KSEEB SSLC Solutions for Class 10 Maths – Real Numbers

KSEEB SSLC Solutions for Class 10 Maths – Real Numbers (English Medium)

Exercise 1.1:

Question 1(i):

Use Euclid’s division algorithm to find the HCF of the following numbers.
65 and 117

Solution :

Question 1(ii):

Use Euclid’s division algorithm to find the HCF of the following numbers.
237 and 81

Solution :

Question 1(iii):

Use Euclid’s division algorithm to find the HCF of the following numbers
55 and 210

Solution :

Question 1(iv):

Use Euclid’s division algorithm to find the HCF of the following numbers
305 and 793

Solution :

Question 2:

Show that any positive even integer is of the form 4q or 4q + 2, where q is a whole number.

Solution :

Let a be an even positive integer.
Apply division algorithm with a and b, where b = 4 is a divisor.
∴ a = ( b × q ) + r
a = (4 × q ) + r
∴ a = 4q + r where 0 ≤ r < 4
when r = 0
a = 4q is an even positive integer
If r = 1,
a = 4q + 1 is not an even positive integer
If r = 2
a = 4q + 2 is an even positive integer
If r = 3
then a = 4q + 3 is not an even positive integer
∴ Any even positive integer is of the form 4q or 4q + 2.

Question 3:

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, but not of the form 3m+2.

Solution :

Let ‘a’ and ‘b’ be two positive integers
∴ a = (b x q) + r 0 ≤ r < b
Let b = 3
∴ a = 3q + r, where 0 ≤ r < 3
∴ r = 0, r = 1 or r = 2

(i) when r = 0,
a= 3q
∴ a² = (3q)²
    a²= 9q²⟹ 3 (3q²) ⟹ 3m, where m = 3q²
   ∴ a² = 3m

(ii) when r = 1
a = 3q + 1
∴ a²= (3q + 1)²
     a²= 9q² + 6q + 1
a²= 3(3q²+ 2q) + 1
∴ a²= 3(m) + 1, where m = 3q² + 2q
a² = 3m + 1

(iii) when r = 2,
Then a = 3q + 2
∴ a² = (3q + 2)²
      a²= 9q² + 12q + 4
a²= 3(3q²+ 4q) + 4
∴ a² = 3m + 4

Square of any positive integer is not of the form 3m + 2.

Question 4:

Prove that the product of three consecutive positive integers is divisible by 6.

Solution :

Let ‘n’ be any positive integer. Any positive integer is of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5.
Let the three consecutive positive integers be n, (n + 1) and (n + 2)
If n = 6q
then n(n + 1) (n + 2) = 6q (6q + 1)(6q + 2)
= 6m (∵ m = q(6q + 1)(6q + 2)
This is divisible by 6
If n = 6q + 1
then n(n + 1) (n + 2) = (6q + 1) (6q + 2)(6q + 3)
= (6q + 1) 2(3q + 1) 3(2q + 1)
= 6 (6q + 1)(3q + 1)(2q + 1)
= 6m, which is divisible by 6
where m = (6q + 1)(3q + 1)(2q + 1)
If n = 6q + 2
then n(n + 1) (n + 2) = (6q + 2) (6q + 3) (6q + 4)
= 2(3q + 1) 3 (2q + 1) 2 (3q + 2)
= 12 (3q + 1) (2q + 1) (3q + 2)
= 12m, which is divisible by 6
where m = (3q + 1) (2q + 1) (3q + 2)
Similarly,
n(n + 1) (n + 2) is divisible by 6 if n = 6q + 3
or n = 6q + 4
or n = 6q + 5
∴ n(n + 1)(n + 2) is divisible by 6.

Question 5:

For any positive integer n, prove that n³-n is divisible by 6.

Solution :

We know that any positive integer is of the form 6q, 6q + 1 or 6q + 2, or 6q + 3 or 6q + 4 or 6q + 5 where q is some integer n³ – n = (n – 1)(n)(n + 1)
(i) When, n  = 6q
then n³ – n  = (n – 1) (n) (n + 1)
= (6q – 1) (6q) (6q + 1)
= 6q(36q² – 1) which is divisible by 6
(ii) When, n = 6q + 1
(n – 1) (n) (n + 1) = (6q) (6q + 1) (6q + 2)
= 6q(6q + 1)(6q + 2), which is divisible by 6
(iii) when n = 6q + 2
(n – 1) (n) (n + 1) = (6q + 1) (6q + 2) (6q + 3)
= (6q + 1) 2 (3q + 1) 3 (2q + 1)
= 6(6q + 1)(3q + 1)(2q + 1), which is divisible by 6
Similerly it is applicable to 6q + 3, 6q + 4 and 6q + 5
∴ n³ – n is divisible by 6

Question 6:

There are 75 roses and 45 lily flowers. These are to be made into bouquets containing both the flowers. All the bouquets should contain the same number of flowers. Find the number of bouquets that can be formed and the number of flowers in them.

Solution :

Question 7:

The length and breadth of a rectangular field is 110m and 30m respectively. Calculate the length of the longest rod which can measure the length and breadth of the field exactly.

Solution :

Exercise 1.2:

Question 1:

Express each number as a product of prime factor
i. 120
ii. 3825
iii. 6762
iv. 32844

Solution :

Question 2:

Solution :

Question 3(i):

Find the L.C.M. and H.C.F. of the following integers by expressing them as product of primes.
12, 15 and 30

Solution :

Question 3(ii):

Find the LCM and HCF of the following integers by expressing them as product of primes.
18,81 and 108

Solution :

Question 4(i):

Find the HCF and LCM of the pairs of integers and verify that LCM (a, b) × HCF (a, b) = a × b.
16 and 80

Solution :

Question 4(ii):

Find the HCF and LCM of the pairs of integers and verify that LCM (a, b) × HCF (a, b) = a × b.
125 and 55

Solution :

Question 5:

If HCF of 52 and 182 is 26, find their LCM?

Solution :

Question 6:

Find the HCF of 105 and 1515 by prime factorisation method and hence find its LCM.

Solution :

Question 7:

Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

Solution :

Question 8:

A rectangular hall is 18m 72cm long and 13m 20cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles.

Solution :

The length and breadth of the hall are 18 m 72 cm and 13 m 20 cm respectively.
The length and breadth of the hall in cm are 1872 cm and 1320 cm respectively.
In order to find the number of square tiles of the same size, we need to find the H.C.F. (1872, 1320).

Question 9:

In a school, the strength in 8^th, 9^th and 10^th standards are respectively 48,42 and 60. Find the least number of books required to be distributed equally among the students of 8^th, 9^th or 10^th Standard.

Solution :

Question 10:

x, y and z start at the same time in the same direction to run around a circular stadium. x completes a round in 126 seconds, y in 154 seconds and z in 231 seconds, all starting at the same point. After what time will they meet again at the starting point? How many rounds would have x, y and z completed by this time?

Solution :

Exercise 1.3:

Question 1:

Solution :

Question 2(i):

Solution :

Question 2(ii):

Solution :

Question 2(iii):

Solution :

Question 2(iv):

Solution :

Question 2(v):

Solution :