10th Class math’s solution Chapter 1.1

10th Class math’s solution Chapter -1

NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers Ex1.1

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 1.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 1.1

Access Answers of Math's NCERT Class 10 Chapter 1 – Real Number Exercise 1.1

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

i. 135 and 225

ii. 196 and 38220

iii. 867 and 255

Solution:

(i) Here, 225 > 125

By Euclid lemma method,

225 = 135 × 1 + 90

Again 135 = 90 × 1 + 45

Again 90 = 45 × 2 + 0

So the remainder is now zero.

Since, the HCF of (135 and 225) is 45.

(ii) Here 38220 > 196

By applying Euclid’s division algorithm,

38220 = 196 × 195 + 0

We have already got the remainder as 0 here.

Therefore, HCF (196, 38220) = 196.

(iii) Here 867 > 255

By Euclid’s division algorithm,

867 = 255 × 3 + 102

Again,255 = 102 × 2 + 51

Again102 = 51 × 2 + 0

So, the remainder is now zero.

Since, HCF (867,255) = 51.

2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Solution:

Let “a” be any positive integer and b = 6.

Then, possible value or r = 0,1,2,3,4,5.

By Euclid lemma method,

a = bq+r,

So, a = 6q+r …………….(1)

Now, put r = 1 in equation (1).

So, a = 6q+1.

Again, put r = 3 in equation (1)

So, a = 6q+3.

Now, put r = 5 in equation (1)

So, a = 6q+5.

If a = 6q, 6q+2, 6q+4, then “a” is an even number and divisible by 2.

Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.

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?

SOLUTIONS:

here, 616>32, therefore,

616 = 32 × 19 + 8

Since, 8 ≠ 0, therefore, taking 32 as new divisor, we have,

32 = 8 × 4 + 0

Now we have got remainder as 0, therefore, HCF (616, 32) = 8.

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

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.

SOLUTION:

Let “a” be any positive integer and b=3.

Then possible value of r= 0, 1, 2.

By Euclid lemma method,

a= bq+r

So, a=3q+r……….(1)

Now put the value of r=0 in equation (1)

a=3q

squaring both side,

a² = (3q)²

a² = 3 ×3q²

So, a²=3m (assume 3q²=m)

Again put the value of r=1 in equation (1)

a=3q+1

squaring both side,

a² = (3q+1)²

a² = (3q + 1)²

a² = (3q)²+1²+2×3q×1

a² = 9q² + 1 +6q

a²= 3(3q²+2q) +1

a²= 3m+1 (assume 3q²+2q=m).

Thus, the square of any positive integer id either of the form 3m or 3m+1.

5. Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8.

Solution:

Let a be any positive integer and b = 3.

Now possible value of r=0,1,2,3,4,5,6,7,8.

By Euclid’s division algorithm,

Then, a=bq+r

So, a=9q+r……..(1)

Therefore, putting the value of r, in equation (1)

a= 3q

a = 3q + 1

a = 3q + 2

Case (i): When r = 0, then,

a³= (3q)³ = 27q³= 9(3q³)= 9m; where m = 3q³

Case (ii): When r = 1, then,

a³ = (3q+1)³ = (3q)³+1³+3×3q×1(3q+1) = 27q³+1+27q²+9q

a³= 9(3q³+3q²+q)+1

Putting = m, we get,

Putting (3q³+3q²⁺q) = m, we get ,

a³ = 9m+1

Case (iii): When r = 2, then,

a³ = (3q+2)³= (3q)³+2³+3×3q×2(3q+2) = 27q³+54q²+36q+8

a³=9(3q³+6q²+4q)+8

Putting (3q³+6q²+4q) = m, we get ,

a³ = 9m+8

Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

10 class maths solution chapter1.2

https://rjeduc.blogspot.com/2021/10/10th-class-maths-solution-chapter-12.html

10 class maths solution chapter1.3

https://rjeduc.blogspot.com/2021/10/10thclass-maths-solution-chapter-1_23.html

10 class maths solution chapter1.4

https://rjeduc.blogspot.com/2021/10/10th-class-maths-solution-chapter-14.html

Class 10 Maths Real Numbers

Rational numbers and irrational numbers are taken together form the set of real numbers. The set of real numbers is denoted by R. Thus every real number is either a rational number or an irrational number. In either case, it has a non–terminating decimal representation. In the case of rational numbers, the decimal representation is repeating (including repeating zeroes) and if the decimal representation is non–repeating, it is an irrational number. For every real number, there corresponds a unique point on the number line ‘l’ or we may say that every point on the line ‘l’ corresponds to a real number (rational or irrational).

From the above discussion we may conclude that:
To every real number there corresponds a unique point on the number line and conversely, to every point on the number line there corresponds a real number. Thus we see that there is one–to–one correspondence between the real numbers and points on the number line ‘l’, that is why the number line is called the ‘real number line’.

rjeduc.com