Rs Aggarwal 2021 2022 for Class 10 Maths Chapter 1 - Real Numbers

Answer:

Euclid's division algorithm states that for any two positive integers a and b, there exist unique integers q and r, such that a = bq + r, where 0 ≤ r < b.

Answer:

We know, Dividend = Divisor $\times$ Quotient + Remainder
 Given: Divisor = 61, Quotient = 27, Remainder = 32
Let the Dividend be x.
∴ x = $61 \times 27 + 32$
        = 1679
Hence, the required number is 1679.

Answer:

Given: Dividend = 1365, Quotient = 31, Remainder = 32
Let the divisor be x.
Dividend = Divisor $\times$ Quotient + Remainder
                       1365 =  x $\times$ 31 + 32
   ⇒                1365 − 32 = 31x
   ⇒                      1333 = 31x

   ⇒                        x = $\frac{1333}{31}$ = 43
Hence, 1365 should be divided by 43 to get 31 as quotient and 32 as remainder.

Answer:

(i) 612 and 1314

612 < 1314
Thus, we divide 1314 by 612 by using Euclid's division lemma

1314 = 612 × 2 + 90

∵ Remainder is not zero,
∴ we divide 612 by 90 by using Euclid's division lemma

612 = 90 × 6 + 72

∵ Remainder is not zero,
∴ we divide 90 by 72 by using Euclid's division lemma

90 = 72 × 1 + 18

∵ Remainder is not zero,
∴ we divide 72 by 18 by using Euclid's division lemma

72 = 18 × 4 + 0

Since, Remainder is zero,

Hence, HCF of 612 and 1314 is 18.

(ii) 1260 and 7344

1260 < 7344
Thus, we divide 7344 by 1260 by using Euclid's division lemma

7344 = 1260 × 5 + 1044

∵ Remainder is not zero,
∴ we divide 1260 by 1044 by using Euclid's division lemma

1260 = 1044 × 1 + 216

∵ Remainder is not zero,
∴ we divide 1044 by 216 by using Euclid's division lemma

1044 = 216 × 4 + 180

∵ Remainder is not zero,
∴ we divide 216 by 180 by using Euclid's division lemma

216 = 180 × 1 + 36

∵ Remainder is not zero,
∴ we divide 180 by 36 by using Euclid's division lemma

180 = 36 × 5 + 0

Since, Remainder is zero,

Hence, HCF of 1260 and 7344 is 36.

(iii) 4052 and 12576

4052 < 12576
Thus, we divide 12576 by 4052 by using Euclid's division lemma

12576 = 4052 × 3 + 420

∵ Remainder is not zero,
∴ we divide 4052 by 420 by using Euclid's division lemma

4052 = 420 × 9 + 272

∵ Remainder is not zero,
∴ we divide 420 by 272 by using Euclid's division lemma

420 = 272 × 1 + 148

∵ Remainder is not zero,
∴ we divide 272 by 148 by using Euclid's division lemma

272 = 148 × 1 + 124

∵ Remainder is not zero,
∴ we divide 148 by 124 by using Euclid's division lemma

148 = 124 × 1 + 24

∵ Remainder is not zero,
∴ we divide 124 by 24 by using Euclid's division lemma

​124 = 24 × 5 + 4

∵ Remainder is not zero,
∴ we divide 24 by 4 by using Euclid's division lemma

24 = 4 × 6 + 0

Since, Remainder is zero,

Hence, HCF of 4052 and 12576 is 4.









 

Answer:

650 and 1170

650 < 1170
Thus, we divide 1170 by 650 by using Euclid's division lemma

1170 = 650 × 1 + 520

∵ Remainder is not zero,
∴ we divide 650 by 520 by using Euclid's division lemma

650 = 520 × 1 + 130

∵ Remainder is not zero,
∴ we divide 520 by 130 by using Euclid's division lemma

520 = 130 × 4 + 0

Since, Remainder is zero,

Therefore, HCF of 650 and 1170 is 130.

Hence, the largest number which divides 650 and 1170 is 130.

Answer:

Smallest prime number is 2.
Smallest composite number is 4.

HCF (2, 4) = 2

Hence, the HCF of the smallest prime number and the smallest composite number is 2.

Answer:

Euclid's division lemma states that for given positive integers a and b, there exists unique integers q and r satisfying $a=bq+r, 0\le r<b$
Applying Euclid's division lemma om n and 6, we have 
$n=6q+r, 0\le r<6$
Therefore, n can have six values, i.e.
$$\begin{gathered} n=6q \\ n=6q+1 \\ n=6q+2 \\ n=6q+3 \\ n=6q+4 \\ n=6q+5 \end{gathered}$$
Case I: When $n=6q$
$$\begin{gathered} n^3=(6q{)}^3 \\ {n}^3-n=(6q{)}^3-6q \\ =6q(36q^2-1) \\ =6m \left[\mathrm{where} m=q(36q^2-1)\right] \end{gathered}$$
Hence, $\forall n=6q, n^3-n$ is divisible by 6

Case II:
When $n=6q+1$
$$\begin{gathered} n^3=(6q+1{)}^3 \\ {n}^3-n=(6q+1{)}^3-(6q+1) \\ =(6q+1)\left[(6q+1{)}^2-1\right] \\ =(6q+1)\left[36q^2+1+12q-1\right] \\ =(6q+1)\left[36q^2+12q\right] \\ =\left[216q^3+72q^2+36q^2+12q\right] \\ =6\left[36q^3+18q^2+2q\right] \\ =6m (\mathrm{where} m=36q^2+18q+2q) \\ Hence, \forall n=6q+1, n^3-n \mathrm{is} \mathrm{divisible} \mathrm{by} 6 \end{gathered}$$

Case III: When $n=6q+2$
$$\begin{gathered} n^3=(6q+2{)}^3 \\ {n}^3-n=(6q+2{)}^3-(6q+2) \\ =(6q+2)\left[(6q+2{)}^2-1\right] \\ =(6q+2)\left[36q^2+4+24q-1\right] \\ =(6q+2)\left[36q^2+24q+3\right] \\ =\left[216q^3+144q^2+18q+72q^2+48q+6\right] \\ =\left[216q^3+216q^2+66q+6\right] \\ =6\left[36q^3+36q^2+11q+1\right] \\ =6m (\mathrm{where} m=36q^3+36q^2+11q+1) \\ Hence, \forall n=6q+1, n^3-n \mathrm{is} \mathrm{divisible} \mathrm{by} 6 \end{gathered}$$

Case IV: When $n=6q+3$
$$\begin{gathered} n^3=(6q+3{)}^3 \\ {n}^3-n=(6q+3{)}^3-(6q+3) \\ =(6q+3)\left[(6q+3{)}^2-1\right] \\ =(6q+3)\left[36q^2+9+36q-1\right] \\ =(6q+3)\left[36q^2+36q+8\right] \\ =\left[216q^3+216q^2+48q+108q^2+108q+24\right] \\ =\left[216q^3+324q^2+156q+24\right] \\ =6\left[36q^3+54q^2+26q+4\right] \\ =6m \end{gathered}$$
Hence, $\forall n=6q+3, n^3-n$ is divisible by 6.

Case V: When $n=6q+4$
$$\begin{gathered} n^3=(6q+4{)}^3 \\ {n}^3-n=(6q+4{)}^3-(6q+4) \\ =(6q+4)\left[(6q+4{)}^2-1\right] \\ =(6q+4)\left[36q^2+16+48q-1\right] \\ =(6q+4)\left[36q^2+48q+15\right] \\ =\left[216q^3+288q^2+90q+144q^2+192q+60\right] \\ =\left[216q^3+432q^2+282q+60\right] \\ =6\left[36q^3+72q^2+47q+10\right] \\ =6m (\mathrm{where} m=36q^3+72q^2+47q+10) \end{gathered}$$
Hence, $\forall n=6q+4, n^3-n$ is divisible by 6.

Case VI: When $n=6q+5$
$$\begin{gathered} n^3=(6q+5{)}^3 \\ {n}^3-n=(6q+5{)}^3-(6q+5) \\ =(6q+5)\left[(6q+5{)}^2-1\right] \\ =(6q+5)\left[36q^2+25+60q-1\right] \\ =(6q+5)\left[36q^2+60q+24\right] \\ =\left[216q^3+360q^2+144q+180q^2+300q+120\right] \\ =\left[216q^3+540q^2+444q+120\right] \\ =6\left[36q^3+90q^2+74q+120\right] \\ =6m (\mathrm{where} m=36q^3+90q^2+74q+120) \end{gathered}$$
Hence, $\forall n=6q+5, n^3-n$ is divisible by 6.

Answer:

Let, n be any positive odd integer and let $x=n \mathrm{and} y=n+2$.
So, $x^2+y^2=(n{)}^2+(n+2{)}^2$
$\mathrm{Or}, x^2+y^2=n^2+(n^2+4+4n)$
$$\begin{gathered} \Rightarrow x^2+y^2=2n^2+4+4n \\ \Rightarrow x^2+y^2=2(n^2+2+2n) \end{gathered}$$
$\Rightarrow x^2+y^2=2m$ (where $m=n^2+2n+2$)
Because $x^2+y^2$ has 2 as a factor, so the value is an even number.
Also, because it does not have any multiple of 4 as a factor, therefore, it is not divisible by 4.

Answer:

Using Euclid's division algorithm, we have


Since 1445 > 1190, we apply Euclid's division lemma to 1445 and 1190 to get;
$1445=1190\times 1+255$
Since the remainder is not zero, we again apply division lemma to 1190 and 255 and get;
$1190=255\times 4+170$
Again, the remainder is not zero, so we apply division lemma to 255 and 170 to get;
$255=170\times 1+85$
Now we finally apply division lemma to 170 and 85 to get;
$170=85\times 2+0$
Since, in this step, 85 completely divides 170 leaving zero remainder, we stop the procedure.
Hence, the HCF is 85. 
Now, using the above division, we have
$$\begin{gathered} 170\times 1+85=255 \\ \Rightarrow 85=255-170\times 1 \\ \Rightarrow 85=(1445-1190\times 1)-(1190-255\times 4) \\ \Rightarrow 85=(1445-1190)-\left[1190-(1445-1190)\times 4\right] \\ \Rightarrow 85=(1445-1190)-\left[1190-1445\times 4+1190\times 4\right] \\ \Rightarrow 85=1445-1190-\left[1190\times 5-1445\times 4\right] \\ \Rightarrow 85=1445-1190-1190\times 5+1445\times 4 \\ \Rightarrow 85=1445\times 5-1190\times 6 \end{gathered}$$
Or, $85=1190(-6)+1445\left(5\right)$
Hence, $m=-6, n=5$

Answer:

Let us first find the HCF of 441 and 567 using Euclid's division lemma.

441 < 567
Thus, we divide 567 by 441 by using Euclid's division lemma

567 = 441 × 1 + 126

∵ Remainder is not zero,
∴ we divide 441 by 126 by using Euclid's division lemma

441 = 126 × 3 + 63

∵ Remainder is not zero,
∴ we divide 126 by 63 by using Euclid's division lemma

126 = 63 × 2 + 0

Since, Remainder is zero,

Therefore, HCF of 441 and 567 is 63.

Now, let us find the HCF of 693 and 63 using Euclid's division lemma.

693 > 63
Thus, we divide 693 by 63 by using Euclid's division lemma

693 = 63 × 11 + 0

Since, Remainder is zero,

Therefore, HCF of 693 and 63 is 63.

Hence, the HCF of 441, 567 and 693 is 63.



 

Answer:

On subtracting 1, 2, and 3 from 1251, 9377 and 15628 respectively, we get 1250, 9375 and 15625.


Now we find the HCF of 1250 and 9375 using Euclid's division lemma

1250 < 9375
Thus, we divide 9375 by 1250 by using Euclid's division lemma

9375 = 1250 × 7 + 625

∵ Remainder is not zero,
∴ we divide 1250 by 625 by using Euclid's division lemma

1250 = 625 × 2 + 0

Since, Remainder is zero,

Therefore, HCF of 1250 and 9375 is 625.

Now, we find the HCF of 15625 and 625 using Euclid's division lemma.

15625 > 625
Thus, we divide 15625 by 625 by using Euclid's division lemma

15625 = 625 × 25 + 0

Since, Remainder is zero,

Therefore, HCF of 15625 and 625 is 625.

Hence, the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2, and 3 respectively is 625.



 

Answer:

Prime factorisation of 429 is:
429 = 3 × 11 × 13

Hence, 429 as a product of its prime factors can be expressed as 3 × 11 × 13.

 

Answer:

Prime Factorisation of 5005 is:
5005 = 5 × 7 × 11 × 13

​Hence, 5005 as a product of its prime factors can be expressed as 5 × 7 × 11 × 13.
 

Answer:

Let the two numbers be a and b.
​Let the value of a be 161.
Given: HCF = 23 and LCM = 1449
we know,        a × b = HCF ​× LCM 
            ⇒     161 × b = 23 × 1449
            ⇒              ∴ b =   23 × 1449   =   33327  = 207
                                          161                  161
   Hence, the other number b is 207.

Answer:

HCF of two numbers = 145
LCM of two numbers = 2175
Let one of the two numbers be 725 and other be x.

Using the formula, Product of two numbers = HCF × LCM
we conclude that

725 × x = 145 × 2175
x = $\frac{145\times 2175}{725}$
   = 435

Hence, the other number is 435.

Answer:

Prime Factorisation of 2431 is:
2431 = 11 × 13 ×  17

​Hence, 2431 as a product of its prime factors can be expressed as 11 × 13 ×  17.

Answer:

(i) 36, 84
   Prime factorisation:
   36 = 2² ⨯ 3²
   84 = 2² ⨯ 3 ⨯ 7
​ HCF = product of smallest power of each common prime factor in the numbers = 2² ⨯ 3 = 12
 LCM = product of greatest power of each prime factor involved in the numbers = 2² ⨯ 3² ⨯ 7 = 252

(ii) 23, 31
   Prime factorisation:
   23 = 23
   31 = 31
​ HCF = product of smallest power of each common prime factor in the numbers = 1
 LCM = product of greatest power of each prime factor involved in the numbers = 23 ⨯ 31 = 713

(iii) 96, 404
   Prime factorisation:
   96 = 2⁵ ⨯ 3
   404 = 2² ⨯ 101
​ HCF = product of smallest power of each common prime factor in the numbers = 2² = 4
 LCM = product of greatest power of each prime factor involved in the numbers = 2⁵ ⨯ 3 ⨯ 101 = 9696

(iv) 144, 198
   Prime factorisation:
   144 = $2^4 \times 3^2$
  198 = $2 \times 3^2 \times 11$
​ HCF = product of smallest power of each common prime factor in the numbers = $2 \times 3^2$ = 18
 LCM = product of greatest power of each prime factor involved in the numbers = $2^{4 }\times 3^{2 }\times 11$ = 1584

(v) 396, 1080
    Prime factorisation:
   396 = $2^2 \times 3^{2 } \times 11$
  1080 = $2^3 \times 3^3 \times 5$
 ​ HCF = product of smallest power of each common prime factor in the numbers = $2^2 \times 3^2$ = 36
LCM = product of greatest power of each prime factor involved in the numbers = $2^3 \times 3^{3 }\times 5 \times 11$ = 11880

(vi) 1152 , 1664
    Prime factorisation:
   1152 = $2^7 \times 3^2$
  1664 = $2^7 \times 13$
HCF = product of smallest power of each common prime factor involved in the numbers = $2^7$ = 128
 LCM = product of greatest power of each prime factor involved in the numbers = $2^7 \times 3^2 \times 13$ = 14976

Answer:

(i) 8, 9, 25

Prime factorisation:
8 = 2 × 2 × 2
9 = 3 × 3
25 = 5 × 5

HCF (8, 9, 25) = 1

LCM (8, 9, 25) = 2 × 2 × 2 × 3 × 3 × 5 × 5
                         = 1800

(ii) 12, 15, 21

Prime factorisation:
12 = 2 × 2 × 3
15 = 3 × 5
21 = 3 × 7

HCF (12, 15, 21) = 3

LCM (12, 15, 21) = 2 × 2 × 3 × 5 × 7
                             = 420

(iii) 17, 23, 29

Prime factorisation:
17 = 17
23 = 23
29 = 29

HCF (17, 23, 29) = 1

LCM (17, 23, 29) = 17 × 23 × 29
                             = 11339

(iv) 24, 36, 40

Prime factorisation:
24 = 2 × 2 × 2 × 3
36 = 2 × 2 × 3 × 3
40 = 2 × 2 × 2 × 5

HCF (24, 36, 40) = 2 × 2
                            = 4

LCM (24, 36, 40) = 2 × 2 × 2 × 3 × 3 × 5
                             = 360

(v) 30, 72, 432

Prime factorisation:
30 = 2 × 3 × 5
72 = 2 × 2 × 2 × 3 × 3
432 = 2 × 2 × 2 × 2 × 3 × 3 × 3

HCF (30, 72, 432) = 2 × 3
                              = 6

LCM (30, 72, 432) = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5
                               = 2160

(vi) 21, 28, 36

Prime factorisation:
21 = 3 × 7
28 = 2 × 2 × 7
36 = 2 × 2 × 3 × 3

HCF (21, 28, 36) = 1

LCM (21, 28, 36) = 2 × 2 × 3 × 3 × 7
                             = 252
 

Answer:

Prime factorisation:
404 = 2 × 2 × 101
96 = 2 × 2 × 2 × 2 × 2 × 3

HCF (404, 96) = 2 × 2
                        = 4

LCM (404, 96) = 2 × 2 × 2 × 2 × 2 × 3 × 101
                         = 9696

Now, LCM × HCF = 9696 × 4
                               = 38784

Product of 404 and 96 = 404 × 96
                                    = 38784

Hence, HCF × LCM = product of two given numbers.

Answer:

It is given that, a = x³y² and b = xy³, where x and y are prime numbers.

$$\begin{gathered} \mathrm{LCM}\left(a, b\right)=\mathrm{LCM}\left(x^3{y}^2, x{y}^3\right) \\ =\mathrm{The} \mathrm{highest} \mathrm{of} \mathrm{indices} \mathrm{of} x \mathrm{and} y \\ =x^3{y}^3 \\ \mathrm{HCF}\left(a, b\right)=\mathrm{HCF}\left(x^3{y}^2, x{y}^3\right) \\ =\mathrm{The} \mathrm{lowest} \mathrm{of} \mathrm{indices} \mathrm{of} x \mathrm{and} y \\ =x{y}^2 \end{gathered}$$

Hence, HCF(a, b) = xy² and LCM(a, b) = x³y³.

Answer:

Let the two numbers be a and b.

Product of two numbers = HCF × LCM
⇒ ab = 5 × 200
⇒ ab = 1000

Hence, the product ab is 1000.

Answer:

Let the HCF of two numbers be x.
Then, LCM = 9x

According to the question,

$$\begin{gathered} \mathrm{LCM}+\mathrm{HCF}=500 \\ \Rightarrow 9x+x=500 \\ \Rightarrow 10x=500 \\ \Rightarrow x=50 \end{gathered}$$

Hence, the HCF of two numbers is 50.

Answer:

​HCF of two numbers = 18
Product of two numbers = 12960
Let their LCM be x.

Using the formula, Product of two numbers = HCF × LCM
we conclude that

12960 = 18 × x
x = $\frac{12960}{18}$
   = 720

Hence, their LCM is 720.

Answer:

No,

Since, LCM is always a multiple of HCF.
But 175 is not a multiple of 15.

Hence, two numbers cannot have 15 as their HCF and 175 as their LCM.

Answer:

(i) Prime factorisation of 69 and 92 is:

69 = 3 × 23
92 = 2² × 23

Therefore, $\frac{69}{92}=\frac{3\times 23}{2^2\times 23}=\frac{3}{2^2}=\frac{3}{4}$
Thus, simplest form of $\frac{69}{92}$ is $\frac{3}{4}$.

(ii) Prime factorisation of 473 and 645 is:

473 = 11 × 43
645 = 3 × 5 × 43

Therefore, $\frac{473}{645}=\frac{11\times 43}{3\times 5\times 43}=\frac{11}{15}$
Thus, simplest form of $\frac{473}{645}$ is $\frac{11}{15}$.

(iii) Prime factorisation of 1095 and 1168 is:

1095 = 3 × 5 × 73
1168 = 2⁴ × 73

Therefore, $\frac{1095}{1168}=\frac{3\times 5\times 73}{2^4\times 73}=\frac{15}{16}$
Thus, simplest form of $\frac{1095}{1168}$ is $\frac{15}{16}$.

(iv) Prime factorisation of 368 and 496 is:

368 = 2⁴ × 23
496 = 2⁴ × 31

Therefore, $\frac{368}{496}=\frac{2^4\times 23}{2^4\times 31}=\frac{23}{31}$
Thus, simplest form of $\frac{368}{496}$ is $\frac{23}{31}$.

Answer:

Largest number which divides 438 and 606, leaving remainder 6 is actually the largest number which divides 438 − 6 = 432 and 606 − 6 = 600, leaving remainder 0.

Therefore, HCF of 432 and 600 gives the largest number.

Now, prime factors of 432 and 600 are:
432 = 2⁴ × 3³ 
600 = 2³ × 3 × 5²

HCF = product of smallest power of each common prime factor in the numbers = 2³ × 3 = 24

Thus, the largest number which divides 438 and 606, leaving remainder 6 is 24.

Answer:

We know that the required number divides 315 (320 − 5) and 450 (457 − 7).
∴ Required number = HCF (315, 450)
On applying Euclid's lemma, we get:
                              315) 450 (1
                                     −​ 315  
                                       135) 315 (2
                                           −   270 
                                                 45) 135 (3
                                                     −​ 135 
                                                          0
Therefore, the HCF of 315 and 450 is 45.
Hence, the required number is 45.

Answer:

Least number which can be divided by 35, 56 and 91 is LCM of 35, 56 and 91.

Prime factorization of 35, 56 and 91 is:

35 = 5 × 7
56 = 2³ × 7
91 = 7 × 13

LCM = product of greatest power of each prime factor involved in the numbers = 2³ × 5 × 7 × 13 = 3640

Least number which can be divided by 35, 56 and 91 is 3640.

Least number which when divided by 35, 56 and 91 leaves the same remainder 7 is 3640 + 7 = 3647.

Thus, the required number is 3647.

Answer:

Let the required number be x.

Using Euclid's lemma,
x = 28p + 8 and x = 32q + 12, where p and q are the quotients
⇒ 28p + 8 = 32q + 12
⇒ 28p = 32q + 4
⇒ 7p = 8q + 1 ..... (1)

Here p = 8n − 1 and q = 7n − 1 satisfies (1), where n is a natural number
On putting n = 1, we get
p = 8 − 1 = 7 and q = 7 − 1 = 6

Thus, x = 28p + 8
             = 28 × 7 + 8
             = 204

Hence, the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 is 204.

Answer:

The smallest number which when increased by 17 is exactly divisible by both 468 and 520 is obtained by subtracting 17 from the LCM of 468 and 520.

Prime factorization of 468 and 520 is:
468 = 2² × 3² × 13
520 = 2³ × 5 × 13

LCM = product of greatest power of each prime factor involved in the numbers = 2³ × 3² × 5 × 13 = 4680

The required number is 4680 − 17 = 4663.

Hence, the smallest number which when increased by 17 is exactly divisible by both 468 and 520 is 4663.

Answer:

Prime factorization:

15 = 3 × 5
24 = 2³ × 3
36 = 2² × 3²

LCM = product of greatest power of each prime factor involved in the numbers = 2³ × 3² × 5 = 360

Now, the greatest four digit number is 9999.
On dividing 9999 by 360 we get 279 as remainder.
Thus, 9999 − 279 = 9720 is exactly divisible by 360.

Hence, the greatest number of four digits which is exactly divisible by 15, 24 and 36 is 9720.

Answer:

Largest 4 digit number is 9999
To find the largest 4 digit number divisible by 4, 7 and 13, we find the LCM of 4, 7 and 13 first. 
LCM(4, 7, 13) = $4\times 7\times 13$ = 364
Now, to we divide 9999 by 364 and subtract the remainder from 9999 to get the number completely divisible by 4, 7 and 13.

$9999-171=9828$ 
Because the number leaves the remainder 3, so we add 3 to 9828. 
Therefore, 9828 + 3 = 9831 is the required number.

Answer:

We find the LCM of 5, 6, 4 and 3 first.​

So, $\mathrm{LCM}(5,6,4,3)=2\times 2\times 3\times 5=60$
Now, divide 2497 by 60, we get

To make the number completely divisible by 60, we must add a number that would make the remainder equal to 60.
Therefore, the number that must be added is 60 $-$ 37 = 23
Hence, 23 must be added to 2497. 
So, the number exactly divisible by 5, 6, 4 and 3 is 2497 + 23 = 2520

Answer:

We need to find the greatest number that would divide 43, 91 and 183 leaving the same remainder every time.
We first find the difference of the numbers and then find the HCF of the got numbers.
$$\begin{gathered} 183-91=92 \\ 183-43=140 \\ 91-43=48 \end{gathered}$$
Now find HCF of 92, 140 and 48, we get
$$\begin{gathered} 92=2\times 2\times 23 \\ 140=2\times 2\times 5\times 7 \\ 48=2\times 2\times 2\times 2\times 3 \end{gathered}$$
HCF(92, 140, 48) = 4
Therefore, 4 is the required number. 

Answer:

First find the LCM of 20, 25, 35 and 40.

$\mathrm{LCM}(20, 25, 35, 40)=2\times 2\times 2\times 5\times 5\times 7=1400$
Now, we can see that
$$\begin{gathered} 20-14=6 \\ 25-19=6 \\ 35-29=6 \\ 40-34=6 \end{gathered}$$
​So, the required number would be$\mathrm{LCM}(20, 25, 35, 40)-6$ 
$$\begin{gathered} =1400-6 \\ =1394 \end{gathered}$$

Answer:

Minimum number of rooms required = $\frac{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{participants}}{\mathrm{HCF}(60,84,108)}$

Prime factorization of 60, 84 and 108 is:

60 = 2² × 3 × 5
84 = 2² × 3 × 7
108 = 2² × 3³

HCF = product of smallest power of each common prime factor in the numbers = 2² × 3 = 12

Total number of paricipants = 60 + 84 + 108 = 252

Therefore, minimum number of rooms required = $\frac{252}{12}=21$

Thus, minimum number of rooms required is 21.

Answer:

Total number of English books = 336
Total number of mathematics books = 240
Total number of science books = 96
∴ Number of books stored in each stack = HCF (336, 240, 96) 
Prime factorisation:
336 = $2^4 \times 3 \times 7$
240 = $2^4 \times 3 \times 5$
96 = $2^5 \times 3$
∴ HCF = Product of the smallest power of each common prime factor involved in the numbers = $2^4 \times 3$ = 48
Hence, we made stacks of 48 books each.

∴ Number of stacks = $\frac{336}{48} + \frac{240}{48} + \frac{ 96}{48}$ = (7 + 5 + 2) = 14

Answer:

The lengths of three pieces of timber are 42 m, 49 m and 63 m, respectively.
We have to divide the timber into equal length of planks.
∴ Greatest possible length of each plank = HCF(42, 49, 63)
Prime factorisation:
42 = $2 \times 3 \times 7$
49 = $7 \times 7$
63 = $3 \times 3 \times 7$
∴ HCF = Product of smallest power of each common prime factor in the numbers = 7
Therefore, the greatest possible length of each plank is 7 m.
Now, to find the total number of planks formed by each of the piece, we divide the length of each piece by the HCF, i.e. by 7.
We know that;
$$\begin{gathered} 7\times 6=42 \\ 7\times 7=49 \\ 7\times 9=63 \end{gathered}$$
Therefore, total number of planks formed$=6+7+9=22$
Hence, total 22 planks will be formed.

Answer:

The three given lengths are 7 m (700 cm), 3 m 85 cm  (385 cm) and 12 m 95 cm (1295 cm).   (∵ 1 m = 100 cm)
∴ Required length = HCF (700, 385, 1295)
Prime factorisation:
700 = $2 \times 2 \times 5 \times 5 \times 7 = 2^2 \times 5^2 \times 7$
385 = $5 \times 7 \times 11$
1295 = $5 \times 7 \times 37$
∴ HCF = $5 \times 7$ = 35
Hence, the greatest possible length is 35 cm.

Answer:

Total number of pens = 1001
Total number of pencils = 910
∴​ Maximum number of students who get the same number of pens and pencils = HCF (1001, 910)
Prime factorisation:
1001 = $11\times 91$
  910 = $10\times 91$
∴ HCF = 91
Hence, 91 students receive same number of pens and pencils.

Answer:

It is given that:
Length of a tile = 15 m 17 cm = 1517 cm                  [∵ 1 m = 100 cm]
Breadth of a tile = 9 m 2 cm = 902 cm
∴ Side of each square tile = HCF (1517, 902)
Prime factorisation:
1517 = $37 \times 41$                                                                                                             
902 = $22 \times 41$                                                                                          
∴ HCF = Product of smallest power of each common prime factor in the numbers = 41             
∴ Required number of tiles = $\frac{\mathrm{Area} \mathrm{of} \mathrm{ceiling}}{\mathrm{Area} \mathrm{of} \mathrm{one} \mathrm{tile}}$ = $\frac{1517 \times 902}{41 \times 41}$ = $37 \times 22$ = 814

Answer:

Length of the three measuring rods are 64 cm, 80 cm and 96 cm, respectively.
∴ Length of cloth that can be measured an exact number of times = LCM (64, 80, 96) 
Prime factorisation:
64 = $2^6$
80 = $2^4 \times 5$
96 = $2^5 \times 3$
∴ LCM = Product of greatest power of  each prime factor involved in the numbers = $2^{6 }\times 3 \times 5$ = 960 cm = 9.6 m
Hence, the required length of cloth is 9.6 m.

Answer:

Beep duration of first device = 60 seconds
Beep duration of second device = 62 seconds
∴ Interval of beeping together = LCM (60, 62) 
Prime factorisation:
60 = $2^2 \times 3 \times 5$
62 = $2 \times 31$
∴ LCM = $2^{2 }\times 3 \times 5 \times 31 = 1860$ seconds = $\frac{1860}{60}$ = 31 min
Hence, they will beep together again at 10 : 31 a.m.

Answer:

We find the LCM of 48, 72 and 108 first to get the time after which they will blink together again. 

Hence, LCM = $2\times 2\times 2\times 2\times 3\times 3\times 3=432$
So, they will blink again at 432 seconds past 8:00 am
or, $\frac{432}{60}=$7 minutes and 12 seconds past 8:00 am 
So, the time will be 08:07:12 hrs

Answer:

Six bells toll together at intervals of 2, 4, 6, 8, 10 and 12 minutes, respectively.
Prime factorisation:
$$\begin{gathered} 2 = 2 \\ 4 = 2 \times 2 \\ 6 = 2 \times 3 \\ 8 = 2 \times 2 \times 2 \\ 10 = 2 \times 5 \\ 12 = 2 \times 2 \times 3 \end{gathered}$$

∴ ​$LCM ( 2, 4, 6, 8, 10, 12 )$ = $2^3 \times 3 \times 5 = 120$
Hence, after every 120 minutes (i.e. 2 hours), they will toll together.
∴ Required number of times = $(\frac{30}{2} + 1) = 16$

Answer:

(i) $\frac{23}{2^3\times 5^2}$ = $\frac{23\times 5}{2^3\times 5^3}=\frac{115}{1000} = 0.115$
We know either 2 or 5 is not a factor of 23, so it is in its simplest form.
Moreover, it is in the form of $(2^m\times 5^n)$.
Hence, the given rational is terminating.

(ii) $\frac{24}{125} = \frac{24}{5^3} = \frac{24 \times 2^3}{5^3\times 2^3} = \frac{192}{1000} = 0.192$
 We know 5 is not a factor of 23, so it is in its simplest form.
Moreover, it is in the form of $(2^m \times 5^n)$.
Hence, the given rational is terminating.

(iii) $\frac{171}{800}$ = $\frac{171 }{2^5 \times 5^2} = \frac{171 \times 5^3}{2^5 \times 5^5} = \frac{21375}{100000} = 0.21375$
     We know either 2 or 5 is not a factor of 171, so it is in its simplest form.
     Moreover, it is in the form of $(2^m \times 5^n)$.
    Hence, the given rational is terminating.

(iv)  $\frac{15}{1600} =$$\frac{15}{2^6 \times 5^2} = \frac{15 \times 5^4}{2^6 \times 5^6} = \frac{9375}{1000000} = 0.009375$
    We know either 2 or 5 is not a factor of 15, so it is in its simplest form.
    Moreover, it is in the form of $(2^m \times 5^n)$.
   Hence, the given rational is terminating.

(v)  $\frac{17}{320} = \frac{17}{2^6 \times 5}$ = $\frac{17 \times 5^5}{2^6 \times 5^6}$ = $\frac{53125}{1000000} = 0.053125$
     We know either 2 or 5 is not a factor of 17, so it is in its simplest form.
     Moreover, it is in the form of $(2^m \times 5^n)$.
    Hence, the given rational is terminating.

(vi) $\frac{19}{3125}$ = $\frac{19}{5^5} = \frac{19 \times 2^5}{5^5 \times 2^5}= \frac{608}{100000} = 0.00608$
    We know either 2 or 5 is not a factor of 19, so it is in its simplest form.
    Moreover, it is in the form of $(2^m \times 5^n)$.
   Hence, the given rational is terminating. 

Answer:

(i) $\frac{11}{2^3 \times 3}$
    We know either 2 or 3 is not a factor of 11, so it is in its simplest form.
   Moreover, $(2^3 \times 3)$≠ $(2^m \times 5^n)$
  Hence, the given rational is non-terminating repeating decimal.

(ii) $\frac{73}{2^2 \times 3^3 \times 5}$
   We know 2, 3 or 5 is not a factor of 73, so it is in its simplest form.
   Moreover, $(2^2 \times 3^3 \times 5)$ ≠ $(2^m \times 5^n)$
  Hence, the given rational is non-terminating repeating decimal.

(iii) $\frac{129}{2^2 \times 5^7 \times 7^5}$
      We know 2, 5 or 7 is not a factor of 129, so it is in its simplest form.
      Moreover, $(2^2 \times 5^7 \times 7^5)$ ≠ $(2^m \times 5^n)$
      Hence, the given rational is non-terminating repeating decimal.

 (iv) $\frac{9}{35} = \frac{9}{5 \times 7}$
    We know either 5 or 7 is not a factor of 9, so it is in its simplest form.
   Moreover, ($5 \times 7$) ≠ $(2^m \times 5^n)$
   Hence, the given rational is non-terminating repeating decimal.


(v) $\frac{77}{210} = \frac{77 ÷ 7}{210 ÷ 7} = \frac{11}{30} = \frac{11}{2 \times 3 \times 5}$
     We know 2, 3 or 5 is not a factor of 11, so $\frac{11}{30}$ is in its simplest form.
    Moreover, $(2 \times 3 \times 5)$ ≠ $(2^m \times 5^n)$
    Hence, the given rational is non-terminating repeating decimal.

(vi) $\frac{32}{147} = \frac{32}{3 \times 7^2}$
     We know either 3 or 7 is not a factor of 32, so it is in its simplest form.
     Moreover, $(3 \times 7^2)$ ≠ $(2^m \times 5^n)$
     Hence, the given rational is non-terminating repeating decimal.

(vii) $\frac{29}{343} = \frac{29}{7^3}$
      We know 7 is not a factor of 29, so it is in its simplest form.
      Moreover, $7^3$ ≠ $(2^m \times 5^n)$
     Hence, the given rational is non-terminating repeating decimal.

(viii) $\frac{64}{455} = \frac{64}{5 \times 7 \times 13}$
     We know 5, 7 or 13 is not a factor of 64, so it is in its simplest form.
     Moreover, $(5 \times 7 \times 13)$ ≠ $(2^m \times 5^n)$
     Hence, the given rational is non-terminating repeating decimal.

Answer:

                  
(i) Let x =  $0.\overline{8}$
∴ x = 0.888                                             ...(1)
10x = 8.888                                            ...(2)
On subtracting equation (1) from (2), we get
9x = 8  ⇒ x = $\frac{8}{9}$ 
               
∴ $0.\overline{8}$ = $\frac{8}{9}$
                   
(ii) Let x = $2.\overline{4}$
∴ x = 2.444                                         ...(1)
10x = 24.444                                      ...(2)
On subtracting equation (1) from (2), we get
 9x = 22 ⇒ x = $\frac{22}{9}$
             
∴ $2.\overline{4}$ = $\frac{22}{9}$
                      
(iii) Let x = $0.\overline{24}$
∴​ x = 0.2424                                   ...(1)
100x = 24.2424                              ...(2)
 On subtracting equation (1) from (2), we get
 99x = 24   ⇒ x = $\frac{8}{33}$
       
 ∴ $0.\overline{24}$  = $\frac{8}{33}$
                     
(iv) Let $x=0.1\overline{2}$
$$\begin{gathered} 10x=1.22222... ...\left(1\right) \\ 100x=12.22222... ...\left(2\right) \end{gathered}$$

On subtracting equation (1) from (2), we get
$$\begin{gathered} 100x-10x=(12.22222...)-(1.22222...) \\ \Rightarrow 90x=11 \\ \Rightarrow x=\frac{11}{90} \end{gathered}$$
                      
(v) Let x = $2.2\overline{4}$  
∴  x = 2.2444                               ...(1)
10x = 22.444                              ...(2)
100x = 224.444                          ...(3)       
On subtracting equation (2) from (3), we get
90x = 202  ⇒ x =$\frac{202}{90}$= $\frac{101}{45}$
                 
Hence, $2.2\overline{4}$ = $\frac{101}{45}$
                       
(vi) Let x = $0.\overline{365}$
∴ x = 0.3656565                          ...(1)
10x = 3.656565                       ...(2)
1000x = 365.656565                 ...(3)
On subtracting (2) from (3), we get
990x = 362  ⇒ x = $\frac{362 }{990}= \frac{181}{495}$
                    
    Hence, $0.\overline{365}$ = $\frac{181}{495}$

Answer:

​​Let x = $5-2\sqrt{3}$ be a rational number.

$$\begin{gathered} x=5-2\sqrt{3} \\ \Rightarrow x^2={\left(5-2\sqrt{3}\right)}^2 \\ \Rightarrow x^2={\left(5\right)}^2+{\left(2\sqrt{3}\right)}^2-2\left(5\right)\left(2\sqrt{3}\right) \\ \Rightarrow x^2=25+12-20\sqrt{3} \\ \Rightarrow x^2-37=-20\sqrt{3} \\ \Rightarrow \frac{37-x^2}{20}=\sqrt{3} \end{gathered}$$
Since x is a rational number, x² is also a rational number.
⇒ 37 − x² is a rational number
⇒ $\frac{37-x^2}{20}$ is a rational number
⇒ $\sqrt{3}$ is a rational number
But $\sqrt{3}$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.

Thus,  ($5-2\sqrt{3}$) is an irrational number.

Answer:

Let $5\sqrt{2}$ is a rational number.

∴ $5\sqrt{2}=\frac{p}{q}$, where p and q are some integers and HCF(p, q) = 1    ....(1)
$$\begin{gathered} \Rightarrow 5\sqrt{2}q=p \\ \Rightarrow {\left(5\sqrt{2}q\right)}^2=p^2 \\ \Rightarrow 2\left(25q^2\right)=p^2 \end{gathered}$$
⇒ p² is divisible by 2
⇒ p is divisible by 2  .....(2)

Let p = 2m, where m is some integer.

∴ $5\sqrt{2}q=2m$
$$\begin{gathered} \Rightarrow {\left(5\sqrt{2}q\right)}^2={\left(2m\right)}^2 \\ \Rightarrow 2\left(25q^2\right)=4m^2 \\ \Rightarrow 25q^2=2m^2 \end{gathered}$$
⇒ q² is divisible by 2
⇒ q is divisible by 2   .....(3)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.

Thus, $5\sqrt{2}$ is irrational.

Answer:

Let $\frac{1}{\sqrt{3}}$ be rational.
∴ $\frac{1}{\sqrt{3}}$ = $\frac{a}{b}$, where $a, b$ are positive integers having no common factor other than 1
∴ $\sqrt{3} = \frac{b}{a}$                           ...(1)
Since $a, b$ are non-zero integers, $\frac{b}{a}$ is rational.
Thus, equation (1)  shows that $\sqrt{3}$ is rational.
This contradicts the fact that $\sqrt{3}$ is rational.
The contradiction arises by assuming $\sqrt{3}$ is rational.
Hence, $\frac{1}{\sqrt{3}}$ is irrational.

Answer:

 $\frac{2}{\sqrt{7}}=\frac{2}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}}=\frac{2}{7}\sqrt{7}$
​Let $\frac{2}{7}\sqrt{7}$ is a rational number.

∴ $\frac{2}{7}\sqrt{7}=\frac{p}{q}$, where p and q are some integers and HCF(p, q) = 1    ....(1)
$$\begin{gathered} \Rightarrow 2\sqrt{7}q=7p \\ \Rightarrow {\left(2\sqrt{7}q\right)}^2={\left(7p\right)}^2 \\ \Rightarrow 7\left(4q^2\right)=49p^2 \\ \Rightarrow 4q^2=7p^2 \end{gathered}$$
⇒ q² is divisible by 7
⇒ q is divisible by 7  .....(2)

Let q = 7m, where m is some integer.

∴ $2\sqrt{7}q=7p$
$$\begin{gathered} \Rightarrow {\left[2\sqrt{7}\left(7m\right)\right]}^2={\left(7p\right)}^2 \\ \Rightarrow 343\left(4m^2\right)=49p^2 \\ \Rightarrow 7\left(4m^2\right)=p^2 \end{gathered}$$
⇒ p² is divisible by 7
⇒ p is divisible by 7   .....(3)

From (2) and (3), 7 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.

Thus, $\frac{2}{\sqrt{7}}$ is irrational.

Answer:

(i) Let $(2 + \sqrt{3)}, ( 2 - \sqrt{3})$ be two irrationals. 
  ∴ $(2 + \sqrt{3}) + ( 2 - \sqrt{3}) = 4$ = rational number

(ii) Let $2\sqrt{3} , 3\sqrt{3}$ be two irrationals. 
   ∴  $2\sqrt{3} \times 3\sqrt{3}$ = 18 = rational number

Answer:

(i) True
(ii) True
(iii) False 
Counter example: $2 + \sqrt{3} \mathrm{and} 2 - \sqrt{3}$ are two irrational numbers. But their sum is 4, which is a rational number.
(iv) False
Counter example: $2\sqrt{3} \mathrm{and} 4\sqrt{3} \mathrm{are} \mathrm{two} \mathrm{irrational} \mathrm{number}s. \text{B}\mathrm{ut} \mathrm{their} \mathrm{product} \mathrm{is} 24, \mathrm{which} \mathrm{is} \text{a} \mathrm{rational} \mathrm{number}.$
(v) True
(vi) True

Answer:

Rational numbers: The numbers of the form $\frac{p}{q}$ where $p , q$ are integers and $q$ ≠ 0 are called rational numbers.
     Example: $\frac{2}{3}$
Irrational numbers: The numbers which when expressed in decimal form are expressible as non-terminating and non-repeating decimals are called irrational numbers.
     Example: $\sqrt{2}$
Real numbers: The numbers which are positive or negative, whole numbers or decimal numbers and rational number or irrational number are called real numbers.
     Example: 2, $\frac{1}{3}$, $\sqrt{2}$, −3 etc.

Answer:

(i) $\frac{22}{7}$ is a rational number because it is of the form of $\frac{p}{q} , q$≠ 0.

(ii) 3.1416 is a rational number because it is a terminating decimal.

(iii) $\mathrm{\pi }$ is an irrational number because it is a non-repeating and non-terminating decimal.
                     
(iv) 3.142857  is a rational number because it is a repeating decimal.

(v) 5.636363... is a rational number because it is a non-terminating, repeating decimal.

(vi) 2.040040004... is an irrational number because it is a non-terminating and non-repeating decimal.

(vii) 1.535335333... is an irrational number because it is a non-terminating and non-repeating decimal.

(viii) 3.121221222... is an irrational number because it is a non-terminating and non-repeating decimal.

(ix) $\sqrt{21 }= \sqrt{3} \times \sqrt{7}$ is an irrational number because $\sqrt{3} \text{and} \sqrt{7}$ are irrational and prime numbers.

(x) $\sqrt[3]{3}$ is an irrational number because 3 is a prime number. So, $\sqrt{3}$ is an irrational number.

Answer:

$$\begin{gathered} \sqrt{2}=1.41421.. \\ \sqrt{3}=1.73205.. \\ \mathrm{One} \mathrm{of} \mathrm{the} \mathrm{rational} \mathrm{number} \mathrm{between} \mathrm{them} \mathrm{is} 1.5=\frac{3}{2} \end{gathered}$$

Hence, a rational number between $\sqrt{2} \mathrm{and} \sqrt{3}$ is $\frac{3}{2}.$

Answer:

Ans

Answer:

Let us assume that $\left(2+\sqrt{3}\right)$ is a rational number.

Thus, $\left(2+\sqrt{3}\right)$ can be represented in the form of $\frac{p}{q},$ where p and q are integers, q ≠ 0, p and q are co-prime numbers.

$$\begin{gathered} 2+\sqrt{3}=\frac{p}{q} \\ \Rightarrow \sqrt{3}=\frac{p}{q}-2 \\ \Rightarrow \sqrt{3}=\frac{p-2q}{q} \\ \mathrm{Since}, \frac{p-2q}{q} \mathrm{is} \mathrm{rational}\Rightarrow \sqrt{3} \mathrm{is} \mathrm{rational} \\ \mathrm{But}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \sqrt{3} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \\ \mathrm{Therefore}, \mathrm{our} \mathrm{assumption} \mathrm{is} \mathrm{wrong}. \\ \mathrm{Hence}, 2+\sqrt{3} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \end{gathered}$$

Answer:

Let us assume that $\left(4-\sqrt{3}\right)$ is a rational number.

Thus, $\left(4-\sqrt{3}\right)$ can be represented in the form of $\frac{p}{q},$ where p and q are integers, q ≠ 0, p and q are co-prime numbers.

$$\begin{gathered} 4-\sqrt{3}=\frac{p}{q} \\ \Rightarrow -\sqrt{3}=\frac{p}{q}-4 \\ \Rightarrow -\sqrt{3}=\frac{p-4q}{q} \\ \Rightarrow \sqrt{3}=\frac{4q-p}{q} \\ \mathrm{Since}, \frac{4q-p}{q} \mathrm{is} \mathrm{rational}\Rightarrow \sqrt{3} \mathrm{is} \mathrm{rational} \\ \mathrm{But}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \sqrt{3} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \\ \mathrm{Therefore}, \mathrm{our} \mathrm{assumption} \mathrm{is} \mathrm{wrong}. \\ \mathrm{Hence}, 4-\sqrt{3} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \end{gathered}$$

Answer:

Let us assume that $\left(3+5\sqrt{2}\right)$ is a rational number.

Thus, $\left(3+5\sqrt{2}\right)$ can be represented in the form of $\frac{p}{q},$ where p and q are integers, q ≠ 0, p and q are co-prime numbers.

$$\begin{gathered} 3+5\sqrt{2}=\frac{p}{q} \\ \Rightarrow 5\sqrt{2}=\frac{p}{q}-3 \\ \Rightarrow 5\sqrt{2}=\frac{p-3q}{q} \\ \Rightarrow \sqrt{2}=\frac{p-3q}{5q} \\ \mathrm{Since}, \frac{p-3q}{5q} \mathrm{is} \mathrm{rational}\Rightarrow \sqrt{2} \mathrm{is} \mathrm{rational} \\ \mathrm{But}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \sqrt{2} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \\ \mathrm{Therefore}, \mathrm{our} \mathrm{assumption} \mathrm{is} \mathrm{wrong}. \\ \mathrm{Hence}, 3+5\sqrt{2} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \end{gathered}$$

Answer:

Let us assume that $\left(2+3\sqrt{5}\right)$ is a rational number.

Thus, $\left(2+3\sqrt{5}\right)$ can be represented in the form of $\frac{p}{q},$ where p and q are integers, q ≠ 0, p and q are co-prime numbers.

$$\begin{gathered} 2+3\sqrt{5}=\frac{p}{q} \\ \Rightarrow 3\sqrt{5}=\frac{p}{q}-2 \\ \Rightarrow 3\sqrt{5}=\frac{p-2q}{q} \\ \Rightarrow \sqrt{5}=\frac{p-2q}{3q} \\ \mathrm{Since}, \frac{p-2q}{3q} \mathrm{is} \mathrm{rational}\Rightarrow \sqrt{5} \mathrm{is} \mathrm{rational} \\ \mathrm{But}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \sqrt{5} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \\ \mathrm{Therefore}, \mathrm{our} \mathrm{assumption} \mathrm{is} \mathrm{wrong}. \\ \mathrm{Hence}, 2+3\sqrt{5} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \end{gathered}$$

Answer:

Let us assume that $\frac{\left(3-4\sqrt{2}\right)}{7}$ is a rational number.

Thus, $\frac{\left(3-4\sqrt{2}\right)}{7}$ can be represented in the form of $\frac{p}{q},$ where p and q are integers, q ≠ 0, p and q are co-prime numbers.

$$\begin{gathered} \frac{3-4\sqrt{2}}{7}=\frac{p}{q} \\ \Rightarrow 3-4\sqrt{2}=\frac{7p}{q} \\ \Rightarrow 4\sqrt{2}=3-\frac{7p}{q} \\ \Rightarrow 4\sqrt{2}=\frac{3q-7p}{q} \\ \Rightarrow \sqrt{2}=\frac{3q-7p}{4q} \\ \mathrm{Since}, \frac{3q-7p}{4q} \mathrm{is} \mathrm{rational}\Rightarrow \sqrt{2} \mathrm{is} \mathrm{rational} \\ \mathrm{But}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \sqrt{2} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \\ \mathrm{Therefore}, \mathrm{our} \mathrm{assumption} \mathrm{is} \mathrm{wrong}. \\ \mathrm{Hence}, \frac{3-4\sqrt{2}}{7} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \end{gathered}$$

Answer:

Let us assume that $\frac{3}{\sqrt{5}}$ is a rational number.

Thus, $\frac{3}{\sqrt{5}}$ can be represented in the form of $\frac{p}{q},$ where p and q are integers, q ≠ 0, p and q are co-prime numbers.

$$\begin{gathered} \frac{3}{\sqrt{5}}=\frac{p}{q} \\ \Rightarrow p\sqrt{5}=3q \\ \Rightarrow \sqrt{5}=\frac{3q}{p} \\ \mathrm{Since}, \frac{3q}{p} \mathrm{is} \mathrm{rational}\Rightarrow \sqrt{5} \mathrm{is} \mathrm{rational} \\ \mathrm{But}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \sqrt{5} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \\ \mathrm{Therefore}, \mathrm{our} \mathrm{assumption} \mathrm{is} \mathrm{wrong}. \\ \mathrm{Hence}, \frac{3}{\sqrt{5}} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \end{gathered}$$

Answer:

Ans

Answer:

Euclid's division lemma, states that for any two positive integers a and b, there exist unique whole numbers q and r, such that
a = b × q + r where 0 ≤ r < b

Answer:

The fundamental theorem of arithmetic, states that every integer greater than 1 either is prime itself or is the product of prime numbers, and this product is unique.

Answer:

Prime factorization:

360 = 2³ × 3² × 5

Answer:

Prime factorization:
a = a
b = b

HCF = product of smallest power of each common prime factor in the numbers = 1

Thus, HCF(a, b) = 1

Answer:

Prime factorization:
a = a
b = b

LCM = product of greatest power of each prime factor involved in the numbers = a × b

Thus, LCM(a, b) = ab.

Answer:

HCF of two numbers = 25
Product of two numbers = 1050
Let their LCM be x.

Using the formula, Product of two numbers = HCF × LCM
we conclude that,

1050 = 25 × x
x = $\frac{1050}{25}$
   = 42

Hence, their LCM is 42.

Answer:

A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself).

Answer:

If two numbers are relatively prime then their greatest common factor will be 1.

Thus, HCF(a, b) = 1.

Answer:

Let x be a rational number whose decimal expansion terminates.
Then, we can express x in the form $\frac{a}{b}$, where a and b are coprime, and prime factorization of b is of the form (2^m × 5ⁿ), where m and n are non negative integers.

Answer:

$$\begin{gathered} \frac{2\sqrt{45}+3\sqrt{20}}{2\sqrt{5}}=\frac{2\sqrt{3\times 3\times 5}+3\sqrt{2\times 2\times 5}}{2\sqrt{5}} \\ =\frac{2\times 3\sqrt{5}+3\times 2\sqrt{5}}{2\sqrt{5}} \\ =\frac{6\sqrt{5}+6\sqrt{5}}{2\sqrt{5}} \\ =\frac{12\sqrt{5}}{2\sqrt{5}} \\ =6 \end{gathered}$$

Hence, simplified form of $\frac{2\sqrt{45}+3\sqrt{20}}{2\sqrt{5}}$ is 6.

Answer:

Decimal expansion:
$$\begin{gathered} \frac{73}{\left(2^4\times 5^3\right)}=\frac{73\times 5}{2^4\times 5^4} \\ =\frac{365}{{\left(2\times 5\right)}^4} \\ =\frac{365}{{\left(10\right)}^4} \\ =\frac{365}{10000} \\ =0.0365 \end{gathered}$$

Thus, the decimal expansion of $\frac{73}{\left(2^4\times 5^3\right)}$ is 0.0365.

Answer:

We can write:
(2ⁿ × 5ⁿ) = (2 × 5)ⁿ
              = 10ⁿ

For any value of n, we get 0 in the end.

Thus, there is no value of n for which (2ⁿ × 5ⁿ) ends in 5.

Answer:

​No, it is not possible to have two numbers whose HCF is 25 and LCM is 520.

Since, HCF must be a factor of LCM, but 25 is not a factor of 520.

Answer:

Let the two irrationals be $4-\sqrt{5} \mathrm{and} 4+\sqrt{5}$.

$\left(4-\sqrt{5}\right)+\left(4+\sqrt{5}\right)=8$

​Thus, sum (i.e., 8) is a rational number.

Answer:

​​Let the two irrationals be $4\sqrt{5} \mathrm{and} 3\sqrt{5}$.

$\left(4\sqrt{5}\right)\times \left(3\sqrt{5}\right)=60$

​Thus, product (i.e., 60) is a rational number.

Answer:

If two numbers are relatively prime then their greatest common factor will be 1.

∴ HCF(a, b) = 1

Using the formula, Product of two numbers = HCF × LCM
we conclude that,

a × b = 1 × LCM
∴ LCM = ab

Thus, LCM(a, b) is ab.

Answer:

If the LCM of two numbers is 1200 then, it is not possible to have their HCF equals to 500.

Since, HCF must be a factor of LCM, but 500 is not a factor of 1200.

Answer:

Let x be $0.\overline{4}$.

$x=0.\overline{4}$  .....(1)
Multiplying both sides by 10, we get
$10x=4.\overline{4}$  .....(2)

Subtracting (1) from (2), we get
$$\begin{gathered} 10x-x=4.\overline{4}-0.\overline{4} \\ \Rightarrow 9x=4 \\ \Rightarrow x=\frac{4}{9} \end{gathered}$$

Thus, simplest form of $0.\overline{4}$ as a rational number is $\frac{4}{9}$.

Answer:

​Let x be $0.\overline{23}$.

$x=0.\overline{23}$  .....(1)
Multiplying both sides by 100, we get
$100x=23.\overline{23}$  .....(2)

Subtracting (1) from (2), we get
$$\begin{gathered} 100x-x=23.\overline{23}-0.\overline{23} \\ \Rightarrow 99x=23 \\ \Rightarrow x=\frac{23}{99} \end{gathered}$$

Thus, simplest form of $0.\overline{23}$ as a rational number is $\frac{23}{99}$.

Answer:

Irrational numbers are non-terminating non-recurring decimals.

Thus, 0.15015001500015 ... is an irrational number.

Answer:

Let $\frac{\sqrt{2}}{3}$ is a rational number.

∴ $\frac{\sqrt{2}}{3}=\frac{p}{q}$, where p and q are some integers and HCF(p, q) = 1   ....(1)

$$\begin{gathered} \Rightarrow \sqrt{2}q=3p \\ \Rightarrow {\left(\sqrt{2}q\right)}^2={\left(3p\right)}^2 \\ \Rightarrow 2q^2=9p^2 \end{gathered}$$
⇒ p² is divisible by 2
⇒ p is divisible by 2   ....(2)

Let p = 2m, where m is some integer.

∴ $\sqrt{2}q=3p$
$$\begin{gathered} \Rightarrow \sqrt{2}q=3\left(2m\right) \\ \Rightarrow {\left(\sqrt{2}q\right)}^2={\left(3\left(2m\right)\right)}^2 \\ \Rightarrow 2q^2=4\left(9p^2\right) \\ \Rightarrow q^2=2\left(9p^2\right) \end{gathered}$$
⇒ q² is divisible by 2
⇒ q is divisible by 2   ....(3)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.

Thus, $\frac{\sqrt{2}}{3}$ is irrational.

Answer:

Since, $\sqrt{3}$ = 1.732....
So, we may take 1.8 as the required rational number between $\sqrt{3} \mathrm{and} 2.$

Thus, the required rational number is 1.8

Answer:

Since, $3.\overline{1416}$ is a non-terminating repeating decimal.

Hence, is a rational number.

Answer:

The numbers that do not share any common factor other than 1 are called co-primes. 
Clearly in option (b), 
factors of 18 are: 1, 2, 3, 6, 9 and 18
factors of 25 are: 1, 5, 25
The two numbers do not share any common factor other than 1. 
They are co-primes to each other.

Answer:

(b) 180
It is given that:
a = $(2^2 \times 3^3 \times 5^4)$ and b = ($2^3 \times 3^2 \times 5$)
∴ HCF (a, b) = Product of smallest power of each common prime factor in the numbers
                       = $2^2 \times 3^2 \times 5$
                       = 180

Answer:

(c) 60

HCF = ($2^3 \times 3^{2 }\times 5$, $2^2 \times 3^{3 }\times 5^2$, $2^{4 }\times 3 \times 5^3 \times 7$)
HCF = Product of smallest power of each common prime factor in the numbers
         = $2^{ 2} \times 3 \times 5$
         = 60

Answer:

(c) 1680

LCM ($2^3 \times 3 \times 5$, $2^4 \times 5 \times 7$) 
∴​ LCM = Product of greatest power of each prime factor involved in the numbers
             = $2^4 \times 3 \times 5 \times 7$
             = $16 \times 3 \times 5 \times 7$
             = 1680

Answer:

(d) 81
Let the two numbers be x and y.
It is given that:
x = 54
​HCF = 27
LCM = 162
We know,
  x $\times$ y = HCF $\times$ LCM
$\Rightarrow$54 $\times$ y = 27 $\times$ 162
$\Rightarrow$  54y = 4374
     $\Rightarrow$   ∴​ y = $\frac{4374}{54}$ = 81

Answer:

(c) 320
Let the two numbers be x and y.
It is given that:
        x $\times$ y = 1600
          HCF = 5
We know,
                 HCF $\times$ LCM = x $\times$ y
             ⇒       5 $\times$ LCM = 1600
             ⇒              ∴  LCM = $\frac{1600}{5}$ = 320
            

Answer:

(c) 128
Largest number that divides each one of 1152 and 1664 = HCF (1152, 1664)
We know, 
               1152 = $2^7 \times 3^2$
               1164 = $2^7 \times 13$
∴ HCF = $2^7$ = 128

Answer:

(a) 13

We know the required number divides 65 (70 − 5) and 117 (125 − 8).
∴ Required number = HCF (65, 117)
we know,
                65 = $13 \times 5$
              117 = $13 \times 3 \times 3$
∴ HCF = 13

Answer:

(b) 16

We know that the required number divides 240 (245 − 5) and 1024 (1029 − 5).
∴ Required number = HCF (240, 1024)
                   240 = $2 \times 2 \times 2 \times 2 \times 3 \times 5$
                 1024 = $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
∴ HCF = $2 \times 2 \times 2 \times 2$ = 16

Answer:

(d) $\frac{15}{16}$

$\frac{1095}{1168} =\frac{1095 ÷73}{1168 ÷ 73} = \frac{15}{16}$

Hence, HCF of 1095 and 1168 is 73.

Answer:

(c) 0 ≤ r < b

Euclid's division lemma states that for any positive integers a and b, there exist unique integers q and r such that a = bq + r,
where r​ must satisfy 0 ≤ r < b

Answer:

(d) 5

We know,
           Dividend = Divisor $\times$ Quotient + Remainder.
It is given that:
Divisor = 143
Remainder = 13
So, the given number is in the form of 143x + 31, where x is the quotient.
∴ 143x + 31 = 13 (11x) + (13 $\times 2$) + 5 = 13 (11x + 2) + 5
Thus, the remainder will be 5 when the same number is divided by 13.

Answer:

(d) 3.141141114...


3.141141114 is an irrational number because it is a non-repeating and non-terminating decimal.

Answer:

(c) an irrational number

$\mathrm{\pi }$ is an irrational number because it is a non-repeating and non-terminating decimal.

Answer:

(b) a rational number
       
2.35 is a rational number because it is a repeating decimal.

Answer:

(c) an irrational number

It is an irrational number because it is a non-terminating and non-repeating decimal.

Answer:

(b) a rational number

It is a rational number because it is a repeating decimal.

Answer:

(c) $\frac{2027}{625}$

 $\frac{124}{165}=\frac{124}{5 \times 33}$; we know 5 and 33 are not the factors of 124. It is in its simplest form and it cannot be expressed as the product of ^{$(2^{m }\times 5^n)$} for some non-negative integers $m , n$.
       
     So, it cannot be expressed as a terminating decimal.


$\frac{131}{30}$ = $\frac{131}{5 \times 6}$; we know 5 and 6 are not the factors of 131. Its is in its simplest form and it cannot be expressed as the product of ( $2^{m }\times 5^n$) for some non-negative integers $m , n$.
  
   So, it cannot be expressed as a terminating decimal.

 $\frac{2027}{625} = \frac{2027 \times 2^4}{5^4 \times 2^4} = \frac{32432}{10000} = 3.2432$; as it is of the form $(2^{m }\times 5^n)$, where $m , n$ are non-negative integers.
  So, it is a terminating decimal.


 $\frac{1625}{462} = \frac{1625}{2 \times 7 \times 33}$; we know 2, 7 and 33 are not the factors of 1625. It is in its simplest form and cannot be expressed as the product of $(2^m \times 5^n)$ for some non-negative integers $m,n$.
So, it cannot be expressed as a terminating decimal.