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Page No 9:

Question 1:

What do you mean by Euclid's division algorithm.

Answer:

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

Page No 9:

Question 2:

A number when divided by 61 gives 27 as quotient and 32 as remainder.
Find the number.

Answer:

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

Page No 9:

Question 3:

By what number should 1365 be divided to get 31 as quotient and 32 as remainder?

Answer:

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

   ⇒                        x = 133331 = 43
Hence, 1365 should be divided by 43 to get 31 as quotient and 32 as remainder.

Page No 9:

Question 4:

Using Euclid's division algorithm, find the HCF of
(i) 612 and 1314
(ii) 1260 and 7344
(iii) 4052 and 12576

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.









 

Page No 9:

Question 5:

By using Euclid's algorithm, find the largest number which divides 650 and 1170.

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.

Page No 9:

Question 6:

Find the HCF of the smallest prime number and the smallest composite number.

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.

Page No 9:

Question 7:

For any positive integer n, prove that n3n is divisible by 6.

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, 0r<b
Applying Euclid's division lemma om and 6, we have 
n=6q+r, 0r<6
Therefore, n can have six values, i.e.
n=6qn=6q+1n=6q+2n=6q+3n=6q+4n=6q+5
Case I: When n=6q
n3=(6q)3n3-n=(6q)3-6q=6q(36q2-1)=6m where m=q(36q2-1)
Hence, n=6q, n3-n is divisible by 6

Case II:
When n=6q+1
n3=(6q+1)3n3-n=(6q+1)3-(6q+1)=(6q+1)(6q+1)2-1=(6q+1)36q2+1+12q-1=(6q+1)36q2+12q=216q3+72q2+36q2+12q=636q3+18q2+2q=6m (where m=36q2+18q+2q)Hence,  n=6q+1, n3-n is divisible by 6

Case III: When n=6q+2
n3=(6q+2)3n3-n=(6q+2)3-(6q+2)=(6q+2)(6q+2)2-1=(6q+2)36q2+4+24q-1=(6q+2)36q2+24q+3=216q3+144q2+18q+72q2+48q+6=216q3+216q2+66q+6=636q3+36q2+11q+1=6m (where m=36q3+36q2+11q+1)Hence, n=6q+1, n3-n is divisible by 6

Case IV: When n=6q+3
n3=(6q+3)3n3-n=(6q+3)3-(6q+3)=(6q+3)(6q+3)2-1=(6q+3)36q2+9+36q-1=(6q+3)36q2+36q+8=216q3+216q2+48q+108q2+108q+24=216q3+324q2+156q+24=636q3+54q2+26q+4=6m
Hence,  n=6q+3, n3-n is divisible by 6.

Case V: When n=6q+4
n3=(6q+4)3n3-n=(6q+4)3-(6q+4)=(6q+4)(6q+4)2-1=(6q+4)36q2+16+48q-1=(6q+4)36q2+48q+15=216q3+288q2+90q+144q2+192q+60=216q3+432q2+282q+60=636q3+72q2+47q+10=6m (where m=36q3+72q2+47q+10)
Hence,  n=6q+4, n3-n is divisible by 6.

Case VI: When n=6q+5
n3=(6q+5)3n3-n=(6q+5)3-(6q+5)=(6q+5)(6q+5)2-1=(6q+5)36q2+25+60q-1=(6q+5)36q2+60q+24=216q3+360q2+144q+180q2+300q+120=216q3+540q2+444q+120=636q3+90q2+74q+120=6m  (where m=36q3+90q2+74q+120)
Hence,  n=6q+5, n3-n is divisible by 6.

Page No 9:

Question 8:

Prove that if x and y are both odd positive integers then x2 + y2 is even but not divisible by 4.

Answer:

Let, be any positive odd integer and let x=n and y=n+2.
So, x2+y2=(n)2+(n+2)2
Or, x2+y2=n2+(n2+4+4n)
x2+y2=2n2+4+4nx2+y2=2(n2+2+2n)
x2+y2=2m (where m=n2+2n+2)
Because x2+y2 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.

Page No 9:

Question 9:

Use Euclid's algorithm to find HCF of 1190 and 1445. Express the HCF in the form 1190m + 1445n.

Answer:

Using Euclid's division algorithm, we have


Since 1445 > 1190, we apply Euclid's division lemma to 1445 and 1190 to get;
1445=1190×1+255
Since the remainder is not zero, we again apply division lemma to 1190 and 255 and get;
1190=255×4+170
Again, the remainder is not zero, so we apply division lemma to 255 and 170 to get;
255=170×1+85
Now we finally apply division lemma to 170 and 85 to get;
170=85×2+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
170×1+85=25585=255-170×185=(1445-1190×1)-(1190-255×4)85=(1445-1190)-1190-(1445-1190)×485=(1445-1190)-1190-1445×4+1190×485=1445-1190-1190×5-1445×485=1445-1190-1190×5+1445×485=1445×5-1190×6
Or, 85=1190(-6)+1445(5)
Hence, m=-6, n=5

Page No 9:

Question 10:

Use Euclid's algorithm to find the HCF of 441, 567 and 693.

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.



 

Page No 9:

Question 11:

Using Euclid's algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2, and 3 respectively.

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.



 



Page No 16:

Question 1:

Express 429 as a product of its prime factors.

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.

 

Page No 16:

Question 2:

Express 5005 as a product of its prime factors.

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.
 

Page No 16:

Question 8:

The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other.

Answer:

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

Page No 16:

Question 9:

The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find the other.

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 = 145×2175725
   = 435

Hence, the other number is 435.



Page No 17:

Question 3:

Express 2431 as a product of its prime factors.

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.

Page No 17:

Question 4:

Using prime factorization, find the HCF and LCM of

(i) 36, 84
(ii) 23, 31
(iii) 96, 404
(iv) 144, 198
(v) 396, 1080
(vi) 1152, 1664

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 = 24 × 32
  198 = 2 × 32 × 11
​ HCF = product of smallest power of each common prime factor in the numbers = 2 × 32 = 18
 LCM = product of greatest power of each prime factor involved in the numbers = 24 × 32 × 11 = 1584

(v) 396, 1080
    Prime factorisation:
   396 = 22 × 32  × 11
  1080 = 23 × 33 × 5
 ​ HCF = product of smallest power of each common prime factor in the numbers = 22 × 32 = 36
LCM = product of greatest power of each prime factor involved in the numbers = 23 × 33 × 5 ×11 = 11880

(vi) 1152 , 1664
    Prime factorisation:
   1152 = 27 × 32
  1664 = 27 × 13
HCF = product of smallest power of each common prime factor involved in the numbers = 27 = 128
 LCM = product of greatest power of each prime factor involved in the numbers = 27 × 32 × 13 = 14976

Page No 17:

Question 5:

Using prime factorization, find the HCF and LCM of:

(i) 8, 9, 25
(ii) 12, 15, 21
(iii) 17, 23, 29
(iv) 24, 36, 40
(v) 30, 72, 432
(vi) 21, 28, 36

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
 

Page No 17:

Question 6:

Find HCF and LCM of 404 and 96 and verify that HCF × LCM = product of two given numbers.

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.

Page No 17:

Question 7:

Two positive integers a  and b can be written as a = x3y2 and b = xy3, where x and y are prime numbers. Find HCF(a, b) and LCM(a, b).

Answer:

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

LCMa, b=LCMx3y2, xy3               =The highest of indices of x and y               =x3y3HCFa, b=HCFx3y2, xy3               =The lowest of indices of x and y               =xy2

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

Page No 17:

Question 10:

The HCF of two number a and b is 5 and their LCM is 200. Find the product ab.

Answer:

Let the two numbers be and b.

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

Hence, the product ab is 1000.

Page No 17:

Question 11:

The LCM of two numbers is 9 times their HCF. The sum of LCM and HCF is 500. Find their HCF.

Answer:

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

According to the question,

LCM+HCF=5009x+x=50010x=500x=50

Hence, the HCF of two numbers is 50.

Page No 17:

Question 12:

The HCF of two numbers is 18 and their product is 12960. Find their LCM.
 

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
x1296018
   = 720

Hence, their LCM is 720.

Page No 17:

Question 13:

Can two numbers have 15 as their HCF and 175 as their LCM? Give reason.

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.

Page No 17:

Question 14:

Find the simplest form of:
(i) 6992
(ii) 473645
(iii) 10951168
(iv) 368496

Answer:

(i) Prime factorisation of 69 and 92 is:

69 = 3 × 23
92 = 22 × 23

Therefore, 6992=3×2322×23=322=34
Thus, simplest form of 6992 is 34.

(ii) Prime factorisation of 473 and 645 is:

473 = 11 × 43
645 = 3 × 5 × 43

Therefore, 473645=11×433×5×43=1115
Thus, simplest form of 473645 is 1115.

(iii) Prime factorisation of 1095 and 1168 is:

1095 = 3 × 5 × 73
1168 = 24 × 73

Therefore, 10951168=3×5×7324×73=1516
Thus, simplest form of 10951168 is 1516.

(iv) Prime factorisation of 368 and 496 is:

368 = 24 × 23
496 = 24 × 31

Therefore, 368496=24×2324×31=2331
Thus, simplest form of 368496 is 2331.

Page No 17:

Question 15:

Find the largest number which divides 438 and 606, leaving remainder 6 in each case.

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 = 2× 3 × 52

HCF = product of smallest power of each common prime factor in the numbers = 2× 3 = 24

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

Page No 17:

Question 16:

Find the largest number which divides 320 and 457 leaving remainders 5 and 7 respectively.

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.

Page No 17:

Question 17:

Find the least number which when divided by 35, 56 and 91 leaves the same remainder 7 in each case.

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 = 2× 7
91 = 7 × 13

LCM = product of greatest power of each prime factor involved in the numbers = 2× 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.



Page No 18:

Question 18:

Find the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 respectively.

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.

Page No 18:

Question 19:

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

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 = 2× 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.

Page No 18:

Question 20:

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

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.

Page No 18:

Question 21:

Find  the largest four-digits number which when divided by 4, 7 and 13 leaves a remainder of 3 in each case.

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) = 4×7×13 = 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.

Page No 18:

Question 22:

Find the least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3.

Answer:

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

So, LCM(5,6,4,3)=2×2×3×5=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

Page No 18:

Question 23:

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

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.
183-91=92183-43=14091-43=48
Now find HCF of 92, 140 and 48, we get
92=2×2×23140=2×2×5×748=2×2×2×2×3
HCF(92, 140, 48) = 4
Therefore, 4 is the required number. 

Page No 18:

Question 24:

Find the least number which when divided by 20, 25, 35 and 40 leaves remainders 14, 19, 29 and 34 respectively.

Answer:

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

LCM(20, 25, 35, 40)=2×2×2×5×5×7=1400
Now, we can see that
20-14=625-19=635-29=640-34=6
​So, the required number would beLCM(20, 25, 35, 40)-6 
=1400-6=1394

Page No 18:

Question 25:

In a seminar, the number of participants in Hindi, English and mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject.

Answer:

Minimum number of rooms required = Total number of participantsHCF(60,84,108)

Prime factorization of 60, 84 and 108 is:

60 = 2× 3 × 5
84 = 2× 3 × 7
108 = 2× 33

HCF = product of smallest power of each common prime factor in the numbers = 2× 3 = 12

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

Therefore, minimum number of rooms required = 25212=21

Thus, minimum number of rooms required is 21.

Page No 18:

Question 26:

Three sets of English, Mathematics and Science books containing 336, 240 and 96 books respectively have to be stacked in such a way that all the books are stored subject wise and the higher of each stack is the same. How many stacks will be there?

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 = 24 × 3 × 7
240 = 24 × 3 × 5
96 = 25 × 3
∴ HCF = Product of the smallest power of each common prime factor involved in the numbers = 24 × 3 = 48
Hence, we made stacks of 48 books each.

∴ Number of stacks = 33648 + 24048 +  9648 = (7 + 5 + 2) = 14

Page No 18:

Question 27:

Three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length. What is the greatest possible length of each plank? How many planks are formed?

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 × 3 × 7
49 = 7 × 7
63 = 3 × 3 × 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;
7×6=427×7=497×9=63
Therefore, total number of planks formed=6+7+9=22
Hence, total 22 planks will be formed.

Page No 18:

Question 28:

Find the greatest possible length which can be used to measure exactly the length 7 m, 3 m 85 cm and 12 m 95 cm.

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 × 2 × 5 × 5 × 7 = 22 × 52 × 7
385 = 5 × 7 × 11
1295 = 5 × 7 × 37
∴ HCF = 5 × 7 = 35
Hence, the greatest possible length is 35 cm.

Page No 18:

Question 29:

Find the maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils.

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 = 11×91
  910 = 10×91
∴ HCF = 91
Hence, 91 students receive same number of pens and pencils.

Page No 18:

Question 30:

Find the leash number of square tiles required to pave the ceiling of a room 15 m 17 cm long and 9 m 2 cm broad.

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 × 41                                                                                                             
902 = 22 × 41                                                                                          
∴ HCF = Product of smallest power of each common prime factor in the numbers = 41             
∴ Required number of tiles = Area of ceilingArea of one tile1517 × 90241 × 41 = 37 × 22 = 814

Page No 18:

Question 31:

Three measuring rods are 64 cm, 80 cm and 96 cm in length. Find the least length of cloth that can be measured an exact number of times, using any of the rods.

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 = 26
80 = 24 × 5
96 = 25 × 3
∴ LCM = Product of greatest power of  each prime factor involved in the numbers = 26 × 3 × 5 = 960 cm = 9.6 m
Hence, the required length of cloth is 9.6 m.



Page No 19:

Question 32:

An electronic device makes a beep after every 60 seconds. Another device makes a beep after 62 seconds. They beeped together at 10 a.m. At what time will they beep together at the earliest?

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 = 22 × 3 × 5
62 = 2 × 31
∴ LCM = 22 × 3 × 5 × 31 = 1860 seconds = 186060 = 31 min
Hence, they will beep together again at 10 : 31 a.m.

Page No 19:

Question 33:

The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they all change simultaneously at 8 a.m. then at what time will they again change simultaneously?

Answer:

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

Hence, LCM = 2×2×2×2×3×3×3=432
So, they will blink again at 432 seconds past 8:00 am
or, 43260=7 minutes and 12 seconds past 8:00 am 
So, the time will be 08:07:12 hrs

Page No 19:

Question 34:

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12, minutes respectively. How many times do they toll together in 30 hours?

Answer:

Six bells toll together at intervals of 2, 4, 6, 8, 10 and 12 minutes, respectively.
Prime factorisation:
2 = 24 = 2 × 26 = 2 × 38 = 2 × 2 × 210 = 2 × 512 = 2 × 2 × 3

∴ ​LCM ( 2, 4, 6, 8, 10, 12 ) = 23 × 3 × 5 = 120
Hence, after every 120 minutes (i.e. 2 hours), they will toll together.
∴ Required number of times = (302 + 1) = 16



Page No 26:

Question 1:

Without actual division, show that each of the following rational numbers is a terminating decimal. Express each in decimal form:

(i) 2323×52

(ii) 24125

(iii)171800

(iv) 151600

(v) 17320

(vi) 193125

Answer:

(i) 2323× 52 = 23×523×53=1151000 = 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 (2m×5n).
Hence, the given rational is terminating.

(ii) 24125 = 2453 = 24 × 2353× 23 = 1921000 = 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 (2m × 5n).
Hence, the given rational is terminating.

(iii) 171800 = 171 25 × 52 = 171 × 5325 × 55 = 21375100000 = 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 (2m ×5n).
    Hence, the given rational is terminating.

(iv)  151600 = 1526 × 52 = 15 × 5426 × 56 = 93751000000 = 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 (2m × 5n).
   Hence, the given rational is terminating.

(v)  17320 = 1726 × 5 = 17 × 5526 × 56 = 531251000000 = 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 (2m × 5n).
    Hence, the given rational is terminating.

(vi) 193125 = 1955 = 19 × 2555 × 25= 608100000 = 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 (2m × 5n).
   Hence, the given rational is terminating. 

Page No 26:

Question 2:

Without actual division, show that each of the following rational numbers is a non-terminating repeating decimal:

(i) 1123×3
(ii) 7322×33×5
(iii)12922×57×75
(iv)935
(v)77210
(vi) 32147
(vii) 29343
(viii) 64455

Answer:

(i) 1123 × 3
    We know either 2 or 3 is not a factor of 11, so it is in its simplest form.
   Moreover, (23 × 3) ≠ (2m × 5n)
  Hence, the given rational is non-terminating repeating decimal.

(ii) 7322 × 33 ×5
   We know 2, 3 or 5 is not a factor of 73, so it is in its simplest form.
   Moreover, (22 × 33 × 5) ≠ (2m × 5n)
  Hence, the given rational is non-terminating repeating decimal.

(iii) 12922 × 57 × 75
      We know 2, 5 or 7 is not a factor of 129, so it is in its simplest form.
      Moreover, (22 × 57 × 75) ≠ (2m × 5n)
      Hence, the given rational is non-terminating repeating decimal.

 (iv) 935 = 95 × 7
    We know either 5 or 7 is not a factor of 9, so it is in its simplest form.
   Moreover, (5 × 7) ≠ (2m × 5n)
   Hence, the given rational is non-terminating repeating decimal.


(v) 77210 = 77 ÷ 7210 ÷ 7 = 1130 = 112 × 3 × 5
     We know 2, 3 or 5 is not a factor of 11, so 1130 is in its simplest form.
    Moreover, (2 × 3 × 5) ≠ (2m × 5n)
    Hence, the given rational is non-terminating repeating decimal.

(vi) 32147 = 323 × 72
     We know either 3 or 7 is not a factor of 32, so it is in its simplest form.
     Moreover, (3 × 72) ≠ (2m × 5n)
     Hence, the given rational is non-terminating repeating decimal.

(vii) 29343 = 2973
      We know 7 is not a factor of 29, so it is in its simplest form.
      Moreover, 73 ≠ (2m × 5n)
     Hence, the given rational is non-terminating repeating decimal.

(viii) 64455 = 645 × 7 × 13
     We know 5, 7 or 13 is not a factor of 64, so it is in its simplest form.
     Moreover, (5 × 7 × 13) ≠ (2m × 5n)
     Hence, the given rational is non-terminating repeating decimal.

Page No 26:

Question 3:

Express each of the following as a fraction in simplest form:

(i) 0.8
(ii) 2.4
(iii) 0.24
(iv) 0.12
(v) 2.24
(vi) 0.365

Answer:

                  
(i) Let x =  0.8
∴ x = 0.888                                             ...(1)
10x = 8.888                                            ...(2)
On subtracting equation (1) from (2), we get
9x = 8  ⇒ x = 89 
               
0.8 = 89
                   
(ii) Let x = 2.4
∴ x = 2.444                                         ...(1)
10x24.444                                      ...(2)
On subtracting equation (1) from (2), we get
 9x22 ⇒ x = 229
             
∴ 2.4229
                      
(iii) Let x = 0.24
∴​ x = 0.2424                                   ...(1)
100x24.2424                              ...(2)
 On subtracting equation (1) from (2), we get
 99x24   ⇒ x833
       
 ∴ 0.24  = 833
                     
(iv) Let x=0.12¯
10x=1.22222...                       ...(1)100x=12.22222...                  ...(2)

On subtracting equation (1) from (2), we get
100x-10x=(12.22222...)-(1.22222...)90x=11x=1190
                      
(v) Let x = 2.24  
∴  x = 2.2444                               ...(1)
10x = 22.444                              ...(2)
100x224.444                          ...(3)       
On subtracting equation (2) from (3), we get
90x = 202  ⇒ x =2029010145
                 
Hence, 2.2410145
                       
(vi) Let x = 0.365
∴ x = 0.3656565                          ...(1)
10x3.656565                       ...(2)
1000x365.656565                 ...(3)
On subtracting (2) from (3), we get
990x362  ⇒ x = 362 990= 181495
                    
    Hence, 0.365181495



Page No 34:

Question 10:

Prove that (5-23) is an irrational number.

Answer:

​​Let x = 5-23 be a rational number.

x=5-23x2=5-232x2=52+232-2523x2=25+12-203x2-37=-20337-x220=3
Since x is a rational number, x2 is also a rational number.
⇒ 37 − xis a rational number
⇒ 37-x220 is a rational number
⇒ 3 is a rational number
But 3 is an irrational number, which is a contradiction.
Hence, our assumption is wrong.

Thus,  (5-23) is an irrational number.

Page No 34:

Question 11:

Prove that 52 is irrational.

Answer:

Let 52 is a rational number.

∴ 52=pq, where p and q are some integers and HCF(pq) = 1    ....(1)
52q=p52q2=p2225q2=p2
p2 is divisible by 2
p is divisible by 2  .....(2)

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

∴ 52q=2m
52q2=2m2225q2=4m225q2=2m2
⇒ 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, 52 is irrational.

Page No 34:

Question 12:

Prove that 13 is irrational.

Answer:

Let 13 be rational.
∴ 13 = ab, where a, b are positive integers having no common factor other than 1
∴ 3 = ba                           ...(1)
Since a, b are non-zero integers, ba is rational.
Thus, equation (1)  shows that 3 is rational.
This contradicts the fact that 3 is rational.
The contradiction arises by assuming 3 is rational.
Hence, 13 is irrational.

Page No 34:

Question 13:

Prove that 27 is irrational.

Answer:

 27=27×77=277
​Let 277 is a rational number.

∴ 277=pq, where p and are some integers and HCF(pq) = 1    ....(1)
27q=7p27q2=7p274q2=49p24q2=7p2
⇒ q2 is divisible by 7
⇒ q is divisible by 7  .....(2)

Let q = 7m, where m is some integer.

∴ 27q=7p
277m2=7p23434m2=49p274m2=p2
⇒ 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, 27 is irrational.

Page No 34:

Question 16:

(i) Give an example of two irrationals whose sum is rational.
(ii) Give an examples of two irrationals whose product is rational.

Answer:

(i) Let (2 + 3), ( 2 - 3)  be two irrationals. 
  ∴ (2 + 3) + ( 2 - 3) = 4 = rational number

(ii) Let 23 , 33 be two irrationals. 
   ∴  23 × 33 = 18 = rational number

Page No 34:

Question 17:

State whether the given statement is true of false:

(i) The sum of two rationals is always rational.
(ii) The product of two rationals is always rational.
(iii) The sum of two irrationals is an irrational.
(iv) The product of two irrationals is an irrational.
(v) The sum of a rational and and irrational is irrational.
(vi) The product of a rational and an irrational is irrational.

Answer:

(i) True
(ii) True
(iii) False 
Counter example: 2 + 3 and 2 - 3 are two irrational numbers. But their sum is 4, which is a rational number.
(iv) False
Counter example: 23 and  43 are two irrational numbers. But their product is 24, which is a rational number.
(v) True
(vi) True



Page No 36:

Question 1:

Define (i) rational numbers (ii) irrational numbers (iii) real numbers.

Answer:

Rational numbers: The numbers of the form pq where p , q are integers and q ≠ 0 are called rational numbers.
     Example: 23
Irrational numbers: The numbers which when expressed in decimal form are expressible as non-terminating and non-repeating decimals are called irrational numbers.
     Example: 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, 132, −3 etc.

Page No 36:

Question 2:

Classify the following numbers as rational or irrational:

(i) 227
(ii) 3.1416
(iii) π
(iv) 3.142857
(v) 5.636363...
(vi) 2.040040004...
(vii) 1.535335333...
(viii) 3.121221222...
(ix) 21
(x) 33

Answer:

(i) 227 is a rational number because it is of the form of pq , q≠ 0.

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

(iii) π 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) 21 = 3 × 7 is an irrational number because 3 and 7 are irrational and prime numbers.

(x) 33 is an irrational number because 3 is a prime number. So, 3 is an irrational number.

Page No 36:

Question 3:

Find a rational number between 2 and 3.

Answer:

2=1.41421..3=1.73205..One of the rational number between them is 1.5=32

Hence, a rational number between 2 and 3 is 32.

Page No 36:

Question 4:

Prove that 6 is an irrational number.

Answer:

Ans

Page No 36:

Question 5:

Prove that 2+3 is an irrational number, given that 3 is an irrational number.

Answer:

Let us assume that 2+3 is a rational number.

Thus, 2+3 can be represented in the form of pq, where p and q are integers, q ≠ 0, p and q are co-prime numbers.

2+3=pq3=pq-23=p-2qqSince, p-2qq is rational3 is rationalBut, it is given that 3 is an irrational number.Therefore, our assumption is wrong.Hence, 2+3 is an irrational number. 

Page No 36:

Question 6:

Prove that 4-3 is an irrational number, given that 3 is an irrational number.

Answer:

Let us assume that 4-3 is a rational number.

Thus, 4-3 can be represented in the form of pq, where p and q are integers, q ≠ 0, p and q are co-prime numbers.

4-3=pq-3=pq-4-3=p-4qq3=4q-pqSince, 4q-pq is rational3 is rationalBut, it is given that 3 is an irrational number.Therefore, our assumption is wrong.Hence, 4-3 is an irrational number. 

Page No 36:

Question 7:

Prove that 3+52 is an irrational number, given that 2 is an irrational number.

Answer:

Let us assume that 3+52 is a rational number.

Thus, 3+52 can be represented in the form of pq, where p and q are integers, q ≠ 0, p and q are co-prime numbers.

3+52=pq52=pq-352=p-3qq2=p-3q5qSince, p-3q5q is rational2 is rationalBut, it is given that 2 is an irrational number.Therefore, our assumption is wrong.Hence, 3+52 is an irrational number. 

Page No 36:

Question 8:

Prove that 2+35 is an irrational number, given that 5 is an irrational number.

Answer:

Let us assume that 2+35 is a rational number.

Thus, 2+35 can be represented in the form of pq, where p and q are integers, q ≠ 0, p and q are co-prime numbers.

2+35=pq35=pq-235=p-2qq5=p-2q3qSince, p-2q3q is rational5 is rationalBut, it is given that 5 is an irrational number.Therefore, our assumption is wrong.Hence, 2+35 is an irrational number. 

Page No 36:

Question 9:

Prove that 3-427 is an irrational number, given that 2 is an irrational number.

Answer:

Let us assume that 3-427 is a rational number.

Thus, 3-427 can be represented in the form of pq, where p and q are integers, q ≠ 0, p and q are co-prime numbers.

3-427=pq3-42=7pq42=3-7pq42=3q-7pq2=3q-7p4qSince, 3q-7p4q is rational2 is rationalBut, it is given that 2 is an irrational number.Therefore, our assumption is wrong.Hence, 3-427 is an irrational number. 

Page No 36:

Question 14:

Prove that 35 is irrational, given that 5 is irrational.

Answer:

Let us assume that 35 is a rational number.

Thus, 35 can be represented in the form of pq, where p and q are integers, q ≠ 0, p and q are co-prime numbers.

35=pqp5=3q5=3qpSince, 3qp is rational5 is rationalBut, it is given that 5 is an irrational number.Therefore, our assumption is wrong.Hence, 35 is an irrational number. 

Page No 36:

Question 15:

Prove that 3+5 is irrational, given that each none of 3 and 5 is irrational.

Answer:

Ans



Page No 37:

Question 1:

State Euclid's division lemma.

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

Page No 37:

Question 2:

State fundamental theorem of arithmatic.

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.



Page No 38:

Question 3:

Express 360 as product of its prime factors.

Answer:

Prime factorization:

360 = 23 × 32 × 5

Page No 38:

Question 4:

If a and b are two prime numbers then find HCF(a, b).

Answer:

Prime factorization:
a = a
b = b

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

Thus, HCF(ab) = 1

Page No 38:

Question 5:

If a and b are two prime numbers then find LCM(ab).

Answer:

Prime factorization:
a = a
b = b

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

Thus, LCM(ab) = ab.

Page No 38:

Question 6:

If the product of two numbers is 1050 and their HCF is 25, find their LCM.

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
x105025
   = 42

Hence, their LCM is 42.

Page No 38:

Question 7:

What is a composite number?

Answer:

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

Page No 38:

Question 8:

If a and b are relatively prime then what is their HCF?

Answer:

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

Thus, HCF(a, b) = 1.

Page No 38:

Question 9:

If the rational number ab has a terminating decimal expansion, what is the condition to be satisfied by b?

Answer:

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

Page No 38:

Question 10:

Simplify: 245+32025.

Answer:

245+32025=23×3×5+32×2×525                        =2×35+3×2525                        =65+6525                        =12525                        =6

Hence, simplified form of 245+32025 is 6.

Page No 38:

Question 11:

Write the decimal expansion of 7324×53.

Answer:

Decimal expansion:
7324×53=73×524×54              =3652×54              =365104              =36510000              =0.0365

Thus, the decimal expansion of 7324×53 is 0.0365.

Page No 38:

Question 12:

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

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.

Page No 38:

Question 13:

Is it possible to have two numbers whose HCF is 25 and LCM is 520?

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.

Page No 38:

Question 14:

Give an example of two irrationals whose sum is rational.

Answer:

Let the two irrationals be 4-5 and 4+5.

4-5+4+5=8

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

Page No 38:

Question 15:

Give an example of two irrationals whose product is rational.

Answer:

​​Let the two irrationals be 45 and 35.

45×35=60

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

Page No 38:

Question 16:

If a and b are relatively prime, what is their LCM?

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(ab) is ab.

Page No 38:

Question 17:

The LCM of two numbers is 1200. Show that the HCF of these numbers cannot be 500. Why?

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.

Page No 38:

Question 18:

Express 0.4¯ as a rational number in simplest form.

Answer:

Let x be 0.4¯.

x=0.4¯  .....(1)
Multiplying both sides by 10, we get
10x=4.4¯  .....(2)

Subtracting (1) from (2), we get
10x-x=4.4¯-0.4¯9x=4x=49

Thus, simplest form of 0.4¯ as a rational number is 49.

Page No 38:

Question 19:

Express 0.23¯ as a rational number in simplest form.

Answer:

​Let be 0.23¯.

x=0.23¯  .....(1)
Multiplying both sides by 100, we get
100x=23.23¯  .....(2)

Subtracting (1) from (2), we get
100x-x=23.23¯-0.23¯99x=23x=2399

Thus, simplest form of 0.23¯ as a rational number is 2399.

Page No 38:

Question 20:

Explain why 0.15015001500015 ... is an irrational number.

Answer:

Irrational numbers are non-terminating non-recurring decimals.

Thus, 0.15015001500015 ... is an irrational number.

Page No 38:

Question 21:

Show that 23 is irrational.

Answer:

Let 23 is a rational number.

∴ 23=pq, where p and q are some integers and HCF(p, q) = 1   ....(1)

2q=3p2q2=3p22q2=9p2
p2 is divisible by 2
⇒ p is divisible by 2   ....(2)

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

∴ 2q=3p
2q=3(2m)2q2=3(2m)22q2=49p2q2=29p2
⇒ 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, 23 is irrational.

Page No 38:

Question 22:

Write a rational number between 3 and 2.

Answer:

Since, 3 = 1.732....
So, we may take 1.8 as the required rational number between 3 and 2.

Thus, the required rational number is 1.8

Page No 38:

Question 23:

Explain why 3.1416¯ is a rational number.

Answer:

Since, 3.1416¯ is a non-terminating repeating decimal.

Hence, is a rational number.



Page No 39:

Question 1:

Which of the following is a pair of co-primes?
(a) (14, 35)
(b) (18, 25)
(c) (31, 93)
(d) (32, 62)

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.

Page No 39:

Question 2:

If a = (22 × 33 × 54) and b = (23 × 32 × 5), then HCF (a, b) = ?

(a) 90
(b) 180
(c) 360
(d) 540

Answer:

(b) 180
It is given that:
a(22 × 33 × 54) and b = (23 × 32 × 5)
∴ HCF (ab) = Product of smallest power of each common prime factor in the numbers
                       = 22 × 32 × 5
                       = 180

Page No 39:

Question 3:

HCF of (23 × 32 × 5), (22 × 33 ×52) and (24 ×3 × 53 × 7) is

(a) 30
(b) 48
(c) 60
(d) 105

Answer:

(c) 60

HCF = (23 × 32 × 522 × 33 × 5224 ×3 × 53 × 7)
HCF = Product of smallest power of each common prime factor in the numbers
         = 2 2 × 3 × 5
         = 60

Page No 39:

Question 4:

LCM of (23 × 3 × 5) and (24 × 5 × 7) is

(a) 40
(b) 560
(c) 1680
(d) 1120

Answer:

(c) 1680

LCM (23 × 3 × 524 × 5 × 7
∴​ LCM = Product of greatest power of each prime factor involved in the numbers
             = 24 × 3 × 5 × 7
             = 16 × 3 × 5 × 7
             = 1680

Page No 39:

Question 5:

The HCF of two numbers is 27 and their LCM is 162. If one of the numbers is 54, what is the other number?

(a) 36
(b) 45
(c) 9
(d) 81

Answer:

(d) 81
Let the two numbers be x and y.
It is given that:
x = 54
​HCF = 27
LCM = 162
We know,
  × = HCF × LCM
 54 × y = 27 × 162
   54y4374
         ∴​ y = 437454 = 81

Page No 39:

Question 6:

The product of two numbers is 1600 and their HCF is 5. The LCM of the numbers is

(a) 8000
(b) 1600
(c) 320
(d) 1605

Answer:

(c) 320
Let the two numbers be and y.
It is given that:
        × y = 1600
          HCF = 5
We know,
                 HCF × LCM = × y
             ⇒       5 × LCM = 1600
             ⇒              ∴  LCM = 16005 = 320
            

Page No 39:

Question 7:

What is the largest number that divides each one of 1152 and 1664 exactly?

(a) 32
(b) 64
(c) 128
(d) 256

Answer:

(c) 128
Largest number that divides each one of 1152 and 1664 = HCF (1152, 1664)
We know, 
               1152 = 27 × 32
               1164 = 27 × 13
∴ HCF = 27 = 128

Page No 39:

Question 8:

What is the largest number that divides 70 and 125, leaving remainders 5 and 8 respectively?

(a) 13
(b) 9
(c) 3
(d) 585

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 × 5
              117 = 13 × 3 × 3
∴ HCF = 13



Page No 40:

Question 9:

What is the largest number that divides 245 and 1029, leaving remainder 5 in each case?

(a) 15
(b) 16
(c) 9
(d) 5

Answer:

(b) 16

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

Page No 40:

Question 10:

The simplest form of 10951168 is

(a) 1726
(b) 2526
(c) 1316
(d) 1516

Answer:

(d) 1516

10951168 =1095 ÷731168 ÷ 73 = 1516

Hence, HCF of 1095 and 1168 is 73.

Page No 40:

Question 11:

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

(a) 1 < r < b
(b) 0 < rb
(c) 0 ≤ r < b
(d) 0 < r < b

Answer:

(c) 0 ≤ r < b

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

Page No 40:

Question 12:

A number when divided by 143 leaves 31 as remainder. What will be the remainder when the same number is divided by 13?

(a) 0
(b) 1
(c) 3
(d) 5

Answer:

(d) 5

We know,
           Dividend = Divisor × 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 × 2) + 5 = 13 (11x + 2) + 5
Thus, the remainder will be 5 when the same number is divided by 13.

Page No 40:

Question 13:

Which of the following is an irrational number?

(a) 227
(b) 3.1416
(c) 3.1416
(d) 3.141141114...

Answer:

(d) 3.141141114...


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

Page No 40:

Question 14:

π is

(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

Answer:

(c) an irrational number

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

Page No 40:

Question 15:

2.35 is
(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

Answer:

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

Page No 40:

Question 16:

2.13113111311113...is

(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

Answer:

(c) an irrational number

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

Page No 40:

Question 17:

3.24636363...is

(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

Answer:

(b) a rational number

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

Page No 40:

Question 18:

Which of the following rational numbers is expressible as a terminating decimal?

(a) 124165
(b) 13130
(c) 2027625
(d) 1625462

Answer:

(c) 2027625

 124165=1245 × 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 (2m × 5n) for some non-negative integers m , n.
       
     So, it cannot be expressed as a terminating decimal.


13130 = 1315 × 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 ( 2m × 5n) for some non-negative integers m , n.
  
   So, it cannot be expressed as a terminating decimal.

 2027625 = 2027 × 2454 × 24 = 3243210000 = 3.2432; as it is of the form (2m × 5n), where m , n are non-negative integers.
  So, it is a terminating decimal.


 1625462 = 16252 × 7 × 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 (2m × 5n) for some non-negative integers m,n.
So, it cannot be expressed as a terminating decimal.



Page No 41:

Question 19:

The decimal expansion of the rational number 3722×5 will terminate after

(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places

Answer:

(b) two decimal places

3722 × 5 = 37 × 522 × 52 = 185100 = 1.85

So, the decimal expansion of the rational number will terminate after two decimal places.

Page No 41:

Question 20:

The decimal expansion of the number 147531250 will terminate after

(a) one decimal place
(b) two decimal place
(c) three decimal place
(d) four decimal place

Answer:

(d) four decimal places

147531250=  1475354 × 2  = 14753 × 2354 × 24 = 11802410000 = 11.8024

So, the decimal expansion of the number will terminate after four decimal places.

Page No 41:

Question 21:

The number 1.732 is
(a) an irrational number
(b) a rational number
(c) an integer
(d) a whole number

Answer:

​Clearly, 1.732 is a terminating decimal.

Hence, a rational number.

Hence, the correct answer is option (b).

Page No 41:

Question 22:

a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. Then, the least prime factor of (a + b) is

(a) 2
(b) 3
(c) 5
(d) 8

Answer:

(a) 2

Since 5 + 3 = 8, the least prime factor of a + b has to be 2, unless a + b is a prime number greater than 2.
If a + b is a prime number greater than 2, then a + must be an odd number. So, either a or b must be an even number. If a is even, then the least prime factor of is 2, which is not 3 or 5. So, neither a nor b can be an even number. Hence, a + b cannot be a prime number greater than 2 if the least prime factor of a is 3 or 5.

Page No 41:

Question 23:

2 is
(a) a rational number
(b) an irrational number
(c) a terminating decimal
(d) a nonterminating repeating decimal

Answer:

Let 2 is a rational number.

∴ 2=pq,  where p and q are some integers and HCF(p, q) = 1    .... (1)
2q=p2q2=p22q2=p2
p2 is divisible by 2
⇒ p is divisible by 2  .... (2)

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

∴ 2q=p
2q=2m2q2=2m22q2=4m2q2=2m2
⇒ q2 is divisible by 2
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, 2 is an irrational number.

Hence, the correct answer is option (b).

Page No 41:

Question 24:

12is

(a) a fraction
(b) a rational number
(c) an irrational number
(d) none of these

Answer:

(c) an irrational number

12 is an irrational number.

Page No 41:

Question 25:

2+2 is
(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

Answer:

(c) an irrational number

2 + 2 is an irrational number.
if it is rational, then the difference of two rational is rational
∴​ (2 + 2 ) - 2 = 2 = irrational

Page No 41:

Question 26:

What is the least number that divisible by all the natural numbers from 1 to 10 (both inclusive)?

(a) 100
(b) 1260
(c) 2520
(d) 5040

Answer:

(c) 2520

We have to find the least number that is divisible by all numbers from 1 to 10.
∴ LCM (1 to 10) =  23 × 32 × 5 × 7 = 2520  
Thus, 2520 is the least number that is divisible by every element and is equal to the least common multiple.



Page No 44:

Question 1:

The decimal representation of 71150is
(a) a terminating decimal
(b) a non-terminating, repeating decimal
(b) a non-terminating and non-repeating decimal
(d) none of these

Answer:

(b) a non-terminating, repeating decimal

71150 = 712 × 3 × 52
We know that 2, 3 or 5 are not factors of 71.
So, it is in its simplest form.
And, (2 × 3 × 52)  ≠ (2m × 5n)
∴  71150  = 0.473¯
 Hence, it is a non-terminating, repeating decimal.

Page No 44:

Question 2:

Which of the following has terminating decimal expansion?

(a) 3291
(b) 1980
(c) 2345
(d) 2542

Answer:

(b) 1980


1980 = 1924 ×5
We know 2 and 5 are not factors of 19, so it is in its simplest form.
And (24 × 5) = (2m × 5n)
Hence, 1980 is a terminating decimal.

Page No 44:

Question 3:

On dividing a positive integer n by 9, we get 7 as remainder. What will be the remainder if (3n − 1) is divided by 9?

(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(b) 2

Let be the quotient.
It is given that:
remainder = 7
On applying Euclid's algorithm, i.e. dividing by 9, we have
         n = 9q + 7
⇒    3n27q21
⇒ 3n − 1 = 27q20
⇒ 3n − 1 = 9 × 3q× 2 + 2
⇒ 3n − 1 = 9 × (3q + 2) + 2
So, when (3n − 1) is divided by 9, we get the remainder 2.

Page No 44:

Question 4:

0.68+0.73=?
(a) 1.41
(b) 1.42
(c) 0.141
(d) None of these

Answer:

(b) 1.42

Page No 44:

Question 5:

Show that any number of the form 4n, nN can never end with the digit 0.

Answer:

If 4n ends with 0, then it must have 5 as a factor.
But we know the only prime factor of 4n is 2.
Also we know from the fundamental theorem of arithmetic that prime factorisation of each number is unique.
Hence, 4n can never end with the digit 0.



Page No 45:

Question 6:

The HCF  of two numbers is 27 and their LCM is 162. If one of the number is 81, find the other.

Answer:

Let the two numbers be x , y.
It is given that:
x = 81
HCF = 27 and  LCM = 162
We know,    Product of two numbers = HCF × LCM
                ⇒                         x × y  = 27 × 162
                ⇒                        81 × y  = 4374
               ⇒                                  y = 437481 = 54
Hence, the other number y is 54.

Page No 45:

Question 7:

Examine whether 1730 is a terminating decimal.

Answer:

1730 = 172 × 3 × 5

We know that 2, 3 and 5 are not the factors of 17.
So, 1730 is in its simplest form.
Also, 30 = 2 × 3 × 5 ≠ ( 2m × 5n)
Hence, 1730 is a non-terminating decimal.

Page No 45:

Question 8:

Find the simplest form of 148185.

Answer:

148185=148 ÷ 37185 ÷ 37 =45 (∵ HCF of 148 and 185 is 37)

Hence, the simplest form is 45.

Page No 45:

Question 9:

Which of the following numbers are irrational?

(a) 2
(b) 63
(c) 3.142857
(d) 2.3
(e) π
(f) 227
(g) 0.232332333...
(h) 5.2741

Answer:

(a) 2 is irrational (∵ if p is prime, then p is irrational).

(b) 63 = 23 × 33 is irrational.

(c) 3.142857 is rational because it is a terminating decimal.
           
(d) 2.3 is rational because it is a non-terminating, repeating decimal.

(e) π is irrational because it is a non-repeating, non-terminating decimal.

(f) 227 is rational because it is in the form of pq , q ≠ 0.

(g) 0.232332333...  is irrational because it is a non-terminating, non-repeating decimal.
                 
(h) 5.2741 is rational because it is a non-terminating, repeating decimal.

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Question 10:

Prove that 2+3 is irrational.

Answer:

Let (2 + 3)  be rational.
Then, both (2 + 3) and 2 are rational.
∴ { (2 + 3) - 2 } is rational [∵ Difference of two rational is rational]
⇒ 3 is rational.
This contradicts the fact that 3 is irrational.
The contradiction arises by assuming (2 + 3) is rational.
Hence, (2 + 3) is irrational.

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Question 11:

Find the HCF and LCM of 12, 15, 18, 27.

Answer:

Prime factorisation:
12 = 2 × 2 × 3 = 22 × 3
15 = 3 × 5
18 = 2 × 3 × 3 = 2 × 32
27 = 3 × 3 × 3 = 33
Now,
HCF = Product of smallest power of each common prime factor in the number
         = 3
LCM = Product of greatest power of each prime factor involved in the number
          =  22 × 33 × 5 = 540

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Question 12:

Give an example of two irrationals whose sum is rational.

Answer:

Let (2 + 2) and (2 - 2) be two irrational numbers.
Sum = (2 + 2) + (2 - 2) = 2 + 2 + 2 - 2 = 4, which is a rational number.

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Question 13:

Give prime factorisation of 4620.

Answer:

Prime factorisation:
4620 = 2 × 2 × 3× 5 × 7 × 11  = 22 × 3× 5 × 7 × 11 

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Question 14:

Find the HCF of 1008 and 1080 by prime factorization method.

Answer:

Prime factorisation:
1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7 = 24 × 32 × 7
1080 = 2 × 2 × 2 × 3 × 3 × 3 × 5 = 23 × 33 × 5
HCF = Product of  smallest power of each common prime factor in the number
         = 23× 32 = 72

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Question 15:

Find the HCF and LCM of 89,1027 and 1681.

Answer:

HCF of fractions = HCF of NumeratorsLCM of Denominators

LCM of fractions= LCM of NumeratorsHCF of Denominators
Prime factorisation of the numbers given in the numerators are as follows:
8 = 2 × 2 × 210 = 2 × 516 = 2 × 2 × 2 × 2

HCF of Numerators = 2
LCM of Numerators = 24 × 5 = 80

Prime factorisation of numbers given in the denominators are as follows:
9 = 3 × 327 = 3 × 3 × 381 = 3 × 3 × 3 × 3

HCF of Denominators = 3 × 3 = 9
LCM of Denominators = 34 = 81


∴ HCF of fractions = HCF of NumeratorLCMof Denominator = 281

              ∴ LCMof fractions = LCMof NumeratorHCF of Denominator = 809

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Question 16:

Find the largest number which divides 546 and 764, leaving remainders 6 and 8 respectively.

Answer:

We know the required number divides 540 (546 − 6) and 756 (764 − 8), respectively.
∴ Required largest number = HCF (540, 756)
Prime factorisation:
     540 = 2×2×3×3×3×5 = 22×32×5
     756 = 2×2×3×3×3×7 = 22 × 33×7
∴ HCF = 22×33=108
Hence, the largest number is 108.

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Question 17:

Prove that 3 is an irrational number.

Answer:

Let 3 be rational and its simplest form be ab.
Then, a, b are integers with no common factors other than 1 and b ≠ 0.
Now 3 = ab ⇒  3 = a2b2                    [on squaring both sides]
                          ⇒ 3b2 = a2           ... (1)
                       
                         ⇒ 3 divides a2                    [since 3 divides 3b2]
                         ⇒ 3 divides a                     [since 3 is prime, 3 divides a2 ⇒ 3 divides a]
Let a = 3c for some integer c.
Putting a = 3c in equation (1), we get
   3b2 = 9c2 ⇒ b = 3c2
                     ⇒ 3 divides b2               [since 3 divides 3c2]
                    ⇒ 3 divides b                 [since 3 is prime, 3 divides b2 ⇒ 3 divides b]
Thus, 3 is a common factor of both a, b.
But this contradicts the fact that a, b have no common factor other than 1.
The contradiction arises by assuming 3 is rational.
Hence, 3 is rational.

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Question 18:

Show that every positive odd integer is of the form (4q + 1) or (4q + 3) for some integer q.

Answer:

Let be the given positive odd integer.
On dividing a by 4,let q be the quotient and r the remainder.
Therefore,by Euclid's algorithm we have
          a = 4q + r           0 ≤ < 4
⇒      a = 4q + r             r​ = 0,1,2,3
⇒      a = 4qa = 4q1,  a = 4q2,  a = 4q + 3
But, 4q  and  4q + 2 = 2 (2q1) = even
Thus, when is odd, it is of the form (4q + 1) or (4q3) for some integer q.

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Question 19:

Show that one and only one out of n, (n + 2) and (n + 4) is divisible by 3, where n is any positive integer.

Answer:

 Let be quotient and be the remainder.
On applying Euclid's algorithm, i.e. dividing by 3, we have
      n = 3q r       0 ≤ < 3
⇒  n = 3q + r       r = 0, 1 or 2
⇒  n = 3q  or  n = (3q1) or n = (3q2)
Case 1​: If n = 3q, then is divisible by 3.
Case 2: If n = (3q1), then (n + 2) = 3q3 = 3(3q1), which is clearly divisible by 3.
             In this case, (n + 2) is divisible by 3.
Case 3 : If n = (3q2), then (n + 4) = 3q + 6 = 3(q + 2), which is clearly divisible by 3.
              In this case, (n + 4) is divisible by 3.
Hence, one and only one out of n, (+ 1) and (n + 2) is divisible by 3.

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Question 20:

Show that 4+32 is irrational.

Answer:

Let (4 + 3√2) be a rational number.
Then both (4 + 32) and 4 are rational.
⇒ ( 4 + 32 − 4) = 32 = rational   [∵ Difference of two rational numbers is rational]
⇒ 32 is rational.
⇒ 13 (32) is rational.         [∵ Product of two rational numbers is rational]
⇒ 2 is rational.
This contradicts the fact that 2 is irrational (when 2 is prime, 2 is irrational).
Hence, (4 + 32 ) is irrational.



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