Category: Class 11 Mathematics

NCERT Exemplar Class 11 Maths Chapter 14 Mathematical Reasoning are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 14 Mathematical Reasoning. https://www.cbselabs.com/ncert-exemplar-problems-class-11-mathematics-chapter-14-mathematical-reasoning/

NCERT Exemplar Class 11 Maths Chapter 14 Mathematical Reasoning

Q1. Which of the following sentences are statements? Justify.
(i) A triangle has three sides.
(ii) O is a complex number.
(iii) Sky is red,
(iv) Every set is an infinite set.
(V) 15 +8>23
(vi) y+9=7
(vii) Where is your bag?
(viii) EVery square is a rectangle.
(ix) Sum of opposite angles of a cyclic quadrilateral is 1800.
(x) sin² x+cos² x=O
Sol: As we know, a statement is a sentence which is either true or false but not
both simultaneously.
(j) It is true statement.
(ii) It is true statement.
(iii) It is false statement.
(iv) It is false statement.
(y) It is false statement.
(vi) y +9 = 7
It is not considered as a statement, since the value of y is not given.
(vii) It is a question, so it is not a statement.
(viii) It is a true statement.
(ix) It is a true statement.
(x) It is a false statement.

Q2. Find the component statements of the following compound statements.
(i) Number 7 is prime and odd.
(ii) Chennai is in India and is the capital of Tamil Nadu.
(iii) The number loo is divisibLe by 3, 11 and 5.
(iv) Chandigarh is the capital of Irlaryana and U.P.
(y) √7 is a rational number or an irrational number.
(vi) O is less than every positive integer and every negative integer.
(vii) Plants use sunlight, water and carbon dioxide for photosynthesis.
(viii) Two lines in a plane either intersect at one point or they are parallel.
(ix) A rectangle is a quadrilateral or a 5-sided polygon.
Sol: (i) p: Number 7 is prime.
q: Number 7 is odd.
(ii) P: Chennai is in India.
q: Chennai is capital of Tamil Nadu.
(iii) p. 100 is divisible by 3.
q: 100 isdivisibleby 11.
r: 100 ¡s divisible by 5.
(iv) p. Chandigarh is capital of Haryana.
q: Chandigarb is capital of UP
(v) p: √7 is a rational number.
q: √7 is an irrational number.
(vi) p: 0 is less than every positive integer.
q: O is less than every negative integer.
(vii) p: Plants use sunlight for photosynthesis.
q: Plants use water for photosynthesis.
q:- Plants use carbon dioxide for photosynthesis.
(viii) p: Two lines in a plane intersect at one point.
q: Two lines ¡n a plane are parallel.
(ix) p: A rectangle is a quadrilateral.
q. A rectangle is a 5-sided polygon.

Q3. Write the component statements of the following compound statements and
check whether the compound statement is true or false.
(i) 57 is divisible by 2 or 3.
(ii) 24 is a multiple of 4 and 6.
(iii) All living things have two eyes and two legs.
(iv) 2 is an even number and a prime number.
Sol: (i) Here component statements are:
p: 57 is divisible by 2. [false]
q: 57 is divisible by 3. [true]
Given compound statement is of the form ‘pvq’.
Since, the statement ‘pvq’ has the truth value T whenever either p or q
or both have the truth value T.
So, it is true statement as 57 is divisible by 3.
(ii) Here component statements are: p: 24 is multiple of 4. q: 24 is multiple of 6.
Given compound statement is of the form ‘p ^ q’
Since, the statement ‘p A q’ has the truth value T whenever bothp and q have the truth value T.
So, it is a true statement as 24 is divisible by 4 and 6.
(iii) Here component statements are:
p: All living things have two eyes. [false]
q: All living things have two legs. [false]
Given compound statement is of the form ‘p ^q’
It is a false statement. Since ‘p ^ q’ has truth value F whenever either p or q or both have the truth value F
(iv) Here component statements are:
p: 2 is an even number. [true]
q: 2 is a prime number. [true]
Given compound statement is of the form ‘p ^ q’.
It is a true statement. Since ‘p ^ q’ has truth value T whenever both p and q or both have the truth value T.

Q4. Write the negative on the following simple statements.
(i) The number 17 is prime.
(ii) 2 + 7 = 6.
(iii) Violets are blue.
(iv) √5 is a rational number.
(v) 2 is not a prime number.
(vi) Every real number is an irrational number.
(vii) Cow has four legs.
(viii) A leap year has 366 days.
(ix) All similar triangles are congruent.
(x) Area of a circle is same as the perimeter of the circle.
Sol: (i) The number 17 is not prime.
(ii) 2 + 7≠6.
(iii) Violets are not blue.
(iv) √5 is not a rational number.
(v) 2 is a prime number.
(vi) Every real number is not an irrational number.
(vii) Cow does not have four legs.
(viii) A leap year does not have 366 days.
(ix) There exist similar triangles which are not congruent.
(x) Area of a circle is not same as the perimeter of the circle

Q5. Translate the following statements into symbolic form.
(i) Rahul passed in Hindi and English.
(ii) x and y are even integers.
(iii) 2, 3 and 6 are factors of 12.
(iv) Either x or x + 1 is an odd integer.
(v) A number is either divisible by 2 or 3.
(vi) Either x = 2 or x = 3 is a root of 3x² – x – 10 = 0
(vii) Students can take Hindi or English as an optional paper.
Sol: (i) p: Rahul passed in Hindi.
q: Rahul passed in English. p ∧ q: Rahul passed in Hindi and English.
(ii) p: x is even integers. . q: y is even integers.
p∧q: x andy are even integers.
(iii) p: 2 is factor of 12. q: 3 is factor of 12. r: 6 is factor of 12.
p ∧ q ∧ r: 2, 3 and 6 are factors of 12
(iv) p: x is an odd integer.
q: (x + 1) is an odd integer. p v q: Either x or (x + 1) is an odd integer.
(v) p: A number is divisible by 2. q: A number is divisible by 3.
pv q: A number is either divisible by 2 or 3.
(vi) p: x = 2 is a root of 3×2 – x – 10 = 0. q: x = 3 is a root of 3×2 – x – 10 = 0
p v q: Either x = 2orx = 3isa root of 3×2 – x – 10 = 0
(vii) p: Students can take Hindi as an optional paper. q: Students can take English as an optional paper.
p v q: Students can take Hindi or English as an optional paper.

Q6. Write down the negation of following compound statements.
(i) All rational numbers are real and complex.
(ii) All real numbers are rationals or irrationals.
(iii) x = 2 and x = 3 are roots of the quadratic equation x² -5x +6 = 0
(iv) A triangle has either 3-sides or 4-sides.
(v) 35 is a prime number or a composite number.
(vi) All prime integers are either even or odd.
(vii) |x| is equal to either x or -x.
(viii) 6 is divisible by 2 and 3.
Sol. (i) Let p: All rational numbers are real.
q: All rational numbers are complex.
~ p: All rational numbers are not real.
~ q ; All rational numbers are not complex.
Then, the negation of the given compound statement is:
~ (p ∧ q): All rational numbers are not real or not complex.
[~(p ∧ q) = ~p v ~q]
(ii) Let p: All real numbers are rationals. q: All real numbers are irrationals.
Then, the negation of the given compound statement is:
~ (p v q): All real numbers are not rational and all real numbers are not irrational. [~(p v q) = ~p ∧ ~ q]
(iii) Let p ; x = 2 is root of quadratic equation x² – 5x + 6 = 0. q: x = 3 is root of quadratic equation x² – 5x + 6 = 0.
Then, the negation of the given compound statement is:
~ (p ∧  q) : x = 2 is not a root of quadratic equation x²– 5x + 6 = 0 or x = 3 is not a root of the quadratic equationx² – 5x + 6 = 0.
(iv) Let p: A triangle has 3-sides. q: A triangle has 4-sides.
Then, the negation of the given compound statement is:
~ (p v q): A triangle has neither 3-sides nor 4-sides.
(v) Let p: 35 is a prime number. q: 35 is a composite number.
Then, the negation of the given compound statement is:
~ (p v q): 35 is not a prime number and it is not a composite number.
(vi) Let p: All prime integers are even. q: All prime integers are odd.
Then, the negation of the given compound statement is given by
~(p v q): All prime integers are not even and all prime integers are not odd.
(vii) Let p:|x| is equal to x. q: |x| is equal to —x.
Then, the negation of the given compound statement is:
~ (p v q): |x| is not equal to JC and it is not equal to —x.
(viii) Let p: 6 is divisible by 2. q: 6 is divisible by 3.
Then, the negation of the given compound statement is:
~ (p∧q): 6 is not divisible by 2 or it is not divisible by 3

Q7. Rewrite each of the following statements in the form of conditional statements.
(i) The square of an odd number is odd.
(ii) You will get a sweet dish after the dinner.
(iii) You will fail, if you will not study.
(iv) The unit digit of an integer is 0 or 5, if it is divisible by 5.
(v) The square of a prime number is not prime.
(vi) 2b = a + c, if a, b and c are in AP.
Sol: (i) If the number is odd number, then its square is odd number.
(ii) It you take the dinner, then you will get sweet dish.
(iii If you will not study, then you will fail.
(iv) If an integer is divisible by 5, then its unit digits are 0 or 5.
(v) If the number is prime, then its square is not prime.
(vi) If a, b and c are in AP, then 2b = a + c.

Q8. Form the biconditional statement p⟷q, where
(i) p: The unit digits of an integer is zero.
q: It is divisible by 5.
(ii) p: A natural number is odd.
q: Natural number is not divisible by 2.
(iii) p: A triangle is an equilateral triangle.
q: All three sides of a triangle are equal.
Sol:(i) p ⟷ q: The unit digit of on integer is zero, if and only if it is divisible by 5.
(ii) p ⟷ q: A natural number is odd if and only if it is not divisible by 2.
(iii) p ⟷q: A triangle is an equilateral triangle if and only if all three sides of triangle are equal.

Q9. Write down the contra positive of the following statements.
(î) lf x =y and y=3,then x = 3.
(ii) If n is a natural number, then n is an integer.
(iii) If all three sides ola triangle are equal, then the triangle is equilateral.
(iv) If x andy are negative integers, then .ty is positive.
(v) If natural number n is divisible by 6, then n is divisible by 2 and 3.
(vi) Jf it snows, then the weather will be cold.
(vii) lix is a real number such that O <x < 1, then x2 < 1
Sol: (i) If x ≠ 3, then x ≠y or y ≠ 3.
(ii) 1f n is not an integer, then it is not a natural number.
(iii) If the triangle is not equilateral, then all three sides of the triangle are not equal.
(iv) If xv is not positive integer, then either x or y is not negative integer.
(v) If natural number n is not divisible by 2 or 3, then n is not divisible by 6.
(vi) The weather will not be cold, if it does not snow.
(vii) lf x² is not less than I, thenx is not a real number such that O <x <1.

Q10. Write down the converse of following statements.
(i) If a rectangle lR’ is a square, then R is a rhombus.
(ii) If today is Monday, then tomorrow is Tuesday.
(iii) If you go to Agra, then you must visit Taj Mahal.
(iv) If sum of squares of two sides of a triangle is equal to the square of third side of a triangle, then the triangle is right angled.
(v) If all three angles of a triangle are equal, then the triangle is equilateral.
(vi) If x : y = 3 : 2, then 2x = 3y.
(vii) If S’ is a cyclic quadrilateral, then the opposite angles of S are supplementary.
(viii) If x is zero, then x is neither positive nor negative.
(ix) If two triangles are similar, then the ratio of their corresponding sides are equal.
Sol: (i) If the rectangle ‘S’ is rhombus, then it is square.
(ii) It tomorrow is Tuesday, then today is Monday.
(iii) If you must visit Taj Mahal, you go to Agra.
(iv) If the triangle is right angle, then the sum of squares of two sides of a triangle is equal to the square of third side.
(v) If the triangle is equilateral, then all three angles of triangle are equal.
(vi) If 2x = 3y, thenx :y = 3:2
(vii) If the opposite angles of a quadrilateral are supplementary, then S is cyclic.
(viii) If x is neither positive nor negative, then x is 0.
(ix) If the ratio of corresponding sides of two triangles are equal, then triangles are similar

Q11. Identify the quantifiers in the following statements.
(i) There exists a triangle which is not equilateral.
(ii) For all real numbers x and y, xy= yx.
(iii) There exists a real number which is not a rational number.
(iv) For every natural number x, x + 1 is also a natural number.
(v) For all real numbers x with x > 3, x² is greater than 9.
(vi) There exists a triangle which is not an isosceles triangle.
(vii) For all negative integers x, x³ is also a negative integers.
(viii) There exists a statement in above statements which is not true.
(ix) There exists an even prime number other than 2.
(x) There exists a real number x such that x² + 1 = 0.
Sol: Quantifier are the phrases like ‘There exists’ and ‘For every1, ‘For all’ etc.
(i) There exists
(ii) For all
(iii) There exists
(iv) For every
(v) For all
(vi) There exists
(vii) For all (viii) There exists
(ix) There exists
(x) There exists

Q12. Prove by direct method that for any integer ‘n’ ,n³ – n is always even.

Q13. Check validity of the following statements.
(i) p: 125 is divisible by 5 and 7.
(ii) q: 131 is a multiple of 3 or 11.
Sol: (i) We have,P : 125 is divisible by 5 and 7.
Let q: 125 is divisible by 5.
r: 125 is divisible by 7. q is true, r is false.
=> q ⋀ r is false.
[since, p ⋀ q has the truth value F (false) whenever eitherp or q or both have the truth value F]
Hence, p is not valid.
(ii) We have,  p: 131 is a multiple of 3 or 11.
Let q: 131 is multiple of 3.
r: 131 is a multiple of 11.
p is true, r is false.
=> p v r is true.
[since, p v q has the truth value T (true) whenever either p or q or both have the truth value T]
Hence, q is valid.

Q14. Prove the following statement by contradiction method.
p: The sum of an irrational number and a rational number is irrational.

Q15. Prove by direct method that for any real number x, y if x = y, then x ²=y²

Sol: Let p: x = y; x, y∈ R
On squaring both sides, we get
x² =y² :q
p ⟹q
Hence, proved.

Q16. Using contra positive method prove that, if n² is an even integer, then n is also an even integer.
Sol: Let p: n² is an even integer. q: n is also an even integer.
Let ~p is true i.e., n is not an even integer.
=> n² is not an even integer. [Since square of an odd integer is odd]
=> ~ p is true.
Therefore, ~q is true which provides that ~p is true.
Hence proved.

Objective Type Questions
Q17. Which of the following is a statement?
(a) x is a real number (b) Switch off the fan
(c) 6 is a natural number (d) Let me go
Sol: (c) As we know that a statement is a sentence which is either true or false.
6 is a natural number; this is true.
Hence, it is a statement.

Q18. Which of the following is not a statement.
(a) Smoking is injurious to health
(b) 2 + 2 = 4
(c) 2 is the only even prime number
(d) Come here
Sol: (d) No sentence can be called a statement, if it is an order. So, ‘Come here’ is not a statement.

Q19. The connective in the statement ‘2 + 7>9or2 + 7<9’is
(a) and
(b) or
(c) >
(d) <
Sol: (b) In ‘2 + 7 > 9 or 2 + 7 < 9’, or is the connective.

Q20. The connective in the statement “Earth revolves round the Sun and Moon is a satellite of earth” is
(a) or
(b) Earth
(c) Sun
(d) and
Sol: (d) Connective word is ‘and’.

Q21. The negation of the statement “A circle is an ellipse” is
(a) An ellipse is a circle
(b) An ellipse is not a circle
(c) A circle is not an ellipse
(d) A circle is an ellipse
Sol: (c) Let p: A circle is an ellipse.
~p: A circle is not an ellipse.

Q22. The negation of the statement “7 is greater than 8” is
(a) 7 is equal to 8
(b) 7 is not greater than 8
(c) 8 is less than 7
(d) None of these
Sol: (b) Letp: 7 is greater than 8.
~p: 7 is not greater than 8

Q23. The negation of the statement “72 is divisible by 2 and 3” is
(a) 72 is not divisible by 2 or 72 is not divisible by 3
(b) 72 is not divisible by 2 and 72 is not divisible by 3
(c) 72 is divisible by 2 and 72 is not divisible by 3
(d) 72 is not divisible by 2 and 72 is divisible by 3
Sol: (a) We have, p: 72 is divisible by 2 and 3.
Let q: 72 is divisible by 2.
r: 72 is divisible by 3.
~q: 72 is not divisible by 2.
~r. 72 is not divisible by 3.
~{q ⋀f ) -~q v ~r
⟹ 72 is not divisible by 2 or 72 is not divisible by 3.

Q24. The negation of the statement “Plants take in C0₂ and give out 0₂” is
(a) Plants do not take in C0₂ and do not given out 0₂
(b) Plants do not take in C0₂ or do not give out 0₂
(c) Plants take is C0₂ and do not give out 0₂
(d) Plants take in C0₂ or do not give out 0₂
Sol: (b) Now, p: Plants take in C0₂ and give out 0₂.
Let q: Plants take in C0₂.
r: Plants give out 0₂.
~q: Plants do not take in C0₂.
~r: Plants do not give out 0₂.
~(q ∧ r): Plants do not take in C0₂ or do not give out 0₂.

Q25. The negative of the statement “Rajesh or Rajni lived in Bangalore” is
(a) Rajesh lives in Bangalore and Rajni did not live in Bangalore
(b) Rajesh did not live in Bangalore and Rajni did not live in Bangalore
(c) Rajesh did not live in Bangalore or Rajni did not live in Bangalore
Sol: (c) We have, p: Rajesh or Rajni lived in Bangalore.
and                q: Rajesh lived in Bangalore.
r: Rajni lived in Bangalore.
~q: Rajesh did not live in Bangalore.
~r. Rajni did not live in Bangalore.
~ (q v r): Rajesh did not live in Bangalore and Rajni did not live in Bangalore.

Q26. The negation of the statement “101 is not a multiple of 3” is
(a) 101 is a multiple of 3
(b) 101 is a multiple of 2
(c) 101 is an odd number                     
(d) 101 is an even number
Sol: (a) Let p: 101 is not a multiple of 3.
~p: 101 is a multiple of 3,

Q27. The contra positive of the statement
“If 7 is greater than 5, then 8 is greater than 6” is
(a) If 8 is greater than 6, then 7 is greater than 5
(b) If 8 is not greater than 6, then 7 is greater than 5
(c) If 8 is not greater than 6, then 7 is not greater than 5
(d) If 8 is greater than 6, then 7 is not greater than 5
Sol: (c) Letp: 7 is greater than 5.
and q: 8 is greater than 6.
∴P→ q
~p: 7 is not greater than 5.
~q: 8 is not greater than 6.
(~q) → (~p) i.e., if 8 is not greater than 6, then 7 is not greater than 5.

Q28. The converse of the statement “If x > y, then x + a > y + a” is
(a)    If x <y, then x + a <y + a      
(b) If x + a >y + a, then x>y
(c)    If x <y, then x+ a <y + a        
(d) If x >y, then x + a<y+ a
Sol: (b) Let p: x >y
q:x + a>y + a
P ⟶ q
Converse of the above statement is:
q⟶P
i.e., If x + a > y + a, then x>y

Q29. The converse of the statement “If sun is not shining, then sky is filled with clouds” is
(a) If sky is filled with clouds, then the Sun is not shining
(b) If Sun is shining, then sky is filled with clouds
(c) If sky is clear, then Sun is shining
(d) If Sun is not shining, then sky is not filled with clouds
Sol: (a) Let p: Sun is not shining.
and q:Sky is filled with clouds.
Converse of the above statement p → q is q → p.
If sky is filled with clouds, then the Sun is not shining.

Q30. The contra positive of the statement “If p, then q” is
(a) if q, then p                                        
(b) if p, then ~q
(c) if ~q, then ~p                                  
(d) if ~p, then ~q
Sol:(c) p → q                                           ‘
If p, then q
Contra positive of the statement p → q is (~q) →(~ p).
If ~q, then ~p.

Q31. The statement “If x² is not even, then x is not even” is converse of the statement
(a) If x² is odd, then x is even
(b) If x is not even, then x² is not even
(c) If x is even, then x² is even
(d) If x is odd, then x² is even
Sol: (b) Let p: x² is not even.
and q: x is not even.
Converse of the statement p →q is q → p. i.e.,
If x is not even, then x² is not even.

Q32. The contra positive of statement ‘If Chandigarh is capital of Punjab, then Chandigarh is in India’ is
(a) If Chandigarh is not in India, then Chandigarh is not the capital of Punjab
(b) If Chandigarh is in India, then Chandigarh is Capital of Punjab
(c) If Chandigarh is not capital of Punjab, then Chandigarh is not capital of India
(d) If Chandigarh is capital of Punjab, then Chandigarh is not is India
Sol: (a) Let p: Chandigarh is capital of Punjab.
and q: Chandigarh is in India.
~ p: Chandigarh is not capital of Punjab.
~q: Chandigarh is not in India.
Contra positive of the statement p → q
if (~q), then (~p).
It Chandigarh is not in India, then Chandigarh is not the capital of Punjab.

Q33. Which of the following is the conditional p → q?
(a) q is sufficient for p                         
(b) p is necessary for q
(c) p only if q                                         
(d) if q then p
Sol: (c) ‘p → q is same as ‘p only if q’.

Q34. The negation of the statement “The product of 3 and 4 is 9” is
(a) it is false that the product of 3 and 4 is 9
(b) the product of 3 and 4 is 12
(c) the product of 3 and 4 is not 12
(d) it is false that the product of 3 and 4 is not 9
Sol: (a) The negation of the above statement is ‘It is false that the product of 3 and 4 is 9’.

Q35. Which of the following is not a negation of “A nature number is greater than zero”
(a) A natural number is not greater than zero
(b) It is false that a natural number is greater than zero
(c) It is false that a natural number is not greater than zero
(d) None of the above
Sol: (c) The false negation of the given statement is “It is false that a natural number is not greater than zero”.

Q36. Which of the following statement is a conjunction?
(a) Ram and Shyam are friends
(b) Both Ram and Shyam are tall
(c) Both Ram and Shyam are enemies
(d) None of the above
Sol: (d) If two simple statements p and q are connected by the word ‘and’, then the resulting compound statement p and q is called a conjunction ofp and q. Here, none of the given statement is conjunction.

Q37. State whether the following sentences are statements or not.
(i) The angles opposite to equal sides of a triangle are equal.
(ii) The moon is a satellites of Earth.
(iii) May God bless you.
(iv) Asia is a continent.
(v) How are you? ,
Sol: (i) It is a statement.
(ii) It is a statement,
(iii) It is not a statement, since it is an exclamations.
(iv) It is a statement.
(v) It is not a statement, since it is a question.

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NCERT Exemplar Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry. https://www.cbselabs.com/ncert-exemplar-problems-class-11-mathematics-chapter-12-introduction-three-dimensional-geometry/

NCERT Exemplar Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

Short Answer Type Questions

Q1. Locate the following points:

Q1. Locate the following points:

(i) (1,-1, 3),
(ii) (-1,2,4)               
(iii) (-2, -4, -7)
(iv) (-4,2, -5)
Sol: Given, coordinates
(i) (1,-1, 3),
(ii) (-1,2,4)
(iii) (-2, -4, -7)
(iv) (-4,2, -5)

Q2. Name the octant in which each of the following points lies.
(i) (1,2,3)             
(ii) (4,-2, 3)             
(iii) (4,-2,-5)              
(iv)(4,2,-5)
(v) (-4,2,5)           
(vi) (-3,-1,6)           
(vii) (2,-4,-7)
(viii) (-4, 2,-5)

Q3. Let A, B, C be the feet of perpendiculars from a point P on the x, y,z-axes respectively. Find the coordinates of A, B and C in each of the following where the point P is:
(i) (3,4,2)             
(ii) (-5,3,7)              
(iii) (4,-3,-5)
Sol: We know that, on x-axis, y, z = 0, on y-axis, x, z = 0 and on z-axis, x,y = 0. Thus, the feet of perpendiculars from given point P on the axis are as follows.

(i) A(3,0,0),5(0,4,0),C(0,0,2)
(ii) A(-5, 0, 0), B(0, 3, 0), C(0, 0, 7)
(iii) A(4, 0, 0), 5(0, -3, 0), C(0,0, -5)

Q4. Let A, B, C be the feet of perpendiculars from a point P on the xy, yz and zx- planes respectively. Find the coordinates of A, B, C in each of the following where the point P is
(i) (3,4,5)
(ii) (-5,3,7)
(iii) (4,-3,-5).
Sol: We know that, on xy-plane z = 0, on yz-plane, x = 0 and on zx-plane, y = 0. Thus, the coordinates of feet of perpendicular on the xy, yz and zx-planes from the given point are as follows:
(i) A(3,4,0), 5(0,4, 5), C(3,0,5)
(ii) A(-5, 3,0), 5(0, 3, 7), C(-5, 0, 7)
(iii) A(4, -3, 0), 5(0, -3, -5), C(4,0, -5)

Q5. How far apart are the points (2,0, 0) and (-3, 0, 0)?
Sol: Given points are A (2, 0, 0) and 5(-3,0, 0).
AB = |2 – (-3)| = 5

Q6. Find the distance from the origin to (6, 6, 7).

Q8. Show that the point ,4(1, -1, 3), 6(2, -4, 5) and (5, -13, 11) are collinear.

Q9. Three consecutive vertices of a parallelogram ABCD are .4(6, -2,4), 6(2,4, -8), C(-2, 2, 4). Find the coordinates of the fourth vertex.

Q10 .Show that the triangle ABC with vertices .4(0,4,1), 6(2,3, -1) and C(4, 5,0) is right angled.
Sol: The vertices of ∆ABC are A(0,4, 1), 5(2, 3, -1) and C(4, 5, 0).

Q11. Find the third vertex of triangle whose centroid is origin and two vertices are (2,4,6) and (0, -2, -5).
Sol: Let the third or unknown vertex of ∆ABC be A(x, y, z).
Other vertices of triangle are 5(2,4, 6) and C(0, -2, -5).
The centroid is G(0, 0, 0).

Q12. Find the centroid of a triangle, the mid-point of whose sides are D (1,2, – 3), E(3,0, l)and F(-l, 1,-4).
Sol: Given that, mid-points of sides of AABC are D(l, 2, -3), E(3, 0, 1) and F(-l, 1,-4).

Q14. Three vertices of a Parallelogram ABCD are A(\, 2, 3), B(-A, -2, -1) and C(2, 3, 2). Find the fourth vertex
Sol: Let the fourth vertex of the parallelogram D(x, y, z).
Mid-point of BD

Q15. Find the coordinate of the points which trisect the line segment joining the points .A(2, 1, -3) and B(5, -8, 3).

Q16. If the origin is the centroid of a triangle ABC having vertices A(a, 1, 3), B(-2, b, -5) and C(4, 7, c), find the values of a, b, c.
Sol: Vertices of AABC are A(a, 1, 3), B(-2, b, -5) and C(4, 7, c).
Also, the centroid is G(0, 0, 0).

Q17. Let A(2, 2, -3), 5(5, 6, 9) and C(2, 7, 9) be the vertices of a triangle. The internal bisector of the angle A meets BC at the point Find the coordinates of D.

Long Answer Type Questions

Q18. Show that the three points A(2, 3, 4), 5(-l, 2, -3) and C(-4, 1, -10) are collinear and find the ratio in which Cdivides
Sol: Given points are A(2, 3, 4), B(-1, 2, -3) and C(-4,1,-10)

Q19. The mid-point of the sides of a triangle are (1, 5, -1), (0,4, -2) and (2, 3,4). Find its vertices. Also, find the centroid of the triangle.

Q20. Prove that the points (0, -1, -7), (2, 1, -9) and (6, 5, -13) are collinear. Find the ratio in which the first point divides the join of the other two.
Sol: Given points are 4(0, -1, -7), 8(2, 1, -9) and C(6, 5, -13).

Q21. What are the coordinates of the vertices of a cube whose edge is 2 units, one of whose vertices coincides with the origin and the three edges passing through the origin, coincides with the positive direction of the axes through the origin?
Sol: The coordinate of the cube whose edge is 2 units, are:
(2, 0, 0), (2,2, 0), (0, 2, 0), (0, 2,2), (0, 0,2), (2,0, 2), (0, 0, 0) and (2,2, 2)

Objective Type Questions

Q22. The distance of point P(3,4, 5) from the yz-plane is
(a) 3 units
(b) 4 units
(c) 5 units
(d) 550
Sol: (a) Given point is P{3,4, 5).
Distance of P from yz-plane = |x coordinate of P| = 3

Q23. What is the length of foot of perpendicular drawn from the point P(3,4, 5) on y-axis?

Q24. Distance of the point (3,4, 5) from the origin (0, 0, 0) is

Q25. If the distance between the points (a,0,1) and (0,1,2) is √27, then the value of a is
(a)     5                      
(b)     ± 5                   
(c)     -5                    
(d)   none of these

Q26. x-axis is the intersection of two planes
(a) xy and xz                                            
(b) yz and zx
(c) xy and yz                                            
(d) none of these
Sol: (a) We know that, on the xy and xz-planes, the line of intersection is x-axis.

Q27. Equation of Y-axis is considered as
(a) x = 0, y = 0                                         
(b) y = 0, z = 0
(c) z = 0, x = 0                                         
(d) none of these
Sol:(c) On the j-axis, x = 0 and z = 0.

Q28. The point (-2, -3, -4) lies in the
(a) First octant                                        
(b) Seventh octant
(c) Second octant                                   
(d) Eighth octant
Sol: (b) The point (-2, -3, -4) lies in seventh octant.

Q29. A plane is parallel to yz-plane so it is perpendicular to
(a) x-axis               
(b) y-axis                 
(c) z-axis                 
(d) none of these
Sol: (a) A plane parallel to yz-plane is perpendicular to x-axis.

Q30. The locus of a point for which y = 0, z = 0 is
(a)    equation of x-axis                         
(b)    equation of y-axis
(c)     equation at z-axis                         
(d)    none of these
Sol: (a) We know that, equation of the x-axis is: y = 0, z = 0 So, the locus of the point is equation of x-axis.

Q31. The locus of a point for which x = 0 is
(a)    xy-plane                                          
(b)    yz-plane
(c)     zx-plane                                        
 (d)    none of these
Sol: (b) On the yz-plane, x = 0, hence the locus of the point is yz-plane.

Q32. If a parallelepiped is formed by planes drawn through the points (5,8,10) and (3, 6, 8) parallel to the coordinate planes, then the length of diagonal of the parallelepiped is

Q33. L is the foot of the perpendicular drawn from a point P(3, 4, 5) on the xy-plane. The coordinates of point L are
(a)    (3,0,0)                                              
(b)    (0,4,5)
(c)     (3, 0, 5)                                            
(d)    none of these
Sol: (d) We know that on the xy-plane, z = 0.
Hence, the coordinates of the points L are (3,4, 0).

Q34. L is the foot of the perpendicular drawn from a point (3, 4, 5) on x-axis. The coordinates of L are
(a)    (3,0,0)                                              
(b)    (0,4,0)
(c)     (0, 0, 5)                                            
(d)    none of these
Sol: (a) On the x-axis, y = 0 and z = 0.
Hence, the required coordinates are (3, 0,0).

Fill in the Blanks Type Questions
Q35. The three axes OX, OY, OZ determine______ .
Sol: The three axes OX, OY and OZ determine three coordinate planes.

Q36. The three planes determine a rectangular parallelepiped which has____ of rectangular faces.

Q37. The coordinates of a point are the perpendicular distance from the _____ on the respective axes.
Sol: Given points

Q38. The three coordinate planes divide the space into _________parts.
Sol: Eight

Q39. If a point P lies in yz-plane, then the coordinates of a point on yz-plane is of the form_______.
Sol: We know that, on yz-plane, x = 0.So, the coordinates of the required point are (0, y, z).

Q40. The equation of yz-plane is ______ .
Sol: On yz-plane for any point x-coordinate is zero.
So, yz-plane is locus of point such that x = 0, which is its equation.

Q41. If the point P lies on z-axis, then coordinates of P are of the form_____.
Sol: On the z-axis, x = 0 and y = 0.
So, the required coordinates are of the form (0, 0, z).

Q42. The equation of z-axis, are ______.
Sol: Any point on the z-axis is taken as (0, 0, z).
So, for any point on z-axis, we have x = 0 and y = 0, which together represents its equation.
Q43. A line is parallel to xy-plane if all the points on the line have equal_________.
Sol: A line is parallel to xy-plane if each point P(x, y, z) on it is at same distance from xy-plane.
Distance of point P from xy plane is ‘z’
So, line is parallel to xy-plane if all the points on the line have equal z-coordinate.

Q44. A line is parallel to x-axis if all the points on the line have equal ______.
Sol: A line is parallel to x-axis if each point on it maintains constant distance from y-axis and z-axis.
So, each point has equal y and z-coordinates. .

Q45. x = a represents a plane parallel to .
Sol: Locus of point P(x, y, z) is x = a.
Therefore, each point P has constant x-coordinate.
Now, x is distance of point P from yz-plane.
So, here plane x = a is at constant distance ‘a’ from yz-plane and parallel to _yz-plane.

Q46. The plane parallel to yz-plane is perpendicular to_____ .
Sol: The plane parallel to yz-plane is perpendicular to x-axis.

Q47. The length of the longest piece of a string that can be stretched straight in a  rectangular room whose dimensions are 10, 13 and 8 units are______ .
Sol: Given dimensions are: a = 10, 6=13 andc = 8.
Required length of the string = yja² + b² + c² = ^100 + 169 + 64 = -7333

Q48. If the distance between the points (a, 2,1) and (1,-1,1) is 5, then a_______ .
Sol: Given points are (a, 2,1) and (1,-1,1).

Q49. If the mid-points of the sides of a triangle AB; BC; CA are D(l, 2, – 3), E( 3, 0, 1) and F(-l, 1, -4), then the centroid of the triangle ABC is________ .
Sol: Given that, mid-points of sides of AABC are D( 1, 2, -3), E(3, 0, 1) and F(-l, 1,-4).

Matching Column Type Questions

Q50. Match each item given under the column C₁ to its correct answer given under column C₂.

Column C,		Column C2
(a)	In xy-plane	(i)	1st octant
(b)	Point (2, 3,4) lies in the	(ii)	vz-plane
(c)	Locus of the points having x coordinate 0 is	(iii)	z-coordinate is zero
(d)	A line is parallel to x-axis if and only	(iv)	z-axis                      .
(e)	If x = 0, y = 0 taken together will represent the	(v)	plane parallel to xy-plane
(f)	z = c represent the plane	(vi)	if all the points on the line have equal y and z-coordinates.
(g)	Planes x = a, y = b represent the line	(vii)	from the point on the respective axis.
00	Coordinates of a point are the distances from the origin to the feet of perpendiculars	(viii)	parallel to z-axis
(i)	A ball is the solid region in the space	(ix)	disc
G)	Region in the plane enclosed by a circle is known as a	00	sphere

 

Sol: (a) In xy-plane, z-coordinate is zero.
(b) The point (2, 3,4) lies in 1st octant.
(c) Locus of the points having x-coordinate zero is yz-plane.
(d) A line is parallel to x-axis if and only if all the points on the line have equal y and z-coordinates.
(e)x = 0, y = 0 represent z-axis
(f) z = c represents the plane parallel to xy-plane.

(g) The plane x = a is parallel to yz-plane.
Plane y = b is parallel to xz-plane.
So,    planes x = a and y = b is line of intersection of these planes.
Now, line of intersection of yz-plane and xz-plane is z-axis.
So, line of intersection of planes x = a andy = b is line parallel to z-axis.
(h) Coordinates of a point are the distances from the origin to the feet of perpendicular from the point on the respective axis.
(i) A ball is the solid region in the space enclosed by a sphere.
(j) The region in the plane enclosed by a circle is known as a disc.
Hence, the correct matches are:
(a) – (iii), (b) – (i), (c) – (ii), (d) – (vi), (e) – (iv),
(f) – (v), (g) – (viii), (h) – (vii), (i) – (x), (j) – (ix),

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NCERT Exemplar Class 11 Maths Chapter 11 Conic Sections of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 11 Conic Sections.

NCERT Exemplar Class 11 Maths Chapter 11 Conic Sections

Short Answer Type Questions

Conic Sections Class 11 Important Questions NCERT

Q1. Find the equation of the circle which touches the both axes in first quadrant and whose radius is a.
Sol:

Q2. Show that the point (x, y) given by  \(x=\frac { 2at }{ 1+{ t }^{ 2 } }   \) and \(y=\frac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } }  \) lies on a circle .

NCERT Exemplar Class 11 Maths Conic Sections

Q3. If a circle passes through the point (0, 0) (a, 0), (0, b) then find the coordinates of its centre.
Sol: We have circle through the point A(0, 0), B(a, 0) and C(0, b).
Clearly triangle is right angled at vertex A.

So, centre of the circle is the mid point of hypotenuse BC which is (a/2, b/2)

Conic Sections Class 11 Exemplar NCERT

Q4. Find the equation of the circle which touches x-axis and whose centre is (1,2).
Sol: Given that, circle with centre (1,2) touches x-axis.
Radius of the circle is, r = 2
So, the equation of the required circle is:
(x – l)² + (y – 2)² = 2²
=>x²-2x + 1 + y²-4y + 4 = 4
=> x² + y² – 2x-4y + 1 = 0

Q5. If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle.
Sol: Given lines are 6x – 8y + 8 = 0 and 6x – 8y – 7 = 0.
These parallel lines are tangent to a circle.

Q6. Find the equation of a circle which touches both the axes and the line 3x – 4y + 8 = 0 and lies in the third quadrant.

NCERT Exemplar Class 11 Conic Sections Solutions

Q7. If one end of a diameter of the circle x² + y² -4x -6y + 11 = 0 is (3,4), then find the coordinate of the other end of the diameter.
Sol: Given equation of the circle is:

Q8. Find the equation of the circle having (1, -2) as its centre and passing through 3x +y= 14, 2x + 5y = 18.

Important Questions Of Conic Sections Class 11

Q9. If the line y= √3 x + k touches the circle x² + y² = 16, then find the value of
Sol: Given line is y = √3 x + k and the circle is x² + y² = 16.

Q10. Find the equation of a circle concentric with the circle x² +y² – 6x + 12y + 15 = 0 and has double of its area.

Conic Sections Class 11 Extra Questions NCERT

Q11. If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.

Q12. Given the ellipse with equation 9X² + 25y² = 225, find the eccentricity and foci.

NCERT Exemplar Class 11 Maths Chapter 11 Solutions

Q13. If the eccentricity of an ellipse is 5/8 and the distance between its foci is 10, then find latus rectum of the ellipse.

Class 11 Conic Sections Extra Questions NCERT

Q14. Find the equation of ellipse whose eccentricity is 2/3, latus rectum is 5 and thecentre is (0, 0).

Class 11 Maths Chapter 11 Extra Questions NCERT

Q15. Find the distance between the directrices of the ellipse \(\frac { { x }^{ 2 } }{ 36 } +\quad \frac { { y }^{ 2 } }{ 20 } \quad =\quad 1  \)

Q16. Find the coordinates of a point on the parabola y² = 8x whose focal distance is 4.

Q17. Find the length of the line-segment joining the vertex of the parabola y² = 4ax and a point on the parabola where the line-segment makes an angle 6 to the x-axis.
Sol: Given equation of the parabola isy² = 4ax.
Let the point on the parabola be P(x₁,,y₁).

Q18. If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.

Q19. If the line y = mx + 1 is tangent to the parabola y² = 4x then find the value of m.

Q20. If the distance between the foci of a hyperbola is 16 and its eccentricity is √2, then obtain the equation of the hyperbola.

Q21. Find the eccentricity of the hyperbola 9y² – 4x² =36
Sol: We have the hyperbola:9y² – 4x² = 36

Q22. Find the equation of the hyperbola with eccentricity 3/2 and foci at (±2, 0).

Long Answer Type Questions

Q23. If the lines 2x – 3y = 5 and 3x-4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.
Sol: Given that lines 2x – 3y – 5 = 0 and 3x – 4y -1 = 0 are diameters of the circle. Solving these lines we get point of intersection as (1, -1), which is centre of the circle.

Q24. Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y – 4x + 3 = 0.

Q25. Find the equation of a circle whose centre is (3, -1) and which cuts off a chord of length 6 units on the line 2x — 5y+ 18 = 0.

Sol: Given centre of the circle 0(3, -1)
Chord of the circle is AB.

Q26. Find the equation of a circle of radius 5 which is touching another circle x² + y² – 2x – 4y – 20 = 0 at (5, 5).

Q27. Find the equation of a circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line y = x -1.
Sol: Given that circle passes through the point A(7, 3) and its radius is 3.

Q28. Find the equation of each of the following parabolas.
(i) Directrix, x = 0, focus at (6, 0)
(ii) Vertex at (0,4), focus at (0, 2)
(iii) Focus at (-1, -2), directrix x – 2y + 3 = 0
Sol: We know that the distance of any point on the parabola from its focus and its directrix is same.
(i) Given that, directrix, x = 0 and focus = (6, 0)
So, for any point P(x, y) on the parabola
Distance of P from directrix = Distance of P from focus =>  x² = (x — 6)² + y²
=>         y²– 12x + 36 = 0
(ii) Given that, vertex = (0,4) and focus = (0, 2)
Now distance between the vertex and directrix is same as the distance between the vertex and focus.
Directrix is y – 6 = 0
For any point of P(x, y) on the parabola
Distance of P from directrix = Distance of P from focus

Q29. Find the equation of the set of all points the sum of whose distances from the points (3, 0) and (9, 0) is 12.

Q30. Find the equation of the set of all points whose distance from (0,4) are 2/3 of their distance from the line y = 9.

Q31. Show that the set of all points such that the difference of their distances from (4, 0)and (-4, 0) is always equal to 2 represent a hyperbola.

True/False Type Questions

Q33. The line x + 3y = 0 is a diameter of the circle x² + y² + 6x + 2y = 0.
Sol: False
Given equation of the circle is x² + y² + 6x + 2y = 0
Centre = (-3, -1)
Clearly, it does not lie on the line x + 3y = 0 as -3 + 3(-l) = -6.
So, this line is not diameter of the circle.

Q34. The shortest distance from the point (2, -7) to the circle x +y² – 1 4jc – lOy- 151 = 0 is equal to 5.
Sol: False

Q35. If the line lx + my = 1 is a tangent to the circle x² + y² = a², then the point (1, m) lies on a circle.

Q36. The point (1,2) lies inside the circle x² + y² – 2x + 6y + 1 = 0.
Sol: False

Q37. The line lx+ my + n = 0 will touch the parabola^² = 4 ax if In = am².
Sol: True




Fill in the Blanks Type Questions

Q41. The equation of the circle having centre at (3, -4) and touching the line 5x + 12y- 12 = 0 is ______.

Q42. The equation of the circle circumscribing the triangle whose sides are the lines y = x + 2, 3y = 4x, 2y = 3x is   _______.
Given equation of line are:

Q43. An ellipse is described by using an endless string which is passed over two pins. If the axes are 6 cm and 4 cm, the length of the string and distance between the pins are _____ .

Q44. The equation of the ellipse having foci (0,1), (0, -1) and minor axis of length 1 is ___ .

Q45. The equation of the parabola having focus at (-1, -2) and the directrix x – 2y + 3 = 0 is______ .
Sol: Given that, focus at S(-l, -2) and directrix is x – 2y + 3 = 0

Q46. The equation of the hyperbola with vertices at (0, ±6) and eccentricity 5/3 ________ and its foci are _____    .

Objective Type Questions

Q47. The area of the circle centred at (1,2) and passing through (4, 6) is

Q48. Equation of a circle which passes through (3, 6) and touches the axes is

Q49. Equation of the circle with centre on the j-axis and passing through the origin and the point (2, 3) is

Q50. The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is

Q51. If the focus of a parabola is (0, -3) and its directrix is y = 3, then its equation is

Q52. If the parabola y² = 4ax passes through the point (3, 2), then the length of its latus rectum is

Q53. If the vertex of the parabola is the point (-3, 0) and the directrix is the line x + 5 = 0, then its equation is
(a) y² = 8(x + 3)
(b) x² = 8(y + 3)
(c) y² = -8(x + 3)
(d) y² = 8(x + 5)

Q54. The equation of the ellipse whose focus is (1, -1), the directrix the line x-y-3 = 0 and eccentricity 1/2 is

Q55. The length of the latus rectum of the ellipse 3x² +y² = 12 is
(a) 4       
(b) 3       
(c) 8       
(d) 4/√3

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NCERT Exemplar Class 11 Maths Chapter 10 Straight Lines are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 10 Straight Lines.

NCERT Exemplar Class 11 Maths Chapter 10 Straight Lines

Short Answer Type Questions

Q1. Find the equation of the straight line which passes through the point (1, -2)

Q2. Find the equation of the line passing through the point (5,2) and perpendicular to the line joining the points (2, 3) and (3, -1)

Q3. Find the angle between the lines y = (2 -√3) (x + 5) and y = (2 + -√3) {x – 7)
Sol: Slope of the line = (2 -√3)(x + 5) is: m_l = (2 -√3 )

Q4. Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.

Q5. Find the points on the line x+y = 4 which lie at a unit distance from the line 4x + 3y= 10
Sol. Let the required point be (h, k) lies on the line x + y = 4

Q7. Find the equation of lines passing through (1,2) and making angle 30° with y-axis.

Q8. Find the equation of the line passing through the point of intersection of lx + y = 5 and x + 3 y +8 = 0 and parallel to the line 3x + 4y = 1.

Q9. For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes.
Sol: Given line is:

Q10. If the intercept of a line between the coordinate axes is divided by the point (-5,4) in the ratio 1 : 2, then find the equation of the line.
Sol: Let the line through the point P(-5, 4) meets axis at A(h, 0) and B(0, k)

According to the question, we have AP: BP =1:2


Q11. Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of x-axis.
Sol: Given that the line makes and angle 120° with positive direction of x-axis.

Q12. Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x + Ay = 4 and the opposite vertex of the hypotenuse is (2, 2).

Long Answer Type Questions

Q13. If the equation of the base of an equilateral triangle is x + y – 2 and the vertex is (2, -1), then find the length of the side of the triangle.

Q14. A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2) and (1, 1) on the line is zero. Find the coordinates of the point P

Q15. In what direction should a line be drawn through the point (1, 2) so that its point of intersection with the line x+y = 4 is at a distance √6/3 from the given point.

Q16. A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point.

Q17. Find the equation of the line which passes through the point (-4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point.
Sol: Let the line through the point P(-A, 3) meets axis at A(h, 0) and 0(0, k)
Now according to the question AP : BP =5:3

Q18. Find the equations of the lines through the point of intersection of the lines x-y+ 1=0 and 2x – 3y + 5 = 0 and whose distance from the point (3, 2) is 7/5

Q19. If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point.

Q20. P_l, P₂ are points on either of the two lines y — √3 |x| = 2 at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from P₁, P₂ on the bisector of the angle between the given lines.


Q21. If p is the length of perpendicular from the origin on the line \(\frac { x }{ a } +\quad \frac { y }{ b }    \)  and a²,p² and are in the A.P , then show that a⁴+b⁴ = 0

Objective Type Questions

Q22. A line cutting off intercept -3 from the y-axis and the tangent of angle to the x-axis is 3/5, its equation is

Q23. Slope of a line which cuts off intercepts of equal lengths on the axes is
(a) -1
(b) 0       
(c) 2       
(d) √3
Sol: (a) Equation of the according to the question is \(\frac { x }{ a } +\quad \frac { y }{ a }     \)
=> x+y = a
Required slope = -1

Q24. The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x is
(a) x-y = 5           
(b) x+y = 5
(c)x+y=l
(d)x-y=1
Sol:(b) Slope of the given line y = x is 1.
Thus, slope of line perpendicular to y = x is -1.
Line passes through the point (3, 2).
So, equation of the required line is:y-2=-l (x – 3) => x + y = 5

Q25. The equation of the line passing through the point (1,2) and perpendicular to the line x +y + 1 = 0 is
(a) y-x +1=0           
(b) y — x—1=0
(c) y-x + 2 = 0                                      
(d) y — x — 2=0
Sol: (b) Slope of the given line +1=0 is-1.
So, slope of line perpendicular to above line is 1.
Line passes through the point (1,2).
Therefore, equation of the required linens:
y-2 = 1(x- 1) => y-x-1=0.

Q26. The tangent of angle between the lines whose intercepts on the axes are a, -b and b, -a, respectively, is

Q27. If the line \(\frac { x }{ a } +\frac { y }{ b }  \)  passes through the points (2, -3) and (4, -5), then (a, b) a b is
(a) (1,1)                  
(b) (-1,1)                  
(c) (1,-1)                  
(d) (-1,-1)

Q28. The distance of the point of intersection of the lines 2x – 3y + 5 = 0 and 3x + 4y = 0 from the line 5x – 2y = 0 is

Q29. The equations of the lines which pass through the point (3, -2) and are inclined at 60° to the line √3 x + y = 1 is

Q30. The equations of the lines passing through the point (1,0) and at a distance √3/2 from the origin, are

Q33. If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is (3, 2), then the equation of the line will be
(a) 2x + 3y = 12                                     
(b) 3x + 2y=l2
(c) 4x-3y = 6                                        
(d) 5x- 2y=10
Sol:

Q34. Equation of the line passing through (1,2) and parallel to the line y = 3x – 1 is
(a)y + 2=x+l                                    
(b) y + 2 = 3(x + 1)
(c) y -2 =  3(x — 1)                             
(d) y-2=x-l
Sol: (c) Line is parallel to the line y = 3x – 1.
So, slope of the line is‘3’.
Also, line passes through the point (1,2).
So, equation of the line is: y – 2 = 3(x – 1)

Q35. Equations of diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y = 1 are

Q37. The point (4, 1) undergoes the following two successive transformations:
(i) Reflection about the line y = x
(ii) Translation through a distance 2 units along the positive x-axis Then the final coordinates of the point are
(a)  (4,3)               
(b) (3,4)                
(c)  (1,4)              
(d) (7/2,7/2)
Sol: (b) Reflection of A (4, 1) in y = x is 5(1,4).
Now translation of point B through a distance ‘2’ units along the positive x-axis shifts B to C( 1 + 2,4) or C(3,4).

Q38. A point equidistant from the lines 4x + 3y+ 10 = 0, 5x – 12y + 26 = 0 and lx + 24y – 50 = 0 is
(a)    (1,-1)              
(b)    (1, 1)               
(c)    (0,0)              
(d)   (0, 1)
Sol: (c) Clearly distance of each of three lines from (0, 0) is 2 units.

Q39. A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is

Q40. The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is
(a) 1:2 (b) 3:7 (c) 2:3 (d) 2:5

Q41. One vertex of the equilateral triangle with centroid at the origin and one side asx + y- 2 = 0is
(a) (-1,-1) (b) (2,2) (c) (-2,-2) (d) (2,-2)

Fill in the Blanks Type Questions

Q42. If a, b, c are in A.P., then the straight lines ax + by + c = 0 will always pass through __________    

Q43.The line which cuts off equal intercept from the axes and pass through the point (1, -2) is

Q44. Equations of the lines through the point (3, 2) and making an angle of 45° with the line x – 2y = 3 are _____

Q45. The points (3,4) and (2, -6) are situated on the_____ of the line 3x – 4y – 8= 0.

Sol: Given line is 3x – 4y – 8 = 0
For point (3, 4), 3(3) – 4(4) – 8 = -15 < 0
For point (2, -6), 3(2) – 4(—6) – 8 = 22 > 0
Hence, the points (3,4) and (2, -6) lies on opposite side of the line.

Q46. A point moves so that square of its distance from the point (3, -2) is numerically equal to its distance from the line 5x – 12y = 3. The equation of its locus is______ .
Sol:

Q47. Locus of the mid-points of the portion of the line x sin θ + y cos θ = p intercepted between the axes is ______

True/False Type Questions

Q48. If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral.

This is a contradiction to the fact that the area is a rational number. Hence, the triangle cannot be equilateral.

Q49. The points A(-2, 1), B(0, 5), C(-l, 2) are collinear.
Sol: False

Q50. Equation of the line passing through the point (a cos³ , a sin³ ) and perpendicular to the line
x sec + y cosec = a isx cos -y sin = a sin 2

Q51. The straight line 5x + 4y = 0 passes through the point of intersection of the straight lines x + 2y— 10 = 0 and 2x +y + 5 = 0.

Q53. The equation of the line joining the point (3, 5) to the point of intersection of the lines 4x +y – 1 = 0 and lx – 3v – 35 = 0 is equidistant from the points (0, 0) and (8, 34).

Sol: True

Q55. The line ax + 2y + 1 = 0, bx + 2y + 1 = 0 and cx + 4y + 1 = 0 are concurrent, if a, b and c are in GP.
Sol: False

Q56. Line joining the points (3, -4) and (-2, 6) is perpendicular to the line joining the points (-3, 6) and (9, -18).

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NCERT Exemplar  Class 11 Maths Chapter 7 Permutations and Combinations are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 7 Permutations and Combinations. https://www.cbselabs.com/ncert-exemplar-problems-class-11-mathematics-chapter-7-permutations-combinations/

NCERT Exemplar Class 11 Maths Chapter 7 Permutations and Combinations

NCERT Exemplar Class 11 Maths Permutations And Combinations

Short Answer Type Questions

Q1. Eight chairs are numbered 1 to 8. Two women and 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the total number of possible arrangements.
Sol. First the women choose the chairs from amongst the chairs numbered 1 to 4.

Permutations And Combinations Class 11 Extra Questions NCERT

Q2. If the letters of the word RACHIT are arranged in all possible ways as listed in dictionary, then what is the rank of the word RACHIT?

Sol: The alphabetical order of the letters of the word RACHIT is: A, C, H, I, R, T. Number of words beginning with A = 5!
Number of words beginning with C = 5!
Number of words beginning with H = 5!
Number of words beginning with 1 = 5!
Clearly, the first word beginning with R is RACHIT.
.•. Rank of the word RACHIT in dictionary = 4×5! + 1= 4 x120+1= 481

Permutation And Combination Class 11 Extra Questions With Answers

Q3. A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. Find the number of different ways of doing questions.

Sol: Since the candidate cannot attempt more than 5 questions from either group, he is able to attempt minimum two questions from either group.
The possible number of questions attempted from each group will be as given in the following table:

Group I	5	4	3	2
Group II	2	3	4	5

NCERT Exemplar Class 11 Maths Chapter 7 Solutions

Q4. Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the point.
Sol: There are 18 point in a plane, of which 5 points are collinear.

Q5. We wish to select 6 persons from 8 but, if the person A is chosen, then B must be chosen. In how many ways can selections be made?
Sol:

Permutation And Combination Class 11 Important Questions NCERT 

Q6. How many committee of five persons with a chairperson can be selected form 12 persons?
Sol: Total number of persons =12
Number of persons to be selected = 5

Q7. How many automobile license plates can be made, if each plate contains two different letters followed by three different digits?
Sol: There are 26 English alphabets and 10 digits (0 to 9).
It is given that each plate contains two different letters followed by three different digits.

Class 11 Maths Chapter 7 Extra Questions NCERT

Q8. A bag contains 5 black and 6 red balls. Determine the number of ways in • which 2 black and 3 red balls can be selected from the lot.

Q9. Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.

Important Questions Of Permutation And Combination Class 11 NCERT

Q10. Find the number of different words that can be formed from the letters of the word TRIANGLE, so that no vowels are together.
Sol: Given word is: TRIANGLE Consonants are: T, R, N, G, L Vowels are: I, A, E
Since we have to form words in such a way that no two vowels are together, we first arrange consonants.
Five consonants can be arranged in 5! ways.

Q11. Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated.
Sol: We have to form 4-digit numbers which are greater than 6000 and less than 7000.
We know that a number is divisible by 5, if at the unit place of the number there is 0 or 5.
So, unit digit can be filled in 2 ways.
The thousandth place can be filled by ‘6’ only.
The hundredth place and tenth place can be filled together in 8 x 7 = 56 ways. So, total number of ways = 56 x 2 = 112

Class 11 Permutations And Combinations Important Questions NCERT

Q12. Thereare 10persons named P₁,P₂,P₃,…,P_!0.Outof 10 persons, 5 persons are to be arranged in a line such that in each arrangement P, must occur whereas P₄ and P₅ do not occur. Find the number of such possible arrangements.
Sol. Given that, P₁, P₂, …, P₁₀, are 10 persons, out of which 5 persons are to be arranged but P, must occur whereas P₄ and P₅ never occurs.
As P, is already occurring w’e have to select now 4 out of 7 persons.
.•. Number of selections = ⁷C₄ = 35 Number of arrangements of 5 persons = 35 x 5! = 35 x 120 = 4200

Q13. There are 10 lamps in a hall each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.
Sol: There are 10 lamps in a hall.
The hall can be illuminated if at least one lamp is switched.
.•. Total number of ways =        ¹⁰C₁+ ¹⁰C₂ + ^l0C₃… + ¹⁰C_]0
= 2¹⁰– 1 = 1024- 1 = 1023

Permutation And Combination Class 11 Questions NCERT

Q14. A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?
Sol: There are two white, three black and four red balls.
We have to draw 3 balls, out of these 9 balls in which at least one black ball is included.
So we have following possibilities:

Black balls	1	2	3
Other than black	2	1	0

.’. Number of selections = ³C₁ x ⁶C₂ + ³C₂ x ⁶C, + ³C₃ x ⁶C₀
= 3×15+ 3×6+1= 45+ 18 + 1= 64

Q15. If ⁿC_r-1=  36 ⁿC_r = 84 and ⁿC_r+1= 126, then find the value of ^rC₂.

Permutation And Combination Questions Class 11 NCERT

Q16. Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8 and 9 where no digits are repeated.
Sol: We have to find the number of integers greater than 7000 with the digits 3,5, 7, 8 and 9.
So, with these digits, we can make maximum five-digit numbers because repetition is not allowed.
Since all the five-digit numbers are greater than 7000, we have Number of five-digit integers = 5x4x3x2x1 = 120 A four-digit integer is greater than 7000 if thousandth place has any one of 7, 8 and 9.
Thus, thousandth place can be filled in 3 ways. The remaining three places can be filled from remaining four digits in ⁴P₃ ways.
So, total number of four-digit integers = 3x ⁴P₃ = 3x4x3x2 = 72 Total number of integers = 120 + 72 = 192

 

Q17. If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other?
Sol: It is given that no two lines are parallel which means that all the lines are intersecting and no three lines are concurrent.
One point of intersection is created by two straight lines.
Number of points of intersection = Number of combinations of 20 straight lines taken two at a time

Permutation And Combination Extra Questions NCERT

Q18. In a certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64. How many telephone numbers have all six digits distinct?
Sol: If first two digit is 41, the remaining 4 digits can be arranged in ⁸P₄ = 8 x 7 x 6×5 = 1680 ways.
Similarly, if first two digit is 42, 46, 62, or 64, the remaining 4 digits can be arranged in ⁸P₄ ways i.e., 1680 ways.
.’. Total number of telephone numbers having all six digits distinct = 5x 1680 = 8400

Q19. In an examination, a student has to answer 4 questions out of 5 questions, questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.
Sol: It is given that 2 questions are compulsory out of 5 questions.
So, the other 2 questions can be selected from the remaining 3 questions in ³C₂ = 3 ways.

Q20. A convex polygon has 44 diagonals. Find the number of its sides.
[Hint: Polygon of n sides has (ⁿC₂ – n) number of diagonals.]
Sol: Let the convex polygon has n sides.
Number of diagonals=Number of ways of selecting two vertices – Number of sides = ⁿC₂ – n
It is given that polygon has 44 diagonals.

Long Answer Type Questions
Q21. 18 mice were placed in two experimental groups and one control group with all groups equally large. In how many ways can the mice be placed into three groups?
Sol: It is given that 18 mice were placed equally in two experimental groups and one control group i.e., three groups.
Each group is of 6 mice.

Q22. A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag, if (i) they can be of any colour (ii) two must be white and two red and (iii) they must all be of the same colour.
Sol:Total number of marbles = 6 white +- 5 red = 11 marbles

(a) If they can be of any colour means we have to select 4 marbles out of 11
∴ Required number of ways = ¹¹C₄
(b) Two white marbles can be selected in ⁶C₂
Two red marbles can be selected in ⁵C₂ ways.
∴ Total number of ways = ⁶C₂ x ⁵C₂ = 15 x 10 = 150
(c) If they all must be of same colour,
Four white marbles out of 6 can be selected in ⁶C₄ ways.
And 4 red marbles out of 5 can be selected in ⁵C₄ ways.
∴ Required number of ways = ⁶C₄ + ⁵C₄ = 15 + 5 = 20

Q23. In how many ways can a football team of 11 players be selected from 16 players? How many of them will

• include 2 particular players?
• exclude 2 particular players?

Sol: Total number of players = 16
We have to select a team of 11 players
So, number of ways = ¹⁶C₁₁
(i) If two particular players are included then more 9 players can be selected from remaining 14 players in ¹⁴C₉
(ii) If two particular players are excluded then all 11 players can be selected from remaining 14 players in ¹⁴C₁₁
Q24.  sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII. If there are 20 students in each of these classes, in how many ways can the team be constituted?
Sol: Total number of students in each class = 20
We have to select at least 5 students from each class.
So we can select either 5 students from class XI and 6 students from class XII or 6 students from class XI and 5 students from class XII.
∴ Total number of ways of selecting a team of 11 players = ²⁰C₅ x ²⁰C₆ + ²⁰C₆ x ²⁰C₅ = 2 x ²⁰C₅ x ²⁰C₆

Q25. A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected, if the team has
(i) no girls
(ii) at least one boy and one girl
(iii) at least three girls
Sol: Number of girls = 4;
Number of boys = 7
We have to select a team of 5 members provided that

Objective Type Questions
Q26. If ⁿC₁₂ = ⁿC₈, then n is equal to 
(a) 20     
(b) 12
(c) 6
(d) 30

Q27. The number of possible outcomes when a coin is tossed 6 times is
(a) 36                    
(b) 64                       
(c) 12                       
(d) 32
Sol: (b) Number of outcomes when a coin tossed = 2 (Head or Tail)
∴Total possible outcomes when a coin tossed 6 times = 2x2x2x2x2x 2 = 64

Q28. The number of different four-digit numbers that can be formed with the digits
2, 3, 4, 7 and using each digit only once is
(a) 120                  
(b) 96                       
(c) 24                       
(d) 100
Sol: (c) Given digits 2,3,4 and 7, we have to form four-digit numbers using these digits.
∴Required number of ways = ⁴P₄ = 4!=4x3x2x1 = 24

Q29. The sum of the digits in unit place of all the numbers formed with the help of 3,4, 5 and 6 taken all at a time is
(a)    432               
(b)    108                  
(c)      36                   
(c)    18
Sol: (b) If the unit place is ‘3’ then remaining three places can be filled in 3! ways.
Thus ‘3’ appears in unit place in 3! times.
Similarly each digit appear in unit place 3! times.
So,    sum of digits in unit place = 3!(3 + 4 + 5 + 6) = 18 x 6 = 108

Q30. The total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is
(a) 60                 
(b) 120                  
(c) 7200              
(d)  720
Sol: (c) Given number of vowels = 4 and number of consonants = 5 We have to form words by 2 vowels and 3 consonants.
So, lets first select 2 vowels and 3 consonants.
Number of ways of selection = ⁴C₂ x ⁵C₃ = 6 x 10 = 60 Now, these letters can be arranged in 5! ways.
So, total number of words = 60 x 5! = 60 x 120 = 7200

 

Q31. A five-digit number divisible by 3 is to be formed using the numbers 0, 1,2,4, and 5 without repetitions. The total number of ways this can be done is
(a)  216               
(b)  600                  
(c)  240                 
(d) 3125
[Hint: 5 digit numbers can be formed using digits 0, 1, 2, 4, 5 or by using digits 1, 2, 3, 4, 5 since sum of digits in these cases is divisible by 3.]

Sol:(a) We know that a number is divisible by 3 if the sum of its digits is divisible by 3.
Now sum of the given six digits is 15 which is divisible by 3. So to form a number of five-digit which is divisible by 3 we can remove either ‘O’ or ‘3’. If digits 1, 2, 3,4, 5 are used then number of required numbers = 5!
If digits 0, 1,2,4, 5 are used then first place from left can be filled in 4 ways and remaining 4 places can be filled in 4! ways. So in this case required numbers are 4 x 4! ways.
So, total number of numbers = 120 + 96 = 216

Q32. Everybody in a room shakes hands with everybody else. If the total number of hand shakes is 66, then the total number of persons in the room is
(a)    11                 
(b)     12                    
(c)     13                   
(d)   14
Sol: (b) Between any two person there is one hand shake.

Q33. The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line is
(a)    105                (b)     15                     (c)     175                  (d)  185
Sol: (d) Number of ways of selecting 3 points from given 12 points = ¹²C₃
But any three points selected from given seven collinear points does not form triangle.
Number of ways of selecting three points from seven collinear points = ⁷C₃Required number of triangles = ¹²C₃ – ⁷C₃ = 220 -35 = 185

Q34. The number of parallelograms that can be formed form a set of four parallel lines intersecting another set of three parallel lines is
(a)    6                     
(b)   18                    
(c)   12                   
(d)   9
Sol: (b) To form parallelogram we required a pair of line from a set of 4 lines and another pair of line from another set of 3 lines.
Required number of parallelograms = ⁴C₂ x ³C₂ = 6×3 = 18

Q35. The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them is
(a)   ¹⁶C₁₁               
(b)    ¹⁶C₅                    
(c)     ¹⁶C₉                  
(d)   ²⁰C₉

Sol: (c) Total number of players = 22
We have to select a team of 11 players.
We have to exclude 4 particular of them, so only 18 players are now available. Also from these 2 particular players are always included. Therefore we have to select 9 more players from the remaining 16 players.
So, required number of ways = ¹⁶C₉

Q36. The number of 5-digit telephone numbers having at least one of their digits repeated is
(a) 900000
(b) 10000                 
(c) 30240                
(d) 69760
Sol: (d) Total number of telephone numbers when there is no restriction = 10⁵ Also number of telephone numbers having all digits different = ^l0P₅ Required number of ways = 10⁵ – ^l0P₅ = 1000000 -10x9x8x7x6 = 1000000-30240 = 69760

Q37. The number of ways in which we.can choose a committee from four men and six women, so that the committee includes at least two men and exactly twice as many women as men is
(a) 94                    
(b) 126                     
(c) 128                    
(d) none of these
Sol: (a) Number of men = 4; Number of women = 6
It is given that committee includes at least two men and exactly twice as many women as men.
So, we can select either 2 men and 4 women or 3 men and 6 women.
∴ Required number of committee formed = ⁴C₂ x ⁶C₄ + ⁴C₃ x ⁶C₆
= 6×15 + 4×1=94

Q38. The total number of 9-digit numbers which have all different digits is
(a) 10!                   
(b) 9!                        
(c) 9×9!               
(d) 10×10!
Sol: (c) We have to form 9-digit number which has all different digit.
First digit from the left can be filled in 9 ways (excluding ‘0’).
Now nine digits are left including ‘O’.
So remaining eight places can be filled with these nine digits in ⁹P_S ways.
So, total number of numbers = 9 x ⁹P₈ = 9×9!

Q39. The number of words which can be formed out of the letters of the word ARTICLE, so that vowels occupy the even place is
(a) 1440               
(b) 144                     
(c) 7!                        
(d) ⁴C₄ x ³C₃
Sol: (b) We have word ARTICLE.
Vowels are A, I, E and consonants are R, T, C, L.
Now vowels occupy three even places (2^nd, 4^th and 6^th) in 3! ways.
In remaining four places four consonants can be arranged in 4! ways.
So, total number of words = 3! x4! = 6×24= 144

Q40. Given five different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is
(a) 3600               
(b) 3720                   
(c) 3800                  
(d) 3600
[Hint: Possible numbers of choosing or not choosing 5 green dyes, 4 blue dyes and 3 red dyes are 2⁵, 2⁴ and 2³, respectively.]

Fill in the Blanks Type Questions

True/False Type Questions

Q51. There are 12 points in a plane of which 5 points are collinear, then the number of lines obtained by joining these points in pairs is ¹²C₂ – ⁵C₂.
Sol: False
Required number of lines = ¹²C₂ – ⁵C₂.+ 1

Q52. Three letters can be posted in fiv.e letter boxes in 3⁵
Sol: False
Each letter can be posted in any one of the five letter boxes.
So, total number of ways of posting three letters = 5x5x5 = 125

Q53. In the permutations of n things r taken together, the number of permutations in which m particular things occur together is

Q54. In a steamer there are stalls for 12 animals and there are horses, cows and calves (not less than 12 each) ready to be shipped. They can be loaded in 3¹² ways.
Sol:True
In each stall any one of the three animals can be shipped.
So total number of ways of loading = 3x3x3x…xl2 times = 3¹²

Q55. If some or all of n objects are taken at a time, then the number of combinations is 2ⁿ– 1.
Sol: True
If some or all objects taken at a time, then number of combinations would be ⁿC₁ + ⁿC₂ + ⁿC₃ + … + ⁿC_n = 2ⁿ – 1

Q56. There will be only 24 selections containing at least one red ball out of a bag containing 4 red and 5 black balls. It is being given that the balls of the same colour are identical.
Sol: False
Number of ways of selecting any number of objects from given n identical objects is 1.
Now selecting zero or more red ball from 4 identical red balls = 1 + 1 + 1 + 1 + 1=5
Selecting at least 1 black ball from 5 identical black balls =1 + T+1 + 1 + 1= 5 So, total number of ways = 5 x 5 = 25

Q58. A candidate is required to answer 7 questions, out of 12 questions which are divided into two groups, each containing 6 He is not permitted to attempt more than 5 questions from either group. He can choose the seven questions in 650 ways.
Sol: False
A candidate can attempt questions in following maimer

Q59. To fill 12 vacancies there are 25 candidates of which 5 are from scheduled castes. If 3 of the vacancies are reserved for scheduled caste candidates while the rest are open to all, the number of ways in which the selection can be made is ⁵C₃ x ²²C₉.
Sol:True
We can select 3 scheduled caste candidate out of 5 in ⁵C₃ ways.
And we can select 9 other candidates out of 22 in ²²C₉ways.
.’. Total number of selections = ⁵C₃ x ²²C₉

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