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Page No 9:

Question 1:

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:

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.

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:

(i) 57


(ii) 83
83=223


(iii) -236=-356


(iv) 1.3
1.3=1310=1310


(v) – 2.4
-2.4=-2410=-125=-225

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:

(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

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:

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.
​
​
 

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:

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.

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:

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
 

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:

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

 

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:

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

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:

(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. 

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:

(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.

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:

(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.

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:

(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
 

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:

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

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:

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
 

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:

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...

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:

(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.

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:

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. 

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:

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. 

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:

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.

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:

(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.

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:

(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.

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:

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.
 

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:

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...

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:

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.

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:

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.

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:

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.

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:

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.

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:

(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

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:

(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

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:

(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

Answer:

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

Page No 28:

Question 4:

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

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:

(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

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:

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

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:

(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.

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:

(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.

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:

(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

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:


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

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:


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

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:


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

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:


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

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:


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

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:


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

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:


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

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:


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

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:

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.

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:

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.

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:


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.

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:

(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

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:


(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

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:


(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

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:


(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

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:


(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

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:

(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

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:


(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

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:


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.

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:


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

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:


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.

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:


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.

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:


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.

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:


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.

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:


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.

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:


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.

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:


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.

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:

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

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:

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.

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:

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.

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:

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.

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:

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.

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:


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.

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:

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.

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:

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 .

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:

(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

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:

(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

Answer:

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

Page No 53:

Question 4:

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

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:

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

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:

(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

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:

(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

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:

(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

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:

(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

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:

(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

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:

(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

Answer:

x234=x21314            =x2314            =x212            =x16

Hence, the result in the exponential form is x16.

Page No 54:

Question 12:

x234=x21314            =x2314            =x212            =x16

Hence, the result in the exponential form is x16.

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:

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

(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

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:

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.

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:

(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.

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:

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

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:

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.

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:

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.

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:

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).

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:

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).

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:

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).

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:

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).

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:

(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.

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:

(d) a real number

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

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:

(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.

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:

(b) either terminating or repeating

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

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:

(d) neither terminating nor repeating

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

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:

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).

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:

(d) 3.141141114...

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

Answer:

Since, 719=7×319×3=2157

Hence, the correct option is (d).

Page No 58:

Question 13:

Since, 719=7×319×3=2157

Hence, the correct option is (d).

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:

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).

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:

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

So, π is an irrational number.

Hence, the correct option is (c).

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:

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

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

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:

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).

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:



 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).

 

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:

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).

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:

(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.

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:

(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.

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:

(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.

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:

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).

 

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:

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).

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:

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

Hence, the correct option is (a).

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:


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).

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:

(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

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:

(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

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:

(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

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:

(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

Answer:

(c) 5×6

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

Page No 60:

Question 31:

(c) 5×6

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

Answer:

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

Page No 60:

Question 32:

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

Answer:

(c) 17×27

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

Page No 60:

Question 33:

(c) 17×27

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

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:

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).

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:

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).

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:


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

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:


-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).

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:


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).

Answer:


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

Page No 60:

Question 39:


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

Answer:


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

Page No 60:

Question 40:


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

Answer:


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



Page No 61:

Question 41:


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

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:


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

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:


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).

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:


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

Answer:


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

Page No 61:

Question 45:


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

Answer:


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

Page No 61:

Question 46:


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

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:

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).

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:

20+7050=1+11              =21              =2

The value of 20+7050 is 2.

Hence, the correct option is (b).

Answer:

24315=3515            =31            =3

The value of 24315 is 3.

Hence, the correct option is (a).

Page No 61:

Question 49:

24315=3515            =31            =3

The value of 24315 is 3.

Hence, the correct option is (a).

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:

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).

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:

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).

Answer:

2234=221314            =22314            =2212            =216

The value of  2234 is 216.

Hence, the correct option is (c).

Page No 61:

Question 52:

2234=221314            =22314            =2212            =216

The value of  2234 is 216.

Hence, the correct option is (c).

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:

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).

Answer:

811212=341212                 =3212                 =3

The value of 811212 is 3.

Hence, the correct option is (a).

Page No 61:

Question 54:

811212=341212                 =3212                 =3

The value of 811212 is 3.

Hence, the correct option is (a).

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:

(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).

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:

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

Hence, the correct option is (b).

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:

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).

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:

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).

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:

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).

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:

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).

Answer:

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

Page No 62:

Question 61:

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

Answer:

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

Page No 62:

Question 62:

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

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:

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). 

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:

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). 

Answer:

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

Page No 62:

Question 65:

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

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

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:

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

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:

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). 

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:

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

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:

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). 

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:

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

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:

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

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:

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). 
 

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:

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). 

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:

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).

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:

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). 

 

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.



Page No 64:

Question 77:

(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.

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:

(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

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:

(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.

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:

(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.

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

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Question 81:

(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

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                                                                                   



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Question 1:

(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                                                                                   

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. 

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Question 2:

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. 

Answer:

3-11 3+11=32-112=9-11=-2

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Question 3:

3-11 3+11=32-112=9-11=-2

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:

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

Answer:

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

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Question 5:

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

Answer:

636+512=6×6+54×3=36+103
 

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Question 6:

636+512=6×6+54×3=36+103
 

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:

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....

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:

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

Hence, the value of 21121027 is 75.

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:

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

Hence, the rationalised form is 3-2.

Answer:

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

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Question 10:

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

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:

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

Hence, 3215+-70+6412 = 11.

Answer:

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

Hence, 8149-32=343729.

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Question 12:

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

Hence, 8149-32=343729.

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
 

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Question 13:


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
 

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.

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Question 14:


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.

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

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Question 15:


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

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
 

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Question 16:


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
 

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.

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Question 17:


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.

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

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Question 18:


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

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

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Question 19:

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

Answer:

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

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Question 20:

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

Answer:

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

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Question 21:

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

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

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Question 22:

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

Answer:

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

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Question 23:

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

Answer:

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



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