{"id":12976,"date":"2021-07-30T09:33:42","date_gmt":"2021-07-30T04:03:42","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=12976"},"modified":"2023-03-29T17:54:55","modified_gmt":"2023-03-29T12:24:55","slug":"ap-10th-class-maths-important-questions-chapter-1","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-10th-class-maths-important-questions-chapter-1\/","title":{"rendered":"AP 10th Class Maths Important Questions Chapter 1 Real Numbers"},"content":{"rendered":"

These AP 10th Class Maths Chapter Wise Important Questions<\/a> 1st Lesson Real Numbers will help students prepare well for the exams.<\/p>\n

AP State Syllabus 10th Class Maths 1st Lesson Important Questions and Answers Real Numbers<\/h2>\n

Question 1.
\nFind the HCF of 60 and 100 by using Euclid division lemma.
\nSolution:
\n\"AP
\n\u2234 HCF of 60 and 100 is 20<\/p>\n

\"AP<\/p>\n

Question 2.
\nFind the value of log5<\/sub> \\(\\sqrt{625}\\) Mar. 19
\nSoluton:
\nGiven that log5<\/sub> \\(\\sqrt{625}\\)
\n= log5<\/sub>25 = logs5<\/sub>52<\/sup>
\n= 2 log5<\/sub> 5 { \u2235 loga<\/sub> xn<\/sup> = n loga<\/sub> x}
\n= 2 .1 = 2<\/p>\n

Question 3.
\nExpress \\(\\frac{36}{99}\\) as a decimal number.
\nSolution:
\n\\(\\frac{36}{99}=\\frac{4 \\times 9}{9 \\times 11}=\\frac{4}{11}=0 . \\overline{36}\\)<\/p>\n

Question 4.
\nWrite the Expand Form of log \\(\\frac{\\mathbf{a}^{\\mathrm{m}} \\mathbf{b}^{\\mathrm{n}}}{\\mathrm{c}^{\\mathrm{z}}}\\)
\nSolution:
\nlog \\(\\frac{\\mathbf{a}^{\\mathrm{m}} \\mathbf{b}^{\\mathrm{n}}}{\\mathrm{c}^{\\mathrm{z}}}\\)
\n= log am<\/sup>bn<\/sup> – log cz<\/sup>
\n[ \u2235 log \\(\\frac{\\mathrm{x}}{\\mathrm{y}}\\) = logx – logy].
\n= log am<\/sup> + log bn<\/sup> – log cz<\/sup>
\n[ \u2235 logxy = logx + logy]
\n= mloga + nlogb – zlogc.<\/p>\n

Question 5.
\nIf x2<\/sup> + y2<\/sup> = 7xy, then prove that \\(\\log \\left(\\frac{x+y}{3}\\right)=\\frac{1}{2}(\\log x+\\log y)\\)
\nSol.
\nGiven x2<\/sup> + y2<\/sup> = 7xy now add (2xy) on both sides. We get
\nx2<\/sup> + y2<\/sup> + 2xy = 7xy + 2xy = 9xy
\nConsidering square root on both sides we get
\n\"AP<\/p>\n

Question 6.
\nFind the HCF of 1260 and 1440 by us\u00acing Euclid’s division lemma.
\nSolution:
\nGiven numbers 1260 and 1440
\n1440 – 1260 x 1 + 180
\n1260 = 180 x 7 + 0
\n\u2234 HCF of 1260 and 1440 is 180.<\/p>\n

Question 7.
\nUsing the Euclid\u2019s division lemma, find the HCF of 4830 and 759.
\nSolution:
\nGiven numbers are 4830 and 759
\n4830 = 759 x 6 + 276
\n759 = 276 x 2 + 207
\n276 = 207 x 1 + 69
\n207 = 69 x 3 + 0
\n\u2234 HCF of 4830 and 759 = 69<\/p>\n

\"AP<\/p>\n

Question 8.
\nIf x2<\/sup> + y2<\/sup> = lOxy, then prove that 2log (x – y) = log x + log y + 3 log 2.
\nSolution:
\nGiven x2<\/sup> + y2<\/sup> = 10xy
\nNow subtract ‘2xy’ on both sides
\nx2<\/sup> + y2<\/sup> – 2xy = 10xy – 2xy = 8xy
\n(x – y)2<\/sup> = 8xy
\ntaking “logarithm” on both sides
\nlog (x – y)2<\/sup> = log 8xy = log 8 + log x + logy
\n\u21d2 2 log (x – y) = 3 log 2 + log x + log y<\/p>\n

Question 9.
\nIs log10<\/sub>5 rational number or irrational number ? Justify your answer.
\nSolution:
\nLet log10<\/sub>5 = x
\nThen 10x<\/sup> = 5
\nBut 2 can\u2019t be written as 10x<\/sup> for any value of x.
\n\u2234 log 5 is irrational.<\/p>\n

Question 10.
\nExpress 1.2333333 In \\(\\frac{\\mathbf{p}}{\\mathbf{q}}\\) form where p and q are co-primes.
\nSolution:
\nLet x = 1.2333333
\n= \\(1.2 \\overline{3}\\)
\n\"AP<\/p>\n

Question 11.
\nProve that \\(\\sqrt{5}+\\sqrt{7}\\) is irrational.
\nSolution:
\nTo show that \\(\\sqrt{5}+\\sqrt{7}\\) is an irrational. Let us prove this by contradiction.
\nSo let us assume \\(\\sqrt{5}+\\sqrt{7}\\) = a
\nWhere ‘a’ is a rational number.
\n\u21d2 \\(\\sqrt{7}\\) = a – \\(\\sqrt{5}\\) now squaring on both sides we get
\n(\\(\\sqrt{7}\\))2<\/sup> = (a-\\(\\sqrt{5}\\))2<\/sup>
\n\u21d2 7 = a2<\/sup> + 5 – 2a\\(\\sqrt{5}\\)
\n2a\\(\\sqrt{5}\\) = a2<\/sup> + 5 – 7 = a2<\/sup> – 2
\n\u21d2 \\(\\sqrt{5}=\\frac{\\mathrm{a}^{2}-2}{2 \\mathrm{a}}\\) in which L.H.S part
\n\\(\\sqrt{5}\\) is an irrational and RHS part \\(\\left(\\frac{a^{2}-2}{2 a}\\right)\\)
\n(\u2235 ‘a’ is a rational)
\nSo from the above a rational number and an irrational numbers are equal, which is impossible, which is wrong, So our assumption is wrong.
\nHence given \\(\\sqrt{5}+\\sqrt{7}\\) won’t be a rational.
\nTherefore \\(\\sqrt{5}+\\sqrt{7}\\) is an irrational.<\/p>\n

Question 12.
\nShow that \\(\\sqrt{3}\\) is irrational.
\nSolution :
\nLet \\(\\sqrt{3}\\) be rational then \\(\\sqrt{3}\\) = a\/b
\n(where a, b are coprimes and b \u2260 0) \u21d2 b\\(\\sqrt{3}\\) = a
\nSquaring on both sides, we get
\n3b2<\/sup> = a2<\/sup>
\n3 divides a2<\/sup> => 3 divides a (1)
\nThere exists an integer c, such that
\na = 3c
\n\u21d2 a2<\/sup> = 9c2<\/sup>
\n\u21d2 3b2<\/sup> = 9c2<\/sup> (\u2235 a2<\/sup> = 3b2<\/sup>)
\n\u21d2 b2<\/sup> = 3c2<\/sup>
\n3 divides b2<\/sup> => 3 divides b (2)
\nFrom (1) & (2)
\n3 is a common factor for a and b This contradicts the fact that a and b are coprimes
\n\u2234 Our assumption that \\(\\sqrt{3}\\) is rational is wrong.
\n\u2234 \\(\\sqrt{3}\\) is irrational.<\/p>\n

\"AP<\/p>\n

Question 13.
\nShow that \\(2+5 \\sqrt{3}\\) is irrational.
\nSolution:
\nGiven that \\(2+5 \\sqrt{3}\\) is an irrational. On the contrary, let us assume that \\(2+5 \\sqrt{3}\\) is a rational number.
\n\"AP
\nHere p, q being integers we can say that \\(\\frac{p-2 q}{5 q}\\) is a rational number. This contradicts the fact that \\(\\sqrt{3}\\) is an irra-tional number. This is due to our as-sumption “2 + 5\\(\\sqrt{3}\\) is a rational num-ber”.
\nHence our assumption is wrong.
\n\u2234 2 + 5\\(\\sqrt{3}\\) is an irrational number.<\/p>\n

Question 14.
\nProve that \\(\\sqrt{5}+\\sqrt{11}\\) is irrational.
\nSolution:
\nLet us assume that \\(\\sqrt{5}+\\sqrt{11}\\) is ratio\u00acnal number.
\nLet us \\(\\sqrt{5}+\\sqrt{11}=\\frac{\\mathrm{a}}{\\mathrm{b}}\\) where a, b are b
\nco – primes b \u2260 0
\n\u2234 \\(\\sqrt{5}=\\frac{a}{b}-\\sqrt{11}\\)
\nSquaring on both sides, we get
\n\"AP
\nSince, a, b are integers \\(\\frac{a^{2}+6 b^{2}}{2 a b}\\) is rational number.
\nSo \\(\\sqrt{11}\\) is rational number.
\nThis contradicts the fact that \\(\\sqrt{11}\\) is irrational number.
\nHence \\(\\sqrt{5}+\\sqrt{11}\\) is irrational number.<\/p>\n

Question 15.
\nShow that \\(\\sqrt{5}+\\sqrt{3}\\) is an irrational number.
\nSolution:
\nSuppose \\(\\sqrt{5}+\\sqrt{3}\\) is not an irrational number. \\(\\sqrt{5}+\\sqrt{3}\\) is a rational number.
\nLet \\(\\sqrt{5}+\\sqrt{3}\\) = p\/q where q \u2260 0 and
\np, q \u2208 Z squaring on both sides
\n5 + 3 – 2\\(\\sqrt{15}\\) = \\(\\frac{\\mathrm{p}^{2}}{\\mathrm{q}^{2}}\\)
\n\\(\\sqrt{15}=\\frac{8 q^{2}-p^{2}}{2 q^{2}}\\)
\np, q \u2208 Z & q \u2260 0
\n8q2<\/sup> – p2<\/sup> & 2q2<\/sup> \u2208 Z and also 2q2<\/sup> \u2260 0.
\nSo \\(\\frac{8 q^{2}-p^{2}}{2 q^{2}}\\) is a rational number.
\nbut \\(\\sqrt{15}\\) is an irrational number.
\nAn irrational number never be equal to a rational number.
\nSo our supposition that \\(\\sqrt{5}+\\sqrt{3}\\) is not an irrational number is false.
\n\u2234 \\(\\sqrt{5}+\\sqrt{3}\\) is an irrational number.<\/p>\n

Question 16.
\nProve that \\(\\sqrt{7}+\\sqrt{11}\\) is irrational.
\nJane 2019
\nSolution:
\nLet us assume that \\(\\sqrt{7}+\\sqrt{11}\\) is ratio\u00acnal number.
\nLet us \\(\\sqrt{7}+\\sqrt{11}\\) = a\/b
\nco – primes b \u2260 0.
\n\u2234 \\(\\sqrt{7}=\\frac{a}{b}-\\sqrt{11}\\)
\nSquaring on both sides, we get
\n\\(7=\\frac{a^{2}}{b^{2}}+11-2 \\frac{a}{b} \\sqrt{11}\\)
\nRearranging
\n\"AP
\nSince, a, b are integers \\(\\frac{a^{2}+4 b^{2}}{2 a b}\\) is ra- tional number.
\nSo \\(\\sqrt{11}\\) is rational number.
\nThis contradicts the fact that \\(\\sqrt{11}\\) is irrational number.
\nHence \\(\\sqrt{7}+\\sqrt{11}\\) is irrational number.<\/p>\n

\"AP<\/p>\n

Question 17.
\nIf (2)x + 1<\/sup> = (3)1 -x<\/sup>, find the value of x.
\nSolution:
\n(2)x + 1<\/sup> = (3) 1 – x<\/sup>
\n(x+ l)log10<\/sub>2 = (1-x)log10<\/sub>3
\nx log10<\/sub>2 + 1 log10<\/sub>2 = 1 log10<\/sub> 3 – x log10<\/sub> 3
\nx log10<\/sub> 2 + x log10<\/sub> 3 = log10<\/sub> 3 – log10<\/sub> 2
\nx (log10<\/sub> 2 + log10<\/sub> 3) – log10<\/sub>3 – log10<\/sub> 2
\nx = \\(\\frac{\\log _{10} 3-\\log _{10} 2}{\\log _{10} 2+\\log _{10} 3}\\)<\/p>\n

Fundamental Theorem of Arithmetic :<\/span><\/p>\n

Question 18.
\nSupport the Fundamental theorem of Arithmetic by considering some examples.
\nSolution:
\nEvery composite number can be ex-pressed as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
\nEx: 1) Now let us consider a com\u00acposite number 186.
\n\u2234 186 can be factorised 2 x 3 x 31 where 2, 3 and 31 are primes.
\nEx: 2) Now consider another com\u00acposite number 384.
\n\u2234 384 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 = 27<\/sup> x 3 where these 2 and 3 are primes.
\nThus we can express any composite number as a product of some primes. Thus the theorem is valid.<\/p>\n

Question 19.
\nWrite 210 as product of prime factors.
\nSolution:
\n210 = 3 x 70
\n= 3 x 7 x 10
\n= 3 x 7 x 5 x 2
\n\u2234 way of 2 x 3 x 5 x 7.
\nSo the composite number 210 can be written as product of primes 2, 3, 5 and 7 in the way of 2 x 3 x 5 x 7.<\/p>\n

Question 20.
\nIs the units digit in the value of 6n where \u2018n\u2019 is a natural number is zero? Justify your answer.
\nSolution:
\nNo, the units number in the value of 6n where ‘n’ is a natural number will not be zero.
\nWhy because, if a (composite) number is written as product of prime factors having 2 and 5 then the units digit becomes zero.
\nIn this case
\n6 = 2 x 3 that means 6 has only 2 in its factors but not 5.
\nHence in the value of 6n<\/sup>, the units digit will not be equal to zero.<\/p>\n

Question 21.
\nIf the HCF of 306 and 657 is 9, then find their LCM.
\nSolution:
\nThe given two numbers are 306, and 657.
\nHCF of given numbers = 9
\nFrom the formula
\nLCM x HCF = Product of given two numbers
\n(LCM) (9) = 657 x 306
\n\u2234 LCM = \\(\\frac{657 \\times 306}{9}\\) = 22, 338
\n\u2234 LCM of 306 and 657 is 22,338.<\/p>\n

\"AP<\/p>\n

Question 22.
\nFind the HCF of 45,75 from Euclid’s division lemma.
\nSolution:
\nLet us write 45 and 75 in the form of a = bq + r where a = 75, b = 45
\n\u2234 75 = 45 (1) + 30
\nAgain using lemma a = bq + r, where
\na = 45, and b = 30,
\nwe get, 45 = 30 (1) + 15
\nthen 30 = 15 (2) + 0
\nHere we get remainder \u20180\u2019 when we
\ndivide it by 15.
\nSo 15 is HCF of 45 and 75.<\/p>\n

Question 23.
\nHow do you say 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are complex numbers ? Explain it.
\nSolution:
\nThe given
\n7 x 11 x 13 + 13= 13 [(7 x 11) + 1] = 13 [77 + 1] =13 x 78
\nSo we can express 7 x 11 x 13 + 13 as product of 13 and 78 that means 13 and 78 are factors of given 7 x 11 x 13 + 13. Hence it is a composite number.
\nSimilarly
\n7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5[(7 x 6 x 4 x 3 x 2 x 1) + 1]
\n= 5 [1008 + 1] = 5 x 1009
\nThat means 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 can be written as product of two factors 5 and 1009. Hence (7 x 6 x 5 x 4 x 3 x 2 x 1) + 5 is a complex number because it has more than 2 factors.<\/p>\n

Logarithms :<\/span><\/p>\n

Question 24.
\nExpress the following in log form,
\ni) 25<\/sup> = 32
\nii) 33<\/sup> = 27
\niii) 53<\/sup> = 125
\nSolution:
\nIf ax<\/sup> = N then loga<\/sub>N = x(by definition)
\ni) if 25<\/sup> = 32 then log2<\/sub>32 =5
\nii) if 33<\/sup> = 27 then log3<\/sub>27 = 3 j log forms
\niii) if 53<\/sup> = 125 then log5<\/sub>125 = 3<\/p>\n

Question 25.
\nCalculate the value of log2<\/sub>512.
\nSolution:
\nWe can write 512 as 29<\/sup>.
\nlog2<\/sub>512 = loga<\/sub>29<\/sup> = 9 log 22 = 9 (1)
\n= 9 (\u2235 loga<\/sub>a = 1)
\nlog2<\/sub>512 = 9<\/p>\n

Question 26.
\nCalculate the value of log10<\/sub> 0.01.
\nSolution:
\nWe can write 0.01 = \\(\\frac{1}{100}=\\frac{1}{10^{2}}\\) = 10-2<\/sup>
\nThen log10<\/sub>0.01= log10<\/sub> 10-2<\/sup>
\n= -2log10<\/sub>10
\n= – 2
\n\u2234 log10<\/sub>0.01 = -2( \u2235 log10<\/sub>10= 1)<\/p>\n

Question 27.
\nExpress log5<\/sub>125 = 3 in its exponential form.
\nSolution:
\nWe know that
\nif loga<\/sub>x = N then aN<\/sup> = x
\nSo, log5<\/sub>125 = 3 => 53<\/sup> = 125
\n[\u2235 x = 125, a = 5, N = 3]
\n\u2235 53<\/sup> = 125 is the exponential form of log5<\/sub>125 = 3<\/p>\n

\"AP<\/p>\n

Question 28.
\nProve that loga<\/sub>xy = loga<\/sub>x + loga<\/sub>y.
\nSolution:
\nLet loga<\/sub>xy = p => ap<\/sup> = xy …………(1)
\nLet loga<\/sub>x = q => aq<\/sup> = x ……….. (2)
\nSimilarly, let loga<\/sub>y = r then ar<\/sup> = y …………… (3)
\nNow put the values of \u2018x\u2019 and \u2018y\u2019 in the equation (1)
\nap<\/sup> = xy
\n= aq<\/sup>. ar<\/sup> [\u2235 x = aq<\/sup>, y = ar<\/sup>]
\nap<\/sup> = aq + r <\/sup>[ \u2235 am<\/sup> . an<\/sup> = am + n<\/sup>]
\n\u21d2 p = q + r
\nNow reproduce the values of p, q and r.
\nWe get, loga<\/sub>xy = loga<\/sub>x + loga<\/sub>y Hence proved.<\/p>\n

Question 29.
\n[H+<\/sup>] ion concentration in a soap used by Sankar is 9.2 x 10-22<\/sup>. Then find its pH using logarithm.
\nSolution:
\nThe [H+<\/sup>] ion concentration in the soap is 9.2 x 10-22<\/sup>.
\nThen the pH = – log10<\/sub> [H+<\/sup>]
\n= -log10<\/sub> (9.2 x 10-22<\/sup>)
\n= -[log10<\/sub>9.2 + log10<\/sub>10-22<\/sup>] = -log109.2-(-22).
\nlogio10
\n= -0.96 + 22 = 21.04
\nHence pH = 21.04 (Actually it is not possible, pH is always less than 14.)<\/p>\n

Irrational Numbers :<\/span><\/p>\n

Question 30.
\nChoose the irrational numbers from the following given
\n\\(\\begin{aligned}
\n&\\sqrt{2}, \\sqrt{3}, 5,5.75,1.735, \\ldots \\ldots \\ldots \\ldots, \\\\
\n&\\sqrt{8}, \\sqrt{4}, \\sqrt{16}
\n\\end{aligned}\\) ……………
\nSolution:
\n\\(\\sqrt{2}, \\sqrt{3}, \\sqrt{8}\\), ……………… are irrational numbers.<\/p>\n

Question 31.
\nDefine the irrational numbers. Give examples.
\nSolution:
\nDefinition of irrational numbers :
\nA number which can be expressed nei-ther as a terminating decimal nor as a non-terminating decimal is called an irrational number.
\nThese are denoted with \u2018S’ or Q’
\n(OR)
\nA number which cannot be repre\u00acsented in the form of p\/q is called an irrational.
\nEx : \\(\\sqrt{2}, \\sqrt{3}, \\sqrt{5}, \\sqrt{6}, \\sqrt{7}, \\sqrt{8}, \\sqrt{11}, \\sqrt{12}, \\ldots \\ldots \\ldots\\)<\/p>\n

\"AP<\/p>\n

Question 32.
\nWrite the differences between rational and irrational numbers.
\nSolution:<\/p>\n\n\n\n\n\n\n\n\n
Irrational Numbers<\/td>\nRational Numbers<\/td>\n<\/tr>\n
1. These are neither in the form of terminating decimal nor as a non-terminating, recurring decimals.<\/td>\n1. These are either in the form of termina
\nting or as a non-terminating, recurring decimal.<\/td>\n<\/tr>\n
2. We cannot express them in the form of p\/q (q \u2260 0).<\/td>\n2. We can express them in the form of p\/q (q \u2260 0)<\/td>\n<\/tr>\n
3. \\(\\sqrt{\\mathrm{p}}\\) is an<\/sup> irrational when p is a prime.<\/td>\n3. All natural numbers\/integers are rational numbers.<\/td>\n<\/tr>\n
4. We denote them with \u2018Q\u2019\u2019 or \u2018S\u2019.<\/td>\n4. We denote them with \u2018Q\u2019.<\/td>\n<\/tr>\n
5. We cannot find their square root value exact.
\nEx : \\(\\sqrt{2}, \\sqrt{3}, \\sqrt{5}, \\ldots \\ldots \\ldots\\) …………….<\/td>\n		5. Ex :\u00a0 \\(\\frac{5}{8}, \\frac{6}{7}\\)1, – 3,…………..<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 33.
\nPlot \\(\\sqrt{2}, \\sqrt{3}, \\sqrt{5}\\) on number line.
\nSolution:
\nNow consider a right angle triangle with 1 cm each leg as shown in figure. AB = AC = 1 cm, then
\nhypotenuse = BC = \\(\\sqrt{1^{2}+1^{2}}=\\sqrt{2}\\)
\nNow plot a point \u2018D’ on the line AB such that AD = BC = \\(\\sqrt{2}\\)
\n\"AP
\nNow plot a point \u2018E\u2019 on the line AB
\nsuch that AE = CD = \\(\\sqrt{3}\\)
\nNow plot a point \u2018F\u2019 on the line AB
\nsuch that AF = CE = 2
\nNow plot another point \u2018G\u2019 on the line
\nAB such that
\nAG = CG = \\(\\sqrt{1^{2}+2^{2}}=\\sqrt{1+4}=\\sqrt{5}\\)
\nThe point \u2018D\u2019 is \\(\\sqrt{2}\\) , ‘E\u2019 is \\(\\sqrt{3}\\) , ‘F’ is \\(\\sqrt{4}\\) , and the point \u2018G’ is \\(\\sqrt{5}\\) on num\u00acber line.<\/p>\n

Question 34.
\nFind the value of \\(\\sqrt{2}\\) upto 4 decimals.
\nSolution:
\n\"AP<\/p>\n

Question 35.
\nWhere do we use irrational numbers in our day to day life ?
\nSolution:
\n1. We use the value of \u03c0 in finding perimeter and area of a circle (n is an irrational).
\n2. To find the length of the diagonal of a square we use \\(\\sqrt{2}\\) which is an irrational one.
\n3. To find the T.S.A., curved surface area and volumes of solids viz cyl-inder, sphere, and cone etc. We use K in above case which is an irratio\u00acnal one.
\n4. For calculation of compound inter-est, E.M.I. (equal monthly install-ments) in Banks they use \u2018e\u2019 value which is an irrational one.
\n5. We use these numbers in space and economics related matters thor-oughly.<\/p>\n

Question 36.
\nEstablish the relation between diago-nal and side of a square.
\nSolution:
\nLet the side of square ABCD = a
\nThen its diagonal AC = Hypotenuse of \u0394ABC.
\n\"AP
\nSo the diagonal of a square is a\\(\\sqrt{2}\\) when its side is \u2018a\u2019.<\/p>\n

Question 37.
\nIs \u03c0 a rational or irrational ? Give reasons.
\nSolution:
\n\u03c0 is an irrational number. The value of
\n\u03c0 is approximately \\(\\frac{22}{7}\\)
\nBut not exactly,
\n\\(\\frac{22}{7}\\) = 3.14285714286. eventhough 22
\n\\(\\frac{22}{7}\\) is in the form of p\/q.
\n\u03c0 is an irrational.<\/p>\n","protected":false},"excerpt":{"rendered":"

These AP 10th Class Maths Chapter Wise Important Questions 1st Lesson Real Numbers will help students prepare well for the exams. AP State Syllabus 10th Class Maths 1st Lesson Important Questions and Answers Real Numbers Question 1. Find the HCF of 60 and 100 by using Euclid division lemma. Solution: \u2234 HCF of 60 and … Read more<\/a><\/p>\n","protected":false},"author":6,"featured_media":35687,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[12],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/07\/Chapter-1-Real-Numbers.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/12976"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=12976"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/12976\/revisions"}],"predecessor-version":[{"id":35688,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/12976\/revisions\/35688"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/35687"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=12976"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=12976"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=12976"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}