NCERT Solution for Class 10 Mathematics Chapter 1 - Real Numbers 

Exercise 1.1

QUESTION-1

Use Euclid’s division algorithm to find the HCF of:

1. 135 and 225
2. 196 and 38220
3. 867 and 225

Solution:

1) We start with the larger number i.e 225

By Euclid’s division algorithm,we have

225=1×135+90

135=1×90+45

90=2×45+0

Hence, HCF(225,135)=HCF(135,90)=HCF(90,45)=45

Therefore,the HCF of 135 and 225 is 45

2) We start with the larger number i.e 38220

By Euclid’s division algorithm,we have

38220=196×195+0

196=196×1+0

Hence , HCF(196,38220)=196

Therefore,the HCF of 196 and 38220 is 196

3) We start with the larger number i.e 867

By Euclid’s division algorithm,we have

867=225×3+102

225=102×2+51

102=51×2+0

Hence, HCF(867,225)=HCF(225,102)=HCF(102,51)=51

Therefore,the HCF of   867 and 225 is 51

Question 2:

Show that any positive odd integer is of form 6q+1,6q+3 & 6q+5

where q is some integer.

Solution :

Using Euclid’s division algorithm,we have

x=bq+r {0≤r<b}…..(1)

Substituting b=6in equation(1)

So, x=6q+r,where r=0,1,2,3,4,5

If r=0, x=6q+0(divisible by 2)…..even

r=1, x=6q+1(not divisible by 2)…..odd

r=2, x=6q+2(divisible by 2)…..even

r=3, x=6q+3(not divisible by 2)…..odd

r=4, x=6q+4(divisible by 2)…..even

r=5, x=6q+5(not divisible by 2)…..odd

Therefore,the number 6q,6q+1,6q+2,6q+3,6q+4,6q+5are either even or odd.Hence ,any positive odd integer is of the form 6q+1,6q+3 & 6q+5Where q is some integer.

Question 3:

An army contingent of 616 members is to march behind an army band of 32 members in a 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?

Solution :

The maximum number of columns would be the HCF of (616,32)

We can find the HCF of 616 and 32 by using the Euclid Division Algorithm.

Therefore,

616=19×32+8

32=4×8+0

8=8×1+0

So, HCF of 6161 and 32 is 8

Hence, the maximum number of columns in which they can march is 8.

Question 4:

Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m, 3m+1 for some integer m.

Solution :

According to Euclid Algorithm ,

We have x=bq+r …..(1)

Substituting , b=3

We get,x=3q+r (where,0≤r<3

i.e r=0,1,2 )

When ,r=0 ,x=3q…….(A)

r=1 ,x=3q+1…(B)

r=2 ,x=3q+2…(C)

So, Squaring eqn(A),(B) and (C)

We get,

From eqn (A),

x2=9q2 x2=3×3q2 x2=3×m (where,m=3q2)

From eqn (B),

x2=(3q+1)2 =9q2+1+6q =9q2+6q+1 =3(3q2+2q)+1 =3m+1 (where,m=3q2+2q)

From eqn (C),

x2=(3q+2)2 =9q2+4+12q =9q2+12q+3+1 =3(3q2+4q+1)+1 =3m+1 (where,m=3q2+4q+1)

 Hence, any positive integer is either of the form 3m,3m+1 for some integer m.

Exercise 4.4

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