{"id":118910,"date":"2022-04-29T16:47:16","date_gmt":"2022-04-29T11:17:16","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=118910"},"modified":"2022-04-29T16:53:31","modified_gmt":"2022-04-29T11:23:31","slug":"chapter-1-real-numbers-questions-and-answers-ncert-solutions-for-class-10-mathematics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-1-real-numbers-questions-and-answers-ncert-solutions-for-class-10-mathematics","title":{"rendered":"Chapter 1 – Real Numbers Questions and Answers: NCERT Solutions for Class 10 Mathematics"},"content":{"rendered":"

Exercise 1.1<\/h2>\n

1. Use Euclid\u2019s division algorithm to find the HCF of:
\ni. 135 and 225
\nii. 196 and 38220
\niii. 867 and 255<\/h2>\n

Solutions:
\ni. 135 and 225
\nAs you can see, from the question 225 is greater than 135. Therefore, by Euclid\u2019s division algorithm, we have,
\n225 = 135 \u00d7 1 + 90
\nNow, remainder 90 \u2260 0, thus again using division lemma for 90, we get,
\n135 = 90 \u00d7 1 + 45
\nAgain, 45 \u2260 0, repeating the above step for 45, we get,
\n90 = 45 \u00d7 2 + 0
\nThe remainder is now zero, so our method stops here. Since, in the last step, the divisor is 45, therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45.
\nHence, the HCF of 225 and 135 is 45.<\/h3>\n

ii. 196 and 38220
\nIn this given question, 38220>196, therefore the by applying Euclid\u2019s division algorithm and taking 38220 as divisor, we get,
\n38220 = 196 \u00d7 195 + 0
\nWe have already got the remainder as 0 here. Therefore, HCF(196, 38220) = 196.
\nHence, the HCF of 196 and 38220 is 196.<\/h3>\n

iii. 867 and 255
\nAs we know, 867 is greater than 255. Let us apply now Euclid\u2019s division algorithm on 867, to get,
\n867 = 255 \u00d7 3 + 102
\nRemainder 102 \u2260 0, therefore taking 255 as divisor and applying the division lemma method, we get,
\n255 = 102 \u00d7 2 + 51
\nAgain, 51 \u2260 0. Now 102 is the new divisor, so repeating the same step we get,
\n102 = 51 \u00d7 2 + 0
\nThe remainder is now zero, so our procedure stops here. Since, in the last step, the divisor is 51, therefore, HCF (867,255) = HCF(255,102) = HCF(102,51) = 51.
\nHence, the HCF of 867 and 255 is 51.<\/h3>\n

2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.<\/h2>\n

Solution:
\nLet a be any positive integer and b = 6. Then, by Euclid\u2019s algorithm, a = 6q + r, for some integer q \u2265 0, and r = 0, 1, 2, 3, 4, 5, because 0\u2264r<6. Now substituting the value of r, we get, If r = 0, then a = 6q Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively. If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.<\/h3>\n

3. An army contingent of 616 members is to march behind an army band 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?<\/h2>\n

Solution: Given, Number of army contingent members=616 Number of army band members = 32 If the two groups have to march in the same column, we have to find out the highest common factor between the two groups. HCF(616, 32), gives the maximum number of columns in which they can march. By Using Euclid\u2019s algorithm to find their HCF, we get, Since, 616>32, therefore,
\n616 = 32 \u00d7 19 + 8
\nSince, 8 \u2260 0, therefore, taking 32 as new divisor, we have,
\n32 = 8 \u00d7 4 + 0
\nNow we have got remainder as 0, therefore, HCF (616, 32) = 8.
\nHence, the maximum number of columns in which they can march is 8.<\/h3>\n

4. Use Euclid\u2019s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.<\/h2>\n

Solutions:
\nLet x be any positive integer and y = 3.
\nBy Euclid\u2019s division algorithm, then,
\nx = 3q + r for some integer q\u22650 and r = 0, 1, 2, as r \u2265 0 and r < 3.
\nTherefore, x = 3q, 3q+1 and 3q+2
\nNow as per the question given, by squaring both the sides, we get,
\nx2\u00a0= (3q)2\u00a0= 9q2\u00a0= 3 \u00d7 3q2
\nLet 3q2\u00a0= m
\nTherefore, x2= 3m \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(1)
\nx2\u00a0= (3q + 1)2\u00a0= (3q)2+12+2\u00d73q\u00d71 = 9q2\u00a0+ 1 +6q = 3(3q2+2q) +1
\nSubstitute, 3q2+2q = m, to get,
\nx2= 3m + 1 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (2)
\nx2= (3q + 2)2\u00a0= (3q)2+22+2\u00d73q\u00d72 = 9q2\u00a0+ 4 + 12q = 3 (3q2\u00a0+ 4q + 1)+1
\nAgain, substitute, 3q2+4q+1 = m, to get,
\nx2= 3m + 1\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (3)
\nHence, from equation 1, 2 and 3, we can say that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.<\/h3>\n

5. Use Euclid\u2019s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.<\/h2>\n

Solution:
\nLet x be any positive integer and y = 3.
\nBy Euclid\u2019s division algorithm, then,
\nx = 3q+r, where q\u22650 and r = 0, 1, 2, as r \u2265 0 and r < 3.
\nTherefore, putting the value of r, we get,
\nx = 3q
\nor
\nx = 3q + 1
\nor
\nx = 3q + 2
\nNow, by taking the cube of all the three above expressions, we get,
\nCase (i):\u00a0When r = 0, then,
\nx2= (3q)3\u00a0= 27q3= 9(3q3)= 9m; where m = 3q3
\nCase (ii):\u00a0When r = 1, then,
\nx3\u00a0= (3q+1)3\u00a0= (3q)3\u00a0+13+3\u00d73q\u00d71(3q+1) = 27q3+1+27q2+9q
\nTaking 9 as common factor, we get,
\nx3\u00a0= 9(3q3+3q2+q)+1
\nPutting = m, we get,
\nPutting (3q3+3q2+q) = m, we get ,
\nx3\u00a0= 9m+1
\nCase (iii): When r = 2, then,
\nx3\u00a0= (3q+2)3= (3q)3+23+3\u00d73q\u00d72(3q+2) = 27q3+54q2+36q+8
\nTaking 9 as common factor, we get,
\nx3=9(3q3+6q2+4q)+8
\nPutting (3q3+6q2+4q) = m, we get ,
\nx3\u00a0= 9m+8
\nTherefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.<\/h3>\n

Exercise 1.2<\/h2>\n

1. Express each number as a product of its prime factors:
\n(i) 140
\n(ii) 156
\n(iii) 3825
\n(iv) 5005
\n(v) 7429<\/h2>\n

Solutions:
\n(i) 140
\nBy Taking the\u00a0LCM of 140, we will get the product of its prime factor.
\nTherefore, 140 = 2 \u00d7 2 \u00d7 5 \u00d7 7 \u00d7 1 = 22\u00d75\u00d77
\n(ii) 156
\nBy Taking the\u00a0LCM of 156, we will get the product of its prime factor.
\nHence, 156 = 2 \u00d7 2 \u00d7 13 \u00d7 3 \u00d7 1 = 22\u00d7 13 \u00d7 3
\n(iii) 3825
\nBy Taking the\u00a0LCM of 3825, we will get the product of its prime factor.
\nHence, 3825 = 3 \u00d7 3 \u00d7 5 \u00d7 5 \u00d7 17 \u00d7 1 = 32\u00d752\u00d717
\n(iv) 5005
\nBy Taking the\u00a0LCM of 5005, we will get the product of its prime factor.
\nHence, 5005 = 5 \u00d7 7 \u00d7 11 \u00d7 13 \u00d7 1 = 5 \u00d7 7 \u00d7 11 \u00d7 13
\n(v) 7429
\nBy Taking the\u00a0LCM of 7429, we will get the product of its prime factor.
\nHence, 7429 = 17 \u00d7 19 \u00d7 23 \u00d7 1 = 17 \u00d7 19 \u00d7 23<\/h3>\n

2. Find the LCM and HCF of the following pairs of integers and verify that LCM \u00d7 HCF = product of the two numbers.
\n(i) 26 and 91
\n(ii) 510 and 92
\n(iii) 336 and 54<\/h2>\n

Solutions:
\n(i) 26 and 91
\nExpressing 26 and 91 as product of its prime factors, we get,
\n26 = 2 \u00d7 13 \u00d7 1
\n91 = 7 \u00d7 13 \u00d7 1
\nTherefore, LCM (26, 91) = 2 \u00d7 7 \u00d7 13 \u00d7 1 = 182
\nAnd HCF (26, 91) = 13
\nVerification
\nNow, product of 26 and 91 = 26 \u00d7 91 = 2366
\nAnd Product of LCM and HCF = 182 \u00d7 13 = 2366
\nHence, LCM \u00d7 HCF = product of the 26 and 91.<\/h3>\n

(ii) 510 and 92
\nExpressing 510 and 92 as product of its prime factors, we get,
\n510 = 2 \u00d7 3 \u00d7 17 \u00d7 5 \u00d7 1
\n92 = 2 \u00d7 2 \u00d7 23 \u00d7 1
\nTherefore, LCM(510, 92) = 2 \u00d7 2 \u00d7 3 \u00d7 5 \u00d7 17 \u00d7 23 = 23460
\nAnd HCF (510, 92) = 2
\nVerification
\nNow, product of 510 and 92 = 510 \u00d7 92 = 46920
\nAnd Product of LCM and HCF = 23460 \u00d7 2 = 46920
\nHence, LCM \u00d7 HCF = product of the 510 and 92.<\/h3>\n

(iii) 336 and 54
\nExpressing 336 and 54 as product of its prime factors, we get,
\n336 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 7 \u00d7 3 \u00d7 1
\n54 = 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 1
\nTherefore, LCM(336, 54) = = 3024
\nAnd HCF(336, 54) = 2\u00d73 = 6
\nVerification
\nNow, product of 336 and 54 = 336 \u00d7 54 = 18,144
\nAnd Product of LCM and HCF = 3024 \u00d7 6 = 18,144
\nHence, LCM \u00d7 HCF = product of the 336 and 54.<\/h3>\n

3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
\n(i) 12, 15 and 21
\n(ii) 17, 23 and 29
\n(iii) 8, 9 and 25<\/h2>\n

Solutions:
\n(i) 12, 15 and 21
\nWriting the product of prime factors for all the three numbers, we get,
\n12=2\u00d72\u00d73
\n15=5\u00d73
\n21=7\u00d73
\nTherefore,
\nHCF(12,15,21) = 3
\nLCM(12,15,21) = 2 \u00d7 2 \u00d7 3 \u00d7 5 \u00d7 7 = 420<\/h3>\n

(ii) 17, 23 and 29
\nWriting the product of prime factors for all the three numbers, we get,
\n17=17\u00d71
\n23=23\u00d71
\n29=29\u00d71
\nTherefore,
\nHCF(17,23,29) = 1
\nLCM(17,23,29) = 17 \u00d7 23 \u00d7 29 = 11339<\/h3>\n

(iii) 8, 9 and 25
\nWriting the product of prime factors for all the three numbers, we get,
\n8=2\u00d72\u00d72\u00d71
\n9=3\u00d73\u00d71
\n25=5\u00d75\u00d71
\nTherefore,
\nHCF(8,9,25)=1
\nLCM(8,9,25) = 2\u00d72\u00d72\u00d73\u00d73\u00d75\u00d75 = 1800<\/h3>\n

4. Given that HCF (306, 657) = 9, find LCM (306, 657).<\/h2>\n

Solution:\u00a0As we know that,
\nHCF\u00d7LCM=Product of the two given numbers
\nTherefore,
\n9 \u00d7 LCM = 306 \u00d7 657
\nLCM = (306\u00d7657)\/9 = 22338
\nHence, LCM(306,657) = 22338<\/h3>\n

5. Check whether 6n\u00a0can end with the digit 0 for any natural number n.<\/h2>\n

Solution:\u00a0If the number\u00a06n\u00a0ends with the digit zero (0), then it should be divisible by 5, as we know any number with unit place as 0 or 5 is divisible by 5.
\nPrime factorization of 6n\u00a0= (2\u00d73)n
\nTherefore, the prime factorization of\u00a06n\u00a0doesn\u2019t contain prime number 5.
\nHence, it is clear that for any natural number n, 6n\u00a0is not divisible by 5 and thus it proves that 6n\u00a0cannot end with the digit 0 for any natural number n.<\/h3>\n

6. Explain why 7 \u00d7 11 \u00d7 13 + 13 and 7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 5 are composite numbers.<\/h2>\n

Solution:\u00a0By the definition of composite number, we know, if a number is composite, then it means it has factors other than 1 and itself. Therefore, for the given expression;
\n7 \u00d7 11 \u00d7 13 + 13
\nTaking 13 as common factor, we get,
\n=13(7\u00d711\u00d71+1) = 13(77+1) = 13\u00d778 = 13\u00d73\u00d72\u00d713
\nHence, 7 \u00d7 11 \u00d7 13 + 13 is a composite number.
\nNow let\u2019s take the other number,
\n7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 5
\nTaking 5 as a common factor, we get,
\n=5(7\u00d76\u00d74\u00d73\u00d72\u00d71+1) = 5(1008+1) = 5\u00d71009
\nHence, 7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 5 is a composite number.<\/h3>\n

7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?<\/h2>\n

Solution:\u00a0Since, Both Sonia and Ravi move in the same direction and at the same time, the method to find the time when they will be meeting again at the starting point is LCM of 18 and 12.
\nTherefore, LCM(18,12) = 2\u00d73\u00d73\u00d72\u00d71=36
\nHence, Sonia and Ravi will meet again at the starting point after 36 minutes.<\/h3>\n

Exercise 1.3<\/h2>\n

1. Prove that \u221a5\u00a0is irrational.<\/h2>\n

Solutions:\u00a0Let us assume, that\u00a0\u221a5 is rational number.
\ni.e.\u00a0\u221a5 = x\/y (where, x and y are co-primes)
\ny\u221a5= x
\nSquaring both the sides, we get,
\n(y\u221a5)2\u00a0= x2
\n\u21d25y2\u00a0= x2\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (1)
\nThus, x2\u00a0is divisible by 5, so x is also divisible by 5.
\nLet us say, x = 5k, for some value of k and substituting the value of x in equation (1), we get,
\n5y2\u00a0= (5k)2
\n\u21d2y2\u00a0= 5k2
\nis divisible by 5 it means y is divisible by 5.
\nClearly, x and y are not co-primes. Thus, our assumption about\u00a0\u221a5 is rational is incorrect.
\nHence,\u00a0\u221a5 is an irrational number.<\/h3>\n

2. Prove that 3 + 2\u221a5 + is irrational.<\/h2>\n

Solutions:\u00a0Let us assume 3 + 2\u221a5 is rational.
\nThen we can find co-prime x and y (y \u2260 0) such that 3 + 2\u221a5 = x\/y
\nRearranging, we get,<\/h3>\n

\"\"
\nSince, x and y are integers, thus,<\/h3>\n

\"\"
\nis a rational number.
\nTherefore,\u00a0\u221a5 is also a rational number. But this contradicts the fact that\u00a0\u221a5 is irrational.
\nSo, we conclude that 3 + 2\u221a5 is irrational.<\/h3>\n

3. Prove that the following are irrationals:
\n(i) 1\/\u221a2
\n(ii) 7\u221a5
\n(iii) 6 +\u00a0\u221a2<\/h2>\n

Solutions:
\n(i) 1\/\u221a2
\nLet us assume 1\/\u221a2 is rational.
\nThen we can find co-prime x and y (y \u2260 0) such that 1\/\u221a2 = x\/y
\nRearranging, we get,
\n\u221a2 = y\/x
\nSince, x and y are integers, thus, \u221a2 is a rational number, which contradicts the fact that \u221a2 is irrational.
\nHence, we can conclude that 1\/\u221a2 is irrational.<\/h3>\n

(ii) 7\u221a5
\nLet us assume 7\u221a5 is a rational number.
\nThen we can find co-prime a and b (b \u2260 0) such that 7\u221a5 = x\/y
\nRearranging, we get,
\n\u221a5 = x\/7y
\nSince, x and y are integers, thus, \u221a5 is a rational number, which contradicts the fact that \u221a5 is irrational.
\nHence, we can conclude that 7\u221a5 is irrational.<\/h3>\n

(iii) 6 +\u221a2
\nLet us assume 6 +\u221a2 is a rational number.
\nThen we can find co-primes x and y (y \u2260 0) such that 6 +\u221a2 = x\/y\u22c5
\nRearranging, we get,
\n\u221a2 = (x\/y) \u2013 6
\nSince, x and y are integers, thus (x\/y) \u2013 6 is a rational number and therefore, \u221a2 is rational. This contradicts the fact that \u221a2 is an irrational number.
\nHence, we can conclude that 6 +\u221a2 is irrational.<\/h3>\n

Exercise 1.4<\/h2>\n

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:<\/h2>\n

(i)\u00a013\/3125 (ii) 17\/8 (iii) 64\/455 (iv) 15\/1600 (v) 29\/343 (vi) 23\/(2352) (vii) 129\/(225775) (viii) 6\/15 (ix) 35\/50 (x) 77\/210<\/h2>\n

Solutions:
\nNote:\u00a0If the denominator has only factors of 2 and 5 or in the form of 2m\u00a0\u00d75n\u00a0then it has terminating decimal expansion.
\nIf the denominator has factors other than 2 and 5 then it has a non-terminating decimal expansion.<\/h3>\n

(i) 13\/3125
\nFactorizing the denominator, we get,
\n3125 = 5 \u00d7 5 \u00d7 5 \u00d7 5 \u00d7 5 = 55
\nSince, the denominator has only 5 as its factor, 13\/3125 has a terminating decimal expansion.<\/h3>\n

(ii) 17\/8
\nFactorizing the denominator, we get,
\n8 = 2\u00d72\u00d72 = 23
\nSince, the denominator has only 2 as its factor, 17\/8 has a terminating decimal expansion.<\/h3>\n

(iii) 64\/455
\nFactorizing the denominator, we get,
\n455 = 5\u00d77\u00d713
\nSince, the denominator is not in the form of 2m\u00a0\u00d7 5n, thus 64\/455 has a non-terminating decimal expansion.<\/h3>\n

(iv) 15\/ 1600
\nFactorizing the denominator, we get,
\n1600 = 26\u00d752
\nSince, the denominator is in the form of 2m\u00a0\u00d7 5n, thus 15\/1600 has a terminating decimal expansion.<\/h3>\n

(v) 29\/343
\nFactorizing the denominator, we get,
\n343 = 7\u00d77\u00d77 = 73\u00a0Since, the denominator is not in the form of 2m\u00a0\u00d7 5n\u00a0thus 29\/343 has a non-terminating decimal expansion.<\/h3>\n

(vi)23\/(2352)
\nClearly, the denominator is in the form of 2m\u00a0\u00d7 5n.
\nHence, 23\/ (2352) has a terminating decimal expansion.<\/h3>\n

(vii) 129\/(225775)
\nAs you can see, the denominator is not in the form of 2m\u00a0\u00d7 5n.
\nHence, 129\/ (225775) has a non-terminating decimal expansion.<\/h3>\n

(viii) 6\/15
\n6\/15 = 2\/5
\nSince, the denominator has only 5 as its factor, thus, 6\/15 has a terminating decimal expansion.<\/h3>\n

(ix) 35\/50
\n35\/50 = 7\/10
\nFactorising the denominator, we get,
\n10 = 2 \u00d7 5
\nSince, the denominator is in the form of 2m\u00a0\u00d7 5n\u00a0thus, 35\/50 has a terminating decimal expansion.
\n(x) 77\/210
\n77\/210 = (7\u00d7 11)\/ (30 \u00d7 7) = 11\/30
\nFactorising the denominator, we get,
\n30 = 2 \u00d7 3 \u00d7 5
\nAs you can see, the denominator is not in the form of 2m\u00a0\u00d7 5n\u00a0.Hence, 77\/210 has a non-terminating decimal expansion.<\/h3>\n

2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.<\/h2>\n

Solutions:
\n(i) 13\/3125<\/h3>\n

\"\"
\n13\/3125 = 0.00416
\n(ii) 17\/8<\/h3>\n

\"\"<\/p>\n

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17\/8 = 2.125
\n(iii) 64\/455\u00a0has a Non terminating decimal expansion<\/h3>\n

(iv)15\/ 1600<\/h3>\n

\"\"
\n15\/1600 = 0.009375
\n(v) 29\/ 343\u00a0has a Non terminating decimal expansion<\/h3>\n

(vi)23\/ (2352)\u00a0= 23\/(8\u00d725)= 23\/200<\/h3>\n

\"\"
\n23\/ (2352) = 0.115
\n(vii) 129\/ (225775)\u00a0has a Non terminating decimal expansion<\/h3>\n

(viii) 6\/15\u00a0= 2\/5<\/h3>\n

\"\"
\n(ix) 35\/50\u00a0= 7\/10<\/h3>\n

\"\"
\n35\/50 = 0.7
\n(x) 77\/210\u00a0has a non-terminating decimal expansion.<\/h3>\n

3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form, p q what can you say about the prime factors of q?
\n(i) 43.123456789
\n(ii) 0.120120012000120000. . .<\/h2>\n

\"\"<\/p>\n

Solutions:
\n(i) 43.123456789
\nSince it has a terminating decimal expansion, it is a rational number in the form of p\/q and q has factors of 2 and 5 only.
\n(ii) 0.120120012000120000. . .
\nSince, it has non-terminating and non- repeating decimal expansion, it is an irrational number.<\/h3>\n

\"\"
\nSince it has non-terminating but repeating decimal expansion, it is a rational number in the form of p\/q and q has factors other than 2 and 5.<\/h3>\n","protected":false},"excerpt":{"rendered":"

Class 10 Mathematics NCERT book solutions for Chapter 1 – Real Numbers Questions and Answers.<\/p>\n","protected":false},"author":21830,"featured_media":117421,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-118910","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/118910","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=118910"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/118910\/revisions"}],"predecessor-version":[{"id":118921,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/118910\/revisions\/118921"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/117421"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=118910"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=118910"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=118910"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}