NCERT Solutions Class 11 Informatics Practices Chapter 5 2026-27

Strategic angle. Imagine one finished output and check whether the stored data can create it. Concept used. A record groups related attribute values. A unique identifier distinguishes otherwise similar records.

1. A certificate must identify its student and examination. Therefore, collect a student ID, name, class, roll number, session, subjects, and marks with their maximum values. Passing rules turn these raw marks into a result. Certificate number and issue date identify the generated document. Two students with the same name remain distinguishable through their IDs.
2. An exhibition card must identify the registered participant and support the planned identity check. Collect name, photograph, organisation, contact details, registration category, and participant ID. Associate the permitted biometric template with that ID. Store validity dates so an expired card can be recognised. A photograph alone does not supply the fingerprint template needed by a fingerprint-based check.
3. An image search result needs both something searchable and somewhere to retrieve the image. Store descriptive tags, title, relevant page text, and the image's URL or file reference. Format, dimensions, and file size help users filter results. The actual picture remains image data, separate from these fields.
4. An appointment confirmation needs patient identity, department, date, and time. Collect patient details and the requested department, then compare the request with doctor schedules and free slots. Save the appointment ID and status after allocation. Updating availability makes the booking usable for later requests.

Structural observation. Eligibility is a record-selection problem. The school does not need an average across all students. Concept used. A logical AND is true only if every linked condition is true.

1. Define a consistent record before collecting data. Use student ID, name, class, year-one percentage, year-two percentage, and annual family income. Retain the year labels to confirm that the results belong to consecutive years. If starting from marks, let M_y be obtained marks and T_y be maximum marks for the same year y. Require T_y>0 and check 0≤ M_y≤ T_y. Then calculate that year's percentage: P_y=100×M_yT_y. Apply this formula separately to each of the two years, using its own obtained and maximum marks.
2. Collect the records for all 500 students. Compare entered values with the supporting school and income records. Flag missing or doubtful values for checking. Missing income is not the same as zero income. Use one currency unit and one annual period throughout.
3. Evaluate each student separately. First check whether the first percentage is greater than 75. Then check the second percentage against the same limit. Finally check whether annual income is below Rs. 5,00,000. The tests are independent requirements; high marks cannot compensate for failing the income requirement.
4. Combine the three true-or-false results using AND. A student with one percentage at 75 does not qualify, even if the other percentage is higher. Similarly, income at the stated upper boundary does not qualify because the question says “less than”.
5. Copy eligible students into the output list, keeping their identifiers and supporting values. Review the selected records, save the final list, and retain the original dataset for correction or later checking. The count of beneficiaries depends on these records; the question does not supply enough data to calculate it.

Strategic angle. Decide what “popular” means before collecting the records. Concept used. A proportion compares a part with its matching whole; a mean summarises numerical observations. Neither measure alone captures every reason for choosing a bank.

1. Use household reach as one clear measure: the fraction of surveyed families with at least one xyz account. For a percentage, divide that family count by all surveyed families and multiply by 100. Include households without bank accounts in the denominator when describing reach among all households.
2. Prepare the survey with family IDs, person IDs, bank names, account counts, and balance fields. Keep the account unit explicit. A joint account belongs to two people but remains one distinct bank account. This distinction prevents inflated account counts.
3. Choose one reference month and a common balance definition. For example, add daily closing balances and divide by the number of calendar days in that month. If the study compares balances per person, first combine the person's included accounts consistently. Do not mix a per-account average for one bank with a per-person average for another.
4. Collect responses from families across city localities using a representative method. Get permission to collect the needed values, check incomplete entries, and store validated records in a spreadsheet or database. Surveying only existing xyz users would exclude many relevant residents.
5. Calculate accounts per family, the fraction of accounts at xyz, and the selected average balance measure. Compare xyz with other banks using the same definitions and survey group. State that sample results estimate the city's pattern, while a complete census describes the covered population directly.

Quick reading. Imagine storing many examples together. Ask whether each uses the same meaningful column headings. Concept used. Structure describes the organisation of data, rather than whether the data are stored electronically.

1. A video contains frames and sound that change with the recording. There is no fixed set of ordinary data columns describing every scene's meaning. Classify its main multimedia content as unstructured. File size and resolution remain useful metadata about that content.
2. An attendance record repeats a stable pattern: which student, on which date, and what attendance status. These values fit predefined columns. Therefore, ordinary attendance-register entries are structured and can be counted by date or student.
3. A tweet's words need not follow a fixed form, and media may also be included. Treat the message body as unstructured. The platform can still store a structured timestamp and user ID beside it. This does not change the classification of the message content asked about here.
4. An online form sets its fields before users fill them in. Every applicant supplies values for the same named attributes, so the form-field data are structured. If the form includes a photograph or essay, distinguish that attachment from the surrounding field values.

Structural observation. Repeated temperatures allow a frequency table to shorten the arithmetic without dropping any day. Concept used. Frequency is the number of occurrences of a value. Multiplying a value by its frequency gives its contribution to the total.

1. Count the observations at each temperature: 27 occurs twice, 28 once, and 34 four times. Check the frequencies: 2+1+4=7, so every day is represented. The weighted sum is ∑ f x=2(27)+1(28)+4(34)
   =54+28+136
   =218. Here f is frequency and x is a distinct temperature. The mean is total divided by count: x̄=∑ f x∑ f
   =2187 ^∘C
   ≈ 31.14 ^∘C.
2. Among the given temperatures, the maximum is 34 ^∘C and the minimum is 27 ^∘C. Subtract them to obtain the range: R=34-27=7 ^∘C. Frequency does not affect this subtraction, although it affects the mean and standard deviation.
3. For an independent arithmetic check, expand the squared-deviation formula. It gives variance as the mean of the squares minus the square of the mean: σ²=∑ f x²n-x̄ ². Calculate the squared-value total explicitly: ∑ f x²=2(27²)+28²+4(34²)
   =1458+784+4624
   =6866. Now substitute the exact mean: σ²=68667-(2187)²
   =48062-4752449
   =53849 (^∘C)². Taking the positive square root returns the original temperature unit: σ=√53849≈ 3.31 ^∘C.

Strategic angle. First identify what is being compared: category counts or numerical test scores. Concept used. A frequency distribution shows how observations are distributed among categories. A mean is suitable when adding and dividing the recorded numerical measurements makes sense.

1. For XII A and XII B, list the same division categories in the same order. Tally each student's division in the correct class column. Check that each column totals the shared class strength. Compare corresponding counts, or convert them to percentages using that common denominator.
2. If asked for one central-tendency measure of these division categories, use the mode. Find the largest frequency and report its division label. If categories tie for the largest count, report the tie. The mode is a label, whereas the frequency tells how many students received it.
3. Keep the full frequency table for the stronger comparison. Two classes might have the same most common division but very different numbers in the lowest division. Assigning numeric labels to divisions does not turn those labels into measured marks with equal intervals. Therefore, a standard deviation of those codes would not answer the stated question reliably.
4. For July to November, work with scores rather than division labels. Convert each score to a percentage if maximum marks differ. For each test, add the included percentages and divide by their count. These five means summarise the class's average performance in the five months.
5. Use the same policy for absence and check whether tests assess comparable content at comparable difficulty. A higher mean describes higher recorded average performance; it does not alone prove that teaching caused improvement.

Structural observation. The two parts concern different units: students in (a), and parents' ages in (b). Concept used. Counting gives the size of a selected group. Dispersion describes how numerical measurements vary.

1. Make one record per student for part (a). Record whether the first parent attended this school and whether the second parent attended it. The four possible pairs are Yes/Yes, Yes/No, No/Yes, and No/No.
2. Build a frequency table for those pairs, or filter directly for Yes/Yes. The frequency of Yes/Yes is the requested number of students. Even if No/No is the most common pair, that mode tells nothing about the exact Yes/Yes count. Do not substitute the label of a category for its frequency.
3. For part (b), first decide that each parent is counted once. Join sibling records using a parent identifier where needed, so the same parent's age is not unintentionally repeated. Measure all ages in years on the same date. Missing ages need checking before the calculation.
4. Add the recorded ages and divide by the number of parents to obtain mean age. For each age, calculate its difference from that mean. Square the difference so negative and positive values cannot cancel. Average those squared differences and take the positive square root. This gives population standard deviation for the full recorded group, using the textbook denominator n.
5. Interpret the result in years. A smaller value means ages are closer to the group's mean. The oldest-minus-youngest range supplies an easy extra comparison, but ignores all other ages. Since no actual records are given, report the techniques without inventing a student count or age result.

Strategic angle. Selection needs evidence of performance, so collect demonstrations of each required skill. Concept used. Structured data use the same fields for every observation; here each observation concerns one student.

1. Part (a): data collection. Arrange a common set of tasks for the 42 students. The singing task tests the stated singing skill. A short script or announcement tests writing. A simulated event sequence tests monitoring, meaning keeping track of the planned sequence and cues in this assessment.
2. Part (a): common collection criteria. State the criteria and rating scale before evaluating anyone. Use the same task length and conditions, and record observations promptly. A questionnaire about interests can supplement the process, but self-reported interest alone does not demonstrate the required skills.
3. Part (b): representation. Create a table with Student ID, Singing rating, Writing rating, Monitoring rating, and Evidence columns. Every student gets one row. These fixed headings make the result structured data. The evidence column can hold a short free-text comment alongside the structured ratings.
4. Part (b): rating definitions. Use an ordered rating scale with descriptions, such as Needs practice through Very good. If numbers are assigned, explain what each level means. The numbers are ratings, not physical measurements of skill. Do not average unexplained labels or let different assessors use different meanings.
5. Compare students' full profiles against the anchor's required tasks. If the school chooses a weighted total, decide the weights and tie-breaking rule before selection. Retain the individual skill ratings so one total does not hide a weakness in a necessary skill.

Quick reading. Compare how a second observation would be added to each example. Concept used. An attribute names a characteristic, and a record stores values for a defined set of attributes.

1. Add another student-day to an attendance table. The existing headings still work: Student ID identifies the student, Date identifies the day, and Status records attendance. Because the new row uses the same fields as earlier rows, the data are structured.
2. Now add another school-event video. Its scenes may have a different length, different speakers, and different visual content. There is no fixed table of ordinary attributes containing all of that meaning, so the main content is unstructured.
3. Separate the content from descriptions about it. A table of video filenames and durations is structured metadata. The corresponding recorded scenes remain unstructured content. Likewise, storing a video in a database does not automatically make its audiovisual content tabular.

Strategic angle. Separate the three tasks: unit-price comparison, sales comparison, and price-spread comparison. Concept used. A mean divides a total by its count; revenue multiplies unit selling price by quantity sold.

Illustrative dataset for this route. These invented classroom records cover one shared day; assume sufficient stock and no returns. Buy and Sell are rupees per unit; Sold is the number of units sold that day. Each juice unit is a 200 ml pack, each biscuit unit is one fixed-size packet, and each samosa unit is one piece.

Formulas used. For a numerical dataset, let x_i be its ith value and let n>0 be the number of observations. The arithmetic mean is sum divided by count, and the range is maximum minus minimum: x̄=∑ _i=1ⁿx_in. R=x_max-x_min. Population variance V is the mean of the squared deviations from x̄; population standard deviation σ is its nonnegative square root: V=∑ _i=1ⁿ(x_i-x̄)²n. σ=√V. Thus V=σ². Price variance has square-rupee units; mean, range and standard deviation have rupee units. For juice prices, use n=3 equally weighted brand observations, all for 200 ml. Apply the mean separately to purchase prices and selling prices, and use selling prices for the requested variation.

1. Check the measurement units first. The three juice brands all supply 200 ml packs. The other products are counted as packets or pieces, and every sales count covers the same day. These choices make the requested comparisons clear and prevent mixing daily sales with weekly sales.
2. For juice, compare each brand's selling price with its purchase price. The differences are 25-20=5, 30-24=6, and 35-28=7 rupees per pack. Their equally weighted mean is 5+6+73=183=6 rupees. This equals the difference between the mean selling price, 30, and mean purchase price, 24. For biscuits, the unit difference is 15-10=5 rupees. These are price differences before other costs, rather than net profits.
3. To compare sales counts, combine the three juice rows: 20+30+10=60 juice packs. Compare 60 juice packs, 80 biscuit packets, and 60 samosas. Biscuits rank first by unit count. Juice and samosas tie in count, though their units are different products.
4. Revenue equals unit selling price multiplied by units sold. For the same illustrative day, the juice revenues are: Juice A: 25× 20=500 rupees,
   Juice B: 30× 30=900 rupees,
   Juice C: 35× 10=350 rupees. Add the three revenues to obtain the juice total: 500+900+350=1750 rupees. Biscuits contribute 15× 80=1200 rupees. Samosas contribute 12× 60=720 rupees. Revenue ranking is juice, biscuits, then samosas. This differs from the quantity ranking because prices differ.
5. For juice price variation, use one observation per brand: 25, 30, and 35 rupees. Their mean is 30. The deviations are -5, 0, and 5; the squared deviations are 25, 0, and 25. Their sum is 50, so variance is 50/3 square rupees. Its positive square root is approximately 4.08 rupees. The range is 35-25=10 rupees.