CBSE Class 10 Mathematics Question Bank Set 04

Chapter (1) : Real Numbers

Question 1. By Euclid division Algorithm find the H.C.F. of (i) 44 and 128 (ii) 4052 and 12576
Answer: (i) For 44 and 128:
\(128 = 44 \times 2 + 40\)
\(44 = 40 \times 1 + 4\)
\(40 = 4 \times 10 + 0\)
Thus, the H.C.F. of 44 and 128 is 4.
(ii) For 4052 and 12576:
\(12576 = 4052 \times 3 + 420\)
\(4052 = 420 \times 9 + 272\)
\(420 = 272 \times 1 + 148\)
\(272 = 148 \times 1 + 124\)
\(148 = 124 \times 1 + 24\)
\(124 = 24 \times 5 + 4\)
\(24 = 4 \times 6 + 0\)
Thus, the H.C.F. of 4052 and 12576 is 4.
In simple words: Euclid's division algorithm uses successive division until the remainder becomes zero, and the last non-zero divisor is the H.C.F.

Exam Tip: Always write each step clearly in the form \(a = bq + r\) where \(0 \le r < b\) to secure full marks.

Question 2. Using fundamental theorem of Arithmetic, find the H.C.F. and L.C.M. of :- (i) 570 and 1425 (ii) 30, 72 and 432
Answer: (i) Prime factorization of 570 and 1425:
\(570 = 2 \times 3 \times 5 \times 19\)
\(1425 = 3 \times 5^2 \times 19\)
H.C.F. = product of smallest power of each common prime factor = \(3^1 \times 5^1 \times 19^1 = 285\)
L.C.M. = product of greatest power of each prime factor = \(2^1 \times 3^1 \times 5^2 \times 19^1 = 2850\)
(ii) Prime factorization of 30, 72, and 432:
\(30 = 2 \times 3 \times 5\)
\(72 = 2^3 \times 3^2\)
\(432 = 2^4 \times 3^3\)
H.C.F. = \(2^1 \times 3^1 = 6\)
L.C.M. = \(2^4 \times 3^3 \times 5^1 = 16 \times 27 \times 5 = 2160\)
In simple words: Find prime factors, use lowest exponents for H.C.F. and highest exponents for L.C.M.

Exam Tip: Verify your answer using \(\text{HCF} \times \text{LCM} = \text{Product of two numbers}\) when applicable.

Question 3. Using fundamental thm of arithmetic. Find prime factorizations of :- (i) 884 (ii) 2400
Answer: (i) Prime factorization of 884:
\(884 = 2 \times 442 = 2 \times 2 \times 221 = 2^2 \times 13 \times 17\)
(ii) Prime factorization of 2400:
\(2400 = 24 \times 100 = (2^3 \times 3) \times (2^2 \times 5^2) = 2^5 \times 3^1 \times 5^2\)
In simple words: Divide the number repeatedly by prime numbers starting from the smallest until you are left with only primes.

Exam Tip: Present the final prime factorization using exponents in ascending order of bases.

Question 4. By applying the fundamental thm of arithmetic, find the HCF of 8624 and 21658. By using HCF \times LCM = product of two numbers, find LCM of the above numbers.
Answer: Prime factorizations:
\(8624 = 2^4 \times 7^3 = 16 \times 343 = 5432 \times \dots\) Let's find prime factors properly:
\(8624 = 2^4 \times 7^3\)
\(21658 = 2 \times 7^2 \times 11 \times 19\) (or checking factors: 21658 / 2 = 10829; 10829 / 7 = 1547; 1547 / 7 = 221; 221 / 13 = 17)
Wait, let's re-factorize 21658:
\(21658 = 2 \times 7^2 \times 221\) - actually \(7^2 = 49\), \(21658 / 49 = 442\), \(442 = 2 \times 13 \times 17\). So \(21658 = 2 \times 7^2 \times 13 \times 17\).
Common factors for HCF: \(2^1 \times 7^2 = 2 \times 49 = 98\).
Now, using \(\text{LCM} = \frac{\text{Product of numbers}}{\text{HCF}}\):
\(\text{LCM} = \frac{8624 \times 21658}{98} = 88 \times 21658 = 1905904\).
In simple words: First find the HCF using prime factorization, then divide the product of the two numbers by this HCF to find the LCM.

Exam Tip: Using the formula relating HCF and LCM is much faster than finding the LCM directly via prime factorization for large numbers.

Question 5. Prove that the square of any positive integer is of the form \(5q\), \(5q+1\), \(5q+4\) for any integer \(q\).
Answer: Let \(a\) be any positive integer. By Euclid's Division Lemma with divisor \(b = 5\), \(a\) can be expressed as:
\(a = 5k + r\), where \(r = 0, 1, 2, 3, 4\) and \(k\) is an integer.
Squaring both sides for each case:
Case 1: If \(a = 5k\), then \(a^2 = (5k)^2 = 25k^2 = 5(5k^2) = 5q\) (where \(q = 5k^2\)).
Case 2: If \(a = 5k + 1\), then \(a^2 = (5k + 1)^2 = 25k^2 + 10k + 1 = 5(5k^2 + 2k) + 1 = 5q + 1\) (where \(q = 5k^2 + 2k\)).
Case 3: If \(a = 5k + 2\), then \(a^2 = (5k + 2)^2 = 25k^2 + 20k + 4 = 5(5k^2 + 4k) + 4 = 5q + 4\) (where \(q = 5k^2 + 4k\)).
Case 4: If \(a = 5k + 3\), then \(a^2 = (5k + 3)^2 = 25k^2 + 30k + 9 = 5(5k^2 + 6k + 1) + 4 = 5q + 4\) (where \(q = 5k^2 + 6k + 1\)).
Case 5: If \(a = 5k + 4\), then \(a^2 = (5k + 4)^2 = 25k^2 + 40k + 16 = 5(5k^2 + 8k + 3) + 1 = 5q + 1\) (where \(q = 5k^2 + 8k + 3\)).
Thus, the square of any positive integer is always of the form \(5q\), \(5q+1\), or \(5q+4\).
In simple words: Represent any number as \(5k + r\) with remainders from 0 to 4, square them, and rearrange to show they fit the required forms.

Exam Tip: Clearly define the substitution for \(q\) in each case to ensure full credit.

Question 6. Show that \(5 + \sqrt{3}\) is irrational.
Answer: Let us assume to the contrary that \(5 + \sqrt{3}\) is rational.
Then, there exist co-prime integers \(a\) and \(b\) (\(b \neq 0\)) such that:
\(5 + \sqrt{3} = \frac{a}{b}\)
Rearranging the equation:
\(\sqrt{3} = \frac{a}{b} - 5\)
\(\sqrt{3} = \frac{a - 5b}{b}\)
Since \(a\) and \(b\) are integers, \(\frac{a - 5b}{b}\) is a rational number.
This implies that \(\sqrt{3}\) is rational, which contradicts the fact that \(\sqrt{3}\) is irrational.
Therefore, our assumption is false, and \(5 + \sqrt{3}\) is irrational.
In simple words: Assume it's rational, isolate the known irrational part (\(\sqrt{3}\)), and reach a contradiction.

Exam Tip: You can directly use the known fact that \(\sqrt{3}\) is irrational without re-proving it unless specified.

Question 7. Show that the sum and product of two irrational numbers \(5 + \sqrt{3}\) and \(5 - \sqrt{3}\) are rational numbers.
Answer: Let \(x = 5 + \sqrt{3}\) and \(y = 5 - \sqrt{3}\).
Sum: \(x + y = (5 + \sqrt{3}) + (5 - \sqrt{3}) = 5 + 5 + \sqrt{3} - \sqrt{3} = 10\), which is a rational number.
Product: \(xy = (5 + \sqrt{3})(5 - \sqrt{3}) = 5^2 - (\sqrt{3})^2 = 25 - 3 = 22\), which is a rational number.
In simple words: Add and multiply the two given expressions using algebraic identities to show the results have no radicals left.

Exam Tip: Mention the algebraic identity \((a+b)(a-b) = a^2 - b^2\) explicitly during multiplication.

Question 8. If the H.C.F. of 408 and 1032 is expressible in the form \(1032x - 408 \times 5\), find the value of \(x\).
Answer: First, find the H.C.F. of 408 and 1032 using Euclid's division algorithm:
\(1032 = 408 \times 2 + 216$
\(408 = 216 \times 1 + 192$
\(216 = 192 \times 1 + 24$
\(192 = 24 \times 8 + 0$
So, the H.C.F. of 408 and 1032 is 24.
According to the given condition:
\(1032x - 408 \times 5 = 24$
\(1032x - 2040 = 24$
\(1032x = 24 + 2040$
\(1032x = 2064$
\(x = \frac{2064}{1032} = 2$
In simple words: Calculate the HCF normally, set it equal to the given linear expression, and solve for \(x\).

Exam Tip: Double-check your long division steps when finding the HCF to avoid errors in the linear equation.

Question 9. Classify as rational and irrational numbers : (i) \(\sqrt{7}\) (ii) \(\sqrt{4}\) (iii) \(\frac{\pi}{2}\) (iv) \(5.7\) (v) \(0.73457456...\) (vi) Reciprocal of \(\sqrt{361}\) (vii) \(0.\overline{66}\) (viii) \(\sqrt[3]{5}\)
Answer: (i) \(\sqrt{7}\): Irrational (non-terminating, non-repeating square root)
(ii) \(\sqrt{4} = 2\): Rational (equals integer 2)
(iii) \(\frac{\pi}{2}\): Irrational (\(\pi\) is irrational)
(iv) \(5.7\): Rational (terminating decimal)
(v) \(0.73457456...\): Irrational (non-terminating, non-repeating pattern implied by dots)
(vi) Reciprocal of \(\sqrt{361} = \frac{1}{19}\): Rational (fraction of integers)
(vii) \(0.\overline{66}\): Rational (recurring decimal can be expressed as a fraction \(\frac{2}{3}\))
(viii) \(\sqrt[3]{5}\): Irrational (cube root of a non-perfect cube)
In simple words: Numbers that can be written as fractions of integers are rational; all others (non-terminating decimals, non-perfect roots) are irrational.

Exam Tip: Always simplify roots first (like \(\sqrt{4} = 2\)) before classifying them.

Question 10. Without long division state whether the following expressions are terminating or non-terminating decimals :- (i) \(\frac{201}{25}\) (ii) \(\frac{243}{5^2 \times 2^3 \times 7^2}\) (iii) \(\frac{9}{27}\) (iv) \(\frac{1}{8}\) (v) \(\frac{59}{4}\) (vi) \(\frac{5000}{70}\)
Answer: A rational number \(\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of \(q\) is of the form \(2^n 5^m\), where \(n\) and \(m\) are non-negative integers.
(i) \(\frac{201}{25} = \frac{201}{5^2}\): Denominator has only factor 5. Hence, Terminating.
(ii) \(\frac{243}{5^2 \times 2^3 \times 7^2}\): Denominator contains factor 7 along with 2 and 5. Hence, Non-terminating repeating.
(iii) \(\frac{9}{27} = \frac{1}{3}\): Denominator has factor 3. Hence, Non-terminating repeating.
(iv) \(\frac{1}{8} = \frac{1}{2^3}\): Denominator has only factor 2. Hence, Terminating.
(v) \(\frac{59}{4} = \frac{59}{2^2}\): Denominator has only factor 2. Hence, Terminating.
(vi) \(\frac{5000}{70} = \frac{500}{7}\): Denominator has factor 7. Hence, Non-terminating repeating.
In simple words: Check the prime factors of the denominator after writing the fraction in simplest form; if they are only 2s and/or 5s, it terminates.

Exam Tip: Always reduce the fraction to its simplest form (like \(\frac{9}{27}\) to \(\frac{1}{3}\)) before analyzing the denominator.