NCERT Grade 9-Number Systems-Answers – MySchoolPage

NCERT Grade 9-Number Systems-Answers

NCERT Solutions for Class 9 Maths

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1.    Consider the definition of a rational number.

A rational number is the one that can be written in the form of , where p and q are integers and q ≠ 0.

Zero can be written as 

So, we arrive at the conclusion that 0 can be written in the form of , where q is any integer.

Therefore, zero is a rational number.

2.    We know that there are infinite rational numbers between any two numbers.

A rational number is the one that can be written in the form of , where p and q are integers and q ≠ 0 .

We know that the numbers 3.1, 3.2, 3.3, 3.4, 3.5 and 3.6 all lie between 3 and 4.

We need to rewrite the numbers 3.1, 3.2, 3.3, 3.4, 3.5 and 3.6 in  form to get the rational numbers between 3 and 4.

So, after converting, we get 

We can further convert the rational numbers  into lowest fractions.

On converting the fractions into lowest fractions, we get 

Therefore, six rational numbers between 3 and 4 are 

3.    We know that there are infinite rational numbers between any two numbers. A rational number is the one that can be written in the form of , where p and q are q integers and q ≠ 0.

We know that the numbers  can also be written as 0.6 and 0.8.

We can conclude that the numbers 0.61, 0.62, 0.63, 0.64 and 0.65 all lie between 0.6 and 0.8.

We need to rewrite the numbers 0.61,0.62,0.63,0.64 and 0.65 in  form to get the rational numbers between 3 and 4.

So after converting, we get 

We can further convert the rotational numbers  into lowest fractions.

On converting the fractions, we get 

Therefore, six rational numbers between 3 and 4 are

4.    (i) Consider the whole numbers and natural numbers separately. We know that whole number series is 0, 1, 2, 3, 4, 5……..

We know that natural number series is 1, 2, 3, 4, 5…..

So, we can conclude that every number of the natural number series lie in the whole number series.

Therefore, we conclude that, yes every natural number is a whole number.

(ii) Consider the integers and whole numbers separately.

We know that integers are those numbers that can be written in the form off ,

where q = 1.

Now, considering the series of integers, we have …. – 4, -3, -2, -1, 0, 1, 2, 3, 4…..

We know that whole number series is 0, 1, 2, 3, 4, 5…………………. .

We can conclude that all the numbers of whole number series lie in the series of integers. But every number of series of integers does not appear in the whole number series.

Therefore, we conclude that every integer is not a whole number.

(iii) Consider the rational numbers and whole numbers separately.

We know that rational numbers are the numbers that can be written in the form 

where q ≠ 0 .

We know that whole number series is 0, 1, 2, 3, 4, 5

We know that every number of whole number series can be written in the form of  as 

We conclude that every number of the whole number series is a rational number.

But, every rational number does not appear in the whole number series.

Therefore, we conclude that every rational number is not a whole number.

5.     (i) Consider the irrational numbers and the real numbers separately.

We know that irrational numbers are the numbers that cannot be converted in the

form , where p and q are integers and q ≠ 0.

We know that a real number is the collection of rational numbers and irrational numbers.

Therefore, we conclude that, yes every irrational number is a real number.

(ii) Consider a number line. We know that on a number line, we can represent negative as well as positive numbers.

We know that we cannot get a negative number after taking square root of any number.

Therefore, we conclude that not every number point on the number line is of the form √m, where m is a natural number.

(iii) Consider the irrational numbers and the real numbers separately.

We know that irrational numbers are the numbers that cannot be converted in the

form , where p and q are integers and q ≠ 0.

We know that a real number is the collection of rational numbers and irrational numbers.

So, we can conclude that every irrational number is a real number. But every real number is not an irrational number.

Therefore, we conclude that, every real number is not a rational number.

6.    We know that square root of every positive integer will not yield an integer.

We know that √4 is 2, which is an integer. But, √7 or √10) will give an irrational number.

Therefore, we conclude that square root of every positive integer is not an irrational number.

7.    According to the Pythagoras theorem, we can conclude that

(√5)2 = (2)2 + (1)2

We need to draw a line segment AB of 1 unit on the number line. Then draw a straight line segment BC of 2 units. Then join the points C and A, to form a line segment BC.

Then draw the arc ACD, to get the number √5 on the number line.

8.    (i) 

On dividing 36 by 100, we get

Therefore, we conclude that = 0.36, which is a terminating decimal.

(ii) 

On dividing 1 by 11, we get

We can observe that while dividing 1 by 11, we got the remainder as 1, which will continue to be 1.

Therefore, we conclude that  = 0.0909…… or  which is a non-terminating decimal and recurring decimal.

(iii) 

On dividing 33 by 8, we get

We can observe that while dividing 33 by 8, we got the remainder as 0.

Therefore, we conclude that = 4.125 , which is a terminating decimal.

(iv) 

On dividing 3 by 13, we get

We can observe that while dividing 3 by 13 we got the remainder as 3, which will continue to be 3 after carrying out 6 continuous divisions.

Therefore, we conclude that  = 0.230769……. or  which is a non-terminating decimal and recurring decimal.

(v)   

On dividing 319 by 300, we get

We can observe that while dividing 329 by 400, we got the remainder as 0.

Therefore, we conclude that  = 0.8225, which is a terminating decimal.

9.     We given that

We need to find the values of without performing long division.

we know that, can be rewritten as 

On substituting value of  as 0.142857……, we get

Therefore, we conclude that, we can predict the values of without performing long division to get

10.  Solution: 

(i) 

We need to multiply both sides by 10 to get

10x = 6.6666 ………………………………………………………….. (b)

We need to subtract (a) from (b), to get

We can also write 9x = 6 as 

Therefore, on converting  form, we get the answer as

(ii)   

We need to multiply both sides by 10 to get

10x = 4.7777 ………………………………………….. . . . .(b)

We need to subtract (a) from (b) , to get

Therefore, on converting    form, we get the answer as 

(iii) 

We need to multiply both sides by 1000 to get

1000x =1.001001 ………..  (b)

We need to subtract (a) from (b), to get

Therefore, on converting  form, we get the answer as

11.    Let x = 0.99999       ….….(a)

We need to multiply both sides by 10 to get 10x = 9.9999 ….….(b)

We need to subtract (a) from (b), to get

Therefore, on converting 0.99999…. in the  form, we get the answer as 1.

Yes at a glance we are surprised at our answer.

But the answer makes sense when we observe that 0.9999…….. goes on forever. SO there is not gap between 1 and 0.9999…… and hence they are equal.

12.   We need to find the number of digits in the recurring block of .

Let us perform the long division to get the recurring block of .

We need to divide 1 by 17, to get

We can observe that while dividing 1 by 17 we got the remainder as 1, which will continue to be 1 after carrying out 16 continuous divisions.

Therefore, we conclude that 

which is a non-terminating decimal and recurring decimal.

13.   Solution:

Let us consider the examples of the form  that are terminating decimals.

We can observe that the denominators of the above rational numbers have powers of 2, 5 or both.

Therefore, we can conclude that the property, which q must satisfy in , so that the rational number  is a terminating decimal is that q must have powers of 2, 5 or both.

14.   The three numbers that have their expansions as non terminating non recurring decimal are given below.

0.04004000400004….

0.07007000700007….

0.013001300013000013….

15.    Let us convert  into decimal form, to get

Three irrational numbers that lie between 0.714285…. and 0.818181…. are:

0.73073007300073….

0.74074007400074….

0.76076007600076….

16.  (i) √23

We know that on finding the square root of 23, we will not get an integer. Therefore, we conclude that √23 an irrational number.

(ii) √225

We know that on finding the square root of √225, we get 15, which is an integer. Therefore, we conclude that √225 is a rational number.

(iii) 0.3796

We know that 0.3796 can be converted into .

While, converting 0.3796 into  form, we get

0.3796 = .

The rational number  can be converted into lowest fractions, to get 

We can observe that 0.3796 can be converted into a rational number.

Therefore, we conclude that 0.3796 is a rational number.

(iv) 7.478478….

We know that 7.478478…. is a non-terminating recurring decimal, which can be converted into  form.

While, converting 7,478478…. into  form, we get

x = 7.478478….                                                   …. (a)

1000x = 7478.478478….                                   …. (b)

While, subtracting (a) from (b), we get

We know that 999x = 7471 can also be written as x = 

Therefore, we conclude that 7.478478…. is a rational number.

(v) 1.101001000100001….

We can observe that the number 1.101001000100001…. is a non terminating non recurring decimal.

We know that non terminating and non recurring decimals cannot be converted into  form.

Therefore, we conclude that 1.101001000100001…. is an irrational number.

17.   We know that the number 3.765 will lie between 3.764 and 3.766.

We know that the numbers 3.764 and 3.766 will lie between 3.76 and 3.77.

We know that the numbers 3.76 and 3.77 will lie between 3.7 and 3.8.

We know that the numbers 3.7 and 3.8 will lie between 3 and 4.

Therefore, we can conclude that we need to use the successive magnification, after locating numbers 3 and 4 on the number line.

18.   We know that the number  can also be written as 4.262…..

We know that the number 4.262…. will lie between 4.261 and 4.263.

We know that the numbers 4.261 and 4.263 will lie between 4.26 and 4.27.

We know that the numbers 4.26 and 4.27 will lie between 4.2 and 4.3.

We know that the numbers 4.2 and 4.3 will lie between 4 and 5.

Therefore, we can conclude that we need to use the successive magnification, after locating numbers 4 and 5 on the number line.

19.   (i) 2 – √5

We know that √5 = 2.236….., which is an irrational number.

2 – √5 = 2 – 2.236……

 = -0.236…., which is also an irrational number

Therefore, we conclude that 2 – √5 is an irrational number.

(ii)  (3 + √23) – √23

(3 + √23) – √23 =3 + √23 – √23

=3.

Therefore, we conclude that (3 + √23) √23 is a rational number.

(iii) 

We can cancel √7 in the numerator and denominator, as √7 is the common number in numerator as well as denominator, to get

Therefore, we conclude that  is a rational number.

(iv) 

We know that √2 = 1.414…., which is an irrational number .

We can conclude that, when 1 is divided by √2 , we will get an irrational number.

Therefore, we conclude that  is an irrational number.

(v) 2π

We know that π = 3.1415…., which is an irrational number.

We can conclude that 2π will also be an irrational number.

Therefore, we conclude that 2π is an irrational number.

20.  (i) (3+ √3) (2 + √2)

We need to apply distributive law to find value of (3 + √3) (2 + √2).

(3 + √3) (2 + √2) = 3 (2 + √2) + √3(2 + √2)

= 6 + 3 √2 + 2√3 + √6

Therefore, on simplifying (3 + √3) (2 + √2) , we get 6 + 3√2 + 2√3 + √6.

(ii) (3 + √3) (3 – √3)

We need to apply distributive law to find value of (3 + √3) (3 – √3)

(3 + √3) (3 – √3) = 3 (3 – √3) +  √3 (3 – √3)

= 9 – 3√3 +  3√3 – 3

= 6

Therefore, on simplifying (3 + √3) (3 – √3), we get 6.

(iii) (√5 + √2)2

 We need to apply the formula (a + b)2 = a2 + 2ab + b2 to find value of (√5 + √2)2

(√5 + √2)= (√5)2 + 2 × √5 × √2 + (√2)

= 5 + 2√10 + 2

= 7 +2√10.

Therefore, on simplifying (√5 + √2)2, we get 7 +2√10.

(iv)   (√5 – √2)  (√5 + √2)

We need to apply the formula (a – b) (a + b) = a2 – b2 to find value of (√5 + √2)2

(√5 – √2)  (√5 + √2) = (√5)2 – (√2)2

 = 5 – 2

 = 3.

Therefore, on simplifying, (√5 – √2)  (√5 + √2), we get 3.

21.   We know that when we measure the length of a line or a figure by using a scale or any device, we do not get an exact measurement. In fact, we get an approximate rational value. So, we are not able to realize that either circumference (c) or diameter (d) of a circle is irrational.

Therefore, we can conclude that as such there is not any contradiction regarding the value of it and we realize that the value of it is irrational.

22.  Mark the distance 9.3 units from a fixed point A on a given line to obtain a point B such that AB = 9.3 units. From B mark a distance of 1 unit and call the new point as C. Find the mid-point of AC and call that point as O. Draw a semi-circle with centre 0 and radius OC = 5.15 units. Draw a line perpendicular to AC passing through B cutting the semi-circle at D. Then BD = √9.3.

23.  (i)

We need to multiply the numerator and denominator of   by √7, to get

Therefore, we conclude that on rationalizing the denominator of , we get 

(ii)  

We need to multiply the numerator and denominator of  get

We need to apply the formula (a — b)(a +b) = a2 – b2 in the denominator to get

Therefore, we conclude that on rationalizing the denominator of , we get √7 + √6.

(iii) 

We need to multiply the numerator and denominator of by √5 – √2, to get

We need to apply the formula (a —b) (a +b) = a2 – b2 in the denominator to get

Therefore, we conclude that on rationalizing the denominator of , we get

(iv)  

We need to multiply the numerator and denominator of  by √7 + 2, to get

We need to apply the formula (a —b) (a +b) = a2 – b2 in the denominator to get

Therefore, we conclude that on rationalizing the denominator of , we get 

24.   (i) 

We know that  where a > 0.

We conclude that  can also be written as 

= 8

Therefore the value of will be 8.

(ii)  

We know that   where a > 0.

We conclude that   can also be written as

= 2

Therefore the value of  will be 2.

(iii) 

We know that  where a > 0.

We conclude that  can also be written as 

= 5

Therefore the value of  will be 5.

25.   (i)

We know that   where a > 0.

We conclude that can also be written as 

 = 3 × 3 × 3

= 27.

Therefore the value of  will be 27.

(ii) 

We know that   where a > 0.

We conclude that can also be written as

= 2 × 2

= 4.

Therefore the value of   will be 4.

(iii) 

26.

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