{"id":16678,"date":"2019-12-31T14:43:21","date_gmt":"2019-12-31T09:13:21","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=16678"},"modified":"2021-07-02T15:33:05","modified_gmt":"2021-07-02T10:03:05","slug":"1st-puc-maths-question-bank-chapter-15","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-maths-question-bank-chapter-15\/","title":{"rendered":"1st PUC Maths Question Bank Chapter 15 Statistics"},"content":{"rendered":"

Students can Download Maths Chapter 15 Statistics Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Maths Question Bank Chapter 15 Statistics<\/h2>\n

Question 1.
\nFind the mean deviation about the mean for the following data
\n(i) 6, 7, 10, 12, 13, 4, 8, 12
\n(ii) 4, 7, 8, 9, 10, 12,13,17
\n(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
\nAnswer:
\n\"1st
\n\"1st<\/p>\n

Question 2.
\nFind the mean deviation about the mean for the data.
\n12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5.
\nAnswer:
\nTry your self.
\n\\(N=20, \\bar{x}=10, \\Sigma\\left|x_{i}-\\bar{x}\\right|=124, \\mathrm{M} \\cdot \\mathrm{D} \\cdot(\\bar{x})=6 \\cdot 2\\)<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nFind the mean deviation about the median for the data:
\n(i) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
\n(ii) 36, 72, 46, 60, 45, 53, 46, 51, 49, 42
\n(iii) 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21
\nAnswer:
\n(i) Arrange the data in ascending order as
\n\"1st
\n\"1st<\/p>\n

Question 4.
\nFind the mean deviation about the mean for the following data.
\n\"1st
\nAnswer:
\n\"1st
\n\"1st<\/p>\n

\"1st
\n\"KSEEB<\/p>\n

Question 5.
\nFind the mean deviation about the median for the data,
\n\"1st
\n\"1st
\nAnswer:
\n\"1st
\n\"1st
\n\"1st
\nQuestion 6.
\nFind the mean deviation about the mean for the following data.
\n(i)<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
Income per day<\/strong><\/td>\nNumber of persons<\/strong><\/td>\n<\/tr>\n
0 – 100<\/td>\n4<\/td>\n<\/tr>\n
100 – 200<\/td>\n8<\/td>\n<\/tr>\n
200 – 300<\/td>\n9<\/td>\n<\/tr>\n
300 – 400<\/td>\n10<\/td>\n<\/tr>\n
400 – 500<\/td>\n7<\/td>\n<\/tr>\n
500 – 600<\/td>\n5<\/td>\n<\/tr>\n
600 – 700<\/td>\n4<\/td>\n<\/tr>\n
700 – 800<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(ii)<\/p>\n\n\n\n\n\n\n\n\n\n
Height (cms)<\/strong><\/td>\nNumber of boys<\/strong><\/td>\n<\/tr>\n
95 – 105<\/td>\n9<\/td>\n<\/tr>\n
105 – 115<\/td>\n13<\/td>\n<\/tr>\n
115 – 125<\/td>\n26<\/td>\n<\/tr>\n
125 – 135<\/td>\n30<\/td>\n<\/tr>\n
135 – 145<\/td>\n12<\/td>\n<\/tr>\n
145 – 155<\/td>\n10<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB<\/p>\n

(iii)<\/p>\n\n\n\n\n\n\n\n\n\n
Marks obtained<\/strong><\/td>\nNumber of students<\/strong><\/td>\n<\/tr>\n
0 – 10<\/td>\n12<\/td>\n<\/tr>\n
10 – 20<\/td>\n18<\/td>\n<\/tr>\n
20 – 30<\/td>\n27<\/td>\n<\/tr>\n
30 – 40<\/td>\n20<\/td>\n<\/tr>\n
40 – 50<\/td>\n17<\/td>\n<\/tr>\n
50 – 60<\/td>\n6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(iv)<\/p>\n\n\n\n\n\n\n\n\n\n\n
Marks obtained<\/strong><\/td>\nNumber of students<\/strong><\/td>\n<\/tr>\n
10 – 20<\/td>\n2<\/td>\n<\/tr>\n
20 – 30<\/td>\n3<\/td>\n<\/tr>\n
30 – 40<\/td>\n8<\/td>\n<\/tr>\n
40 – 50<\/td>\n14<\/td>\n<\/tr>\n
50 – 60<\/td>\n8<\/td>\n<\/tr>\n
60 – 70<\/td>\n3<\/td>\n<\/tr>\n
70 – 80<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\n\"1st<\/p>\n

Method (II)
\n\"1st<\/p>\n

Method (I)
\n\"1st<\/p>\n

Method (ii)
\nHere a = 130, h = 10
\n\"1st<\/p>\n

(iii) Method (I)
\n\"1st<\/p>\n

Question 7.
\nFind the mean deviation about the median for the data:
\n(i)<\/p>\n\n\n\n\n\n\n\n\n\n
Class<\/strong><\/td>\nFrequency<\/strong><\/td>\n<\/tr>\n
0 – 10<\/td>\n6<\/td>\n<\/tr>\n
10 – 20<\/td>\n7<\/td>\n<\/tr>\n
20 – 30<\/td>\n15<\/td>\n<\/tr>\n
30 – 40<\/td>\n16<\/td>\n<\/tr>\n
40 – 50<\/td>\n4<\/td>\n<\/tr>\n
50 – 60<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(ii)<\/p>\n\n\n\n\n\n\n\n\n\n
Marks<\/strong><\/td>\nNumber of Girls<\/strong><\/td>\n<\/tr>\n
0 – 10<\/td>\n6<\/td>\n<\/tr>\n
10 – 20<\/td>\n8<\/td>\n<\/tr>\n
20 – 30<\/td>\n14<\/td>\n<\/tr>\n
30 – 40<\/td>\n16<\/td>\n<\/tr>\n
40 – 50<\/td>\n4<\/td>\n<\/tr>\n
50 – 60<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\n\"1st
\n(ii)
\n\"1st
\n\"1st
\n\"KSEEB<\/p>\n

Question 8.
\nCalculate the mean deviation about median age for the age distribution of 100 persons given below.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
Age<\/strong><\/td>\nNumber<\/strong><\/td>\n<\/tr>\n
16 – 20<\/td>\n5<\/td>\n<\/tr>\n
21 – 25<\/td>\n6<\/td>\n<\/tr>\n
26 – 30<\/td>\n12<\/td>\n<\/tr>\n
31 – 35<\/td>\n14<\/td>\n<\/tr>\n
36 – 40<\/td>\n26<\/td>\n<\/tr>\n
41- 45<\/td>\n12<\/td>\n<\/tr>\n
46 – 50<\/td>\n16<\/td>\n<\/tr>\n
51 – 55<\/td>\n9<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\nFirst, modify the classes to make the data. continuous by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit.
\n\"1st
\n\"1st<\/p>\n

Question 9.
\nFind Mean and variance for each data:
\n(i) 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
\n(ii) 6, 7, 10, 12, 13, 4, 8, 12
\n(iii) First n natural numbers.
\n(iv) First 10 multiples of 3.
\nAnswer:
\n\"1st
\n\"1st
\n\"1st
\n\"1st
\nQuestion 10.
\nFind mean and variance.
\n(i)<\/p>\n\n\n\n\n
xi<\/sub><\/td>\n6<\/td>\n10<\/td>\n14<\/td>\n18<\/td>\n24<\/td>\n28<\/td>\n30<\/td>\n<\/tr>\n
fi<\/sub><\/td>\n2<\/td>\n4<\/td>\n7<\/td>\n12<\/td>\n8<\/td>\n4<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(ii)<\/p>\n\n\n\n\n
xi<\/sub><\/td>\n92<\/td>\n93<\/td>\n97<\/td>\n98<\/td>\n102<\/td>\n104<\/td>\n109<\/td>\n<\/tr>\n
fi<\/sub><\/td>\n3<\/td>\n2<\/td>\n3<\/td>\n2<\/td>\n6<\/td>\n3<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\n(i)
\n\"1st
\n(ii)
\n\"1st<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nFind the mean and standard deviation using short-cut method.<\/p>\n\n\n\n\n
xi<\/sub><\/td>\n60<\/td>\n61<\/td>\n62<\/td>\n63<\/td>\n64<\/td>\n65<\/td>\n66<\/td>\n67<\/td>\n68<\/td>\n<\/tr>\n
fi<\/sub><\/td>\n2<\/td>\n1<\/td>\n12<\/td>\n29<\/td>\n25<\/td>\n12<\/td>\n10<\/td>\n4<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\n\"1st
\n\"1st<\/p>\n

Question 12.
\nFind the mean and variance
\n(i)<\/p>\n\n\n\n\n\n\n\n\n\n\n
Class<\/strong><\/td>\nFrequency<\/strong><\/td>\n<\/tr>\n
0 – 30<\/td>\n2<\/td>\n<\/tr>\n
30 – 60<\/td>\n3<\/td>\n<\/tr>\n
60 – 90<\/td>\n5<\/td>\n<\/tr>\n
90 – 120<\/td>\n10<\/td>\n<\/tr>\n
120 – 150<\/td>\n3<\/td>\n<\/tr>\n
150 – 180<\/td>\n5<\/td>\n<\/tr>\n
180 – 210<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(ii)<\/p>\n\n\n\n\n\n\n\n\n
Class<\/strong><\/td>\nFrequency<\/strong><\/td>\n<\/tr>\n
0 – 10<\/td>\n5<\/td>\n<\/tr>\n
10 – 20<\/td>\n8<\/td>\n<\/tr>\n
20 – 30<\/td>\n15<\/td>\n<\/tr>\n
30 – 40<\/td>\n16<\/td>\n<\/tr>\n
40 – 50<\/td>\n6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\n\"1st
\n\"1st<\/p>\n

Question 13.
\nFind the mean, variance and standard deviation using short-cut method.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n
Height (cms)<\/strong><\/td>\nNo of Children<\/strong><\/td>\n<\/tr>\n
70 – 75<\/td>\n3<\/td>\n<\/tr>\n
75 – 80<\/td>\n4<\/td>\n<\/tr>\n
80 – 85<\/td>\n7<\/td>\n<\/tr>\n
85 – 90<\/td>\n7<\/td>\n<\/tr>\n
90 – 95<\/td>\n15<\/td>\n<\/tr>\n
95 – 100<\/td>\n9<\/td>\n<\/tr>\n
100 – 105<\/td>\n6<\/td>\n<\/tr>\n
105 – 110<\/td>\n6<\/td>\n<\/tr>\n
110 – 115<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\n\"1st
\n\"1st<\/p>\n

Question 14.
\nThe diameters of circles (in mm ) drawn in a design are given below calculate S.D and mean diameter.<\/p>\n\n\n\n\n\n\n\n\n
Diameter<\/strong><\/td>\nCircles<\/strong><\/td>\n<\/tr>\n
70 – 75<\/td>\n15<\/td>\n<\/tr>\n
75 – 80<\/td>\n17<\/td>\n<\/tr>\n
80 – 85<\/td>\n21<\/td>\n<\/tr>\n
85 – 90<\/td>\n22<\/td>\n<\/tr>\n
90 – 95<\/td>\n25<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\n\"1st
\n\"1st<\/p>\n

\"KSEEB<\/p>\n

Question 15.
\nFrom the data given below state which group is more variable, A or B?<\/p>\n\n\n\n\n\n\n\n\n\n\n
Marks<\/strong><\/td>\nGroup A<\/strong><\/td>\nGroup B<\/strong><\/td>\n<\/tr>\n
10 – 20<\/td>\n9<\/td>\n10<\/td>\n<\/tr>\n
20 – 30<\/td>\n17<\/td>\n20<\/td>\n<\/tr>\n
30 – 40<\/td>\n32<\/td>\n30<\/td>\n<\/tr>\n
40 – 50<\/td>\n33<\/td>\n25<\/td>\n<\/tr>\n
50 – 60<\/td>\n40<\/td>\n143<\/td>\n<\/tr>\n
60 – 70<\/td>\n10<\/td>\n15<\/td>\n<\/tr>\n
70 – 80<\/td>\n9<\/td>\n7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\nWe know that, the group having greater C.V.(co-efficient of variation) is said w be more variable than other. So, we compute mean and S.D. for groups A and B.
\nFirst we compute mean and S.D. for A.
\n\"1st
\n\"1st
\n\"1st
\nClearly, B having greater C. V. than A. Therefore B is more variable.<\/p>\n

Note:
\nFor two frequency distributions with equal means, the series with greater S.D. is more variable than the other.<\/p>\n

\"KSEEB<\/p>\n

Question 16.
\nFrom the prices of shares X and Y below, find out which is more stable in value.
\n\"1st
\nAnswer:
\nHere, number of items for both = 10.
\nFirst we compute mean and S.D. for x, taking a = 56, N = 10.
\n\"1st
\n\"1st
\n\"1st
\nQuestion 17.
\nAn analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results.<\/p>\n\n\n\n\n\n\n
<\/td>\nA<\/td>\nB<\/td>\n<\/tr>\n
No. of wage earners<\/td>\n586<\/td>\n648<\/td>\n<\/tr>\n
Mean of monthly wages (Rs )<\/td>\n5253<\/td>\n5253<\/td>\n<\/tr>\n
Variance of the distribution of wages<\/td>\n100<\/td>\n121<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(i) Which firm A or B pays larger amount as monthly wages?
\n(ii) Which firm A or B shows greater variability in individual wages?
\nAnswer:
\n(i) Amount of monthly wages paid by firm A.
\n= 586 x mean wages
\n= 586 x 5253 = Rs.3078258
\nAnd amount of monthly wages paid by form B
\n= 648 x mean wages
\n= 648 x 5253 = Rs. 3403944.
\nClearly, firm B pays more wages.
\nAlternatively,
\nThe variance of wages in A is 100.
\n\u2234 S.D. of distribution of wage in A = 10
\nAlso the variance of distribution of wages in firm B is 121.
\n\u2234 S.D. of distribution of wages in B = 11.
\nsince the average monthly wages in both firms is same i.e., Rs. 5253, therefore the firm with greater S.D. will have more variability.
\n\u200d\u2234 Firm B pays more wages.<\/p>\n

(ii) Firm B with greater S.D. shows greater variability in individual wages.<\/p>\n

\"KSEEB<\/p>\n

Question 18.
\nTwo plants A and B of a factory show following results about the number of workers and the wages paid to them.<\/p>\n\n\n\n\n\n\n
<\/td>\nA<\/td>\nB<\/td>\n<\/tr>\n
No – of workers<\/td>\n5000<\/td>\n6000<\/td>\n<\/tr>\n
Average monthly wages<\/td>\n2500<\/td>\n2500<\/td>\n<\/tr>\n
Variance distribution of wages<\/td>\n81<\/td>\n100<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

In which plant A or B is there greater variability in individual wages?
\nAnswer:
\nB. (Try yourself) (Ex. 13 Text book)<\/p>\n

Question 19.
\nThe following is the record of goals scored by team A in a football session:
\nFor the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?
\nAnswer:
\nFirst, compute mean number of goals and S.D. for team A.
\n\"1st
\nGiven : Mean number of goals for the team B is 2 and S.D. is 1.25
\nSince S.D. For A is less than S.D for B.
\n\u2234 A is more consistent.<\/p>\n

Question 20.
\nThe sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
\n\"1st
\nWhich is more varying, the length or weight?
\nAnswer:
\n\"1st<\/p>\n

\"KSEEB<\/p>\n

Question 21.
\nCo-efficient of variation of two distributions are 60 and 70, and the standard deviations are 21 and 16, respectively. What are their arithmetic means.
\nAnswer:
\nLet \\(\\bar{x}_{1} \\text { and } \\bar{x}_{2} \\) be arithmetic means of first and second distribution respectively. Then
\n\"1st<\/p>\n

Question 22.
\nThe following values are calculated in respect of heights and weight of the students of a section of class XI.<\/p>\n\n\n\n\n\n
<\/td>\nHeight<\/strong><\/td>\nWeight<\/strong><\/td>\n<\/tr>\n
Mean<\/td>\n162 – 6 cm<\/td>\n52-36 kg<\/td>\n<\/tr>\n
Variance<\/td>\n127 – 69 cm2<\/sup><\/td>\n23. 1361kg2<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Can we say that the weight show greater variation than the height.
\nAnswer:
\n\"1st
\nClearly C.V. (weights) is greater than C.V. (heights). Weights show more variability than heights.<\/p>\n

Miscellaneous Examples<\/strong><\/p>\n

Question 1.
\nThe mean and variance of eight observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13. Find the remaining two observations.
\nAnswer:
\nLet the remaining two observation be x and y.
\n\"1st<\/p>\n

\"1st<\/p>\n

Question 2.
\nThe mean of 5 observations is 4.4 and their variance is 8.24. If three observations are 1, 2 and 6, find the remaining two observations.
\nAnswer:
\nLet the remaining two observations be x and
\n\u2234 Five observations are 1, 2, 6, x, y.
\n\"1st
\n\"1st<\/p>\n

Question 3.
\nThe mean and variance of 7 observations are 8 and 16 respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.
\nAnswer:
\n6 and 8 (Try yourself)
\nHere, x + y = 14, x2<\/sup> + y2<\/sup> = 100<\/p>\n

Question 4.
\nThe mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation.
\nAnswer:
\nLet six observations be
\n\"1st<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nGiven that \\(\\bar{x} \\)is the mean and \u03c32<\/sup> is the variance of n observations x1<\/sub> , x2<\/sub>,……….. ,xn<\/sub> . Prove that the mean and variance of the observations ax1<\/sub>,ax2<\/sub>,………………,axn<\/sub> are \\(a \\bar{x} \\) and a2<\/sup>a1<\/sup> respectively (a \u2260 0)
\nAnswer:
\n\"1st<\/p>\n

Question 6.
\nIf each of the observation x1<\/sub>,x2<\/sub>,………. ,xn<\/sub> is increased by \u2018a\u2019, where a is positive or negative number, show that the variance remains unchanged.
\nAnswer:
\n\"1st
\n\"1st<\/p>\n

Question 7.
\nThe mean and standard deviation 20 observations are found to be 10 and 2 respectively. On rechecking, it was fond that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases :
\n(i) If wrong item is omitted,
\n(ii) If it is replaced by 12.
\nAnswer:
\n\"1st
\n\"1st
\n\"1st<\/p>\n

Question 8.
\nThe mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?
\nAnswer:
\nGiven: N = 100
\nIncorrect mean = 40 = \\(a \\bar{x} \\)
\nIncorrect S.D. = 5.1
\nWe have, \\(\\bar{x}=\\frac{1}{N} \\Sigma x_{i}\\)
\n\u21d2\\(\\Sigma x_{i}=N \\bar{x}=100 \\times 40=4000\\)
\n\u2234 Incorrect sum of observations = 4000.
\n\u2234 Correct sum of observations = 4000 – 50 + 40 = 3990.
\n\"1st<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nThe mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find and standard deviation if the incorrect observations are omitted.
\nAnswer:
\nGiven N = 100,New N = 97
\n\"1st
\n\"1st
\nQuestion 10.
\nThe mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below.<\/p>\n\n\n\n\n\n
Subject<\/strong><\/td>\nMathematics<\/strong><\/td>\nPhysics<\/strong><\/td>\nChemistry<\/strong><\/td>\n<\/tr>\n
Mean<\/td>\n42<\/td>\n32<\/td>\n40 – 9<\/td>\n<\/tr>\n
S D<\/td>\n12<\/td>\n15<\/td>\n20<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Which of three subjects shows the highest variability in marks and which shows the lowest?
\nAnswer:
\n\"1st
\nChemistry shows high variability and Mathematics shows lowest variability.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 15 Statistics Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 1st PUC Maths Question Bank Chapter 15 Statistics Question 1. Find the mean deviation about the mean for …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/16678"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=16678"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/16678\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=16678"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=16678"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=16678"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}