Category: OP Malhotra Solutions

Students can cross-reference their work with ISC Class 11 Maths Solutions S Chand Chapter 28 Statistics Ex 28(a) to ensure accuracy.

S Chand Class 11 ICSE Maths Solutions Chapter 28 Statistics Ex 28(a)

Question 1.
One set of 100 observations has the mean 15 and another set of 150 observations has the mean 16. Find the mean of 250 observations by combining the two sets of given observations.
Solution:
Here, n₁ = 100 ; \(\bar{x}_1\) = 15 and n₂ – 150 ; \(\bar{x}_2\) = 16
∴ combined Mean \(\bar{x}\)₁₂ = \(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\) = \(\frac{100 \times 15+150 \times 16}{100+150}\) = \(\frac{1500+2400}{250}\) = \(\frac{3900}{250}\) = 15.6

Question 2.
The mean age of 40 students is 6 years and the mean age of another group of 60 students is 20 years. Find out the mean age of the 100 students combined together.
Solution:
Here n₁ = 40; \(\bar{x}_1\) = 16 and n₂ = 60; \(\bar{x}_2\) = 20
∴ combined Mean \(\bar{x}\)₁₂ = \(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\) = \(\frac{40 \times 16+60 \times 20}{40+60}\) = \(\frac{640+1200}{100}\) = \(\frac{1840}{100}\) = 18.4
Thus required mean age be 18.4 years.

Question 3.
The mean of marks obtained in an examination by a group of 100 students is found to be 49.46. The mean of the marks obtained in the same examination by another group of 200 students was 52.32. Find the mean of the marks obtained by both the groups of students taken together.
Solution:
Here, n₁ = 100; \(\bar{x}_1\) = 49. 46 and n₂ = 200; \(\bar{x}_2\) = 52.32
Thus marks obtained by first group of students = n₁\(\bar{x}_1\)
and marks obtained by second group of students = n₂\(\bar{x}_2\)
Hence, required mean marks obtained by both groups of students taken together = \(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\) = \(\frac{100 \times 49.46+200 \times 52.32}{100+200}\) = \(\frac{4946+10464}{300}\) = 51.366

Question 4.
The number of students in section X A and X B are 30 and 35 respectively. The mean scores of students in the mathematics test are as follows:

X A	X B	X A and X B combined
70	?	62

Find the mean score of XB.
Solution:
Given n₁ = 30; n₂ = 35; \(\bar{x}_1\) = 70; \(\bar{x}\)₁₂ = 62; \(\bar{x}_2\) = ?
We know that \(\bar{x}\)₁₂ = \(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\)
⇒ 62 = \(\frac{30 \times 70+35 \times \bar{x}_2}{30+35}\)
⇒ 4030 – 2100 = 35\(\bar{x}_2\)
⇒ \(\bar{x}_2\) = \(\frac { 1930 }{ 35 }\)
= 55.14

Question 5.
Two samples of sizes 50 and 100 are given. The mean of these samples respectively are 56 and 50. Find the mean of size 150 by combining.
Solution:
Here n₁ = 50; n₂ = 100; \(\bar{x}_1\) = 56; \(\bar{x}_2\) = 50
Then \(\bar{x}\)₁₂ = \(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\)
⇒ \(\bar{x}\)₁₂ = \(\frac{50 \times 56+100 \times 50}{50+100}\)
⇒ \(\bar{x}\)₁₂ = \(\frac{2800+5000}{150}\) = \(\frac{7800}{150}\) = 52

Question 6.
The mean and standard deviation of the marks obtained by the groups of students, consisting of 50 each, are given below. Calculate the mean and standard deviation of the marks obtained by all the 100 students.

Groups	Mean	Standard deviation
1	60	8
2	55	7

Solution:
Here n₁ = n₂ = 50; \(\bar{x}_1\) = 60; \(\bar{x}_2\) = 55; σ₁ = 8; σ₁ = 7
Then combined mean \(\bar{x}\)₁₂ = \(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\) = \(\frac{50 \times 30+50 \times 44}{50+50}\) = \(\frac{3000+2750}{100}\) = \(\frac{5750}{100}\) = 57.5
∴ d₁ = \(\bar{x}\)₁₂ – \(\bar{x}_1\) = 57.5 – 60 = – 2.5 and d₂ = \(\bar{x}\)₁₂ – \(\bar{x}_2\) = 57.5 – 55 = 2.5
Therefore, combined S.D = σ₁₂ =

Question 7.
The mean and standard deviation of distribution of 100 and 150 items are 50, 5 and 40, 6 respectively. Find the mean and standard deviation of all the 250 items taken together.
Solution:
Here n₁ = 100; n₂ = 150; \(\bar{x}_1\) = 50; \(\bar{x}_2\) = 40
σ₁ = 5; σ₂ = 6
∴ Combined Mean \(\bar{x}\)₁₂ = \(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\) = \(\frac{100 \times 50+150 \times 40}{100+150}\) = \(\frac{11000}{250}\) = 44
∴ d₁ = \(\bar{x}\)₁₂ – \(\bar{x}_1\) = 44 – 50 = – 6 and d₂ = \(\bar{x}\)₁₂ – \(\bar{x}_2\) = 44 – 40 = 4
∴ Combined S.D = σ₂ =

Access to comprehensive ISC Mathematics Class 11 OP Malhotra Solutions Chapter 27 Mathematical Reasoning Chapter Test encourages independent learning.

S Chand Class 11 ICSE Maths Solutions Chapter 27 Mathematical Reasoning Chapter Test

Question 1.
for each of the following compound statements, first identify the connective words and then break it into component statements.
(i) All rational numbers are real and all real numbers are not complex.
(ii) Square of an integer is positive or negative.
(iii) x = 2 and x = 3 are the roots of the equation 3x² – x – 10 = 0.
Solution:
(i) The connective word be ‘AND’ and component statements are :
p : all rational numbers are real.
q : all real numbers are not complex.

(ii) The connective word is ‘OR’ and component statements are ;
p : square of an integer is positive
q : square of an integer is negative

(iii) The connective word is ‘AND’ and component statements are ;
p : x = 2 be a root of eqn. 3x² – x – 10 = 0
q : x = 3 be a root of eqn. 3x² – x – 10 = 0

Question 2.
Identify the quantifier in the following statements and write the negation of the statements.
(i) There exists a number which is equal to its square.
(ii) For every real numbers x, x is less than x + 1.
(iii) There exists a capital for every state in India.
Solution:
(i) Here, the quantifier be ‘there exists’ Negation of given statement: There does not exists a number which is equal to its square.
(ii) Here, the quantifier be “For every.” Negation of given statement be given by It is false that for every real number x, x is less than x + 1.
(iii) Here, the quantifier be “There exists.” Negation of given statement be given by There exists a state in India which does not have a capital.

Question 3.
For any statement ‘p’ prove that ~(~p) ≡ p.
Solution:

p	~p	~(~p)
T	F	T
F	T	F

From truth tables we observe that, p ≡ ~ (~p)

Question 4.
Write the converse, contradiction and contrapositive of the statement
‘If x + 3 = 9, then x = 6.’
Solution:
The component statements are ;
let p : x + 3 = 9
q : x = 6
Then given statement be p ⇒ q
Its converse be q ⇒ p
i.e. if x = 6 Then x + 3 = 9
Contradiction of given statement is x = 6 ⇒ x + 3 ≠ 9
Contrapositive of given statement is ~q ⇒ ~p i.e. if x ≠ 6 then x + 3 ≠ 9

Question 5.
For any statements, p and q, prove that p ⇒ q = (~p ∨ q).
Solution:
The truth table for p ⇒ q and (~p ∨ q) is given as under:

I	II	III	IV	V
p	q	p ⇒ q	~p	~p ∨ q
T	T	T	F	T
T	F	F	F	F
F	T	T	T	T

From column III and V, we observe that
P ⇒ q ≡ (~p ∨ q)

Question 6.
Write the following implications (p ⇒ q) in the form (~ p ∨ q) and write its negation. ‘If △ABC is isosceles then the base angles A andB are equal.’
Solution:
The truth table for p ⇒ q and (~p ∨ q) is given as under :

I	II	III	IV	V
p	q	p ⇒ q	~p	~p ∨ q
T	T	T	F	T
T	F	F	F	F
F	T	T	T	T
F	F	T	T	T

From column III and V, we observe that p ⇒ q ≡ (~ p ∨ q)
∴ ~(~p ∨ q) = ~ (~p) ∧ (~q) = p ∧ ~q
Let p : △ABC is isosceles
q : The base angles A and B are equal.
Then ~ p ∨ q : either △ABC is not an isosceles triangle or the base angles A and B are equal.
Thus the negation of given statement be (p ∧ ~ q)
i.e. △ABC be an isosceles triangle and the base angles A and B are not equal.

Students appreciate clear and concise ISC Mathematics Class 11 OP Malhotra Solutions Chapter 27 Mathematical Reasoning Ex 27(g) that guide them through exercises.

S Chand Class 11 ICSE Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(g)

Question 1.
Show that the statement p : ‘If x is a real number such that x³ + 4x = 0, then x = 0’ is true by
(i) Direct method,
(ii) Method of contradiction,
(iii) Method of contrapositive.
Solution:
Let q : x ∈ R s.t x³ + 4x = 0
r : x = 0
Then p :q → r
(i) Then q be true ⇒ x³ + 4x = 0
⇒ x (x² + 4x) = 0 ⇒ x = 0 (∵ x ∈ R)
⇒ r be true
Thus, q ⇒ r is true ⇒ p is true statement.

(ii) Let x ≠ 0 and let x = k, where k e R and k be a root of x² + 4x = 0
⇒ k³ + 4k = 0 ⇒ k (k² + 4) = 0
⇒ k = 0 (∵ k ∈ R)
since k² + 4 ≠ 0
∴ which contradict the our assumption that k ≠ 0
∴ our supposition is wrong.
∴ x = 0
Thus q ⇒ r is true.

(iii) Let r be not true ∴ x ≠ 0
Let x = p ≠ 0 ∈ R
∴ x³ + 4x = p³ + 4p – p (p² + 4) ≠ 0
[∵ p ≠ 0, p² + 4
⇒ q be not true
∴ ~ r ⇒ ~ q is true
Hence q ⇒ r is true.

Question 2.
Show that the statement ‘For any real numbers a and b, a² = b² implies that a = b’ is not true by giving counter examples.
Solution:
Let us take a = 2 and b = – 2
Here a² = b² = 4 but a ≠ b

Question 3.
Show that the following statement is true by the method of contrapositive.
p : If x is an integer and x² is even, then x is also even.
Solution:
Let q : x is an integer and x² is even r : x is also even
Then p : q ⇒ r be the given statement.
We want to check, whether the statement q ⇒ r is true by contrapositive method.
i.e. we want to check the validity of ~ r ⇒ ~ q
Let ~ r be true ⇒ r be not true
⇒ x is not even
⇒ x = 2 k + 1, where k ∈ I
⇒ x² = (2k + 1)² = 4k² + 4k + 1
=4k(k + 1) + 1 = 8k’ + 1,
where k’ ∈ I [∵ product of n consecutive integers is divisible by n!]
⇒ x² is not even integer
Thus ~ r ⇒ ~ q be true
Hence q ⇒ r be a true statement.

Question 4.
Given below are two statements :
p : 30 is a multiple of 5.
q : 30 is a multiple of 9.
Write the compound statement, connecting these two statements with ‘and’ and ‘or’. In both cases, check the validity of the compound statement.
Solution:
Given p : 30 is a multiple of 5 (True)
q : 30 is a multiple of 9 (False)
Then compound statement p ∧ q : 30 is a multiple of 5 and 9
Its truth value be false since one of the component statement is false and hence p ∧ q is not valid.
The compound statement p ∨ q : 30 is a multiple of 5 or 9
Its truth value be true and it is a valid statement.

Question 5.
Verify by the method of contradiction that √7 is irrational.
Solution:
Let p be the given statement
i.e. p : √7 is irrational
Let if possible p be not true i.e. √7 is a rational number ⇒ √7 = \(\frac { a }{ b }\)
where a, b ∈ l and (a, b) = 1 i.e. have no common factor.
⇒ 7 = \(\frac{a^2}{b^2}\)
⇒ a² = 7b²
⇒ 7 divides a² ⇒ 7 divides a
⇒ a = 7k, where k ∈ I
⇒ (7k)² = 7b² ⇒ 7k² = b²
⇒ 7 divides b² ⇒ 7 divides b
⇒ 7 is a common factor of both a and b. which contradicts the fact that a and b have no common factor and hence our supposition is wrong. Thus p is true. Hence, √7 be an irrational number.

Interactive ISC Mathematics Class 11 OP Malhotra Solutions Chapter 27 Mathematical Reasoning Ex 27(f) engage students in active learning and exploration.

S Chand Class 11 ICSE Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(f)

Question 1.
Write the converse, inverse and contra-positive of the following statements.
(i) If you do not drink your milk, you will not be strong.
(ii) If you drink milk, you will be strong.
(iii) You will be strong only if you drink your milk.
Solution:
(i) Converse : q ⇒ p : If you are not strong then you do not drink your milk.
Inverse : ~ p ⇒ ~ q : If you drink your milk then you will be strong.
Contrapositive : ~ q ⇒ ~ p : If you are strong then you drink your milk.

(ii) Converse : If you are strong then you drink your milk.
Inverse : If you do not drink your milk then you are not strong.
Contrapositive : If you are not strong then you do not drink your milk.

(iii) Converse (q ⇒ p): If you drink your milk then you are strong.
Inverse (~ p ⇒ ~ q) : If you are strong then you do not drink your milk.
Contrapositive (~ q ⇒ ~ p) : If you do not drink your milk then you are not strong.

Question 2.
Write the converse of each of the following statements. In which cases is the converse true?
(i) If an integer is even, then its square is divisible by 4.
(ii) If it is raining, then there are clouds in the sky.
(iii) In order to get this job, I must be a graduate.
(iv) If Mr. Sexena is elected to office, then all our problems are over.
Solution:
(i) Converse of given statement is q ⇒ p
If square of an integer is divisible by 4 then the integer is even (True)

(ii) Let p : it is raining
q : There are clouds in the sky
Then converse of p ⇒ q be q ⇒ p
i.e. If there are clouds in the sky then it is raining (False)

(iii) Let p : I get this job
q : I must be a graduate Then the converse of p ⇒ q is q ⇒ p
i.e. If I am a graduate then I can get this job, which is false.

(iv) Let p : Mr. Sexena is elected to office
q : all our problems are over
Then converse of p ⇒ q is q ⇒ p
i.e. If all our problems are over then Mr. Sexena is elected to office (False)

Question 3.
Consider the statements :
p : You will work hard
q : You will become wealthy.
Translate each of the symbolic statements into an English sentence.
(i) P ⇒ q
(ii) q ⇒ p
(iii) (~p) ⇒ (~ q)
(iv) (~q) ⇒ (~p)
Solution:
Given statements are ;
p : You will work hard
q : You will become wealthy
(i) p ⇒ q means if you will work hard then you will become wealthy.
(ii) q ⇒ p: If you will become wealthy then you will work hard
(iii) ~ p ⇒ ~ q : If you will not weak hard then you will not become wealthy
(iv) ~ q ⇒ ~ p : If you will not become wealthy then you will not work hard.

Question 4.
Compare the following statements :
(i) P ⇒ q
(ii) If p, then q
(iii) p is a sufficient condition for q.
(iv) q is a necessary condition for p.
(v) p, only if q.
Solution:
All the given five statements are equivalent and all are equivalent to p ⇒ q.

Question 5.
Construct truth tables for each of the following :
(i) (p ⇒ q) ∧ (q ⇒ p)
(ii) q ⇒ [(~p) q]
(iii) [(~p) ∧ q] ⇒ (p ∨ q)
Solution:
(i) The truth table for (p ⇒ q) ∧ (q ⇒ p) is given are under:

p	q	P ⇒ q	q ⇒ p	(p ⇒ q) ∧ (q ⇒ p)
T	T	T	T	T
T	F	F	T	F
F	T	T	F	F
F	F	T	T	T

(ii) The truth tale for q ⇒ [(~ p) ∨ q] is given as under:

s	q	~ p	~ p ∨ q	q ⇒ [(~p) ∨ q]
T	T	F	T	T
T	F	F	F	T
F	T	T	T	T
F	F	T	T	T

(iii) The truth table for [(~ p) ∧ q] ⇒ (p ∨ q) is given as under :

p	q	~ p	~ p ∧ q	p ∨ q	[~p ∧ q] ⇒ p ∨ q
T	T	F	F	T	T
T	F	F	F	T	T
F	T	T	T	T	T
F	F	T	F	F	T
T	T	F	F	T	T

Question 6.
Write the converse, inverse and contra-positive for the statement (~ p) ⇒ q.
Solution:
Given statement is ~p ⇒ q
Converse :q ⇒ ~p
Inverse : ~ (~p) ⇒ ~ q i.e. p ⇒ ~ q
Contrapositive : ~ q ⇒ ~ (~p) i.e. ~ q ⇒ p

Question 7.
Write the inverse of the converse of p ⇒ q.
Solution:
Given statement be p ⇒ q
∴ its converse : q ⇒ p
Thus required inverse be ~q ⇒ ~p

Question 8.
Write the converse of the inverse of p ⇒ q.
Solution:
The inverse of p ⇒ q be ~p ⇒ ~ q
Then its converse be ~ q ⇒ ~ p

Question 9.
Write the contrapositive of the inverse of p ⇒ q.
Solution:
Given statement be p ⇒ q
Inverse : ~ p ⇒ ~q
Contrapositive : ~ (~ q) ⇒ ~ (~ p)
i.e. q ⇒ p

Question 10.
Write the converse of the contrapositive of p ⇒ q.
Solution:
The contrapositive of p ⇒ q be ~q ⇒ ~p
∴ its converse be ~ p ⇒ ~ q

Question 11.
Write the contrapositive of the contrapositive of p ⇒ q.
Solution:
The contrapositive of p ⇒ q be ~ q ⇒ ~p
Then the contrapositive of ~ q
⇒ ~p be ~ (~p) ⇒ ~ (~ q) i.e. p ⇒ q.

Question 12.
Does completing each of the problems 6 through 10 result in a conditional? What is the relationship of each resulting condition to the original conditional P ⇒ q?
Solution:
Yes, contrapositive (# 7) ; contrapositive (# 8); converse (# 9); Increase (# 10) and original condition (#11)

Question 13.
If p and q are any two propositions then prepare the truth table for p ⇒ q, ~q ⇒ ~p and show that the above statements are equivalent.
Hence, or otherwise determine which of the following two arguments is valid?
(i) Given : If you work hard, then you pass the course.
Given : You did not work hard.
Conclusion : You did not pass the course.
(ii) Given : If you work hard, then you pass the course.
Given : You did not pass the course.
Conclusion : You did not work hard.
Solution:
The truth table is given as under :

I	II	III	IV	V	VI	VII	VIII
p	q	p ⇒ q	q ⇒ p	~P	~q	~p ⇒ ~q	~q ⇒ ~p
T	T	T	T	F	F	T	T
T	F	F	T	F	T	T	F
F	T	T	F	T	F	F	T
F	F	T	T	T	T	T	T

From column III and VIII; we have p ⇒ q ≡ ~q ⇒ ~p
From column IVth and VII; we have q ⇒ p ≡ ~p ⇒ ~q

(i) Let p : You work hard, q : You pass the course,
~ p : You did not work hard, ~ q : You did not pass the course
Now p ⇒ q and ~p ⇒ ~q are not equivalent statement
∴ p ⇒ q Then ~p ⇒ ~ q is not valid.

(ii) p ⇒ q and ~q ⇒ ~p are equivalent statements.
Thus, p ⇒ q then ~ q ⇒ ~ p is valid.

Utilizing ISC Mathematics Class 11 OP Malhotra Solutions Chapter 27 Mathematical Reasoning Ex 27(e) as a study aid can enhance exam preparation.

S Chand Class 11 ICSE Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(e)

Question 1.
Write each sentence in the “If ……….. then” form.
(i) Roses are vegetables if carrots are flowers.
(ii) All ducks are birds.
(iii) Vertical angles are equal.
(iv) Freezing water expands.
(v) A set with no elements is called the empty set.
(vi) A racer wins the race only if he runs fast.
(vii) Any two parallel lines are coplanar.
Solution:
(i) If carrots are flowers then roses are vegetables.
(ii) If the creature is a duck then it is a bird.
(iii) If two angles are vertical angles then they are equal.
(iv) If the water is freezing then it expands.
(v) If the set has no element then it is called the empty set.
(vi) If the racer runs fast then the racer wins the race.
(vii) If two lines are parallel then they are coplanar.

Question 2.
Let p be “I will marry her,” and let q be “she is beautiful.” Translate into symbolic form.
(i) If she is beautiful, then I will marry her.
(ii) If I will marry her, then she is beautiful.
(iii) If I will marry her, then she is beautiful.
(iv) If she is not beautiful, then I will not marry her.
(v) If I will not marry her, then she is not beautiful.
(vi) If she is beautiful, then I will not marry her.
Solution:
Given statements are ;
p : I will marry her
q : she is beautiful
(i) q ⇒ p
(ii) p ⇒ q
(iii) q ⇔ p
(iv) ~q ⇒ ~p
(v) ~p ⇒ ~ q
(vi) q ⇒ ~p

Question 3.
Determine whether p, q and ‘If p, then q’ are true or false in each case given below :

(i) 3 is a prime number	3 is an even number
(ii) 5 <7	5 is odd
(iii) 3 > 2	2 × 7 = 14
(iv) 1 > 4	2 is even
(v) 5 × 3 = 16	2 + 7 = 6
(vi) 3 (5 ÷ 6) < 1	8 – 3 ÷ 6 > 9

Solution:
(i) Given p : 3 is a prime number (True)
q : 3 is an even number (False)
Then p ⇒ q is false

(ii) p : 5 < 7 (True) q : 5 is odd (True) Then p ⇒ q is true (iii) p : 3 > 2 (True)
q : 2 × 7 = 14 (True)
Then p ⇒ q is true.

(iv) p : 1 > 5 (False)
q : 2 is even (True)
Then p ⇒ q is true

(v) p : 5 × 3 = 16 (False)
q : 2 + 7 = 6 (False)
Then p ⇒ q is true

(vi) p : 3 (5 ÷ 6) < 1 (False) q : 8 – 3 ÷ 6 > 9 (False)

Then p ⇒ q is true.

Question 4.
Write T before each true statement and write F before each false statement. Then give the truth value of the conditional expressed.
(i) If Asia is a continent, then Delhi is in Japan.
(ii) If monkeys climb trees, then 6 is divisible by 2.
(iii) If oxygen is a gas, then gold is a compound.
(iv) Water is dry ⇒ show is hot.
(v) Show is cold ⇒ water is wet.
(vi) If 3 = 5, then 7 is a prime number.
(vii) 5 × 6 – 4 = 21 ⇒ 2 (5 ÷ 15 + 3) = \(\frac { 20 }{ 3 }\).
(viii) If a triangle is a rectangle, then a circle is a rhombus.
(ix) If 51 is the product of + 17 and – 3, then lions can fly in the air.
(x) If √5 is an integer, then 3 is an integer.
Solution:
(i) Let p : Asia is a continent (T)
q : Delhi is in Japan (F)
Then p ⇒ q is false (F)

(ii) Let p : Monkeys climb Trees (T)
q : 6 is divisible by 2 (T)
Then given statement p
⇒ q is True (T)

(iii) Let p : Oxygen is a gas (T)
q : Gold is a compound (F)
Then given statement p
⇒ q is false (F)

(iv) Let p : water is dry (F)
q : snow is hot (F)
Then p ⇒ q is true (T)

(v) Let p : Snow is cold (T)
and q : water is wet (T)
Then p ⇒ q is True (T)

(vi) Let p : 3 = 5 (F)
q : 7 is a prime number (T)
Then p ⇒ q is True (T)

(vii) Let p : 5 × 6 – 4 = 21 (F)
q : 2(5 ÷ 15 + 3) = 2 \(\left(\frac{1}{3}+3\right)\) = \(\frac { 20 }{ 3 }\) (T)
Then p ⇒ q is true

(viii) Let p : triangle is a rectangle (F)
q : Circle is a rhombus (F)
Then p ⇒ q is true (T)

(ix) Let p : 51 is the product of + 17 and (-3) (F)
q : lions can fly in the air (F)
Then p ⇒ q is true (T)

(x) Let P : √5 is an integer (F)
q : 3 is an integer (T)
Then p ⇒ q is true (T)

The availability of step-by-step ISC Mathematics Class 11 OP Malhotra Solutions Chapter 27 Mathematical Reasoning Ex 27(d) can make challenging problems more manageable.

S Chand Class 11 ICSE Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(d)

Question 1.
Let p, q, r and s represent simple statements. Assume that p is false, q is true, r is false, and s is true. Determine the truth value of each statement expressed below :
(i) q ∧ r
(ii) r ∨ p
(iii) p ∧ S
(iv) p ∨ s
(v) ~q
(vi) q ∨ s
(vii) ~ r
(viii) s ∧ q
(ix) r ∧ p
Solution:
Given p is false, q is true, r is false and s is true.
(i) The truth value of statement is false as one of the component statement is false.
(ii) The truth value of statement is false since both component statements are false.
(iii) The truth value of compound statement is false since one of the component statement is false.
(iv) The truth value of compound statement is true since one of the component statement is true.
(v) The truth value of statement is false,
(vi) The truth value of q ∨ s is true since both component statements are true.
(vii) The truth value of compound statement is true.
(viii) The compound statement has truth value is true.
since, both component statements are true.
(ix) The truth value of compound statement is false.
since both component statements are false.

Question 2.
Let a, b, c and d represent simple statements. Assume that a ∧ d is true, b ∧ C is false, and ~ c is false.
(i) What is the truth value of a?
(ii) What is the truth value of d?
(iii) What is truth value of c?
(iv) What is the truth value of b?
Solution:
Given a ∧ d is true ⇒ a and d both statements are true.
and b ∧ C is false. …(1)
Since ~ c is false ⇒ c is true
∴ from (1); b is false (∵ F A T = F)
(i) The truth value of a be true.
(ii) The truth value of d be true.
(iii) The truth value of c be true.
(iv) The truth value of b be false.

Question 3.
Assume that two given statements p and q are both true and indicate whether or not you would expect each of the following statements to be true :
(i) P ∧ q
(ii) p ∨ q
(iii) p ∨ (~q)
(iv) (~p) ∨ (~ q)
Solution:
Given statements p and q are both true.
(i) Thus, the statement p ∧ q = T ∧ I = T is also true.
(ii) The statement p ∨ q is also statement.
(iii) p ∨ (~ q) = T ∨ F = T
(iv) (~p) ∨ (~ q) = F ∨ F = F
Construct truth tables for :

Question 4.
(~p) ∧ q
Solution:

p	q	~ p	~ P ∧ q
T	F	F	F
T	T	F	F
F	T	T	T
F	F	T	F

Question 5.
(~p) ∧ (~q)
Solution:

P	q	~ p	~ q	(~ p) ∧ (~ q)
T	T	F	F	F
T	F	F	T	F
F	T	T	F	F
F	F	T	T	T

Question 6.
~(p ∧ q)
Solution:
The truth table for ~ (p ∧ q) is given as under:

p	q	P ∧ q	~ (P ∧ q)
T	T	T	F
T	F	F	T
F	T	F	T
F	F	F	T

Question 7.
p ∧ (~ q)
Solution:
The truth table for statement p ∨ (~ q) is given as under:

P	q	~q	P ∨ (~q)
T	T	F	T
T	F	T	T
F	T	F	F
F	F	T	T

Question 8.
~ [p ∨ (~ q)]
Solution:
The truth table for ~[p ∨ (~ q)] is given as under:

P	q	~ q	P ∨ (~ q)	~[p ∨ (~q)]
T	T	F	T	F
T	F	T	T	F
F	T	F	F	T
F	F	T	T	F

Question 9.
~ (~ p ∧ ~ q)
Solution:
The truth table for ~ (~p ∧ ~ q) is given as under:

P	q	~ P	~ q	( ~ p ∧ ~ q)	~ (~ p ∧ ~ q)
T	T	F	F	F	T
T	F	F	T	F	T
F	T	T	F	F	T
F	F	T	T	T	F

Question 10.
(p ∧ q) ∨ (~p ∧ q)
Solution:
The truth table for (p ∧ q) ∨ (~ p ∧ q) is given as under:

p	q	P ∧ q	~ p	~ p  ∧ q	(p  ∧ q) ∨  (~ p  ∧ q)
T	T	T	F	F	T
T	F	F	F	F	F
F	T	F	T	T	T
F	F	F	T	F	F

Question 11.
p ∧ (q ∨ r)
Solution:
The truth table for p ∧ (q ∨ r) is given as under:

p	q	r	q ∧ r	P ∧ (q ∨ r)
T	T	T	T	T
T	T	F	T	T
T	F	T	T	T
T	F	F	F	F
F	T	T	T	F
F	T	F	T	F
F	F	T	T	F
F	F	F	F	F

Question 12.
(~p ∧ ~ q) ∨ (p ∧ ~ q)
Solution:
The truth table for (~ p ∧ ~ q) ∨ (p ∧ ~ q) is given as under:

p	q	~ p	~ q	(~ p ∧ ~ q)	P ∧ ~ q	(~ p ∧ ~ q) ∨ (p ∧ ~ q)
T	T	F	F	F	F	F
T	F	F	T	F	T	T
F	T	T	F	F	F	F
F	F	T	T	T	F	T

Question 13.
(p ∨ q) ∨ (r ∧ ~ q)
Solution:
The truth table for (p ∨ q) ∨ (r ∧ ~ q) is given as under:

p	q	r	~q	P v q	r ∨ ~ q	(p ∨ q) ∨ (r ∧ ~ q)
T	T	T	F	T	F	T
T	T	F	F	T	F	T
T	F	T	T	T	T	T
T	F	F	T	T	F	T
F	T	T	F	T	F	T
F	T	F	F	T	F	T
F	F	T	T	F	T	T
F	F	F	T	F	F	F

Question 14.
Let p be “Ananya is beautiful,” and let q be “Ananya is 165 centimetres tall.”
(i) Under what conditions is the statement, “Ananya is beautiful and 165 centimetres tall.” true?
(ii) Under what conditions is the statement, “Ananya is beautiful and 165 centimetres tall,” false?
(iii) Under what conditions is the statement, “Ananya is beautiful or 165 centimetres tall,” true?
(iv) Under what conditions is the statement, “Ananya is beautiful or 165 centimetres tall,” false?
Solution:
Given p : Ananya is beautiful
q: Ananya is 165 centimetres tall
(i) Only if Ananya is both beautiful and 165 cm tall.
(ii) If Ananya is not beautiful and/or not 165 cm tall.
(iii) If Ananya is beautiful and/or 165 cm tall.
(iv) Only if Ananya is neither beautiful nor 165 cm tall.

Students often turn to ISC Mathematics Class 11 OP Malhotra Solutions Chapter 27 Mathematical Reasoning Ex 27(c) to clarify doubts and improve problem-solving skills.

S Chand Class 11 ICSE Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(c)

Question 1.
Write the given statement in symbolic form using the letter in parentheses to represent the corresponding component.
(i) This is April (p) and income tax returns must be filed (q).
(ii) Accountancy is a required subject for Chartered Accountants (m) but not for engineers (n).
(iii) Mukesh Patel is a teacher (t) or a lawyer (u).
(iv) Jack went up the hill (c) and Jill went up the hill (d).
(v) I plan to take science (a) or commerce (c) in class 11.
(vi) I will not drive to Jaipur (~ d) but I shall go by train (i) or by plane (p).
Solution:
(i) p ∧ q
(ii) m ∧ ~ n
(iii) t ∨ u
(iv) c ∧ d
(v) Compound statement is the disjunction of a and c i.e. a ∨ c
(vi) The compound statement is the conjunction of ~ d and (disjunction of t and p) i.e. (~ d) A (t ∨ p)

Question 2.
Let p be “Shruti can type”, and let q be “Shruti takes shorthand.” Write the following statements in symbolic form :
(i) Shruti can type and take shorthand.
(ii) Shruti can type but she does not take shorthand.
(iii) Shruti can neither type nor take shorthand.
(iv) It is not true that Shruti can type and take shorthand.
Solution:
Given p : Shruti can type
q : Shruti takes shorthand
(i) Compound statement is the conjunction of p and q i.e. p ∧ q.
(ii) ~ q : she does not takes shorthand The given statement is the conjunction ofp and ~ q i.e. p ∧ q.
(iii) ~ p : Shruti can’t type
∴ given statement is the conjunction of ~p and ~ q i.e. ~p ∧ ~ q.
(iv) p ∧ q : Shruti can type and takes shorthand
∴ ~(p ∧ q): It is not true that Shruti can type and take shorthand.

Question 3.
Use p : Ramesh is rich ; q : Pradeep is poor. Think of “poor” as “not rich”, and write each of these statements in symbolic form.
(i) Ramesh is poor and Pradeep is rich.
(ii) Pradeep and Ramesh are both rich.
(iii) Neither Ramesh nor Pradeep is rich.
(iv) Ramesh is not rich and Pradeep is poor.
(v) It is not true that Ramesh and Pradeep both are rich.
(vi) Either Ramesh is poor or Pradeep is poor.
(vii) Either Ramesh or Pradeep is rich.
Solution:
Given statements are : p : Ramesh is rich q : Pradeep is poor ~p : Ramesh is not rich i.e. poor ~ q : Pradeep is not poor i.e. rich.
(i) (~P) ∧ (~ q)
(ii) (~ q) ∧ p
(iii) (~ p) ∧ q
(iv) ~ p ∧ q
(v) ~ [p ∧ (~ q)]
(vi) ~p ∨ q
(vii) p ∨ (~q)

Question 4.
Use p : I like this school ; q : I like Mr. Sexena. Express each of the following statements in words.
(i) p ∧ q
(ii) ~ q
(iii) ~p
(iv) (~p) ∧ (~ q)
(v) (~p) ∧ q
(vi) p ∨ q
(vii) ~ (p ∧ q)
(viii) ~ [(~ p) ∧ q]
Solution:
Given statements are ;
p : I like this school
q : I like Mr. Sexena.
(i) p ∧ q : I like this school and I like Mr. Sexena.
(ii) ~ q : I do not like Mr. Sexena.
(iii) ~p : I do not like this school.
(iv) ~p ∧ ~ q : I do not like this school and I do not like Mr. Sexena.
(v) ~ p ∧ q : I donot like this school but I like Mr. Sexena.
(vi) p v q : I like this school or I like Mr. Sexena.
(vii) ~ (p ∧ q) : It is not true that I like this school and I like Mr. Sexena.
(viii) ~ [(~ p) ∧ q] : It is not true that I donot like this school and I like Mr. Sexena.

Question 5.
Give the negation of each of the following statements.
(i) Either he is bald or he is tali.
(ii) Nobody does not like Madhuri.
(iii) It is not true that the set of prime numbers is finite.
(iv) All circles are round.
(v) Some students passed this course.
Solution:
(i) Let p : he is bald
q : he is tall.
∴ p ∨ q: either he is bald or he is tall.
Then ~ (p ∨ q) : He is not bald and he is not tall.
(ii) Negation of given statement is that somebody does not like Madhuri.
(iii) Let p : It is not true that set of prime numbers is finite.
∴ ~p : set of prime numbers is finite.
(iv) p : all circles are sound
~ p : all circles are not round or it is not the case that all circles are round.
(v) p : Some students passed this course
~ p : It is not true that, some students passed this course or No student passed this course.

Accessing ISC Mathematics Class 11 OP Malhotra Solutions Chapter 27 Mathematical Reasoning Ex 27(b) can be a valuable tool for students seeking extra practice.

S Chand Class 11 ICSE Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(b)

Question 1.
Identify the quantifier in the following statements.
(i) There exists a capital city for every state of India.
(ii) For every real numbers, x is less than x + 1.
(iii) At least one natural number is not a prime number.
Solution:
(i) There exists.
(ii) For every.
(iii) Atleast.

Question 2.
Symbolise the following statements.
(i) There is at least one number in the set of natural numbers which is equal to ‘its’ cube.
(ii) The square of every real number is positive.
(iii) There exists at least one number in A = {5, 7, 8, 9, 10} which is an even number.
(iv) For every real number x, x < x + 1.
(v) The square roots of all prime numbers are irrational numbers. (Let P denote the set of prime numbers and Q that of irrational numbers).
Solution:
(i) ∃ x ∈ N s.t x = x³
since 1 e N and 1 = 1³
(ii) ∀ x ∈ R s.tx² > 0
(iii) ∃ x ∈ A s.t x is even, since 8, 10 ∈ A
(iv) ∀ x ∈ R, x < x + 1, since successor of every real number is greater than itself.
(v) ∀ x ∈ P, √x ∈ Q

Practicing ISC Mathematics Class 11 OP Malhotra Solutions Chapter 27 Mathematical Reasoning Ex 27(a) is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 11 ICSE Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(a)

Question 1.
Open the door.
Solution:
It is an imperative sentence so it does not represents a statement.

Question 2.
5 is a prime number.
Solution:
It is a statement.

Question 3.
Do you like mathematics ?
Solution:
It is not a statement as it is an Interrogative sentence.

Question 4.
Every rectangle is a square.
Solution:
It is a statement.

Question 5.
Today is Sunday and tomorrow is Monday.
Solution:
It is a compound statement as its component statement are;
(i) Today is Sunday
(ii) Tomorrow is Monday.

Question 6.
May you live long!
Solution:
It is not a statement. It is an exclamatory sentence.

Question 7.
Rekha is studying in class eleven and she has to offer 5 subjects.
Solution:
It is a compound statement and its components are ;
(i) Rekha is studying in class eleven.
(ii) She has to offer 5 subjects.

Question 8.
The earth revolves around the moon.
Solution: It is a simple statement.

Question 9.
New Delhi is a big city and it is the capital of India.
Solution:
It is a compound statement and its components are ;
(i) New Delhi is a big city.
(ii) It is the capital of India.

Question 10.
20 is a prime number and 20 is less than 21.
Solution:
It represents a compound statement and its components are ;
(i) 20 is a prime number
(ii) 20 is less than 21.
Write down whether the following statements are true (T) or false (F).

Question 11.
8 is a prime number.
Solution:
It is a false statement.

Question 12.
Every square is a rectangle.
Solution:
It is a true statement.

Question 13.
The earth revolves around the moon.
Solution:
It is a false statement.

Question 14.
The set of whole numbers is a finite set.
Solution:
It is a false statement as the set of whole numbers is an infinite set.

Question 15.
32 is a multiple of 8.
Solution:
It is a true statement.

Question 16.
3 + 4i is a complex number.
Solution:
It is a true statement.

The availability of step-by-step ISC Mathematics Class 11 OP Malhotra Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Chapter Test can make challenging problems more manageable.

S Chand Class 11 ICSE Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Chapter Test

Question 1.
Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).
Solution:
Let P be any point on y-axis and its coordinates are (0, y, 0) and given points are A(3, 1, 2) and B (5, 5, 2).
According to given condition; we have

On squaring both sides; we have
9 + (1 – y)² + 4 = 25 + (5 – y)² + 4
⇒ 14 + y² – 2y = 54 + y² – 10y
⇒ 8y = 40
⇒ y = 5
Thus required point on y-axis be P(0, 5, 0).

Question 2.
Show that the points (a, b, c),(b, c, a) and (c, a, b) are the vertices of an equilateral trinagle.
Solution:
Let (a, b, c); (b, c, a) and (c, a, b) are the given vertices A, B and C of △ABC.

Thus, △ABC be an equilateral triangle.

Question 3.
Find out whether the points (0, 7, -10), (1, 6, 6) and (4, 9, -6) are the vertices of a right angled triangle.
Solution:
Let A(0,7,-10) ; B(1, 6, 6) and C(4, 9, -6) are the vertices of △ABC respectively.
AB² = (1 – 0)² + (6 – 7)² + (6 + 10)² = 1 + 1 +256 = 258
BC² = (4 – 1)² + (9 – 6)² + (- 6 – 6)² = 9 + 9 + 144 = 162
CA² = (4 – 0)² + (9 – 7)² + (- 6 + 10)² = 16 + 4 + 16 = 36
Thus,
AB² ≠ BC² + AC²
BC² ≠ AB² + AC²
AC² ≠ AB² + BC²
Thus, pythagoras theorem does not holds good. ∴ △ABC is not a right angled Δ.

Question 4.
Show that the points A(-2, 3, 5), B(1, 2, 3) and C(7, 0, -1) are collinear.
Solution:
Given points are A(-2, 3, 5); B(1, 2, 3) and C(7, 0, -1)
Here AB = \(\sqrt{(1+2)^2+(2-3)^2+(3-5)^2}\) = \(\sqrt{9+1+4}\) = \(\sqrt{14}\)
BC = \(\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}\) = \(\sqrt{36+4+16}\) = \(\sqrt{56}\)
and CA = \(\sqrt{(-2-7)^2+(3-0)^2+(5+1)^2}\) = \(\sqrt{81+9+36}\) = \(\sqrt{126}\) = \(3 \sqrt{14}\)
Thus AB + BC + CA
Therefore all the given points A, B and C lies on same straight line and hence given points are collinear.

Question 5.
Find the lengths of the medians of the tri- angle A(0, 0, 6), B(0, 4, 0) and C(6, 0, 0)
Solution:
Let D, E and F are the mid-points of sides BC, CA and AB of △ABC.
Thus coordinates of D, E and F using mid-point formula be given by D(3, 2, 0); E(3, 0, 3) and F(0, 2, 3).

Question 6.
Using section formula, show that the points (2, -3, 4), (-1, 2, 1) and (0, \(\frac { 1 }{ 3 }\), 2) are collinear.
Solution:
Let A, B and C are the given points (2, -3, 4), (-1, 2, 1) and (0, \(\frac { 1 }{ 3 }\), 2) respectively.
Thus the coordinates of the point divides the straight line joining the points B and C in the ratio k : 1 internally are

∴ 2 = –\(\frac{1}{k+1}\) ⇒ k + 1 = –\(\frac{1}{2}\) ⇒ k = –\(\frac{3}{2}\)
\(\frac{\frac{k}{3}+2}{k+1}\) = – 3 ⇒ \(\frac{k}{3}\) + 2 = – 3k – 3
⇒ \(\frac{10k}{3}\) = -5 ⇒ k = –\(\frac{3}{2}\)
and \(\frac{2 k+1}{k+1}\) ⇒ 2k + 1 = 4k + 4
⇒ 2k = – 3 ⇒ k = –\(\frac{3}{2}\)
Thus from all equations, we get same value of k.
Hence, the point A lies on the straight line joining B and C. Thus, points A, B and C are collinear.

Question 7.
Find the ratio in which the yz-plane divides the line segment formed by joining the point (-2, 4, 7) and (3, -5, 8).
Solution:
Let the point P divides the line segment joining A(-2, 4, 7) and B(3, -5, 8) in the ratio k : 1.

Clearly the line segment AB is divided by yz-plane so x-coordinate of point P is 0.
∴ \(\frac{3 k-2}{k+1}\) = 0 ⇒ k = \(\frac{2}{3}\)
Thus, required ratio be k : 1 i.e. 2 : 3

Students often turn to ISC Mathematics Class 11 OP Malhotra Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) to clarify doubts and improve problem-solving skills.

S Chand Class 11 ICSE Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b)

Question 1.
Find the coordinates of the points which divide the join of the points (2, -1, 3) and (4, 3, 1) in the ratio 3 : 4 internally.
Solution:
Let R be the point which divides the join of points A(2, -1, 3) and B(4, 3, 1) in the ratio 3 : 4 internally.

Question 2.
Find the coordinates of the points which divide the line joining the points (2, -4, 3), (-4, 5, -6) in the ratio
(i) 1 : -4
(ii) 2 : 1
Solution:
(i) Let the point P divides the line segment AB in ratio 1 :-4
Then coordinates of P are

Question 3.
Find the ratio in which the line joining the points (2, 4, 5),(3, 5, -4) is divided by the yz-plane.
Solution:
Let the point P divides the line segment joining A(2, 4, 5) and B(3, 5, -4) in the ratio k : 1

Since the line joining AB is divided by yz – plane
i.e. x-coordinates of point P is 0 .
\(\frac{3 k+2}{k+1}\) = 0
⇒ k = –\(\frac{2}{3}\)
Thus required ratio be k : 1 i.e. –\(\frac{2}{3}\) : 1 i.e. – 2 : 3.

Question 4.
The three points A(0, 0, 0), B(2, -3, 3), C(-2, 3, -3) are collinear. Find in what ratio each point divides the segment joining the other two.
Solution:
Let the point B (2, -3, 3) divides the line segment AC in the ratio k : 1 internally.
Then by section formula, we have
The coordinates of B are

∴ \( \frac{-2 k}{k+1}\) = 2
⇒ – 2k = 2k + 2
⇒ 4k = – 2
⇒ k = \(\frac{-1}{2}\)
and \(\frac{3 k}{k+1}\) = – 3
⇒ 3k = -3 k – 3
⇒ 6 k = – 3
⇒ k =-\(\frac{1}{2}\)
and \(\frac{-3 k}{k+1}\) = 3
⇒ – 3k = 3k + 3
⇒ k = –\(\frac{1}{2}\)
Thus the required ratio be k : 1 i.e. – 1 : 2.
Let the point C(-2, 3, -3) divides AB in the ratio λ : 1.
Then coordinates of C are

\(\frac{2 \lambda}{\lambda+1}\) = – 2 ⇒ 2λ = – 2λ – 2 ⇒ λ = –\(\frac { 1 }{ 2 }\)
\(\frac{-3 \lambda}{\lambda+1}\) = 3 ⇒ -3λ = 3λ + 3 ⇒ λ = –\(\frac { 1 }{ 2 }\)
and \(\frac{3 \lambda}{\lambda+1}\) = – 3 ⇒ 6λ = – 3
Thus required ratio be λ : 1 i.e. -1 : 2. Let the point A(0, 0, 0) divides line segment BC in the ratio p : 1.

Question 5.
Find the coordinates of the points which trisect AB given that A(2, 1, -3) and B (5, -8, 3).
Solution:
Let P and Q be the point of trisection of line segment AB.

Thus point P divides the line segment AB in the ratio 1 : 2.
Then coordinates of P are
\(\left(\frac{5+4}{1+2}, \frac{-8+2}{1+2}, \frac{3-6}{1+2}\right)\) i.e. P(3, -2, -1)
Also, point Q divides the line segment AB in the ratio 2 : 1.
Then coordinates of $\mathrm{Q}$ are
\(\left(\frac{10+2}{2+1}, \frac{-16+1}{2+1}, \frac{6-3}{2+1}\right)\) i.e. Q(4, -5, 1).

Question 6.
Find the coordinates of the point which is three-fifths of the way from (3, 4, 5) to (-2, -1, 0).
Solution:
Let the given points are A(3, 4, 5) and E(-2, -1, 0) and let point P is at \(\frac{3}{5}\)th of the way from A.
∴ P is at a \(\frac{2}{5}\)th of the way from B i.e. p divides line segment AB in the ratio 3 : 2.
Then by section formula, we have

Thus, the coordinates of point P are (0, 1, 2).

Question 7.
Show that the point (1, -1, 2) is common to the lines which join (6, -7), 0) to (16, -19, -4) and (0, 3, -6) to (2, -5, 10).
Solution:
Any point on line segment AB be

If AB and CD have a common point. Then P and Q coincide for some values of k and k’.

Thus eqns. (1), (2) and (3) are satisfied or consistent for k = \(\frac{-1}{3}\) and k’ = 1
putting k = \(\frac{-1}{3}\) in coordinates of P
we get, the required point be (1, -1, 2). Hence, the point P(1, -1, 2) is common to lines which join A(6, – 7, 0)
and B(16, -19, -4) and C(0, 3, -6) and D(2, -5, 10).

Question 8.
Find the lengths of the medians of the triangle whose vertices are A(2, -3, 1), B (-6, 5, 3), C (8, 7, – 7).
Solution:
Let D, E and F are the mid-points of sides BC, CA and AB of △ABC.

Question 9.
Find the point of intersection of the medians of the triangle with vertices (-1, -3, -4), (4, -2, -7), (2, 3, -8).
Solution:
Let the vertices of △ABC are A(-1, -3, -4); B(4, -2, -7) and C(2, 3, -8)
We know that the point of intersection of all medians of a triangle is called centroid of triangle.

Question 10.
Find the ratio in which the join of A(2, 1, 5) and B(3, 4, 3) is divided by the plane 2x + 2y – 2z = 1. Also, find the coordinates of the point of division.
Solution:
Let the point R divides the line segment PQ in the ratio k : 1 internally.

Then by section formula, we have coordinates of R are
\(\left(\frac{3 k+2}{k+1}, \frac{4 k+1}{k+1}, \frac{3 k+5}{k+1}\right)\)
Clearly it is given that, line segment PQ is divided by the plane
2x + 2y – 2z = 1 …(1)
Thus the point R lies on eqn. (1); we have
\(2\left(\frac{3 k+2}{k+1}\right)+2\left(\frac{4 k+1}{k+1}\right)-2\left(\frac{3 k+5}{k+1}\right)\) = 1
⇒ 6k + 4 + 8k + 2 – 6k – 10 = k + 1
⇒ 7k = 5 ⇒ k = \(\frac{5}{7}\)
Thus the required ratio be k : 1
i.e. \(\frac{5}{7}\) : 1 i.e. 5 : 7
Thus, required point of division be

Question 11.
The mid-points of the sides of astriangle are (1, 5, -1),(0, 4, -2) and (2, 3, 4). Find its vertices.
Solution:
Let the vertices of △ABC are A (x₁, y₁, z₁);
B (x₂, y₂, z₂) and
C (x₃, y₃, z₃).
It is given that D (1, 5, -1) ; E(0, 4, – 2) and F(2, 3, 4) are the mid-points of sides BC, CA and AB of △ABC.
Now D be the mid-point of BC.

On adding eqn. (1), (4) and (7); we have
x₁ + x₂ + x₃ = 3 …(10)
From eqn. (1) and eqn. (10); x₁ = 1
From (4) and (10); x₂ = 3
and From eqn. (7) and (10); x₃ = -1
On adding eqn. (2), (5) and (8); we have
y₁ + y₂ + y₃ = 12 …(11)
From (2) and (11); y₁ = 2
From (5) and (11); y₂ = 4
From (8) and (11); y₃ = 6
On adding eqn. (3), (6) and (9); we have
z₁ + z₂ + z₃ = 1 …(12)
From (3) and (12); z₁ = 3
From (6) and (12); z₂ = 5
From (9) and (12); z₃ = -7
Thus the required vertices of △ABC are (1, 2, 3) ;(3, 4, 5) and (-1, 6, -7).

Question 12.
Three vertices of a parallelogram ABCD are A(3, -1, 2), B(1, 2, -4) and C(-1, 1, 2). Find the coordinates of the fourth vertex D.
Solution:
Given vertices of parallelogram ABCD are A(3, -1, 2); B(1, 2, -4); C(-1, 1, 2) and let the coordinates of fourth vertex D are (α, β ,γ).
Mid-point of AC = \(\left(\frac{3-1}{2}, \frac{-1+1}{2}, \frac{2+2}{2}\right)\)
and Mid-point of BD = \(\left(\frac{1+\alpha}{2}, \frac{2+\beta}{2}, \frac{-4+\gamma}{2}\right)\)
Since ABCD be a parallelogram.
∴ diagonals of || gm ABCD bisect each other.
Thus, mid-point of AC = mid-point of BD

Thus the coordinates of fourth vertex D are $(1,-2,8)$.

Question 13.
What is the locus of a point for which
(i) x = 0
(ii) y = 0
(iii) z = 0
(iv) x =a
(v) y = b
(vi) z = c?
Solution:
(i) The point for which x = 0 is of the form (0, y, z) and lies in yoz plane. Thus required locus of point be yz-plane.

(ii) The point for which y = 0 is of the form (x, 0, z) and lies in the xoz plane.
Thus, required locus of point be xz-plane.

(iii) The point for which z = 0 is of the form (x, y, 0) and lies in xoy plane. Thus required locus of a point be xy-plane.

(iv) Since x = a be the plane parallel to x = 0 i.e. yz plane. Thus locus of a point for which x = a be a plane parallel to yz plane at a distance of a units from it.

(v) Since y = b be the plane parallel to y = 0 i.e. xz-plane. Thus, locus of a point for which y = b be a plane parallel to xz plane at a distance b units from it.

(vi) Since z = c be a plane parallel to plane z = 0 i.e. xy plane. Thus locus of a point for which z = c be a plane || to xy plane at a distance c units from it.

Question 14.
What is the locus of a point for which
(i) x = 0, y = 0
(ii) y = 0, z = 0
(iii) z = 0, x = 0
(iv) x = a, y = b
(v) y = b, z = c
(vi) z = c, x = a?
Solution:
(i) We know that on z-axis, x = 0 = y Thus required locus be z-axis.
(ii) We know that on x-axis, we have y = 0 = z Thus required locus be x-axis.
(iii) We know that on y-axis, we have x = 0 = z Thus, required locus be y-axis.
(iv) x = a be the line || to y-axis and y = b be the line || to x-axis
Thus locus of a point for which x = a, y = b is the line of intersection of given planes x = a and y = b
(v) Clearly the locus of a point for which y = b, z = c is the line of intersection of given planes y = b and z = c
(vi) Given planes are z = c and x = a
Required locus is the line of intersection of planes z = c and x = a.

The availability of step-by-step ISC Mathematics Class 11 OP Malhotra Solutions Chapter 25 Hyperbola Chapter Test can make challenging problems more manageable.

S Chand Class 11 ICSE Maths Solutions Chapter 25 Hyperbola Chapter Test

Question 1.
Find the eccentricity and the coordinate of foci of the hyperbola 25x² – 9y² = 225.
Solution:
Given eqn. of hyperbola can be written as ;
\(\frac{x^2}{9}\) – \(\frac{y^2}{25}\) = 1 …(1)
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
we have a² = 9 and b² = 25
We know that b² = a² (e² – 1)
⇒ 25 = 9(e² – 1)
⇒ \(\frac{25}{9}\) + 1 = e²
⇒ e = \(\frac{\sqrt{34}}{3}\) (∵ e > 0)
Thus the required eccentricity of given eqn. (1) be \(\frac{\sqrt{34}}{3}\)
The coordinates of foci are ( ± ae, 0) i.e. \(\left( \pm 3 \times \frac{\sqrt{34}}{3}, 0\right)\) i.e. \(( \pm \sqrt{34}, 0)\).

Question 2.
Find the value (s) of k so that the line 2x + y + k = 0 may touch the hyperbola 3x² – y² = 3.
Solution:
eqn. of given hyperbola be 3x² – y² = 3 ⇒ \(\frac{x^2}{1}\) – \(\frac{y^2}{3}\) = 1
On comparing eqn. (1) with\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
we have a² = 1; b² = 3
eqn. of given line be
2x + y + k = 0 ⇒ y = – 2x – k
We know that the line y = mx + c touches hyperbola (1) if c = ±\(\sqrt{a^2 m^2-b^2}\)
Here m = – 2 and c = – k
Thus, eqn. (2) touches hyperbola (1)
if – k = ± \(\sqrt{1 \times(-2)^2-3}\)
⇒ – k = ± 1
⇒ k = ± 1

Question 3.
From the following information, find the equation of the hyperbola and the equation of the transverse axis.
Focus (-2, 1), Directrix : 2x – 3y + 1 = 0, e = \(\frac{2}{\sqrt{3}}\)
Solution:
Let P (x, y) be any point on the hyperbola.
Then by definition, we have | PF | = | PM |

On squaring both sides ; we have
(x + 2)² + (y – 1)² =\frac{4}{39}(2x – 3y + 1)²
⇒ 39 [x² + 4x + 4 + y² – 2y + 1]
= 4 [4x² + 9y² – 12xy – 6y + 1+ 4x]
which is the required eqn. of hyperbola.
Axis of hyperbola is a line ⊥ to the directrix and pass through the focus (- 2, 1).
Thus eqn. of line ⊥ to 2x – 3y + 1 = 0 be given by
3x + 2y + k = 0 …(1)
eqn. (1) pass through the point (- 2, 1).
-6 + 2 + k = 0
⇒ k = 4
Thus eqn. (1) reduces to ; 3x + 2y + 4 = 0 be the required eqn. of axis of hyperbola.

Question 4.
Find the equation of the hyperbola whose eccentricity is √5 and the sum of whose semi-axes is 9.
Solution:
Let the required eqn. of hyperbola be
\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
given eccentricity of hyperbola be √5
∴ e = √5
Let a be length of semi-major and b be the length of semi-minor axes of hyperbola.
According to given condition, we have
a + b = 9 …(2)
We know that b² = a² (e² – 1)
⇒ b² = a²(5 – 1) = 4a²
⇒ (9 – a)² = 4a² [using eqn. (2)]
⇒ 3a² + 18a – 81 = 0
⇒ (a – 3)(3a + 27) = 0
⇒ a = 3 (∵ a > 0)
∴ from (2); b = 9 – 3 = 6
Thus eqn. (1) reduces to ;
\(\frac{x^2}{9}\) – \(\frac{y^2}{36}\) = 1
which is the required eqn. of hyperbola.

Question 5.
Find the equation of the hyperbola whose foci are (4, 1),(8, 1) and whose eccentricity is 2.
Solution:
We know that centre is the mid-point of line joining the two foci (4, 1) and (8, 1).
∴ Coordinates of centre of hyperbola are \(\left(\frac{4+8}{2}, \frac{1+1}{2}\right)\) i.e. (6, 1).
Since ordinate of both foci are identical
∴ Transverse axis of the hyperbola is parallel to x-axis.
Thus eqn. of hyperbola can be taken as
\(\frac{(x-6)^2}{a^2}\) – \(\frac{(y-1)^2}{b^2}\) = 1
Given eccentricity of eqn. (1) be 2 i.e. e = 2
∴ distance between foci
= \(\sqrt{(8-4)^2+(1-1)^2}\) = 4
⇒ 2ae = 4 ⇒ ae = 2 ⇒ a = 1
∴ b² = a² (e² – 1) = 1²(4 – 1) = 3
Thus, eqn. (1) reduces to ;
\(\frac{(x-6)^2}{1}\) – \(\frac{(y-1)^2}{3}\) = 1
⇒ 3 (x – 6)² – (y – 1)² = 3
which is the required eqn. of hyperbola.

Question 6.
Show that the line y = x + √7 touches the hyperbola 9x² – 16y² = 144.
Solution:
Given eqn. of hyperbola be
9x² – 16y² = 144
and eqn. of given line be
y = x + √7 …(2)
From (2); y = x + √7, putting in eqn. (1); we get
9x² – 16 (x + √7)² = 144
⇒ 9x² – 16 (x² + 7 + 2√7 x) – 144 = 0
⇒ -7x² – 32√7x – 256 = 0
⇒ 7x² + 32√7x + 256 = 0
⇒ (√7x +1 6)² = 0
eqn. (3) is quadratic in x and have equal roots
∴ from (3) ; x = \(\frac{-16}{\sqrt{7}}\)
Thus line (2) touches hyperbola eqn. (1).

Question 7.
Find the equation of the hyperbola whose foci are (0, ± 13) and the length of the conjugate axis is 20.
Solution:
Coordinates of foci are (0, ± 13) which are lies on y-axis and y-axis be the transverse axis of hyperbola.
Let the eqn. of hyperbola be,
\(\frac{y^2}{a^2}\) – \(\frac{x^2}{b^2}\) = 1
Here ae = 13
and 2b = 20 and b = 10
We know that, b² = a² (e² – 1)
⇒ 100 = 13² – a²
⇒ a² = 69
Thus eqn. (1) reduces to ; \(\frac{y^2}{69}\) – \(\frac{x^2}{100}\) = 1,
which is the required eqn. of hyperbola.

Question 8.
Find the equation of the hyperbola whose transverse and conjugate axes are the x and y axes respectively, given that the length of conjugate axis is 5 and distance between the foci is 13 .
Solution:
Let the eqn. of hyperbola be
\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1 …(1)
Given length of conjugate axis be 5
∴ 2b = 5 ⇒ b = \(\frac{5}{2}\)
and distance between foci = 13
⇒ 2 ae = 13 …(2)
We know that b² = a² (e² – 1)
⇒ \(\left(\frac{5}{2}\right)^2\) = \(\left(\frac{13}{2}\right)^2\) – a²
⇒ a² = \(\frac{169}{4}\) – \(\frac{25}{4}\) = \(\frac{144}{4}\) = 36
Thus, eqn. (1) reduces to;
\(\frac{x^2}{36}\) – \(\frac{y^2}{25 / 4}\) = 1
⇒ \(\frac{x^2}{36}\) – \(\frac{4 y^2}{25}\) = 1
⇒ 25x² – 144y² = 900
which is the required eqn. of hyperbola.

Question 9.
Find the equation to the conic whose focus is (1, -1), eccentricity is \(\frac{1}{2}\) and the directrix is the line x – y = 3. Is the conic section an ellipse?
Solution:
Given focus of ellipse be F (1, -1) and eqn. of directrix be x – y -3 = 0 and e = \(\frac{1}{2}\)
Let P (x, y) be any point on ellipse. Then by def. of ellipse, we have | PF | = e | PM |
where PM be the ⊥ drawn from P on given directrix.

On squaring both sides, we have
8[(x – 1)² + (y + 1)²] = (x – y – 3)²
⇒ 8 (x² – 2x + y² + 2y + 2) = x² + y² + 9 – 2xy + 6y – 6x
⇒ 7x² + 7y² + 2xy – 10x + 10y + 7 = 0
which is the required eqn. of ellipse.