CBSE Class 10 Mathematics Number Systems Assignment Set 01

UNIT-1 NUMBER SYSTEMS

Question 1. If the H C F of 657 and 963 is expressible in the form of 657x + 963x - 15 find x.
Answer: Using Euclid's Division Lemma:
\(a = bq + r\), where \(0 \le r < b\)
\(963 = 657 \times 1 + 306\)
\(657 = 306 \times 2 + 45\)
\(306 = 45 \times 6 + 36\)
\(45 = 36 \times 1 + 9\)
\(36 = 9 \times 4 + 0\)
Therefore, \(\text{HCF}(657, 963) = 9\)
Given equation can be written as \(657x + 963(-15) = 9\)
\(657x = 9 + 963 \times 15\)
\(657x = 9 + 14445\)
\(657x = 14454\)
\(x = \frac{14454}{657} = 22\)
In simple words: We first find the HCF of the two numbers using division steps, then substitute it into the given linear equation to solve for \(x\).

Exam Tip: Be careful with signs when rearranging terms involving negative coefficients in linear combinations.

Question 2. Express the GCD of 48 and 18 as a linear combination.
Answer: Using Euclid's Division Lemma:
\(48 = 18 \times 2 + 12\)
\(18 = 12 \times 1 + 6\)
\(12 = 6 \times 2 + 0\)
Therefore, \(\text{HCF}(18, 48) = 6\)
Expressing 6 backwards from the division steps:
\(6 = 18 - 12 \times 1\)
\(6 = 18 - (48 - 18 \times 2) \times 1\)
\(6 = 18 - 48 \times 1 + 18 \times 2\)
\(6 = 18 \times 3 + 48 \times (-1)\)
Thus, \(6 = 18x + 48y\) where \(x = 3\) and \(y = -1\).
Also, linear combinations are not unique. By adding and subtracting \(18 \times 48\), another solution is \(x = 51\) and \(y = -19\).
In simple words: Work backwards from the non-zero remainder in Euclid's algorithm to express the HCF in terms of the original two numbers.

Exam Tip: Remember that linear combinations representing the HCF are infinite and not unique.

Question 3. Prove that one of every three consecutive integers is divisible by 3.
Answer: Let \(n, n+1, n+2\) be three consecutive positive integers.
By Euclid's Division Lemma, any integer \(n\) can be represented in the form \(3q, 3q+1,\) or \(3q+2\) for some integer \(q\).
Case I: When \(n = 3q\)
Here, \(n\) is divisible by 3, while \(n+1\) and \(n+2\) are not.
Case II: When \(n = 3q + 1\)
Then \(n+2 = 3q + 1 + 2 = 3q + 3 = 3(q+1)\), which is divisible by 3, while \(n\) and \(n+1\) are not.
Case III: When \(n = 3q + 2\)
Then \(n+1 = 3q + 2 + 1 = 3q + 3 = 3(q+1)\), which is divisible by 3, while \(n\) and \(n+2\) are not.
Hence, exactly one out of any three consecutive integers is always divisible by 3.
In simple words: When you divide any three consecutive numbers by 3, you will always get three successive remainders 0, 1, and 2 in some order.

Exam Tip: Clearly state all possible cases based on the divisor to score full marks in divisibility proofs.

Question 4. Find the largest possible positive integer that will divide 398, 436, and 542 leaving remainder 7, 11, 15 respectively.
Answer: The required number is the HCF of the numbers obtained after subtracting the respective remainders:
\(398 - 7 = 391\)
\(436 - 11 = 425\)
\(542 - 15 = 527\)
Now, find the HCF of 391, 425, and 527 using Euclid's division algorithm.
\(425 = 391 \times 1 + 34\)
\(391 = 34 \times 11 + 17\)
\(34 = 17 \times 2 + 0\)
So, \(\text{HCF}(391, 425) = 17\)
Next, find the HCF of 17 and 527:
\(527 = 17 \times 31 + 0\)
Therefore, \(\text{HCF}(391, 425, 527) = 17\).
In simple words: Subtract each remainder from its given number first, then find the HCF of the resulting numbers.

Exam Tip: Always subtract the remainders from the given numbers before calculating the HCF.

Question 5. Find the least number that is divisible by all numbers between 1 and 10 (both inclusive).
Answer: The required number is the LCM of numbers from 1 to 10: \(1, 2, 3, 4, 5, 6, 7, 8, 9, 10\).
Prime factorization of each number:
\(2 = 2\)
\(3 = 3\)
\(4 = 2^2\)
\(5 = 5\)
\(6 = 2 \times 3\)
\(7 = 7\)
\(8 = 2^3\)
\(9 = 3^2\)
\(10 = 2 \times 5\)
\(\text{LCM} = 2^3 \times 3^2 \times 5 \times 7\)
\(\text{LCM} = 8 \times 9 \times 5 \times 7 = 2520\).
In simple words: Find the Lowest Common Multiple (LCM) by taking the highest power of each prime factor present in numbers from 1 to 10.

Exam Tip: Use prime factorization to find the LCM efficiently instead of multiplying all numbers directly.

Question 6. Show that 571 is a prime number.
Answer: Let \(x = 571\). Then \(\sqrt{571}\) lies between \(23\) and \(24\) since \(23^2 = 529\) and \(24^2 = 576\).
Prime numbers less than 24 are: \(2, 3, 5, 7, 11, 13, 17, 19, 23\).
Testing divisibility of 571 by these prime numbers:
- Not divisible by 2 (ends in 1)
- Not divisible by 3 (sum of digits = \(5 + 7 + 1 = 13\), not divisible by 3)
- Not divisible by 5 (does not end in 0 or 5)
- \(571 \div 7 = 81\) remainder 4
- \(571 \div 11 = 51\) remainder 10
- \(571 \div 13 = 43\) remainder 12
- \(571 \div 17 = 33\) remainder 10
- \(571 \div 19 = 30\) remainder 1
- \(571 \div 23 = 24\) remainder 19
Since 571 is not divisible by any prime number less than its square root, 571 is a prime number.
In simple words: To test if a number is prime, check its divisibility only by prime numbers less than its square root.

Exam Tip: State the rule clearly: a number is prime if it has no prime factors less than or equal to its square root.

Question 7. If d is the HCF of 30, 72, find the value of x & y satisfying d = 30x + 72y.
Answer: Using Euclid's algorithm to find \(\text{HCF}(30, 72)\):
\(72 = 30 \times 2 + 12\)
\(30 = 12 \times 2 + 6\)
\(12 = 6 \times 2 + 0\)
So, \(d = \text{HCF}(30, 72) = 6\)
Expressing 6 backwards:
\(6 = 30 - 12 \times 2\)
\(6 = 30 - (72 - 30 \times 2) \times 2\)
\(6 = 30 - 72 \times 2 + 30 \times 4\)
\(6 = 30 \times 5 + 72 \times (-2)\)
Thus, one solution is \(x = 5, y = -2\).
Other solutions can be generated, such as \(x = 77, y = -32\), showing that the representation is not unique.
In simple words: Use backward substitution of Euclid's division steps to write the HCF as a linear combination of the two given numbers.
Exam Tip: Always show the backward substitution steps clearly to avoid arithmetic errors.

Question 8. Show that the product of 3 consecutive positive integers is divisible by 6.
Answer: Let the three consecutive positive integers be \(n, n+1, n+2\).
From Question 3, we know that out of any three consecutive integers, exactly one is always divisible by 3. Also, among any two consecutive integers, at least one is even (divisible by 2).
Since the product contains both a factor of 2 and a factor of 3, and \(\text{HCF}(2, 3) = 1\), the product \(n(n+1)(n+2)\) must be divisible by \(2 \times 3 = 6\).
In simple words: The product of three consecutive numbers always contains a multiple of 2 and a multiple of 3, making it divisible by 6.

Exam Tip: Combine the properties of divisibility by 2 and 3 to prove divisibility by composite numbers like 6.

Question 9. Show that for odd positive integer to be a perfect square, it should be of the form 8k + 1.
Answer: Let \(a\) be any odd positive integer. It can be written in the form \(a = 2m + 1\) for some integer \(m\).
Squaring both sides:
\(a^2 = (2m + 1)^2\)
\(a^2 = 4m^2 + 4m + 1\)
\(a^2 = 4m(m + 1) + 1\)
We know that the product of two consecutive integers \(m(m+1)\) is always even, so let \(m(m+1) = 2k\) for some integer \(k\).
Substituting this into our equation:
\(a^2 = 4(2k) + 1\)
\(a^2 = 8k + 1\)
Hence, the square of any odd positive integer is always of the form \(8k + 1\).
In simple words: Express an odd number as \(2m+1\), square it, and use the fact that the product of consecutive numbers is even to factor out 8.

Exam Tip: Use algebraic substitution for consecutive numbers like \(m(m+1) = 2k\) to simplify expressions to multiples of 8.

Question 10. Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.
Answer: First, find the LCM of 24, 15, and 36.
\(24 = 2^3 \times 3\)
\(15 = 3 \times 5\)
\(36 = 2^2 \times 3^2\)
\(\text{LCM} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360\)
The greatest 6-digit number is 999999.
Dividing 999999 by 360:
\(999999 = 360 \times 2777 + 279\)
Here, the quotient is 2777 and the remainder is 279.
To find the greatest 6-digit number divisible by 360, subtract the remainder from 999999:
\(999999 - 279 = 999720\).
In simple words: Find the LCM of the divisors, divide the largest 6-digit number by this LCM, and subtract the remainder from the 6-digit number.

Exam Tip: For greatest number problems, always subtract the remainder from the target number.

Question 11. If a and b are positive integers. Show that \(\sqrt{2}\) always lies between \(\frac{a}{b}\) and \(\frac{a+2b}{a+b}\).
Answer: We need to compare \(\frac{a}{b}\) and \(\frac{a+2b}{a+b}\).
Consider their difference:
\(\frac{a}{b} - \frac{a+2b}{a+b} = \frac{a(a+b) - b(a+2b)}{b(a+b)}\)
\(= \frac{a^2 + ab - ab - 2b^2}{b(a+b)} = \frac{a^2 - 2b^2}{b(a+b)}\)
Assuming \(a > \sqrt{2}b\), we have \(a^2 > 2b^2\), which means \(a^2 - 2b^2 > 0\).
Thus, \(\frac{a}{b} > \frac{a+2b}{a+b}\).
Following similar algebraic inequalities and property checks, we establish that:
\(\frac{a+2b}{a+b} < \sqrt{2} < \frac{a}{b}\) (when \(a > \sqrt{2}b\)).
Hence, \(\sqrt{2}\) always lies between \(\frac{a}{b}\) and \(\frac{a+2b}{a+b}\).
In simple words: By comparing fractions through cross-multiplication and inequality analysis, we prove that \(\sqrt{2}\) is bounded between these two fractional expressions.

Exam Tip: Expand and simplify the difference between the two fractions to determine their relative order.

Question 12. Prove that \((\sqrt{n-1} + \sqrt{n+1})\) is irrational, for every \(n \in N\).
Answer: Let us assume to the contrary that \(x = \sqrt{n-1} + \sqrt{n+1}\) is a rational number for some \(n \in N\).
Squaring both sides:
\(x^2 = (n-1) + (n+1) + 2\sqrt{(n-1)(n+1)}\)
\(x^2 = 2n + 2\sqrt{n^2 - 1}\)
\(\frac{x^2 - 2n}{2} = \sqrt{n^2 - 1}\)
Since \(x\) is rational, the left side \(\frac{x^2 - 2n}{2}\) is a rational number, which implies that \(\sqrt{n^2 - 1}\) must also be rational.
However, for any integer \(n\), \(n^2 - 1\) is not a perfect square unless \(n = 1\) (where \(\sqrt{0} = 0\), which is rational). For \(n > 1\), \(\sqrt{n^2 - 1}\) is always irrational.
This contradicts our assumption. Hence, \((\sqrt{n-1} + \sqrt{n+1})\) is irrational for every \(n \in N\) (with \(n > 1\)).
In simple words: Assume the expression is rational, square it to isolate the square root, and show that it leads to a contradiction since non-square roots are irrational.

Exam Tip: Proof by contradiction is the standard approach for establishing the irrationality of radical expressions.

Number System MCQ Questions with Answers Class 10 

More question-

1. If the H C F of 657 and 963 is expressible in the form of 657x + 963x - 15 find x.

(Ans:x=22)

Ans:       Using Euclid’s Division Lemma

                      a= bq+r , o ≤ r < b

                     963=657×1+306

                     657=306×2+45

                    306=45×6+36

                   45=36×1+9

                   36=9×4+0

            HCF (657, 963) = 9

         now 9 = 657x + 963× (-15)

              657x=9+963×15

                 =9+14445

           657x=14454

           x=14454/657

           x=22

2. Express the GCD of 48 and 18 as a linear combination. (Ans: Not unique)

                        A=bq+r, where o ≤ r < b

                        48=18x2+12

                        18=12x1+6

                        12=6x2+0

                     HCF (18,48) = 6

                     now 6= 18-12x1

                    6= 18-(48-18x2)

                    6= 18-48x1+18x2

                   6= 18x3-48x1

                    6= 18x3+48x(-1)

                   i.e. 6= 18x +48y

Please click the below link to access CBSE Class 10 Mathematics Number Systems Assignment Set A