MCQ Questions

Students who are searching for NCERT MCQ Questions for Class 12 Business Studies Chapter 9 Financial Management with Answers Pdf free download are compiled here to get good practice on all fundamentals. Know your preparation level on MCQ Questions for Class 12 Business Studies with Answers. You can also verify your answers from our provided Financial Management Class 12 MCQs Questions with Answers. So, ace up your preparation with MCQ on Financial Management Class 12 Objective Questions.

Financial Management Class 12 MCQs Questions with Answers

Question 1.
Higher debt-equity ratio results in:
(a) a Higher degree of financial risk
(b) a Higher degree of operating risk
(c) Higher EPS
(d) Lower financial risk

Answer

Answer: (a) a Higher degree of financial risk

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Question 2.
Cost of advertising and printing prospectus is called__________
(a) Floatation cost
(b) Debt cost
(c) Equity cost
(d) Dividend cost

Answer

Answer: (a) Floatation cost

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Question 3.
The primary goal of the financial management is ____________
(a) To maximize the return
(b) To minimize the risk
(c) To maximize the wealth of owners
(d) To maximize profit

Answer

Answer: (c) To maximize the wealth of owners

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Question 4.
Which of the following affects the Dividend Decision of a company?
(a) Taxation Policy
(b) Cash Flow Position
(c) Earnings
(d) All of the above

Answer

Answer: (d) All of the above

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Question 5.
Which of the following affects capital budgeting decision?
(a) Investment Criteria and interest rate
(b) Rate of Return
(c) Cash Flow of the Project
(d) All of the above

Answer

Answer: (d) All of the above

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Question 6.
Higher working capital usually results in:
(a) Higher equity, lower risk, and lower profits
(b) Lower current ratio, higher risk, and profits
(c) Lower equity, lower risk, and higher profits
(d) Higher current ratio, higher risk, and higher profits

Answer

Answer: (d) Higher current ratio, higher risk, and higher profits

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Question 7.
Which of the following is not concerned with the Long term investment decision
(a) Management of fixed capital
(b) Inventory management
(c) Research and Development Programme
(d) Opening a new branch

Answer

Answer: (b) Inventory management

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Question 8.
Favourable financial leverage is a situation where _____
(a) ROI is higher than the rate of interest on debt
(b) ROI is Equal to the Rate of interest on debt
(c) ROI is lower than the rate of interest on debt
(d) None of the above

Answer

Answer: (a) ROI is higher than the rate of interest on debt

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Question 9.
Other things remaining the same, an increase in the tax rate on corporate profits will:
(a) Make the debt relatively cheaper
(b) Make the debt relatively the dearer
(c) Have no impact on the cost of debt
(d) None of the above

Answer

Answer: (a) Make the debt relatively cheaper

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Question 10.
Higher dividend per share is associated with:
(a) High earnings, high cash flows, stable earnings, and high growth opportunities
(b) High earnings, high cash flows, stable earnings, and lower growth opportunities
(c) High earnings, low cash flows, stable earnings, and lower growth opportunities
(d) High earning, high cash flows, unstable earnings, and higher growth opportunities

Answer

Answer: (b) High earnings, high cash flows, stable earnings, and lower growth opportunities

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Question 11.
The main objective of financial planning is to ensure that_________
(a) Enough funds are available at the right time
(b) Dividend is paid to shareholders at the right time
(c) Purchase of raw material
(d) Purchase of fixed assets

Answer

Answer: (a) Enough funds are available at the right time

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Question 12.
Financial planning arrives at:
(a) Doing only what is possible with the funds that the firms have at its disposal
(b) Entering that the firm always have significantly more funds than required so that there is no paucity of funds
(c) Minimising the external borrowing by resorting to equity issues
(d) Ensuring that the firm faces neither a shortage nor a glut of unusable funds

Answer

Answer: (d) Ensuring that the firm faces neither a shortage nor a glut of unusable funds

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Question 13.
Which of the following is not a financial Decision?
(a) Financing Decision
(b) Investment Decision
(c) Staffing Decision
(d) Dividend Decision

Answer

Answer: (c) Staffing Decision

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Question 14.
The cheapest source of finance is:
(a) Preference share
(b) Retained earning
(c) Equity share capital
(d) Debenture

Answer

Answer: (b) Retained earning

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Question 15.
Financial leverage is called favourable if:
(a) Return on Investment is lower than the cost of debt
(b) If the degree of existing financial leverage is low
(c) Debt is easily available
(d) ROI is higher than the cost of debt

Answer

Answer: (d) ROI is higher than the cost of debt

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Question 16.
Short-term Investment Decision is also known as ____
(a) Working capital
(b) Dividend Decision
(c) Capital Budgeting
(d) None of the above

Answer

Answer: (a) Working capital

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Question 17.
Which of the following affects the Dividend Decision of a company?
(a) Earnings
(b) Cash Flow Position
(c) Taxation Policy
(d) All of the above

Answer

Answer: (d) All of the above

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Question 18.
Short term investment decisions affect the ___________
(a) Purchase of fixed assets
(b) Long term profitability
(c) Day to Day working of the business
(d) Large amount of funds for future

Answer

Answer: (c) Day to Day working of the business

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Question 19.
Financial Management is mainly concerned with ______________
(a) All aspects of acquiring and utilizing financial resources for firms activities
(b) Arrangement of funds
(c) Efficient Management of every business
(d) Profit maximization

Answer

Answer: (a) All aspects of acquiring and utilizing financial resources for firms activities

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Question 20.
A fixed asset should be financed through:
(a) A long-term liability
(b) A short-term liability
(c) A mix of long and short-term liabilities
(d) None of the above

Answer

Answer: (a) A long-term liability

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Question 21.
Long term investment decision is also known as _____________
(a) Capital Budgeting
(b) Working Capital
(c) Dividend Decision
(d) None of the above

Answer

Answer: (a) Capital Budgeting

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Question 22.
Current assets of a business firm should be financed through:
(a) Current liability only
(b) Long-term liability only
(c) Both of the above
(d) None of the above

Answer

Answer: (c) Both of the above

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Question 23.
Companies with a higher growth pattern are likely to:
(a) Dividends are not affected by growth considerations
(b) Pay higher dividends
(c) Pay lower dividends
(d) None of the above

Answer

Answer: (c) Pay lower dividends

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Use the above-provided NCERT MCQ Questions for Class 12 Business Studies Chapter 9 Financial Management with Answers Pdf free download and get a good grip on the fundamentals. Need any support from our end during the preparation of CBSE Class 12 Business Studies Financial Management MCQs Multiple Choice Questions with Answers then leave your comments below. We’ll revert back to you soon.

Students who are searching for NCERT MCQ Questions for Class 12 Business Studies Chapter 10 Financial Markets with Answers Pdf free download are compiled here to get good practice on all fundamentals. Know your preparation level on MCQ Questions for Class 12 Business Studies with Answers. You can also verify your answers from our provided Financial Markets Class 12 MCQs Questions with Answers. So, ace up your preparation with MCQ on Financial Markets Class 12 Objective Questions.

Financial Markets Class 12 MCQs Questions with Answers

Question 1.
Treasury bills are also known as _____________
(a) Fixed interest Bonds
(b) Low-Interest Bonds
(c) Flat Rate Bonds
(d) Zero-Coupon Bonds

Answer

Answer: (d) Zero-Coupon Bonds

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Question 2.
Which of the following is a method of floatation?
(a) Offer for sale
(b) Private Placement
(c) Offer through the prospectus
(d) All of the above

Answer

Answer: (d) All of the above

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Question 3.
Primary and secondary markets:
(a) Control each other
(b) Complement each other
(c) Compete with each other
(d) Function independently

Answer

Answer: (b) Complement each other

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Question 4.
The total number of Stock Exchanges in India is:
(a) 22
(b) 21
(c) 23
(d) 20

Answer

Answer: (b) 21

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Question 5.
Money market deals in ________
(a) Long term Securities
(b) Medium-term securities
(c) Short term Securities
(d) None of the above

Answer

Answer: (c) Short term Securities

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Question 6.
A Treasury Bill is basically:
(a) An instrument to borrow short-term funds
(b) An instrument to borrow long-term funds
(c) An instrument of capital market
(d) None of the above

Answer

Answer: (a) An instrument to borrow short-term funds

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Question 7.
What is meant by Demat Account?
(a) Development Market Account
(b) Depository Participant Account
(c) Dematerialisation of Securities
(d) Demand Depository Account

Answer

Answer: (c) Dematerialisation of Securities

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Question 8.
To be listed on OTCEI, the minimum capital requirement for a company is:
(a) Rs. 10 crores
(b) Rs. 3 crores
(c) Rs. 25 crores
(d) Rs. 5 crores

Answer

Answer: (b) Rs. 3 crores

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Question 9.
Clearing and settlement operations of NSE are carried out by:
(a) NSDL
(b) SBI
(c) NSCCL
(d) CDSL

Answer

Answer: (c) NSCCL

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Question 10.
Instruments traded in capital market are__________
(a) Bonds
(b) Equity Shares and Preference Shares
(c) Debentures
(d) All of the above

Answer

Answer: (d) All of the above

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Question 11.
OTCEI was started on the lines of:
(a) NSE
(b) NYSE
(c) NASDAQ
(d) NASAQ

Answer

Answer: (c) NASDAQ

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Question 12.
_________ is not a participant in money market
(a) SEBI
(b) NBFCs
(c) RBI
(d) Mutual Funds

Answer

Answer: (a) SEBI

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Question 13.
When a trade bill is accepted by a commercial bank, it is known as a _____
(a) Certificate of deposit
(b) Commercial Bill
(c) Call money
(d) None of the above

Answer

Answer: (b) Commercial Bill

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Question 14.
What type of instruments are traded in a Money Market?
(a) Treasury bills
(b) Commercial bills
(c) Call money
(d) All of the above

Answer

Answer: (d) All of the above

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Question 15.
Capital market deals in ___________
(a) Medium and long term securities
(b) Very short term securities
(c) Short term securities
(d) None of the above

Answer

Answer: (a) Medium and long term securities

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Question 16.
What is the meaning of the right issue?
(a) Company sells the securities to some selected institutions
(b) Company offers new shares to its existing shareholders
(c) Securities are not issued to existing shareholders at all
(d) None of the above

Answer

Answer: (b) Company offers new shares to its existing shareholders

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Question 17.
_______ Market instruments enjoy a higher degree of liquidity
(a) Both money and capital market
(b) Money market
(c) Capital market
(d) None of the above

Answer

Answer: (b) Money market

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Question 18.
At present only two depositories are registered with SEBI _____
(a) NSDL and HDFC
(b) NSDL and CDSL
(c) NSDL and ABSL
(d) NEFT and TDS

Answer

Answer: (b) NSDL and CDSL

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Question 19.
Which of the following is not a part of the capital market?
(a) Banks
(b) Financial Institutions
(c) Stock Exchanges
(d) RBI

Answer

Answer: (d) RBI

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Question 20.
Securities Exchange Board of India (SEBI) was established in ________
(a) 1992
(b) 1956
(c) 2001
(d) 1984

Answer

Answer: (a) 1992

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Question 21.
A Treasury bill is an instrument of _____________
(a) Long term debt
(b) Short term debt
(c) Interest
(d) Dividend

Answer

Answer: (b) Short term debt

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Question 22.
The settlement cycle in NSE is:
(a) T + 2
(b) T + 5
(c) T + 3
(d) T + 1

Answer

Answer: (a) T + 2

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Question 23.
Only institutional investors can participate in __________
(a) Loan Market
(b) Money Market
(c) Foreign Market
(d) Capital Market

Answer

Answer: (b) Money Market

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Use the above-provided NCERT MCQ Questions for Class 12 Business Studies Chapter 10 Financial Markets with Answers Pdf free download and get a good grip on the fundamentals. Need any support from our end during the preparation of CBSE Class 12 Business Studies Financial Markets MCQs Multiple Choice Questions with Answers then leave your comments below. We’ll revert back to you soon.

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Complex Numbers and Quadratic Equations Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Objective Questions.

Complex Numbers and Quadratic Equations Class 11 MCQs Questions with Answers

Students are advised to solve the Complex Numbers and Quadratic Equations Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Complex Numbers and Quadratic Equations Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Complex Numbers and Quadratic Equations Class 11 with answers provided with detailed solutions by looking below.

Question 1.
Let z₁ and z₂ be two roots of the equation z² + az + b = 0, z being complex. Further assume that the origin, z₁ and z₂ form an equilateral triangle. Then
(a) a² = b
(b) a² = 2b
(c) a² = 3b
(d) a² = 4b

Answer

Answer: (c) a² = 3b
Given, z₁ and z₁ 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

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Question 2.
The value of iⁱ is
(a) 0
(b) e^-π
(c) 2e^-π/2
(d) e^-π/2

Answer

Answer: (d) e^-π/2
Let A = iⁱ
⇒ log A = i log i
⇒ log A = i log(0 + i)
⇒ log A = i [log 1 + i tan^-1 ∞]
⇒ log A = i [0 + i π/2]
⇒ log A = -π/2
⇒ A = e^-π/2

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Question 3.
The value of √(-25) + 3√(-4) + 2√(-9) is
(a) 13 i
(b) -13 i
(c) 17 i
(d) -17 i

Answer

Answer: (c) 17 i
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
= 17 i
So, √(-25) + 3√(-4) + 2√(-9) = 17 i

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Question 4.
If the cube roots of unity are 1, ω and ω², then the value of (1 + ω / ω²)³ is
(a) 1
(b) -1
(c) ω
(d) ω²

Answer

Answer: (b) -1
Given, the cube roots of unity are 1, ω and ω²
So, 1 + ω + ω² = 0
and ω³ = 1
Now, {(1 + ω)/ ω²}³ = {-ω²/ ω²}³ = {-1}³ = -1

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Question 5.
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
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

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Question 6.
The value of [i¹⁹ + (1/i)²⁵]² is
(a) -1
(b) -2
(c) -3
(d) -4

Answer

Answer: (d) -4
Given, [i¹⁹ + (1/i)²⁵]²
= [i¹⁹ + 1/i²⁵]²
= [i¹⁶ × i³ + 1/(i²⁴ × i)]²
= [1 × i³ + 1/(1 × i)]² {since i⁴ = 1}
= [i³ + 1/i]²
= [i² × i + 1/i]²
= [(-1) × i + 1/i]² {since i² = -1}
= [-i + 1/i]²
= [-i + i⁴ /i]²
= [-i + i³]²
= [-i + i² × i]²
= [-i + (-1) × i]²
= [-i – i]²
= [-2i]²
= 4i²
= 4 × (-1)
= -4
So, [i¹⁹ + (1/i)²⁵]² = -4

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Question 7.
If z and w be two complex numbers such that |z| ≤ 1, |w| ≤ 1 and |z + iw| = |z – iw| = 2, then z equals {w is congugate of w}
(a) 1 or i
(b) i or – i
(c) 1 or – 1
(d) i or – 1

Answer

Answer: (c) 1 or – 1
Given |z + iw| = |z – iw| = 2 {w is congugate of w}
⇒ |z – (-iw)| = |z – (iw)| = 2
⇒ |z – (-iw)| = |z – (-iw)|
So, z lies on the perpendicular bisector of the line joining -iw and -iw.
Since, -iw is the mirror in the x-axis, the locus of z is the x-axis.
Let z = x + iy and y = 0
⇒ |z| < 1 and x² + 0² < 0
⇒ -1 ≤ x ≤ 1
So, z may take value 1 or -1

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Question 8.
The value of {-√(-1)}⁴ⁿ⁺³, n ∈ N is
(a) i
(b) -i
(c) 1
(d) -1

Answer

Answer: (a) i
Given, {-√(-1)}⁴ⁿ⁺³
= {-i}⁴ⁿ⁺³ {since √(-1) = i}
= {-i}⁴ⁿ × {-i}³
= {(-i)⁴}ⁿ × (-i³) {since i⁴ = 1}
= 1ⁿ ×(-i × i²)
= -i × (-1) {since i² = -1}
= i

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Question 9.
Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is real
(a) π
(b) nπ
(c) nπ/2
(d) 2nπ

Answer

Answer: (b) nπ
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 real if sin θ = 0
⇒ sin θ = sin nπ
⇒ θ = nπ

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Question 10.
If i = √(-1) then 4 + 5(-1/2 + i√3/2)³³⁴ + 3(-1/2 + i√3/2)³⁶⁵ is equals to
(a) 1 – i√3
(b) -1 + i√3
(c) i√3
(d) -i√3

Answer

Answer: (c) i√3
Given, 4 + 5(-1/2 + i√3/2)³³⁴ + 3(-1/2 + i√3/2)³⁶⁵
= 4 + 5w³³⁴ + 3w³⁶⁵ {since w = -1/2 + i√3/2}
= 4 + 5w + 3w² {since w³ = 1}
= 4 + 5(-1/2 + i√3/2) + 3(-1/2 – i√3/2) {since w² = (-1/2 – i√3/2)}
= i√3

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Question 11.
The real part of the complex number √9 + √(-16) is
(a) 3
(b) -3
(c) 4
(d) -4

Answer

Answer: (a) 3
Given, √9 + √(-16) = √9 + √(16) × √(-1)
= 3 + 4i {since i = √(-1)}
So, the real part of the complex number is 3

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Question 12.
The modulus of 5 + 4i is
(a) 41
(b) -41
(c) √41
(d) -√41

Answer

Answer: (c) √41
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

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Question 13.
The modulus of 1 + i√3 is
(a) 1
(b) 2
(c) 3
(d) None of these

Answer

Answer: (b) 2
Let Z = 1 + i√3
Now modulus of Z is calculated as
|Z| = √{1² + (√3)²}
⇒ |Z| = √(1 + 3)
⇒ |Z| = √4
⇒ |Z| = 2
So, the modulus of 1 + i√3 is 2

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Question 14.
The value of {-√(-1)}⁴ⁿ⁺³, n ∈ N is
(a) i
(b) -i
(c) 1
(d) -1

Answer

Answer: (a) i
Given, {-√(-1)}⁴ⁿ⁺³
= {-i}⁴ⁿ⁺³ {since √(-1) = i}
= {-i}⁴ⁿ × {-i}³
= {(-i)⁴}ⁿ ×(-i³) {since i⁴ = 1}
= 1ⁿ × (-i × i²)
= -i × (-1) {since i² = -1}
= i

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Question 15.
If ω is cube root of unity (ω ≠ 1) , then the least value of n where n is a positive integer such that (1 + ω²)ⁿ = (1 + ω⁴)ⁿ is
(a) 2
(b) 3
(c) 5
(d) 6

Answer

Answer: (b) 3
Given ω is an imaginary cube root of unity.
So 1 + ω + ω² = 0 and ω³ = 1
Now, (1 + ω²)ⁿ = (1 + ω⁴)ⁿ
⇒ (-1)ⁿ ×(ω)ⁿ = (1 + ω × ω³)ⁿ
⇒ (-1)ⁿ × (ω)ⁿ = (1 + ω)ⁿ
⇒ (-1)ⁿ × (ω)ⁿ = (-ω²)ⁿ
⇒ (-1)ⁿ × (ω)ⁿ = (-1)ⁿ × ω²ⁿ
⇒ ωⁿ = ω²ⁿ
Since ω³ = 1, So the least value of n is 3

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Question 16.
The value of i⁹ + i¹⁰ + i¹¹ + i¹² is
(a) i
(b) 2i
(c) 0
(d) 1

Answer

Answer: (c) 0
Given, i⁹ + i¹⁰ + i¹¹ + i¹²
= i9 (1 + i + i2 + i3 )
= i9 (1 + i – 1 – i ) {since i2 = (-1) and i4 = 1}
= i9 × 0
= 0

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Question 17.
If a = cos α + i sin α and b = cos β + i sin β , then the value of 1/2(ab + 1/ ab) is
(a) sin (α + β)
(b) cos (α + β)
(c) sin (α – β)
(d) cos (α – β)

Answer

Answer: (b) cos (α + β)
Given a = cos α + i sin α and b = cos β + i sin β
Now, 1/a = 1/(cos α + i sin α)
⇒ 1/a = {1 × (cos α – i sin α)/{(cos α + i sin α) × (cos α + i sin α)}
⇒ 1/a = (cos α – i sin α)/(cos² α + i sin² α)
⇒ 1/a = (cos α – i sin α)
Again, 1/b = 1/(cos β + i sin β)
⇒ 1/b = {1 × (cos β – i sin β)/{(cos β + i sin β) × (cos β + i sin β)}
⇒ 1/b = (cos β – i sin β)/(cos² β + i sin² β)
⇒ 1/b = (cos β – i sin β)
Now, ab = (cos α + i sin α) × (cos β + i sin β)
⇒ ab = cos α × cos β + i cos α × sin β + i sin α × cos β – sin α × sin β
Again, 1/ab = (cos α – i sin α) × (cos β – i sin β)
⇒ 1/ab = cos α × cos β – i cos α × sin β – i sin α × cos β – sin α × sin β
Now, ab + 1/ab = cos α × cos β + i cos α × sin β + i sin α × cos β – sin α × sin β + cos α × cos β – i cos α × sin β – i sin α × cos β – sin α × sin β
⇒ ab + 1/ab = 2(cos α × cos β – sin α × sin β)
⇒ 1/2(ab + 1/ ab) = 2(cos α × cos β – sin α × sin β)/2
⇒ 1/2(ab + 1/ ab) = cos α × cos β – sin α × sin β
⇒ 1/2(ab + 1/ ab) = cos(α + β)

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Question 18.
The polar form of -1 + i is
(a) √2(cos π/2 + i × sin π/2)
(b) √2(cos π/4 + i × sin π/4)
(c) √2(cos 3π/2 + i × sin 3π/2)
(d) √2(cos 3π/4 + i × sin 3π/4)

Answer

Answer: (d) √2(cos 3π/4 + i × sin 3π/4)
The polar form of a com plex number = r(cos θ + i × sin θ)
Given, complex number = -1 + i
Let x + iy = -1 + i
Now, x = -1, y = 1
Now, r = √{(-1)² + 1²} = √(1 + 1) = √2
and tan θ = y/x
⇒ tan θ = 1/(-1)
⇒ tan θ = -1
⇒ θ = 3π/4
Now, polar form is √2(cos 3π/4 + i × sin 3π/4)

---

Question 19.
For all complex numbers z₁, z₂ satisfying |z₁| = 12 and |z₂ – 3 – 4i| = 5, the minimum value of |z₁ – z₂| is
(a) 0
(b) 2
(c) 7
(d) 17

Answer

Answer: (b) 2
Given For all complex numbers z₁, z₂ satisfying |z₁| = 12 and |z₂ – 3 – 4i| = 5
Now, mod(z₁) = 12 represents a circle centred at 0 and radius 12
mod(z₂ – 3 – 4i) = 5 represents a circle centred at (3, 4) and radius 5
This circle passes through the origin. Distance of diametrically opposite end is 10
So, the minimum value (z₁ – z₂) = 2

---

Question 20.
The value of (1 – i)² is
(a) i
(b) -i
(c) 2i
(d) -2i

Answer

Answer: (d) -2i
Given, (1 – i)² = 1 + i² – 2i
= 1 + (-1) – 2i
= 1 – 1 – 2i
= -2i

---

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Linear Inequalities Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 6 Linear Inequalities Objective Questions.

Linear Inequalities Class 11 MCQs Questions with Answers

Students are advised to solve the Linear Inequalities Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Linear Inequalities Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Linear Inequalities Class 11 with answers provided with detailed solutions by looking below.

Question 1.
If -2 < 2x – 1 < 2 then the value of x lies in the interval
(a) (1/2, 3/2)
(b) (-1/2, 3/2)
(c) (3/2, 1/2)
(d) (3/2, -1/2)

Answer

Answer: (b) (-1/2, 3/2)
Given, -2 < 2x – 1 < 2
⇒ -2 + 1 < 2x < 2 + 1
⇒ -1 < 2x < 3
⇒ -1/2 < x < 3/2
⇒ x ∈ (-1/2, 3/2)

---

Question 2.
If x² < -4 then the value of x is
(a) (-2, 2)
(b) (2, ∞)
(c) (-2, ∞)
(d) No solution

Answer

Answer: (d) No solution
Given, x² < -4
⇒ x² + 4 < 0
Which is not possible.
So, there is no solution.

---

Question 3.
If |x| < -5 then the value of x lies in the interval
(a) (-∞, -5)
(b) (∞, 5)
(c) (-5, ∞)
(d) No Solution

Answer

Answer: (d) No Solution
Given, |x| < -5
Now, LHS ≥ 0 and RHS < 0
Since LHS is non-negative and RHS is negative
So, |x| < -5 does not posses any solution

---

Question 4.
The graph of the inequations x ≤ 0 , y ≤ 0, and 2x + y + 6 ≥ 0 is
(a) exterior of a triangle
(b) a triangular region in the 3rd quadrant
(c) in the 1st quadrant
(d) none of these

Answer

Answer: (b) a triangular region in the 3rd quadrant
Given inequalities x ≥ 0 , y ≥ 0 , 2x + y + 6 ≥ 0
Now take x = 0, y = 0 and 2x + y + 6 = 0
when x = 0, y = -6
when y = 0, x = -3
So, the points are A(0, 0), B(0, -6) and C(-3, 0)

So, the graph of the inequations x ≤ 0 , y ≤ 0 , and 2x + y + 6 ≥ 0 is a triangular region in the 3rd quadrant.

---

Question 5.
The graph of the inequalities x ≥ 0, y ≥ 0, 2x + y + 6 ≤ 0 is
(a) a square
(b) a triangle
(c) { }
(d) none of these

Answer

Answer: (c) { }
Given inequalities x ≥ 0, y ≥ 0, 2x + y + 6 ≤ 0
Now take x = 0, y = 0 and 2x + y + 6 = 0
when x = 0, y = -6
when y = 0, x = -3
So, the points are A(0, 0), B(0, -6) and C(-3, 0)

Since region is outside from the line 2x + y + 6 = 0
So, it does not represent any figure.

---

Question 6.
Solve: 2x + 1 > 3
(a) [-1, ∞]
(b) (1, ∞)
(c) (∞, ∞)
(d) (∞, 1)

Answer

Answer: (b) (1, ∞)
Given, 2x + 1 > 3
⇒ 2x > 3 – 1
⇒ 2x > 2
⇒ x > 1
⇒ x ∈ (1, ∞)

---

Question 7.
The solution of the inequality 3(x – 2)/5 ≥ 5(2 – x)/3 is
(a) x ∈ (2, ∞)
(b) x ∈ [-2, ∞)
(c) x ∈ [∞, 2)
(d) x ∈ [2, ∞)

Answer

Answer: (d) x ∈ [2, ∞)
Given, 3(x – 2)/5 ≥ 5(2 – x)/3
⇒ 3(x – 2) × 3 ≥ 5(2 – x) × 5
⇒ 9(x – 2) ≥ 25(2 – x)
⇒ 9x – 18 ≥ 50 – 25x
⇒ 9x – 18 + 25x ≥ 50
⇒ 34x – 18 ≥ 50
⇒ 34x ≥ 50 + 18
⇒ 34x ≥ 68
⇒ x ≥ 68/34
⇒ x ≥ 2
⇒ x ∈ [2, ∞)

---

Question 8.
Solve: 1 ≤ |x – 1| ≤ 3
(a) [-2, 0]
(b) [2, 4]
(c) [-2, 0] ∪ [2, 4]
(d) None of these

Answer

Answer: (c) [-2, 0] ∪ [2, 4]
Given, 1 ≤ |x – 1| ≤ 3
⇒ -3 ≤ (x – 1) ≤ -1 or 1 ≤ (x – 1) ≤ 3
i.e. the distance covered is between 1 unit to 3 units
⇒ -2 ≤ x ≤ 0 or 2 ≤ x ≤ 4
Hence, the solution set of the given inequality is
x ∈ [-2, 0] ∪ [2, 4]

---

Question 9.
Solve: -1/(|x| – 2) ≥ 1 where x ∈ R, x ≠ ±2
(a) (-2, -1)
(b) (-2, 2)
(c) (-2, -1) ∪ (1, 2)
(d) None of these

Answer

Answer: (c) (-2, -1) ∪ (1, 2)
Given, -1/(|x| – 2) ≥ 1
⇒ -1/(|x| – 2) – 1 ≥ 0
⇒ {-1 – (|x| – 2)}/(|x| – 2) ≥ 0
⇒ {1 – |x|}/(|x| – 2) ≥ 0
⇒ -(|x| – 1)/(|x| – 2) ≥ 0

Using number line rule:
1 ≤ |x| < 2
⇒ x ∈ (-2, -1) ∪ (1, 2)

---

Question 10.
If x² < 4 then the value of x is
(a) (0, 2)
(b) (-2, 2)
(c) (-2, 0)
(d) None of these

Answer

Answer: (b) (-2, 2)
Given, x² < 4
⇒ x² – 4 < 0
⇒ (x – 2) × (x + 2) < 0
⇒ -2 < x < 2
⇒ x ∈ (-2, 2)

---

Question 11.
Solve: 2x + 1 > 3
(a) [1, 1)
(b) (1, ∞)
(c) (∞, ∞)
(d) (∞, 1)

Answer

Answer: (b) (1, ∞)
Given, 2x + 1 > 3
⇒ 2x > 3 – 1
⇒ 2x > 2
⇒ x > 1
⇒ x ∈ (1, ∞)

---

Question 12.
If a is an irrational number which is divisible by b then the number b
(a) must be rational
(b) must be irrational
(c) may be rational or irrational
(d) None of these

Answer

Answer: (b) must be irrational
If a is an irrational number which is divisible by b then the number b must be irrational.
Ex: Let the two irrational numbers are √2 and √3
Now, √2/√3 = √(2/3)

---

Question 13.
Sum of two rational numbers is ______ number.
(a) rational
(b) irrational
(c) Integer
(d) Both 1, 2 and 3

Answer

Answer: (a) rational
The sum of two rational numbers is a rational number.
Ex: Let two rational numbers are 1/2 and 1/3
Now, 1/2 + 1/3 = 5/6 which is a rational number.

---

Question 14.
If |x| = -5 then the value of x lies in the interval
(a) (-5, ∞)
(b) (5, ∞)
(c) (∞, -5)
(d) No solution

Answer

Answer: (d) No solution
Given, |x| = -5
Since |x| is always positive or zero
So, it can not be negative
Hence, given inequality has no solution.

---

Question 15.
The value of x for which |x + 1| + √(x – 1) = 0
(a) 0
(b) 1
(c) -1
(d) No value of x

Answer

Answer: (d) No value of x
Given, |x + 1| + √(x – 1) = 0, where each term is non-negative.
So, |x + 1| = 0 and √(x – 1) = 0 should be zero simultaneously.
i.e. x = -1 and x = 1, which is not possible.
So, there is no value of x for which each term is zero simultaneously.

---

Question 16.
If x² < -4 then the value of x is
(a) (-2, 2)
(b) (2, ∞)
(c) (-2, ∞)
(d) No solution

Answer

Answer: (d) No solution
Given, x² < -4
⇒ x² + 4 < 0
Which is not possible.
So, there is no solution.

---

Question 17.
The solution of |2/(x – 4)| > 1 where x ≠ 4 is
(a) (2, 6)
(b) (2, 4) ∪ (4, 6)
(c) (2, 4) ∪ (4, ∞)
(d) (-∞, 4) ∪ (4, 6)

Answer

Answer: (b) (2, 4) ∪ (4, 6)
Given, |2/(x – 4)| > 1
⇒ 2/|x – 4| > 1
⇒ 2 > |x – 4|
⇒ |x – 4| < 2
⇒ -2 < x – 4 < 2
⇒ -2 + 4 < x < 2 + 4
⇒ 2 < x < 6
⇒ x ∈ (2, 6), where x ≠ 4
⇒ x ∈ (2, 4) ∪ (4, 6)

---

Question 18.
The solution of the function f(x) = |x| > 0 is
(a) R
(b) R – {0}
(c) R – {1}
(d) R – {-1}

Answer

Answer: (b) R – {0}
Given, f(x) = |x| > 0
We know that modulus is non negative quantity.
So, x ∈ R except that x = 0
⇒ x ∈ R – {0}
This is the required solution

---

Question 19.
Solve: |x – 1| ≤ 5, |x| ≥ 2
(a) [2, 6]
(b) [-4, -2]
(c) [-4, -2] ∪ [2, 6]
(d) None of these

Answer

Answer: (c) [-4, -2] ∪ [2, 6]
Given, |x – 1| ≤ 5, |x| ≥ 2
⇒ -(5 ≤ (x – 1) ≤ 5), (x ≤ -2 or x ≥ 2)
⇒ -(4 ≤ x ≤ 6), (x ≤ -2 or x ≥ 2)
Now, required solution is
x ∈ [-4, -2] ∪ [2, 6]

---

Question 20.
The solution of the 15 < 3(x – 2)/5 < 0 is
(a) 27 < x < 2
(b) 27 < x < -2
(c) -27 < x < 2
(d) -27 < x < -2

Answer

Answer: (a) 27 < x < 2
Given inequality is:
15 < 3(x – 2)/5 < 0
⇒ 15 × 5 < 3(x – 2) < 0 × 5
⇒ 75 < 3(x – 2) < 0
⇒ 75/3 < x – 2 < 0
⇒ 25 < x – 2 < 0
⇒ 25 + 2 < x < 0 + 2
⇒ 27 < x < 2

---

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Permutations and Combinations Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 7 Permutations and Combinations Objective Questions.

Permutations and Combinations Class 11 MCQs Questions with Answers

Students are advised to solve the Permutations and Combinations Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Permutations and Combinations Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Permutations and Combinations Class 11 with answers provided with detailed solutions by looking below.

Question 1.
It is required to seat 5 men and 4 women in a row so that the women occupy the even places. The number of ways such arrangements are possible are
(a) 8820
(b) 2880
(c) 2088
(d) 2808

Answer

Answer: (b) 2880
Total number of persons are 9 in which there are 5 men and 4 women
So total number of place = 9
Now women seat in even place
So total number of arrangement = 4! (_W_W_W_W_) (W-Woman)
Men sit in odd place
So total number of arrangement = 5! (MWMWMWMWM) (M-Man)
Now Total number of arrangement = 5! × 4! = 120 × 24 = 2880

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Question 2.
Six boys and six girls sit along a line alternately in x ways and along a circle (again alternatively in y ways), then
(a) x = y
(b) y = 12x
(c) x = 10y
(d) x = 12y

Answer

Answer: (d) x = 12y
Given, six boys and six girls sit along a line alternately in x ways and along a circle
(again alternatively in y ways).
Now, x = 6! × 6! + 6! × 6!
⇒ x = 2 × (6!)2
and y = 5! × 6!
Now, x/y = {2 × (6!)2}/(5! × 6!)
⇒ x/y = {2 × 6! × 6! }/(5! × 6!)
⇒ x/y = {2 × 6!}/5!
⇒ x/y = {2 × 6 × 5!}/5!
⇒ x/y = 12
⇒ x = 12y

---

Question 3.
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
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

---

Question 4.
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of at least 3 girls
(a) 588
(b) 885
(c) 858
(d) None of these

Answer

Answer: (a) 588
Given number of boys = 9
Number of girls = 4
Now, A committee of 7 has to be formed from 9 boys and 4 girls.
Now, the committee consists of atleast 3 girls:
⁴C₃ × ⁹C₄ + ⁴C₄ × ⁹C₃
= [{4! / (3! × 1!)} × {9! / (4! × 5!)}] + 9C₃
= [{(4 × 3!) /3!} × {(9 × 8 × 7 × 6 × 5!) / (4! × 5!)}] + 9! /(3! × 6!)
= [4 × {(9 × 8 × 7 × 6) / 4!}] + (9×8×7×6!)/(3! × 6!)
= [{4 × (9 × 8 × 7 × 6)} / (4 × 3 × 2 × 1)] + (9 × 8 × 7)/3!
= (9 × 8 × 7) + (9 × 8 × 7)/(3 × 2 × 1)
= 504 + (504/6)
= 504 + 84
= 588

---

Question 5.
In how many ways can 12 people be divided into 3 groups where 4 persons must be there in each group?
(a) none of these
(b) 12!/(4!)³
(c) Insufficient data
(d) 12!/{3! × (4!)³}

Answer

Answer: (d) 12!/{3! × (4!)³}
Number of ways in which
m × n”>
m × n distinct things can be divided equally into n
n”> groups
= (mn)!/{n! × (m!)n }
Given, 12(3 × 4) people needs to be divided into 3 groups where 4 persons must be there in each group.
So, the required number of ways = (12)!/{3! × (4!)n}

---

Question 6.
How many factors are 2⁵ × 3⁶ × 5² are perfect squares
(a) 24
(b) 12
(c) 16
(d) 22

Answer

Answer: (a) 24
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

---

Question 7.
If ⁿC₁₅ = ⁿC₆ then the value of ⁿC₂₁ is
(a) 0
(b) 1
(c) 21
(d) None of these

Answer

Answer: (b) 1
We know that
if ⁿC_r1 = ⁿC_r2
⇒ n = r₁ + r₂
Given, ⁿC₁₅ = ⁿC₆
⇒ n = 15 + 6
⇒ n = 21
Now, ²¹C₂₁ = 1

---

Question 8.
If ⁿ⁺¹C₃ = 2 ⁿC₂, then the value of n is
(a) 3
(b) 4
(c) 5
(d) 6

Answer

Answer: (d) 6
Given, ⁿ⁺¹C₃ = 2 ⁿC₂
⇒ [(n + 1)!/{(n + 1 – 3) × 3!}] = 2n!/{(n – 2) × 2!}
⇒ [{n × n!}/{(n – 2) × 3!}] = 2n!/{(n – 2) × 2}
⇒ n/3! = 1
⇒ n/6 = 1
⇒ n = 6

---

Question 9.
There are 15 points in a plane, no two of which are in a straight line except 4, all of which are in a straight line. The number of triangle that can be formed by using these 15 points is
(a) ¹⁵C₃
(b) 490
(c) 451
(d) 415

Answer

Answer: (c) 451
The required number of triangle = ¹⁵C₃ – ⁴C₃ = 455 – 4 = 451

---

Question 10.
In how many ways in which 8 students can be sated in a circle is
(a) 40302
(b) 40320
(c) 5040
(d) 50040

Answer

Answer: (c) 5040
The number of ways in which 8 students can be sated in a circle = ( 8 – 1)!
= 7!
= 5040

---

Question 11.
Let R = {a, b, c, d} and S = {1, 2, 3}, then the number of functions f, from R to S, which are onto is
(a) 80
(b) 16
(c) 24
(d) 36

Answer

Answer: (d) 36
Total number of functions = 3⁴ = 81
All the four elements can be mapped to exactly one element in 3 ways, and exactly 3
elements in 3(2⁴ – 2) = 3(16 – 2) = 3 × 14 = 42
Thus the number of onto functions = 81 – 42 -3 = 81 – 45 = 36

---

Question 12.
If (1 + x)ⁿ = C₀ + C₁ x + C₂ x² + …………..+ C_n xⁿ, then the value of C₀² + C₁² + C₂² + …………..+ C_nⁿ = ²ⁿC_n is
(a) (2n)!/(n!)
(b) (2n)!/(n! × n!)
(c) (2n)!/(n! × n!)2
(d) None of these

Answer

Answer: (b) (2n)!/(n! × n!)
Given, (1 + x)ⁿ = C₀ + C₁ x + C₂ x² + ………….. + C_n xⁿ ………. 1
and (1 + x)ⁿ = C₀ xⁿ + C₁ x^n-1 + C₂ x^n-2 + ………….. C_r x^n-r + ………. + C_n-1 x + C_n ……….. 2
Multiply 1 and 2, we get
(1 + x)²ⁿ = (C₀ + C₁ x + C₂ x² + …………..+ C_n xⁿ) × (C₀ xⁿ + C₁ x^n-1 + C₂ x^n-2 + ………….. C_r x^n-r + ………. + C_n-1 x + C_n)
Now, equating the coefficient of xn on both side, we get
C₀² + C₁² + C₂² + …………..+ C_nⁿ = ²ⁿC_n = (2n)!/(n! × n!)

---

Question 13.
The total number of 9 digit numbers of different digits is
(a) 99!
(b) 9!
(c) 8 × 9!
(d) 9 × 9!

Answer

Answer: (d) 9 × 9!
Given digit in the number = 9
1st place can be filled = 9 ways = 9 (from 1-9 any number can be placed at first position)
2nd place can be filled = 9 ways (from 0-9 any number can be placed except the number which is placed at the first position)
3rd place can be filled = 8 ways
4th place can be filled = 7 ways
5th place can be filled = 6 ways
6th place can be filled = 5 ways
7th place can be filled = 4 ways
8th place can be filled = 3 ways
9th place can be filled = 2 ways
So total number of ways = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2
= 9 × 9!

---

Question 14.
The number of ways in which 6 men add 5 women can dine at a round table, if no two women are to sit together, is given by
(a) 30
(b) 5 ! × 5 !
(c) 5 ! × 4 !
(d) 7 ! × 5 !

Answer

Answer: (b) 5 ! × 5 !
Again, 6 girls can be arranged among themselves in 5! ways in a circle.
So, the number of arrangements where boys and girls sit attentively in a circle = 5! × 5!

---

Question 15.
There are 15 points in a plane, no two of which are in a straight line except 4, all of which are in a straight line. The number of triangle that can be formed by using these 15 points is
(a) ¹⁵C₃
(b) 490
(c) 451
(d) 415

Answer

Answer: (c) 451
The required number of triangle = ¹⁵C₃ – ⁴C₃ = 455 – 4 = 451

---

Question 16.
The number of 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated are
(a) 110
(b) 120
(c) 130
(d) 140

Answer

Answer: (b) 120
A number is divisible by 10 if the unit digit of the number is 0.
Given digits are 0, 1, 3, 5, 7, 9
Now we fix digit 0 at unit place of the number.
Remaining 5 digits can be arranged in 5! ways
So, total 6-digit numbers which are divisible by 10 = 5! = 120

---

Question 17.
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
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

---

Question 18.
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of exactly 3 girls
(a) 540
(b) 405
(c) 504
(d) None of these

Answer

Answer: (c) 504
Given number of boys = 9
Number of girls = 4
Now, A committee of 7 has to be formed from 9 boys and 4 girls.
Now, If in committee consist of exactly 3 girls:
⁴C₃ × ⁹C₄
= {4! / (3! × 1!)} × {9! / (4! × 5!)}
= {(4×3!) /3!} × {(9 × 8 × 7 × 6 × 5!) / (4! × 5!)}
= 4 × {(9 × 8 × 7 × 6) / 4!}
= {4 × (9 × 8 × 7 × 6)} / (4 × 3 × 2 × 1)
= 9 × 8 × 7
= 504

---

Question 19.
How many factors are 2⁵ × 3⁶ × 5² are perfect squares
(a) 24
(b) 12
(c) 16
(d) 22

Answer

Answer: (a) 24
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

---

Question 20.
The value of 2 × P(n, n-2) is
(a) n
(b) 2n
(c) n!
(d) 2n!

Answer

Answer: (c) n!
Given, 2 × P(n, n – 2)
= 2 × {n!/(n – (n – 2))}
= 2 × {n!/(n – n + 2)}
= 2 × (n!/2)
= n!
So, 2 × P(n, n – 2) = n!

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 10 Straight Lines with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Straight Lines Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 10 Straight Lines Objective Questions.

Straight Lines Class 11 MCQs Questions with Answers

Students are advised to solve the Straight Lines Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Straight Lines Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Straight Lines Class 11 with answers provided with detailed solutions by looking below.

Question 1.
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)
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 coordiantes 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 cooridnates of B are (7, -2)

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Question 2.
The equation of straight line passing through the point (1, 2) and perpendicular to the line x + y + 1 = 0
(a) y – x + 1 = 0
(b) y – x – 1 = 0
(c) y – x + 2 = 0
(d) y – x – 2 = 0

Answer

Answer: (b) y – x – 1 = 0
Given straight line is: x + y + 1 = 0
⇒ y = -x – 1
Slope = -1
Now, required line is perpendicular to this line.
So, slope = -1/-1 = 1
Hence, the line is
y – 2 = 1 × (x – 1)
⇒ y – 2 = x – 1
⇒ y – 2 – x + 1 = 0
⇒ y – x – 1 = 0

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Question 3.
The points (-a, -b), (0, 0), (a, b) and (a², ab) are
(a) vertices of a square
(b) vertices of a parallelogram
(c) collinear
(d) vertices of a rectangle

Answer

Answer: (c) collinear
Let the four points are P(-a, -b), O(0, 0), Q(a, b) and R(a², ab)
Now,
m₁ = slope of OP = b/a
m₂ = slope of OQ = b/a
m₃ = slope of OR = b/a
Since m₁ = m₂ = m₃
So, the points O, P, Q, R are collinear.

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Question 4.
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
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

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Question 5.
The slope of a line which passes through points (3, 2) and (-1, 5) is
(a) 3/4
(b) -3/4
(c) 4/3
(d) -4/3

Answer

Answer: (b) -3/4
Given, points are (3, 2) and (-1, 5)
Now, slope m = (5 – 2)/(-1 – 3)
⇒ m = -3/4
So, the slope of the line is -3/4

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Question 6.
The ratio of the 7th to the ( n – 1)th mean between 1 and 31, when n arithmetic means are inserted between them, is 5 : 9. The value of n is
(a) 15
(b) 12
(c) 13
(d) 14

Answer

Answer: (d) 14
Let the A.P. are 1, A₁, A₂, A₃ …… A_m, 31
a = 1, a_n = 31 and n = m + 2
Now, a_n = a + (n – 1)d
⇒ 31 = 1 + (m + 2 – 1)d
⇒ 30 = (m + 1)d
⇒ d = 30/(m + 1)
Again, A₇ = a + 7d = 1 + 7[30/(m + 1)] …………….. 1
and A_m-1 = a + (m – 1)d = 1 + (m – 1)[30/(m + 1)] ………. 2
From equation 1 and 2, we get
A₇/A_m-1 = 5/9
⇒ 1 + 7[30/(m + 1) / 1 + (m – 1)[30/(m + 1)] = 5/9
⇒ [m + 1 + 7(30)] / [m + 1 + 30 m – 30] = 5/9
⇒ [m + 211] / [31 m – 29] = 5/9
⇒ 9[m + 211] = 5[31 m – 29]
⇒ 9 m + 1899 = 155 m – 145
⇒ 146 m = 2044
⇒ m = 2044/146
⇒ m = 14
So, the value of m is 14

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Question 7.
The ortho centre of the triangle formed by lines xy = 0 and x + y = 1 is :
(a) (0, 0)
(b) none of these
(c) ( 1/2, 1/2)
(d) ( 1/3, 1/3)

Answer

Answer: (a) (0, 0)
Given lines are:
xy = 0 and x + y = 1
⇒ x = 0, y = 0 and x + y = 1

In a triangle OAB, OA and OB are the altitudes which intersect at O.
So, the required orthocentre is (0, 0)

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Question 8.
Two lines a₁ x + b₁ y + c₁ = 0 and a₂ x + b₂ y + 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
Two lines a₁ x + b₁ y + c₁ = 0 and a₂ x + b₂ y + c₂ = 0 are parallel if
a₁ /a₂ = b₁ /b₂ ≠ c₁ /c₂

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Question 9.
If the line x/a + y/b = 1 passes through the points (2, -3) and (4, -5), then (a, b) is
(a) a = 1 and b = 1
(b) a = 1 and b = −1
(c) a = −1 and b = 1
(d) a = −1 and b = -1

Answer

Answer: (d) a = −1 and b = -1
Given equation of the line is x/a + y/b = 1
⇒ bx + ay = ab
It is given that this line passes through (2, -3)
⇒ b(2) + a(-3) = ab
⇒ 2b – 3a = ab ——– (1)
It also passes through (4, -5)
⇒ 4b – 5a = ab ——– (2)
On solving equation (1) and (2), we get
a = -1 and b = -1

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Question 10.
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)
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)

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Question 11.
The points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units is
(a) (0, 32/3) and (0, 8/3)
(b) (0, -32/3) and (0, 8/3)
(c) (0, -32/3) and (0, -8/3)
(d) (0, 32/3) and (0, -8/3)

Answer

Answer: (d) (0, 32/3) and (0, -8/3)
Given equation of line is (x/3) + (y/4) = 1
⇒ 4x + 3y = 12
⇒ 4x + 3y – 12 = 0 ……………. 1
Let (0, b) is the point of the y-axis whose distance from given line is 4 unit.
When we compare equation 1 with general form of the equation Ax + By + C = 0, we get
A = 4, B = 3, C = -12
Now perpendicular distance of a line Ax + By + C = 0 from a point (x₁, y₁) is
d = |Ax₁ + By₁ + C|/√(A² + B²)
So perpendicular distance of a line 4x + 3y – 12 = 0 from a point (0 ,b) is
4 = |4×0 + 3×b – 12|/√(4² + 3²)
⇒ 4 = |3b – 12|/√(16 + 9)
⇒ 4 = |3b – 12|/√25
⇒ 4 = |3b – 12|/5
⇒ 4 × 5 = |3b – 12|
⇒ |3b – 12| = 20
Now
3b – 12 = 20 and 3b – 12 = -20
⇒ 3b = 20 12 and 3b = -20 + 12
⇒ 3b = 32 and 3b = -8
⇒ b = 32/3 and b = -8/3
So the points are (0, 32/3) and (0, -8/3)

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Question 12.
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
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

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Question 13.
The distance between the lines 3x + 4y = 9 and 6x + 8y = 15 is
(a) 3/10
(b) 2/3
(c) 3/2
(d) 7/10

Answer

Answer: (a) 3/10
Given equations are:
3x + 4y = 9
⇒ 3x + 4y – 9 = 0 and
6x + 8y = 15
⇒ 6x + 8y – 15 = 0
⇒ 3x + 4y – 15/2 = 0
Now, compare these lines with a₁ x + b₁ y + c₁ = 0 and a₂ x + b₂ y + c₂ = 0, we get
a₁ = 3, b₁ = 4, c₁ = -9 and
a₂ = 3, b₂ = 4, c₂ = -15/2
Now, distance between two parallel line = |c₁ – c₂|/√(a₁² + b₁²)
= |-9 + 15/2|/√(3² + 4²)
= |(-18 + 15)/2|/√25
= |(-3/2)|/5
= (3/2)/5
= 3/10

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Question 14.
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
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

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Question 15.
Two lines a₁ x + b₁ y + c₁ = 0 and a₂ x + b₂ y + 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
Two lines a₁ x + b₁ y + c₁ = 0 and a₂ x + b₂ y + c₂ = 0 are parallel if
a₁/a₂ = b₁/b₂ ≠ c₁/c₂

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Question 16.
The slope of a line making inclination of 30° with the positive direction of x-axis is
(a) 1/2
(b) √3
(c) √3/2
(d) 1/√3

Answer

Answer: (d) 1/√3
Here inclination of the line is 30°
So, slope of the line m = tan 30° = 1/√3

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Question 17.
The perpendicular distance of a line 4x + 3y + 5 = 0 from the point (-1, 2) is
(a) 5
(b) 4
(c) 2
(d) 1

Answer

Answer: (c) 2
The perpendicular distance of a line 4x + 3y + 5 = 0 from the point (-1, 2)
d = |4 × (-1) + 3 × 3 + 5|/√(4² + 3²)
⇒ d = |-4 + 9 + 5|/√(16 + 9)
⇒ d = 10/√(25)
⇒ d = 10/5
⇒ d = 2

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Question 18.
The inclination of the line 5x – 5y + 8 = 0 is
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer

Answer: (b) 45°
Given line is: 5x – 5y + 8 = 0
⇒ 5y = 5x + 8
⇒ y = (5/5)x + 8/5
⇒ y = x + 8/5
Now tan θ = 1
⇒ tan θ = tan 45°
⇒ θ = 45°
So, the inclination of the line is 45°

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Question 19.
The points (-a, -b), (0 , 0), (a, b) and (a², ab) are
(a) vertices of a square
(b) vertices of a parallelogram
(c) collinear
(d) vertices of a rectangle

Answer

Answer: (c) collinear
Let the four points are P(-a, -b), O(0, 0), Q(a, b) and R(a², ab)
Now,
m₁ = slope of OP = b/a
m₂ = slope of OQ = b/a
m₃ = slope of OR = b/a
Since m₁ = m₂ = m₃
So, the points O, P, Q, R are collinear.

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Question 20.
Given the three straight lines with equations 5x + 4y = 0, x + 2y – 10 = 0 and 2x + y + 5 = 0, then these lines are
(a) none of these
(b) the sides of a right angled triangle
(c) concurrent
(d) the sides of an equilateral triangle

Answer

Answer: (c) concurrent
Since the determinant of these lines is equal to zero
i.e.
|5 4 0|
|1 2 -10| = 0
|2 1 -5|
So, these three lines are concurrent.

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We believe the knowledge shared regarding NCERT MCQ Questions for Class 11 Maths Chapter 10 Straight Lines with Answers Pdf free download has been useful to the possible extent. If you have any other queries regarding CBSE Class 11 Maths Straight Lines MCQs Multiple Choice Questions with Answers, feel free to reach us via the comment section and we will guide you with the possible solution.