{"id":11105,"date":"2020-10-03T17:20:49","date_gmt":"2020-10-03T11:50:49","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=11105"},"modified":"2022-03-02T11:24:13","modified_gmt":"2022-03-02T05:54:13","slug":"real-numbers-class-10-extra-questions","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/real-numbers-class-10-extra-questions\/","title":{"rendered":"Real Numbers Class 10 Extra Questions Maths Chapter 1 with Solutions Answers"},"content":{"rendered":"

Here you will find Real Numbers Class 10 Extra Questions Maths Chapter 1 with Answers Solutions, Extra Questions for Class 10 Maths<\/a> are solved by experts and will guide students in the right direction.<\/p>\n

Extra Questions for Class 10 Maths Real Numbers\u00a0with Answers Solutions<\/h2>\n

Extra Questions for Class 10 Maths Chapter 1 Real Numbers with Solutions Answers<\/strong><\/p>\n

Real Numbers Class 10 Extra Questions Objective Type<\/h3>\n

Question 1.
\nThe graph of y = f(x) is given. The no. of zeros of f(x) will be:
\n\"Real
\n(a) 1
\n(b) 3
\n(c) 2
\n(d) 4
\nAnswer:
\n(b) 3
\nSolution.
\nAs graph of\/(x) cut the x and at 3 points
\n\u2234 Zeros of fix) are 3.
\nChoice (b) is correct.<\/p>\n

Question 2.
\nFor some integer m, the form of every positive even integer will be:
\n(a) m
\n(b) m + 1
\n(c) 2m
\n(d) 2m + 1
\nAnswer:
\n(c) 2m
\nSolution.
\n2m is even integer
\nHence, Choice (c) is correct.<\/p>\n

\"Real<\/p>\n

Question 3.
\nThe rational number which can be expressed as a terminating decimal number will be:
\n(a) \\(\\frac {77}{210}\\)
\n(b) \"Real
\n(c) \\(\\frac {13}{3125}\\)
\n(d) \\(\\frac {8}{17}\\)
\nAnswer:
\n(c) \\(\\frac {13}{3125}\\)
\nSolution.
\nAs the prime factorization of denominators of \\(\\frac {13}{3125}\\) is only at the ninth place of 2 and 5.
\nHence, choice (c) is correct.<\/p>\n

Question 4.
\nFor some integer n, the form of every positive odd integer will be:
\n(a) n
\n(b) n + 1
\n(c) 2n
\n(d) 2n + 1
\nAnswer:
\n(d) 2n + 1
\nSolution.
\nFor some integer n, the odd positive integer will be (2n + 1)
\nHence, choice (d) is correct.<\/p>\n

Question 5.
\nWhich of the following graph is not of a quadratic polynomial?
\n(a) \"Real
\n(b) \"Real
\n(c) \"Real
\n(d) \"Real
\nAnswer:
\n(d)\"Real<\/p>\n

Solution.
\nChoice (d) is correct, (d)<\/p>\n

\"Real<\/p>\n

Question 6.
\nDecimal expansion of rational number \\(\\frac {14587}{1250}\\) will end after the following decimal places:
\n(a) one
\n(b) two
\n(c) three
\n(d) four
\nAnswer:
\n(d) four
\nSolution.
\nGiven Rational Number \\(\\frac {14587}{1250}\\)
\nThe factors of the denominator 1250 is 5 \u00d7 2
\nSo the Number is terminating decimal expansion
\n\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-7\"
\n\u2234 Number will end after 4 is decimal places.
\nHence choice (d) is correct. Answer:<\/p>\n

Question 7.
\nThe decimal expansion of \\(\\frac {131}{120}\\) will terminate after how many places of decimals?
\n(a) 3
\n(b) 4
\n(c) 1
\n(d) 2
\nAnswer:
\n(a) 3<\/p>\n

\"Real<\/p>\n

Question 8.
\nThe decimal expansion of the rational number \\(\\frac{11}{2^{3} \\cdot 5^{2}}\\) will terminate after:
\n(a) one decimal place
\n(b) two decimal places
\n(c) three decimal places
\n(d) more than 3 decimal places.
\nAnswer:
\n(c) three decimal places<\/p>\n

Question 9.
\nThe decimal expansion of the rational number \\(\\frac{43}{2^{4} \\times 5^{3}}\\)will terminate after:
\n(a) 3 places
\n(b) 4 places
\n(c) 5 places
\n(d) 1 place.
\nAnswer:
\n(b) 4 places<\/p>\n

Question 10.
\nThe decimal expansion of the rational numbers \\(\\frac{23}{2^{2} \\cdot 5}\\) will terminate after:
\n(a) one decimal place
\n(b) two decimal places
\n(c) three decimal places
\n(d) more than three decimal places.
\nAnswer:
\n(b) two decimal places<\/p>\n

Real Numbers Class 10 Extra Questions Very Short Answer Type<\/h3>\n

Question 1.
\nUse Euclid\u2019s division algorithm to find the HCF of:
\n(i) 135 and 225
\n(ii) 196 and 38220
\n(iii) 867 and 255
\nSolution:
\n(i) We start with the larger number 225.
\nBy Euclid’s division algorithm, we have Dividend = (Divisor \u00d7 Quotient + Remainder)
\n225 = 135 \u00d7 1 + 90
\n\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-8\"
\nWe apply Euclid’s division algorithm on divisor 135 and the remainder 90.
\nDividend = (Divisor \u00d7 Quotient + Remainder)
\n135 = 90 \u00d7 1 + 45 1
\n\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-9\"
\nAgain, we apply Euclid’s division algorithm on divisor 90 and remainder 45.
\n90 = 45 \u00d7 2 + 0
\n\u2234 HCF (135,225) =45<\/p>\n

(ii) We have,
\nDividend = 38220 and Divisor = 196
\nDividend = (Divisor \u00d7 Quotient + Remainder)
\n38220 = 196 \u00d7 195 + 0
\nHence, HCF (196, 38220) = 196<\/p>\n

(iii) By Euclid’s division algorithm,
\nwe have
\nDividend = (Divisor \u00d7 Quotient + Remainder)
\n867 = 255 + 3 + 102
\nWe apply Euclid’s division algorithm on the divisor 255 and the remainder 102.
\nDividend = (Divisor \u00d7 Quotient + Remainder)
\n255 = 102 \u00d7 2 + 51
\nAgain, we apply Euclid’s division algorithm on the divisor 102 and the remainder 51.
\nDividend = (Divisor \u00d7 Quotient + Remainder)
\n102 = 51 \u00d7 2 + 0
\n\u2234 HCF (867, 255) = 51<\/p>\n

\"Real<\/p>\n

Question 2.
\nUsing Euclid’s division algorithm prove that: 847, 2160 are co-primes\/relatively prime.
\nSolution.
\nDefinition of co-primes or relatively primes : Two numbers are said to be co-prime or relatively prime. If their HCF is 1. Hence to prove 847 and 2160 as co-prime numbers we will find their HCF and which should be 1.
\nNow steps to find HCF will be as under
\n2160 = 847 \u00d7 2 + 466
\n847 = 466 \u00d7 1 + 381
\n466 = 381 \u00d7 1 + 85
\n381 = 85 \u00d7 4 + 41
\n85 = 41 \u00d7 2 + 3
\n41 = 3 \u00d7 13 + 2
\n3 = 2 \u00d7 1 + 1 .
\n2 = 1 \u00d7 2 + 0
\nTherefore, the HCF = 1.
\nHence, the numbers are co-prime relatively prime.<\/p>\n

Question 3.
\nFind the largest number that divides 628, 3129 and 15630 to leave remainders 3, 4 and 5 respectively.
\nSolution.
\nRequired number = HCF of (628 – 3), (3129 – 4) and (15630 – 6) i.e., required number
\n= HCF of 625, 3125,15625
\nTo find HCF of 625, 3125 and 15625 first we will find HCF of 625 and 3125 and then we will find HCF of HCF of (625 & 3125) and 15625.
\nStep (1): HCF of 625 and 3125 by using Euclid’s division algorithm is
\n3125 = 625 \u00d7 5 + 0
\n\u21d2 HCF of 625 and 3125 is 625.
\nStep (2): Now HCF of625 and 15625 will be : 15625 = 625 \u00d7 25 + 0
\n\u21d2 HCF of 625 and 15625 is 625.
\nHence HCF of 625, 3125 and 15625 is 625.
\nHence required number is 625.<\/p>\n

\"Real<\/p>\n

Question 4.
\nAn Army contingent of 616 members is to march behind an army bond 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?
\nSolution.
\nTo find the maximum number of columns, we have to find the HCF of 616 and 32. By Euclid’s division algorithm 19 4
\n\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-10\"
\ni.e., 616 = 32 \u00d7 19 + 8 i.e., 32 – 8 \u00d7 4 + 0
\n\u2234 The HCF of 616 and 32 is 8.
\nHence, maximum number of columns is 8.<\/p>\n

Question 5.
\nShow that the square of an odd positive integer is of the form 8m + 1, for some whole number m.
\nSolution.
\nAny positive odd integer is of the form 2q + 1, where q is a whole number.
\nTherefore,
\n(2q + 12<\/sup> = 4q2<\/sup> + 4q + 1
\n= 4 q(q + 1) + 1 …(1)
\nq(Question + 1) is either 0 or even because out of q and q + 1 one of the number is even. So, it is 2m, where m is a whole number.
\nTherefore, (2q + 1)2<\/sup>
\n= 4.2 m + 1 = 8m + 1. [From (1)]<\/p>\n

\"Real<\/p>\n

Question 6.
\nGiven that HCF (306, 657) = 9, find LCM (306, 657).
\nSolution.
\nWe have,
\nHCF (306, 657) = 9
\nWe know that,
\nProduct of LCM and HCF
\n= Product of two numbers
\n\u21d2 LCM \u00d7 9 = 306 \u00d7 657
\n\u21d2 \"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-11\"
\nHence, LCM (306, 657) = 22338<\/p>\n

Real Numbers Class 10 Extra Questions Short Answer Type<\/h3>\n

Question 1.
\nUse 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.
\nSolution:
\nBy Euclid’s division algorithm, we have
\na – bq + r …(i)
\nOn putting b = 3 in Eq (i), we get a = 3q + r, [0 \u2264 r < 3, i.e. , r = 0, 1, 2]
\nIf r = 0 \u21d2 a – 3q
\n\u21d2 a2<\/sup> = 9q2<\/sup> …(ii)
\nIf r = 1 \u21d2 a = 3q + 1
\n\u21d2 a2<\/sup> = 9q2<\/sup> + 6q + 1 …(iii)
\nIf r = 2 \u21d2 a = 3q + 2
\n\u21d2 a2<\/sup> = 9q2<\/sup> + 12g + 4 …(iv)
\nFrom Eq (ii), 9q2<\/sup> is a square of the form 3m, where m 3q2<\/sup>
\nFrom Eq (iii), 9q2<\/sup> + 6q + 1 i.e., 3(3q2<\/sup> + 2q) + 1 is a square which is of the form
\n3m + 1, where m = 3q2<\/sup> + 2q
\nFrom Eq (iv) 9q2<\/sup> + 12q + 4 i.e.,3(3q2<\/sup> + 4q + 1) + 1 is a square which is of the form 3m + 1, where m – 3q2<\/sup> + 4q + 1.
\n\u2234 The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.<\/p>\n

\"Real<\/p>\n

Question 2.
\nShow that that square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.
\nSolution:
\nWe know that any positive integer can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5, for some integer m.
\nThus, an odd positive integer can be of the form 6m + 1, 6m + 3, or 6m + 5.
\nThus we have :
\n(6m + 1)2<\/sup> – 36m2<\/sup> + 12m + 1
\n= 6(6m2<\/sup> + 2m) + 1 = 6q + 1,
\nq = 6m2<\/sup> + 2m is an integer.
\n(6m + 3)2<\/sup> – 36m2<\/sup> + 36m + 9
\n6(6m2<\/sup> + 6m + 1) + 3
\n= 6q + 3,
\nq = 6m2<\/sup> + 6m + 1 is an integer.
\n(6m + 5)2<\/sup> 36m2<\/sup> + 60m + 25
\n= 6(6m2<\/sup> + 10m + 4) + 1 = 6q+ 1, q
\n= 6m2<\/sup> + 10m + 4 is an integer.
\nThus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.<\/p>\n

Question 3.
\nFind the LCM and HCF of the fo11owing pairs of integers and verify that LCM \u00d7 HCF = product of the two numbers.
\n(i) 26 and 91 (ii) 510 and 92
\nSol:
\n(i) 26 and 91
\n26 = 2 \u00d7 13 and 91 = 7 \u00d7 13
\nLCM of 26 and 91
\n= 2 \u00d7 7 \u00d7 13 = 182
\nand HCF of 26 and 91 = 13<\/p>\n

\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-12\"
\nNow, 182 \u00d7 13 = 2366 and 26 \u00d7 91 = 2366
\nHence, 182 \u00d7 13 = 26 \u00d7 91<\/p>\n

(ii) 510 and 92
\n510 = 2 \u00d7 3 \u00d7 5 \u00d7 17
\nand 92 = 2 \u00d7 2 \u00d7 23<\/p>\n

\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-13\"
\n\u2234 LCM of 510 and 92
\n= 2 \u00d7 2 \u00d7 3 \u00d7 5 \u00d7 17 \u00d7 23 = 23460
\nand HCF of 510 and 92 = 2
\nNow, 23460 \u00d7 2 = 46920
\nand 510 \u00d7 92 = 46920
\nHence, 23460 \u00d7 2 = 510 \u00d7 92<\/p>\n

\"Real<\/p>\n

Question 4.
\nProve that 5 + \u221a3 is an irrational number.
\nSolution:
\nLet us assume, to the contrary, that 5 + \u221a3 is rational.
\nSo that we can find integers p and q(q \u2260 0) such that
\n5 + \u221a3 = P\/q,
\nWhere p and q are coprime.
\n\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-14\"
\nSince, p and q are integers, we get \\(\\frac {p}{q}\\) – \\(\\frac {5q}{q}\\) is a rational. So \u221a3 is rational.
\nBut this contradict the fact that \u221a3 is irrational.
\nSo, we conc1ude that 5 + \u221a3 is irrational.
\nProved<\/p>\n

Question 5.
\nWith out performing 1ong division, show whether the rational number is terminating decimal or non-terminating decimal. Find a1so its decimal expansion without performing actual division.
\nSolution:
\nGiven Number is \\(\\frac {17}{8}\\) in this 17 and 8 are coprime and the denominator is on1y the mu1tip1e of 2.
\n\u2234 \\(\\frac {17}{8}\\) is a terminating decimal.
\nNow,
\n\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-15\"<\/p>\n

\"Real<\/p>\n

Question 6.
\nWithout performing 1ong division procidure, show that the rational Number \\(\\frac {17}{90}\\) is terminating decimal or non terminating repeating decimal.
\nSolution:
\nGiven, Number
\n\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-16\"
\nIn denominator it contains 32 therefore this number is non-terminating repeating decimal.<\/p>\n

Question 7.
\nExp1ain, Why
\n(17 \u00d7 5 \u00d7 11 \u00d7 3 \u00d7 2 + 2 \u00d7 11) is a composite number.
\nSolution:
\n17 \u00d7 5 \u00d7 11 \u00d7 3 \u00d7 2 + 2 \u00d7 11
\n= 2 \u00d7 11 [17 \u00d7 5 \u00d7 3 + 1]
\n= 22 \u00d7 256 = 2 \u00d7 11 \u00d7 4 \u00d7 4 \u00d7 4 \u00d7 4
\nWhich is a product of more than two prime numbers i.e., 2 and 11. Hence it is a composite number.<\/p>\n

Question 8.
\nWithout performing the 1ong division process, find whether the rational Number \\(\\frac {987}{10500}\\) is terminating or non terminating decimal. Give reason of your answer.
\nSolution:
\n\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-17\"
\n\u2235 The prime factor of denominator is of the form 53 \u00d7 22
\n\u2234 The given rational number is terminating decimal.<\/p>\n

\"Real<\/p>\n

Real Numbers Class 10 Extra Questions Long Answer Type<\/h3>\n

Question 1.
\nProve that \u221a2 is irrational.
\nSolution:
\nIf possible, let \u221a2 be rational and let its simp1est form be \\(\\frac {a}{b}\\) .
\nThen, a and b are integers having no common factor other than 1, and b \u2260 0.
\nNow,
\n\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-18\"
\n[on squaring both sides]
\n\u21d2 2 b2<\/sup> = a2<\/sup> …(i)
\n\u21d2 2 divides a2<\/sup> [\u2235 2 divides 2b2<\/sup>]
\n\u21d2 2 divides a
\n[ \u2235 2 is prime and divides c2<\/sup> \u21d2 2 divides a]
\nLet a = 2c for some integer c.
\nPutting a = 2c in (i), we get
\n2b2<\/sup> = 4c2<\/sup>
\n\u21d2 b2<\/sup> = 2c2<\/sup>
\n\u21d2 2 divides b2<\/sup> [ \u2235 2 divides 2c2<\/sup>]
\n\u21d2 2 divides b[ \u2235 2 is prime and 2 divides b2<\/sup> \u21d2 2 divides b]
\nThus, 2 is a common factor of a and b.
\nBut, this contradicts the fact that a and b have no common factor other than 1.
\nThe contradiction arises by assuming that \u221a2 is rational.
\nHence, \u221a2 is irrational.<\/p>\n

\"Real<\/p>\n

Question 2.
\nWhy n is an irrational number.
\nSolution:
\nWe know that the ratio of circumference of a circle to 1ength of its corresponding diameter is known as ‘\u03c0’ (pie).
\ni.e., \u03c0 = \\(\\frac {c}{d}\\)
\nNow, read this to understand why ‘\u03c0’ is an irrational number.
\nIf we take circumference of a circle as an integer (natural number) then we find length of its corresponding diameter comes in decimal not an integer (natura1 number)] so ratio of \\(\\frac {c}{d}\\) is not in the form of \\(\\frac {p}{q}\\), i.e., p and q bothare not integers.<\/p>\n

Simi1arly if we take diameter of circle as an integer (natura1 number) then circumference of its corresponding circle comes out to be in the decimal form so again ratio \\(\\frac {c}{d}\\) = \u03c0 does not comes in the form of \\(\\frac {p}{q}\\) because diameter, i.e., ‘q’ is an integer and ‘p’ is in decimal form.<\/p>\n

So it is clear that for any circle if length of diameter is natura1 number then circumference is the decimal form or if circumference is a natura1 number then diameter is a decimal number. So we can say that the ratio \\(\\frac{\\pi}{d}\\) can never come in the form of \\(\\frac {p}{q}\\) where p and q both are integers and q \u2260 0. So it is clear that the ratio \\(\\frac{c}{d}\\), i.e., \u03c0 can never be rational hence ‘\u03c0’ is a1ways an irrational number.<\/p>\n

\"Real<\/p>\n

Question 3.
\nProve that 3 + 2\u221a5 is irrational.
\nSolution:
\nlet us assume, to the contrary, that 3 + 2\u221a5 is a rational number.
\nNow, let 3 + 2\u221a5 = x, where a and b are coprime and b \u2260 0.
\n\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-19\"
\nSince, a and b are integers, therefore
\n\\(\\frac {a}{2b}\\) – \\(\\frac {3}{2}\\) is a rational number.
\n\u2234 \u221a5 is a rational number.
\nBut \u221a5 is an irrational number.
\nThis shows that our assumption is wrong.
\nSo, 3 + 2 \u221a5 is an irrational number.
\nHence proved.
\nExample 4. Prove that \u221a3 is irrational.
\nSolution:
\nIf possible, let \u221a3 be rational and let its simplest form be \\(\\frac{a}{b}\\)
\nThen, a and b are integers having no common factor other than 1, and b \u2260 0.
\n\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-20\" [on squaring both sides]
\n\u21d2 3b2<\/sup> = a2<\/sup> …(i)
\n\u21d2 3 divides a2<\/sup> [ \u2235 3 divides 3b2<\/sup>]
\n\u21d2 3 divides a [\u2235 3 is prime and divides a2<\/sup> \u21d2 3 divides a]
\nLet a = 3c for some integer c.
\nPutting a = 3c in (i), we get
\n3b2<\/sup> = 9c2<\/sup>
\n\u21d2 b2<\/sup> = 3c2<\/sup>
\n\u21d2 3 divides b2<\/sup> [ \u2235 3 divides 3c2<\/sup>]
\n\u21d2 3 divides b [ \u2235 3 is prime and 3 divides b2<\/sup> \u21d2 3 divides b]
\nThus, 3 is a common factor of a and b. But, this contradicts the fact that a and b have no common factor other than 1.
\nThe contradiction arises by assuming that \u221a3 is rational.
\nHence, \u221a3 is irrational.<\/p>\n

\"Real<\/p>\n

Question 3.
\nThe fo11owing rea1 numbers have decimal expansions as given be1ow. In each case, decide whether they are rational or not. If they are rational and of the form \\(\\frac{p}{q}\\), what can you say about the prime factors of q?
\n(i) 43.123456789
\n(ii) 0.120120012000120000….
\n(iii) \\(43 . \\overline{123456789}\\)
\nSolution:
\n(i) 43.123456789 is terminating. So, it represents a rational number.
\nThus,
\n\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-21\"
\nThus, q = 109<\/sup>
\n(ii) 0.120120012000120000…. is non-terminating and non-repeating. So, it is an irrational number.
\n(iii) \\(43 . \\overline{123456789}\\) is non-terminating but repeating. So it is a rational.
\n\"Real-Numbers-Class-10-Extra-Questions-Maths-Chapter-1-with-Solutions-Answers-22\"
\nThus q = 999999999<\/p>\n

\"Real<\/p>\n","protected":false},"excerpt":{"rendered":"

Here you will find Real Numbers Class 10 Extra Questions Maths Chapter 1 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction. Extra Questions for Class 10 Maths Real Numbers\u00a0with Answers Solutions Extra Questions for Class 10 Maths Chapter 1 Real Numbers with …<\/p>\n

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