Rs Aggarwal (2015) Solutions for Class 10 Math Chapter 1 Real Numbers are provided here with simple step-by-step explanations. These solutions for Real Numbers are extremely popular among Class 10 students for Math Real Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal (2015) Book of Class 10 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal (2015) Solutions. All Rs Aggarwal (2015) Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 11:

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 11:

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 11:

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 11:

Question 4:

Using Euclid's algortihm, find the HCF of

(i) 405 and 2520
(ii) 504 and 1188
(iii) 960 and 1575

Answer:

(i)

On applying Euclid's algorithm, i.e. dividing 2520 by 405, we get:
         Quotient = 6, Remainder = 90
         ∴ 2520 = 405 × 6 + 90
   Again on applying Euclid's algorithm, i.e. dividing 405 by 90, we get:
          Quotient = 4, Remainder = 45
          ∴ 405 = 90 × 4 + 45
      Again on applying Euclid's algorithm, i.e. dividing 90 by 45, we get:
         ∴ 90 = 45 × 2 + 0                                                                      
                                                                                                                      
Hence, the HCF of 2520 and 405 is 45.

(ii)

On applying Euclid's algorithm, i.e. dividing 1188 by 504, we get:
       Quotient = 2, Remainder = 180
         ∴ 1188 = 504 × 2 +180                                 
    Again on applying Euclid's algorithm, i.e. dividing 504 by 180, we get:                           
        Quotient = 2, Remainder = 144                             
            ∴ 504 = 180× 2 + 144                                       
   Again on applying Euclid's algorithm, i.e. dividing 180 by 144, we get:                        
        Quotient = 1, Remainder = 36                                        
              ∴ 180 = 144 × 1 + 36                                          
 Again on applying Euclid's algorithm, i.e. dividing 144 by 36, we get:                         
             ∴ 144 = 36 × 4 + 0                                               

Hence, the HCF of 1188 and 504 is 36.  
                                                                                                 
(iii)

On applying Euclid's algorithm, i.e. dividing 1575 by 960, we get:
         Quotient = 1, Remainder = 615           
        ∴ 1575 = 960 × 1 + 615                          
    Again on applying Euclid's algorithm, i.e. dividing 960 by 615, we get:                         
       Quotient = 1, Remainder = 345                   
        ∴ 960 = 615 × 1 + 345                         
  Again on applying Euclid's algorithm, i.e. dividing 615 by 345, we get:                     
     Quotient = 1, Remainder = 270                                       
       ∴ 615 = 345 × 1 + 270                     
 Again on applying Euclid's algorithm, i.e. dividing 345 by 270, we get:                   
    Quotient = 1, Remainder = 75        
      ∴ 345 = 270 × 1 + 75                        
Again on applying Euclid's algorithm, i.e. dividing 270 by 75, we get:                                       
   Quotient = 3, Remainder = 45                                                                                   
      ∴270 = 75 × 3 + 45                                                                 
 Again on applying Euclid's algorithm, i.e. dividing 75 by 45, we get:                          
   Quotient = 1,  Remainder = 30                                                                                                
     ∴ 75  = 45 × 1 + 30                                                                                             
Again on applying Euclid's algorithm, i.e. dividing 45 by 30, we get:
  Quotient = 1, Remainder = 15
     ∴ 45  = 30 × 1 + 15
Again on applying Euclid's algorithm, i.e. dividing 30 by 15, we get:
 Quotient = 2, Remainder = 0
   ∴ 30 = 15 × 2 + 0

Hence, the HCF of 960 and 1575 is 15.                                                                              

Page No 11:

Question 5:

Using prime factorisation, find the HCF and LCM of

(i) 144, 198
(ii) 396, 1080
(iii) 1152, 1664

Answer:

(i) 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

(ii) 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

(iii) 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 11:

Question 6:

Using prime factorisation, find the HCF and LCM of

(i) 24, 36, 40
(ii) 30, 72, 432
(iii) 21, 28, 36, 45

Answer:

(i) 24 = 2× 2 × 2 ×3 = 23 × 3
    36 = 2 × 2 ×3 × 3 = 22 × 32
    40 =  2 × 2 ×2 × 5 = 23 × 5
  ∴ ​HCF = Product of smallest power of each common prime factor in the numbers = 22 = 4
 ∴​ LCM = Product of the greatest power of each prime factor involved in the numbers =  23×32×5 = 360

(ii) 30 = 2 × 3 × 5
     72 = 2 × 2 × 2 × 3 × 3 = 23 × 32
    432 = 2 × 2 × 2 × 2 × 3 × 3× 3 = 24 × 33
  ∴ HCF = Product of smallest power of each common prime factor in the numbers = 2 × 3 = 6
  ∴ ​LCM = Product of the greatest power of each prime factor involved in the numbers = 24 × 33 × 5 = 2160

(iii) 21 = 3 × 7
     28  = 28 = 2 × 2 × 7 = 22 × 7
     36 = 2 × 2 × 3 × 3 = 22 × 32
     45 = 5 × 3 × 3 = 5 × 32
   ∴​ HCF = Product of smallest power of each common prime factor in the numbers = 1
   ∴​ LCM = Product of the greatest power of each prime factor involved in the numbers = 22 × 32 × 5 × 7 = 1260

Page No 11:

Question 7:

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 11:

Question 8:

The HCF of two numbers is 11 and their LCM is 7700. If one of the numbers is 275, find the other.

Answer:

Let the two numbers be x and y.
Given: HCF = 11   and    LCM = 7700
Let the value of x be 275.
We know,         × y = 11 ​× 7700

              ⇒  ∴ 275 ​× y = 84700
              ⇒             ∴​ y =   84700 = 308
                                             275
Hence, the other number y is 308.

Page No 11:

Question 9:

Find the missing numbers in the following factorisation:

Answer:

660 = 2 × 2 × 3 × 5 × 11

Page No 11:

Question 10:

Find the largest number which divides 378 and 510 leaving remainder 6 in each case.

Answer:

We know that the required number divides 372 (378 − 6) and 504 (510 − 6).
∴ Required number = HCF (372, 504)
On applying Euclid's algorithm, we get:
                        372) 504 (1
                             −​ 372  
                                132) 372 (2
                                      −​ 264  
                                        108) 132 (1
                                              −​ 108  
                                                  24) 108 (4
                                                      −​   96  
                                                          12) 24 (2
                                                             −​  24  
                                                                  0
Therefore, the HCF of  372 and 504 is 12.
Hence, the required largest number is 12.                                 
                                 



Page No 12:

Question 11:

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 12:

Question 12:

Find the simplest form of

(i) 6992
(ii) 561748
(iii) 10951168

Answer:

(i) 6992= 69 ÷ 2392 ÷ 23= 34;  ∴​ HCF (69, 92) = 23

(ii) 561748 = 561 ÷ 187748 ÷187 = 34; ∴​ HCF (561, 748) = 187

(iii) 10951168= 1095 ÷ 731168 ÷ 73 = 1516; ∴​ HCF (1095, 1168) = 73

Page No 12:

Question 13:

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?

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
Hence, the greatest possible length of each plank is 7 m.

Page No 12:

Question 14:

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 12:

Question 15:

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 12:

Question 16:

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 12:

Question 17:

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 12:

Question 18:

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 12:

Question 19:

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

Answer:

The traffic lights at three different places changes after every 48 seconds, 72 seconds and 108 seconds, respectively.
∴ Interval of change = LCM (48, 72, 108)
Prime factorisation:
48 = 24 × 3
72 = 23 × 32
108 = 22 × 33
LCM = Product of greatest power of each prime factor involved in the numbers
          = 24 × 33 = 432 seconds = 7 min 12 seconds             [∵ 1 min = 60 seconds]
∴ Required next time of simultaneous change = 8 : 7 : 12 hours

Page No 12:

Question 20:

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 12:

Question 21:

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 18:

Question 1:

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 18:

Question 2:

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 18:

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) 0.24
(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
                 
∴ 2.4 = 229
                          
(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
     ∴ x = 0.1212                               ...(1)
         100x12.1212                        ...(2)
  On subtracting equation (1) from (2), we get
          99x12  ⇒ x = 433
                  
∴ 0.12 = 433
                          
(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, 2.24  = 10145
                             
(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, 0.365 = 181495

Page No 18:

Question 4:

Decide whether the given number is rational or not:

(i) 53.123456789
(ii) 31.123456789
(iii) 0.12012001200012...
Get reason to support your answer.

Answer:

(i) 53.123456789  is rational because it is terminating.
                           _
(ii) 31.123456789  is rational because it is a repeating decimal.

(iii) 0.12012001200012... is not rational because it is non-terminating and non-repeating decimal.



Page No 26:

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 26:

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 27:

Question 3:

Prove that each of the following numbers is irrational:

(i) 6
(ii) 2-3
(iii) 3+2
(iv) 2+5
(v) 5+32
(vi) 37
(vii) 35
(viii) 2-35
(ix) 3+5

Answer:

(i) Let 6 = 2 × 3 be rational.
   Hence,  2, 3 are both rational.
   This contradicts the fact that 2, 3 are irrational.
   The contradiction arises by assuming 6 is rational.
  Hence, 6 is irrational.

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

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

(iv) Let 2 + 5 be rational.
  Hence, 2 + 5 and 5 are rational.  
     ∴   (2 + 5 ) -2= 2   + 5 - 2 = 5 = rational         [∵ Difference of two rational is rational]
       This contradicts the fact that 5 is irrational.
    The contradiction arises by assuming 2 -5 is rational.
   Hence, 2 - 5 is irrational.

(v) Let, 5 + 32 be rational.
      Hence, 5 and 5 + 32 are rational.
     ∴ (5 + 32 - 5)  = 32 = rational       [∵ Difference of two rational is rational]
       ∴ 13× 32 = 2 = rational            [∵ Product of two rational is rational]
      This contradicts the fact that 2 is irrational.
      The contradiction arises by assuming 5 + 32 is rational.
     Hence,  5 + 32 is irrational.

(vi) Let 37 be rational.
      ∴ 13× 37 = 7 = rational           [∵ Product of two rational is rational]
    This contradicts the fact that 7 is irrational.
    The contradiction arises by assuming 37 is rational.
    Hence, 37 is irrational.

(vii) Let 35 be rational.
     ∴ 13× 35 = 15 = rational         [∵ Product of two rational is rational]
   This contradicts the fact that 15 is irrational.
∴ 1×55×5  = 155
So, if 15 is rational, then 155 is rational.
∴ 5 ×155 = 5 = rational           [∵ Product of two rational is rational]
  Hence, 15 is irrational.
The contradiction arises by assuming 35 is rational.
Hence, 35 is irrational.

(viii) Let 2 - 35 be rational.
      Hence 2 and 2 - 35 are rational.
 ∴ 2 -(2 - 35 ) = 2 - 2 + 35 = 35 = rational   [∵ Difference of two rational is rational]
 ∴ 13×35 = 5 = rational                                      [∵ Product of two rational is rational]
This contradicts the fact that 5 is irrational.
The contradiction arises by assuming 2 - 35 is rational.
Hence, 2 - 35 is irrational.

(ix) Let 3 + 5 be rational.
    ∴ 3 + 5 = a, where a is rational
                    ∴  3 = a - 5                                ... (1)
   On squaring both sides of equation (1), we get
                              3 = ( a - 5)2 = a2 + 5 - 25a 
  ⇒                       5 = a2+22a     
This is impossible because right-hand side is rational, whereas the left-hand side is irrational.
This is a contradiction.
Hence, 3 + 5 is irrational.                                                                                              

Page No 27:

Question 4:

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 27:

Question 5:

(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 27:

Question 6:

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 27:

Question 1:

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 28:

Question 2:

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 28:

Question 3:

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 28:

Question 4:

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 28:

Question 5:

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 28:

Question 6:

If the HCF of 65 and 117 is of the form (65m − 117), then m = ?

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

Answer:

(b) 2
It is given that:
HCF (65, 117 ) = ( 65m − 117)
We know,
           65 = 13 × 5
         117 = 13 × 9
∴ HCF = 13
∴ 65m − 117 = 13
65m = 13 + 117 = 130
  m = 13065 = 2
∴ m = 2

Page No 28:

Question 7:

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 28:

Question 8:

For some positive integer m, every positive even integer is of the form

(a) m − 1
(b) m + 1
(c) 2m
(d) 2m + 1

Answer:

(c) 2m

Every positive even integer is in the form of 2m.
Proof: 
 Let m be a positive even  integer.
On dividing m by 2, let q be the quotient and r be the remainder.
     ∴ m = 2q + r, where 0 ≤ r ≤ 2
m = 2q + r, where r = 0,1
m = 2q and m = 2q + 1
But,
      m = 2q + 1 = odd
∴ m = 2q = even

Page No 28:

Question 9:

For some positive integer n, every positive odd integer is of the form

(a) n
(b) n + 1
(c) 2n
(d) 2n + 1

Answer:

(d) 2n + 1

Every positive odd integer is of the form 2n + 1
Proof: 
            Let n be the given positive odd integer.
On dividing n by 2, let q be the quotient and r be the remainder.
On applying Euclid's algorithm, we get:
          n = 2q + r, where 0 ≤ r ≤ 2
    ⇒  n = 2q + r, where r = 0, 1
    ⇒  n = 2q and n = 2q + 1
But,
        n = 2q = even
Thus, when n is odd, it is in the form of n = 2q + 1.

Page No 28:

Question 10:

(n2 − 1) is divisible by 8, if n is

(a) any natural number
(b) any integer
(c) any odd positive integer
(d) any even positive integer

Answer:

(c) any odd positive integer

(n2 - 1) is divisible by 8 if n is an odd positive integer.
We know that an odd positive integer n is of the form (4q + 1) or (4q3) for some integer q.
Case 1: When n = (4q + 1), we have:
             (n2 - 1) ={ ( 4q + 1)2 - 1} = {16q2 + 1 + 8q -1} = 16q2 + 8q = 8q( 2q + 1),
which is clearly divisible by 8.
Case 2: When n = 4q + 3, we have:
          (n2 - 1) = {(4q + 3)2 - 1} = { 16q2 + 9 + 24q - 1} = 16q2 + 24q + 8 = 8(2q2 + 3q + 1),
which is clearly divisible by 8.

Page No 28:

Question 11:

If a and b are positive integers such that a = x3y2 and b = xy3, where x, y are prime numbers, then HCF (a, b) = ?

(a) xy
(b) xy2
(c) x2y2
(d) x3y3

Answer:

(b) xy2
It is given that:
a = x3y2   and   xy3, where x and y are prime numbers
∴ HCF (a, b​) = HCF (x3y2xy3)
we know,
     ​  ​          HCF = Product of smallest power of each common prime factor in the numbers
                          = xy​2

Page No 28:

Question 12:

If p and q are positive integers such that p = ab2 and q = a3b, where a, b are prime numbers, then LCM (p, q) = ?

(a) ab
(b) a2b2
(c) a3b2
(d) a3b3

Answer:

(c) a3b2
We know and are positive integers.
It is given that:
 p = ab2 and q = a3b, where a and are prime numbers
∴ LCM (​p, q) = LCM (ab2a3b)
We know,
                LCM = Product of greatest power of each prime factor involved in the numbers
                         = a3b2

Page No 28:

Question 13:

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 28:

Question 14:

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 29:

Question 15:

Which of the following is a pair of co-primes?

(a) (14, 35)
(b) (18, 25)
(c) (31, 93)
(d) (32, 62)

Answer:

(b) (18, 25)

(14, 35) is not a pair of co-primes because it has a common factor 7 other than 1.
(18, 25) is a co-primes because it has only common factor 1.
(31, 93) is not a pair of co-primes because it has a common factor 31 other than 1.
(32, 62) is  not a pair of co-primes because it has a common factor 2 other than 1.

Page No 29:

Question 16:

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 29:

Question 17:

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 29:

Question 18:

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 29:

Question 19:

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 29:

Question 20:

A positive integer n, when divided by 9, gives 7 as remainder. What will be the remainder when (3n − 1) is divided by 9?

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

Answer:

(b) 2

Let q be the quotient.
It is given that:
Remainder = 7 and Divisor = 9 
On applying Euclid's algorithm, we get:
                  n = 9q + 7          ...(1)
By multiplying both sides by 3, equation (1) becomes:
              3n27q21        ...(2)
By subtracting 1 from both sides, equation (2) becomes:
       3n − 1 = 27q20
 ⇒ 3n − 1 = 9 × 3q× 2 + 2
 ⇒ 3n − 1 = 9 × (3q2) + 2
So, when (3n − 1) is divided by 9, the remainder is 2.

Page No 29:

Question 21:

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 29:

Question 22:

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 29:

Question 23:

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 29:

Question 24:

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 29:

Question 25:

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

(a) 13511250
(b) 2017250
(c) 32191800
(d) 1723625

Answer:

(c) 32191800

 13511250 = 135154 × 2
   We know 2 and 5 are not the factors of 1351.
   So, the given rational is in its simplest form.
And it is of the form (2m  × 5n) for some integers m , n.
  So, the given number is a terminating decimal.
   ∴ 135154 × 2= 1351 × 254 × 24 = 1080810000 = 1.0808

 2017250 = 201753 × 2
     We know 2 and 5 are not the factors of 2017.
    So, the given rational number is in its simplest form.
And it is of the form (2m × 5n) for some integers m , n.
     So, the given rational number is a terminating decimal.
    ∴ 201753 × 2 = 2017 × 2253 × 23 = 80681000 = 8.068

 32191800 = 321923 × 52 × 32
     We know 2, 3 and 5 are not the factors of 3219.
     So, the given rational number is in its simplest form.
  ∴ (23 × 52 × 32)  ≠ (2m × 5n)
  Hence, 32191800 is not a terminating decimal.
32191800= 1.78833333.....
Thus, it is a repeating decimal.

 1723625  = 172354
    We know 5 is not a factor of 1723.
    So, the given rational number is in its simplest form.
   And it is not of the form (2m × 5n ).
   Hence, 1723650 is not a terminating decimal.

Page No 29:

Question 26:

The product of a non-zero rational and an irrational number is

(a) always rational
(b) always irrational
(c) rational of irrational
(d) 1

Answer:

(b) always irrational
Product of non-zero rational and an irrational number is always an irrational.
For example, let 2 and 2 be rational and an irrational number such that their product is 2 × 2 = 22, which is an irrational number.

Page No 29:

Question 27:

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 29:

Question 28:

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 29:

Question 29:

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 29:

Question 30:

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 29:

Question 31:

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 29:

Question 32:

1.2348 is
(a) an integer
(b) an irrational number
(c) a rational number
(d) none of these

Answer:

(c) a rational number
           
1.2348  = 1.234848....
It is a repeating decimal, hence a rational number.

Page No 29:

Question 33:

3 is

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

Answer:

(c) an irrational number

3 = 1.73205080.....
Since it is non-repeating and non-terminating, it is an irrational number.

Page No 29:

Question 34:

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

Answer:

(b) an irrational number

25 is an irrational number.
Proof: Let 25 be rational.
12 × 25 = 5 [∵ Product of two rational is rational]
    ∴ 5 is rational, which is not true (As 5 is prime, 5 is an irrational number)
Hence, 5 is an irrational number.

Page No 29:

Question 35:

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 30:

Question 36:

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

Answer:

(c) an irrational number

12 = 2 × 2 × 3 = 23
It is an irrational number.
Proof:  Let 23 be rational.
∴ 12× 23  = 3 = rational (false)
Hence, 3 is an irrational number.

Page No 30:

Question 37:

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 31:

Question 38:

Assertion (A)
1236250 is a terminating decimal.
Reason (R)
The rational number pq is a terminating decimal, if q= (2m × 5n) for some whole numbers m and n.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (R) is a correct explanation of Assertion (A).
(b) Both Assertion of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (R) is a correct explanation of Assertion (A).


1236250 = 12355 × 2 = terminating decimal
It is given that:
p = 123
q = (2m × 5n), where m , n are whole numbers.
Hence, both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

Page No 31:

Question 39:

Assertion (A)
124249is a non-terminating, repeating decimal.
Reason (R)
The rational number pq is a terminating decimal if q = (2m × 5n) for some whole numbers m and n.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

124249 = 124272 = a non-terminating, repeating decimal
We know 7 is not a factor of 1242; so, it is in its simplest form.
Moreover, it is not of the form (2m × 5n).

Page No 31:

Question 40:

Assertion (A)
8 is an irrational number.
Reason (R)
If m is a natural number which is not a perfect square, then m is irrational.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

8 = 2 × 2 × 2 = 22 = an irrational number and not a perfect square
So, Reason (R) is a correct explanation of Assertion (A).



Page No 32:

Question 41:

Assertion (A)
The HCF of two number is 9 and their LCM is 2016. If one of the numbers is 54, then the other is 306.
Reason (R)
For any positive integers a and b, we have:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(d) Assertion (A) is false and Reason (R) is true.
It is given that:
HCF = 9 and LCM = 2016
Let the two positive integers be x and y.
It is given that:
x = 54
We know,
               Product of two numbers = HCF × LCM
                                      54 × y = 9 × 2016
                                              ∴ y = 1814454  =336 ≠ 306
Hence, Assertion is false and the Reason (R) is true.

Page No 32:

Question 42:

Match the following columns:

Column I Column II
(a) 63 (p) terminates after 4 places of decimal
(b) 0.3465 (q) is irrational
(c) 1234625 (r) terminates after 3 places of decimal
(d) 3416125 (s) is a non-terminating and repeating decimal

Answer:

     Column I Column II
(a) 63 (q) is irrational
                 
(b) 0.3465
(s) is a non-terminating and repeating decimal
(c) 1234625 (p) terminates after four places of decimal
(d) 3416125 (r) terminates after three places of decimal



Page No 35:

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 36:

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 36:

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 36:

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 36:

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 36:

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 36:

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 36:

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 36:

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.

Page No 36:

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.

Page No 36:

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

Page No 36:

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.

Page No 36:

Question 13:

Give prime factorisation of 4620.

Answer:

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

Page No 36:

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

Page No 36:

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

Page No 36:

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.

Page No 36:

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.



Page No 37:

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.

Page No 37:

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.

Page No 37:

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.



View NCERT Solutions for all chapters of Class 10