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Complete 10th Class Maths Solution
- NCERT Solutions For Class 10 Maths – Chapter 1 Exercise 1.2 Real Number
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CHAPTER 1 – REAL NUMBER – EXERCISE 1.1
Q.1) Use Euclid’s division algorithm to find the HCF of: (1.1) 135 and 225
Solution:
135 and 225 Since 225>135 ,
Therefore, by Euclid’s division algorithm 225= 135X1+90(since remainder 90 is not equal to 0 and we apply algorithm until we get remainder 0)
135= 90×1+45
90 = 45×2+0 (since remainder is now 0 and the divisor is 45).
Therefore the HCF of 135 and 225 is 45.
(1.2) 196 and 38220
Solution: Since 38220>196 ,
Therefore, by Euclid’s division algorithm 38220= 196×195+0 (since remainder is 0 and the divisor is 196) Therefore the HCF of 196 and 38220 is 196.
(1.3) 867 and 225
solution: 867 and 225 since 867>255 ,
Therefore, by Euclid’s division algorithm 867= 255X3+192 (since remainder 192 is not equal to 0 and we apply algorithm until we get remainder 0)
255= 192×1+33
192= 33×5+27
33= 27×1+6
27= 6×4+3 6= 3×2+0 (since remainder is now 0 and the divisor is 3)
Therefore the HCF of 867 and 225 is 3.
Q2). 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 is a positive odd integer and b= 6.
Then, by Euclid’s algorithm
a = 6q + r, for some integer q>=0,
Since 0<=r<6, the possible remainders are 0, 1, 2, 3, 4,5 .
That is, a can be 6q, 6q+1, 6q+2, 6q+3, 6q+4, 6q+5, where q is quotient.
However, since a is odd, a cannot be 6q, 6q + 2, 6q+4 (since they are even and divisible by 2).
Therefore, any positive odd integer is of the form 6q+1, or 6q+3, or 6q+5.
Q3). 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: Given:-number of army contingent of members=616 number of army band of member =32.
If the two groups are to march in the same number of columns.we have to find the HCFof (616,32) By using Euclid’s division algorithm
since 616>32 616=32×19+8 (since remainder 8=! 0 and we apply algorithm until we get remainder 0) 32=8×4+0 (since remainder is now 0 and the divisor is 8)
The HCF of 616 and 32 is 8 Thereforr, the maximum number of columns in which they can march is 8.
Q4). 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 x is a positive odd integer and y= 3.
Then, by using Euclid’s algorithm, x=3q+r for some integer q>=0
Since 0<=r<3, the possible remainders are 0, 1, 2
Therefore x=3q, 3q+1, 3q+2
according to question, by squaring both sides
x²= (3q)²=9q²=3X3q² let m= 3q²
Therefore x²= 3m …………….eq 1
x²=(3q+1)²=(3q)²+1²+2X3qX1 =9q²+1+6q =3(3q²+2q)+1 substitute, m= 3q²+2q x²=3m+1 ……………………eq 2
x²=(3q+2)² = (3q)²+2²+2X3qX2 =9q²+4+12q =3(3q²+4q+1)+1 again, substitute, m=3q²+4q+1 X²=3m+1……eq 3
Therefore, from equation 1,2 and 3 we can say that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Q5). 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: let x is a positive odd integer and y= 3.
Then, by using Euclid’s algorithm, x=3q+r for some integer q>=0
Since 0<=r<3, the possible remainders are 0, 1, 2
Therefore x=3q, 3q+1, 3q+2 according to question, by taking cube of both sides of all above
x³=(3q)³=27q³=9X(3q³) =9m; where m= 3q³ x³=9m ………………………………….eq 1
x³=(3q+1)³=(3q)³+1³+3X3qX1(3q+1) =27q³+1+27q²+9q =9(3q³+3q²+q)+1 take m =3q³+3q²+q x³=9m+1………eq 2
x³=(3q+2)³=(3q)³+2³+3X3qX2(3q+2) =27q³+54q²+36q+8 x³=9(3q³+6q²+4q)+8 ( by taking 9 common) take m =3q³+6q²+4q
x³= 9m+8 ………………………….eq 3
Therefore, from equation 1,2 and 3 we can say that, the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
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