CBSE Class 10 Mathematics Number Systems Assignment Set B

 

More question-

 

1. If the H C F of 657 and 963 is expressible in the form of 657x + 963x - 15 find x.

(Ans:x=22)

Ans:       Using Euclid’s Division Lemma

                      a= bq+r , o ≤ r < b

                     963=657×1+306

                     657=306×2+45

                    306=45×6+36

                   45=36×1+9

                   36=9×4+0

            HCF (657, 963) = 9

         now 9 = 657x + 963× (-15)

              657x=9+963×15

                 =9+14445

              657x=14454

           x=14454/657

              x=22

 

2. Express the GCD of 48 and 18 as a linear combination. (Ans: Not unique)

                        A=bq+r, where o ≤ r < b

                        48=18x2+12                     

                       18=12x1+6

                        12=6x2+0

                     HCF (18,48) = 6

                     now 6= 18-12x1

                    6= 18-(48-18x2)

                    6= 18-48x1+18x2

                   6= 18x3-48x1                  

                   6= 18x3+48x(-1)

                   i.e. 6= 18x +48y 

 6= 18×3 +48×(-1)

 

=18×3 +48×(-1) + 18×48-18×48

=18(3+48)+48(-1-18)

=18×51+48×(-19)

6=18x+48y

Hence, x and y are not unique.

 

3. Prove that one of every three consecutive integers is divisible by 3.

Ans:
n,n+1,n+2 be three consecutive positive integers

We know that n is of the form 3q, 3q +1, 3q + 2

So we have the following cases

Case – I when n = 3q

In the this case, n is divisible by 3 but n + 1 and n + 2 are not divisible by 3

Case - II When n = 3q + 1

Sub n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. but n and n+1 are not

divisible by 3

 

Case – III When n = 3q +2

Sub n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. but n and n+1 are not  divisible by 3

Hence one of n, n + 1 and n + 2 is divisible by 3

4. Find the largest possible positive integer that will divide 398, 436, and 542 leaving remainder 7, 11, 15 respectively.
(Ans: 17)

Ans: The required number is the HCF of the numbers

Find the HCF of 391, 425 and 527 by Euclid’s algorithm

 HCF (425, 391) = 17

Now we have to find the HCF of 17 and 527

527 = 17 X 31 +0