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NCERT exemplar Class 10 Maths solutions chapter 1 is very important for all students who want to pursue mathematics in higher classes. This chapter provides a detailed understanding of real numbers. NCERT exemplar Class 10 Maths chapter 1 solutions can help the students build a strong base which will provide an extra edge while solving questions of NCERT Class 10 Maths.
These Class 10 Maths NCERT exemplar chapter 1 solutions are complete and descriptive which makes the learning of concepts of real numbers extremely effective. CBSE Syllabus for Class 10 has been kept in vision while preparing the NCERT exemplar Class 10 Maths solutions chapter 1.
Question:1
Choose the correct answer from the given four options in the following questions For some integer m, every even integer is of the form
(A) m
(B) m + 1
(C) 2m
(D) 2m + 1
Answer:
If the value of m is odd number than option A give odd number. Therefore, option A is not the answer of given question. When value of m is even, then option B always given odd number.
For example let m = 2, then m + 1 = 2 + 1 = 3
Which is odd hence B is not the answer.
When value of m is odd/even then option C gives always even number.
For example let m = 2, then 2 m = 4;
let m = 3, then 2 m = 6
Hence C is correct option.
Option D always gives odd integer for every value of m, hence option D is not the answer.
Question:2
For some integer q, every odd integer is of the form
(A) q
(B) q + 1
(C) 2q
(d) 2q + 1
Answer:
If the value of q is even number than option A gives even number. Therefore, option A is not the answer of given question.
When value of q is odd, then option B always gives even number.
For example let q = 3, then q + 1 = 3 + 1 = 4
Which is even hence B is not the answer.
When value of q is odd/even then option C gives always even number.
For example let m = 2, then 2 m = 4;
let m = 3, then 2 m = 6
Hence C is incorrect option.
Option D always gives odd integer for every value of q, hence option D is the answer.
For example let q = 2, then 2q + 1 = 2 × 2 + 1 5
Question:3
n^{2} – 1 divisible by 8, if n is
(A) an integer
(B) a natural number
(C) an odd integer
(D) an even integer
Answer:
let n = 2 then n^{2} – 1 = 4 – 1 = 3, which is not divisible by 8
When n is a natural number,
let n = 4 then n^{2 }– 1 = 16 – 1 = 15, which is not divisible by 8
When n is an odd integer then n^{2} – 1 is always divisible by 8,
let n = 3, n^{2} – 1 = 9 – 1 = 8 which is divisible by 8.
Proof: let n = 2q + 1
n^{2}– 1 = (2q + 1)^{2} – 1
= 4q^{2} + 1 + 4q – 1
= 4q (q + 1)
Product of two consecutive number is divisible by 2, hence number n^{2} – 1 is divisible by 8.
When n is an even integer then n^{2} – 1 is not divisible by 8.
let n = 2,
n^{2} – 1 = 4 – 1 = 3
Question:4
If the HCF of 65 and 117 is expressible in the form 65 m – 117, then the value of m is
(A) 4 (B) 2 (C) 1 (D) 3
Answer:
Answer. [B]
Solution. First of all let us find the HCF of 65 and 117
65 = 1 × 5 × 13 117 = 1 × 3 × 13 × 3
HCF = 13 ….(1)
According to question HCF = 65 m – 117
Using (1)
13 = 65 m – 117
13 + 117 = 65 m
m = 2
Therefore, option (b) is correct.
Question:5
The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is
(A) 13 (B) 65 (C) 875 (D) 1750
Answer:
Answer. [A]
Solution. First of all subtract remainder from the given number.
(70 – 5) = 65
(125 – 8) = 117
Require number is HCF of 65 and 117
65 = 5 × 13 × 1
117 = 3 × 3 × 13 × 1
HCF = 13
Hence 13 is the largest number which is divides by 70 and 120 & leaves remainder 5 and 8.
Therefore, option (A) is correct.
Question:6
If two positive integers a and b are written as a = x^{3} y^{2} and b = xy^{3}; x, y are prime numbers then HCF (a, b)
(A) xy (B) xy^{2} (C) x^{3}y^{3} (D) x^{2}y^{2}
Answer:
For HCF, we have to take all common prime numbers from both the numbers.
Hence,
Hence option B is correct.
Question:7
If two positive integers p and q can be expressed as p = ab^{2} and q = a^{3}b; a, b being prime numbers then LCM (p, q) is
(A) ab (B) a^{2}b^{2} (C) a^{3}b^{2} (D) a^{3}b^{3}
Answer:
Answer. [C]
Solution.
For LCM, we have to take all the primes from prime factorization of two number.
The common prime in both numbers has to be taken only once.
Therefore,
Hence option C is correct.
Question:8
The product of a non-zero rational and an irrational number is
(A) always irrational
(B) always rational
(C) Rational or irrational
(D) one
Answer:
Rational number is the number, which can be written:
in the form of p/q where q not equal to zero. (Here P and Q has to be an integer)
Irrational numbers cannot be written in the form of p/q.
Consider a rational number = 1/2
Therefore, option (A) is correct.
Question:9
The least number that is divisible by all the numbers from 1 to 10 is:
(A) 10 (B) 504 (C) 100 (D) 2520
Answer:
Option A is not the answer of the question, because 10 is not divisible by 3, 4, 6, 7, 8, 9
Option B is also not the answer of the question, because 504 is not divisible by 5 and 10
Option (C) is also not the answer of the question because 100 is not divisible by 3, 6, 7, 8 and 9
Option (D) is the answer of the question because 2520 is divisible by all the numbers from 1 to 10.
Calculation:
Prime factorization of all the numbers:
1 = 1
2 = 2
3 =3
4 =22
5 =5
6 = 32
7 = 7
8 = 222
+9 = 33
10 = 25
For LCM, we have to take all the primes from prime factorization of two number.
The common prime in both numbers has to be taken only once.
Therefore,
LCM = 22233357 =2520
Therefore, option (D) is correct.
Question:10
The decimal expansion of the rational numbers will terminate after:
(A) one decimal place
(B) two decimal places
(C) there decimal places
(D) four decimal places
Answer:
Answer. [D]
Hence the given rational number will terminate after four decimal places. (D) four decimal places
Question:1
Answer:
The above statement is not true in all the conditions. For example, if we take the value of q as – 1. Then 4q + 2 = – 4 + 2 = – 2 which is not a positive integer.
And moreover, if q is positive then 4q + 2 will be always even. Therefore, we can never write an odd positive number in this form. Hence given statement is false
Question:2
Answer:
The above statement is true in all the conditions. Consider a number P = q (q + 1)
If q is odd then q+ 1 will be even, hence P will be divisible by 2.
If q is even then q+ 1 will be odd, but P will be divisible by 2.
For example:
let us take numbers 2 and 3:
product is 2 x 3 = 6 which is divisible by 2
Hence given statement is true
Question:3
Answer:
The given statement is true in all the conditions.
Let three consecutive positive integers be, x, x + 1 and x + 2.
Divisibility by 3: Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
n = 3p or 3p + 1 or 3p + 2, where p is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 3.
x (x + 1) (x + 2) is divisible by 3.
Divisibility by 2: Whenever a number is divided 2, the remainder obtained is 0 or 1.
n = 2q or 2q + 1, where q is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 2.
x (x + 1) (x + 2) is divisible by 2.
Since, x (x + 1) (x + 2) is divisible by 2 and 3.
x (x + 1) (x + 2) is divisible by 6.
Question:4
Answer:
Any positive integer can be written in the form of 3q or 3q+1 or 3q + 2.
Since square of 3q = 9q² = 3(3q²) = 3m (where 3q² = m)
Square of (3q+1) = (3q+1) ² = 9q²+1+6q = 3(3q²+2q) +1 = 3m + 1
(where, m = 3q² + 2q)
Square of (3q+2) = (3q+2) ² = 9q²+4+12q = 3(3q²+4q) +4 = 3m + 4 (where, m = 3q²+2q)
Thus, there is not any square of a positive integer, which can be written in the form of 3m + 2
Hence the given statement is false.
Question:5
Answer:
Let us take the square of (3q + 1)
(3q + 1)2 = (3q) 2 + (1)2 + 2 × 3q × 1
= 9q2 + 1 + 6q
= 9q2 + 6q + 1
= 3(3q2 + 2q) + 1
= 3m + 1
{where m = 3q2 + 2q}
Hence, the square of 3q + 1 (where q is a positive integer) cannot be
written in any form other than 3m + 1
Hence the given statement is false.
Question:6
Answer:
The highest common divisor of any two numbers is called HCF.
Here 75 is the highest common divisor.
We can also find HCF:
Applying Euclid’s division algorithm on 525 and 3000 we get
Hence 75 is the HCF of numbers 525 and 3000
Question:7
Explain why 3 5 7 + 7 is a composite number
Answer:
Composite number – the number which is divisible by a number other than one and itself. Here, 3 x 5 x 7 + 7 = 112 and 112 is divisible by 2, 4, 7, 8, 14, 16, 28 and 56. Hence it is a composite number.
Question:8
Can two numbers have 18 as their HCF and 380 as then LCM? Give reasons.
Answer:
As we know, LCM is a multiple of HCF
has to be an integer.
If the given statement is true:
380 = 18n where n is any integer.
But there is no such type of n exist as we can say that 380 is not divisible by 18.
Hence two numbers, with 18 as their HCF and 380 as their LCM, do not exist.
Question:9
Answer:
We can also write this number as
:
=
=
Here, we have a finite number of digits after the decimal point.
Hence it is terminating decimal expansion.
Question:10
Answer:
The given decimal number is 327.7881.
The q is 10000, therefore its prime factorization, can be written as:
q = 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5
= 24 × 54 = (2 × 5)4
Question:1
Show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q
Answer:
Any positive integer can be written in the form of 4m or 4m + 1 or 4m + 2 or 4m + 3.
Case 1:
A = 4m
(4m)^{2} = 16 m^{2}
=4(4m^{2})
=4q (Here q = 4m^{2})
Case 2:
A = 4m + 1
(4m + 1)^{2}= (4m)^{2} + 1 + 8m
=16m^{2} + 8m + 1
=4(4m^{2} + 2m) + 1
=4q + 1 (here q = 4m^{2} + 2m)
Case 3:
A = 4m + 2
(4m + 2) ^{2}
= 16 m^{2} + 4 + 16 m
=4 × (4m^{2} + 1) + 4 m
= 4q
Here q = (4m^{2} + 1 + 4m)
Case 4:
A = 4m + 3
(4m + 3)2 = 16 m2 + 9 + 24 m
4(4m^{2} + 6m + 2) + 1
4q + 1
here q = (4m^{2} + 6m +2)
Hence square of any positive integer is either of the form 4q or 4q + 1.
Question:2
Show that cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer m.
Answer:
By Euclid’s division –
Any positive integer can be written as:
A = bq + r
Here b = 4
r is remainder when we divide A by 4, therefore:
A = 4q + r …(1)
Case 1:
A = 4q
A3 = (4q)^{3} = 64q^{3} = 4(16q^{3})
A^{3} = 4m
(m = 16 q3)
Case 2:
A = 4q + 1
(4q + 1)^{3} = (4q)^{3} + (1)^{3} + 3(4q)^{2} (1) + 3(4q) (1)^{2}
(Using (a + b)^{3} = a^{3}+ b^{3} + 3a2b + 3ab^{2})
= 64q^{3} + 1 + 48q^{2} + 12q
= 4(16q^{3} + 12q^{2} + 3q) + 1
A^{3} = 4m + 1
(m = 16q^{3} + 12q^{2} + 3q)
Case 3:
A = 4q + 2
= (4q + 2)^{3} = (4q)^{3} + (2)^{3} + 3(4q)^{2} (2) + 3(4q) (2)^{2}
(using (a + b)^{3} = a^{3} + b3 + 3a2b + 3ab^{2})
= 64q^{3} + 8 + 96q^{2} + 48q
= 4(16q^{3} + 2 + 24q^{2}+ 12q)
= 4m
(m = 16q^{3} + 2 + 24q^{2} + 12 q)
Case 4:
A = 4q + 3
A3 = (4q + 3)^{2} = (4q)^{3} + (3)^{3} + 3(4q)^{2} (3) + 3(4q) (3)^{2}
(Using (a + b)^{3} = a^{3} + b^{3} + 3a2b + 3ab^{2})
= 64q^{3} + 27 + 144q^{2} + 108q
= 64q^{3} + 24 + 144 q^{2} + 108q + 3
= 4(16q3 + 6 + 36q2 + 27q) + 3
= 4m + 3
(m = 16q3 + 6 + 36q2 + 27q)
Hence any positive integer’s cube can be written in the form of
4m, 4m + 1, 4m + 3.
Question:3
Answer:
By Euclid’s division –
Any positive integer can be written as:
A = bm + r
Here b = 5
r is remainder when we divide A by 5, therefore:
0 r 5, r = 0, 1, 2, 3.4
A = 5m + r …(1)
Case 1:
A = 5m
(5m)^{2} = 25 m^{2}
= 5(5m^{2})
= 5q
(Here q = 5 m^{2})
Case 2:
A = 5m + 1
(5m + 1)^{2} = 25 m^{2} + 1 + 10 m
25m^{2} + 10 m + 1
5(5m^{2} + 2m) + 1
5q + 1
(Here q = 5m^{2} + 2m)
Similarly, we can verify it for 5m + 2, 5m + 3, 5m + 4
Here, square of any positive integer is in the form of 5q, 5q + 1
Hence, square of any positive integer cannot be of the form 5q + 2 or 5q + 3.
Question:4
Answer:
By Euclid’s division –
Any positive integer can be written as:
A = bq + r
Here b = 6
r is remainder when we divide A by 5, therefore:
0 r 6, r = 0, 1, 2, 3,4,5
A = 6q + r …(1)
Case 1:
A = 6q
A^{2} = (6q)^{2} = 36q^{2} = 6(6q^{2})
A^{2} = 6m
(m = 6q^{2})
Case 2:
A = 6q + 1
A^{2} = (6q + 1)^{2}
A^{2} = 36q^{2} + 1 + 2(6q)
(Using (a + b)^{2} = a^{2} + b^{2} + 2ab)
A^{2} = 36q^{2} + 1 + 12q
A^{2} = 6(6q^{2} + 2q) + 1
A^{2} = 6m + 1
(m = 6q^{2} + 2q)
Similarly, we can verify it for 6q + 2, 6q + 3, 6q + 4, 6q + 5
Now, it is clear that the square of any positive integer cannot be of the form 6m + 2, 6m + 5.
Question:5
Show that the square of any odd integer is of the form 4q + 1 for some integer q.
Answer:
By Euclid’s division –
Any positive integer can be written as:
A = bq + r
Here b = 4
r is remainder when we divide A by 4, therefore:
0 r 4, r = 0, 1, 2, 3
Even: A = 4q and 4q + 2
Odd: A = 4q + 1 and 4q + 3
Case 1:
Now squaring the odd terms, we get
(4q + 1)^{2} = (4q) ^{2} + (1)^{2} + 8q
= 16q2 + 8q + 1
= 4(4q2 + 2q) + 1
= 4m + 1
Here m = 4q^{2} + 2q
Case 2:
(4q + 3)^{2} = (4q)^{2} + (3)^{2} +24q
= 16q^{2} + 8 + 1 + 24q
=16q^{2} + 24q + 8 + 1
=4(4q^{2} + 6q + 2) + 1
=4m + 1
Here m = 4q^{2} + 6q + 2
Hence we can say that square of any odd integer is of the form 4q + 1 for some integer q.
Question:6
If n is an odd integer, then show that n^{2} – 1 is divisible by 8.
Answer:
We can represent any odd positive integer in the form of (4q + 1) or (4q + 3) for some integer q.
Case 1:
If n = 4q + 1
n^{2} – 1 = (4q + 1)^{2} –1
n^{2}– 1 = (4q)^{2} + (1)^{2} + 2(4q) (1) – 1
Using (a + b)^{2} = a^{2} + b^{2} + 2ab
n^{2} – 1 = 16q^{2} + 1 + 8q – 1
n^{2} – 1 = 16q^{2} + 8q = 8q (2q + 1)
Which is divisible by 8.
Case 2:
If n = 4q + 3
n^{2} – 1 = (4q + 3)^{2} – 1
n^{2} – 1 = (4q)^{2} + (3)^{2} + 2(4q) (3) – 1
(Using (a + b)^{2} = a^{2} + b^{2} + 2ab)
n^{2} – 1 =16q^{2} + 9 + 24q –1
n^{2} – 1 = 16q^{2} + 8 + 24q = 8(2q^{2} + 1+3q)
Which is divisible by 8
Hence any odd positive integer n when written in n2 – 1 from is divisible by 8.
Question:7
Answer:
We can represent any positive integer in the form of (2q +1) for any integer q.
Let x = 2q + 1; y = 2p + 1
x^{2} + y^{2} = (2q + 1)^{2} + (2p + 1)^{2}
x^{2} + y^{2} = (2q)^{2} + (1)^{2} + 2(2q) (1) + (2p)^{2} + (1)^{2} + 2(2p) (1)
(Using (a + b) ^{2} = a^{2} + b + 2ab)
x^{2} + y^{2} = 4q^{2} + 1 + 4q + 4p^{2} + 1 + 4p
x^{2} + y^{2} = 2(2q^{2} + 2q + 2p^{2} + 2p + 1)
(2q^{2} + 2q + 2p^{2} + 2p + 1) is odd
Hence, x^{2} + y^{2} is divisible by 2 but not divisible by 4.
Question:8
Use Euclid’s division algorithm to find the HCF of 441, 567, 693
Answer:
First, we find the HCF of 441 and 567 by using Euclid division algorithm.
Here HCF of (567, 41) = 63 Now, we will find the HCF of 63 and 693
693 = 63 11 + 0
Therefore HCF of (63, 693) = 63
Thus HCF of (441, 467, 693) = 63.
Question:9
Answer:
As we know that 1, 2 and 3 are the remainders of given number 1251, 9377 and 15628
After subtracting remainders, we get
1251 – 1 = 1250
9377 – 2 = 9375
15628 – 3 = 15625
Now HCF of 1250, 9375, 15625 can be calculated as:
First, we will calculate the HCF of 9375 and 15625;
HCF (15625, 9375) = 3125 Now we will calculate the HCF of 3125 and 1250.
HCF of 3125 and 1250 is 625. Here HCF of (1250, 9375, 15625) = 625
Hence largest number is 625.
Question:10
Answer:
We will do it by the method of contradiction: We will assume is a rational number. If it leads to some absurd outcome then it is a wrong assumption.
Let is a rational number
Squaring both side.
(Using (a – b)^{2} = a^{2} + b^{2} = 2ab)
LHS is a rational number but RHS is an irrational number.
This is not possible, hence our assumption was wrong.
Hence is a irrational number.
Question:11
Show that 12^{n} cannot end with the digit 0 or 5 for any natural number n.
Answer:
We know that, if any number ends up with digit 0 or 5 then it must be divisible by 5.
Here 12n = (2 × 2 × 3)^{n}
12^{n} = (2^{2} × 3)^{n}
12^{n} = (2^{2n} × 3^{n})
5 is not there, in the prime factorization form.
Hence 12^{n} can’t end with the digit 0 or 5.
Question:12
Answer:
Step-length of first person = 40 cm
Step-length of second person = 42 cm
Step-length of third person = 45 cm
For finding minimum distance which will be multiple of all step-lengths. we have to find the LCM of (40, 42, 45)
Prime factorization forms: 40 = 2 × 2 × 2 × 5 × 1
42 = 2 × 3 × 7 × 1
45 = 5 × 3 × 3 × 1
LCM = 2 × 3 × 5 × 2 × 2 × 7 × 3 = 2520
Hence minimum distance is 2520 cm.
Question:13
Answer:
Here, denominator = 5000
It can be written as:
5000 = 2 × 2 × 2 × 5 × 5 × 5 × 5
Hence the given fraction can be written as:
(On dividing & multiply by 21)
Hence, it is a terminating decimal.
Question:14
Prove that is irrational, where p, q are primes.
Answer:
We will do it by method of contradiction: We will assume is a rational number. If it leads to some absurd outcome then it is wrong assumption.
Let is a rational number
Squaring both sides
(Using (a – b)^{2} = a^{2} + b^{2} – 2ab)
RHS is a rational number but LHS is an irrational number.
This is not possible, hence our assumption was wrong.
Here has to be irrational.
Question:1
Answer:
By Euclid’s division –
Any positive integer can be written as:
A = bq + r
Here b = 6
r is remainder when we divide A by 5, therefore:
0 r 6, r = 0, 1, 2, 3,4,5
A = 6q + r
Here A = 6q + r
A^{3} = (6q + r)^{3} = 216q^{3} + r^{3} + 3.6q.r (6q + r)
[ (a + b)^{3} = a^{3} + b^{3} + 3ab (a + b)]
A^{3} = (216q^{3} + 108q^{2}r + 18qr^{2}) + r^{3} …(1)
Put r = 0, 1, 2, 3, 4, 5
Case 1:
A = 6q
A^{2} = 216q^{3}
A^{3} =6 (36q^{3})
=6 (m )
Case 2:
A = 6q + 1
A^{3} = (216q^{3} + 108q^{2} + 18q) + 1
=6(36q^{3} + 18q^{2} + 3q) + 1
a^{3} = 6m + 1
{m = 36q^{3} + 18q^{2} + 3q}
Case 3:
A = 6q + 2
A^{3} = (216q^{3} + 216q^{3} + 72q) + 8
= (216q^{3} + 216q^{2} + 72q + 6) + 2
= 6m + 2
Case 4:
A = 6q + 3
A^{3}=216q^{3} + 324q^{2} + 162q + 24 + 3
=6(36q^{3} + 54q^{2} + 27q + 4) + 3
= 6m + 3
{m = 36 q^{3} + 54q^{2} + 27q + 4}
Case 5:
A = 6q + 4
A^{3} = (216 q^{3} + 432 q^{2} + 288q) + 64
=6(36q^{3} + 72q^{2} + 48 q) + 60 + 4
=6(36q^{3} + 72q^{2} + 48q + 10) + 4
=6m + 4
{m = 36q^{3} + 72q^{2} + 48q + 10}
Case 6:
A = 6q + 5
A^{3} = (216 q^{3} + 540 q^{2} + 450 q) + 125
= 216 q^{3} + 540 q^{2} + 450 q + 120 + 5
= 6 (36q^{3} + 90q^{2} + 75 q + 20) + 5
= 6m + 5
(where m = 36 q^{3} + 90q^{2} + 75q + 20)
Hence, cube of a positive integer of the form 6q + r, q is a integer can be express in the form 6m + r where r = 0, 1, 2, 3, 4 and 5.
Question:2
Answer:
Any positive integer can be written in form of 3m, 3m + 1 or 3m + 2
Case-1:
n = 3m
3m is divisible by 3
Adding two on both sides
n + 2 = 3m + 2
Here when n + 2 is divided by 3 it leaves remainder 2
Adding four on both sides
n + 4 = 3m + 4
= 3m + 3 + 1
= 3(m + 1) + 1
Here on dividing by 3, it leaves remainder 1
In this case n is divisible by 3 but (n + 2) and (n + 4) are not divisible by 3
Case-2 :
n = 3m + 1
on dividing by 3, it leaves remainder 1
adding two on both sides
n + 2 = 3m + 1 + 2
n + 2 = 3m + 3
= 3(m + 1)
on dividing by 3 it leaves remainder 0
Hence it is divisible by 3
adding four on both sides
x + 4 = 3m + 1 + 4
x + 4 = 3m + 3 + 2
= 3(m + 1) + 2
on dividing by 3, it leaves remainder 2
In this (n + 2) is divisible by 3 but n and (n + 4) are not divisible by 3
Hence through both the cases we conclude that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer.
Similarly, we can show for n = 3m + 2 form.
Question:3
Prove that one of any three consecutive positive integers must be divisible by 3.
Answer:
Let three consecutive integers are n, n + 1, n + 2 when n is a natural number
i.e., n = 1, 2, 3, 4, ….
for n = 1, chosen numbers :(1, 2, 3) and 3 is divisible by 3
for n = 2, chosen numbers :(2, 3,4) and 3 is divisible by 3
for n = 3, chosen numbers :( 3, 4, 5) and 3 is divisible by 3
for n = 4, chosen numbers :(4, 5, 6) and 6 is divisible by 3
for n = 5, (chosen numbers :(5, 6, 7) and 6 is divisible by 3
Proof:
Case 1 : Let n is divisible by 3, it means n can be written as :
n = 3m,
Now, n +1 = 3m + 1 ; it gives remainder 1, when divided by 3
Now, n +2 = 3m + 2; it gives remainder 2, when divided by 3
Case 2: Let n + 1 is divisible by 3, it means n +1 can be written as :
n +1 = 3m,
Now, n = 3m – 1= 3(m – 1) + 2 ; it gives remainder 2, when divided by 3
Now, n +2 = 3m + 1; it gives remainder 1, when divided by 3
Case 3: Let n + 2 is divisible by 3, it means n +2 can be written as :
n +2 = 3m,
Now, n = 3m – 2= 3(m – 1) + 1 ; it gives remainder 1, when divided by 3
Now, n +1 = 3m – 1=3(m – 1) + 2 ; it gives remainder 2, when divided by 3
Question:4
For any positive integer n, prove that n^{3} – n is divisible by 6.
Answer:
Let three consecutive positive integers be, x, x + 1 and x + 2.
Divisibility by 3: Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2
.n = 3p or 3p + 1 or 3p + 2, where p is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 3.
x (x + 1) (x + 2) is divisible by 3.
Divisibility by 2: Whenever a number is divided 2, the remainder obtained is 0 or 1.
n = 2q or 2q + 1, where q is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 2.
x (x + 1) (x + 2) is divisible by 2.
Since, x (x + 1) (x + 2) is divisible by 2 and 3.
x (x + 1) (x + 2) is divisible by 6.
Now the given number is:
P = n^{3} – n
P = n (n^{2} – 1)
P = n (n – 1) (n + 1)
P = (n – 1) n (n + 1)
Therefore, P is the product of three consecutive numbers.
Now P can be written as:
P = x (x + 1) (x + 2) which is is divisible by 6.
Hence P is divisible by 6.
Hence proved
Question:5
Answer:
By Euclid’s division –
Any positive integer can be written as:
n = bm + r
Here b = 5
r is remainder when we divide n by 5, therefore:
0 r 5, r = 0, 1, 2, 3.4
n = 5m + r, therefore n can have values:
n = 5m, 5m + 1, 5m + 2, 5m + 3, 5m + 4.
Here m is natural number
Case 1 : Let n is divisible by 5, it means n can be written as :
n = 5m,
Now, n +4 = 5m + 4;it gives remainder 4, when divided by 5
Now, n + 8 = 5m + 8= 5(m +1) +3; it gives remainder 3, when divided by 5
Now, n + 12 = 5m + 12= 5(m +2) +2; it gives remainder 2, when divided by 5
Now, n + 16 = 5m + 16= 5(m +3) +1; it gives remainder 1, when divided by 5
Case 2: Let n + 4 is divisible by 5, it means n + 4 can be written as :
n + 4 = 5m,
Now, n = 5m – 4 = 5(m – 1) + 1 ; it gives remainder 1, when divided by 5
Now, n + 8 = 5m + 4; it gives remainder 4, when divided by 5
Now, n + 12 = 5m + 8 = 5(m +1) +3; it gives remainder 3, when divided by 5
Now, n + 16 = 5m + 12 = 5(m +2) +2; it gives remainder 2, when divided by 5
Similarly, we can show for other cases.
These Class 10 Maths NCERT exemplar chapter 1 solutions provide a basic knowledge of real numbers which has great importance in higher classes.
The questions based on Real Numbers can be practiced in a better way along with these solutions.
The NCERT exemplar Class 10 Maths chapter 1 solution Real Numbers has a good amount of problems for practice and are sufficient for a student to easily sail through other books such as NCERT Class 10 Maths, RD Sharma Class 10 Maths, RS Aggarwal Class 10 Maths.
NCERT exemplar Class 10 Maths solutions chapter 1 pdf download are available through online tools for the students to access this content in an offline version so that no breaks in continuity are faced while practicing NCERT exemplar Class 10 Maths chapter 1.
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Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
If the sum of digits is divisible by three then the number will be divisible by three.
The chapter on Real Numbers is important for competitive examinations like JEE Advanced. A better understanding of the concepts will help the student solve complex problems in higher classes and competitive examinations.
The chapter on Real Numbers holds around 7-9% of the total marks of the paper. NCERT exemplar Class 10 Maths solutions chapter 1 can help the students learn the concepts and provide ample amount of practice questions.
The chapter of Real Numbers is extremely important for Board examinations as it holds around 7-8% weightage of the whole paper.
Application Date:20 November,2023 - 19 December,2023
Application Date:20 November,2023 - 19 December,2023
The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.
According to CBSE norms, a student must be 14 years old by the end of the year in which the exam will be held in order to sit for the 10th board exam. If your age is greater than 14, however, there are no limits. Therefore, based on your DOB, you will be 12 years and 6 months old by the end of December 2022. After the actual 16 June 2024, you'll be qualified to take your 10th board.
Hello aspirant,
Central Board Of Secondary Education(CBSE) is likely to declare class 10 and 12 terms 2 board result 2022 by July 15. The evaluation process is underway. Students are demanding good results and don't want to lack behind. They are requesting the board to use their best scores in Term 1 and Term 2 exams to prepare for the results.
CBSE concluded board exams 2022 for 10, and 12 on June 15 and May 24. Exams for both classes began on April 26. A total of 35 lakh students including 21 lakh class 10 students and 14 lakh class 12 students appeared in exams and are awaiting their results.
You can look for your results on websites- cbse.gov (//cbse.gov) .in, cbseresults.nic.in
Thank you
Hello SIR CBSE board 9th class admission 2022 Kara Raha hu entrance ki problem hai please show the Entrance for Baal vidya mandir school sambhal
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