Find what sum will amount to ₹ 73810 in two years at 10% per annum compound interest

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Solution:

We have,

Rate = 5 % per annum

Compound Interest (CI) = ₹164

Time (t) = 2 years

By using the formula,

Let P be ‘x’

CI = A – P

164 = P (1 + R/100) n – P

Substituting the values, we have

= P [(1 + R/100)n – 1]

= x [(1 + 5/100)2 – 1]

= x [(105/100)2 – 1]

164 = x ((1.05)2 – 1)

x = 164 / ((1.05)2 – 1)

= 164/0.1025

= ₹1600

Therefore 

The required sum is ₹1600.

Question 2. Find the principal if the interest compounded annually at the rate of 10% for two years is ₹210.

Solution:

We have,

Rate = 10 % per annum

Compound Interest (CI) = ₹210

Time (t) = 2 years

By using the formula,

Let P be ‘x’

CI = A – P

210 = P (1 + R/100)n – P

Substituting the values, we have

= P [(1 + R/100)n – 1]

= x [(1 + 10/100)2 – 1]

= x [(110/100)2 – 1]

210 = x ((1.1)2 – 1)

x = 210 / ((1.1)2 – 1)

= 210/0.21

= ₹1000

Therefore,

The required sum is ₹1000.

Question 3. A sum amounts to ₹756.25 at 10% per annum in 2 years, compounded annually. Find the sum.

Solution:

We have,

Rate = 10 % per annum

Amount = ₹756.25

Time (t) = 2 years

By using the formula,

A = P (1 + R/100)n

Substituting the values, we have

756.25 = P (1 + 10/100)2

P = 756.25/(1 + 10/100)2

= 756.25/1.21

= 625

Therefore, 

The principal amount is ₹625.

Question 4.  What sum will amount to ₹4913 in 18 months, if the rate of interest is 12 ½ % per annum, compounded half-yearly?

Solution:

We have,

Rate = 12 ½% per annum = 25/2% = 25/2/2 = 25/4% half yearly

Amount = ₹4913

Time (t) = 18months = 18/12years = 3/2 × 2 = 3 half years

By using the formula,

A = P (1 + R/100)n

Substituting the values, we have

4913 = P (1 + 25/4 ×100)3

P = 4913 / (1 + 25/400)3

= 4913/1.19946

= 4096

Therefore, 

The principal amount is ₹4096.

Question 5. The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is ₹283.50. Find the sum.

Solution:

We have,

Rate = 15 % per annum

Compound Interest (CI) – Simple Interest (SI)= ₹283.50

Time (t) = 3 years

By using the formula,

CI – SI = 283.50

P [(1 + R/100)n – 1] – (PTR)/100 = 283.50

Substituting the values, we have

P [(1 + 15/100)3 – 1] – (P(3)(15))/100 = 283.50

P[1.520 – 1] – (45P)/100 = 283.50

0.52P – 0.45P = 283.50

0.07P = 283.50

P = 283.50/0.07

= 4000

Therefore,

The sum is ₹4000.

Question 6. Rachana borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years ₹1290 as interest compounded annually, find the sum she borrowed.

Solution:

We have,

Rate = 15 % per annum

Time = 2 years

CI = Rs 1290

By using the formula,

CI = P [(1 + R/100)n – 1]

Substituting the values, we have

1290 = P [(1 + 15/100)2 – 1]

1290 = P [0.3225]

P = 1290/0.3225

= 4000

Therefore,

The sum is ₹4000.

Question 7.  The interest on a sum of ₹2000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is ₹163.20.

Solution:

We have,

Rate = 4 % per annum

CI = ₹163.20

Principal (P) = Rs 2000

By using the formula,

CI = P [(1 + R/100)n – 1]

Substituting the values, we have

163.20 = 2000[(1 + 4/100)n – 1]

163.20 = 2000[(1.04)n -1]

163.20 = 2000 × (1.04)n – 2000

163.20 + 2000 = 2000 × (1.04)n

2163.2 = 2000 × (1.04)n

(1.04)n = 2163.2/2000

(1.04)n = 1.0816

(1.04)n = (1.04)2

So on comparing both the sides, n = 2

Therefore,

Time required is 2 years.

Question 8. In how much time would ₹5000 amount to ₹6655 at 10% per annum compound interest?

Solution:

We have,

Rate = 10% per annum

A = ₹6655

Principal (P) = ₹5000

By using the formula,

A = P (1 + R/100)n

Substituting the values, we have

6655 = 5000 (1 + 10/100)n

6655 = 5000 (11/10)n

(11/10)n = 6655/5000

(11/10)n = 1331/1000

(11/10)n = (11/10)3

So on comparing both the sides, n = 3

Therefore,

Time required is 3 years.

Question 9. In what time will ₹4400 become ₹4576 at 8% per annum interest compounded half-yearly?

Solution:

We have,

Rate = 8% per annum = 8/2 = 4% (half yearly)

A = Rs 4576

Principal (P) = ₹4400

Let n be ‘2T’

By using the formula,

A = P (1 + R/100)n

Substituting the values, we have

4576 = 4400 (1 + 4/100)2T

4576 = 4400 (104/100)2T

(104/100)2T = 4576/4400

(104/100)2T= 26/25

(26/25)2T = (26/25)1

So on comparing both the sides, n = 2T = 1

Therefore, 

Time required is 1/2 year.

Question 10. The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is ₹20. Find the sum.

Solution:

We have,

Rate = 4 % per annum

Time = 2 years

Compound Interest (CI) – Simple Interest (SI)= ₹20

By using the formula,

CI – SI = 20

P [(1 + R/100)n – 1] – (PTR)/100 = 20

Substituting the values, we have

P [(1 + 4/100)2 – 1] – (P(2)(4))/100 = 20

P[51/625] – (2P)/25 = 20

51/625P – 2/25P = 20

(51P-50P)/625 = 20

P = 20 × 625

P = 20/7.918

= 12500

Therefore 

The sum is ₹12500.

Question 11. In what time will Rs. 1000 amount to Rs. 1331 at 10% per annum, compound interest?

Solution:

We have,

Principal = Rs 1000

Amount = Rs 1331

Rate = 10% per annum

Let time = T years

By using the formula,

A = P (1 + R/100)n

Substituting the values, we have

1331 = 1000 (1 + 10/100)T

1331 = 1000 (110/100)T

(11/10)T = 1331/1000

(11/10)T = (11/10)3

So on comparing both the sides, n = T = 3

Therefore,

Time required is 3 years.

Question 12. At what rate percent compound interest per annum will Rs. 640 amount to Rs. 774.40 in 2 years?

Solution:

We have,

Principal = Rs 640

Amount = Rs 774.40

Time = 2 years

Let rate = R%

By using the formula,

A = P (1 + R/100)n

Substituting the values, we have

774.40 = 640 (1 + R/100)2

(1 + R/100)2 = 774.40/640

(1 + R/100)2 = 484/400

(1 + R/100)2 = (22/20)2

By canceling the powers on both sides,

(1 + R/100) = (22/20)

R/100 = 22/20 – 1

= (22-20)/20

= 2/20

= 1/10

R = 100/10

= 10%

Therefore,

Required Rate is 10% per annum.

Question 13. Find the rate percent per annum if Rs. 2000 amount to Rs. 2662 in 1 ½ years, interest being compounded half-yearly?

Solution:

We have,

Principal = Rs 2000

Amount = Rs 2662

Time = 1 ½ years = 3/2 × 2 = 3 half years

Let rate be = R% per annum = R/2 % half yearly

By using the formula,

A = P (1 + R/100)n

Substituting the values, we have

2662 = 2000 (1 + R/2×100)3

(1 + R/200)3 = 1331/1000

(1 + R/100)3 = (11/10)3

By canceling the powers on both sides,

(1 + R/200) = (11/10)

R/200 = 11/10 – 1

= (11-10)/10

= 1/10

R = 200/10

= 20%

Therefore,

Required Rate is 20% per annum.

Question 14. Kamala borrowed from Ratan a certain sum at a certain rate for two years simple interest. She lent this sum at the same rate to Hari for two years compound interest. At the end of two years she received Rs. 210 as compound interest, but paid Rs. 200 only as simple interest. Find the sum and the rate of interest.

Solution:

We have,

C.I that Kamala receives = Rs 210

S.I that Kamala paid = Rs 200

Time = 2 years

So,

We know, SI = PTR/100

= P×2×R/100

P×R = 10000 ………….. Equation 1

CI = A – P

CI = P [(1 + R/100)n – 1]

Substituting the values, we have

210 = P [(1 + R/100)2 – 1]

210 = P (12 + R2/1002 + 2(1)(R/100) – 1) (by using the formula (a+b)2)

210 = P (1 + R2/10000 + R/50 – 1)

210 = P (R2/10000 + R/50)

210 = PR2/10000 + PR/50

We know PR = 10000 from Equation 1

210 = 10000R/10000 + 10000/50

210 = R + 200

R = 210 – 200

= 10%

In Equation 1, PR = 10000

P = 10000/R

= 10000/10

= 1000

Therefore,

Required sum is Rs 1000.

Chapter 14 Compound Interest – Exercise 14.3 | Set 2

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Solution:

We have,

Principal = Rs 2000

Amount = Rs 2315.25

Time = 1 ½ years = 3/2 years

Let rate be = R % per annum

By using the formula,

A = P (1 + 

)n

Substituting the values, we have

2315.25 = 2000 (1 + 

)3/2

(1 + 

)3/2 = 2315.25/2000

(1 + 

)3/2 = (1.1576)

(1 + 

) = 1.1025

 = 1.1025 – 1

= 0.1025 × 100

= 10.25

Therefore,

Required Rate is 10.25% per annum.

Question 16. Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.

Solution:

We have,

Time = 3 years

Let rate be = R %

Also principal be = P

So, amount becomes = 2P

By using the formula,

A = P (1 + 

)n

Substituting the values, we have

2P = P (1 + 

)3

(1 + 

)3 = 2

(1 + 

) = 21/3

1 + 

 = 1.2599

 = 1.2599-1

= 0.2599

R = 0.2599 × 100

= 25.99

Therefore,

Required Rate is 25.99% per annum.

Question 17. Find the rate at which a sum of money will become four times the original amount in 2 years, if the interest is compounded half-yearly

Solution:

We have,

Time = 2 years = 2×2 = 4 half years

Let rate = R % per annum = R/2% half years

Let principal be = P

So, Amount becomes = 4P

By using the formula,

A = P (1 + 

)n

Substituting the values, we have

4P = P (1 + 

)4

(1 + 

)4 = 4

(1 + 

) = 41/4

1 + 

 = 1.4142

 = 1.4142-1

= 0.4142

R = 0.4142 × 200

= 82.84%

Therefore,

Required Rate is 82.84% per annum.

Question 18. A certain sum amounts to Rs. 5832 in 2 years at 8% compounded interest. Find the sum.

Solution:

We have,

Amount = Rs 5832

Time = 2 years

Rate = 8%

Let principal be = P

By using the formula,

A = P (1 + 

)n

Substituting the values, we have

5832 = P (1 + 

)2

5832 = P (1.1664)

P = 5832/1.1664

= 5000

Therefore,

Required sum is Rs 5000.

Question 19. The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs. 360. Find the sum.

Solution:

We have,

Time = 2 years

Rate = 7.5 % per annum

Let principal = Rs P

Compound Interest (CI) – Simple Interest (SI) = Rs 360

C.I – S.I = Rs 360

By using the formula,

P [(1 + 

)n – 1] – (PTR)/100 = 360

Substituting the values, we have

P [(1 + 

)2 – 1] – (P(2)(7.5))/100 = 360

P[249/1600] – (3P)/20 = 360

249/1600P – 3/20P = 360

(249P-240P)/1600 = 360

9P = 360 × 1600

P = 576000/9

= 64000

Therefore,

The sum is Rs 64000.

Question 20. The difference in simple interest and compound interest on a certain sum of money at 623 % per annum for 3 years in Rs. 46. Determine the sum.

Solution:

We have,

Time = 3 years

Rate = 6

 % per annum = 20/3%

Let principal = Rs P

Compound Interest (CI) – Simple Interest (SI) = Rs 46

C.I – S.I = Rs 46

By using the formula,

P [(1 + 

)n – 1] – (PTR)/100 = 46

Substituting the values, we have

P [(1 + 

)3 – 1] – (P(3)(20/3))/100 = 46

P[(1 + 

)3 – 1] – P/5 = 46

P[721/3375] – P/5 = 46

721/3375P – 1/5P = 46

(721P-675P)/3375 = 46

46P = 46 × 3375

46P = 46 × 3375/46

= 3375

Therefore, 

The sum is Rs 3375.

Question 21. Ishita invested a sum of Rs. 12000 at 5% per annum compound interest. She received an amount of Rs. 13230 after n years. Find the value of n.

Solution:

We have,

Principal = Rs 12000

Amount = Rs 13230

Rate = 5% per annum

Let time = T years

By using the formula,

A = P (1 + 

)n

Substituting the values, we have

13230 = 12000 (1 + 

)T

13230 = 12000 (

)T

(21/20)T = 13230/12000

(21/20)T = 441/400

(21/20)T = (21/20)2

So on comparing both the sides, n = T = 2

Therefore, 

Time required is 2 years.

Question 22. At what rate percent per annum will a sum of Rs. 4000 yield compound interest of Rs. 410 in 2 years?

Solution:

We have,

Principal = Rs 4000

Time = 2 years

CI = Rs 410

Rate be = R% per annum

By using the formula,

CI = P [(1 + 

)n – 1]

Substituting the values, we have

410 = 4000 [(1 + 

)2 – 1]

410 = 4000 (1 + 

)2 – 4000

410 + 4000 = 4000 (1 + 

)2

(1 + 

)2 = 4410/4000

(1 + 

)2 = 441/400

(1 + 

)2 = (21/20)2

By canceling the powers on both the sides,

1 + 

 = 21/20

 = 21/20 – 1

= (21-20)/20

= 1/20

R = 100/20

= 5

Therefore,

Required Rate is 5% per annum.

Question 23. A sum of money deposited at 2% per annum compounded annually becomes Rs. 10404 at the end of 2 years. Find the sum deposited.

Solution:

We have,

Time = 2years

Amount = Rs 10404

Rate be = 2% per annum

Let principal be = Rs P

By using the formula,

A = P [(1 + 

)n

Substituting the values, we have

10404 = P [(1 + 

)2]

10404 = P [1.0404]

P = 10404/1.0404

= 10000

Therefore, 

Required sum is Rs 10000.

Question 24. In how much time will a sum of Rs. 1600 amount to Rs. 1852.20 at 5% per annum compound interest?

Solution:

We have,

Principal = Rs 1600

Amount = Rs 1852.20

Rate = 5% per annum

Let time = T years

By using the formula,

A = P (1 + 

)n

Substituting the values, we have

1852.20 = 1600 (1 + 

)T

1852.20 = 1600 (

)T

(21/20)T = 1852.20/1600

(21/20)T = 9261/8000

(21/20)T = (21/20)3

So on comparing both the sides, n = T = 3

Therefore,

Time required is 3 years.

Question 25. At what rate percent will a sum of Rs. 1000 amount to Rs. 1102.50 in 2 years at compound interest?

Solution:

We have,

Principal = Rs 1000

Amount = Rs 1102.50

Rate = R% per annum

Let time = 2 years

By using the formula,

A = P (1 + 

)n

Substituting the values, we have

1102.50 = 1000 (1 + 

)2

(1 + 

)2 = 1102.50/1000

(1 + 

)2 = 4410/4000

(1 +

)2 = (21/20)2

1 + 

 = 21/20

= 21/20 – 1

= (21-20)/20

= 1/20

R = 100/20

= 5

Therefore,

Required Rate is 5%.

Question 26. The compound interest on Rs. 1800 at 10% per annum for a certain period of time is Rs. 378. Find the time in years.

Solution:

We have,

Principal = Rs 1800

CI = Rs 378

Rate = 10% per annum

Let time = T years

By using the formula,

CI = P [(1 + 

)n – 1]

Substituting the values, we have

378 = 1800 [(1 + 

)T – 1]

378 = 1800 [(

)T – 1]

378 = 1800 [(

)T – 1800

378 + 1800 = 1800 [(

)T

(11/10)T = 2178/1800

(11/10)T = 726/600

(11/10)T = 121/100

(11/10)T = (11/10)2

So on comparing both the sides, n = T = 2

Therefore,

Time required is 2 years.

Question 27. What sum of money will amount to Rs. 45582.25 at 6 ¾ % per annum in two years, interest being compounded annually

Solution:

We have,

Time = 2years

Amount = Rs 45582.25

Rate be = 6 ¾ % per annum = 27/4%

Let principal be = Rs P

By using the formula,

A = P [(1 + 

)n

Substituting the values, we have

45582.25 = P [(1 + 27/4×100)2]

45582.25 = P (1 + 

)2

45582.25 = P (

)2

45582.25 = P × 427/400 × 427/400

P = (45582.25 × 400 × 400) / (427×427)

P = 7293160000/182329

= 40000

Therefore,

Required sum is Rs 40000.

Question 28. Sum of money amounts to Rs. 453690 in 2 years at 6.5% per annum compounded annually. Find the sum.

Solution:

We have,

Time = 2years

Amount = Rs 453690

Rate be = 6.5 % per annum

Let principal be = Rs P

By using the formula,

A = P [(1 + 

)n

Substituting the values, we have

453690 = P [(1 + 

)2]

453690 = P (

)2

453690 = P × 106.5/100 × 106.5/100

P = (453690 × 100 × 100) / (106.5×106.5)

P = 4536900000/11342.25

= 400000

Therefore,

Required sum is Rs 400000.

Page 3

(i) The sky is blue, and the grass is green.

(ii) The earth is round, or the sun is cold.

(iii) All rational numbers are real, and all real numbers are complex.

(iv) 25 is a multiple of 5 and 8.

Solution:

Every compound statement has components .

(i) The component’s of the statement are:

A: The sky is blue.

B: The grass is green.

(ii) The component’s of the statement are:

A: The earth is round.

B: The sun is cold.

(iii) The component’s statement are:

A: All rational number are real.

B: All real number are complex.

(iv) The component’s of the statement are:

A: 25 is multiple of 5.

B: 25 is multiple of 8.

Question 2. For each of the following statements, determine whether an inclusive “OR” o exclusive “OR” is used. Give reasons for your answer.

(i) Students can take Hindi or Sanskrit as their third language.

(ii) To entry a country, you need a passport or a voter registration card.

(iii) A lady gives birth to a baby boy or a baby girl.

(iv) To apply for a driving license, you should have a ration card or a passport.

Solution:

(i) A choice is given to a student to choose either Hindi or Sanskrit as their third language. The student can’t choose both. Therefore, an exclusive “OR” is used because a student cannot take both Hindi and Sanskrit.

(ii) A passport or voter registration card is needed by a person to enter a country. However, a person may have both. Therefore, an inclusive “OR” is used because a person can have both a passport and a voter registration card to enter a country.

(iii) A lady cannot give birth to a baby who is both a boy and a girl, therefore, an exclusive “OR” is used because either a boy or girl is born.

(iv) A ration card  or passport is needed to apply a driving license. However, a person may have both.Therefore, an inclusive “OR” is used because a person can have both a ration card and passport to apply for a driving license.

(i) To enter into a public library children need an identification card from the school or a letter from the school authorities.

(ii) All rational numbers are real and all real numbers are not complex.

(iii) Square of an integer is positive or negative.

(iv) x = 2 and x = 3 are the roots of the equation 3x2 – x – 10 = 0.

(v) The sand heats up quickly in the sun and does not cool down fast at night.

Solution:

(i) The components are:

P: To get into a public library children need an identity card from the school.

Q: To get into a public library children need a letter from the school authorities.

Both of the statements P and Q are true. Therefore, the compound statement is true.

(ii) The components are:

P: All rational numbers are real.

Q: All real numbers are not complex.

The statement P is true and Q is false therefore, P and Q both together are False. Therefore, the compound statement is False

(iii) The components are:

P: Square of an integer is positive.

Q: Square of an integer is negative.

Both statements P and Q are true. Therefore, the compound statement is True.

(iv) The components of the compound statement are:

P: x = 2 is a root of the equation 3x2 – x – 10 = 0

Q: x = 3 is a root of the equation 3x2 – x – 10 = 0

The statement P is true, but Q is false then P and Q both combined are False. Therefore, the compound statement is False.

(v) The components of the compound statement are:

P: The sand heats up quickly in the sun.

Q: The sand does not cool down fast at night.

Both statements P and Q both are False. Therefore, the compound statement is False.

(i) Delhi is in India and 2 + 2 = 4

(ii) Delhi is in England and 2 + 2 = 4

(iii) Delhi is in India and 2 + 2 = 5

(iv) Delhi is in England and 2 + 2 = 5

Solution:

(i) The components are:

A: Delhi is in India.

B: 2 + 2 = 4

Both statements A and B are true. Hence, the compound statement is True.

(ii) The components are:

A: Delhi is in England.

B: 2 + 2 = 4

The statement A is false, and B is true. So, both A and B together are false. Hence, the compound statement is False.

(iii) The components are:

A: Delhi is in India.

B: 2 + 2 = 5

The statement B is false, and A is true. So, both A and B together are false. Hence, the compound statement is False.

(iv) The components are:

A: Delhi is in England.

B: 2 + 2 = 5

Both the statements A and B are false. Hence, the compound statement is False.

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(i) Bangalore is the capital of Karnataka.

(ii) It rained on July 4, 2005.

(iii) Ravish is honest.

(iv) The earth is round.

(v) The sun is cold.

Solution:

(i) Bangalore is not the capital of Karnataka or It is false that “Bangalore is the capital of Karnataka.”

(ii) It did not rain on July 4, 2005 or It is false that it rained on July 4, 2005.

(iii) Ravish is dishonest or It is false that “Ravish is honest”.

(iv) The earth is not round or It is false that “The earth is round.”

(v) The sun is not cold or It is false that “The sun is cold.”

Question 2. Write the negation of the following statement:

(i) All birds sing.

(ii) Some even integers are prime.

(iii) There is a complex number which is not a real number.

(iv) I will not go to school.

(v) Both the diagonals of a rectangle have the same length.

(vi) All policemen are thieves.

Solution:

(i) All birds do not sing or It is false that “All birds sing.”

(ii) Not all even integers are prime or It is false that “even integers are prime.”

(iii) All complex number are real numbers or  It is false that “complex numbers are not a real number.”

(iv) I will go to school.

(v) There is at least one rectangle whose both diagonals of unequal length.

(vi) No policemen are thief.

Question 3. Are the following pairs of statements are a negation of each other:

(i) The number x is not a rational number.

The number x is not an irrational number.

(ii) The number x is not a rational number.

The number x is an irrational number.

Solution:

(i) The number x is not a irrational number, means that the number x is a rational number.

Therefore, the second statement is negation of the first statement. 

(ii) The number x is not a rational number means that the number x is an irrational number.

Therefore, the second statement is similar to the first statement, and therefore, they are not negation of each other. 

Question 4. Write the negation of the following statements:

(i) p: For every positive real number x, the number (x – 1) is also positive.

(ii) q: For every real number x, either x > 1 or x < 1.

(iii) r: There exists a number x such that 0 < x < 1.

Solution:

(i) We have, 

p: For every positive real number x, the number (x – 1) is also positive.

The negation of the statement is, 

~p: There exists at least one positive real number x, such that the number (x – 1) is not positive.

(ii) The negation of the statement:

~q: There exists at least one real number, s.t neither x>1 nor x<1.

(iii) The negation of the statement:

~r: For all real numbers x, such that either x ≤ 0 or x ≥ 1.

Question 5. Check whether the following pair of statements is a negation of each other. Give reasons for your answer.

(i) a + b = b + a is true for every real number a and b.

(ii) There exist real numbers a and b for which a + b = b + a.

Solution:

The negation of the (i) statement:

There exist real numbers are ‘a’ and ‘b’ for which a + b ≠ b + a.

So, (ii) is not negation of (i). Hence, these statements are not a negation of each other.

Page 5

(i) You can access the website only if you pay a subscription fee.

(ii) There is traffic jam whenever it rains.

(iii) It is necessary to have a passport to log on to the server.

(iv) It is necessary to be rich in order to be happy.

(v) The game is canceled only if it is raining.

(vi) It rains only if it is cold.

(vii) Whenever it rains, it is cold.

(viii) It never rains when it is cold.

Solution:

(i) If you pay a subscription fee, then you can access the website .

(ii) If it rains, then there is a traffic jam.

(iii) If you log on the server, then you must have a passport.

(iv) If he is happy, then he is rich.

(v) If it is raining, then the game is canceled.

(vi) If it rains, then it is cold.

(vii) If it rains, then it is cold.

(viii) If it is cold, then it never rains.

Question 2. State the converse and contrapositive of each of the following statements:

(i) If it is hot outside, then you feel thirsty.

(ii) I go to a beach whenever it is a sunny day.

(iii) A positive integer is prime only if it has no divisions other than 1 and itself.

(iv) If you live in Delhi, then you have winter clothes.

(v) If a quadrilateral is a parallelogram, then its diagonals bisect each other.

Solution:

(i) If it is hot outside, then you feel thirsty.

The converse and contrapositive are as follows : 

Converse: If you feel thirsty, then it is hot outside.

Contrapositive: If you do not feel thirsty, then it is not hot outside.

(ii) I go to a beach whenever it is a sunny day.

The converse and contrapositive are as follows : 

Converse: If I go to a beach, then it is a sunny day.

Contrapositive: If I do not go to a beach, then it is not a sunny day.

(iii) A positive integer is prime only if it has no divisions other than 1 and itself.

The converse and contrapositive are as follows : 

Converse: If an integer has no divisor other that 1 and itself, then it is prime.

Contrapositive: If an integer has some divisor other than 1 and itself, then it is prime.

(iv) If you live in Delhi, then you have winter clothes.

The converse and contrapositive are as follows : 

Converse: If you have winter clothes, then you live in Delhi.

Contrapositive: If you do not have winter clothes, then you do not live in Delhi.

(v) If a quadrilateral is a parallelogram, then its diagonals bisect each other.

The converse and contrapositive are as follows : 

Converse: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Contrapositive: If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram.

(i) p: If you watch television, then your mind is free, and if your mind is free, then you watch television.

(ii) q: If a quadrilateral is equiangular, then it is a rectangle, and if a quadrilateral is a rectangle, then it is equiangular.

(iii) r: For you to get an A grade, it is necessary and sufficient that you do all the homework you regularly.

(iv) s: If a tumbler is half empty, then it is half full, and if a tumbler is half full, then it is half empty.

Solution:

(i) You watch television if and only if your mind is free.

(ii) A quadrilateral is a rectangle if and only if it is equiangular.

(iii) You get an A grade if and only if you do all the homework regularly.

(iv) A tumbler is half empty if and only if it is half full.

Question 4. Determine the Contrapositive of each of the following statements:

(i) If Mohan is a poet, then he is poor.

(ii) Only if Max studies will he pass the test.

(iii) If she works, she will earn money.

(iv) If it snows, then they do not drive the car.

(v) It never rains when it is cold.

(vi) If Ravish skis, then it snowed.

(vii) If x is less than zero, then x is not positive.

(viii) If he has courage he will win.

(ix) It is necessary to be strong in order to be a sailor.

(x) Only if he does not tire will he win.

(xi) If x is an integer and x2 is odd, then x is odd.

Solution:

(i) If Mohan is a poet, then he is poor.

The contrapositive is as follows : 

Contrapositive: If Mohan is not poor, then he is not a poet.

(ii) Only if Max studies will he pass the test.

The contrapositive is as follows :

Contrapositive: If Max does not study, then he will not pass the test.

(iii) If she works, she will earn money.

The contrapositive is as follows :

Contrapositive: If she does not earn money, then she does not work.

(iv) If it snows, then they do not drive the car.

The contrapositive is as follows :

Contrapositive: If then they do not drive the car, then there is no snow.

(v) It never rains when it is cold.

The contrapositive is as follows:

Contrapositive: If it rains, then it is not cold.

(vi) If Ravish skis, then it snowed.

The contrapositive is as follows :

Contrapositive: If it did not snow, then Ravish will not ski.

(vii) If x is less than zero, then x is not positive.

The contrapositive is as follows:

Contrapositive: If x is positive, then x is not less than zero.

(viii) If he has courage he will win.

The contrapositive is as follows :

Contrapositive: If he does not win, then he does not have courage.

(ix) It is necessary to be strong in order to be a sailor.

The contrapositive is as follows :

Contrapositive: If he is not strong, then he is not a sailor

(x) Only if he does not tire will he win.

The contrapositive is as follows :

Contrapositive: If he tries, then he will not win.

(xi) If x is an integer and x2 is odd, then x is odd.

The contrapositive is as follows :

Contrapositive: If x is even, then x2 is even.