Ganita Manjari Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data End-of-Chapter Exercises Solutions

Runs so far: 6 + 12 = 18 in 2 overs.

A run rate of 6 after 19 overs means 6 × 19 = 114 runs. Add the 12 runs of the last over.

This is a weighted mean of 6 (weight 19 overs) and 12 (weight 1 over), so it stays close to 6.

Count the prices by type. Nylon: 3 + 4 + 3 + 2 + 1 = 13 prices. Feather: 3 + 0 + 2 + 2 + 3 = 10 prices.

The two expressions give the same answer, because both are the total of all 23 prices divided by 23. Only the way of grouping the prices is different.

Old money spent: 25 × 150 = ₹3750. New purchase: 10 × 30 = ₹300.

The result lies between 30 and 150, and closer to 150 because he has many more old shares.

Old money: 5 × 150 = ₹750. Let n be the new shares bought.

Check: 750 + 30015 = 105015 = 70.

This is Brahmagupta's way of finding the mean depth of an uneven pit.

The new average always lies between 15 and the new price, and it moves towards the new price when more gold is bought. So (i) and (ii) are above 15, with (ii) nearer 30. (iii) and (iv) are below 15, with (iv) very near 10. (vi) buys at the same price, so nothing changes.

The weights are the grams of gold: 1 g at 15 plus the new grams at the new price.

Take the values 10 and 20 with weights 1 and 3.

Double the weights to 2 and 6.

The average is the same.

The factor 2 is in every term of the top and of the bottom. It cancels, and the original weighted mean remains. The same is true for any factor, not just 2: what matters is the ratio of the weights, not their size.

In 2003 the column reaches a little below 10,000, about 9,800 objects. Double is about 19,600. The columns first pass about 19,600 around 2021.

The 2025 column is about 33,000 tall, of which payload objects are about 17,000 and other objects about 16,000.

• Number of payload objects: low and slowly rising for decades, then a very steep rise in the last few years.
• Number of other objects: a steady climb for decades with a sudden jump around 2007, and then growth again.
• Share of payload objects: about a quarter of all objects around 2000, but about half now. Its share has gone up.
• Share of other objects: was about three quarters, now about half. Its share has come down, even though its number has gone up.

So the numbers of both kinds keep rising, but the shares tell a different story: payload objects are now catching up.

A percentage is a share, not a count. We can only use it to find a number if we know the total number of schools.

• (a) Haryana is 90% and Uttarakhand is 78%, but we do not know how many schools each state has. Uttarakhand could have more schools. Not an inference.
• (b) Arunachal Pradesh is about 68%, which is close to 23 ≈ 66.7%. A correct inference.
• (c) Punjab (98%) has a high percentage, not necessarily the most schools, and Dadra and Nagar Haveli and Daman and Diu is higher at 99.8%. Not an inference.
• (d) Maharashtra has 93% of a larger number of schools, while Telangana has 74% of a smaller number. Both factors favour Maharashtra, so Maharashtra has more schools with a playground. A correct inference.

So the correct inferences are (b) and (d).

No. The national percentage is a weighted mean of the state percentages, and the weights are the number of schools in each state.

We would need the number of schools in each state or union territory. Averaging the state percentages directly would treat a small state like Goa as equal to a huge state like Uttar Pradesh.

From the book

↔ Swipe sideways to see the whole chart

Plays A and B have about the same share of high ratings and very few low ratings. Play C has the largest share of 5★ (about 40%), but a third of its ratings are 1★. People either love or dislike it. A cautious choice is Play A or Play B; Play B is slightly better with fewer 1★.

From the book

↔ Swipe sideways to see the whole chart

The count chart shows how many ratings each play has: about 210 for Play A, 920 for Play B and 380 for Play C.

Play B has many more ratings than Play A. Its good score (about 71% high ratings) is based on about 920 people, so it is much more reliable than the same score from only about 210 people.

A 100% stacked bar shows shares but hides the totals. The stacked bar shows the totals. Using both, Play B is the safest decision, and the decision does not change. It becomes more certain.

• (a) Who finished first? No. All bars have equal length, so total times are hidden.
• (b) Fraction of time Athlete 1 spent cycling? Yes. This is a share of the athlete's own total. The exact value is 300563 ≈ 0.53, using 5 h = 300 min and 9 h 23 = 563 min, so a little more than half.
• (c) Who took the longest for running? No. The chart shows shares of each athlete's own total, not actual times.

• Each bar is a 100% bar, so it shows the share of each type of disability within an age group.
• Seeing and hearing together take a large share in every age group, about 33% to 44%. The share of seeing alone rises from about 18% (0–19) to 25% (60 and above).
• The share of movement disability is about 13% in the 0–19 group, but about 25% in the 60 and above group. It goes up with age.
• The shares of speech (about 9% down to 4%) and mental retardation (about 8% down to 2%) are larger among the young and smaller among those aged 60 and above.
• The share of 'any other' disability is about 21% in 0–19 but about 11% in 60 and above, while multiple disability is largest at 60 and above (about 12%).

• How many persons are in each age group? The chart cannot say, because every bar has the same length.
• Why does movement disability take a larger share at older ages? Could it be due to age-related causes, such as injuries or arthritis?
• Do the shares change in newer census data?

We cannot say that there are more persons with movement disability in the 60 and above group than in the 0–19 group. The graph gives shares within each age group, not the number of persons.

1. Write what you did hour by hour from waking up to sleeping again, and tag each hour as lying down, sitting or standing or moving.
2. Add the hours for each tag. The three numbers must add up to 24.
3. Draw one stacked bar per person on a scale from 0 to 24 hours, with the three segments one after another.
4. Write how you decided each hour, for example whether travelling in a bus counts as sitting.

1. Categories: list the headings in your family budget, such as housing, food, education, transport, health, recreation and others.
2. Collect: note the amount for each category in at least 3 months.
3. Convert: find each share = category amountmonthly total × 100.
4. Draw: one bar per month, all from 0% to 100%.
5. Observe: write what changed from month to month.

Example for one month with a total of ₹30,000:

Check: 30 + 25 + 15 + 10 + 8 + 5 + 7 = 100%. If the shares do not add up to 100, there is an error in the amounts.

The total weight is 5 + 4 + 3 + 2 + 1 = 15.

One rider rates safety 4, punctuality 3, cleanliness 5, comfort 4 and fare 5.

Do this for each of the 10 people. The overall average is the mean of their 10 individual ratings.

For each aspect, count how many of the 10 people gave 5★, 4★, 3★, 2★ and 1★. Turn the counts into shares of 10 and draw one bar per aspect from 0% to 100%.

• What it captures well: the overall rating reflects what matters most, because safety counts five times as much as fare.
• What it misses: the weights are our opinion, so another group may choose differently, and the average hides how much people disagree.

The average of the family averages would be wrong, because families have different sizes. A family with more members must count more. So the family size is the weight.

Each student calculates: average age × number of members = the total age of that family. The class then adds all the totals and divides by the total number of members.

A simple average of 30, 25 and 40 would give 31.67, which is not correct.

Take values 10 and 20 with weights 1 and 3.

The average changed. It moved towards 15, the simple average of 10 and 20.

The top has gained the sum of all the values, and the bottom has gained n. That is the same as adding n more equal weights of 1 to the data. So the new weighted mean is pulled towards the simple average of the values.

It stays unchanged only when the weighted mean already equals the simple mean, for example when all the weights are equal.

Shares always add up to 100%, so they cannot all go down or all go up. This kills (i) and (ii).

Let each population be multiplied by a factor smaller than 1: a for cows, b for sheep and d for chickens. The whole farm changes by a factor F, and F is a weighted mean of a, b and d with weights 60, 25 and 15.

A share goes up if its factor is above F, down if it is below F, and stays the same if it equals F.

• (iii) a = b = d. Every factor equals F, so all shares stay. Possible. Example: each population falls to 90%.
• (iv) Need a < d < b. Take a = 0.80, b = 0.99. Then d = 60×0.80 + 25×0.9985 = 72.7585 ≈ 0.856, which lies between them. Possible. Check: 60 × 0.80 = 48 cows, 25 × 0.99 = 24.75 sheep, 15 × 0.856 = 12.84 chickens, total 85.59. Shares: cows 56.1% (down), sheep 28.9% (up), chickens 15.0% (same).
• (v) Need a and b above F and d below F. Take a = b = 0.95 and d = 0.5. Then F = 57 + 23.75 + 7.5100 = 0.8825, so a and b are above F and d is below. Possible. Shares: cows 64.6% (up), sheep 26.9% (up), chickens 8.5% (down).