Debenture holders are called of the company.
(a) Creditors
(b) Debtors
(c) Owners
(d) Bankers

Answer

Answer: (a) Creditors

Question 60.
A company issued ₹ 1,00,000 12% debentures of ₹ 100 each. The amount of interest on debentures will be:
(a) ₹ 12,000
(b) ₹ 1,20,000
(c) ₹ 12,00,000
(d) None of these

Answer

Answer: (c) ₹ 12,00,000

Question 61.
Premium on redemption of debentures account is :
(a) A real account
(b) A nominal account
(c) A personal account
(d) None of these

Answer

Answer: (c) A personal account

Question 62.
Debenture premium can be used to :
(a) Write off the discount on issue of shares or debentures
(b) Write off the premium on redemption of shares or debentures
(c) Write off capital loss
(d) All of the above

Answer

Answer: (d) All of the above

Question 63.
Which of the following is false ?
(a) A company can issue redeemable debentures
(b) A company can issue debentures with voting rights
(c) A company can buy its own shares
(d) A company can buy its own debentures

Answer

Answer: (b) A company can issue debentures with voting rights

Question 64.
Debenture is the part of:
(a) Share Capital
(b) Long-term Borrowings
(c) Owned Capital
(d) None, of these

Answer

Answer: (b) Long-term Borrowings

Question 65.
Consideration of Debenture is:
(a) Profit
(b) Dividend
(c) Interest
(d) None of these

Answer

Answer: (c) Interest

Question 66.
Premium on Redemption of Debenture A/c is.:
(a) Asset
(b) Liability
(c) Expense
(d) Revenue

Answer

Answer: (b) Liability

Question 67.
Debenture is a :
(a) Loan Certificate
(b) Cash Certificate
(c) Credit Certificate
(d) None of these

Answer

Answer: (a) Loan Certificate

Question 68.
Loss on issue of debentures is recorded as :
(a) Intangible Asset
(b) Current Asset
(c) Current Liability
(d) Miscellaneous Expenditure

Answer

Answer: (d) Miscellaneous Expenditure

We hope the given NCERT MCQ Questions for Class 12 Accountancy Chapter 7 Issue and Redemption of Debentures with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 12 Accountancy Issue and Redemption of Debentures MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

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MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations with Answers

Permutations and Combinations Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below Online Education NCERT MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Permutations and Combinations Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 7 MCQ With Answers

Maths Class 11 Chapter 7 MCQs On Permutations and Combinations

MCQ Questions On Permutation And Combination Class 11 Question 1.
There are 12 points in a plane out of which 5 are collinear. The number of triangles formed by the points as vertices is
(a) 185
(b) 210
(c) 220
(d) 175

Answer

Answer: (b) 210
Hint:
Total number of triangles that can be formed with 12 points (if none of them are collinear)
= ¹²C₃
(this is because we can select any three points and form the triangle if they are not collinear)
With collinear points, we cannot make any triangle (as they are in straight line).
Here 5 points are collinear. Therefore we need to subtract ⁵C₃ triangles from the above count.
Hence, required number of triangles = ¹²C₃ – ⁵C₃ = 220 – 10 = 210

Permutation And Combination Class 11 MCQ Question 2.
The number of combination of n distinct objects taken r at a time be x is given by
(a) ^n/2C_r
(b) ^n/2C_r/2
(c) ⁿC_r/2
(d) ⁿC_r

Answer

Answer: (d) ⁿC_r
Hint:
The number of combination of n distinct objects taken r at a time be x is given by
ⁿC_r = n!/{(n – r)! × r!}
Let the number of combination of n distinct objects taken r at a time be x.
Now consider one of these n ways. There are e objects in this selection which can be arranged in r! ways.
So, each of the x combinations gives rise to r! permutations. So, x combinations will give rise to x×(r!).
Consequently, the number of permutations of n things, taken r at a time is x×(r!) and it is equal to ⁿP_r
So, x×(r!) = ⁿP_r
⇒ x×(r!) = n!/(n – r)!
⇒ x = n!/{(n – r)! × r!}
⇒ ⁿC_r = n!/{(n – r)! × r!}

MCQ On Permutation And Combination Class 11 Question 3.
Four dice are rolled. The number of possible outcomes in which at least one dice show 2 is
(a) 1296
(b) 671
(c) 625
(d) 585

Answer

Answer: (b) 671
Hint:
No. of ways in which any number appearing in one dice = 6
No. of ways in which 2 appear in one dice = 1
No. of ways in which 2 does not appear in one dice = 5
There are 4 dice.
Getting 2 in at least one dice = Getting any number in all the 4 dice – Getting not 2 in any of the 4 dice.
= (6×6×6×6) – (5×5×5×5)
= 1296 – 625
= 671

Permutation And Combination MCQ Question 4.
If repetition of the digits is allowed, then the number of even natural numbers having three digits is
(a) 250
(b) 350
(c) 450
(d) 550

Answer

Answer: (c) 450
Hint:
In a 3 digit number, 1st place can be filled in 5 different ways with (0, 2, 4, 6, 8)
10th place can be filled in 10 different ways.
100th place can be filled in 9 different ways.
So, the total number of ways = 5 × 10 × 9 = 450

MCQ On Permutation And Combination Question 5.
The number of ways in which 8 distinct toys can be distributed among 5 children is
(a) 5⁸
(b) 8⁵
(c) ⁸P₅
(d) ⁵P₅

Answer

Answer: (a) 5⁸
Hint:
Total number of toys = 8
Total number of children = 5
Now, each toy can be distributed in 5 ways.
So, total number of ways = 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5
= 5⁸

Permutation And Combination MCQ Class 11 Question 6.
The value of P(n, n – 1) is
(a) n
(b) 2n
(c) n!
(d) 2n!

Answer

Answer: (c) n!
Hint:
Given,
Given, P(n, n – 1)
= n!/{(n – (n – 1)}
= n!/(n – n + 1)}
= n!
So, P(n, n – 1) = n!

Class 11 Maths Chapter 7 MCQ With Answers Question 7.
In how many ways can 4 different balls be distributed among 5 different boxes when any box can have any number of balls?
(a) 5⁴ – 1
(b) 5⁴
(c) 4⁵ – 1
(d) 4⁵

Answer

Answer: (b) 5⁴
Hint:
Here, both balls and boxes are different.
Now, 1st ball can be placed into any of the 5 boxes.
2nd ball can be placed into any of the 5 boxes.
3rd ball can be placed into any of the 5 boxes.
4th ball can be placed into any of the 5 boxes.
So, the required number of ways = 5 × 5 × 5 × 5 = 5⁴

Permutations And Combinations MCQ Question 8.
The number of ways of painting the faces of a cube with six different colors is
(a) 1
(b) 6
(c) 6!
(d) None of these

Answer

Answer: (a) 1
Hint:
Since the number of faces is same as the number of colors,
therefore the number of ways of painting them is 1

MCQ Of Chapter 7 Maths Class 11 Question 9.
Out of 5 apples, 10 mangoes and 13 oranges, any 15 fruits are to be distributed among 2 persons. Then the total number of ways of distribution is
(a) 1800
(b) 1080
(c) 1008
(d) 8001

Answer

Answer: (c) 1008
Hint:
Given there are 5 apples, 10 mangoes and 13 oranges.
Let x₁ is for apple, x₂ is for mango and x₃ is for orange.
Now, first we have to select total 15 fruits out of them.
x₁ + x₂ + x₃ = 15 (where 0 ⇐ x₁ ⇐ 5, 0 ⇐ x₂ ⇐ 10, 0 ⇐ x₃ ⇐ 13)
= (x⁰ + x¹ + x² +………+ x⁵)×(x⁰ + x¹ + x² +………+ x1¹⁰)×(x⁰ + x¹ + x² +………+ x¹³)
= {(1- x⁶)/(1 – x)}×{(1- x¹¹)/(1 – x)}×{(1- x¹⁴)/(1 – x)}
= {(1- x⁶)×(1- x¹¹)×{(1- x¹⁴)}/(1 – x)³
= {(1- x⁶)×(1- x¹¹)×{(1- x¹⁴)} × ∑^3+r+1C_r × x^r
= {(1- x¹¹ – x⁶ + x¹⁷)×{(1- x¹⁴)} × ∑^3+r+1C_r × x^r
= {(1- x¹¹ – x⁶ + x¹⁷ – x¹⁴ + x²⁵ + x²⁰ – x³¹)} × ∑^2+rC_r × x^r
= 1 × ∑^2+rC_r × x^r – x¹¹ × ∑^2+rC_r × x^r – x⁶ × ∑^2+rC_r × x^r + x¹⁷ × ∑^2+rC_r × x^r – x¹⁴ × ∑^2+rC_r × x^r + x²⁵ × ∑^2+rC_r × x^r + x²⁰ × ∑^2+rC_r × x^r – x³¹ × ∑^2+rC_r × x^r
= ∑^2+rC_r × x^r – ∑^2+rC_r × x^r+11 – ∑^2+rC_r × x^r+6 + ∑^2+rC_r × x^r+17 – ∑^2+rC_r × x^r+14 + ∑^2+rC_r × x^r+25 + ∑^2+rC_r × x^r+20 – ∑^2+rC^r × x^r+25
Now we have to find co-efficeient of x¹⁵
= ²⁺¹⁵C₁₅ – ²⁺⁴C₄ – ²⁺⁹C⁹ – ²⁺¹C¹ (rest all terms have greater than x¹⁵, so its coefficients are 0)
= ¹⁷C₁₅ – ⁶C₄ – ¹¹C₉ – ³C₁
= ¹⁷C₂ – ⁶C₂ – ¹¹C₂ – ³C₁
= {(17×16)/2} – {(6×5)/2} – {(11×10)/2} – 3
= (17×8) – (3×5) – (11×5) – 3
= 136 – 15 – 55 – 3
= 136 – 73
= 63
Again we have to distribute 15 fruits between 2 persons.
So x₁ + x₂ = 15
= ^2-1+15C₁₅
= ¹⁶C₁₅
= ¹⁶C₁
= 16
Now total number of ways of distribution = 16 × 63 = 1008

Permutation And Combination MCQs With Answers Question 10.
6 men and 4 women are to be seated in a row so that no two women sit together. The number of ways they can be seated is
(a) 604800
(b) 17280
(c) 120960
(d) 518400

Answer

Answer: (a) 604800
Hint:
6 men can be sit as
× M × M × M × M × M × M ×
Now, there are 7 spaces and 4 women can be sit as ⁷P₄ = ⁷P₃ = 7!/3! = (7 × 6 × 5 × 4 × 3!)/3!
= 7 × 6 × 5 × 4 = 840
Now, total number of arrangement = 6! × 840
= 720 × 840
= 604800

MCQ Of Permutation And Combination Class 11 Question 11.
The number of ways can the letters of the word ASSASSINATION be arranged so that all the S are together is
(a) 152100
(b) 1512
(c) 15120
(d) 151200

Answer

Answer: (d) 151200
Hint:
Given word is : ASSASSINATION
Total number of words = 13
Number of A : 3
Number of S : 4
Number of I : 2
Number of N : 2
Number of T : 1
Number of O : 1
Now all S are taken together. So it forms a single letter.
Now total number of words = 10
Now number of ways so that all S are together = 10!/(3!×2!×2!)
= (10×9×8×7×6×5×4×3!)/(3! × 2×2)
= (10×9×8×7×6×5×4)/(2×2)
= 10×9×8×7×6×5
= 151200
So total number of ways = 151200

MCQs On Permutation And Combination Question 12.
If repetition of the digits is allowed, then the number of even natural numbers having three digits is
(a) 250
(b) 350
(c) 450
(d) 550

Answer

MCQs On Permutations And Combinations Class 11 Question 13.
Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon on n sides. If T_n+1 – T_n = 21, then n equals
(a) 5
(b) 7
(c) 6
(d) 4

Answer

Answer: (b) 7
Hint:
The number of triangles that can be formed using the vertices of a regular polygon = ⁿC₃
Given, T_n+1 – T_n = 21
⇒ ⁿ⁺¹C₃ – ⁿC₃ = 21
⇒ ⁿC₂ + ⁿC₃ – ⁿC₃ = 21 {since ⁿ⁺¹C_r = ⁿC_r-1 + ⁿC_r}
⇒ ⁿC₂ = 21
⇒ n(n – 1)/2 = 21
⇒ n(n – 1) = 21×2
⇒ n² – n = 42
⇒ n² – n – 42 = 0
⇒ (n – 7)×(n + 6) = 0
⇒ n = 7, -6
Since n can not be negative,
So, n = 7

Permutations And Combinations MCQ Class 11 Question 14.
How many ways are here to arrange the letters in the word GARDEN with the vowels in alphabetical order?
(a) 120
(b) 240
(c) 360
(d) 480

Answer

Answer: (c) 360
Hint:
Given word is GARDEN.
Total number of ways in which all letters can be arranged in alphabetical order = 6!
There are 2 vowels in the word GARDEN A and E.
So, the total number of ways in which these two vowels can be arranged = 2!
Hence, required number of ways = 6!/2! = 720/2 = 360

Permutations And Combinations Class 11 MCQ Questions Question 15.
How many factors are 2⁵ × 3⁶ × 5² are perfect squares
(a) 24
(b) 12
(c) 16
(d) 22

Answer

Answer: (a) 24
Hint:
Any factors of 2⁵ × 3⁶ × 5² which is a perfect square will be of the form 2^a × 3^b × 5^c
where a can be 0 or 2 or 4, So there are 3 ways
b can be 0 or 2 or 4 or 6, So there are 4 ways
a can be 0 or 2, So there are 2 ways
So, the required number of factors = 3 × 4 × 2 = 24

Permutation And Combination Class 11 Extra Questions With Answers Question 16.
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
(a) 40
(b) 196
(c) 280
(d) 346

Answer

Answer: (b) 196
Hint:
There are two cases
1. When 4 is selected from the first 5 and rest 6 from remaining 8
Total arrangement = ⁵C₄ × ⁸C₆
= ⁵C₁ × ⁸C₂
= 5 × (8×7)/(2×1)
= 5 × 4 × 7
= 140
2. When all 5 is selected from the first 5 and rest 5 from remaining 8
Total arrangement = ⁵C₅ × ⁸C₅
= 1 × ⁸C₃
= (8×7×6)/(3×2×1)
= 8×7
= 56
Now, total number of choices available = 140 + 56 = 196

Permutation And Combination MCQs With Answers Pdf Question 17.
Four dice are rolled. The number of possible outcomes in which at least one dice show 2 is
(a) 1296
(b) 671
(c) 625
(d) 585

Answer

Question 18.
In how many ways in which 8 students can be sated in a line is
(a) 40230
(b) 40320
(c) 5040
(d) 50400

Answer

Answer: (b) 40320
Hint:
The number of ways in which 8 students can be sated in a line = ⁸P₈
= 8!
= 40320

Question 19.
The number of squares that can be formed on a chess board is
(a) 64
(b) 160
(c) 224
(d) 204

Answer

Answer: (d) 204
Hint:
A chess board contains 9 lines horizontal and 9 lines perpendicular to them.
To obtain a square, we select 2 lines from each set lying at equal distance and this equal
distance may be 1, 2, 3, …… 8 units, which will be the length of the corresponding square.
Now, two lines from either set lying at 1 unit distance can be selected in ⁸C₁ = 8 ways.
Hence, the number of squares with 1 unit side = 8²
Similarly, the number of squares with 2, 3, ….. 8 unit side will be 7², 6², …… 1²
Hence, total number of square = 8² + 7² + ……+ 1² = 204

Question 20.
How many 3-letter words with or without meaning, can be formed out of the letters of the word, LOGARITHMS, if repetition of letters is not allowed
(a) 720
(b) 420
(c) none of these
(d) 5040

Answer

Answer: (a) 720
Hint:
The word LOGARITHMS has 10 different letters.
Hence, the number of 3-letter words(with or without meaning) formed by using these letters
= ¹⁰P₃
= 10 ×9 ×8
= 720

We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Permutations and Combinations MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

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MCQ Questions for Class 11 Maths Chapter 16 Probability with Answers

Probability Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 16 Probability with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Probability Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 16 MCQ With Answers

Maths Class 11 Chapter 16 MCQs On Probability

MCQ On Probability Class 11 Question 1.
Two cards from a pack of 52 cards are lost. One card is drawn from the remaining cards. If drawn card is diamond then the probability that the lost cards were both hearts is
(a) 143/1176
(b) 143/11760
(c) 143/11706
(d) 134/11760

Answer

Answer: (b) 143/11760
Hint:
Total number of cards = 52
Two cards are lost.
So remaining cards = 50
Now one card is drawn.
Probability that it is a diamond card = 13/50
Now probability that both lost cards are heart = 13/50 ×(¹¹C₂ / ⁴⁹C₂)
= 13/50 ×[{(11×10)/2}/{(49×48/2)}]
= 13/50 ×{(11×10)/(49×48)}
= {(13×11×10)/(50×49×48)}
= {(13×11)/(5×49×48)}
= 143/11760
So probability that both lost card are heart = 143/11760

Probability Class 11 MCQ Question 2.
If four whole numbers taken at random are multiplied together, then the chance that the last digit in the product is 1, 3, 5, 7 is
(a) 16/25
(b) 16/125
(c) 16/625
(d) none of these

Answer

Answer: (c) 16/625
Hint:
The last digit of the four whole number can be
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
The chance that any of the four numbers is divisible by 2 or 5 = 6/10 = 3/5
Hence, the chance that any of the four numbers is not divisible by 2 or 5 = 1 – 3/5 = 2/5
So, the chance that all of the four numbers are divisible by 2 or 5 = (2/5)×(2/5)×(2/5)×(2/5)
= 16/625
This is the chance that the last digit in the product will not be 0, 2, 4, 5, 6, 8 and this is also the chance that the last digit in the product is 1, 3, 7 or 9

Probability MCQ Class 11 Question 3.
Three identical dice are rolled. The probability that the same number will appear on each of them is
(a) 1/6
(b) 1/36
(c) 1/18
(d) 3/28

Answer

Answer: (b) 1/36
Hint:
Total number of cases = 6³ = 216
The same number can appear on each of the dice in the following ways:
(1, 1, 1), (2, 2, 2), ………….(3, 3, 3)
So, favourable number of cases = 6
Hence, required probability = 6/216 = 1/36

Probability Class 11 Extra Questions Question 4.
There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is
(a) 1/3
(b) 1/6
(c) 1/2
(d) 1/4

Answer

Answer: (b) 1/6
Hint:
First, we choose 1 machine out of given 4.
The probability that it is fault = 2/4 = 1/2
Now, we have to pick the second fault machine.
The probability that it is fault = 1/3
So, required probability = (1/2)×(1/3) = 1/6

Class 11 Probability MCQ Question 5.
Two unbiased dice are thrown. The probability that neither a doublet nor a total of 10 will appear is
(a) 3/5
(b) 2/7
(c) 5/7
(d) 7/9

Answer

Answer: (d) 7/9
Hint:
When two dice are throw, then Total outcome = 36
A doublet: {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Favourable outcome = 6
Sum is 10: {(4, 6), (5, 5), (6, 4)}
Favourable outcome = 3
Again, A doublet and sum is 10: (5, 5)
Favourable outcome = 1
Now, P(either dublet or a sum of 10 appears) = P(A dublet appear) + P(sum is 10) – P(A dublet appear and sum is 10)
⇒ P(either dublet or a sum of 10 appears) = 6/36 + 3/36 – 1/36
= (6 + 3 – 1)/36
= 8/36
= 2/9
So, P(neither dublet nor a sum of 10 appears) = 1 – 2/9 = 7/9

Probability Class 11 MCQ Questions Question 6.
Two dice are thrown the events A, B, C are as follows A: Getting an odd number on the first die. B: Getting a total of 7 on the two dice. C: Getting a total of greater than or equal to 8 on the two dice. Then AUB is equal to
(a) 15
(b) 17
(c) 19
(d) 21

Answer

Answer: (d) 21
Hint:
When two dice are thrown, then total outcome = 6×6 = 36
A: Getting an odd number on the first die.
A = {(1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4),(3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4),(5, 5), (5, 6)}
Total outcome = 18
B: Getting a total of 7 on the two dice.
B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
Total outcome = 6
C: Getting a total of greater than or equal to 8 on the two dice.
C = {(2, 6), (3, 5), (3, 6), (4, 4),(4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6),(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Total outcome = 15
Now n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
⇒ n(A ∪ B) = 18 + 6 – 3
⇒ n(A ∪ B) = 21

Class 11 Maths Probability Extra Questions Question 7.
Two numbers are chosen from {1, 2, 3, 4, 5, 6} one after another without replacement. Find the probability that the smaller of the two is less than 4.
(a) 4/5
(b) 1/15
(c) 1/5
(d) 14/15

Answer

Answer: (a) 4/5
Hint:
Total number of ways of choosing two numbers out of six = ⁶C₂ = (6×5)/2 = 3×5 = 15
If smaller number is chosen as 3 then greater has choice are 4, 5, 6
So, total choices = 3
If smaller number is chosen as 2 then greater has choice are 3, 4, 5, 6
So, total choices = 4
If smaller number is chosen as 1 then greater has choice are 2, 3, 4, 5, 6
So, total choices = 5
Total favourable case = 3 + 4 + 5 = 12
Now, required probability = 12/15 = 4/5

Probability Class 11 Extra Questions With Solutions Question 8.
The probability that when a hand of 7 cards is drawn from a well-shuffled deck of 52 cards, it contains 3 Kings is
(a) 1/221
(b) 5/716
(c) 9/1547
(d) None of these

Answer

Answer: (c) 9/1547
Hint:
Total number of cards = 52
Number of king card = 4
Now, 7 cards are drawn from 52 cards.
P (3 cards are king) = {⁴C₃ × ⁴⁸C₄}/⁵²C₇
= {4×(48×47×46×45)/(4×3×2×1)}/{(52×51×50×49×48×47×46)/(7×6×5×4×3×2×1)}
= {4×(48×47×46×45)×(7×6×5×4×3×2×1)}/{(4×3×2×1)×{(52×51×50×49×48×47×46)}
= (7×6×5×4×45)/(52×51×50×49)
= (6×5×4×45)/(52×51×50×7)
= (6×4×45)/(7×52×51×10)
= (6×45)/(7×13×51×10)
= (6×3)/(7×13×17×2)
= (3×3)/(7×13×17)
= 9/1547

MCQ Of Probability Class 11 Question 9.
A certain company sells tractors which fail at a rate of 1 out of 1000. If 500 tractors are purchased from this company, what is the probability of 2 of them failing within first year
(a) e^-1/2/2
(b) e^-1/2/4
(c) e^-1/2/8
(d) none of these

Answer

Answer: (c) e^-1/2/8
Hint:
This question is based on Poisson distribution.
Now, λ = np = 500×(1/1000) = 500/1000 = 1/2
Now, P(x = 2) = {e^-1/2 × (1/2)²}/2! = e^-1/2/(4×2) = e^-1/2/8

Probability Class 11 Questions Question 10.
The probability that in a random arrangement of the letters of the word INSTITUTION the three T are together is
(a) 0.554
(b) 0.0554
(c) 0.545
(d) 0.0545

Answer

Answer: (d) 0.0545
Hint:
Given word: INSTITUTION
Total letters = 11
The word contains 3 I, 2 N, 1 S, 3 T, 1 U and 1 O
Total number of arrangement = 11!/(3!×2!×3!) = 554400
Now, taken 3 T are together.
So total latter = 9
The number of favorable cases = 9!/(3!×2!) = 30240
Now, P(3 T are together) = 30240/554400 = 0.0545

Class 11 Probability Questions Question 11.
Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is
(a) 2/9
(b) 1/9
(c) 8/9
(d) 7/9

Answer

Answer: (b) 1/9
Hint:
One person can select one house out of 3 = ³C₁ = 3
So, three persons can select one house out of three = 3×3×3 = 27
Thus, probability that all the three can apply for the same house = 3/27 = 1/9

Extra Questions Of Probability Class 11 Question 12.
A bag contains 5 brown and 4 white socks . A man pulls out two socks. The probability that both the socks are of the same colour is
(a) 9/20
(b) 2/9
(c) 3/20
(d) 4/9

Answer

Answer: (d) 4/9
Hint:
Total number of shocks = 5 + 4 = 9
Two shocks are pulled.
Now, P(Both are same color) = (⁵C₂ + ⁴C₂)/⁹C₂
= {(5×4)/(2×1) + (4×3)/(2×1)}/{(9×8)/(2×1)}
= {(5×4) + (4×3)/}/{(9×8)
= (5 + 3)/(9×2)
= 8/18
= 4/9

Probability Extra Questions Class 11 Question 13.
When a coin is tossed 8 times getting a head is a success. Then the probability that at least 2 heads will occur is
(a) 247/265
(b) 73/256
(c) 247/256
(d) 27/256

Answer

Answer: (c) 247/256
Hint:
Let x be number a discrete random variable which denotes the number of heads obtained in n (in this question n = 8)
The general form for probability of random variable x is
P(X = x) = ⁿC_x × p^x × q^n-x
Now, in the question, we want at least two heads
Now, p = q = 1/2
So, P(X ≥ 2) = ⁸C₂ × (1/2)² × (1/2)^8-2
⇒ P(X ≥ 2) = ⁸C₂ × (1/2)² × (1/2)⁶
⇒ 1 – P(X < 2) = ⁸C₀ × (1/2)⁰ × (1/2)⁸ + ⁸C₁ × (1/2)¹ × (1/2)^8-1
⇒ 1 – P(X < 2) = (1/2)⁸ + 8 × (1/2)¹ × (1/2)⁷
⇒ 1 – P(X < 2) = 1/256 + 8 × (1/2)⁸
⇒ 1 – P(X < 2) = 1/256 + 8/256
⇒ 1 – P(X < 2) = 9/256
⇒ P(X < 2) = 1 – 9/256
⇒ P(X < 2) = (256 – 9)/256
⇒ P(X < 2) = 247/256

Probability Class 11 Important Questions Question 14.
A couple has two children. The probability that both children are females if it is known that the elder child is a female is
(a) 0
(b) 1
(c) 1/2
(d) 1/3

Answer

Answer: (c) 1/2
Hint:
Given, a couple has two children.
Let A denotes both children are females i.e. {FF}
Now, P(A) = (1/2)×(1/2) = 1/4
and B denotes elder children is a female i.e. {FF, FM}
P(B) = 1/4 + 1/4 = 1/2
Now, P(A ∩ B) = 1/4
Now, P(Both the children are female if elder child is female)
P(A/B) = P(A ∩ B)/P(B)
⇒ P(A/B) = (1/4)/(1/2)
⇒ P(A/B) = 1/2

Important Questions Of Probability Class 11 Question 15.
A certain company sells tractors which fail at a rate of 1 out of 1000. If 500 tractors are purchased from this company, what is the probability of 2 of them failing within first year
(a) e^-1/2/2
(b) e-^-1/2/4
(c) e^-1/2/8
(d) none of these

Answer

Answer: (c) e^-1/2/8
Hint:
This question is based on Poisson distribution.
Now, λ = np = 500×(1/1000) = 500/1000 = 1/2
Now, P(x = 2) = {e^-1/2 × (1/2)²}/2! = e^-1/2 /(4×2) = e^-1/2/8

Probability Important Questions Class 11 Question 16.
A random variable X has poison distribution with mean 2. Then, P (X > 1.5) equals
(a) 1 – 3/e²
(b) 2/e²
(c) 3/e²
(d) 0

Answer

Answer: (a) 1 – 3/e²
Hint:
Here m = 2
Now, P(X > 1.5) = ∑_r {(e^-2 × 2^r)/r!} {2 ≤ r ≤ ∞}
= e^-2 {2²/2! + 2³/3! + 2⁴/4! + …}
= e^-2 {(1 + 2 /1! + 2²/2! + 2³/3! + …) – 1 – 2}
= e^-2 (e² – 3)
= 1 – 3e^-2
= 1 – 3/e²

Class 11 Probability Important Questions Question 17.
Let A and B are two mutually exclusive events and if P(A) = 0.5 and P(B ̅) = 0.6 then P(A∪B) is
(a) 0
(b) 1
(c) 0.6
(d) 0.9

Answer

Answer: (d) 0.9
Hint:
Given, A and B are two mutually exclusive events.
So, P(A ∩ B) = 0
Again given P(A) = 0.5 and P(B ̅) = 0.6
P(B) = 1 – P(B ̅) = 1 – 0.6 = 0.4
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ P(A ∪ B) = P(A) + P(B)
⇒ P(A ∪ B) = 0.5 + 0.4 = 0.9

Probability Questions Class 11 Question 18.
The probability of getting 53 Sundays in a leap year is
(a) 1/7
(b) 2/7
(c) 3/7
(d) None of these

Answer

Answer: (b) 2/7
Hint:
In a leap year, the total number of days = 366 days.
In 366 days, there are 52 weeks and 2 days.
Now two days may be
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
Now there are total 7 possibilities, So total outcomes = 7
In 7 possibilities, Sunday came two times.
So, favorable case = 2
Hence, the probabilities of getting 53 Sundays in a leap year = 2/7

Questions On Probability Class 11 Question 19.
The probability of getting the number 6 at least once in a regular die if it can roll it 6 times?
(a) 1 – (5/6)⁶
(b) 1 – (1/6)⁶
(c) (5/6)⁶
(d) (1/6)⁶

Answer

Answer: (a) 1 – (5/6)⁶
Hint:
Let A is the event that 6 does not occur at all.
Now, the probability of at least one 6 occur = 1 – P(A)
= 1 – (5/6)⁶

Question 20.
On his vacation, Rahul visits four cities (A, B, C, and D) in a random order. The probability that he visits A first and B last is
(a) 1/2
(b) 1/6
(c) 1/10
(d) 1/12

Answer

Answer: (d) 1/12
Hint:
Total cities are 4 i.e. A, B, C, D
Given, Rahul visit four cities, So, n(S) = 4! = 24
Now, sample space IS:
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BDAC, BDCA, BCAD, BCDA, CABD, CADB, CBDA, CDAD, CDAB,CDBA, DABC, DACB, DBCA, DBAC, DCAB, DCBA}
Let G = Rahul visits A firsta and B last
⇒ G = {ACDB, ADCB}
⇒ n(G) = 2
So, P(G) = n(G)/n(S) = 2/24 = 1/12

We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 16 Probability with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Probability MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Maths MCQ:

MCQ Questions for Class 11 Maths Chapter 15 Statistics with Answers

Statistics Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 15 Statistics with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Statistics Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 15 MCQ With Answers

Maths Class 11 Chapter 15 MCQs On Statistics

Statistics Class 11 MCQ Question 1.
The sum of 10 items is 12 and the sum of their squares is 18. The standard deviation is
(a) 1/5
(b) 2/5
(c) 3/5
(d) 4/5

Answer

Answer: (c) 3/5
Hint:
Given, ∑x = 12 and ∑x² = 18
Now, varience = ∑x²/n – (∑x/n)²
⇒ varience = 18/10 – (12/10)²
⇒ varience = 9/5 – (6/5)²
⇒ varience = 9/5 – 36/25
⇒ varience = (9 × 5 – 36)/25
⇒ varience = (45 – 36)/25
⇒ varience = 9/25
⇒ Standard deviation = √(9/25)
⇒ Standard deviation = 3/5

MCQ On Statistics Class 11 Question 2.
The algebraic sum of the deviation of 20 observations measured from 30 is 2. So, the mean of observations is
(a) 30.0
(b) 30.1
(c) 30.2
(d) 30.3

Answer

Answer: (b) 30.1
Hint:
Given, algebraic sum of of the deviation of 20 observations measured from 30 is 2
⇒ ∑(x_i – 30) = 2 {1 ≤ i ≤ 20}
⇒ ∑x_i – 30 × 20 = 2
⇒ (∑x_i)/20 – (30 × 20)/20 = 2/20
⇒ (∑x_i)/20 – 30 = 0.1
⇒ Mean – 30 = 0.1
⇒ Mean = 30 + 0.1
⇒ Mean = 30.1

Statistics MCQ Class 11 Question 3.
The coefficient of variation is computed by
(a) S.D/.Mean×100
(b) S.D./Mean
(c) Mean./S.D×100
(d) Mean/S.D.

Answer

Answer: (b) S.D./Mean
Hint:
The coefficient of variation = S.D./Mean

Class 11 Statistics MCQ Question 4.
When tested the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623. The mean of the lives of 5 bulbs is
(a) 1445
(b) 1446
(c) 1447
(d) 1448

Answer

Answer: (b) 1446
Hint:
Given, lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623
Now, mean = (1357 + 1090 + 1666 + 1494 + 1623)/5
= 7230/5
= 1446

MCQ Of Statistics Class 11 Question 5.
If mode of a series exceeds its mean by 12, then mode exceeds the median by
(a) 4
(b) 8
(c) 6
(d) 12

Answer

Answer: (b) 8
Hint:
Given, Mode = Mean + 12
⇒ Mode – 12 = Mean
Now, Mode = 3×Median – 2×Mean
⇒ Mode = 3×Median – 2(Mode – 12)
⇒ Mode = 3×Median – 2×Mode + 24
⇒ Mode + 2×Mode = 3×Median + 24
⇒ 3×Mode = 3×Median = 24
⇒ Mode = Median + 8
So, mode exceeds the median by 8

Class 11 Maths Chapter 15 MCQ Question 6.
The median and SD of a distributed are 20 and 4 respectively. If each item is increased by 2, the new median and SD are
(a) 20, 4
(b) 22, 6
(c) 22, 4
(d) 20, 6

Answer

Answer: (c) 22, 4
Hint:
Since each value is increased by 2, therefore the median value is also increased by
2. So, new median = 22
Again, the variance is independent of the change of origin. So it remains the same.

Class 11 Maths Statistics MCQ Question 7.
Range of the data 4, 7, 8, 9, 10, 12, 13 and 17 is
(a) 4
(b) 17
(c) 13
(d) 21

Answer

Answer: (c) 13
Hint:
Give, data are: 4, 7, 8, 9, 10, 12, 13 and 17
Range = Maximum value – Minimum Value
= 17 – 4
= 13

MCQs Statistics Class 11 Question 8.
If Mean = Median = Mode, then it is
(a) Symmetric distribution
(b) Asymmetric distribution
(c) Both symmetric and asymmetric distribution
(d) None of these

Answer

Answer: (a) Symmetric distribution
Hint:
In a symmetric distribution,
Mean = Median = Mode

Class 11 Maths Chapter 15 MCQ With Answers Question 9.
If the difference of mode and median of a data is 24, then the difference of median and mean is
(a) 12
(b) 24
(c) 8
(d) 36

Answer

Answer: (a) 12
Hint:
Given the difference of mode and median of a data is 24
⇒ Mode – Median = 24
⇒ Mode = Median + 24
Now, Mode = 3×Median – 2×Mean
⇒ Median + 24 = 3×Median – 2×Mean
⇒ 24 = 3×Median – 2×Mean – Median
⇒ 24 = 2×Median – 2×Mean
⇒ Median – Mean = 24/2
⇒ Median – Mean = 12

Statistics Class 11 MCQ Questions Question 10.
If r is the correlation coefficient, then
(a) |r| ≤ 1
(b) r ≤ 1
(c) |r| ≥ 1
(d) r ≥ 1

Answer

Answer: (a) |r| ≤ 1
Hint:
If r is the correlation coefficient, then |r| ≤ 1

Class 11 Maths Ch 15 MCQ Question 11.
If the varience of the data is 121 then the standard deviation of the data is
(a) 121
(b) 11
(c) 12
(d) 21

Answer

Answer: (b) 11
Hint:
Given, varience of the data = 121
Now, the standard deviation of the data = √(121) = 11

Ch 15 Maths Class 11 MCQ Question 12.
If the mean of the following data is 20.6, then the value of p is
x = 10 15 p 25 35
f = 3 10 25 7 5
(a) 30
(b) 20
(c) 25
(d) 10

Answer

Answer: (b) 20
Hint:
Mean = ∑ f_i× x_i /∑ f_i
⇒ 20.6 = (10 × 3 + 15 × 10 + p × 25 + 25 × 7 + 35 × 5)/(3 + 10 + 25 + 7 + 5)
⇒ 20.6 = (30 + 150 + 25p + 175 + 175)/50
⇒ 20.6 = (530 + 25p)/50
⇒ 530 + 25p = 20.6 × 50
⇒ 530 + 25p = 1030
⇒ 25p = 1030 – 530
⇒ 25p = 500
⇒ p = 500/25
⇒ p = 20
So, the value of p is 20

MCQs On Statistics Class 11 Question 13.
If the mean of first n natural numbers is 5n/9, then n =
(a) 5
(b) 4
(c) 9
(d) 10

Answer

Answer: (c) 9
Hint:
Given mean of first n natural number is 5n/9
⇒ (n+1)/2 = 5n/9
⇒ n + 1 = (5n×2)/9
⇒ n + 1 = 10n/9
⇒ 9(n + 1) = 10n
⇒ 9n + 9 = 10n
⇒ 10n – 9n = 9
⇒ n = 9

MCQ On Statistics Class 11 Maths Question 14.
If one of the observation is zero then geometric mean is
(a) (Sum of observation)/n
(b) (Multiplication of all observations)ⁿ
(c) (Multiplication of all observations)^1/n
(d) 0

Answer

Answer: (d) 0
Hint:
Let the observations are 0, a, b, c, ……… up to n
Now, geometric mean = (0 × a × b × c × ……… up to n)^1/n
= 0
So, geometric mean is 0

MCQ For Statistics Class 11 Question 15.
Which one is measure of dispersion method
(a) Renge
(b) Quartile deviation
(c) Mean deviation
(d) all of the above

Answer

Answer: (d) all of the above
Hint:
Range, Quartile deviation, Mean deviation all are the measure of dispersions method.

MCQs Of Statistics Class 11 Question 16.
If a variable takes discrete values x + 4, x – 7/2, x – 5/2, x – 3, x – 2, x + 1/2, x – 1/2, x + 5 (x > 0), then the median is
(a) x – 5/4
(b) x – 1/2
(c) x – 2
(d) x + 5/4

Answer

Answer: (a) x – 5/4
Hint:
Given, discrete values x + 4, x – 7/2, x – 5/2, x – 3, x – 2, x + 1/2, x – 1/2, x + 5
Now, arrange them in ascending order, we get
x – 7/2, x – 3, x – 5/2, x – 2, x – 1/2, x + 1/2, x + 4, x + 5
Total observations = 8
Now, median = AM of 4th and 5th observations
= AM of (x – 2) and (x – 1/2) observations
= (x – 2 + x – 1/2)/2
= (2x – 5/2)/2
= x – 5/4

Statistics Class 11 MCQs Question 17.
If covariance between two variables is 0, then the correlation coefficient between them is
(a) nothing can be said
(b) 0
(c) positive
(d) negative

Answer

Answer: (b) 0
Hint:
The relationship between the correlation coefficient and covariance for two variables as shown below:
r_{(x, y)} = COV(x, y)/{s_x × s_y}
r_{(x, y)} = correlation of the variables x and y
COV(x, y) = covariance of the variables x and y
s_x = sample standard deviation of the random variable x
s_x = sample standard deviation of the random variable y
Now given COV(x, y) = 0
Then r_{(x, y)} = 0

Statistics MCQs Class 11 Question 18.
The mean of a group of 100 observations was found to be 20. Later on, it was found that three observations were incorrect, which was recorded as 21, 21 and 18. Then the mean if the incorrect observations are omitted is
(a) 18
(b) 20
(c) 22
(d) 24

Answer

Answer: (b) 20
Hint:
Given mean of 100 observations is 20
Now
∑ xi/100 = 20 (1 = i = 100)
⇒ ∑xi = 100×20
⇒ ∑xi = 2000
3 observations 21, 21 and 18 are recorded in-correctly.
So ∑xi = 2000 – 21 – 21 – 18
⇒ ∑xi = 2000 – 60
⇒ ∑xi = 1940
Now new mean is
∑ xi/100 = 1940/97 = 20
So, the new mean is 20

Statistics Class 11 MCQ Maths Question 19.
Varience is independent of change of
(a) origin only
(b) scale only
(c) origin and scale both
(d) None of these

Answer

Answer: (a) origin only
Hint:
Varience is independent of change of origin only.

MCQ Of Chapter 15 Maths Class 11 Question 20.
Let x₁, x₂, x₃, ……… , x_n, be n observations and X be the arithmetic mean. Then formula for the standard deviation is given by
(a) ∑(x_i – mean)²
(b) ∑(x_i – mean)2 /n
(c) √{∑(x_i – mean)²/n}
(d) None of these

Answer

Answer: (c) √{∑(x_i – mean)²/n}
Hint:
Given, x₁, x₂, x₃, ………. , x_n be n observations and X be the arithmetic mean.
Now standard deviation = √{∑(x_i – mean)²/n}

We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 15 Statistics with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Statistics MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Maths MCQ:

MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives with Answers

Limits and Derivatives Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Limits and Derivatives Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 13 MCQ With Answers

Maths Class 11 Chapter 13 MCQs On Limits and Derivatives

Limits Class 11 MCQ Question 1.
The value of the limit Lim_x→0 (cos x)^{cot2 x} is
(a) 1
(b) e
(c) e^1/2
(d) e^-1/2

Answer

Answer: (d) e^-1/2
Hint:
Given, Lim_x→0 (cos x)^{cot² x}
= Lim_x→0 (1 + cos x – 1)^{cot² x}
= eLim_x→0 (cos x – 1) × cot² x
= eLim_x→0 (cos x – 1)/tan² x
= e^-1/2

Limits And Derivatives Class 11 MCQ Question 2.
The value of limit Lim_x→0 {sin (a + x) – sin (a – x)}/x is
(a) 0
(b) 1
(c) 2 cos a
(d) 2 sin a

Answer

Answer: (c) 2 cos a
Hint:
Given, Lim_x→0 {sin (a + x) – sin (a – x)}/x
= Lim_x→0 {2 × cos a × sin x}/x
= 2 × cos a × Lim_x→0 sin x/x
= 2 cos a

MCQ On Limits Class 11 Question 3.
Lim_x→-1 [1 + x + x² + ……….+ x¹⁰] is
(a) 0
(b) 1
(c) -1
(d) 2

Answer

Answer: (b) 1
Hint:
Given, Lim_x→-1 [1 + x + x² + ……….+ x¹⁰]
= 1 + (-1) + (-1)² + ……….+ (-1)¹⁰
= 1 – 1 + 1 – ……. + 1
= 1

Limits MCQ Class 11 Question 4.
The value of Lim_x→01 (1/x) × sin^-1 {2x/(1 + x²) is
(a) 0
(b) 1
(c) 2
(d) -2

Answer

Answer: (c) 2
Hint:
Given, Lim_x→0 (1/x) × sin^-1 {2x/(1 + x²)
= Lim_x→0 (2× tan^-1 x)/x
= 2 × 1
= 2

Limits And Derivatives Class 11 MCQ Questions Question 5.
Lim_x→0 log(1 – x) is equals to
(a) 0
(b) 1
(c) 1/2
(d) None of these

Answer

Answer: (a) 0
Hint:
We know that
log(1 – x) = -x – x²/2 – x³/3 – ……..
Now,
Lim_x→0 log(1 – x) = Lim_x→0 {-x – x²/2 – x³/3 – ……..}
⇒ Lim_x→0 log(1 – x) = Lim_x→0 {-x} – Lim_x→0 {x²/2} – Lim_x→0 {x³/3} – ……..
⇒ Lim_x→0 log(1 – x) = 0

MCQ On Limits And Derivatives Class 11 Question 6.
Lim_x→0 {(a^x – b^x)/ x} is equal to
(a) log a
(b) log b
(c) log (a/b)
(d) log (a×b)

Answer

Answer: (c) log (a/b)
Hint:
Given, Lim_x→0 {(a^x – b^x)/ x}
= Lim_x→0 {(a^x – b^x – 1 + 1)/ x}
= Lim_x→0 {(a^x – 1) – (b^x – 1)}/ x
= Lim_x→0 {(a^x – 1)/x – (b^x – 1)/x}
= Lim_x→0 (a^x – 1)/x – Lim_x→0 (b^x – 1)/x
= log a – log b
= log (a/b)

Class 11 Limits MCQ Question 7.
The value of lim_y→0 {(x + y) × sec (x + y) – x × sec x}/y is
(a) x × tan x × sec x
(b) x × tan x × sec x + x × sec x
(c) tan x × sec x + sec x
(d) x × tan x × sec x + sec x

Answer

Answer: (d) x × tan x × sec x + sec x
Hint:
Given, lim_y→0 {(x + y) × sec (x + y) – x×sec x}/y
= lim_y→0 {x sec (x + y) + y sec (x + y) – x×sec x}/y
= lim_y→0 [x{ sec (x + y) – sec x} + y sec (x + y)]/y
= lim_y→0 x{ sec (x + y) – sec x}/y + lim_y→0 {y sec (x + y)}/y
= lim_y→0 x{1/cos (x + y) – 1/cos x}/y + lim_y→0 {y sec (x + y)}/y
= lim_y→0 [{cos x – cos (x + y)} × x/{y×cos (x + y)×cos x}] + lim_y→0 {y sec (x + y)}/y
= lim_y→0 [{2sin (x + y/2) × sin(y/2)} × 2x/{2y×cos (x + y)×cos x}] + lim_y→0 {y sec (x + y)}/y
= lim_y→0 {sin (x + y/2) × lim_y→0 {sin(y/2)/(2y/2)} × lim_y→0 { x/{y×cos (x + y)×cos x}] + sec x
= sin x × 1 × x/cos² x + sec x
= x × tan x × sec x + sec x
So, lim_y→0 {(x + y) × sec (x + y) – x×sec x}/y = x × tan x × sec x + sec x

Class 11 Maths Chapter 13 MCQ Question 8.
Lim_y→∞ {(x + 6)/(x + 1)}^(x+4) equals
(a) e
(b) e³
(c) e⁵
(d) e⁶

Answer

Answer: (c) e⁵
Hint:
Given, Lim_y→∞ {(x + 6)/(x + 1)}(x + 4)
= Lim_y→∞ {1 + 5/(x + 1)}(x + 4)
= e^{Lim_y→∞ 5(x + 4)/(x + 1)}
= e^{Lim_y→∞ 5(1 + 4/x)/(1 + 1/x)}
= e^{5(1 + 4/∞)/(1 + 1/∞)}
= e^{5/(1 + 0)}
= e⁵

Limits MCQs With Answers Question 9.
The derivative of [1+(1/x)] /[1-(1/x)] is
(a) 1/(x-1)²
(b) -1/(x-1)²
(c) 2/(x-1)²
(d) -2/(x-1)²

Answer

Answer: (d) A
Hint:
Let y = [1+(1/x)] /[1-(1/x)]
then dy/dx = [{1-(1/x)}*(-1/x²)]/[{1+(1/x)}*(1/x²)]
= (1/x²) [(1/x) -1 – 1 – (1/x)]/[1-(1/x)]²
= [-2/x²]/[(x-1)/x]²
= -2/(x-1)²

Limits And Derivatives MCQ Question 10.
The expansion of log(1 – x) is
(a) x – x²/2 + x³/3 – ……..
(b) x + x²/2 + x³/3 + ……..
(c) -x + x²/2 – x³/3 + ……..
(d) -x – x²/2 – x³/3 – ……..

Answer

Answer: (d) -x – x²/2 – x³/3 – ……..
Hint:
log(1 – x) = -x – x²/2 – x³/3 – ……..

MCQs Of Limits Class 11 Question 11.
If f(x) = x × sin(1/x), x ≠ 0, then Lim_x→0 f(x) is
(a) 1
(b) 0
(c) -1
(d) does not exist

Answer

Answer: (b) 0
Hint:
Given, f(x) = x × sin(1/x)
Now, Lim_x→0 f(x) = Lim_x→0 x × sin(1/x)
⇒ Lim_x→0 f(x) = 0

Ch 13 Maths Class 11 MCQ Question 12.
The value of Lim_n→∞ {1² + 2² + 3² + …… + n²}/n³ is
(a) 0
(b) 1
(c) -1
(d) n

Answer

Answer: (a) 0
Hint:
Given, Lim_n→∞ {1² + 2² + 3² + …… + n²}/n³
= Lim_n→∞ [{n×(n + 1)×(2n + 1)}/6]/{n(n + 1)/2}²
= Lim_n→∞ [{n×n×n ×(1 + 1/n)×(2 + 1/n)}/6]/{n × n ×(1 + 1/n)/2}²
= Lim_n→∞ [{n³ ×(1 + 1/n)×(2 + 1/n)}/6]/{n² ×(1 + 1/n)/2}²
= Lim_n→∞ [{(1 + 1/n)×(2 + 1/n)}/6]/[n⁴ × {(1 + 1/n)/2}²]
⇒ Lim_n→∞ [{(1 + 1/n)×(2 + 1/n)}/6]/[n × {(1 + 1/n)/2}²]
= [{(1 + 1/∞)×(2 + 1/∞)}/6]/[∞×{(1 + 1/∞)/2}²
= [{(1 + 0)×(2 + 0)}/6]/∞ {since 1/∞ = 0}
= {(1 × 2)/6}/∞
= (2/6)/∞
= (1/3)/∞
= 0
So, Lim_n→∞ {1² + 2² + 3² + …… + n²}/n³ = 0

MCQ Of Limits And Derivatives Class 11 Question 13.
The value of Lim_n→∞ (sin x/x) is
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (a) 0
Hint:
Lim_n→∞ (sin x/x) = Lim_y→0 {y × sin (1/y)} = 0

Class 11 Maths Limits MCQ Question 14.
The value of Lim_x→0 ax is
(a) 0
(b) 1
(c) 1/2
(d) 3/2

Answer

Answer: (b) 1
Hint:
We know that
a^x = 1 + x/1! × (log a) + x²/2! × (log a)² + x³/3! × (log a)³ + ………..
Now,
Lim_x→0 a^x = Lim_x→0 {1 + x/1! × (log a) + x²/2! × (log a)² + x³/3! × (log a)³ + …}
⇒ Lim_x→0 a^x = Lim_x→0 1 + Lim_x→0 {x/1! × (log a)} + Lim_x→0 {x² /2! × (log a)²}+ ………
⇒ Lim_x→0 a^x = 1

MCQ Of Limits Class 11 Question 15.
Let f(x) = cos x, when x ≥ 0 and f(x) = x + k, when x < 0 Find the value of k given that Lim_x→0 f(x) exists.
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (b) 1
Hint:
Given, Lim_x→0 f(x) exists
⇒ Lim_x→0 – f(x) = Lim_x→0 + f(x)
⇒ Lim_x→0 (x + k) = Lim_x→0 cos x
⇒ k = cos 0
⇒ k = 1

MCQ On Limits Class 11 Pdf Question 16.
The value of Lim_x→0 (1/x) × sin^-1 {2x/(1 + x²) is
(a) 0
(b) 1
(c) 2
(d) -2

Answer

Answer: (c) 2
Hint:
Given, Lim_x→0 (1/x) × sin^-1 {2x/(1 + x²)
= Lim_x→0 (2 × tan^-1 x)/x
= 2 × 1
= 2

Limit Class 11 MCQ Question 17.
Lim_x→0 sin (ax)/bx is
(a) 0
(b) 1
(c) a/b
(d) b/a

Answer

Answer: (c) a/b
Hint:
Given, Lim_x→0 sin (ax)/bx
= Lim_x→0 [{sin (ax)/ax} × (ax/bx)]
⇒ (a/b) Lim_x→0 sin (ax)/ax
= a/b

Class 11 Maths Ch 13 MCQ Question 18.
The value of the limit Lim_x→0 {log(1 + ax)}/x is
(a) 0
(b) 1
(c) a
(d) 1/a

Answer

Answer: (c) a
Hint:
Given, Lim_x→0 {log(1 + ax)}/x
= Lim_x→0 {ax – (ax)² /2 + (ax)³ /3 – (ax)⁴ /4 + …….}/x
= Lim_x→0 {ax – a² x² /2 + a³ x³ /3 – a⁴ x⁴ /4 + …….}/x
= Lim_x→0 {a – a² x /2 + a³ x² /3 – a⁴ x³ /4 + …….}
= a – 0
= a

MCQs On Limits Class 11 Question 19.
If f(x) = (x + 1)/x then df(x)/dx is
(a) 1/x
(b) -1/x
(c) -1/x²
(d) 1/x²

Answer

Answer: (c) -1/x²
Hint:
Given, f(x) = (x + 1)/x
Now, df(x)/dx = d{(x + 1)/x}/dx
= {1 × x – (x + 1)×1}/x²
= (x – x – 1)/x²
= -1/x²

Class 11 Maths Chapter 13 MCQ With Answers Question 20.
Lim_x→0 (e^x² – cos x)/x² is equals to
(a) 0
(b) 1
(c) 2/3
(d) 3/2

Answer

Answer: (d) 3/2
Hint:
Given, Lim_x→0 (e^x² – cos x)/x²
= Lim_x→0 (e^x² – cos x -1 + 1)/x²
= Lim_x→0 {(e^x² – 1)/x² + (1 – cos x)}/x²
= Lim_x→0 {(e^x² – 1)/x² + Lim_x→0 (1 – cos x)}/x²
= 1 + 1/2
= (2 + 1)/2
= 3/2

We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Limits and Derivatives MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Maths MCQ:

MCQ Questions for Class 11 Maths Chapter 1 Sets with Answers

Sets Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 1 Sets with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Sets Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 1 MCQ With Answers

Maths Class 11 Chapter 1 MCQs On Sets

Sets Class 11 MCQ Question 1.
If A, B and C are any three sets, then A – (B ∪ C) is equal to
(a) (A – B) ∪ (A – C)
(b) (A – B) ∪ C
(c) (A – B) ∩ C
(d) (A – B) ∩ (A – C)

Answer

Answer: (d) (A – B) ∩ (A – C)
Hint:
Given A, B and C are any three sets.
Now, A – (B ∪ C) = (A – B) ∩ (A – C)

Sets MCQ Questions Class 11 Question 2.
(A’)’ = ?
(a) ∪ – A
(b) A’
(c) ∪
(d) A

Answer

Answer: (d) A
Hint:
(A’)’ = A

MCQ On Sets Class 11 Question 3.
A – B is read as?
(a) Difference of A and B of B and A
(b) None of the above
(c) Difference of B and A
(d) Both a and b

Answer

Answer: (a) Difference of A and B of B and A
Hint:
A – B will read as difference of A and B of B and A
Ex: Let A = {1, 2, 3, 4, 5} and B = {1, 3, 5, 7}
Now, A – B = {2, 4}

Class 11 Sets MCQ Questions Question 4.
If A, B and C are any three sets, then A × (B ∪ C) is equal to
(a) (A × B) ∪ (A × C)
(b) (A ∪ B) × (A ∪ C)
(c) None of these
(d) (A × B) ∩ (A × C)

Answer

Answer: (a) (A × B) ∪ (A × C)
Hint:
Given A, B and C are any three sets.
Now, A × (B ∪ C) = (A × B) ∪ (A × C)

MCQ Of Sets Class 11 Question 5.
IF A = [5, 6, 7] and B = [7, 8, 9] then A ∪ B is equal to
(a) [5, 6, 7, 8, 9]
(b) [5, 6, 7]
(c) [7, 8, 9]
(d) None of these

Answer

Answer: (a) [5, 6, 7, 8, 9]
Hint:
Given A = [5, 6, 7] and B = [7, 8, 9]
then A ∪ B = [5, 6, 7, 8, 9]

Class 11 Maths Chapter 1 MCQ Question 6.
Which of the following sets are null sets
(a) {x: |x |< -4, x ?N}
(b) 2 and 3
(c) Set of all prime numbers between 15 and 19
(d) {x: x < 5, x > 6}

Answer

Answer: (b) 2 and 3
Hint:
2 and 3 is the null set.

Class 11 Maths Chapter 1 MCQ With Answers Question 7.
IF R = {(2, 1),(4, 3),(4, 5)}, then range of the function is?
(a) Range R = {2, 4}
(b) Range R = {1, 3, 5}
(c) Range R = {2, 3, 4, 5}
(d) Range R {1, 1, 4, 5}

Answer

Answer: (b) Range R = {1, 3, 5}
Hint:
Given R = {(2, 1),(4, 3),(4, 5)}
then Range(R) = {1, 3, 5}

Class 11 Maths MCQ Chapter 1 Question 8.
The members of the set S = {x | x is the square of an integer and x < 100} is
(a) {0, 2, 4, 5, 9, 58, 49, 56, 99, 12}
(b) {0, 1, 4, 9, 16, 25, 36, 49, 64, 81}
(c) {1, 4, 9, 16, 25, 36, 64, 81, 85, 99}
(d) {0, 1, 4, 9, 16, 25, 36, 49, 64, 121}

Answer

Answer: (b) {0, 1, 4, 9, 16, 25, 36, 49, 64, 81}
Hint:
The set S consists of the square of an integer less than 100
So, S = {0, 1, 4, 9, 16, 25, 36, 49, 64, 81}

Class 11 Maths Ch 1 MCQ Question 9.
In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math. How many opt for none of the three subjects?
(a) 19
(b) 41
(c) 21
(d) 57

Answer

Answer: (b) 41
Hint:
The number of students who took at least one of the three subjects can be found by finding out A ∪ B ∪ C, where A is the set of those who took Physics, B the set of those who took Chemistry and C the set of those who opted for Math.
Now, A ∪ B ∪ C = A + B + C – (A ∩ B + B ∩ C + C ∩ A) + (A ∩ B ∩ C)
A is the set of those who opted for Physics = 120/2 = 60 students
B is the set of those who opted for Chemistry = 120/5 = 24
C is the set of those who opted for Math = 120/7 = 17
The 10th, 20th, 30th….. numbered students would have opted for both Physics and Chemistry.
Therefore, A ∩ B = 120/10 = 12
The 14th, 28th, 42nd….. Numbered students would have opted for Physics and Math.
Therefore, C ∩ A = 120/14 = 8
The 35th, 70th…. numbered students would have opted for Chemistry and Math.
Therefore, B ∩ C = 120/35 = 3
And the 70th numbered student would have opted for all three subjects.
Therefore, A ∪ B ∪ C = 60 + 24 + 17 – (12 + 8 + 3) + 1 = 79
Number of students who opted for none of the three subjects = 120 – 79 = 41

MCQ Questions For Class 11 Maths With Answers Chapter 1 Question 10.
{ (A, B) : A² +B² = 1} on the sets has the following relation
(a) reflexive
(b) symmetric
(c) none
(d) reflexive and transitive

Answer

Answer: (b) symmetric
Hint:
Given {(a, b) : a² + b² = 1} on the set S.
Now a² +b² = b² + a² = 1
So, the given relation is symmetric.

MCQ Class 11 Maths Chapter 1 Question 11.
Two finite sets have N and M elements. The number of elements in the power set of first set is 48 more than the total number of elements in power set of the second test. Then the value of M and N are
(a) 7, 6
(b) 6, 4
(c) 7, 4
(d) 6, 3

Answer

Answer: (b) 6, 4
Hint:
Let A and B be two sets having m and n numbers of elements respectively
Number of subsets of A = 2m
Number of subsets of B = 2n
Now, according to question
2m – 2n = 48
⇒ 2n(2m – n – 1) = 24(22 – 1)
So, n = 4
and m – n = 2
⇒ m – 4 = 2
⇒ m = 2 + 4
⇒ m = 6

MCQ Questions On Sets Class 11 Question 12.
The range of the function f(x) = 3x – 2‚ is
(a) (- ∞, ∞)
(b) R – {3}
(c) (- ∞, 0)
(d) (0, – ∞)

Answer

Answer: (a) (- ∞, ∞)
Hint:
Let the given function is
y = 3x – 2
⇒ y + 2 = 3x
⇒ x = (y + 2)/3
Now x is saisfied by all values.
So, Range{f(x)} = R = (-∞, ∞)

Ch 1 Maths Class 11 MCQ Question 13.
If A, B, C be three sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C, then,
(a) B = C
(b) A = C
(c) A = B = C
(d) A = B

Answer

Answer: (a) B = C
Hint:
Given A, B, C be three sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C then B = C

MCQ On Sets For Class 11 Pdf With Answers Question 14.
In 2nd quadrant?
(a) X < 0, Y < 0
(b) X < 0, Y > 0
(c) X > 0, Y > 0
(d) X > 0, Y < 0

Answer

Answer: (b) X < 0, Y > 0
Hint:
In the second quadrant,
MCQ Questions for Class 11 Maths Chapter 1 Sets with Answers 1
X < 0, Y > 0

MCQ Of Chapter Sets Class 11 Question 15.
How many rational and irrational numbers are possible between 0 and 1?
(a) 0
(b) Finite
(c) Infinite
(d) 1

Answer

Answer: (c) Infinite
Hint:
There are infinite many rational and irrational numbers are possible between 0 and 1
This is because between any two numbers, there are infinite numbers.

Question 16.
Empty set is a?
(a) Finite Set
(b) Invalid Set
(c) None of the above
(d) Infinite Set

Answer

Answer: (a) Finite Set
Hint:
In mathematics, and more specifically set theory, the empty set is the unique set having no elements and its size or cardinality (count of elements in a set) is zero.
So, an empty set is a finite set.

Question 17.
If A = [5, 6, 7] and B = [7, 8, 9] then A U B is equal to
(a) [5, 6, 7, 8, 9]
(b) [5, 6, 7]
(c) [7, 8, 9]
(d) None of these

Answer

Answer: (a) [5, 6, 7, 8, 9]
Hint:
Given A = [5, 6, 7] and B = [7, 8, 9]
then A U B = [5, 6, 7, 8, 9]

Question 18.
Which of the following two sets are equal?
(a) A = {1, 2} and B = {1}
(b) A = {1, 2} and B = {1, 2, 3}
(c) A = {1, 2, 3} and B = {2, 1, 3}
(d) A = {1, 2, 4} and B = {1, 2, 3}

Answer

Answer: (c) A = {1, 2, 3} and B = {2, 1, 3}
Hint:
Two sets are equal if and only if they have the same elements.
So, A = {1, 2, 3} and B = {2, 1, 3} are equal sets.

Question 19.
In a class of 50 students, 10 did not opt for math, 15 did not opt for science and 2 did not opt for either. How many students of the class opted for both math and science.
(a) 24
(b) 25
(c) 26
(d) 27

Answer

Answer: (d) 27
Hint:
Total students = 50
Students who did not opt for math = 10
Students who did not opt for Science = 15
Students who did not opt for either maths or science = 2
Total of 40 students in math and 13 did not opt for science but did for math = 40 – 13 = 27
So, students of the class opted for both math and science is 27

Question 20.
In last quadrant?
(a) X < 0, Y > 0
(b) X < 0, Y < 0
(c) X > 0, Y < 0
(d) X > 0, Y > 0

Answer

Answer: (d) X > 0, Y > 0
Hint:
In the last quadrant,
MCQ Questions for Class 11 Maths Chapter 1 Sets with Answers 2
X > 0, Y > 0

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Class 11 Maths MCQ:

MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers

Trigonometric Functions Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Trigonometric Functions Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 3 MCQ With Answers

Maths Class 11 Chapter 3 MCQs On Trigonometric Functions

MCQ On Trigonometry For Class 11 Pdf Question 1.
The value of cos² x + cos² y – 2cos x × cos y × cos (x + y) is
(a) sin (x + y)
(b) sin² (x + y)
(c) sin³ (x + y)
(d) sin⁴ (x + y)

Answer

Answer: (b) sin² (x + y)
Hint:
cos² x + cos² y – 2cos x × cos y × cos(x + y)
{since cos(x + y) = cos x × cos y – sin x × sin y }
= cos² x + cos² y – 2cos x × cos y × (cos x × cos y – sin x × sin y)
= cos² x + cos² y – 2cos² x × cos² y + 2cos x × cos y × sin x × sin y
= cos² x + cos² y – cos² x × cos² y – cos² x × cos² y + 2cos x × cos y × sin x × sin y
= (cos² x – cos² x × cos² y) + (cos² y – cos² x × cos² y) + 2cos x × cos y × sin x × sin y
= cos² x(1- cos² y) + cos² y(1 – cos² x) + 2cos x × cos y × sin x × sin y
= sin² y × cos² x + sin² x × cos² y + 2cos x × cos y × sin x × sin y (since sin² x + cos² x = 1 )
= sin² x × cos² y + sin² y × cos² x + 2cos x × cos y × sin x × sin y
= (sin x × cos y)² + (sin y × cos x)² + 2cos x × cos y × sin x × sin y
= (sin x × cos y + sin y × cos x)²
= {sin (x + y)}²
= sin² (x + y)

Trigonometry MCQ Class 11 Question 2.
If a×cos x + b × cos x = c, then the value of (a × sin x – b²cos x)² is
(a) a² + b² + c²
(b) a² – b² – c²
(c) a² – b² + c²
(d) a² + b² – c²

Answer

Answer: (d) a² + b² – c²
Hint:
We have
(a×cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b²
⇒ c² + (a × sin x – b × cos x)² = a² + b²
⇒ (a × sin x – b × cos x)² = a² + b² – c²

Trigonometry Class 11 MCQ Question 3.
If cos a + 2cos b + cos c = 2 then a, b, c are in
(a) 2b = a + c
(b) b² = a × c
(c) a = b = c
(d) None of these

Answer

Answer: (a) 2b = a + c
Hint:
Given, cos A + 2 cos B + cos C = 2
⇒ cos A + cos C = 2(1 – cos B)
⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2)
⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2)
⇒ cos((A-C)/2) = 2sin (B/2)
⇒ cos((A-C)/2) = 2cos((A+C)/2)
⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)
⇒ 2sin(A/2)sin(C/2) = sin(B/2)
⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac
⇒ 2(s – b) = b
⇒ a + b + c – 2b = b
⇒ a + c – b = b
⇒ a + c = 2b

Trigonometric Functions Class 11 MCQ Question 4.
The value of cos 5π is
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (c) -1
Hint:
Given, cos 5π = cos (π + 4π) = cos π = -1

Class 11 Trigonometry MCQ Questions Question 5.
In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals
(a) none of these
(b) c/a
(c) 1
(d) a/c

Answer

Answer: (c) 1
Hint:
Given cosec A (sin B cos C + cos B sin C)
= cosec A × sin(B+C)
= cosec A × sin(180 – A)
= cosec A × sin A
= cosec A × 1/cosec A
= 1

Class 11 Maths Chapter 3 MCQ With Answers Question 6.
If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is
(a) 4 : (√5 – 1)
(b) 5 : 4
(c) (√5 – 1) : 4
(d) none of these

Answer

Answer: (a) 4 : (√5 – 1)
Hint:
Given, the angles of a triangle be in the ratio 1 : 4 : 5
⇒ x + 4x + 5x = 180
⇒ 10x = 180
⇒ x = 180/10
⇒ x = 18
So, the angle are: 18, 72, 90
Since a : b : c = sin A : sin B : sin C
⇒ a : b : c = sin 18 : sin 72 : sin 90
⇒ a : b : c = (√5 – 1)/4 : {√(10 + 2√5)}/4 : 1
⇒ a : b : c = (√5 – 1) : {√(10 + 2√5)} : 4
Now, c /a = 4/(√5 – 1)
⇒ c : a = 4 : (√5 – 1)

MCQ On Trigonometry For Class 11 Pdf Download Question 7.
The value of cos 180° is
(a) 0
(b) 1
(c) -1
(d) infinite

Answer

Answer: (c) -1
Hint:
180 is a standard degree generally we all know their values but if we want to go theoretically then
cos(90 + x) = – sin(x)
So, cos 180 = cos(90 + 90)
= -sin 90
= -1 {sin 90 = 1}
So, cos 180 = -1

MCQ Of Trigonometry Class 11 Question 8.
The perimeter of a triangle ABC is 6 times the arithmetic mean of the sines of its angles. If the side b is 2, then the angle B is
(a) 30°
(b) 90°
(c) 60°
(d) 120°

Answer

Answer: (b) 90°
Hint:
Let the lengths of the sides if ∆ABC be a, b and c
Perimeter of the triangle = 2s = a + b + c = 6(sinA + sinB + sinC)/3
⇒ (sinA + sinB + sinC) = ( a + b + c)/2
⇒ (sinA + sinB + sinC)/( a + b + c) = 1/2
From sin formula,Using
sinA/a = sinB/b = sinC/c = (sinA + sinB + sinC)/(a + b + c) = 1/2
Now, sinB/b = 1/2
Given b = 2
So, sinB/2 = 1/2
⇒ sinB = 1
⇒ B = π/2

Trigonometry Objective Questions For Class 11 Question 9:
If 3 × tan(x – 15) = tan(x + 15), then the value of x is
(a) 30
(b) 45
(c) 60
(d) 90

Answer

Answer: (b) 45
Hint:
Given, 3×tan(x – 15) = tan(x + 15)
⇒ tan(x + 15)/tan(x – 15) = 3/1
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = (3 + 1)/(3 – 1)
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 4/2
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 2
⇒ sin(x + 15 + x – 15)/sin(x + 15 – x + 15) = 2
⇒ sin 2x/sin 30 = 2
⇒ sin 2x/(1/2) = 2
⇒ 2 × sin 2x = 2
⇒ sin 2x = 1
⇒ sin 2x = sin 90
⇒ 2x = 90
⇒ x = 45

MCQ Questions On Trigonometry Class 11 Question 10.
If the sides of a triangle are 13, 7, 8 the greatest angle of the triangle is
(a) π/3
(b) π/2
(c) 2π/3
(d) 3π/2

Answer

Answer: (c) 2π/3
Hint:
Given, the sides of a triangle are 13, 7, 8
Since greatest side has greatest angle,
Now Cos A = (b² + c² – a²)/2bc
⇒ Cos A = (7² + 8² – 13²)/(2×7×8)
⇒ Cos A = (49 + 64 – 169)/(2×7×8)
⇒ Cos A = (113 – 169)/(2×7×8)
⇒ Cos A = -56/(2×56)
⇒ Cos A = -1/2
⇒ Cos A = Cos 2π/3
⇒ A = 2π/3
So, the greatest angle is
= 2π/3

MCQ On Trigonometry For Class 11 Question 11.
The value of tan 20 × tan 40 × tan 80 is
(a) tan 30
(b) tan 60
(c) 2 tan 30
(d) 2 tan 60

Answer

Answer: (b) tan 60
Hint:
Given, tan 20 × tan 40 × tan 80
= tan 40 × tan 80 × tan 20
= [{sin 40 × sin 80}/{cos 40 × cos 80}] × (sin 20/cos 20)
= [{2 * sin 40 × sin 80}/{2 × cos 40 × cos 80}] × (sin 20/cos 20)
= [{cos 40 – cos 120}/{cos 120 + cos 40}] × (sin 20/cos 20)
= [{cos 40 – cos (90 + 30)}/{cos (90 + 30) + cos 40}] × (sin 20/cos 20)
= [{cos 40 + sin30}/{-sin30 + cos 40}] × (sin 20/cos 20)
= [{(2 × cos 40 + 1)/2}/{(-1 + cos 40)/2}] × (sin 20/cos 20)
= [{2 × cos 40 + 1}/{-1 + cos 40}] × (sin 20/cos 20)
= [{2 × cos 40 × sin 20 + sin 20}/{-cos 20 + cos 40 × cos 20}]
= (sin 60 – sin 20 + sin 20)/(-cos 20 + cos 60 + cos 20)
= sin 60/cos 60
= tan 60
So, tan 20 × tan 40 × tan 80 = tan 60

MCQ Trigonometry Class 11 Question 12.
If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is
(a) 4 : (√5 – 1)
(b) 5 : 4
(c) (√5 – 1) : 4
(d) none of these

Answer

Class 11 Maths Trigonometry MCQ Questions Question 13.
The general solution of √3 cos x – sin x = 1 is
(a) x = n × π + (-1)n × (π/6)
(b) x = π/3 – n × π + (-1)n × (π/6)
(c) x = π/3 + n × π + (-1)n × (π/6)
(d) x = π/3 – n × π + (π/6)

Answer

Answer: (c) x = π/3 + n × π + (-1)n × (π/6)
Hint:
√3 cos x-sin x=1
⇒ (√3/2)cos x – (1/2)sin x = 1/2
⇒ sin 60 × cos x – cos 60 × sin x = 1/2
⇒ sin (x – 60) = 1/2
⇒ sin (x – π/3) = sin 30
⇒ sin (x – π/3) = sinπ/6
⇒ x – π/3 = n × π + (-1)n × (π/6) {where n ∈ Z}
⇒ x = π/3 + n × π + (-1)n × (π/6)

Class 11 Maths Trigonometry MCQs Question 14.
If tan² θ = 1 – e², then the value of sec θ + tan³ θ × cosec θ is
(a) 2 – e²
(b) (2 – e²)^1/2
(c) (2 – e²)²
(d) (2 – e²)^3/2

Answer

Answer: (d) (2 – e²)^3/2
Hint:
Given, tan² θ = 1 – e²
⇒ tan θ = √(1 – e²)
MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers 1
From the figure and Pythagorus theorem,
AC² = AB² + BC²
⇒ AC² = {√(1 – e²)}² + 12
⇒ AC² = 1 – e² + 1
⇒ AC² = 2 – e²
⇒ AC = √(2 – e²)
Now, sec θ = √(2 – e²)
cosec θ = √(2 – e²)/√(1 – e²)
and tan θ = √(1 – e²)
Given, sec θ + tan³ θ × cosec θ
= √(2 – e²) + {(1 – e²)^3/2 × √(2 – e²)/√(1 – e²)}
= √(2 – e²) + {(1 – e²) × (1 – e²) × √(2 – e²)/√(1 – e²)}
= √(2 – e²) + (1 – e²) × √(2 – e²)
= √(2 – e²) × (1 + 1 – e²)
= √(2 – e²) × (2 – e²)
= (2 – e²)^3/2
So, sec θ + tan³ θ × cosec θ = (2 – e²)^3/2

Trigonometry Class 11 MCQ With Answers Question 15.
The value of cos 20 + 2sin² 55 – √2 sin65 is
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (b) 1
Hint:
Given, cos 20 + 2sin² 55 – √2 sin65
= cos 20 + 1 – cos 110 – √2 sin65 {since cos 2x = 1 – 2sin² x}
= 1 + cos 20 – cos 110 – √2 sin65
= 1 – 2 × sin {(20 + 110)/2 × sin{(20 – 110)/2} – √2 sin65 {Apply cos C – cos D formula}
= 1 – 2 × sin 65 × sin (-45) – √2 sin65
= 1 + 2 × sin 65 × sin 45 – √2 sin65
= 1 + (2 × sin 65)/√2 – √2 sin65
= 1 + √2 ( sin 65 – √2 sin 65
= 1
So, cos 20 + 2sin² 55 – √2 sin65 = 1

Question 16.
If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ ( = PR), then the angle P is
(a) 2π/3
(b) π/3
(c) π/2
(d) π/6

Answer

Answer: (a) 2π/3
Hint:
Let S be the center of the circumcircle of triangle PQR.
So, SP = SQ = SR = PQ = PR, where SP, SQ & SR are radii.
Thus SPQ & SPR are equilateral triangles.
⇒ ∠QSP = 60°;
Similarly ∠RQP = 60°
⇒ Angle at the center QSP = 120°
So, SRPQ is a rhombus, since all the four sides are equal.
Hence, its opposite angles are equal; so ∠P = ∠QSP = 120°

Question 17.
If cos a + 2cos b + cos c = 2 then a, b, c are in
(a) 2b = a + c
(b) b² = a × c
(c) a = b = c
(d) None of these

Answer

Answer: (a) 2b = a + c
Hint:
Given, cos A + 2 cos B + cos C = 2
⇒ cos A + cos C = 2(1 – cos B)
⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin² (B/2)
⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2)
⇒ cos((A-C)/2) = 2sin (B/2)
⇒ cos((A-C)/2) = 2cos((A+C)/2)
⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)
⇒ 2sin(A/2)sin(C/2) = sin(B/2)
⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac
⇒ 2(s – b) = b
⇒ a + b + c – 2b = b
⇒ a + c – b = b
⇒ a + c = 2b

Question 18.
The value of 4 × sin x × sin(x + π/3) × sin(x + 2π/3) is
(a) sin x
(b) sin 2x
(c) sin 3x
(d) sin 4x

Answer

Answer: (c) sin 3x
Hint:
Given, 4 × sin x × sin(x + π/3) × sin(x + 2π/3)
= 4 × sin x × {sin x × cos π/3 + cos x × sin π/3} × {sin x × cos 2π/3 + cos x × sin 2π/3}
= 4 × sin x × {(sin x)/2 + (√3 × cos x)/2} × {-(sin x)/2 + (√3 × cos x)/2}
= 4 × sin x × {-(sin 2x)/4 + (3 × cos 2x)/4}
= sin x × {-sin 2x + 3 × cos 2x}
= sin x × {-sin 2x + 3 × (1 – sin 2x)}
= sin x × {-sin 2x + 3 – 3 × sin 2x}
= sin x × {3 – 4 × sin 2x}
= 3 × sin x – 4 sin 3x
= sin 3x
So, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = sin 3x

Question 19.
If tan A – tan B = x and cot B – cot A = y, then the value of cot (A – B) is
(a) x + y
(b) 1/x + y
(c) x + 1/y
(d) 1/x + 1/y

Answer

Answer: (d) 1/x + 1/y
Hint:
Given,
tan A – tan B = x ……………. 1
and cot B – cot A = y ……………. 2
From equation,
1/cot A – 1/cot B = x
⇒ (cot B – cot A)/(cot A × cot B) = x
⇒ y/(cot A × cot B) = x {from equation 2}
⇒ y = x × (cot A × cot B)
⇒ cot A × cot B = y/x
Now, cot (A – B) = (cot A × cot B + 1)/(cot B – cot A)
⇒ cot (A – B) = (y/x + 1)/y
⇒ cot (A – B) = (y/x) × (1/y) + 1/y
⇒ cot (A – B) = 1/x + 1/y

Question 20.
The value of (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x) is
(a) tan 6x
(b) 2 tan 6x
(c) 3 tan 6x
(d) 4 tan 6x

Answer

Answer: (b) 2 tan 6x
Hint:
Given, (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x)
⇒ [{2 × sin(7x+5x)/2 × cos(7x-5x)/2}/{2 × cos(7x+5x)/2 × cos(7x-5x)/2}] + [{2 × sin(9x+3x)/2 × cos(9x-3x)/2}/{2 × cos(9x+3x)/2 × cos(9x-3x)/2}]
⇒ [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] + [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}]
⇒ (sin 6x/cos 6x) + (sin 6x/cos 6x)
⇒ tan 6x + tan 6x
⇒ 2 tan 6x

We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Trigonometric Functions MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Maths MCQ:

MCQ Questions for Class 11 Maths Chapter 10 Straight Lines with Answers

Straight Lines Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 10 Straight Lines with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Straight Lines Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 10 MCQ With Answers

Maths Class 11 Chapter 10 MCQs On Straight Lines

Straight Lines Class 11 MCQ Question 1.
The locus of a point, whose abscissa and ordinate are always equal is
(a) x + y + 1 = 0
(b) x – y = 0
(c) x + y = 1
(d) none of these.

Answer

Answer: (b) x – y = 0
Hint:
Let the coordinate of the variable point P is (x, y)
Now, the abscissa of this point = x
and its ordinate = y
Given, abscissa = ordinate
⇒ x = y
⇒ x – y = 0
So, the locus of the point is x – y = 0

MCQ On Straight Lines Class 11 Question 2.
The equation of straight line passing through the point (1, 2) and parallel to the line y = 3x + 1 is
(a) y + 2 = x + 1
(b) y + 2 = 3 × (x + 1)
(c) y – 2 = 3 × (x – 1)
(d) y – 2 = x – 1

Answer

Answer: (c) y – 2 = 3 × (x – 1)
Hint:
Given straight line is: y = 3x + 1
Slope = 3
Now, required line is parallel to this line.
So, slope = 3
Hence, the line is
y – 2 = 3 × (x – 1)

Straight Lines Class 11 MCQ Questions Question 3.
What can be said regarding if a line if its slope is negative
(a) θ is an acute angle
(b) θ is an obtuse angle
(c) Either the line is x-axis or it is parallel to the x-axis.
(d) None of these

Answer

Answer: (b) θ is an obtuse angle
Hint:
Let θ be the angle of inclination of the given line with the positive direction of x-axis in the anticlockwise sense.
Then its slope is given by m = tan θ
Given, slope is positive
⇒ tan θ < 0
⇒ θ lies between 0 and 180 degree
⇒ θ is an obtuse angle

Straight Lines MCQ Class 11 Question 4:
The equation of the line which cuts off equal and positive intercepts from the axes and passes through the point (α, β) is
(a) x + y = α + β
(b) x + y = α
(c) x + y = β
(d) None of these

Answer

Answer: (a) x + y = α + β
Hint:
Let the equation of the line be x/a + y/b = 1 which cuts off intercepts a and b with
the coordinate axes.
It is given that a = b, therefore the equation of the line is
x/a + y/a = 1
⇒ x + y = a …..1
But it is passes through (α, β)
So, α + β = a
Put this value in equation 1, we get
x + y = α + β

Straight Line Class 11 MCQ Question 5.
Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are coincedent if
(a) a₁/a₂ = b₁/b₂ ≠ c₁/c₂
(b) a₁/a₂ ≠ b₁/b₂ = c₁/c₂
(c) a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
(d) a₁/a₂ = b₁/b₂ = c₁/c₂

Answer

Answer: (d) a₁/a₂ = b₁/b₂ = c₁/c₂
Hint:
Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are coincedent if
a₁/a₂ = b₁/b₂ = c₁/c₂

MCQ Of Straight Line Class 11 Question 6:
The equation of the line passing through the point (2, 3) with slope 2 is
(a) 2x + y – 1 = 0
(b) 2x – y + 1 = 0
(c) 2x – y – 1 = 0
(d) 2x + y + 1 = 0

Answer

Answer: (c) 2x – y – 1 = 0
Hint:
Given, the point (2, 3) and slope of the line is 2
By, slope-intercept formula,
y – 3 = 2(x – 2)
⇒ y – 3 = 2x – 4
⇒ 2x – 4 – y + 3 = 0
⇒ 2x – y – 1 = 0

Class 11 Maths Chapter 10 MCQ Question 7.
The slope of the line ax + by + c = 0 is
(a) a/b
(b) -a/b
(c) -c/b
(d) c/b

Answer

Answer: (b) -a/b
Hint:
Give, equation of line is ax + by + c = 0
⇒ by = -ax – c
⇒ y = (-a/b)x – c/b
It is in the form of y = mx + c
Now, slope m = -a/b

Class 11 Straight Lines MCQ Question 8.
Equation of the line passing through (0, 0) and slope m is
(a) y = mx + c
(b) x = my + c
(c) y = mx
(d) x = my

Answer

Answer: (c) y = mx
Hint:
Equation of the line passing through (x₁, y₁) and slope m is
(y – y₁) = m(x – x₁)
Now, required line is
(y – 0 ) = m(x – 0)
⇒ y = mx

Class 11 Maths Straight Lines MCQ Question 9.
The angle between the lines x – 2y = y and y – 2x = 5 is
(a) tan^-1 (1/4)
(b) tan^-1 (3/5)
(c) tan^-1 (5/4)
(d) tan^-1 (2/3)

Answer

Answer: (c) tan^-1 (5/4)
Hint:
Given, lines are:
x – 2y = 5 ………. 1
and y – 2x = 5 ………. 2
From equation 1,
x – 5 = 2y
⇒ y = x/2 – 5/2
Here, m₁ = 1/2
From equation 2,
y = 2x + 5
Here. m₂ = 2
Now, tan θ = |(m₁ + m₂)/{1 + m₁ × m₂}|
= |(1/2 + 2)/{1 + (1/2) × 2}|
= |(5/2)/(1 + 1)|
= |(5/2)/2|
= 5/4
⇒ θ = tan^-1 (5/4)

Straight Line MCQ Questions Question 10.
Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel if
(a) a₁/a₂ = b₁/b₂ ≠ c₁/c₂
(b) a₁/a₂ ≠ b₁/b₂ = c₁/c₂
(c) a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
(d) a₁/a₂ = b₁/b₂ = c₁/c₂

Answer

Answer: (a) a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Hint:
Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel if
a₁/a₂ = b₁/b₂ ≠ c₁/c₂

MCQ Questions On Straight Lines For Class 11 Question 11.
The locus of a point, whose abscissa and ordinate are always equal is
(a) x + y + 1 = 0
(b) x – y = 0
(c) x + y = 1
(d) none of these.

Answer

Class 11 Maths Ch 10 MCQ Question 12.
In a ΔABC, if A is the point (1, 2) and equations of the median through B and C are respectively x + y = 5 and x = 4, then B is
(a) (1, 4)
(b) (7, – 2)
(c) none of these
(d) (4, 1)

Answer

Answer: (b) (7, – 2)
Hint:
The equation of median through B is x + y = 5
The point B lies on it.
Let the coordinates of B are (x₁, 5 – x₁)
Now CF is a median through C,
So co-ordiantes of F i.e. mid-point of AB are
((x₁+1)/2, (5 – x₁+ 2)/2)
Now since this lies on x = 4
⇒ (x₁ + 1)/2 = 4
⇒ x₁ + 1 = 8
⇒ x₁ = 7
Hence, the co-oridnates of B are (7, -2)

Ch 10 Maths Class 11 MCQ Question 13.
The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150 degrees with the positive direction of the y-axis. Then the equation of line is
(a) x + y = 14
(b) √3y + x = 14
(c) √3x + y = 14
(d) None of these

Answer

Answer: (c) √3x + y = 14
Hint:
Given, The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150 degrees with the positive direction of the y-axis.
Now, equation of line is
x × cos 30 + y × sin 30 = 7
⇒ √3x/2 + y/2 = 7
⇒ √3x + y = 7×2
⇒ √3x + y = 14

MCQs On Straight Lines Class 11 Question 14.
If two vertices of a triangle are (3, -2) and (-2, 3) and its orthocenter is (-6, 1) then its third vertex is
(a) (5, 3)
(b) (-5, 3)
(c) (5, -3)
(d) (-5, -3)

Answer

Answer: (d) (-5, -3)
Hint:
Let the third vertex of the triangle is C(x, y)
Given, two vertices of a triangle are A(3,-2) and B(-2,3)
Now given orthocentre of the circle = H(-6, 1)
So, AH ⊥ BC and BH ⊥ AC
Since the product of the slope of perpendicular lines equal to -1
Now, AH ⊥ BC
⇒ {(-2 – 1)/(3 + 6)} × {(y + 2)/(x – 3)} = -1
⇒ (-3/9) × {(y + 2)/(x – 3)} = -1
⇒ (-1/3)×{(y – 3)/(x + 2)} = -1
⇒ (y – 3)/{3×(x + 2)} = 1
⇒ (y – 3) = 3×(x + 2)
⇒ y – 3 = 3x + 6
⇒ 3x + 6 – y = -3
⇒ 3x – y = -3 – 6
⇒ 3x – 2y = -9 ………… 1
Again, BH ⊥ AC
⇒ {(3 – 1)/(-2 + 6)} × {(y – 3)/(x + 2)} = -1
⇒ (2/4) × {(y – 3)/(x + 2)} = -1
⇒ (1/2)×{(y – 3)/(x + 2)} = -1
⇒ (y – 3)/{2×(x + 2)} = 1
⇒ (y – 3) = 2×(x + 2)
⇒ y – 3 = 2x + 4
⇒ 2x + 4 – y = -3
⇒ 2x – y = -3 – 4
⇒ 2x – y = -7 ………… 2
Multiply equation 2 by 2, we get
4x – 2y = -14 ……… 3
Subtract equation 1 and we get
-x = 5
⇒ x = -5
From equation 2, we get
2×(-5) – y = -7
⇒ -10 – y = -7
⇒ y = -10 + 7
⇒ y = -3
So, the third vertex of the triangle is (-5, -3)

MCQ Questions On Straight Lines Class 11 Question 15.
The sum of squares of the distances of a moving point from two fixed points (a, 0) and (-a, 0) is equal to 2c² then the equation of its locus is
(a) x² – y² = c² – a²
(b) x² – y² = c² + a²
(c) x² + y² = c² – a²
(d) x² + y² = c² + a²

Answer

Answer: (c) x² + y² = c² – a²
Hint:
Let P(h, k) be any position of the moving point and let A(a, 0) and B(-a, 0) be the given points. Then
PA² + PB² = 2c²
⇒ (h – a)² + (k – 0)² + (h + a)² + (k – 0)² = 2c²
⇒ h² – 2ah + a² + k² + h² + 2ah + a² + k² = 2c²
⇒ 2h² + 2k² + 2a² = 2c²
⇒ h² + k² + a² = c²
⇒ h² + k² = c² – a²
Hence, the locus of (h, k) is x² + y² = c² – a²

Question 16.
The equation of the line through the points (1, 5) and (2, 3) is
(a) 2x – y – 7 = 0
(b) 2x + y + 7 = 0
(c) 2x + y – 7 = 0
(d) x + 2y – 7 = 0

Answer

Answer: (c) 2x + y – 7 = 0
Hint:
Given, points are: (1, 5) and (2, 3)
Now, equation of line is
y – y₁ = {(y₂ – y₁)/(x₂ – x₁)} × (x – x₁)
⇒ y – 5 = {(3 – 5)/(2 – 1)} × (x – 1)
⇒ y – 5 = (-2) × (x – 1)
⇒ y – 5 = -2x + 2
⇒ 2x + y – 5 – 2 = 0
⇒ 2x + y – 7 = 0

Question 17.
What can be said regarding if a line if its slope is zero
(a) θ is an acute angle
(b) θ is an obtuse angle
(c) Either the line is x-axis or it is parallel to the x-axis.
(d) None of these

Answer

Answer: (c) Either the line is x-axis or it is parallel to the x-axis.
Hint:
Let θ be the angle of inclination of the given line with the positive direction of x- axis in the anticlockwise sense.
Then its slope is given by m = tan θ
Given, slope is zero
⇒ tan θ = 0
⇒ θ = 0°
⇒ Either the line is x-axis or it is parallel to the x-axis.

Question 18.
Two lines are perpendicular if the product of their slopes is
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (c) -1
Hint:
Let m₁ is the slope of first line and m₂ is the slope of second line.
Now, two lines are perpendicular if m₁ × m₂ = -1
i.e. the product of their slopes is equals to -1

Question 19.
y-intercept of the line 4x – 3y + 15 = 0 is
(a) -15/4
(b) 15/4
(c) -5
(d) 5

Answer

Answer: (d) 5
Hint:
Given, equation of line is 4x – 3y + 15 = 0
⇒ 4x – 3y = -15
⇒ 4x/(-15) + (-3)y/(-15) = 1
⇒ x/(-15/4) + 3y/15 = 1
⇒ x/(-15/4) + y/(15/3) = 1
⇒ x/(-15/4) + y/5 = 1
Now, compare with x/a + y/b = 1, we get
y-intercept b = 5

Question 20.
The equation of the locus of a point equidistant from the point A(1, 3) and B(-2, 1) is
(a) 6x – 4y = 5
(b) 6x + 4y = 5
(c) 6x + 4y = 7
(d) 6x – 4y = 7

Answer

Answer: (b) 6x + 4y = 5
Hint:
Let P(h, k) be any point on the locus. Then
Given, PA = PB
⇒ PA² = PB²
⇒ (h – 1)² + (k – 3)² = (h + 2)² + (k – 1)²
⇒ h² – 2h + 1 + k² – 6k + 9 = h² + 4h + 4 + k² – 2k + 1
⇒ -2h – 6k + 10 = 4h – 2k + 5
⇒ 6h + 4k = 5
Hence, the locus of (h, k) is 6x + 4y = 5

We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 10 Straight Lines with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Straight Lines MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

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Category: MCQ Questions

Class 11 Accountancy Chapter 8 Bills of Exchange MCQ With Answers

Accountancy Class 11 Chapter 8 MCQs On Bills of Exchange

Class 11 Maths Chapter 5 MCQ With Answers

Maths Class 11 Chapter 5 MCQs On Complex Numbers and Quadratic Equations

Class 11 Business Studies Chapter 2 Forms of Business Organisation MCQ With Answers

Business Studies Class 11 Chapter 2 MCQs On Forms of Business Organisation

Class 12 Accountancy Chapter 7 Issue and Redemption of Debentures MCQ With Answers

Accountancy Class 12 Chapter 7 MCQs On Issue and Redemption of Debentures

Class 11 Maths Chapter 7 MCQ With Answers

Maths Class 11 Chapter 7 MCQs On Permutations and Combinations

Class 11 Maths Chapter 16 MCQ With Answers

Maths Class 11 Chapter 16 MCQs On Probability

Class 11 Maths Chapter 15 MCQ With Answers

Maths Class 11 Chapter 15 MCQs On Statistics

Class 11 Maths Chapter 13 MCQ With Answers

Maths Class 11 Chapter 13 MCQs On Limits and Derivatives

Class 11 Maths Chapter 1 MCQ With Answers

Maths Class 11 Chapter 1 MCQs On Sets

Class 11 Maths Chapter 3 MCQ With Answers

Maths Class 11 Chapter 3 MCQs On Trigonometric Functions

Class 11 Maths Chapter 10 MCQ With Answers

Maths Class 11 Chapter 10 MCQs On Straight Lines