Number Systems

1. Number Systems

Class : IX

Subject: Mathematics

NCERT Text book Solution.

EXERCISE 1.1

1. Is zero a rational number? Can you write it in the form p/q, where p and q are integers and q ≠ 0?

Solution:

Yes zero is a rational number.

It can be written in the form p/q which is 0/1.

2. Find six rational numbers between 3 and 4.

Solution:

Both 3 and 4 are multiplied and divided by (6 + 1 = 7)

3 = 3×7/7 = 21/7

4 = 4×7/7 = 28/7

Required rational numbers are

22/7, 23/7, 24/7, 25/7, 26/7 and 27/7 Ans.

3. Find five rational numbers between ⅗ and ⅘.

Solution:

Both ⅗ and ⅘ multiplied and divided by (5+1=6)

⅗ = 18/30

⅘ = 24/30

Required rational numbers are

19/30, 20/30, 21/30, 22/30 and 23/30 Ans.

4. State whether the following statements are true or false. Give reasons for your answers.

(i) Every natural number is a whole number.

(ii) Every integer is a whole number.

(iii) Every rational number is a whole number.

Solution:

(i)

True, Natural numbers are a part of whole numbers.

(ii)

False, Negative integers are not whole numbers.

(iii)

False, ½ is a rational number but not a whole number.

प्रश्नावली 1.1

1. क्या शून्य एक परिमेय संख्या है? क्या इसे आप p/q के रूप में लिख सकते हैं, जहाँ p और q पूर्णांक हैं और q ≠ 0 है?

हल:

हां, शून्य एक परिमेय संख्या है। इसे p/q के रूप में लिखा जा सकता है जो 0/1 है।

2. 3 और 4 के बीच में छः परिमेय संख्याएँ ज्ञात कीजिए।

हल:

3 और 4 दोनों के अंश एवं हर को 7 से गुना करने पर।

3 = 21/7

4 = 28/7

अतः 6 परिमेय संख्या है

22/7, 23/7, 24/7, 25/7, 26/7 एवं 27/7।

3. ⅗ और ⅘ के बीच पाँच परिमेय संख्याएँ ज्ञात कीजिये ।

हल:

⅗ एवं ⅘ दोनों के अंश एवं हर को 6 से गुना करने पर।

⅗ = 18/30

⅘ = 24/30

अतः 5 परिमेय संख्या है

19/30, 20/30, 21/30, 22/30 एवं 23/30

4. नीचे दिए गए कथन सत्य हैं या असत्य? कारण के साथ अपने उत्तर दीजिए।

(i) प्रत्येक प्राकृत संख्या एक पूर्ण संख्या होती है।

(ii) प्रत्येक पूर्णांक एक पूर्ण संख्या होती है।

(iii) प्रत्येक परिमेय संख्या एक पूर्ण संख्या होती है।

हल:

(i)

सत्य है, क्योंकि पूर्ण संख्या के समूह में प्राकृत संख्या शामिल है।

(ii)

असत्य है, क्योंकि ऋणात्मक पूर्णांक पूर्ण संख्या नही होता है।

(iii)

असत्य है, जैसे ½ परिमेय संख्या है लेकिन पूर्ण संख्या नही है।

EXERCISE 1.2

1. State whether the following statements are true or false. Justify your answers.

(i) Every irrational number is a real number.

Solution:

True, every irrational numbers are represented on a number line.

(ii) Every point on the number line is of the form, √m where m is a natural number.

Solution:

False, every point on the number is of the form of real number.

(iii) Every real number is an irrational number.

Solution:

False, every real number is either rational or irrational.

2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

Solution:

The square root of all positive integers are not irrational.

Example

√4 is a rational number.

3. Show how √5 can be represented on the number line.

Solution:

Following steps are used to show √5 on number line.

Step 1.

Apply Pythagoras theorem to get base for unit perpendicular and for hypotenuse of √5.

b² = (√5)² - 1² = 4 = 2² 

Therefore, 

b = 2.

Step 1.

Draw perpendicular on 2 of unit that is P.

Step 2.

Join O to P.

Step 3.

Draw an arch of radius OP and centre O. The arch intersects the number line at Q. This represent √5 on number line.

Figure is given below.

Quadratic Equations

Let's Solve the Questions of

Class X.

4. Quadratic Equations.

Exercise 4.4 

Subject Mathematics 

Publication : NCERT.

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) 2x² – 3x + 5 = 0 

(ii) 3x² – 4√3x + 4 = 0

(iii) 2x² – 6x + 3 = 0

Solution:

(i)

a = 2, b = -3, c = 5

D = b² -4ac

or, D = -3² -4×2×5

or, D = 9 -40

∴ D = -31

D<0

Hence, roots are not real.

(ii)

a = 3, b =-4√3, c = 4

D = b² -4ac

D = -4√3² -4×3×4

D = 48 - 48

D = 0

D = 0

Hence, roots are real and equal.

x = (-b ±√D)÷2a

or, x = (-(-4√3) ±√0)÷2×3

or, x = 4√3 ÷ 6

x = 2√3/3 Ans.

(iii)

a = 2, b = -6, c = 3

D = b² -4ac

or, D = -6² -4×2×3

or, D = 36 - 24

or, D = 12

D>0

Hence, roots are real and equal.

x = (-b ±√D)÷2a

x = (-(-6) ±√12)÷2×2

x = (6 ±2√3)÷4

x = (3 ±√3)÷2

∴ x = (3 +√3)/2 or (3 -√3)/2 Ans.

2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(i) 2x² + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0

Solution:

2x² + kx + 3 = 0

a = 2, b = k , c = 3

For equal roots 

D = 0

b² - 4ac = 0

k² - 4.2.3 = 0

k² = 24

k = √24

k = ±2√6 Ans.

(ii)

kx(x - 2) + 6 = 0

kx² - 2kx + 6 = 0

a = k, b = -2k, c = 6

For equal roots

D = 0

b² - 4ac = 0

(-2k)² - 4.k.6 = 0

4k² - 24k = 0

4k(k - 6) = 0

Therefore,

k = 0 or 6 Ans.

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

Solution:

Let the breadth be x m.

Length = 2x

Area = Length ×breadth

800 = 2x × x

or, 800 = 2x²

or, 2x² -800 = 0

or, x² -400 = 0

a = 1, b = 0, c = -400

D = b² -4ac

or, D = 0² -4×1×-400

or, D = 1600

Here, D > 0

So, roots are real and unequal.

Hence design is possible.

Therefore, 

x = (-b ±√D)÷2a

or, x = (-0 ±√1600)÷2

or, x = ±40÷2

or, x = ±20

Breadth not be in negative.

Therefore,

x = 20 m.

Length = 2x = 2×20 = 40 m.

Required Answers

40 m & 20 m Ans.

3. क्या एक एेसी आम की बगिया बनाना संभव है जिसकी लंबाई, चौड़ाई से दुगुनी हो और उसका क्षेत्रफल 800 m²  हो? यदि है, तो उसकी लंबाई और चौड़ाई ज्ञात कीजिए।

हल:

मान लिया कि चौडाई = x मीटर।

∴ लंबाई = 2x मीटर।

हम जानते है कि

क्षेत्रफल = लंबाई×चौड़ाई

800 = 2x × x

or, 800 = 2x²

or, 2x² -800 = 0

or, x² -400 = 0

a = 1, b = 0, c = -400

D = b² -4ac

or, D = 0² -4×1×-400

or, D = 1600

D > 0

अतः बगिया बनना संभव है।

x = (-b ±√D)÷2a

or, x = (-0 ±√1600)÷2×1

or, x = ±40÷2

चौडाई ऋणात्मक नही हो सकता है।

अतः x = 20 मीटर।

लंबाई = 2x = 2×20 = 40 मीटर।

अभिष्ट उत्तर

40 मीटर एवं 20 मीटर Ans.

4. Is the following situation possible? If so, determine their present ages.

The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Solution:

Let the present age of one friend be x yrs.

Therefore,

Present age of second friend = 20 -x yrs.

4 yrs ago

(x -4)×(20 -x -4) = 48

or, (x -4)×(16 -x) = 48

or, 20x -x² -64 = 48

or, x² -20x + 112 = 0

a = 1, b = -20, c = 112

D = b² -4ac

or, D = (-20)² -4×1×112

or, D = 400 - 448

or, D = -48

Here D is less than 0

Hence, under this situation  age cannot be determined. 

5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m² ? If so, find its length and breadth.

Solution:

Let the length of the park is = x metre. 

Breadth = 80/2 -x 

or, Breadth = (40 -x)

Area = Length× breadth

400 = x(40 -x)

or, 400 = 40x - x²

ot, x² -40x + 400 = 0

a = 1, b = -40, c = 400

or, D = b² -4ac

or, D = -40² -4×1×400

or, D = 1600 -1600

or, D = 0

It is not possible to design rectangular park but possible to design square park.

x = -b ÷ 2a

or, x = -(-40) ÷ 2×1

or, x = 40 ÷ 2

or, x = 20 meter.

Breadth = 40 -x = 40 -20 = 20 meter

Hence, both Length and breadth are equal. 

So, design of square park is possible instead of rectangular park.