NCERT solutions for Class 10 Maths Chapter 1:Real Numbers EXERCISE 1.1
NCERT solutions for Class 10 Maths Chapter 1 EXERCISE 1.1
In this page we have NCERT solutions for Class 10 Maths Chapter 1:Real Numbers for EXERCISE 1.1. Formula Used Euclid’s Division Lemma
For a and b any two positive integer, we can always find unique integer q and r such that
$a=bq + r \; ,\; \; 0 \leq r < b$
If r =0, then b is divisor of a. HCF (Highest common factor)
HCF of two positive integers can be find using the Euclid’s Division Lemma algorithm
We know that for any two integers a,b. we can write following expression
$a=bq + r \; \; \; 0 \leq r < b$
If r=0 ,then
HCF( a,b) =b
If r≠0 , then
$HCF ( a,b) = HCF ( b,r)$
Again expressing the integer b,r in Euclid’s Division Lemma, we get
$ b=pr + r_1$
$HCF ( b,r)=HCF ( r,r_1)$
Similarly successive Euclid ‘s division can be written until we get the remainder zero, the divisor at that point is called the HCF of the a and b
This exercise contains Questions on finding HCF using Euclid Division Lemma
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Question 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)We should always start with Larger number
So in this case the larger number is 225
Now we can write 135 and 225 in Euclid division algorithm
$225=135 \times 1 +90$
Now $HCF (225,135) = HCF ( 135,90)$
Again writing 135,90 is Euclid division formula
$135=90 \times 1 +45$
Now $HCF (135,90)= HCF (90,45)$
Again writing 90,45 is Euclid division formula
$90=45 \times 2 +0$
Now r =0, 45 is HCF (90,45)
45 is HCF ( 225,135)
(ii) We should always start with Larger number
So in this case the larger number is 38220
Now we can write 38220 and 196 in Euclid division algorithm
$38220=196 \times 195 +0$
Now r =0, 196 is HCF (196,38220)
(iii) We should always start with Larger number
So in this case the larger number is 867
Now we can write 867 and 255 in Euclid division algorithm
$867=255 \times 3 +102$
Now $HCF (867,255) = HCF ( 255,102)$
Again writing 255,102 is Euclid division formula
$255=102 \times 2 +51$
Now $HCF (255,102)= HCF (102,51)$
Again writing 102,51 is Euclid division formula
$102=51 \times 2 +0$
Now r =0, 51 is HCF (102,51)
51 is HCF ( 867,255)
Question 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 By Euclid’s Division Lemma
For a and b any two positive integer, we can always find unique integer q and r such that
$ a=bq + r \; ,\; \; 0 \leq r < b$
Let us take a as any positive integers and b=6 ,Then using Euclid’s algorithm we get,
$a = 6q + r$ as $0 \leq r < 6$,possible reminders are r = 0, 1, 2, 3, 4, 5
So total possible forms will $6q + 0 , 6q + 1 , 6q + 2,6q + 3, 6q + 4, 6q + 5$
$a=6q+ r, 0 \leq r < 6$
$a=6q $, This is a even number
$a=6q+1$ , This is a odd number
$a=6q+2$ ,This is an even number as 6 and 2 are divisible by 2
$a=6q+3$, it is not divisible by 2
$a=6q+4$, it is divisible by 2
$a=6q+5$ , it is not divisible by 2
So $6q,6q+2,6q +4$ are even number
$6q+1,6q+3,6q+5$ are odd numbers
Hence Proved
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
According to the questions, we need to find the maximum number of column
maximum number of column is the HCF of number 32 and 616
So question has reduced to finding the HCF of 32 and 616 using Euclid division algorithm
We should always start with Larger number
So in this case the larger number is 616
Now we can write 616 and 32 in Euclid division algorithm
$616=32 \times 19 +8$
Now HCF of numbers (616,32) = HCF ( 32,8)
Again writing 32,8 is Euclid division formula
$32=8 \times 4+0$
As r=0, 8 is the HCF of 616 and 32
Question 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.
[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]
Solution Euclid’s Division Lemma
For a and b any two positive integer, we can always find unique integer q and r such that
$ a=bq + r \; ,\; 0 \leq r < b$
Let us take a as any positive integers and b=3 ,Then using Euclid’s algorithm we get,
$a = 3q + r$ as $0 \leq r < 3$,possible reminders are r = 0, 1, 2
So total possible forms will $3q + 0 , 3q + 1 , 3q + 2$
$a=3q \;,\; a^2=9q^2 =3m \; \\ where \; m=3q^2$
$a=3q+1\;,\; a^2=9q^2 +6q+1 \\=3(3q^2+2q) +1 =3m+1 \; where \; m=3q^2+2q$
$a=3q+2\;,\; a^2=9q^2 +12q+4 \\=3(3q^2+4q+1) +1 =3m+1 \; where \; m=3q^2+4q+1$
Question 5.
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8
Solution Euclid’s Division Lemma
For a and b any two positive integer, we can always find unique integer q and r such that
$ a=bq + r \; ,\; 0 \leq r < b$
Let us take a as any positive integers and b=3 ,Then using Euclid’s algorithm we get,
$a = 3q + r$ as $0 \leq r < 3$,possible reminders are r = 0, 1, 2
So total possible forms will $3q + 0 , 3q + 1 , 3q + 2$
$a=3q\;,\; a^3=27q^3 =9m \; \\ where \; m=3q^3$
$a=3q+1\;,\; a^3=27q^3 +27q^2 +9q +1 \\=9(3q^3+3q+q) +1 =9m+1 \; \\where \; m=3q^3+3q+q$
$a=3q+2\;,\; a^3=27q^3 +54q^2 +36q +8 \\=9(3q^3+6q+4q) +8 =9m+8 \; \\ where \; m=3q^3+6q+4q$ Download Real Numbers Exercise 1.1 as pdf
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