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Page No 9:

Question 1:

Is zero a rational number? Justify.

Answer:

Yes, 0 is a rational number.

0 can be expressed in the form of the fraction pq, where p=0 and q can be any integer except 0.

Page No 9:

Question 2:

Represent each of the following rational number line:
(i) 57
(ii) 83
(iii) -236
(iv) 1.3
(v) – 2.4

Answer:

(i) 57


(ii) 83
83=223


(iii) -236=-356


(iv) 1.3
1.3=1310=1310


(v) – 2.4
-2.4=-2410=-125=-225

Page No 9:

Question 3:

Find a rational number between
(i) 38 and 25
(ii) 1.3 and 1.4
(iii) -1 and 12
(iv) -34 and-25
(v) 19 and 29

Answer:

(i) 38 and 25
Let:
x = 38 and y = 25
Rational number lying between x and y:
12x + y = 1238 + 25
= 1215+1640 = 3180

(ii) 1.3 and 1.4
Let:
x = 1.3 and y = 1.4
Rational number lying between x and y:
12x + y = 121.3+1.4
= 122.7= 1.35

(iii) -1 and 12
Let:
x = -1 and y = 12
Rational number lying between x and y:
12x + y = 12-1 + 12
= -14

(iv) -34 and-25
Let:
x-34 and y = -25
Rational number lying between x and y:
12x + y = 12-34 - 25
= 12-15-820 = -2340

(v) 19 and 29
A rational number lying between 19 and 29 will be
1219+29=12×13=16

Page No 9:

Question 4:

Find three rational numbers lying between 35 and 78. How many rational numbers can be determined between these two numbers?

Answer:

x=35 and y=78
n = 3
d=y-xn+1=78-353+1=1140×14=11160
Rational numbers between x=35 and y=78 will be
x+d,x+2d,...,x+nd35+11160,35+2×11160,35+3×11160107160,118160,129160107160,5980,129160
There are infinitely many rational numbers between two given rational numbers.


 

Page No 9:

Question 5:

Find four rational numbers between 37 and 57.

Answer:

n = 4
n + 1 = 4 + 1 = 5
37=37×55=1535 57=57×55=2535
Thus, rational numbers between 37 and 57 are 1635,1735,1835,1935.

Page No 9:

Question 6:

Find six rational numbers between 2 and 3.

Answer:

x = 2, y = 3 and n = 6
d=y-xn+1=3-26+1=17
Thus, the required numbers are 
x+d,x+2d,x+3d,...,x+nd=2+17,2+2×17,2+3×17,2+4×17,2+5×17,2+6×17=157,167,177,187,197,207
 

Page No 9:

Question 7:

Find five rational numbers between 35 and 23.

Answer:

n = 5
n + 1 = 6
x=35,y=23
d=y-xn+1=23-356=10-990=190

Thus, rational numbers between 35 and 23 will be 
x+d,x+2d,x+3d,x+4d,x+5d=35+190,35+290,35+390,35+490,35+590=5590,5690,5790,5890,5990=1118,2845,1930,2945,5990

 



Page No 10:

Question 8:

Insert 16 rational numbers between 2.1 and 2.2.

Answer:

Let:
x = 2.1, y = 2.2 and n = 16

We know:
d = y-xn+1=2.2-2.116+1=0.117=1170= 0.005 (approx.)
So, 16 rational numbers between 2.1 and 2.2 are:
(x + d), (x + 2d), ...(x + 16d)
= [2.1 + 0.005], [2.1 + 2(0.005)],...[2.1 + 16(0.005)]
= 2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17, 2.175 and 2.18

Page No 10:

Question 9:

State whether the following statements are true or false. Give reasons for your answer.
(i) Every natural number is a whole number.
(ii) Every whole number is a natural number.
(iii) Every integer is a whole number.
(iv) Every integer is a rational number.
(v) Every rational number is an integer.
(vi) Every rational number is a whole number.

Answer:

(i) Every natural number is a whole number.
True, since natural numbers are counting numbers i.e N = 1, 2,...
Whole numbers are natural numbers together with 0. i.e W = 0, 1, 2,...
So, every natural number is a whole number

(ii) Every whole number is a natural number.
False, as whole numbers contain natural numbers and 0 whereas natural numbers only contain the counting numbers except 0. 

(iii) Every integer is a whole number.
False, whole numbers are natural numbers together with a zero whereas integers include negative numbers also.

(iv) Every integer is a rational number.
True, as rational numbers are of the form pq where q0. All integers can be represented in the form pq where q0.
(v) Every rational number is an integer.
False, as rational numbers are of the form pq where q0. Integers are negative and positive numbers which are not in pq form.
For example, 12 is a rational number but not an integer. 

(vi) Every rational number is a whole number.
False, as rational numbers are of the form pq where q0. Whole numbers are natural numbers together with a zero.
For example, 57 is a rational number but not a whole number. 



Page No 18:

Question 1:

Write actual division, find which of the following rational numbers are terminating decimals.
(i) 1380

(ii) 724

(iii) 512

(iv) 31375

(v) 16125

Answer:

(i) 1380
Denominator of 1380 is 80.
And,
80 = 24×5
Therefore, 80 has no other factors than 2 and 5.
Thus, 1380 is a terminating decimal.

(ii) 724
Denominator of 724 is 24.
And,
24 = 23×3
So, 24 has a prime factor 3, which is other than 2 and 5.
Thus, 724 is not a terminating decimal.

(iii) 512
Denominator of 512 is 12.
And,
12 = 22×3
So, 12 has a prime factor 3, which is other than 2 and 5.
Thus, 512 is not a terminating decimal.
 
(iv) 31375
Denominator of 31375 is 375. 
375=53×3
So, the prime factors of 375 are 5 and 3.
Thus, 31375 is not a terminating decimal. 

(v) 16125
Denominator of 16125 is 125.
And,
125 = 53
Therefore, 125 has no other factors than 2 and 5.
Thus, 16125 is a terminating decimal.



Page No 19:

Question 2:

Write each of the following in decimal form and say what kind of decimal expansion each has.
(i) 58

(ii) 725

(iii) 311

(iv) 513

(v) 1124

(vi) 261400

(vii) 231625

(viii) 2512

Answer:

(i) 58 = 0.625
By actual division, we have:

It is a terminating decimal expansion.

(ii) 725
725 = 0.28
By actual division, we have:

 
It is a terminating decimal expansion.

(iii) 311 = 0.27¯


It is a non-terminating recurring decimal.
(iv) 513 = 0.384615¯

It is a non-terminating recurring decimal.

(v) 1124
1124 =
By actual division, we have:
 
It is nonterminating recurring decimal expansion.

(vi) 261400=0.6525

It is a terminating decimal expansion.

(vii) 231625=0.3696

It is a terminating decimal expansion.

(viii) 2512
2512 = 2912 =
By actual division, we have:

It is non-terminating decimal expansion.

Page No 19:

Question 3:

Express each of the following decimals in the form pq, where p, q are integers and q ≠ 0.
(i) 0.2¯
(ii) 0.53¯
(iii) 2.93¯
(iv) 18.48¯
(v) 0.235¯
(vi) 0.0032¯
(vii) 1.323¯
(viii) 0.3178¯
(ix) 32.1235¯
(x) 0.407¯

Answer:

(i) 0.2¯
Let x = 0.222...                               .....(i)
Only one digit is repeated so, we multiply x by 10.
10x = 2.222...                                 .....(ii) 
Subtracting (i) from (ii) we get
9x=2x=29

(ii) 0.53¯
Let x = 0.5353...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 53.5353...                                 .....(ii) 
Subtracting (i) from (ii) we get
99x=53x=5399

(iii) 2.93¯
Let x = 2.9393...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 293.9393...                                 .....(ii) 
Subtracting (i) from (ii) we get
99x=291x=29199=9733

(iv) 18.48¯
Let x = 18.4848...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 1848.4848...                                 .....(ii) 
Subtracting (i) from (ii) we get
99x=1830x=183099=61033

(v) 0.235¯
Let x = 0.235235...                               .....(i)
Three digits are repeated so, we multiply x by 1000.
1000x = 235.235235...                                 .....(ii) 
Subtracting (i) from (ii) we get
999x=235x=235999

(vi) 0.0032¯
Let x = 0.003232...                               .....(i)
we multiply x by 100.
100x = 0.3232...                                 .....(ii) 
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 32.3232...                           .....(iii)
Subtracting (ii) from (iii) we get
9900x=32x=329900=82475

(vii) 1.323¯
Let x = 1.32323...                               .....(i)
we multiply x by 10.
10x = 13.2323...                                 .....(ii) 
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
1000x = 1323.2323...                           .....(iii)
Subtracting (ii) from (iii) we get
990x=1310x=13199

(viii) 0.3178¯
Let x = 0.3178178...                               .....(i)
we multiply x by 10.
10x = 3.178178...                                 .....(ii) 
Again multiplying by 1000 as there are 3 repeating numbers after decimals we get
10000x = 3178.178178...                           .....(iii)
Subtracting (ii) from (iii) we get
9990x=3175x=31759990=6351998

(ix) 32.1235¯
Let x = 32.123535...                               .....(i)
we multiply x by 100.
100x = 3212.3535...                                 .....(ii) 
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 321235.35...                           .....(iii)
Subtracting (ii) from (iii) we get
9900x=318023x=3180239900

(x) 0.407¯
Let x = 0.40777...                               .....(i)
we multiply x by 100.
100x = 40.7777...                                 .....(ii) 
Again multiplying by 10 as there is 1 repeating number after decimals we get
1000x = 407.777...                           .....(iii)
Subtracting (ii) from (iii) we get
900x=367x=367900
 

Page No 19:

Question 4:

Express 2.36¯+0.23¯ as a fraction in simplest form.

Answer:

Given: 2.36¯+0.23¯
Let 
x=2.36¯                      ...iy=0.23¯                      ...ii
First we take x and convert it into pq
100x = 236.3636...        ...(iii)
Subtracting (i) from (iii) we get
99x=234x=23499
Similarly, multiply y with 100 as there are 2 decimal places which are repeating themselves. 
100y=23.2323...               ...(iv)
Subtracting (ii) from (iv) we get
99y=23y=2399
Adding x and y we get
2.36¯+0.23¯x+y=23499+2399=25799

Page No 19:

Question 5:

Express in the form of pq: 0.38¯+1.27¯.

Answer:

Let 0.38¯=x1.27¯=y
x = 0.3838...                                  ...(i)
Multiply with 100 as there are 2 repeating digits after decimals
100x = 38.3838...                          ...(ii)
Subtracting (i) from (ii) we get
99x = 38
x=3899
Similarly, we take 
y = 1.2727...                                  ...(iii)
Multiply y with 100 as there are 2 repeating digits after decimal.
100y = 127.2727...                       ...(iv)
Subtract (iii) from (iv) we get
99y = 126
y=12699Now, x+y=3899+12699=16499
 



Page No 23:

Question 1:

What are irrationl numbers? How do they differ from rational numbers? Give examples.

Answer:

A number that can neither be expressed as a terminating decimal nor be expressed as a repeating decimal is called an irrational number. A rational number, on the other hand, is always a terminating decimal, and if not, it is a repeating decimal.
Examples of irrational numbers:
0.101001000...
0.232332333...

Page No 23:

Question 2:

Classify the following numbers as rational or irrational. give reasons to support your answer.
(i) 381

(ii) 361

(iii) 21

(iv) 1.44

(v) 236

(vi) 4.1276

(vii) 227

(viii) 1.232332333...

(ix) 3.040040004....

(x) 2.356565656...

(xi) 6.834834...

Answer:

(i) 381
381=127=1313
It is an irrational number.

(ii) 361 = 19
So, it is rational.

(iii) 21
21=3×7=4.58257...It is an irrational number.

(iv) 1.44 = 1.2
So, it is rational. 

(v) 236
It is an irrational number

(vi) 4.1276
It is a terminating decimal. Hence, it is rational.

(vii) 227
227 is a rational number because it can be expressed in the pqform.

(viii) 1.232332333...is an irrational number because it is a non-terminating, non-repeating decimal.

(ix) 3.040040004... is an irrational number because it is a non-terminating, non-repeating decimal.(x) 2.356565656... is a rational number because it is repeating.

(xi) 6.834834... is a rational number because it is repeating.

Page No 23:

Question 3:

Let x be a rational number and y be an irrational number. Is x + y necessarily an irrational number? Give a example in support of your answer.

Answer:

x be a rational number and y be an irrational number then x + y necessarily will be an irrational number.
Example: 5 is a rational number but 2 is irrational. 
So, 5 + 2 will be an irrational number. 

Page No 23:

Question 4:

Let a be a rational number and b be an irrational number. Is ab necessarily an irrational number? Justify your answer with an example.

Answer:

a be a rational number and b be an irrational number then ab necessarily will be an irrational number.
Example: 6 is a rational number but 5 is irrational. And 65 is also an irrational number. 

Page No 23:

Question 5:

Is the product of two irrationals always irrational? Justify your answer.

Answer:

Product of two irrational numbers is not always an irrational number.
Example: 5 is irrational number. And 5×5=5 is a rational number. But the product of another two irrational numbers 2 and 3 is 6 which is also an irrational numbers.

Page No 23:

Question 6:

Give an example of two irrational numbers whose
(i) difference is an irrational number.
(ii) difference is a rational number.
(iii) sum is an irrational number.
(iv) sum is a rational number.
(v) product is an irrational number.
(vi) product is a rational number.
(vii) quotient is an irrational number.
(viii) quotient is a rational number.

Answer:

(i) 2 irrational numbers with difference an irrational number will be 3-5 and 3+5.
(ii) 2 irrational numbers with difference is a rational number will be 5+3 and 2+3
(iii) 2 irrational numbers with sum an irrational number 7+5 and 6-8 
(iv) 2 irrational numbers with sum a rational number is 3-2 and 3+2
(v) 2 irrational numbers with product an irrational number will be 6+3 and 7-3
(vi) 2 irrational numbers with product a rational number will be 5+7 and 5-7
(vii) 2 irrational numbers with quotient an irrational number will be 15 and 5
(viii) 2 irrational numbers with quotient a rational number will be 63 and 7.

Page No 23:

Question 7:

Examine whether the following numbers are rational or irrational.
(i) 3+3

(ii) 7-2

(iii) 53×253

(iv) 7×343

(v) 13117

(vi) 8×2

Answer:

(i) Let us assume, to the contrary, that 3+3 is rational.
Then, 3+3=pq, where p and q are coprime and q0.
3=pq-33=p-3qq
Since, p and q are are integers.
p-3qq is rational.
So, 3 is also rational.
But this contradicts the fact that 3 is irrational.
This contradiction has arisen because of our incorrect assumption that 3+3 is rational.
Hence, 3+3 is irrational.

(ii) Let us assume, to the contrary, that 7-2 is rational.
Then, 7-2=pq, where p and q are coprime and q0.
7=pq+27=p+2qq
Since, p and q are are integers.
p+2qq is rational.
So, 7 is also rational.
But this contradicts the fact that 7 is irrational.
This contradiction has arisen because of our incorrect assumption that 7-2 is rational.
Hence, 7-2 is irrational.

(iii) As, 53×253

=5×253=1253=5, which is an integer
Hence, 53×253 is rational.

(iv) As, 7×343
=7×343=2401=49, which is an integer
Hence, 7×343 is rational.

(v) As, 13117=19=13, which is rational
Hence, 13117 is rational.

(vi) As, 8×2
=8×2=16=4, which is an integer
Hence, 8×2 is rational.

Page No 23:

Question 8:

Insert a rational and an irrational number between 2 and 2.5.

Answer:

As, few rational numbers between 2 and 2.5 are: 2.1, 2.2, 2.3, 2.4, ...
And,
Since, 2=4 and 2.5=6.25
So, irrational number between 2 ans 2.5 are: 4.1, 4.2, ..., 5, ...

Hence, a rational and an irrational number can be 2.1 and 5, respectively.

Disclaimer: There are infinite rational and irrational numbers between any two rational numbers.
 

Page No 23:

Question 9:

How many irrational numbers lie between 2 and 3? Find any three irrational numbers lying between 2 and 3.

Answer:

There are infinite number of irrational numbers lying between 2 and 3.

As, 2=1.414 and 3=1.732
So, the three irrational numbers lying between 2 and 3 are:
1.420420042000..., 1.505005000... and 1.616116111...

Page No 23:

Question 10:

Find two rational and two irrational number between 0.5 and 0.55.

Answer:

The two rational numbers between 0.5 and 0.55 are: 0.51 and 0.52

The two irrational numbers between 0.5 and 0.55 are: 0.505005000... and 0.5101100111000...

Disclaimer: There are infinite number of rational and irrational numbers between 0.5 and 0.55.

Page No 23:

Question 11:

Find three different irrational numbers between the rational numbers 57 and 911.

Answer:

As, 570.714 and 9110.818

So, the three different irrational numbers are: 0.72020020002..., 0.7515511555111... and 0.808008000...

Disclaimer: There are an infinite number of irrational numbers between two rational numbers.



Page No 24:

Question 12:

Find two rational numbers of the form pq between the numbers 0.2121121112... and 0.2020020002... .

Answer:

The rational numbers between the numbers 0.2121121112... and 0.2020020002... are:

0.21=21100 and 0.205=2051000=41200

Disclaimer: There are an infinite number of rational numbers between two irrational numbers.

Page No 24:

Question 13:

Find two irrational numbers between 0.16 and 0.17.

Answer:

The two irrational numbers between 0.16 and 0.17 are 0.161161116... and 0.1606006000...

Disclaimer: There are an infinite number of irrational numbers between two rational numbers.

Page No 24:

Question 14:

State in each case, whether the given statement is true of false.
(i) The sum of two rational numbers is rational.
(ii) The sum of two irrational numbers is irrational.
(iii) The product of two rational numbers is rational.
(iv) The product of two irrational number is irrational.
(v) The sum of a rational number and an irrational number is irrational.
(vi) The product of a nonzero rational number and an irrational number is a rational number.
(vii) Every real number is rational.
(viii) Every real number is either rational or irrational.
(ix) πis irrational and227is rational.

Answer:

(i) True

(ii) False
Example: 2+3+2-3=4Here, 4 is a rational number.

(iii) True

(iv) False
Example: 3×3=3Here, 3 is a rational number.

(v) True

(vi) False
Example: 4×5=45 Here, 45 is an irrational number.

(vii) False 
Real numbers can be divided into rational and irrational numbers.

(viii) True

(ix) True



Page No 27:

Question 1:

Add:
(i) 23-52and3+22
(ii) 22+53-75and33-2+5
(iii) 237-12+611and137+322-11

Answer:

(i) 23-52+3+22=23+3+22-52=33-32(ii) 22+53-75+33-2+5=22-2+53+33+5-75=2+83-65(iii) 237-122+611+137+322-11=237+137-11+611+322-122=7+511+2



Page No 28:

Question 2:

Multiply:
(i) 35 by 25
(ii) 615 by 43
(iii) 26 by 33
(iv) 38 by 32
(v) 10 by 40
(vi) 328 by 27

Answer:

(i) 35×25=3×2×5×5=6×5=30(ii) 615×43=6×4×5×3×3=24×3×5=725(iii) 26×33=2×3×2×3×3=6×3×2=182(iv) 38×32=3×3×2×2×2×2=9×4=36   (v) 10×40=2×5×2×2×2×5=2×2×2×2×5×5=2×2×5=20(vi) 328×27=67×4×7=6×7×4=42×2=84

Page No 28:

Question 3:

Divide:
(i) 166 by 42
(ii) 125 by 43
(iii) 1821 by 67

Answer:

(i) 16642=162342=43(ii) 121543=125×343=35(iii) 182167=187367=33

Page No 28:

Question 4:

Simplify
(i) 3-11 3+11
(ii) -3+5 -3-5
(iii) 3-32
(iv) 5-32
(v) 5+7 2+5
(vi) 5-2 2-3

Answer:

(i) 3-11 3+11

=32-112             [a-ba+b=a2-b2]=9-11=-2

(ii) -3+5 -3-5

=-32-52             [a+ba-b=a2-b2]=9-5=4

(iii) 3-32

=32+32-2×3×3             [a-b2=a2+b2-2ab]=9+3-63=12-63

(iv) 5-32

=52+32-2×53      [a-b2=a2+b2-2ab]=5+3-215=8-215
=5×2-5×3-2×2+2×3    =10-15-2+6

(v) 5+7 2+5

5+72+5=5×2+5×5+7×2+7×5=10+55+27+35

(vi) 5-2 2-3

5-22-3=5×2-5×3-2×2+2×3    =10-15-2+6

Page No 28:

Question 5:

Simplify 3+3 2+22.

Answer:

3+3 2+22=3+3 22+22+2×22=3+3 4+2+42=3+3 6+42

=3×6+3×42+3×6+3×42=18+122+63+46

Page No 28:

Question 6:

Examine whether the following numbers are rational or irrational:

(i) 5-5 5+5

(ii) 3+22

(iii) 213352-4117

(iv) 8+432-62

Answer:

(i) 5-5 5+5

=52-52                  a-ba+b=a2-b2=25-5=20, which is an integer

Hence, 5-5 5+5 is rational.

(ii) 3+22

=32+22+2×3×2                  a+b2=a2+b2+2ab=3+4+43=7+43

Since, the sum and product of rational numbers and an irrational number is always an irrational.

7+43 is irrational.

Hence, 3+22​ is irrational.

(iii) 213352-4117

=213313×4-413×9=2131334-49=23×2-4×2

=26-8=2-2=-1, which is an integer

Hence, 213352-4117 is rational.

(iv) 8+432-62

=22+4×42-62=22+162-62=122

Since, the product of a rational number and an irrational number is always an irrational.

Hence, 8+432-62 is rational.

Page No 28:

Question 7:

On her birthday Reema distributed chocolates in an orphanage. The total number of chocolates she distributed is given by 5+11 5-11.
(i) Find the number of chocolates distributed by her.
(ii) Write the moral values depicted here by Reema.

Answer:

(i) As, 5+11 5-11

=52-112                   a+ba-b=a2-b2=25-11=14

Hence, the number of chocolates distributed by Reema is 14.

(ii) The moral values depicted here by Reema is helpfulness and caring.

Disclaimer: The moral values may vary from person to person.

Page No 28:

Question 8:

Simplify

(i) 345-125+200-50

(ii) 2306-314028+5599

(iii) 72+800-18

Answer:

(i) 345-125+200-50

=39×5-25×5+100×2-25×2=3×35-55+102-52=95-55+52=45+52

(ii) 2306-314028+5599

=26×56-328×528+5×119×11=26×56-328×528+5×119×11=25-35+53

=-5+53=-35+53=-253

(iii) 72+800-18

=36×2+400×2-9×2=62+202-32=232



Page No 35:

Question 1:

Represent 5 on the number line.

Answer:


To represent 5 on the number line, follow the following steps of construction:

(i) Mark points 0 and 2 as O and P, respectively.
(ii) At point A, draw AB  OA such that AB = 1 units.
(iii) Join OB.
(iv) With O as centre and radius OB, draw an arc intersecting the number line at point P.
Thus, point represents 5 on the number line.

Justification:

In right OAB,

Using Pythagoras theorem,

OB=OA2+AB2=22+12=4+1=5

Page No 35:

Question 2:

Locate 3 on the number line.

Answer:


To represent 3
 on the number line, follow the following steps of construction:

(i) Mark points 0 and 1 as O and A, respectively.
(ii) At point A, draw AB  OA such that AB = 1 units.
(iii) Join OB.
(iv) At point B, draw DB  OA such that DB = 1 units.
(v) Join OD.
(vi) 
With O as centre and radius OD, draw an arc intersecting the number line at point Q.
Thus, point Q represents 3 on the number line.

Justification:

In right OAB,

Using Pythagoras theorem,

OB=OA2+AB2=12+12=1+1=2

Again, in right 
ODB,

Using Pythagoras theorem,

OD=OB2+DB2=22+12=2+1=3

Page No 35:

Question 3:

Locate 10 on the number line.

Answer:


To represent 10 
on the number line, follow the following steps of construction:

(i) Mark points 0 and 3 as O and B, respectively.
(ii) At point A, draw AB  OA such that AB = 1 units.
(iii) Join OA.
(iv) With O as centre and radius OA, draw an arc intersecting the number line at point P.
Thus, point P represents 10 on the number line.

Justification:

In right OAB,

Using Pythagoras theorem,

OA=OB2+AB2=32+12=9+1=10

Page No 35:

Question 4:

Locate 8 on the number line.

Answer:


To represent 8 
on the number line, follow the following steps of construction:

(i) Mark points 0 and 2 as O and B, respectively.
(ii) At point B, draw AB  OA such that AB = 2 units.
(iii) Join OA.
(iv) With O as centre and radius OA, draw an arc intersecting the number line at point P.
Thus, point P represents 8 on the number line.

Justification:

In right OAB,

Using Pythagoras theorem,

OA=OB2+AB2=22+22=4+4=8

Page No 35:

Question 5:

Represent 4.7 geometrically on the number line.

Answer:


To represent 4.7 on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 4.7 units.

(ii) From B, mark a point C on the same given line such that BC = 1 unit.

(iii) Find the mid point of AC and mark it as O.

(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.

(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.

(vi)  With B as centre and radius BD, draw an arc intersecting the given line at point E.

Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents 4.7.

Justification:

Here, in semi-circle, radii OA = OC = OD = 4.7+12=5.72=2.85 units

And, OB = AB - AO = 4.7 - 2.85 = 1.85 units

In a right angled triangle OBD,

BD=OD2-OB2=2.852-1.852=2.85+1.852.85-1.85             a2-b2=a+ba-b=4.7×1=4.7

Page No 35:

Question 6:

Represent 10.5 on the number line.

Answer:


To represent 10.5 on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 10.5 units.

(ii) From B, mark a point C on the same given line such that BC = 1 unit.

(iii) Find the mid point of AC and mark it as O.

(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.

(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.

(vi)  With B as centre and radius BD, draw an arc intersecting the given line at point E.

Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents 10.5.


Justification:

Here, in semi-circle, radii OA = OC = OD = 10.5+12=11.52=5.75 units

And, OB = AB - AO = 10.5 - 5.75 = 4.75 units

In a right angled triangle OBD,

BD=OD2-OB2=5.752-4.752=5.75+4.755.75-4.75             a2-b2=a+ba-b=10.5×1=10.5

Page No 35:

Question 7:

Represent 7.28 geometrically on the number line.

Answer:


To represent 7.28 on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 7.28 units.

(ii) From B, mark a point C on the same given line such that BC = 1 unit.

(iii) Find the mid point of AC and mark it as O.

(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.

(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.

(vi)  With B as centre and radius BD, draw an arc intersecting the given line at point E.

Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents 7.28.


Justification:

Here, in semi-circle, radii OA = OC = OD = 7.28+12=8.282=4.14 units

And, OB = AB - AO = 7.28 - 4.14 = 3.14 units

In a right angled triangle OBD,

BD=OD2-OB2=4.142-3.142=4.14+3.144.14-3.14             a2-b2=a+ba-b=7.28×1=7.28

Page No 35:

Question 8:

Represent 1+9.5 on the number line.

Answer:


To represent 1+9.5 on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 9.5 units.

(ii) From B, mark a point C on the same given line such that BC = 1 unit.

(iii) Find the mid point of AC and mark it as O.

(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.

(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.

(vi)  With B as centre and radius BD, draw an arc intersecting the given line at point E.

(vii) From E, mark a point F on the same given line such that EF = 1 unit.

Thus, let us treat the given line as the number line, with B as 0, C as 1, E as 9.5 and so on, then point F represents 1+9.5.


Justification:

Here, in semi-circle, radii OA = OC = OD = 9.5+12=10.52=5.25 units

And, OB = AB - AO = 9.5 - 5.25 = 4.25 units

In a right angled triangle OBD,

BD=OD2-OB2=5.252-4.252=5.25+4.255.25-4.25             a2-b2=a+ba-b=9.5×1=9.5

So, BF=BE+EF=9.5+1                  BD=BE=9.5 Radii

Page No 35:

Question 9:

Visualize the representation of 3.765 on the number line using successive magnification.

Answer:

3 < 3.765 < 4
Divide the gap between 3 and 4 on the number line into 10 equal parts.
Now, 3.7 < 3.765 < 3.8
In order to locate the point 3.765 on the number line, divide the gap between 3.7 and 3.8 into 10 equal parts.
Further, 3.76 < 3.765 < 3.77
So, to locate the point 3.765 on the number line, again divide the gap between 3.76 and 3.77 into 10 equal parts.
Now, the number 3.765 can be located on the number line. This can be shown as follows:


Here, the marked point represents the point 3.765 on the number line.

Page No 35:

Question 10:

Visualize the representation of 4.67¯ on the number line up to 4 decimal places.

Answer:

4.67¯=4.6767  (Upto 4 decimal places)
4 < 4.6767 < 5
Divide the gap between 4 and 5 on the number line into 10 equal parts.
Now, 4.6 < 4.6767 < 4.7
In order to locate the point 4.6767 on the number line, divide the gap between 4.6 and 4.7 into 10 equal parts.
Further, 4.67 < 4.6767 < 4.68
To locate the point 4.6767 on the number line, again divide the gap between 4.67 and 4.68 into 10 equal parts.
Again, 4.676 < 4.6767 < 4.677
To locate the point 4.6767 on the number line, again divide the gap between 4.676 and 4.677 into 10 equal parts.
Now, the number 4.6767 can be located on the number line. This can be shown as follows:


Here, the marked point represents the point 4.67¯ on the number line up to 4 decimal places.



Page No 43:

Question 1:

Write the rationalising factor of the denominator in 12+3.

Answer:


12+3
=13+2×3-23-2=3-232-22
=3-23-2=3-21
Here, the denominator i.e. 1 is a rational number. Thus, the rationalising factor of the denominator in 12+3 is 3-2.

Page No 43:

Question 2:

Rationalise the denominator of each of the following.
(i) 17                     (ii) 523                   (iii) 12+3
(iv) 15-2               (v) 15+32              (vi) 17-6
(vii) 411-7        (viii) 1+22-2              (ix) 3-223+22

Answer:

(i) 17
On multiplying the numerator and denominator of the given number by 7, we get:

 17 = 17×77 = 77

(ii) 523
On multiplying the numerator and denominator of the given number by 3, we get:

 523 = 523×33 = 156

(iii) 12+3
On multiplying the numerator and denominator of the given number by 2-3, we get:
 12+3 = 12+3×2-32-3 =2-322-32= 2-34-3=2-31 = 2-3

(iv) 15-2
On multiplying the numerator and denominator of the given number by 5+2, we get:
 15-2 = 15-2×5+25+2 =5+252-22= 5+25-4=5+21 = 5+2

(v) 15+32
On multiplying the numerator and denominator of the given number by 5-32, we get:
 15+32 = 15+32×5-325-32 =5-3252-322= 5-3225-18=5-327 

(vi) 17-6
Multiplying the numerator and denominator by 7+6, we get
17-6=17-6×7+67+6=7+672-62
=7+67-6=7+6

(vii) 411-7     
Multiplying the numerator and denominator by 11+7, we get
411-7=411-7×11+711+7=411+7112-72 
=411+711-7=411+74=11+7

(viii) 1+22-2 
Multiplying the numerator and denominator by 2+2, we get 
1+22-2=1+22-2×2+22+2=2+2+22+222-22        
=4+324-2=4+322
(ix) 3-223+22
Multiplying the numerator and denominator by 3-22, we get 
3-223+22=3-223+22×3-223-22=3-22232-222
=9+8-1229-8              a-b2=a2+b2-2ab=17-122

Page No 43:

Question 3:

It being given that 2=1.414, 3=1.732, 5=2.236 and 10=3.162, find the value of three places of decimals, of each of the following.
(i) 25

(ii) 2-33

(iii) 10-52

Answer:


(i)
 25=25×55=255
=2×2.2365=0.894
(ii) 
2-33=2-33×33=23-33
=2×1.732-33=0.155
(iii)
 10-52=10-52×22=210-52
=1.414×3.162-2.2362=0.655

Page No 43:

Question 4:

Find rational numbers a and b such that

(i) 2-12+1=a+b2

(ii) 2-52+5=a5+b

(iii) 3+23-2=a+b6

(iv) 5+237+43=a+b3

Answer:


(i)
2-12+1=2-12+1×2-12-1=2-1222-12
=2+1-222-1=3-22
2-12+1=3+-22=a+b2a=3, b=-2

(ii)
2-52+5=2-52+5×2-52-5=2-5222-52
=4+5-454-5=9-45-1=-9+45
2-52+5=45+-9=a5+ba=4, b=-9

(iii)
3+23-2=3+23-2×3+23+2=3+2232-22
=3+2+2×3×23-2=5+26
3+23-2=5+26=a+b6a=5, b=2

(iv)
5+237+43=5+237+43×7-437-43=35-203+143-2472-432
=11-6349-48=11-63
5+237+43=11+-63=a+b3a=11,b=-6

Page No 43:

Question 5:

It being given  that 3=1.732, 5=2.236, 6=2.449  and 10=3.162, find to three places of decimal, the value of each of the following.

(i) 16+5

(ii) 65+3

(iii) 143-35

(iv) 3+53-5

(v) 1+232-3

(vi) 5+25-2

Answer:


(i)
 16+5=16+5×6-56-5=6-562-52
=6-56-5=6-5=2.449-2.236
=0.213
(ii) 
65+3=65+3×5-35-3=65-352-32
=65-35-3=65-32=35-3
=3×2.236-1.732=1.512
(iii)
 143-35=143-35×43+3543+35=43+35432-352
=43+3548-45=4×1.732+3×2.2363=4.545
(iv)
 3+53-5=3+53-5×3+53+5=3+5232-52
=9+5+659-5=14+654=7+352
=7+3×2.2362=6.854
(v)
 1+232-3=1+232-3×2+32+3=2+3+43+622-32
=8+534-3=8+53=8+5×1.732
=16.660
(vi)
 5+25-2=5+25-2×5+25+2=5+2252-22
=5+2+2×5×25-2=7+2103=7+2×3.1623
=4.441



Page No 44:

Question 6:

Simplify by rationalising the denominator.

(i) 73-5248+18

(ii) 26-535-26

Answer:


(i) 
73-5248+18=73-5216×3+9×2=73-5243+32
=73-5243+32×43-3243-32=73×43-73×32-52×43+52×32432-322=84-216-206+3048-18
=114-41630
(ii)
26-535-26=26-535-26×35+2635+26=26×35+26×26-5×35-5×26352-262
=630+24-15-23045-24=9+43021

Page No 44:

Question 7:

Simplify

(i) 4+54-5+4-54+5

(ii) 13+2-25-3-32-5

(iii) 2+32-3+2-32+3+3-13+1

(iv) 262+3+626+3-836+2

Answer:

(i)
4+54-5+4-54+5=4+54-5×4+54+5+4-54+5×4-54-5=4+5242-52+4-5242-52
=16+5+85+16+5-8516-5=4211
(ii)
13+2-25-3-32-5=13+2×3-23-2-25-3×5+35+3-32-5×2+52+5=3-232-22-25+352-32-32+522-52
=3-23-2-25+35-3-32+52-5=3-2-25+32-32+5-3=3-2-5-3+2+5
=0
(iii)
2+32-3+2-32+3+3-13+1=2+32-3×2+32+3+2-32+3×2-32-3+3-13+1×3-13-1=2+3222-32+2-3222-32+3-1232-12
=4+3+434-3+4+3-434-3+3+1-233-1=7+43+7-43+4-232=14+2-3
=16-3
(iv)
262+3+626+3-836+2=263+2×3-23-2+626+3×6-36-3-836+2×6-26-2=26×3-26×232-22+62×6-62×362-32-83×6-83×262-22
=218-2123-2+612-666-3-818-866-2=218-212+612-663-818-864=218-212+212-26-218+26
=0

Page No 44:

Question 8:

Prove that
(i) 13+7+17+5+15+3+13+1=1
(ii) 11+2+12+3+13+4+14+5+15+6+16+7+17+8+18+9=2

Answer:


(i)
13+7+17+5+15+3+13+1=13+7×3-73-7+17+5×7-57-5+15+3×5-35-3+13+1×3-13-1=3-732-72+7-572-52+5-352-32+3-132-12
=3-79-7+7-57-5+5-35-3+3-13-1=3-72+7-52+5-32+3-12=3-7+7-5+5-3+3-12
=22=1
(ii)
11+2+12+3+13+4+14+5+15+6+16+7+17+8+18+9=11+2×1-21-2+12+3×2-32-3+13+4×3-43-4+14+5×4-54-5+15+6×5-65-6+16+7×6-76-7+17+8×7-87-8+18+9×8-98-9
=1-212-22+2-322-32+3-432-42+4-542-52+5-652-62+6-762-72+7-872-82+8-982-92=1-21-2+2-32-3+3-43-4+4-54-5+5-65-6+6-76-7+7-87-8+8-98-9=1-2-1+2-3-1+3-4-1+4-5-1+5-6-1+6-7-1+7-8-1+8-9-1
=2-1+3-2+4-3+5-4+6-5+7-6+8-7+9-8=3-1=2

Page No 44:

Question 9:

Find the values of a and b if
7+353+5-7-353-5=a+b5

Answer:


7+353+5-7-353-5=7+353+5×3-53-5-7-353-5×3+53+5=73-5+353-532-52-73+5-353+532-52
=21-75+95-159-5-21+75-95-159-5=6+254-6-254
=6+25-6+254=454=5
7+353+5-7-353-5=0+1×5
Comparing with the given expression, we get

a = 0 and b = 1

Thus, the values of a and b are 0 and 1, respectively.

Page No 44:

Question 10:

Simplify 13-1113+11+13+1113-11.

Answer:


13-1113+11+13+1113-11=13-1113+11×13-1113-11+13+1113-11×13+1113+11=13-112132-112+13+112132-112
=13+11-2×13×1113-11+13+11+2×13×1113-11=24-21432+24+21432=24-2143+24+21432
=482=24

Page No 44:

Question 11:

If x=3+22, check whether x+1x is rational or irrational.

Answer:


x=3+22                 .....1
1x=13+221x=13+22×3-223-221x=3-2232-222
1x=3-229-81x=3-22              .....2
Adding (1) and (2), we get
x+1x=3+22+3-22=6, which is a rational number
Thus, x+1x is rational.

Page No 44:

Question 12:

If x=2-3, find value of x-1x3.

Answer:


x=2-3                  .....11x=12-31x=12-3×2+32+3
1x=2+322-321x=2+34-31x=2+3                .....2
Subtracting (2) from (1), we get
x-1x=2-3-2+3x-1x=2-3-2-3=-23x-1x3=-233=-243
Thus, the value of x-1x3 is -243.

Page No 44:

Question 13:

If x=9-45, find the value of x2+1x2.

Answer:


x=9-45                .....11x=19-451x=19-45×9+459+45
1x=9+4592-4521x=9+4581-801x=9+45             .....2
Adding (1) and (2), we get
x+1x=9-45+9+45x+1x=18
Squaring on both sides, we get
x+1x2=182x2+1x2+2×x×1x=324x2+1x2=324-2=322
Thus, the value of x2+1x2 is 322.

Page No 44:

Question 14:

If x=5-212, find the value of x+1x.

Answer:


x=5-212                    .....11x=15-2121x=25-21
1x=25-21×5+215+211x=25+2152-2121x=25+2125-21
1x=25+2141x=5+212               .....2
Adding (1) and (2), we get
x+1x=5-212+5+212x+1x=5-21+5+212x+1x=102=5
Thus, the value of x+1x is 5.

Page No 44:

Question 15:

If a=3-22, find the value of a2-1a2.

Answer:


a=3-22a2=3-222a2=9+8-122a2=17-122               .....1
1a2=117-1221a2=117-122×17+12217+1221a2=17+122172-1222
1a2=17+122289-2881a2=17+122              .....2
Subtracting (2) from (1), we get
a2-1a2=17-122-17+122a2-1a2=17-122-17-122a2-1a2=-242
Thus, the value of a2-1a2 is -242.



Page No 45:

Question 16:

If x=13+23, find the value of x-1x.

Answer:


x=13+23                 .....11x=113+231x=113+23×13-2313-23
1x=13-23132-2321x=13-2313-121x=13-23              .....2
Subtracting (2) from (1), we get
x-1x=13+23 -13-23 x-1x=13+23 -13+23x-1x=43 
Thus, the value of x-1x is 43.

Page No 45:

Question 17:

If x=2+3, find the value of x3+1x3.

Answer:


x=2+3                  .....11x=12+31x=12+3×2-32-3
1x=2-322-321x=2-34-31x=2-3             .....2
Adding (1) and (2), we get
x+1x=2+3+2-3=4          .....3
Cubing both sides, we get
x+1x3=43x3+1x3+3×x×1xx+1x=64
x3+1x3+3×4=64             [Using (3)]
x3+1x3=64-12=52
Thus, the value of x3+1x3 is 52.

Page No 45:

Question 18:

If x=5-35+3 and y=5+35-3, show that x2-y2=-10311.

Answer:

Disclaimer: The question is incorrect.

x=5-35+3x=5-35+3×5-35-3x=5-3252-32
x=25+3-10325-3x=28-10322x=14-5311

y=5+35-3y=5+35-3×5+35+3y=5+3252-32
y=25+3+10325-3y=28+10322y=14+5311

x2-y2=14-53112-14+53112=196+75-1403121-196+75+1403121
=271-1403121-271+1403121=271-1403-271-1403121=-2803121
The question is incorrect. Kindly check the question.
The question should have been to show that x-y=-10311.
x-y=14-5311-14+5311=14-53-14-5311=-10311

Page No 45:

Question 19:

If a=5+25-2 and b=5-25+2, show that 3a2+4ab-3b2=4+56310.

Answer:

According to question,
a=5+25-2 and b=5-25+2

a=5+25-2   =5+25-2×5+25+2   =5+2252-22   =52+22+2525-2   =5+2+2103   =7+2103    ...1b=5-25+2   =5-25+2×5-25-2   =5-2252-22   =52+22-2525-2   =5+2-2103   =7-2103  ...2
Now,
    3a2+4ab-3b2=3a2-b2+4ab=3a+ba-b+4ab=37+2103+7-21037+2103-7-2103+47+2103×7-2103=31434103+472-21029=56310+449-409=56310+4
Hence, 3a2+4ab-3b2=4+56103.

Page No 45:

Question 20:

If a=3-23+2 and b=3+23-2, find the value of a2 + b2 – 5ab.

Answer:

According to question,
a=3-23+2 and b=3+23-2

a=3-23+2   =3-23+2×3-23-2   =32+22-23232-22   =3+2-263-2   =5-261   =5-26     ....1b=3+23-2   =3+23-2×3+23+2   =32+22+23232-22   =3+2+263-2   =5+261   =5+26      ....2
Now,
    a2+b2-5ab=a-b2-3ab=5-26-5+262-35-265+26=-462-325-24=96-3=93

Hence, the value of a2 + b2 – 5ab is 93.

Page No 45:

Question 21:

If p=3-53+5 and q=3+53-5, find the value of p2 + q2.

Answer:

According to question,
p=3-53+5 and q=3+53-5

p=3-53+5   =3-53+5×3-53-5   =3-5232-52   =32+52-2359-5   =9+5-654   =14-654    ...(1)q=3+53-5   =3+53-5×3+53+5   =3+5232-52   =32+52+2359-5   =9+5+654   =14+654    ...(2)

Now,
p2+q2=p+q2-2pq            =14-654+14+6542-214-65414+654            =2842-2142-65216            =72-2196-18016            =49-21616            =49-2            =47

Hence, the value of p2 + q2 is 47.

Page No 45:

Question 22:

Rationalise the denominator of each of the following.
(i) 17+6-13                 (ii) 33+5-2             (iii) 42+3+7

Answer:

i17+6-13=17+6-13×7+6+137+6+13=7+6+137+62-132=7+6+1372+62+276-132=7+6+137+6+242-13=7+6+13242=7+6+13242×4242=76+67+134284=76+67+54684
Hence, the rationalised form is 76+67+54684.
ii33+5-2=33-2+5×3-2-53-2-5=33-2-53-22-52=33-2-532+22-232-52=33-2-53+2-26-5=33-2-5-26=33-2-5-26×66=332-23-30-12=30+23-324
Hence, the rationalised form is 30+23-324.
iii42+3+7=42+3+7×2+3-72+3-7=42+3-72+32-72=42+3-722+32+223-72=42+3-74+3+43-7=42+3-743=2+3-73=2+3-73×33=23+3-213
Hence, the rationalised form is 23+3-213.

Page No 45:

Question 23:

Given, 2=1.414 and 6=2.449, find the value of 13-2-1 correct to 3 places of decimal.

Answer:


13-2-1=13-2+1×3+2+13+2+1=3+2+132-2+12=3+2+132-22-221-12=3+2+13-2-22-1=3+2+1-22=3+2+1-22×22=6+2+2-4=2.449+2+1.414-4              2=1.414 and 6=2.449=5.863-4=-1.465

Hence, the value of 13-2-1 correct to 3 places of decimal is −1.465.

Page No 45:

Question 24:

If x=12-3, find the value of x3 – 2x2 – 7x + 5.

Answer:

x=12-3x=12-3×2+32+3x=2+322-32x=2+34-3x=2+3    ...1Now,x2=2+32x2=22+32+223x2=4+3+43x2=7+43    ...2Also,x3=x2.xx3=7+432+3x3=14+73+83+12x3=26+153    ...3

Now,
x3-2x2-7x+5=26+153-27+43-72+3+5    (using 1, 2 and 3)=26+153-14-83-14-73+5=31-28+153-153=3

Hence, the value of x3 – 2x2 – 7x + 5 is 3.

Page No 45:

Question 25:

Evaluate 1510+20+40-5-80, it being given that 5=2.236 and 10=3.162.
Hint
1510+20+40-5-80=1510+25+210-5-45=15310-35=510-5

Answer:

1510+20+40-5-80=1510+25+210-5-45=15310-35=510-5=510-5×10+510+5=510+5102-52=510+510-5=510+55=10+5=3.162+2.236     (given)=5.398

Hence, 1510+20+40-5-80 = 5.398 .



Page No 53:

Question 1:

Simplify
(i) 223×213

(ii) 223×215

(iii) 756×723

(iv) 129614×129612

Answer:

(i) 223×213
223×213=223+13               =22+13               =233               =21               =2

(ii) 223×215
223×215=223+15               =210+315               =21315

(iii) 756×723
756×723=756+23               =75+46               =796               =732

(iv) 129614×129612
129614×129612=129614+12                                 =12961+24                                 =129634                                 =6434                                 =63                                 =216

Page No 53:

Question 2:

Simplify:
(i) 61/461/5
(ii) 81/282/3
(iii) 56/752/3

Answer:

(i) 614615=614-15=65-420= 6120         aman=am-n(ii) 812823=812-23=83-46=8-16(iii) 567523=567-23=518-1421=5421

Page No 53:

Question 3:

Simplify:
(i) 31/4×51/4
(ii) 25/8×35/8
(iii) 61/2×71/2

Answer:

(i) 314×514=1514                            (am×bm)=abm(ii) 258×358=658(iii) 612×712=4212

Page No 53:

Question 4:

Simplify:
(i) (34)1/4
(ii) (31/3)4
(iii) 134

Answer:

(i)  3414=34×14=3                           (a)mn=amn(ii)  3134=313×4=343(iii)  13412=134×12=132=19=3-2

Page No 53:

Question 5:

Evaluate
(i) 12513

(ii) 6416

(iii) 2532

(iv) 8134

(v) 64-12

(vi) 8-13

Answer:

(i) 12513=5313=53×13=5(ii) 6416=2616=26×16=2

(iii) 2532= 52×32= 53 = 125               (am)n= amn(iv) 8134= 3434= 34×34= 33= 27
(v) 64-12=82-12=82×-12=8-1=18(vi) 8-13=23-13=23×-13=2-1=12

Page No 53:

Question 6:

If a = 2, b = 3, find the values of

(i) (ab + ba)–1

(ii) (aa + bb)–1

Answer:

(i) (ab + ba)–1
ab+ba-1=23+32-1                    =8+9-1                    =17-1                    =117



(ii) (aa + bb)–1
aa+bb-1=22+33-1                    =4+27-1                    =31-1                    =131

Page No 53:

Question 7:

Simplify

(i) 8149-32

(ii) (14641)0.25

(iii) 32243-45

(iv) 7776243-35

Answer:

(i) 8149-32
8149-32=972-32                =97-3                =793                =343729

(ii) (14641)0.25
146410.25=1464125100                    =1464114                    =11414                    =11

(iii) 32243-45
32243-45=2433245                  =32545                  =324                  =8116

(iv) 7776243-35
7776243-35=243777635                    =36535                    =123                    =18



Page No 54:

Question 8:

Evaluate
(i) 4216-23+1256-34+2243-15

(ii) 64125-23+256625-14+370

(iii) 8116-34 259-32÷52-3

(iv) 2552×7291312523×2723×843

Answer:

(i) 4216-23+1256-34+2243-15
4216-23+1256-34+2243-15=463-23+144-34+235-15=46-2+14-3+23-1=462+43+23=144+64+6=214


(ii) 64125-23+256625-14+370
64125-23+256625-14+370=453-23+454-14+1=45-2+45-1+1=542+54+1=2516+54+1=25+20+1616=6116


(iii) 8116-34 259-32÷52-3
8116-34 259-32÷52-3=16813492532÷253=2343435232÷8125=233353÷8125=827×271258125=1

(iv) 2552×7291312523×2723×843
2552×7291312523×2723×843=5252×93135323×3323×2343=55×9152×32×24=5×5×5×5×5×95×5×3×3×2×2×2×2=12516

Page No 54:

Question 9:

Evaluate
(i) 13+23+3312

(ii) 5813+2713314

(iii) 20+7050

(iv) 161212

Answer:

(i) 13+23+3312
13+23+3312=1+8+2712                         =3612                         =6212                         =6

(ii) 5813+2713314
5813+2713314=52313+3313314                               =52+3314                               =5414                               =5

(iii) 20+7050
20+7050=1+11              =2

(iv) 161212
161212=421212                =4112                =2212                =2

Page No 54:

Question 10:

Prove that
(i) 8-23×212×25-54÷32-25×125-56=2

(ii) 64125-23+125662514+25643=6516

(iii) 78114+25614144=16807

Answer:

(i) 8-23×212×25-54÷32-25×125-56=2
LHS=8-23×212×25-54÷32-25×125-56        =23-23×212×52-54÷25-25×53-56        =2-2×212×5-52÷2-2×5-52        =2-2×212×5-522-2×5-52        =212        =2        =RHS 8-23×212×25-54÷32-25×125-56=2

(ii) 64125-23+125662514+25643=6516
LHS=64125-23+125662514+25643        =453-23+145414+5×54×4×43        =45-2+145+54        =542+54+54        =2516+54+54        =25+20+2016        =6516        =RHS 64125-23+125662514+25643=6516

(iii) 78114+25614144=16807
LHS=78114+25614144        =73414+4414144        =73+4144        =71×7144        =71+144        =7544        =75        =16807        =RHS78114+25614144=16807

Page No 54:

Question 11:

Simplify x234 and express the result in the exponential form of x.

Answer:

x234=x21314            =x2314            =x212            =x16

Hence, the result in the exponential form is x16.

Page No 54:

Question 12:

Simplify the product 23·24·3212.

Answer:

23·24·3212=213.214.32112                           =213.214.25112                           =213.214.2512                           =213+14+512                           =24+3+512                           =21212                           =21                           =2

Page No 54:

Question 13:

Simplify
(i) 1513914-6

(ii) 1215271552

(iii) 1514312-2

Answer:

(i) 1513914-6
1513914-6=91415136                 =9641563                 =932152                 =3232152                 =33152                 =27225                 =325

(ii) 1215271552
1215271552=121552271552                =12122712                =122712                =4912                =23212                =23

(iii) 1514312-2
1514312-2=31215142                =3221524                =31512                =3312.512                =31-12512                =312512                =35

Page No 54:

Question 14:

Find the value of x in each of the following.

(i) 5x+25=2

(ii) 3x-23=4

(iii) 343 43-7=342x

(iv) 5x-3×32x-8=225

(v) 33x·32x3x=3204

Answer:

i 5x+25=25x+215=25x+2155=255x+2=325x=32-25x=30x=305x=6
Hence, the value of x is 6.

ii 3x-23=43x-213=43x-2133=433x-2=643x=64+23x=66x=663x=22
Hence, the value of x is 22.

iii 343 43-7=342x343 347=342x343+7=342x3410=342x10=2x102=x5=x
Hence, the value of x is 5.

iv 5x-3×32x-8=2255x-3×32x-8=1525x-3×32x-8=52×32x-3=2 and 2x-8=2x=2+3 and 2x=2+8x=5 and 2x=10x=5 and x=102x=5 and x=5x=5
Hence, the value of x is 5.

v 33x·32x3x=320433x+2x3x=3201435x3x=3535x-x=3534x=354x=5x=54
Hence, the value of x is 54.



Page No 55:

Question 15:

Prove that

(i) x-1y·y-1z·z-1x=1.

(ii) x1a-b1a-c·x1b-c1b-a·x1c-a1c-b=1

(iii) xab-cxba-c÷xbxac=1

(iv) xa+b2 xb+c2 xc+a2xaxbxc4=1

Answer:

(i) x-1y·y-1z·z-1x=1
LHS=x-1y·y-1z·z-1x        =x-1y12·y-1z12·z-1x12        =x-12y12·y-12z12·z-12x12        =x-12+12y12-12z12-12        =x0y0z0        =1        =RHS

Hence, x-1y·y-1z·z-1x=1.

(ii) x1a-b1a-c·x1b-c1b-a·x1c-a1c-b=1
LHS=x1a-b1a-c·x1b-c1b-a·x1c-a1c-b        =x1a-b1a-c·x-1c-b1b-a·x1c-a1c-b        =x1a-b1a-c·x-1b-a1c-b·x1c-a1c-b        =x1a-b1a-c·x-1b-a.x1c-a1c-b        =x1a-b1a-c·x1c-a-1b-a1c-b        =x1a-b1a-c·xb-a-c+ac-ab-a1c-b        =x1a-b1a-c·xb-cc-ab-a-1b-c        =x1a-ba-c·x-1c-ab-a        =x1b-ac-a·x-1c-ab-a        =x1b-ac-a-1c-ab-a        =x0        =1        =RHS

Hence, x1a-b1a-c·x1b-c1b-a·x1c-a1c-b=1.

(iii) xab-cxba-c÷xbxac=1
LHS=xab-cxba-c÷xbxac        =xab-cxba-c×xaxbc        =xab-cxba-c×xacxbc        =xab-acxba-bc×xacxbc        =xab-ac-ba+bc.xac-bc        =x-ac+bc.xac-bc        =x-ac+bc+ac-bc        =x0        =1        =RHS

Hence, xab-cxba-c÷xbxac=1.

(iv) xa+b2 xb+c2 xc+a2xaxbxc4=1
LHS=xa+b2 xb+c2 xc+a2xaxbxc4        =x2a+2b x2b+2c x2c+2ax4ax4bx4c        =x2a+2b+2b+2c+2c+2ax4a+4b+4c        =x4a+4b+4cx4a+4b+4c        =1        =RHS
Hence, xa+b2 xb+c2 xc+a2xaxbxc4=1.

Page No 55:

Question 16:

If x is a positive real number and exponents are rational numbers, simplify

xbxcb+c-a·xcxac+a-b·xaxba+b-c

Answer:

xbxcb+c-a·xcxac+a-b·xaxba+b-c=xb-cb+c-a·xc-ac+a-b·xa-ba+b-c=xb-cb.xb-cc-a·xc-ac+a-b·xa-bb-c.xa-ba=xb-cb.xb-cc-a·xc+a-bc-a·xa-bb-c.xa-ba=xb-cb.xb-cc-a·xc+a-bc-a·xa-bb-c.xa-ba=xb-cb.xb-c+c+a-bc-a·xa-bb-c.xa-ba=xb-cb.xac-a·xa-bb-c.xa-ba=xbb-c.xac-a·xa-bb-c.xa-ba=xbb-c·xa-bb-c.xac-a.xa-ba=xb+a-bb-c.xac-a.xa-ba=xab-c.xac-a.xa-ba=xb-ca.xc-aa.xa-ba=xb-c+c-a+a-ba=x0=1

Page No 55:

Question 17:

If 9n×32×3-n2-2-27n33m×23=127, prove that m – n = 1.

Answer:

9n×32×3-n2-2-27n33m×23=12732n×32×3-n-1-33n33m×23=13332n×32×3n-33n33m×23=13332n+2+n-33n33m×23=13333n+2-33n33m×23=13333n×32-33n33m×23=13333n9-133m×8=13333n833m×8=13333n33m=13333n-3m=3-33n-3m=-33n-m=-3n-m=-1m-n=1
Hence, m – n = 1.

Page No 55:

Question 18:

Write the following in ascending order of magnitude.

66, 73, 84.

Answer:

66, 73, 8466=616=6212=62112=36112   ...173=713=7412=74112=2401112  ...284=814=8312=83112=512112  ...3On Comparing 1, 2 and 3, we get36112<512112<240111266<84<73Hence, 66<84<73.



Page No 57:

Question 1:

Which of the following is a rational number?

(a) 1+3

(b) π

(c) 23

(d) 0

Answer:

Since, the sum and product of a rational and an irrational is always irrational.

So, 1+3 and 23 are irrational numbers.

Also, π is an irrational number.

And, 0 is an integer.

So, 0 is a rational number.

Hence, the correct option is (d).

Page No 57:

Question 2:

A rational number between –3 and 3 is
(a) 0
(b) –4.3
(c) –3.4
(d) 1.101100110001...

Answer:

Since, –4.3 < –3.4 < –3 < 0 < 1.101100110001... < 3

But 1.101100110001... is an irrational number

So, the rational number between –3 and 3 is 0.

Hence, the correct option is (a).

Page No 57:

Question 3:

Two rational numbers between 23 and 53 are
(a) 16 and 26

(b) 12 and 21

(c) 56 and 76

(d) 23 and 43

Answer:

We have,

23=2×23×2 =46 and 53=5×23×2=106

And, 12=1×32×3=36 and 21=2×61×6=126

Also, 23=2×23×2 =46 and 43=4×23×2=86

Since, 16<26<3612<46=23<56<76<86=43<106=53<126=21

So, the two rational numbers between 23 and 53 are 56 and 76.

Hence, the correct opion is (c).

Page No 57:

Question 4:

Every point on a number line represents
(a) a rational number
(b) a natural number
(c) an irrational number
(d) a unique number

Answer:

As, all rational numbers, all natural numbers and all irrational numbers can be represented on a nuumber line in an unique way.

So, every point on a number line represents a unique number.

Hence, the correct option is (d).

Page No 57:

Question 5:

Which of the following is a rational number?
(a) 2
(b) 23
(c) 225
(d) 0.1010010001....

Answer:

(c) 225

Because 225 is a square of 15, i.e., 225 = 15, and it can be expressed in the pq form, it is a rational number.

Page No 57:

Question 6:

Every rational number is
(a) a natural number
(b) a whole number
(c) an integer
(d) a real number

Answer:

(d) a real number

Every rational number is a real number, as every rational number can be easily expressed on the real number line.

Page No 57:

Question 7:

Between any two rational numbers there
(a) is no rational number
(b) is exactly one rational numbers
(c) are infinitely many rational numbers
(d) is no irrational number

Answer:

(c) are infinitely many rational numbers

Because the range between any two rational numbers can be easily divided into any number of divisions, there can be an infinite number of rational numbers between any two rational numbers.

Page No 57:

Question 8:

The decimal representation of a rational number is
(a) always terminating
(b) either terminating or repeating
(c) either terminating or non-repeating
(d) neither terminating nor repeating

Answer:

(b) either terminating or repeating

As per the definition of rational numbers, they are either repeating or terminating decimals.



Page No 58:

Question 9:

The decimal representation of an irrational number is
(a) always terminating
(b) either terminating or repeating
(c) either terminating or non-repeating
(d) neither terminating nor repeating

Answer:

(d) neither terminating nor repeating

As per the definition of irrational numbers, these are neither terminating nor repeating decimals.

Page No 58:

Question 10:

The decimal expansion that a rational number cannot have is
(a) 0.25
(b) 0.2528¯
(c) 0.2528¯
(d) 0.5030030003...

Answer:

As, any number which have a terminating or non-terminating recurring decimal expansion is a rational number.

So, 0.5030030003... which is non-termintaing non-recurring decimal expansion is not a rational number.

Hence, the correct option is (d).

Page No 58:

Question 11:

Which of the following is an irrational number?
(a) 3.14
(b) 3.141414...
(c) 3.14444...
(d) 3.141141114...

Answer:

(d) 3.141141114...

Because 3.141141114... is neither a repeating decimal nor a terminating decimal, it is an irrational number.

Page No 58:

Question 12:

A rational number equivalent to 719 is
(a) 17119

(b) 1457

(c) 2138

(d) 2157

Answer:

Since, 719=7×319×3=2157

Hence, the correct option is (d).

Page No 58:

Question 13:

Choose the rational number which does not lie between -23 and -15.
(a) -310

(b) 310

(c) -14

(d) -720

Answer:

We have, -23=-2×203×20=-4060 and -15=-1×125×12=-1260

And, -310=-3×610×6=-1860, 310=3×610×6=1860, -14=-1×154×15=-1560, and -720=-7×320×3=-2160

Since, -4060=-23<-2160=-720<-1860=-310<-1560=-14<-1260=-15<1860=310

So, the rational number which does not lie between -23 and -15 is 310.

Hence, the correct option is (b).

Page No 58:

Question 14:

π is
(a) a rational number
(b) an integer
(c) an irrational number
(d) a whole number

Answer:

Since, π has a non-terminating non-recurring decimal expansion.

So, π is an irrational number.

Hence, the correct option is (c).

Page No 58:

Question 15:

Decimal expansion of 2 is
(a) a finite decimal
(b) 1.4121
(c) non-terminating recurring
(d) non-terminating, non-recurring

Answer:

(c) a non-terminating and non-repeating decimal

Because 2 is an irrational number, its decimal expansion is non-terminating and non-repeating.

Page No 58:

Question 16:

Which of the following is an irrational number?

(a) 23

(b) 225

(c) 0.3799

(d) 7.478¯

Answer:

Since, 225 = 15, which is an integer,

0.3799 is a number with terminating decimal expansion, and

7.478¯ is a number with non-terminating recurring decimal expansion

Also, 23 is a prime number.

So, 23 is an irrational number.                n is always an irrational number, if n is a prime number.

Hence, the correct option is (a).

Page No 58:

Question 17:

How many digits are there in the repeating block of digits in the decimal expansion of 177?
(a) 16
(b) 6
(c) 26
(d) 7

Answer:



 177=2.428571¯

So, there are 6 digits in the repeating block of digits in the decimal expansion of 177.

Hence, the correct option is (b).

 

Page No 58:

Question 18:

Which of the following numbers is irrational?

(a) 49

(b) 12508

(c) 8

(d) 246

Answer:

Since,

49=23, which is a rational number,

12508=12508=6254=252, which is a rational number,

8=22, which is an irrational number, and

246=246=4=2, which is a rational number

Hence, the correct option is (c).



Page No 59:

Question 19:

The product of two irrational number is
(a) always irrational
(b) always rational
(c) always an integer
(d) sometimes rational and sometimes irrational

Answer:

(d) sometimes rational and sometimes irrational

For example:
2 is an irrational number, when it is multiplied with itself  it results into 2, which is a rational number.
2  when multiplied with 3, which is also an irrational number, results into 6, which is an irrational number.

Page No 59:

Question 20:

Which of the following is a true statment?
(a) The sum of two irrational numbers is an irrational number
(b) The product of two irrational numbers is an irrational number
(c) Every real number is always rational
(d) Every real number is either rational or irrational

Answer:

(d) Every real number is either rational or irrational.

Because a real number can be further categorised into either a rational number or an irrational number, every real number is either rational or irrational.

Page No 59:

Question 21:

Which of the following is a true statment?
(a) π and 227are both rationals
(b) π and 227are both irrationals
(c) π is rational and 227is irrational
(d) π is irrational and 227is rational

Answer:

(d) π is irrational and 227 is rational.
Because the value of π is neither repeating nor terminating, it is an irrational number. 227, on the other hand, is of the form pq, so it is a rational number.

Page No 59:

Question 22:

A rational number lying between 2 and 3 is
(a) 2+32
(b) 6
(c) 1.6
(d) 1.9

Answer:

Since, 2+32 and 6 are irrational numbers,

And, 2=1.414 and 3=1.732

So, the rational number lying between 2 and 3 is 1.6 .

Hence, the correct option is (c).

 

Page No 59:

Question 23:

Which of the following is a rational number?
(a) 5
(b) 0.101001000100001...
(c) π
(d) 0.853853853...

Answer:

Since, a number whose decimal expansion is terminating or non-terminating recurring is rational number.

So, 0.853853853... is a rational number.

Hence, the correct option is (d).

Page No 59:

Question 24:

The product of a nonzero rational number with an irrational number is always a/an
(a) irrational number
(b) rational number
(c) whole number
(d) natural number

Answer:

Since, the product of a non-zero rational number with an irrational number is always an irrational number.

Hence, the correct option is (a).

Page No 59:

Question 25:

The value of 0.2¯ in the form pq, where p and q are integers and q ≠ 0, is
(a) 15  

(b) 29

(c) 25

(d) 18

Answer:


Let x=0.2¯=0.222...             .....1
Multiplying both sides by 10, we get
10x=2.2¯        .....2
Subtracting (1) from (2), we get
10x-x=2.2-0.29x=2x=290.2=29
Hence, the correct answer is option (b).

Page No 59:

Question 26:

The simplest for of 1.6 is
(a) 833500
(b) 85
(c) 53
(d) none of these

Answer:

(c) 53
Let x = 1.6666666...       ...(i)
Multiplying by 10 on both sides, we get:
10x = 16.6666666...       ...(ii)
Subtracting (i) from (ii), we get:
9x = 15
x = 159 =53

Page No 59:

Question 27:

The simplest form of 0.54 is
(a) 2750
(b) 611
(c) 47
(d) none of these

Answer:

(b) 611
Let x = 0.545454...               ...(i)
Multiplying both sides by 100, we get:
100x = 54.5454545...           ...(ii)
Subtracting (i) from (ii), we get:
99x = 540
x5499 = 611



Page No 60:

Question 28:

The simplest form of 0.32 is
(a) 1645
(b) 3299
(c) 2990
(d) none of these

Answer:

(c) 2990
Let x = 0.3222222222...          ...(i)
Multiplying by 10 on both sides, we get:
10x = 3.222222222...              ...(ii)
Again, multiplying by 10 on both sides, we get:
100x = 32.222222222...          ...(iii)
On subtracting (ii) from (iii), we get:
90x = 29
x = 2990

Page No 60:

Question 29:

The simplest form of 0.123 is
(a) 41330
(b) 37330
(c) 41333
(d) none of these

Answer:

(d) none of these
Let x = 0.12333333333...         ...(i)
Multiplying by 100 on both sides, we get:
100x = 12.33333333...             ...(ii)
Multiplying by 10 on both sides, we get:
1000x = 123.33333333...         ...(iii)
Subtracting (ii) from (iii), we get:
900x = 111

x = 111900

Page No 60:

Question 30:

An irrational number between 5 and 6 is
(a) 125+6
(b) 5+6
(c) 5×6
(d) none of these

Answer:

(c) 5×6

An irrational number between a and b is given as ab.

Page No 60:

Question 31:

An irrational number between 2 and 3is
(a) 2+3
(b) 2×3
(c) 51/4
(d) 61/4

Answer:

(d) 61/4
An irrational number between 2and 3:2×3=614

Page No 60:

Question 32:

An irrational number between 17and27is
(a) 1217+27
(b) 17×27
(c) 17×27
(d) none of these

Answer:

(c) 17×27

An irrational number between a and b is given as ab.

Page No 60:

Question 33:

The sum of 0.3¯ and 0.4¯ is
(a) 710

(b) 79

(c) 711

(d) 799

Answer:

Let x=0.3¯=0.333...             .....1
Multiplying both sides by 10, we get
10x=3.3¯              .....2
Subtracting (1) from (2), we get
10x-x=3.3-0.39x=3x=390.3=39
Let y=0.4¯=0.444...             .....3
Multiplying both sides by 10, we get
10y=4.4¯              .....4
Subtracting (3) from (4), we get
10y-y=4.4-0.49y=4y=490.4=49
Sum of 0.3¯ and 0.4¯ = 0.3+0.4=39+49=79
Hence, the correct answer is option (b).

Page No 60:

Question 34:

The value of 2.45¯+0.36¯ is

(a) 6733

(b) 2411

(c) 3111

(d) 167110

Answer:

Let x=2.45=2.4545...             .....1
Multiplying both sides by 100, we get
100x=245.45              .....2
Subtracting (1) from (2), we get
100x-x=245.45-2.4599x=245-2=243x=243992.45=24399
Let y=0.36¯=0.3636...             .....3
Multiplying both sides by 100, we get
100y=36.36¯              .....4
Subtracting (3) from (4), we get
100y-y=36.36-0.3699y=36y=36990.36=3699
So, 2.45+0.36=24399+3699=243+3699=27999=3111
Hence, the correct answer is option (c).

Page No 60:

Question 35:

Which of the following is the value of 11-7 11+7?
(a) –4
(b) 4
(c) 11
(d) 7

Answer:


11-711+7=112-72               a-ba+b=a2-b2=11-7=4
Hence, the correct answer is option (b).

Page No 60:

Question 36:

-2-3 -2+3 when simplified is
(a) positive and irrational
(b) positive and rational
(c) negative and irrational
(d) negative and rational

Answer:


-2-3-2+3=-2+3×-2-3=2+32-3
=22-32            a+ba-b=a2-b2=4-3=1
Thus, the given expression when simplified is positive and rational.

Hence, the correct answer is option (b).

Page No 60:

Question 37:

6+27-3+3+1-23 when simplified is
(a) positive and irrational
(b) positive and rational
(c) negative and irrational
(d) negative and rational

Answer:


6+27-3+3+1-23=6+3×3×3-3+3+1-23=6+33-3-3+1-23=4
Thus, the given expression when simplified is positive and rational.

Hence, the correct answer is option (b).

Page No 60:

Question 38:

When 1515 is divided by 33, the quotient is

(a) 53

(b) 35

(c) 55

(d) 33

Answer:


1515÷33=155×333=15×5×333             ab=a×b=55
Hence, the correct answer is option (c).

Page No 60:

Question 39:

The value of 20×5 is

(a) 10

(b) 25

(c) 205

(d) 45

Answer:


20×5=2×2×5×5=25×5
=2×5=10
Hence, the correct answer is option (a).

Page No 60:

Question 40:

The value of 4121227 is
(a) 19

(b) 29

(c) 49

(d) 89

Answer:


4121227=42×2×3123×3×3=4×2312×33=29
Hence, the correct answer is option (b).



Page No 61:

Question 41:

10×15=?
(i) 25
(ii) 56
(iii) 65
(iv) None of these

Answer:


10×15=2×5×3×5=2×5×3×5              ab=a×b
=5×2×3=56
Hence, the correct answer is option (ii).
 

Page No 61:

Question 42:

32+488+12=?
(a) 2
(b) 2
(c) 4
(d) 8

Answer:


32+488+12=16×2+16×34×2+4×3=42+4322+23             ab=a×b
=42+322+3=42=2
Hence, the correct answer is option (b).

Page No 61:

Question 43:

(125)–1/3 = ?
(a) 5
(b) –5
(c) 15
(d) -15

Answer:


125-13=53-13=53×-13                  xab=xab
=5-1=15               x-a=1xa
Hence, the correct answer is option (c).

Page No 61:

Question 44:

The value of 712·812 is
(a) (28)1/2
(b) (56)1/2
(c) (14)1/2
(d) (42)1/2

Answer:


712·812=7×812                xa×ya=x×ya=5612
Hence, the correct answer is option (b).

Page No 61:

Question 45:

After simplification, 13151313 is
(a) 13215

(b) 13815

(c) 1313

(d) 13-215

Answer:


13151313=1315-13                 xaxb=xa-b=133-515=13-215
Hence, the correct answer is option (d).

Page No 61:

Question 46:

The value of 64-24 is

(a) 18

(b) 12

(c) 8

(d) 164

Answer:

64-24=64-214               =26-214               =2-1214               =2-3               =123               =18

The value of 64-24 is 18.

Hence, the correct option is (a).

Page No 61:

Question 47:

The value of 20+7050 is
(a) 0

(b) 2

(c) 95

(d) 15

Answer:

20+7050=1+11              =21              =2

The value of 20+7050 is 2.

Hence, the correct option is (b).

Page No 61:

Question 48:

The value of 24315 is

(a) 3

(b) –3

(c) 5

(d) 13

Answer:

24315=3515            =31            =3

The value of 24315 is 3.

Hence, the correct option is (a).

Page No 61:

Question 49:

93+-33-63 = ?
(a) 432
(b) 270
(c) 486
(d) 540

Answer:

93+-33-63=729+-27-216                         =729-27-216                         =729-27+216                         =729-243                         =486 93+-33-63=486

Hence, the correct option is (c).

Page No 61:

Question 50:

Simplified value of 16-14×164 is
(a) 0
(b) 1
(c) 4
(d) 16

Answer:

16-14×164=24-14×2414                         =2-1×21                         =2-1+1                         =20                         =1

Simplified value of 16-14×164 is 1.

Hence, the correct option is (b).

Page No 61:

Question 51:

The value of  2234 is
(a) 2-16

(b) 2-6

(c) 216

(d) 26

Answer:

2234=221314            =22314            =2212            =216

The value of  2234 is 216.

Hence, the correct option is (c).

Page No 61:

Question 52:

Simplified value of 2513×513 is
(a) 25
(b) 3
(c) 1
(d) 5

Answer:

2513×513=5213×513                   =523×513                   =523+13                   =52+13                   =533                   =5

Simplified value of 2513×513 is 5.

Hence, the correct option is (d).

Page No 61:

Question 53:

The value of 811212 is
(a) 3
(b) –3
(c) 9
(d) 13

Answer:

811212=341212                 =3212                 =3

The value of 811212 is 3.

Hence, the correct option is (a).

Page No 61:

Question 54:

There is a number x such that x2 is irrational but x4 is rational. Then, x can be

(a) 5

(b) 2

(c) 23

(d) 24

Answer:

(a) Let x=5.x2=52=5, which is a rational number.(b) Let x=2.x2=22=2, which is a rational number.(c) Let x=23.x2=232=223, which is an irrational number.x4=234=243, which is also an irrational number.(d) Let x=24.x2=242=224=212, which is an irrational number.x4=244=244=2, which is a rational number.

x can be 24.

Hence, the correct option is (d).



Page No 62:

Question 55:

If x=75 and 5x=p7 then the value of p is
(a) 725

(b) 257

(c) 715

(d) 157

Answer:

5x=p7575=p7     x=755×57=p7257=p7257×7=p257=p The value of p is 257.

Hence, the correct option is (b).

Page No 62:

Question 56:

The value of 256x1681y4-14 is
(a) 3y8x4

(b) 3y4x4

(c) 4y5x4

(d) 4x43y

Answer:

256x1681y4-14=28 x1634 y4-14                       =34 y428 x1614                       =3414 y4142814 x1614                       =3 y22 x4                       =3y4x4 The value of 256x1681y4-14 is 3y4x4.

Hence, the correct option is (b).

Page No 62:

Question 57:

The value of xp-q·xq-r·xr-p is equal to
(a) 0

(b) 1

(c) x

(d) xpqr

Answer:

xp-q·xq-r·xr-p=xp-q+q-r+r-p                              =x0                              =1 The value of xp-q·xq-r·xr-p is equal to 1.

Hence, the correct option is (b).

Page No 62:

Question 58:

The value of p-1q·q-1r·r-1p is
(a) –1

(b) 0

(c) 1

(d) 2

Answer:

p-1q·q-1r·r-1p=p-1q12·q-1r12·r-1p12                                          =p-12q12.q-12r12·r-12p12                                          =p-12+12q12-12r12-12                                          =p0q0r0                                          =1 The value of p-1q·q-1r·r-1p is equal to 1.

Hence, the correct option is (c).

Page No 62:

Question 59:

23×24×3212=?

(a) 2

(b) 2

(c) 22

(d) 42

Answer:

23×24×3212=213×214×32112                              =213×214×25112                              =213×214×2512                              =213+14+512                              =24+3+512                              =21212                              =2 23×24×3212=2.

Hence, the correct option is (a).

Page No 62:

Question 60:

If 23x 322x=8116 then x=?
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

23x 322x=811623x23-2x=811623x-2x=811623-x=342432x=3424x=4
Hence, the correct answer is option (d).

Page No 62:

Question 61:

If 332=9x then 5x=?
(a) 1
(b) 5
(c) 25
(d) 125

Answer:

332=9x 323=32xx=3Now5x=53=125
Hence, the correct answer is option (d).

Page No 62:

Question 62:

On simplification, the expression 5n+2-6×5n+113×5n-2×5n+1 equals
(a) 53

(b) -53

(c) 35

(d) -35

Answer:

5n+2-6×5n+113×5n-2×5n+1=5n×52-6×5n×513×5n-2×5n×5=5n×55-65n13-2×5=-53
Hence, the correct answer is option (b). 

Page No 62:

Question 63:

The simplest rationalisation factor of 5003 is
(a) 5
(b) 3
(c) 53
(d) 23

Answer:

5003=5×5×2×5×23=53×223=543  
So, the simplest rationalisation factor of 5003 is 23.
Hence, the correct answer is option (d). 

Page No 62:

Question 64:

The simplest rationalisation ractor of 22-3 is
(a) 22+3
(b) 22+3
(c) 2+3
(d) 2-3

Answer:

Simplest rationalisation ractor of 22-3 is 22+3.
Hence, the correct answer is option (b). 

Page No 62:

Question 65:

The rationalisation factor of 123-5 is
(a) 5-23
(b) 3+25
(c) 3+5
(d) 12+5

Answer:

Rationalisation factor of 123-5 will be 23+5=4×3+5=12+5.
Hence, the correct answer is option (d). 

Page No 62:

Question 66:

Rationalisation of the denominator of 15+2 gives
(a) 110
(b) 5+2
(c) 5-2
(d) 5-23

Answer:

15+2
Rationalisation of denominator gives
15+2×5-25-2=5-25-2=5-23
Hence, the correct answer is option (d). 

Page No 62:

Question 67:

If x=2+3 then x+1x equals
(a) -23
(b) 2
(c) 4
(d) 4-23

Answer:

x=2+31x=12+3=12+3×2-32-3=2-34-3=2-3
x+1x=2+3+2-3=4
Hence, the correct answer is option (c). 



Page No 63:

Question 68:

13+22=?

(a) 3-2217

(b) 3-2213

(c) 3-22

(d) None of these

Answer:

13+22=13+22×3-223-22=3-229-8=3-22
Hence, the correct answer is option (c). 

Page No 63:

Question 69:

If x=7+43 then x+1x=?
(a) 83
(b) 14
(c) 49
(d) 48

Answer:

Given: x=7+43
1x=17+43=17+43×7-437-43=7-4349-48=7-43
x+1x=7+43+7-43=14
Hence, the correct answer is option (b). 

Page No 63:

Question 70:

If 2=1.41 then 12=?
(a) 0.075
(b) 0.75
(c) 0.705
(d) 7.05

Answer:

12=12×22=22Given: 2=1.41So, 22=1.412=0.705
Hence, the correct answer is option (c). 

Page No 63:

Question 71:

If 7=2.646 then 17 =?
(a) 0.375
(b) 0.378
(c) 0.441
(d) None of these

Answer:

17=17×77=77
Given that 
7=2.646So, 77=2.6467=0.378
Hence, the correct answer is option (b). 

Page No 63:

Question 72:

The value of 3-22 is
(a) 3+2
(b) 3-2
(c) 2+1
(d) 2-1

Answer:

3-22=2+1-2×2×1=22+12-2×2×1
This is of the form
a2+b2-2ab=a-b2
22+12-2×2×1=2-12So, 3-22=2-12=2-1
Hence, the correct answer is option (d). 
 

Page No 63:

Question 73:

The value of 5+26 is
(a) 5+6
(b) 5-6
(c) 3+2
(d) 3-2

Answer:

5+26=2+3+2×3×2=22+32+2×3×2
This is in the form
a2+b2+2ab=a+b2
So, we have 
22+32+2×3×2=2+32
Thus, 5+26=2+32=2+3
Hence, the correct answer is option (c). 

Page No 63:

Question 74:

If 2=1.414 then 2-12+1=?
(a) 0.207
(b) 2.414
(c) 0.414
(d) 0.621

Answer:

2-12+1=2-12+1×2-12-1=2-122-1=2+1-221=3-22=3-2×1.414=3-2.828=0.172=0.414
Hence, the correct answer is option (c).

Page No 63:

Question 75:

If x=3+8 then x2+1x2=?
(a) 34
(b) 56
(c) 28
(d) 63

Answer:

Given: x=3+8
1x=13+8=13+8×3-83-8=3-89-8=3-81=3-8
x+1x=3+8+3-8=6
x+1x2=x2+1x2+2×x×1x=x2+1x2+262=x2+1x2+236=x2+1x2+2x2+1x2=36-2=34
Hence, the correct answer is option (a). 

 

Page No 63:

Question 76:

Assertion: Three rational numbers between 25 and 35are920, 1020and 1120.
Reason: A rational number between two rational numbers p and q is 12p+q.
(a) Both Assertion and Reason are true and Reasom is a correct explanation of Assertion.
(b) Both Assertion and Reason and Reasom are true but Reasom is not a correct explanation of Assertion.
(c) Assertion is true and Reasom is false.
(d) Assertion is false and Reasom is true.

Answer:

(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Rational number between 25 and 35: 25+352=12=1020Rational number between 25 and 1020: 25+10202=1840=920Rational number between 35 and 1020: 35+10202=2240=1120

So, Assertion and Reason are correct (property of rational numbers). Also, Reason is the correct explanation of Assertion.



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Question 77:

Assertion: 3is an irrational number.
Reason: Square root a positive integer which is not a perfect square is an irrational number.
(a) Both Assertion and Reason are true and Reasom is a correct explanation of Assertion.
(b) Both Assertion and Reason and Reasom are true but Reasom is not a correct explanation of Assertion.
(c) Assertion is true and Reasom is false.
(d) Assertion is false and Reasom is true.

Answer:

(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

3is not a perfect square; hence, it is irrational. 3 33 is not a perfect square and hence is irrational and the reason is correct explanation for the assertion thus (a) is correct

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Question 78:

Assertion: e is irrational number.
Reason: π is an irrational number.
(a) Both Assertion and Reason are true and Reasom is a correct explanation of Assertion.
(b) Both Assertion and Reason and Reasom are true but Reasom is not a correct explanation of Assertion.
(c) Assertion is true and Reasom is false.
(d) Assertion is false and Reasom is true.

Answer:

(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.

It is known that e and π are irrational numbers, but Reason is not the correct explanation.

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Question 79:

Assertion: 3is an irrational number.
Reason: The sum of rational number and an irrational number is an irrational number.
(a) Both Assertion and Reason are true and Reasom is a correct explanation of Assertion.
(b) Both Assertion and Reason and Reasom are true but Reasom is not a correct explanation of Assertion.
(c) Assertion is true and Reasom is false.
(d) Assertion is false and Reasom is true.

Answer:

(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.
3 is not a perfect square and is irrational.Reason: Let the sum of a rational number a and an irrational number b be a rational number c.Thus, we have: a +b=c b=c-aNow, c-a is rational because both c and a are rational, but b is irrational; thus, we arrive at a contradiction.Hence, the sum of a rational number and an irrational number is an irrational number.Thus, Reason R is not a correct explanation.

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Question 80:

Match the following columns:

Column I Column II
(a) 6.54 is ....... . (p) 14
(b) π is ...... . (q) 6
(c) The length of period of 17=...... . (r) a rational number
(d) If x=2-3, then x2+1x2....... . (s) an irrational number
(a) ........
(b) ........
(c) ........
(d) ........

Answer:

(a) Because it is a non-terminating and repeating decimal, it is a rational number.

(b) π is an irrational number.

(c) 17=.142857142857...
Hence, its period is 6.
                                                     
(d)
x2+1x2=2-32+12-32=22+33-2×2×3+122+33-2×2×3=4+3-4×3+14+3-4×3=7-4×3+17-4×3=7-4×327-4×3+17-4×3=72+432-2×7×43+17-4×3=49+48-563+17-4×3=98 -5637-4×3=14×7-4×37-4×3=14

Page No 64:

Question 81:

Match the following columns:

Column I Column II
(a) 81-24=...... . (p) 4
(b) If abx-2=bax-4, then x = ........ . (q) 29
(c) If x=9+45, then x-1x= ...... . (r) 19
(d) 8116-3/4×6427-1/3=? (s) 3

(a) ......
(b) ......
(c) ......
(d) ......

Answer:

(a)
81-214=9-414=9-4×14=9-1=19     

(b)

abx-2=bax-4ba2-x=bax-42-x=x-42x=6  x=3

(c)

x=9+45and 1x=19+45×9-459-45=9-4581-80=9-45
Now,x+1x=9+45+9-45=18Thus, we have:x+1x=18We know:x-1x2=x+1x-2×x×1xx-1x2=18-2x-1x2=16Taking the square root of both sides, we get:x-1x=4

(d)

324×-34×433×-13=32-3×43-1=32-3×34=3-32-3×322=3-3×32-3×22=3-22-1=29                                                                                   



Page No 65:

Question 1:

What can you say about the sum of a rational number and an irrational number?

Answer:

Sum of a rational number and an irrational number is an irrational number. 
Example: 4 + 5 represents sum of rational and an irrational number where 4 is rational and 5 is irrational. 

Page No 65:

Question 2:

Solve 3-11 3+11.

Answer:

3-11 3+11=32-112=9-11=-2

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Question 3:

The number 665625 will terminate after how many decimal places?

Answer:

665625=5×19×754=19×753=19×7×2353×23=10641000=1.064
So, 665625 will terminate after 3 decimal places. 



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Question 4:

Find the value of 12960.17×12960.08.

Answer:

12960.17×12960.08=12960.17+0.08=12960.25=129614=12964=6

Page No 66:

Question 5:

Simplify 636+512.

Answer:

636+512=6×6+54×3=36+103
 

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Question 6:

Find an irrational number between 5 and 6.

Answer:

A number which is non terminating and non recurring is known as irrational number.

There are infinitely many irrational numbers between 5 and 6.

One of the example is 5.40430045000460000....

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Question 7:

Find the value of 21121027.

Answer:

21121027=212×2×3103×3×3               =21×2310×33               =3×7×25×2×3               =75

Hence, the value of 21121027 is 75.

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Question 8:

Rationalise 13+2.

Answer:

13+2=13+2×3-23-2                  =3-232-22                  =3-23-2                  =3-21                  =3-2

Hence, the rationalised form is 3-2.

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Question 9:

Solve for x: 252x-2=323125.

Answer:

252x-2=323125252x-2=2555252x-2=2552x-2=52x=5+2x=72
Hence, x=72.

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Question 10:

Simplify 3215+-70+6412.

Answer:

3215+-70+6412=2515+1+2612                                     =2+1+23                                     =2+1+8                                     =11

Hence, 3215+-70+6412 = 11.

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Question 11:

Evaluate 8149-32.

Answer:

8149-32=498132               =729232               =79232               =793               =7393               =343729

Hence, 8149-32=343729.

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Question 12:

Simplify 81x8y4z164.

Answer:


81x8y4z164=34×x8×y4×z1614=3414×x814×y414×z1614                   x×y×z×...a=xa×ya×za×...
=34×14×x8×14×y4×14×z16×14                  xab=xab=3×x2×y×z4=3x2yz4
 

Page No 66:

Question 13:

If a = 1, b = 2 then find the value of (ab + ba)–1.

Answer:


For = 1and b = 2,
ab+ba-1=12+21-1=1+2-1
=3-1=13         x-a=1xa
Thus, the value of (ab + ba)–1 when a = 1 and b = 2 is 13.

Page No 66:

Question 14:

Simplify 312524345.

Answer:


312524345=5×5×5×5×53×3×3×3×345=553545=53545                     xya=xaya
=535×45                       xab=xab=534=5×5×5×53×3×3×3=62581

Page No 66:

Question 15:

Give an example of two irrational numbers whose sum as well as product is rational.

Answer:


Let the two irrational numbers be 2+3 and 2-3.
Sum of these irrational numbers =2+3+2-3=4, which is rational
Product of these irrational numbers =2+32-3=22-32=4-3=1, which is rational
 

Page No 66:

Question 16:

Is the product of a rational and an irrational number always irrational? Give an example.

Answer:


Yes, the product of a rational and an irrational number is always an irrational number.
Example: 
2 is a rational number and 3 is an irrational number.
Now, 2×3=23, which is an irrational number.

Page No 66:

Question 17:

Give an example of a number x such that x2 is an irrational number and x3 is a rational number.

Answer:


The cube roots of natural numbers which are not perfect cubes are all irrational numbers.
Let x=23=213.
Now,
x2=2132=223=2213=413, which is an irrational number
Also,
x3=2133=23×13=2, which is a rational number

Page No 66:

Question 18:

Write the reciprocal of 2+3.

Answer:

The reciprocal of 2+3

=12+3=12+3×2-32-3=2-32+32-3=2-322-32                        a+ba-b=a2-b2

=2-34-3=2-31=2-3

Page No 66:

Question 19:

If 10=3.162, find the value of 110.

Answer:

The value of 110=110×1010=1010=3.16210=0.3162

Page No 66:

Question 20:

Simplify 25+322.

Answer:

25+322=252+322+22532                      a+b2=a2+b2+2ab=20+18+1210=38+1210

Page No 66:

Question 21:

If 10x = 64, find the value of 10x2+1.

Answer:

We have,

10x=64

Taking square root from both sides, we get

10x=6410x12=810x2=8

Multiplying both sides by 10, we get

10x2×10=8×10 10x2+1=80

Page No 66:

Question 22:

Evaluate 2n+2n-12n+1-2n.

Answer:

2n+2n-12n+1-2n=2n1+2-12n2-1=1+121=2+12=32

Page No 66:

Question 23:

Simplify 256-12142.

Answer:

256-12142=256-1212=256-14=44-14=4-1=14



View NCERT Solutions for all chapters of Class 9