NCERT Solutions Class 10 Maths Chapter 1

Q.1 Show that the square of any positive integer is of the form 3m or 3m+1 for some integer m.

Ans- Let a be any positive integer. Then it is of the form 3q or 3q+1 or 3q+2.

Case I – When a = 3q, we have
a² = (3q)² = 9q² = 3(3q²) = 3m ; where m = 3q²

Case II – When a = 3q+1, we have
a² = (3q+1)² = 9q² +6q+1
or, a² = 3q(3q+2)+1 = 3m+1; where m = 3q(3q+2)

Case III – When a = 3q+2, we have
a² = (3q+2)² = 9q²+12q+4=9q²+12q+3+1
or, a² = 3(3q²+4q+1)+1 = 3m+1 ; where m = (3q²+4q+1)

Hence, a is either of the form 3m or 3m+1 for some integer m.

Q.2 Prove that n²-n is divisible by 2 for every positive integer n.

Ans- We know that any positive integer is of the form 2q
or 2q+1 for some integer q.

Case I – When n=2q then we have
n²-n =(2q)² – 2q
or, n²-n =4q²-2q
or, n²-n = 2q(2q-1)
or, n²-n = 2r; where r= q(2q-1)
Thus, n²-n is divisible by 2.

Case II – When n=2q+1, we have
n²-n =(2q+1)² – (2q+1)
or, n²-n =(2q+1)(2q+1 –1)
or, n²-n = 2q(2q+1)= 2k ; where k=q(2q+1)
Thus n²-n is divisible by 2.

Hence, n²-n is divisible by 2 for every positive integer n.

Q.3 Show that any positive integer is of the form 3q or 3q+1 or 3q+2 for some integer q.

$\begin{array}{l}\text{Ans- Let\hspace{0.17em}a\hspace{0.17em}be\hspace{0.17em}any\hspace{0.17em}positive\hspace{0.17em}integer\hspace{0.17em}and\hspace{0.17em}b=3,\hspace{0.17em}then there}\\ \text{\hspace{0.17em}exist\hspace{0.17em}integer\hspace{0.17em}q\hspace{0.17em}and\hspace{0.17em}r\hspace{0.17em}such\hspace{0.17em}that}\end{array}$ $\begin{array}{l}\text{a=3q+r,\hspace{0.17em}where\hspace{0.17em}0}\le \text{r\hspace{0.17em}<\hspace{0.17em}3}\mathrm{\dots }\left(\text{1}\right)\left[\text{by\hspace{0.17em}Euclid’s\hspace{0.17em}Division\hspace{0.17em}Lemma}\right]\end{array}$ $\begin{array}{l}\text{now\hspace{0.17em}0}\le \text{r\hspace{0.17em}<\hspace{0.17em}3}\Rightarrow \text{0}\le \text{r\hspace{0.17em}<\hspace{0.17em}2}\end{array}$ $\begin{array}{l}\text{r=0,1,2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \text{r\hspace{0.17em}is\hspace{0.17em}integer}\right]\end{array}$ $\begin{array}{l}\text{put\hspace{0.17em}r=0\hspace{0.17em}in\hspace{0.17em}}\left(\text{1}\right)\text{\hspace{0.17em}we\hspace{0.17em}get\hspace{0.17em}a=3q}\end{array}$ $\begin{array}{l}\text{put\hspace{0.17em}r=1\hspace{0.17em}in\hspace{0.17em}}\left(\text{1}\right)\text{\hspace{0.17em}we\hspace{0.17em}get\hspace{0.17em}a=3q+1}\end{array}$ $\begin{array}{l}\text{put\hspace{0.17em}r=2\hspace{0.17em}in\hspace{0.17em}}\left(\text{1}\right)\text{\hspace{0.17em}we\hspace{0.17em}get\hspace{0.17em}a=3q+2}\end{array}$ $\begin{array}{l}\text{Hence\hspace{0.17em}any\hspace{0.17em}positive\hspace{0.17em}integer\hspace{0.17em}is\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}form\hspace{0.17em}3q}\\ \text{or,\hspace{0.17em}}\end{array}$ $\begin{array}{l}\text{3q+1,\hspace{0.17em}or\hspace{0.17em}3q+2\hspace{0.17em}for\hspace{0.17em}some\hspace{0.17em}integer\hspace{0.17em}q.}\end{array}$

Q.4 Use Euclid’s division algorithm to find the HCF of 210 and 55.

Ans- Clearly, 210 > 55.
On applying Euclid’s Division Lemma to 210 and 55
we get, 210 = 55×3 + 45
Since remainder 45 ≠ 0, therefore on again applying Euclid’s Division Lemma we get,
55 = 45×1 + 10
Since remainder 10 ≠ 0 therefore, on again applying Euclid’s Division Lemma we get,
45 = 10×4 + 5
Since remainder 5 ≠ 0 therefore, on again applying Euclid’s Division Lemma we get,
10 = 5×2 + 0
Since remainder is 0 therefore, HCF = 5

Q.5 Prove that if x and y are odd positive integers then x²+y² is even but not divisible by 4.

$\begin{array}{l}\text{Ans- We\hspace{0.17em}know\hspace{0.17em}that\hspace{0.17em}any\hspace{0.17em}odd\hspace{0.17em}integer\hspace{0.17em}is\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}form}\\ \text{2q+1\hspace{0.17em}for\hspace{0.17em}some\hspace{0.17em}integer\hspace{0.17em}q.}\\ \text{so,\hspace{0.17em}let\hspace{0.17em}x=2m+1\hspace{0.17em}and\hspace{0.17em}y=2n+1\hspace{0.17em}for\hspace{0.17em}some\hspace{0.17em}integers\hspace{0.17em}m\&n.}\\ \therefore {\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{=}{\left(\text{2m+1}\right)}^{\text{2}}\text{+}{\left(\text{2n+1}\right)}^{\text{2}}\\ \Rightarrow {\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}{\text{=4m}}^{\text{2}}{\text{+4m+1+4n}}^{\text{2}}\text{+4n+1}\\ \Rightarrow {\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}{\text{=4m}}^{\text{2}}{\text{+4n}}^{\text{2}}\text{+4m+4n+2}\mathrm{\dots }\left(\text{1}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2}\left({\text{2m}}^{\text{2}}{\text{+2n}}^{\text{2}}\text{+2m+2n+1}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2k}\left(\text{even\hspace{0.17em}integer}\right)\text{\hspace{0.17em}where\hspace{0.17em}k=}\left({\text{2m}}^{\text{2}}{\text{+2n}}^{\text{2}}\text{+2m+2n+1}\right)\\ {\text{Hence\hspace{0.17em}x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{\hspace{0.17em}is\hspace{0.17em}an\hspace{0.17em}even\hspace{0.17em}integer.}\\ {\text{Now\hspace{0.17em}we\hspace{0.17em}will\hspace{0.17em}prove\hspace{0.17em}x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{\hspace{0.17em}is\hspace{0.17em}an\hspace{0.17em}even\hspace{0.17em}integer.}\\ \text{From\hspace{0.17em}equation\hspace{0.17em}}\left(\text{1}\right)\text{\hspace{0.17em}we\hspace{0.17em}get,}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{=4}\left({\text{m}}^{\text{2}}{\text{+n}}^{\text{2}}\text{+m+n}\right)\text{+2}\\ \Rightarrow {\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{=4q+2\hspace{0.17em}where\hspace{0.17em}q=}\left({\text{m}}^{\text{2}}{\text{+n}}^{\text{2}}\text{+m+n}\right)\\ \text{Which\hspace{0.17em}leaves\hspace{0.17em}remainder\hspace{0.17em}2\hspace{0.17em}when\hspace{0.17em}divided\hspace{0.17em}by\hspace{0.17em}4}\\ \Rightarrow {\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}4.}\end{array}$

Q.6 Prove that one of every three consecutive positive integers is divisible by 3.

Let n, n+1, n+2 be three consecutive positive integers where n can take the form 3q, 3q+1 or 3q+2.

Case I
when n = 3q
Then n is divisible by 3
but neither n+1 nor n+2 is divisible by 3.

Case II
when n = 3q+1
Then n is not divisible by 3.
n+1 = 3q+1+1 = 3q+2,
which is not divisible by 3.
n+2 = 3q+1+2 = 3q+3 = 3(q+1),
which is divisible by3.

Case III
when n = 3q+2
Then n is not divisible by 3.
n+1 = 3q+2+1 =3q+3 = 3(q+1),
which is divisible by 3.
n+2 = 3q+2+2 = 3q+4,
which is not divisible by 3.

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

Q.7 Prove that if a and b are positive integers such that a = bq + r, then every common divisor of a and b is a common divisor of b and r.

Proof: Let c be a common divisor of a and b.
Therefore,
c divides a ⇒ a = cm …(1) for some integer m
& c divides b ⇒ b = cn …(2) for some integer n
Now, a = bq + r given
r = a – bq
r = cm – cnq [using (1) & (2)]
r = c(m – nq)
⇒ c divides r …(3)
from (2) & (3) c divides b and r
⇒ c is common divisor of b and r
Hence, every common divisor of a
and b is a common divisor of b and r.

Q.8 Prove that √5 is irrational number.

Let √5 be a rational number.

Then √5=p/q (where p and q are co-prime numbers and q 0)
Squaring both the sides we get
5=p²/q²
⇒ p²=5q² …(1)
⇒ 5 divides p²
⇒ 5 divides p….(2)
Let p=5r (where r is an integer)
Squaring both the sides, we get
p²=25r² …(3)
From equation (1) & (3), we get
5q²=25r²
⇒ q²=5r²
⇒ 5 divides q²
⇒ 5 divides q…(4)
From equation (2) & (4) we get
HCF of p and q is 5, but co-prime has only 1 as HCF.
This is a contradiction.
Thus, our assumption was wrong.
Therefore, √5 is irrational number.

Q.9 Prove that √3 is irrational number.

Let √3 be a rational number.
Then √3=p/q (where p and q are co-prime numbers and q ≠ 0)
By squaring both the sides, we get
3=p²/q²
⇒ p²=3q² …(1)
⇒ 3 divides p²
⇒ 3 divides p….(2)
Let p=3r (where r is an integer)
Squaring both the sides, we get
p²=9r² …(3)
From equation (1) &(3) , we get
3q²=9r²
⇒ q²=3r²
⇒ 3 divides q²
3 divides q….(4)

From equation (2) & (4) we get
The HCF of p and q is 3, but co-prime has only 1 as HCF.
Hence, this is a contradiction.
Thus, our assumption was wrong.
Therefore, √3 is irrational number.

Q.10 In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required if in each room same numbers of participants are to be seated and all of them being in the same subject.

The number of rooms will be minimum if each room accommodates maximum number of participants. Therefore, the number of participants in each room must be the HCF of 60, 84 and 108.

60 = 2x2x3x5 = 2²x3¹x5

84 = 2x2x3x7 = 2²x3x7

108 = 2x2x3x3x3 = 2²x3³

HCF of 60, 84 and 108 = 2²X3 = 4X3 = 12

Therefore, in each room 12 participants can be seated.

Number of rooms required=Total no. of participants/12

= (60+84+108)/12

=252/12

=21

Express the positive integers 144 & 180 as the product of its prime factors.

$\begin{array}{l}2\overline{)144}\\ 2\overline{)72}\\ 2\overline{)36}\\ 2\overline{)18}\\ 2\overline{)9}\\ 3\\ 144=2\times 2\times 2\times 2\times 3\times 3\\ =2^4\times 3^2\\ 2\overline{)180}\\ 2\overline{)90}\\ 2\overline{)45}\\ 2\overline{)15}\\ 5\\ 180=2\times 2\times 3\times 3\times 5\\ =2^2\times 3^2\times 5\end{array}$

Q.12 Two tankers contain 2340 litres and 3825 litres of petrol respectively. Find the maximum capacity of the container that can measure the petrol of either tanker in exact number of times.

Clearly, the maximum capacity of the container that can measure the capacity of either tanker is the

HCF of 2340 and 3825.

HCF is common prime factors with smallest exponents

2340 = 2×2×3×3×5×13 = 2²×3²×5×13

3825 = 3×3×5×5×17 = 3²×5²×17

HCF = 3²×5
= 9×5
= 45
Hence, capacity of the container must be 45 litres.