RS Aggarwal Solutions Chapter 1 Real Numbers Exercise - 1A Class 10 Maths
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Chapter Name |
RS Aggarwal Chapter 1 Real Numbers Solutions |
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Book Name |
RS Aggarwal Mathematics for Class 10 |
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Other Exercises |
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Related Study |
NCERT Solutions for Class 10 Maths |
Exercise 1A Solutions
1. What do you mean by Euclid’s division algorithm?
Solution
Euclid’s division algorithm states that for any two positive integers a and b, there exit unique integers q and r, such that a = bq + r. where 0 ≤ r ≤ b.
2. A number when divided by 61 gives 27 as quotient and 32 as remainder. Find the number.
Solution
We know, Dividend = Divisor × Quotient + Remainder
Given: Divisor = 61, Quotient = 27, Remainder = 32
Let the Dividend be x.
∴ x = 61×27 + 32
= 1679
Hence, the required number is 1679.
3. By what number should be 1365 be divided to get 31 as quotient and 32 as remainder?
Solution
Given: Dividend = 1365, Quotient = 31, Remainder = 32
Let the divisor be x.
Dividend = Divisor×Quotient + Remainder
1365 = x×31 + 32
⇒ 1365 – 32 = 31 x
⇒ 1333 = 31 x
⇒ x = 1333/31
⇒ x = 43
Hence, 1365 should be divided by 43 to get 31 as quotient and 32 as remainder.
4. Using Euclid’s algorithm, find the HCF of
(i) 405 and 2520
(ii) 504 and 1188
(iii) 960 and 1575
Solution
(i)
On applying Euclid’s algorithm, i.e. dividing 2520 by 405, we get:
Quotient = 6, Remainder = 90
∴ 2520 = 405×6 + 90
Again on applying Euclid’s algorithm, i.e. dividing 405 by 90, we get:
Quotient = 4, Remainder = 45
∴ 405 = 90×4 + 45
Again on applying Euclid’s algorithm, i.e. dividing 90 by 45, we get:
∴ 90 = 45×2 + 0
Hence, the HCF of 2520 and 405 is 45.
(ii)
On applying Euclid’s algorithm, i.e. dividing 1188 by 504, we get:
Quotient = 2, Remainder = 180
∴ 1188 = 504×2 + 180
Again on applying Euclid’s algorithm, i.e. dividing 504 by 180, we get:
Quotient = 2, Remainder = 144
∴ 504 = 180×2 + 144
Again on applying Euclid’s algorithm, i.e. dividing 180 by 144, we get:
Quotient = 1, Remainder = 36
∴ 180 = 144×1 + 36
Again on applying Euclid’s algorithm, i.e. dividing 144 by 36, we get:
∴ 144 = 36×4 + 0
Hence, the HCF of 1188 and 504 is 36.
(iii)
On applying Euclid’s algorithm, i.e. dividing 1575 by 960, we get:
Quotient = 1, Remainder = 615
∴ 1575 = 960×1 + 615
Again on applying Euclid’s algorithm, i.e. dividing 960 by 615, we get:
Quotient = 1, Remainder = 345
∴ 960 = 615×1 + 345
Again on applying Euclid’s algorithm, i.e. dividing 615 by 345, we get:
Quotient = 1, Remainder = 270
∴ 615 = 345×1 + 270
Again on applying Euclid’s algorithm, i.e. dividing 345 by 270, we get:
Quotient = 1, Remainder = 75
∴ 345 = 270 × 1 + 75
Again on applying Euclid’s algorithm, i.e. dividing 270 by 75, we get:
Quotient = 3, Remainder = 45
∴ 270 = 75×3 + 45
Again on applying Euclid’s algorithm, i.e. dividing 75 by 45, we get:
Quotient = 1, Remainder = 30
∴ 75 = 45×1 + 30
Again on applying Euclid’s algorithm, i.e. dividing 45 by 30, we get:
Quotient = 1, Remainder = 15
∴ 45 = 30×1 + 15
Again on applying Euclid’s algorithm, i.e. dividing 30 by 15, we get:
Quotient = 2, Remainder = 0
∴ 30 = 15×2 + 0
Hence, the HCF of 960 and 1575 is 15.
5. Show that every positive integer is either even or odd?
Solution
Let us assume that there exist a smallest positive integer that is neither odd nor even, say n.
Since n is least positive integer which is neither even nor odd, n – 1 must be either odd or even.
Case 1: If n – 1 is even, n – 1 = 2k for some k.
But this implies n = 2k + 1
this implies n is odd.
Case 2: If n – 1 is odd, n – 1 = 2k + 1 for some k.
But this implies n = 2k + 2 (k+1)
this implies n is even.
In both ways we have a contradiction.
Thus, every positive integer is either even or odd.
6. Show that every positive even integer is of the form (6m+1) or (6m+3) or (6m+5) where m is some integer.
Solution
Let n be any arbitrary positive odd integer.
On dividing n by 6, let m be the quotient and r be the remainder. So, by Euclid’s division lemma, we have
n = 6m + r, where 0 ≤ r ˂ 6.
As 0 ≤ r ˂ 6 and r is an integer, r can take values 0, 1, 2, 3, 4, 5.
⇒ n = 6m or n = 6m + 1 or n = 6m + 2 or n = 6m + 3 or n = 6m + 4 or n = 6m + 5
But n ≠ 6m or n ≠ 6m + 2 or n ≠ 6m + 4 (∵ 6m, 6m + 2, 6m + 4 are multiples of 2, so an even integer whereas n is an odd integer)
⇒ n = 6m + 1 or n = 6m + 3 or n = 6m + 5
Thus, any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer.
7. Show that every positive even integer is of the form 4m and that every positive odd integer is of the form 4m + 1 for some integer m.
Solution
Let n be any arbitrary positive odd integer.
On dividing n by 4, let m be the quotient and r be the remainder. So, by Euclid’s division lemma, we have
n = 4m + r, where 0 ≤ r ˂ 4.
As 0 ≤ r ˂ 4 and r is an integer, r can take values 0, 1, 2, 3.
⇒ n = 4m or n = 4m + 1 or n = 4m + 2 or n = 4m + 3
But n ≠ 4m or n ≠ 4m + 2 (∵ 4m, 4m + 2 are multiples of 2, so an even integer whereas n is an odd integer)
⇒ n = 4m + 1 or n = 4m + 3
Thus, any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer.