Q.1.      Use Euclid’s division algorithm to find the HCF of :
(i) 135 and 225        (ii) 196 and 38220        (iii) 867 and 255
Sol.     (i)  135 and 225

Given integers are 135 and 225 clearly 225 > 135 applying Euclid’s division lemma on 135 and 225.

We get

225 = 135 × + 90……………… (i)

Here remainder  So we again apply EDL on divisor 135 and remainder 90

135 = 90 × 1 + 45………………(ii)

Here, remainder  , so we apply  Euclid’s division lemma on divisor 90 and remainder 45

90 = 45 × 2 + 0 ………………….. (iii)

From equation (iii), remainder = 0. So the divisor at this stage and remainder of previous stage

i.e. 45 is HCF ( 135, 225) = 45

(ii) 196 and 38220

Given positive integers are 196 and 38220 and 38220 > 196 so applying EDL,

we get

38220 = 196 × 195 + 0 …………… (i)

Remainder at this stage is zero. So, the divisor of this stage i.e 196 is HCF of 38220 and 196

HCF ( 196 , 38220) = 196

(iii) 867 and 255

Given positive integers are 867 and 255 and 867 > 255 So, applying Euclid’s division algorithm

We get

867 = 255 × 3 + 102  ………………. (i)

Here, remainder . So, we again apply Euclid’s division algorithm on division 255 and remainder 102 .

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Q.2     Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
Sol.       Let a be any positive integer and b = 6.Then, by Euclid’s algorithm a = 6q + r, for some integer q≥0 and where 0≤r<6 the possible remainders are 0, 1, 2, 3, 4, 5 i.e, a can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5,where q is the quotient. If a = 6q or 6q + 2 or 6q + 4, then a is an even integer. Also, an integer can be either even or odd.Therefore, any odd integer is of the form 6a + 1 or 6q + 3 or 6q + 5, where q is some integer.

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Q.3     An army contingent of 616 members is to march behind an army band of 32 members in parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ?
Sol.       To find the maximum number of columns, we have to find the HCF of 616 and 32.

                                                                                                             

⇒616=32×19+8  
             ⇒32=8×4+0
            Therefore,the HCF of 616 and 32 is 8.Hence, maximum number of columns is 8.

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Q.4     Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m +1 for some integer m.
Sol.       Let x be any positive integer, then it is of the form 3q, 3q + 1 or 3q +2. Now, we have to prove that the squre of each of these can be written in the form 3m or 3m +1.
             Now, (3q)2=9q2=3(3q2)=3m, where m=3q2
             (3q+1)2=9q2+6q+1
              =3(3q2+2q)+1
             = 3m + 1, where m=3q2+2q
             and, (3q+2q)2=9q2+12q+4
              =3(3q2+4q+1)+1
              = 3m + 1, where m=3q2+4q+1
             Hence, the result.

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Q.5     Use Euclid’s division lemma to show that cube of any positive integer is either of the form 9q, 9q + 1 or 9q + 8.
Sol.      Let x be any positive integer, then it is of  the form 3m, 3m + 1 or 3m +2. Now, we have prove that the cube of each of these can be rewritten in the form 9q + 1 or 9q + 8.
            Now, (3m)3=27m3=9(3m3)
            = 9q, where q=3m3
            (3m+1)3=(3m)3+3(3m)2.1+3(3m).12+1
            =27m3+27m2+9m+1
            =9(3m3+3m2+m)+1
            = 9q + 1, where q=3m3+3m2+m
            and (3m+2)3=(3m)3+3(3m)2.2+3(3m).22+8
            =27m3+54m2+36m+8
            =9(3m3+6m2+4m)+8
            = 9q + 8, where q=3m3+6m2+4m

 

Q.1     Express each number as product of its prime factors:
(i) 140      (ii) 156      (iii) 3825      (iv) 5005      (vi) 7429
Sol.       (i) We use the division method as shown below :

Therefore, 140 = 2 × 2 × 5 × 7 =22×5×7

(ii) We use the division method as shown below:

Therefore, 156 = 2 × 2 × 3 × 13  =22×3×13

(iii) We use the division method as shown below :

Therefore, 3825 = 3 × 3 × 5 × 5 × 17 =32×52×17

(iv) We use the division method as shown below :

Therefore, 5005 = 5 × 7 × 11 × 13

(v) We use the division method as shown below :

Therefore, 7429 = 17 × 19 × 23

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Q.2     Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91     (ii) 510 and 92     (iii) 336 and 54
Sol.       (i) 26 and 91

                             

26 = 2 × 13 and 91 = 7 × 13
Therefore, LCM of 26 and 91 = 2 × 7 × 13 = 182
               and HCF of 26 and 91 = 13
Now, 182 × 13 = 2366 and 26 × 91 = 2366
Since, 182 × 13 = 26 × 91
Hence verified.

(ii) 510 and 92

                              

              510 = 2 × 3 × 5 × 17 and 92 = 2 × 2 × 23
Therefore, LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23 = 23460
              and HCF of 510 and 92 = 2
              Now, 23460 × 2 = 46920 and 510 × 92 = 46920
Since 23460 × 2 = 510 × 92
Hence verified.

(iii) 336 and 54

                                 

           336 = 2 × 2 × 2 × 2 × 3 × 7
and 54 = 2 × 3 × 3 × 3
Therefore, LCM of 336 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
and HCF of 336 and 54 = 2 × 3 = 6
Now, 3024 × 6 = 18144
and 336 × 54 = 18144
Since, 3024 × 6 = 336 × 54
Hence verified.

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Q.3       Find the LCM and HCF of the following integers by applying the prime factorisation method
(i) 12,15 and 21      (ii) 17, 23 and 29      (iii) 8, 9 and 25
Sol.         (i) First we write the prime factorisation of each of the given numbers.
12 = 2 × 2 × 3 = 22 × 3, 15 = 3 × 5 and 21 = 3 × 7
Therefore, LCM = 22 × 3 × 5 × 7 = 420
and, HCF = 3

(ii) First we write the prime factorisation of each of the given numbers.
17 = 17, 23 = 23 and 29 = 29
Therefore, LCM = 17 × 23 × 29 = 11339
and HCF = 1

(iii)  First we write the prime factorisation of each of the given numbers.
                  8 = 2 × 2 × 2 =23,9=3×3=32, 25 = 5 × 5 = 52
Therefore, LCM =23×32×52=8×9×25=1800
and HCF = 1

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Q.4     Given that HCF (306, 657) = 9, find LCM (306, 657).
Sol.       We know that the product of the HCF and the LCM of two numbers is equal to the product of the given numbers.
Therefor, HCF (306,657) × LCM (306,657) = 306 × 657
⇒ 9 × LCM (306 × 657) = 306 × 657
⇒ LCM (306,657) =306×6579  = 22338

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Q.5.     Check whether 6n can end with the digit 0 for any natureal number n.
Sol.        If the number 6n, for any n ends with the digit zero, then it is divisible by 5. That is, the prime factorisation of 6n contains the prime 5. That is, not possible as the only prime in the factorisation of 6n is 2 and 3 and the uniqueness of the Fundamental Theorem of Arihmetic guarantees that there are no other primes in the factorisation of 6n. So, there is no n∈N for which 6n ends with the digit zero.

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Q.6    Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Sol.     Since, 7 × 11 × 13 + 13 = 13 × (7 × 11 × 1 + 1)
= 13 × (77 + 1)
= 13 × 78
⇒ It is a composite number.
Again, 7 × 6 × 5 × 4 × 3 × 1 × 1 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 1 × 1 + 1)
⇒ It is a composite number.

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Q.7     There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the path, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they again at the starting point ?
Sol.      To find the LCM of 18 and 12, we have

                      

18 = 2 × 3 × 3 and 12 = 2 × 2 × 3
LCM of 18 and 12 = 2 × 2 × 3 × 3 = 36
So, Sonia and Ravi will meet again at the starting point after 36 minutes.

 

Q.1     Prove that 5–√ is irrational.
Sol.       Let us assume, to the contrary, that 5–√ is rational.
Now, let 5–√=ab , where a and b are coprime and b≠0. Squaring on both side, we get
5=a2b2⇒5b2=a2                                                          …(1)
This shows that a2 is divisible by 5
             It follows that a is divisible by 5                                              …(2)
⇒a=5m for some integer m.
Substituting a = 5m in (1), we get
5b2=(5m)2=25m2
or b2=5m2
⇒b2 is divisible by 5
and hence b is divisible by 5                                                   …(3)
From (2) and (3), we can conclude that 5 is a common factor of both a and b.
But this contradicts our supposition that a and b are coprime.
Hence, 5–√ is irrational.

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Q.2     Prove that 3 + 25–√ is irrational.
Sol.       Let us assume, to the contrary, that 3+25–√, is a rational number.
             Now, let  3+25–√=ab, where a and b are coprime and b≠0
⇒25–√=ab−3 or 5–√=a2b−32
Since, a and b are integers.
Therefore, a2b−32 is a rational number
⇒ 5–√ is a rational number.
But 5–√ is an irrational number.
This shows that our assumption is incorrect.
So, 3+25–√ is an irrational number .

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Q.3     Prove that the following are irrationals :
(i) 12√     (ii) 75–√     (iii) 6+2–√
Sol.       (i) Let us assume, to the contrary, that 12√ is rational. That is, we can find co-prime integers p and q(≠0) such that
12√=pq⇒1×2√2√×2√=pq⇒2√2=pq
⇒2–√=2pq
Since p and q are integers, 2pq is rational, and so 2–√ is rational.
But this contradicts the fact that 2–√ is irrational.So, we conclude that 12√ is irrational.

             (ii) Let us assume, to the contrary, that 75–√ is rational. 
             That is, we can find co-prime integers p and q(≠0) such that 75–√=pq
Since p and q are integers, p7q is rational and so is 5–√ 
             But this contradicts the fact that 5–√ is irrational . So, we conclude that 75–√ is irrational.

(iii) Let us assume, to the contrary, that 2–√ is rational. That is, we can find integers p and q(≠0) such that
6+2–√=pq⇒6−pq=2–√
⇒2–√=6−pq
Since p and q are integers, we get 6−pq is rational, and so 2–√ is rational.
But this contradicts the fact that 2–√ is irrational.
So, we conclude that 6 + 2–√ is irrational

 

Q.1     Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non – terminating repeating decimal expansion :
          (i) 133125                                   (ii) 178
          (iii) 64455                                   (iv) 151600
          (v) 29343                                    (iv) 232352
         (vii) 129225775             (viii) 615
         (ix) 3550                                      (x) 77210

Sol.     We know that if the denominator of a rational number has no prime factors other than 2 or 5, then it is expressible as a terminating, otherwise it has non – terminating  repeating decimal representation. Thus, we will have to check the prime factors of the denominators of each of the given rational numbers.  
          (i) In 133125, the denominator is 3125.

We have, 3125 = 5 × 5 × 5 × 5 × 5.
Thus, 3125 has 5 as the only prime factor.
Hence, 133125 must have a terminating decimal representation.

 (ii) In 178, the denominator is 8.

We have, 8 = 2 × 2 × 2
Thus, 8 has 2 as the only prime factor.
Hence, 178 must have a terminating decimal representation.

(iii) In 64455, denominator is 455. We have, 455 = 5 × 7 × 13
Clearly, 455 had prime factors other than 2 and 5. So, it will not have a terminating decimal representation.

(iv) In 151600, the denominator is 1600.

We have, 1600
= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
Thus, 1600 has only 2 and 5 as prime factors.
Hence, 151600 must have a terminating decimal representation.

 (v) In 29343, the denominator is 343.

We have, 343 = 7 × 7 × 7
Clearly, 343 has prime factors other than 2 and 5.
So, it will not have terminating decimal representation.

(vi) In 2323.52 Clearly, the denominator 23.52 has only 2 and 5 as prime factors.
       Hence, 2323.52 must have a terminating decimal representation.
       (vii) In 12922.57.75 Clearly, the denominator 22.57.75 has prime factors other than 2 and 5.So, it will not have terminating decimal representation.

 (viii) In 615, we have 15 = 3 × 5
Clearly, 15 has prime factors other than 2 and 5. So, it will not have terminating decimal representation.

  (ix) In 3550 , we have 50 = 2 × 5 × 5 The denominator has only 2 and 5 as prime factors. Hence, 3550 must have a terminating decimal representation.

(x) In 77210, the denominator is 210.

We have, 210 = 2 × 3 × 5 × 7
Clearly, 210 has prime factors other than 2 and 5.
So, it will not have terminating decimal representation.

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Q.2     Write down the decimal expansion of those rational numbers in Question 1 above which have terminating decimal expansions.
Sol.      (i) 133125=135×5×5×5×5
=13×2×2×2×2×25×2×5×2×5×2×5×2×5×2
=13×3210×10×10×10×10=416100000 = 0.00416

(ii) 178=17×5323×53=17×53103=17×125103
=21251000=2.125

 (iii) Non – terminating.

(iv) 151600=1526×52=1524×22×52
=1524×102=15×5424×54×102
=15×625104×102=93751000000=0.009375

   (v) Non – terminating.

 (vi) 2323.52=232.22.52=232.102=23×52×5×102
=11510×102=1151000=0.115

 (vii) Non – terminating.

 (viii) 615=25=410=0.4

  (ix) 3550=35×250×2=70100=0.70

  (x) Non – terminating.

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Q.3     The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form pq, what can you say about the prime factors of q ?
(i) 43.123456789      (ii) 0.120120012000120000…….      (iii) 43.123456789¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Sol.     (i) 43.123456789 is terminating.
So, it represents a rational number.
Thus, 43.123456789 = pq, where q=109.

(ii) 0.12012001200012000… is non – terminating and non-repeating. So, it is irrational.

 (iii) 43.123456789¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is non – terminating but repeating. So, it is rational.
Thus, 43.123456789¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ =pq, where q = 999999999.