NCERT Solutions Ganita Prakash Chapter 4 – 99Section 4.4 Drawing a Bar Graph — Figure it Out

The largest number is only 10, so the simplest scale works: 1 unit length = 1 insect.

1. Draw a horizontal line and a vertical line meeting at 0.
2. On the horizontal line mark the five kinds of creatures, equally spaced.
3. On the vertical line mark 0, 2, 4, 6, 8, 10 — equal gaps for equal numbers.
4. Draw bars of the same width with the same gap between them: heights 6, 10, 5, 3 and 2 units.
5. Give the graph a title and label both lines.

a. Vidisha → 24 tickets (write 24 just above that bar).

b. Jabalpur → 20 tickets.

c. The scale. The same bar is described in two ways — 6 unit lengths, and 24 tickets. So one unit length must stand for

d. The bar for Sagar.

e. The numbers on the vertical axis must go up in steps of 4, starting from 0:

f. Seoni is 16 tickets = 16 ÷ 4 = 4 units; Indore is 28 tickets = 28 ÷ 4 = 7 units.

• Seoni's bar is correct — on the board it is drawn 4 unit lengths tall, exactly as it should be.
• Indore's bar is wrong. On the board it is drawn only about 4½ unit lengths — barely taller than Seoni's, and shorter than Vidisha's. Indore sold the most tickets, so its bar must be 7 unit lengths and must be the tallest bar of all.

The corrected graph is shown below.

a. Read Chinu's list row by row and put one tally mark for each vehicle.

b. The bike was used the most — 13 out of 50 vehicles, more than a quarter of all the traffic. Two-wheelers together (bike 13 + scooter 9 + bicycle 8) make 30 of the 50 vehicles.

c. How I would collect this data:

1. Decide the place and the exact time first — one spot on the road, from 9:00 to 10:00 a.m.
2. Make a table beforehand with one row for each kind of vehicle, and keep a spare row for “any other”.
3. Sit facing the road and put one tally mark for every vehicle as it passes, in bundles of five. Do not try to write full words — vehicles pass too fast.
4. Count vehicles going in both directions (or fix one direction and say so), and never count the same vehicle twice.
5. At the end of the hour, count the tally marks, write the frequencies, and add them up to check the total.

This is an experiment — roll the die yourself. Here is one set of 30 actual rolls:

For this experiment:

• a. 4 appeared the minimum number of times — only 3 times.
• b. 5 appeared the maximum number of times — 7 times.
• c. 3 and 6 appeared an equal number of times — 5 times each.

a. The table tells us how many matches Bumrah took each number of wickets in. For example, the row “2 — 6” means that in 6 of the matches he took exactly 2 wickets. It does not tell us which match was which, or against whom he played.

b. A good title would be “Wickets taken by Jaspreet Bumrah in his last 30 matches”. (Other titles work too — “Bumrah's wicket-taking record: last 30 matches”.) A title must say whose data it is, what is being counted and over what period.

c. Things worth noticing:

• Taking 3 wickets is his most usual performance — it happened in 8 matches, more than any other.
• He went wicketless in only 2 matches out of 30.
• A haul of 6 or 7 wickets is rare — just 1 match each.
• The frequencies rise up to 3 wickets and then fall — but 5 wickets (5 matches) breaks the fall.
• The numbers of matches add up to 2 + 4 + 6 + 8 + 3 + 5 + 1 + 1 = 30 ✔, which is exactly what the title promises.

d. Bumrah took 4 wickets in 3 matches.

e. No, Mayank's method is wrong.

That would be the answer only if he had played exactly one match of each kind. But he took 3 wickets in 8 different matches, so those 3 wickets must be counted 8 times, not once. Mayank is adding the labels of the rows instead of using the counts.

f. The correct method: multiply each number of wickets by the number of matches, then add.

The key is “one symbol = 1 tractor”, so count the symbols in each row.

a. Village D — only 3 tractors, the shortest row.

b. Village C — 8 tractors, the longest row.

c. Village C has

d. Yes, Komal is right.

Here one symbol stands for 4 girls, so a half symbol stands for 2 girls. Every row must first be turned into a number:

a. Class 8 — its row is the shortest of all eight rows: 1½ symbols, that is 1 × 4 + 2 = 6 girls, the least in the school.

b. Class 5 shows 2½ symbols and Class 6 shows 4 symbols:

c. Class 2's row is 4½ symbols, that is 4 × 4 + 2 = 18 girls. With 2 more girls it becomes

So the half symbol at the end of Class 2's row would become a full symbol, and the row would then show 5 complete symbols. Nothing else in the graph changes.

d. Class 7 shows 3 symbols:

a. Look for a number that divides all six values exactly, and that is big enough to keep the rows short. The numbers are 18, 36, 12, 48, 18, 24 — every one of them is a multiple of 6.

A useful key is “one symbol = 6 dogs”. Every row then comes out in whole symbols — no half or quarter pictures are needed at all. (A key of 12 would give 1½, 3, 1, 4, 1½, 2 — it works but needs half symbols; a key of 2 or 3 would make the rows uncomfortably long.)

b. Village B has 36 dogs, so

c. Yes, Kamini is right.

The picture shows it too: the rows of B and D together have 6 + 8 = 14 symbols, while the other four rows have 3 + 2 + 3 + 4 = 12 symbols.

First turn every frequency into a bar height, using the given scale 1 unit length = 5 students:

Now mark the vertical line 0, 5, 10, 15, … 50 and draw bars of equal width with equal gaps.

Other than playing, reading story books is preferred by the most students — 30 out of 120.

Read the height of each bar against the vertical line, where the marks go up in tens.

Six of the bars end exactly on a marked line. Monday's bar is the only one that does not — it rises a little above the 50 line, at about 52.

a. Wednesday and Thursday together:

b. The whole week:

c. The greatest number of saplings were planted on Saturday (60) and the least on Wednesday (30).

Possible reasons:

• Saturday is usually a half day or a holiday, so more students, teachers and even parents were free to join the plantation drive.
• Wednesday was a full working day with regular classes, so only a short time was left for planting. It may also have rained heavily, or the supply of saplings may have run short that day.
• July is the sowing month — soft, wet soil makes planting easy, which is why the numbers are large on every day.

How to check whether these explanations are right:

• Ask the teachers and students how much time was given for planting on each day, and note it beside the graph.
• Look at the school attendance register — were fewer people present on Wednesday?
• Check the rainfall record for that week, and the record of how many saplings were delivered each day.
• Repeat the drive next month and see whether Saturday is again the biggest day. If the same pattern appears again, the explanation is much more believable.

The horizontal line is marked 0, 1000, 2000, 3000, 4000, so 1 unit length = 1000 tigers. Read each bar against that scale and compare it with the table.

The mistakes: only the 2022 bar is drawn correctly. Of the other four:

• The 2014 and 2018 bars have been interchanged — 2014 has been given the 2018 value (about 3000) and 2018 has been given the 2014 value (about 2200). This is easy to spot without any measuring: the 2018 bar comes out shorter than the 2014 bar, yet the table says the tigers increased from 2200 to 3000 in those years.
• The 2006 bar is far too short — about 800 instead of 1400.
• The 2010 bar is too short — about 1500 instead of 1700.

How to fix them: redraw the four wrong bars to the lengths the table gives — 2006 to 1400 (a little less than 1½ units), 2010 to 1700, 2014 to 2200 and 2018 to exactly 3000 (3 units). Keep every bar starting from 0, of the same width and with equal gaps. After the fix the bars grow steadily longer from 2006 to 2022, as they must.