{"id":13240,"date":"2019-12-19T11:02:33","date_gmt":"2019-12-19T05:32:33","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=13240"},"modified":"2021-07-02T15:35:53","modified_gmt":"2021-07-02T10:05:53","slug":"2nd-puc-statistics-question-bank-chapter-7","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-statistics-question-bank-chapter-7\/","title":{"rendered":"2nd PUC Statistics Question Bank Chapter 7 Statistical Quality control"},"content":{"rendered":"

You can Download Chapter 7 Statistical Quality Control Questions and Answers, Notes, 2nd PUC Statistics Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Statistics Question Bank Chapter 7 Statistical Quality Control<\/h2>\n

2nd PUC Statistical Quality Control One and Two Marks Questions With Answers<\/h3>\n

Question 1.
\nWhat do you mean by Statistical Quality Control?
\nAnswer:
\nThe control of quality of manufactured articles using statistical, techniques is called SQC.<\/p>\n

Question 2.
\nMention two types of causes for variation in a manufacturing process.
\nAnswer:
\nChance causes and Assignable causes.<\/p>\n

Question 3.
\nIn X\u0304 – chart, if one of the sample means lies outside the control lines, what would you conclude?
\nAnswer:
\nWe conclude that the production process is out of control’ or lack of control.<\/p>\n

Question 4.
\nSQC helps in which type of variation?\/ Which type of ’cause of variation’ is detected by SQC.
\nAnswer:
\nSQC helps in detecting Assignable causes.<\/p>\n

Question 5.
\nWhich cause of variation cannot be eliminated from the process control?
\nAnswer:
\nChance cause of variation.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nDistinguish between a defect and a defective\/what are Defects and Defectives?
\nAnswer:<\/p>\n

    \n
  • A defect is a quality characteristic, which does not conform to specifications.<\/li>\n
  • The presence of one or more defects in a product\/article is a defective.<\/li>\n<\/ul>\n

    Question 7.
    \nWrite a merit of ‘Acceptance sampling in “SQC”
    \nAnswer:
    \nIt is useful when the items are of destructive in nature.<\/p>\n

    Question 8.
    \nGive an example for a defect & defective
    \nAnswer:
    \nMisprints, damages in weaving on a cloth, a glass bottle with one or more bubble, excessive thickness of a glass pane, are all defects The cloth, glass bottle, glass pane are all defectives.<\/p>\n

    Question 9.
    \nName the Control Chart for variable
    \nAnswer:
    \nX\u0304 and R charts are used for the control of variables.<\/p>\n

    Question 10.
    \nName the control chart used in case of defective in S.Q.C.
    \nAnswer:
    \nnp\/d-chart<\/p>\n

    Question 11.
    \nWhat are process control and product control?.
    \nAnswer:<\/p>\n

      \n
    • Controlling the quality of the product during the manufacturing process itself is called process control.<\/li>\n
    • Controlling the quality of the finished products\/manufactured products is called product control.<\/li>\n<\/ul>\n

      Question 12.
      \nMention any two uses\/advantages of SQC
      \nAnswer:<\/p>\n

        \n
      • It improve the quality and to decrease the proportion of defectives<\/li>\n
      • It provides greater quality assurance at lower inspection cost.<\/li>\n<\/ul>\n

        Question 13.
        \nWhat are control charts\/What is a control chart?.
        \nAnswer:
        \nControl chart is graphical device, which used to verify whether the production process is in Statistical control or not.<\/p>\n

        Question 14.
        \nWhat is Acceptance-sampling \/ Acceptance sampling plan?
        \nAnswer:
        \nThe process of inspection of manufactured lot for acceptability is called Acceptance sampling plan<\/p>\n

        Question 15.
        \nWhat do you mean by Single and Double Sampling Plans?
        \nAnswer:
        \nIn a single sampling plan, the decision about accepting or rejecting a lot is based on one sample only. Where as in a double sampling plan, the decision is based on two samples drawn from a lot.<\/p>\n

        \"KSEEB<\/p>\n

        Question 16.
        \nWrite Upper and Lower control Limits for ‘number of defects’ chart when standards are given.
        \nAnswer:
        \nNumber of defects use C-chart: U.C.L = \\(\\lambda^{\\prime}+3 \\sqrt{\\lambda^{\\prime}}\\) L.C.L = \\(\\lambda^{1}-3 \\sqrt{\\lambda^{1}}\\)<\/p>\n

        Question 17.
        \nIf c\u0305 = 2 find U.C.L
        \nAnswer:
        \n\"2nd<\/p>\n

        Question 18.
        \nIf p’ = 0.13 and n = 100, find the value of central line(CL) for np-chart.
        \nAnswer:
        \nC.L = np’ = 100 \u00d7 0.13 = 13<\/p>\n

        Question 19.
        \nIf p’ = 0.02 and n = 100, calculate U.C.L for d – chart (np – chart).
        \nAnswer:
        \nnp’ + \\(3 \\sqrt{n P^{\\prime} Q^{\\prime}}\\) =100 \u00d7 0.02 + \\(3 \\sqrt{100 \\times 0.02 \\times 0.98}\\) = 3.4<\/p>\n

        Question 20.
        \nWhat do you mean by statistical quality control?
        \nAnswer:
        \nThe control of quality of manufactured products using statistical techniques is called statistical quality control.<\/p>\n

        Question 21.
        \nWhat are chance causes?
        \nAnswer:
        \nA small amount of variation in the quality of the product for no specific cause\/reason, and such variation is called chance causes.<\/p>\n

        Question 22.
        \nWhat are assignable causes?
        \nAnswer:
        \nVariation in the quality of the product due to use of substandard raw material, improper machine setting, wrong handling of machines, power variation etc., and such variations are called assignable causes.<\/p>\n

        Question 23.
        \nWhat is a control chart?
        \nAnswer:
        \nControl chart is a graphical device which is used to verify whether the production process is in statistical control or not.<\/p>\n

        Question 24.
        \nName the control chart used in case of defects in statistical quality control.
        \nAnswer:
        \nC-chart.<\/p>\n

        Question 25.
        \nWrite down control limits for x-charts when standards are given and not given.
        \nAnswer:
        \n\"2nd<\/p>\n

        \"KSEEB<\/p>\n

        Question 26.
        \nMention any two differences between single sampling plan and double sampling plan.
        \nAnswer:
        \nIn single sampling plan, decision about accepting or rejecting a lot is based on one sample only, whereas in double sampling plan decision is based on two samples.
        \nIn S.S.P. the lot is accepted if:
        \nd \u2264 c , where d = no. of defectives, c-acceptance number and<\/p>\n

        In D.S.P the lot is accepted if:
        \nd1<\/sub> \u2264 c2<\/sub> and (d1<\/sub> + d2<\/sub>) \u2264 c2<\/sub> .
        \nwhere d1<\/sub>,d2<\/sub> – no. of defectives ; c1<\/sub>, c2<\/sub> – acceptance number .<\/p>\n

        Question 27.
        \nWrite down control limits for R-charts when standards are given and not given. Ans
        \nAnswer:
        \n\"2nd
        \nWhere \u03c3’ – Standard given and \\(\\overline{\\mathrm{R}}=\\frac{\\Sigma \\mathrm{R}}{\\mathrm{K}}\\) ; and D1<\/sub> D2<\/sub>, D3<\/sub> and D4<\/sub> table constants.<\/p>\n

        Question 28.
        \nWrite down control limits for number of defectives (np\/d -chart) when standards given and not given.
        \nAnswer:
        \n\"2nd
        \nWrite P’ – given standard, \\(\\overline{\\mathrm{P}}=\\frac{\\Sigma d}{\\mathrm{nK}}\\) ; d – number of defectives, K – number of samples n – sample size<\/p>\n

        Question 29.
        \nWrite down control limits for number of defects (C – chart)
        \nAnswer:
        \n\"2nd
        \nWhere \u03bb’ – given standard, \\(\\overline{\\mathrm{C}}=\\frac{\\Sigma \\mathrm{C}}{\\mathrm{K}}\\) , and C – number of defects,
        \nK – number of samples<\/p>\n

        Question 30.
        \nGive two merits and demerits of single sampling plan and double sampling plan.
        \nAnswer:
        \n(a) Decision and testing procedure of single sampling plan is easier than double sampling plan.
        \n(b) Decision based on two samples is more reliable and has psychological advantage of giving the lot a second chance compared to single sampling plan.<\/p>\n

        Question 31.
        \nMention any two advantages and disadvantages of acceptance sampling plan.
        \nAnswer:
        \nAdvatage: When the items are of distructive in nature, it is less espensive, requires few staff as compared to 100% inspection.
        \nDisadvantages: There is a risk of accepting bad lot and rejecting good lot. Moreover double sampling plans more costly and time consuming.<\/p>\n

        Question 32.
        \nIf P|<\/sup> = 0.02 and n = 100, calculate the upper control limit for d-chart (np-chart).
        \nAnswer: U.C.L when standards given is :
        \nU.C.L = np’+\\(3 \\sqrt{n p^{\\prime} Q^{\\prime}}\\) = 100 \u00d7 0.02 + \\(3 \\sqrt{100 \\times 0.02 \\times 0.98}\\) = 2 + \\(3 \\sqrt{1.96}\\)
        \nU.C.L = 6.2<\/p>\n

        2nd PUC Statistical Quality Control Five Marks Questions With Answers<\/h3>\n

        Question 1.
        \nMeasurements in mean (X\u0304) and Range (R) from 10 samples each of size 5 gave results as X\u033f = 80, R\u0304=3.5 Find 3\u03c3 conttrol limits for drawing X\u0304 chart ( A2 <\/sub>\u00a0is given as 0.557)
        \nAnswer:
        \nGiven : X\u033f = 80, R\u0304 =3.5 n = 5
        \nThe control limits for X\u0304 chart are (standards are not given)
        \nC.L = X\u033f = 80
        \nL.C.L = X\u033f – A2<\/sub>R\u0304 =80-0.557 \u00d7 3.5 = 78.05
        \nU.C.L = X\u033f + A2<\/sub>R\u0304= 80 + 0.557 \u00d7 3.5 = 81.95<\/p>\n

        Question 2.
        \nA drill drills holes with standard deviation of depth 0.05 cms and mean depth of holes is 5.70cms. Find 3 \u03c3 control limits for mean of samples of size 4. Given A = 1.500.
        \nAnswer:
        \nGiven: X\u03041<\/sup> =5.70 cms : \u03c31<\/sup> =0.05 cms, n=4
        \nThe 3\u03c3 – control limits for X\u0304 – chart (standards are given) are
        \nC.L = X\u03041<\/sup> = 5.70
        \nL.C.L = X\u03041<\/sup> – A\u03c31<\/sup> =5.70 – 1,5 \u00d7 0.05 = 5.625
        \nU.C.L = X\u03041<\/sup> + A\u03c31<\/sup> =5-70 + 1.5 \u00d7 0.05 = 5.775<\/p>\n

        Question 3.
        \nIf X\u03041<\/sup> = 35.5 , \u03c31<\/sup> = 3.5 and n = 5. Write down the control limits for X\u0304 and R – charts.
        \nAnswer:
        \nGiven: X\u03041<\/sup> = 35.5, \u03c31<\/sup>=3.5, n = 5, A = 1.342 [From the table]
        \nThe control limits for X\u0304 – chart (standards are given) are:
        \nC.L = X\u03041<\/sup> = 35.5
        \nL.C.L = X\u03041<\/sup> – A\u03c31<\/sup> = 35.5 – 1.342 \u00d7 3.5 = 30.803
        \nU.C.L = X\u03041<\/sup> + A\u03c31<\/sup> = 35.5 + 1.342 \u00d7 3.5 = 40.197
        \nThe control limits for R – chart are (standards are given) : d2<\/sub> = 2.326, D1<\/sub> = 0, D2<\/sub> = 4.918 (From the table)
        \nC.L = d2<\/sub> \u03c31<\/sup> = 2.326 \u00d7 3.5 = 8.141
        \nL.C.L = D1<\/sub> \u03c31<\/sup> = 0 \u00d7 3.5 = 0 .
        \nU.C.L = D2<\/sub>\u03c31<\/sup> = 4.918 \u00d7 3.5 = 17.213<\/p>\n

        Question 4.
        \nIf X\u033f = 112.6, R\u0304 = 4.2, n = 4, write down the control limits for X\u0304 and R- charts.
        \nAnswer:
        \nGiven X\u033f = 112.6, R\u0304 = 4.2, n = 4, and the constants from the table A2<\/sub> = 0.729, D3<\/sub> = 0, D4<\/sub>= 2.115
        \nWhen standards are not given:
        \nThe control limits for R- chart are:
        \nC.L = X\u033f = 112.6
        \nL.C.L = X\u033f – A2<\/sub>R\u0304 =112.6 – 0.729 \u00d7 4.2 = 109.54
        \nU.C.L = X\u033f + A2<\/sub>R\u0304 = 112.6 + 0.729 \u00d7 4.2= 115.66
        \nThe control limits for R – chart are :
        \nC.L = R\u0304 = 4.2
        \nL.C.L =D3<\/sub>R\u0304 = 0 \u00d7 4.2 = 0
        \nU.C.L = D4<\/sub>R\u0304 = 2.115 \u00d7 4.2 = 8.883<\/p>\n

        \"KSEEB<\/p>\n

        Question 5.
        \nGiven fraction defective = 0.05, and sample size 8. Write down the control limits for number of defectives
        \nAnswer:
        \nGiven: P1<\/sup> = 0.05, n = 8,
        \nHere Q1<\/sup> = 1 – P1<\/sup> = 1 – 0.05 = 0.95
        \nThe control limits for number of defectives np\/d – chart (When standards are given) are:
        \nC.L = np1<\/sup> =8 \u00d7 0.05 = 0.4
        \nL.C.L = np1<\/sup> – \\(3 \\sqrt{n p^{\\prime} Q^{\\prime}}\\) = 8 \u00d7 0.05 – \\(3 \\sqrt{8 \\times 0.5 \\times 0.05}\\) =-1.45 = 0 (Taken as 0)
        \nU.C.L = np1<\/sup> + \\(3 \\sqrt{n p^{\\prime} Q^{\\prime}}\\) = 8 \u00d7 0.05 + \\(3 \\sqrt{8 \\times 0.5 \\times 0.05}\\) = 2.25<\/p>\n

        Question 6.
        \nIn a bulb manufacturing company, the proportion of defectives is 0.08. If the process control is based on samples of size 150 each. Write down the control limit for np or d-chart.
        \nAnswer:
        \nGiven: P1<\/sup> = 0.08, n = 150
        \n(\u2234 Q1<\/sup>= 1 -P1<\/sup>)= 1 – 0.08 = 0.92
        \nThe control limits for np – chart are
        \nC.L =np1<\/sup>= 150 \u00d7 0.08 = 12
        \nL.C.L = np1<\/sup> – \\(3 \\sqrt{n p^{\\prime} Q^{\\prime}}\\)= 150 \u00d7 0.08 – \\(3 \\sqrt{150 \\times 0.08 \\times 0.92}\\) x 0.92 = 2.032
        \nU.C.L = np1<\/sup> + \\(3 \\sqrt{n p^{\\prime} Q^{\\prime}}\\) =150 \u00d7 0.08 + \\(3 \\sqrt{150 \\times 0.08 \\times 0.92}\\) =21.96 = 22<\/p>\n

        Question 7.
        \nThe following data show the values of sample mean (X\u0304) and range (R) for samples of size 5 each. Calculate values for central line and control limits for mean chart. (Given A2<\/sub> = 0.577 for n = 5)
        \n\"2nd
        \nAnswer:
        \nHere Standards are not given
        \n\"2nd
        \nThe control limits for X\u0304 – chart are :
        \nC.L = X\u033f = 10.66
        \nL.C.L = X\u033f – A2<\/sub>R\u0304= 10.66 – 0.577 \u00d7 6.3 = 7.025
        \nU.C.L = X\u033f + A2<\/sub>R\u0304 = 10.66 + 0.577 \u00d7 6.3 = 14.30<\/p>\n

        Question 8.
        \nYou are given values of samples means (X\u0304) and Range (R) for 6 samples of the size 4 each. Find control limits for mean and Range charts
        \n\"2nd
        \nyou may use the following control chart constants
        \nFor n = 4, A2<\/sub> = 0.729, D3<\/sub> = 0; D4<\/sub> = 2.282
        \nAnswer:
        \nHere standards are not given and K = 4
        \n\"2nd
        \nThe control limits for X – chart are :
        \nC.L = X\u033f = 63.75
        \nL.C.L = X\u033f – A2<\/sub>R\u0304 = 63.75 – 0.729 \u00d7 8.5 = 57.55
        \nU.C.L = X\u033f + A2<\/sub>R\u0304 = 63.75 + 0.729 \u00d7 8.5 = 69.55
        \nThe control limits for R – chart are:
        \nC.L = R\u0304 = 8.5
        \nL.C.L = D3<\/sub>R\u0304 = 0 \u00d7 8.5 = 0
        \nU.C.L = D4<\/sub>R\u0304 =2.282 \u00d7 8.5 = 19.40<\/p>\n

        Question 9.
        \nThe following are the number of defects noted in the final inspection of balls of woolen cloth 0, 3, 1, 4, 2, 2, 1, 3, 5, 0, 2, 1, 0, 3.
        \nWrite down an appropriate control limits
        \nAnswer:
        \nHere Standards are not given and number of defects are given.
        \nThe chart would be C-chart
        \n\"2nd
        \nC\u0305 = \\(\\frac { 27 }{ 15 }\\) = 1.8 15
        \nC.L = C\u0305 = 1.8
        \n\"2nd<\/p>\n

        Question 10.
        \nIf \u03bb1<\/sup> = 2.5 Write down the control limits for C-Chart.
        \nAnswer: Given Standard is : \u03bb1<\/sup> = 2.5
        \nThe control limits for c-chart are :
        \nC.L = \u03bb1<\/sup> =2.5
        \nL.C.L = \\(\\lambda^{1}-3 \\sqrt{\\lambda^{1}}=2.5-3 \\sqrt{2.5}\\) = -2.24 = 0 (Taken as \u2018o\u2019. )
        \nU.C.L = \\(\\lambda^{1}+3 \\sqrt{\\lambda^{1}}=2.5-3 \\sqrt{2.5}\\)= 7.24<\/p>\n

        \"KSEEB<\/p>\n

        Question 11.
        \nIn a printing industry, at regular intervals, cloth is inspected for defects in printing. If on an average 0.5 defects are expected per square metre, obtain suitable control limits.
        \nAnswer:
        \nGiven : Standard \u03bb1<\/sup> = 0.5
        \nThe control limits for C-chart are :
        \nC.L = \u03bb1<\/sup> = 0.5
        \nL.C.L = \\(\\lambda^{\\prime}-3 \\sqrt{\\lambda^{\\prime}}=0.5-3 \\sqrt{0.5}\\) = -1.62 =0 (Taken as 0)
        \nU.C.L = \\(\\lambda^{\\prime}-3 \\sqrt{\\lambda^{\\prime}}=0.5+3 \\sqrt{0.5}\\) =2.62<\/p>\n

        Question 12.
        \nFrom a soap manufacturing machine, samples consisting of 5 packed soap cakes are drawn and the following observations were recorded: X\u03041<\/sup> = 75 gms ; \u03c31<\/sup> = 3 gms
        \n\"2nd
        \ndraw X\u0304-chart and comment
        \nAnswer:
        \nGiven: n – 5, X\u03041<\/sup> = 75 gms , \u03c31<\/sup> = 3 gms (Standards are given)
        \nThe X\u0304 – control limits are
        \nC.L = X\u03041<\/sup> = 75
        \nL C.L = X\u03041<\/sup> – A\u03c31<\/sup> = 75 – 1.342 \u00d7 3S = 70.97 = 71
        \nU.C.L = X\u03041<\/sup> + A\u03c31<\/sup> = 75 + 1.342 \u00d7 3 = 79.03 = 79
        \nThe X\u0304 – chart and the plotted points are as shown below:
        \n\"2nd
        \nInterpretation: In X\u0304 -chart one of the point lie outside the UCL, the process is out of control.<\/p>\n

        Question 13.
        \nIf the standards are known to be X\u03041<\/sub> =20 and \u03c31<\/sup> = 6, construct X\u0304 chart for the data sample size n = 5 and A = 1.342)
        \n\"2nd
        \nAnswer:
        \nGiven :n = 5, A = 1.342, X\u03041<\/sup> =20, \u03c31<\/sup> =6
        \nStandards are given
        \nThe control limits for X\u0304 chart are:
        \nC.L = X\u03041<\/sup> = 20
        \nL.C.L = X\u03041<\/sup> – A\u03c31<\/sup> =20 – 1.342 \u00d7 6 = 11.948 = 11.95
        \nU.C.L. = X\u03041<\/sup> + A\u03c31<\/sup> =20 + 1.342 \u00d7 6 = 23.052 = 28.05
        \n\"2nd
        \nInterpretation: All the plotted lie with in the control limits. Therefore the production process is under statistical control.<\/p>\n

        Question 14.
        \nConstruct a suitable control chart for the following data and state your conclusion.
        \n\"2nd
        \nAnswer:
        \nHere standards are not given;
        \n\"2nd
        \n\u2234 q\u0304 =l – p\u0304 = 1 – 0.09 = 0.91
        \nwhere n = sample size, = 100
        \nK = number of samples and d = number of defectives
        \nThe control limits for np-chart (d-chart) are:
        \nC.L = np\u0304 =100 \u00d7 0.09 = 9
        \n\"2nd
        \n\"2nd<\/p>\n

        \"KSEEB<\/p>\n

        Question 15.
        \nDuring an examination of equal lengths of cloth, the following are the number of defects observed.
        \n2, 3, 4, 0, 5, 6, 7, 4, 3,2.
        \nDraw a control chart for the number of defects and comment whether the process is under control or not
        \nAnswer:
        \nHere Standards are not given and So,
        \n\"2nd
        \nc\u0304 = \\(\\frac { 36 }{ 10 }\\) = 3.6
        \nThe control limits for c-chart are
        \nC.L = c\u0304 =3.6
        \n\"2nd
        \n\"2nd
        \nInterpretation : The points are with in central limits, the process is under control<\/p>\n

        Question 16.
        \nGiven D3<\/sub> = 0, D4<\/sub>, =2.115, and R\u0304 = 4
        \ndraw R-chart for the following data.
        \n\"2nd
        \nAnswer:
        \nThe control limits for R-chart are :
        \nwhen standards are not given
        \n\"2nd
        \nk = number of samples = 6
        \nC.L. = R\u0304 = 4
        \nL.C.L = D3<\/sub>R\u0304 = 0 \u00d7 4 = 0
        \nU. C. L = D4<\/sub>R\u0304 = 2.115 \u00d7 4 = 8.46
        \n\"2nd
        \nInterpretation : All points are with in control limits. The process is under control<\/p>\n

        Question 17.
        \nA company manufactures flooring titles. samples of 100 tiles are drawn at regular intervals. The number of defective tiles are given below
        \n\"2nd
        \nFind suitable control limits for the above data.
        \nAnswer:
        \nGiven n = 100, k – 12, d = defectives (standards are not given)
        \n\"2nd
        \nq\u0304 = 0.9683
        \nThe control limits for np\/d chart are –
        \nC.L = np\u0304 = 100 \u00d7 0.0317 = 3.17
        \nL.C.L = \\(n \\bar{p}-3 \\sqrt{n \\overline{p q}}\\)
        \n= 100 \u00d7 0.0317 – \\(3 \\sqrt{100 \\times 0.0137 \\times 0.9683}\\)= -2.09 (Taken as 0)
        \nL.C.L =0
        \nU.C.L = \\(n \\bar{p}+3 \\sqrt{n \\overline{p q}}\\)= 100 x 0.0317 – \\(3 \\sqrt{100 \\times 0.0137 \\times 0.9683}\\)
        \nU.C.L =8.43 .<\/p>\n

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

        You can Download Chapter 7 Statistical Quality Control Questions and Answers, Notes, 2nd PUC Statistics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 2nd PUC Statistics Question Bank Chapter 7 Statistical Quality Control 2nd PUC Statistical Quality Control One and Two …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/13240"}],"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=13240"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/13240\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=13240"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=13240"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=13240"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}