Chapter 4: Data Handling and Presentation (Data Handling)

Organise the list: write each game once and put a tally mark against it for every student who chose it, then count the tally marks. The game with the most tally marks is the most popular.

Make a frequency table with tally marks for each game and count; the game with the highest count is the most popular.

Hockey (8 students) is the most popular game. (Cricket, which Navya thought was the most popular, has only 6.)

Hockey, chosen by 8 of the 31 students.

Ask every classmate their favourite game (once each), record the answers with tally marks in a table like the one above, and count. The game with the largest count is the most popular in your class. (Make sure nobody is counted twice and nobody is missed: the total should equal the number of students.)

Collect each classmate's choice, tally and count; the largest count gives the answer.

• a. ✓: she must ask her classmates.
• b. ✗: it is a known fact (15 August 1947).
• c. ✓: she must observe and measure in her locality.
• d. ✗: it is a known fact (New Delhi).

a. ✓ b. ✗ c. ✓ d. ✗

a. 6 b. 3 c. 13 d. 7 e. 9

a. 6 b. 3 c. 13 d. 7 e. 9

No. The table tells only how many students chose each sweet, not which student chose which sweet. To give each child the right sweet, Shri Nilesh needs a list of names with each child's choice, for example a table with columns "Name of student" and "Sweet chosen", or the names grouped under each sweet.

No: it gives only the numbers, not who chose what. A list of each student's name and choice (or names grouped under each sweet) is needed.

From the arranged list (3 three times, 4 nine times, 5 ten times, 6 four times, 7 once: 27 students):

a. 7 b. 3 c. 10 d. sizes 5, 6 and 7: 10 + 4 + 1 = 15

a. 7 b. 3 c. 10 d. 15

In ascending order, the smallest value is at the start and the largest at the end, and equal values are grouped together, so they can be counted easily without searching through the whole jumbled list. Values "larger than 4" are all together at the end of the list.

The smallest and largest are at the two ends, and equal sizes come together, so counting is quick and easy.

Yes:

• in descending order (largest to smallest);
• in a frequency table with tally marks:

• as a pictograph or a bar graph.

Yes: descending order, a frequency table (3, 9, 10, 4, 1 for sizes 3 to 7), a pictograph or a bar graph.

Sample record (yours will be different):

a. Neem (7) b. Banyan (1) c. Yes: mango and ashoka, 4 each.

Answer from your own table; in the sample, neem is the most (7), banyan the least (1), and mango and ashoka are equal (4 each).

a. Usually e. b. Usually x.

c. The order is usually x, c, r, i, e. This happens because in English some letters are used much more often than others: e is the most common letter in English words (the, be, we, here, see…), while x appears in very few words. A news item has many words, so it follows the usual pattern of the language.

d.–f. Process: make a table with the letters; read the article word by word and put a tally mark for each c, e, i, r or x; cross out each word after checking it so it is not counted twice; count the tally marks at the end. With another news item, follow the same steps, or divide the article into parts and count each part separately, then add (or share the counting with a friend, one letter each).

Usually e is the most and x the least, giving the order x, c, r, i, e, because e is the commonest and x one of the rarest letters in English. Count with tally marks, word by word, crossing out words already checked.

With one symbol for 10 and half a symbol for 5, we can show only multiples of 5. For 33 we would need 3 full symbols and $\frac{3}{10}$ of a symbol, and for 27 two full symbols and $\frac{7}{10}$ of a symbol. Such parts of a symbol are very hard to draw and to read accurately. (We could write the number or choose a smaller key, like 1 symbol = 1 student, but that takes a lot of space.)

33 and 27 are not multiples of 5, so we would need to draw 3/10 or 7/10 of a symbol, which is hard to draw and read.

Reading the pictograph: Monday 5, Tuesday 4, Wednesday 2, Thursday 0, Friday 5, Saturday 8.

a. Thursday (no books at all).

b. $5+4+2+0+5+8=$ 24 books.

c. Saturday (8 books). A possible reason: Sunday is a holiday, so students borrow books on Saturday to read at home over the weekend.

a. Thursday b. 24 books c. Saturday, probably because students take books home to read on Sunday.

With 1 kite symbol = 100 kites, half a symbol stands for 50 kites:

a. 3 symbols (300 kites).

b. Poonam Ben (700 kites, 7 symbols).

c. Jasmeet (450 kites) bought more than Chaman (250 kites).

d. Yes. Double of Rani's kites is $2\times 300=600$, and Poonam Ben bought 700, which is more than 600.

a. 3 symbols b. Poonam Ben c. Jasmeet d. Yes: 700 is more than 2 × 300 = 600.

1. 5 students. 2. Class 8 (7 students). 3. Class 5 (0 absent).

1. 5 2. Class 8 3. Class 5

From the bar graph: 6–7: about 150; 7–8: 1200; 8–9: 1000; 9–10: 800; 10–11: 700; 11–12: 600.

$$150+1200+1000+800+700+600=4450$$

About 4450 vehicles.

Early in the morning most people are still at home: offices, schools, shops and markets have not opened yet, so very few people need to travel.

Very early in the morning most people have not started for work, school or shopping yet.

This is the rush hour: children go to school, and many people leave for offices, factories and shops at this time, so the most vehicles are on the road (school buses, cars, scooters, autos).

It is the rush hour, when people leave for school and work.

Most people reach school or work by 9 or 10 a.m., so fewer people are still travelling. Later in the morning only those going for shopping or other work are on the road, so the traffic keeps reducing until noon.

After 8 a.m. most people have already reached school or work, so fewer and fewer vehicles are on the road.

Questions to ask: In which year was the population the least? In which decade did it increase the most? About how many times did the population grow from 1951 to 2001?

Increase over 50 years: $102-36=66$ crores.

Increase in each decade:

The increase became larger each decade.

It increased by 66 crores in 50 years; the increase in each decade was 8, 10, 14, 16 and 18 crores.

1. Most on food (₹3400), second most on house rent (₹3000).

2. Yes. Electricity ₹400 is exactly half of ₹800.

3. Yes. One-fourth of ₹3400 is ₹850, and education costs ₹800, which is less than ₹850.

1. Food, then house rent 2. Yes (400 = half of 800) 3. Yes (800 < 3400 ÷ 4 = 850)

Take 1 unit length = 1 insect, mark the insects equally spaced on the horizontal line, and draw bars of equal width with heights 6, 10, 5, 3 and 2 units:

Bars of heights 6, 10, 5, 3 and 2 units (1 unit = 1 insect); caterpillars are the most and grasshoppers the fewest.

a. 24 b. 20

c. $24÷6=4$ (and $20÷5=4$), so 1 unit length = 4 tickets.

d. Sagar: $16÷4=4$ unit lengths.

e. Mark 0, 4, 8, 12, 16, 20, 24, 28, 32 on the vertical axis.

f. Seoni needs $16÷4=4$ units: its bar is correct. Indore needs $28÷4=7$ units, but its bar is only about 4½ units high, so it is wrong; it must be redrawn 7 units high.

a. 24 b. 20 c. 1 unit length = 4 tickets d. Sagar: 4 units e. 0, 4, 8, …, 32 f. Seoni is correct; Indore is wrong and should be 7 units (28 tickets).

a.

b. Bike (13).

c. Steps: (1) Sit at a safe place where the road is clearly visible, with a watch, a notebook and a pencil. (2) Before starting, make a table with the likely means of transport (and space to add new ones). (3) From exactly 9:00 to 10:00 a.m., put a tally mark against the right row every time a vehicle passes; draw the fifth mark across the four. (4) At 10:00 a.m. stop, count the tally marks and write the frequencies. (Two friends can share the work: one for each direction of the road.)

a. Bike 13, scooter 9, bicycle 8, auto rickshaw 8, car 6, bus 4, bullock cart 2 (total 50) b. Bike c. Watch the road from 9 to 10 a.m. and put a tally mark against each vehicle type as it passes, then count.

A sample record of 30 rolls (your results will be different):

In this sample: a. 5 appeared the minimum number of times (3). b. 4 appeared the maximum number of times (7). c. 3 and 6 appeared an equal number of times (5 each). Check that the frequencies add up to 30.

Answers depend on your rolls; in the sample, 5 is the least (3 times), 4 the most (7 times), and 3 and 6 are equal (5 times each).

a. It tells in how many matches Bumrah took 0, 1, 2, …, 7 wickets in his last 30 matches.

b. "Wickets taken by Jaspreet Bumrah in his last 30 matches".

c. For example: he took 3 wickets most often (in 8 matches); he took 5 wickets in as many as 5 matches; he went without a wicket in only 2 matches; he once took 7 wickets.

d. 3 matches.

e. No. 0 + 1 + 2 + … + 7 = 28 counts each number of wickets only once. But he took, for example, 3 wickets in 8 different matches (that is 24 wickets), not just once. We must take into account how many matches each number happened in.

f. Multiply each number of wickets by its number of matches, and add:

He took 90 wickets in his last 30 matches.

a. The number of matches in which he took 0 to 7 wickets b. "Wickets taken by Jaspreet Bumrah in his last 30 matches" c. e.g. 3 wickets in 8 matches d. 3 e. No: each number of wickets must be counted as many times as it happened f. Multiply and add: 90 wickets.

Reading the pictograph: A: 6, B: 5, C: 8, D: 3, E: 6.

a. Village D (3). b. Village C (8). c. $8-5=$ 3 more. d. Yes: Village D has 3 and Village E has 6, and 3 is half of 6.

a. Village D b. Village C c. 3 d. Yes (3 is half of 6).

Each full symbol is 4 girls, so a half symbol is 2 girls:

a. Class 8 (6 girls).

b. Class 6 has 16 and Class 5 has 10: difference = 6 girls.

c. Class 2 would have 18 + 2 = 20 girls, so its half symbol would become a full symbol: 5 full symbols.

d. 12 girls (3 symbols).

a. Class 8 b. 6 c. The half symbol in Class 2 becomes a full one (5 symbols = 20 girls) d. 12

a. All the numbers are multiples of 6 (18, 36, 12, 48, 18, 24), so a useful key is 1 symbol = 6 dogs; then no part-symbols are needed.

b. $36÷6=$ 6 symbols.

c. Yes. B and D together: $36+48=84$ dogs (14 symbols). The other four: $18+12+18+24=72$ dogs (12 symbols). 84 is more than 72.

a. 1 symbol = 6 dogs b. 6 symbols c. Yes: B + D = 84 dogs, the other four together have 72.

Heights of the bars: playing $45÷5=9$ units, reading $30÷5=6$, TV $20÷5=4$, music $10÷5=2$, painting $15÷5=3$ units.

Other than playing, reading story books (30 students) is the most preferred activity.

Bars of 9, 6, 4, 2 and 3 units; after playing, reading story books is the most preferred.

From the graph: Monday 50, Tuesday 40, Wednesday 30, Thursday 40, Friday 50, Saturday 60, Sunday 40.

a. $30+40=$ 70.

b. $50+40+30+40+50+60+40=$ 310.

c. Greatest on Saturday (60), least on Wednesday (30).

Possible reasons: on Saturday there may be more free time (half day at school, more parents and villagers could join); on Wednesday it may have rained heavily, or fewer students may have come, or there were fewer saplings. To check: ask the teachers and students, look at the attendance register and the weather on those days, and check how many saplings were available each day.

a. 70 b. 310 c. Saturday (60) and Wednesday (30); for example more free time on Saturday and rain or low attendance on Wednesday; check with teachers, attendance and weather records.

Comparing the bars with the table:

• 2006: the bar shows about 800 instead of 1400; too short.
• 2010: the bar shows about 1500 instead of 1700; too short.
• 2014: the bar shows about 2900 instead of 2200; too long.
• 2018: the bar shows about 2200 instead of 3000; too short (the 2014 and 2018 values seem to have been mixed up).
• 2022: the bar (about 3700) is correct.

The corrected graph (each bar exactly as long as the number in the table):

The bars for 2006 (≈ 800), 2010 (≈ 1500), 2014 (≈ 2900) and 2018 (≈ 2200) are wrong; they should be 1400, 1700, 2200 and 3000. Only 2022 (3700) is correct.

Vertical bars (a column graph). Heights are measured upwards from the ground, so bars that rise from the ground look like the persons standing; it is easy to compare them at a glance.

Vertical bars, because heights are measured upward from the ground, like standing people.

Horizontal bars, because the length of a river is measured along the ground, from its source to its mouth.

Africa (Nile) and South America (Amazon) have the longest rivers, followed closely by Asia and North America.

Horizontal bars, because river lengths lie along the ground. Africa (Nile, about 6650 km) and South America (Amazon, about 6400 km) have the longest rivers.