Class 7 Maths Ganita Prakash Part 2 Chapter 5 MCQs – Connecting the Dots

Different students can have different heights. I would measure the students, organise their heights and study the results.
Thus the question expects variation in collected data. The other questions have fixed mathematical answers and do not require such data collection.

The arithmetic mean is the equal share of the total.
Mean = sum of all values ÷ number of values.
Adding the values gives the total amount, and dividing by the number of observations distributes that total equally among them.

Total guavas = 3 + 8 + 10 + 5 + 4 = 30.
Each friend’s share = 30 ÷ 5 = 6 guavas.
The mean represents this equal share even though the friends originally collected different numbers of guavas.

Total flowers = 2 + 7 + 9 + 4 + 3 = 25.
Mean = 25 ÷ 5 = 5 flowers per day.
This does not mean that exactly five flowers bloomed every day; it is the equal daily share of the total.

Sum = 6 + 2 + 9 + 5 + 4 + 6 + 3 + 5 = 40.
There are eight attempts.
Mean = 40 ÷ 8 = 5 bounces per attempt. Each attempt, including the smallest count, contributes to the mean.

Total runs = 23 + 7 + 10 + 52 + 18 = 110.
Mean score = 110 ÷ 5 = 22 runs per match.
Checking: five equal scores of 22 would also give a total of 110 runs.

Shubman’s mean = 110 ÷ 5 = 22 runs.
Yashasvi’s mean = 96 ÷ 4 = 24 runs.
Difference = 24 − 22 = 2 runs per match. Comparing only totals would ignore that the players played different numbers of matches.

Range = maximum value − minimum value.
Yahapur’s range = 59 − 24 = ₹35 per kg.
Wahapur’s range = 60 − 17 = ₹43 per kg.
Since 43 > 35, prices have a greater spread by this measure in Wahapur.

Each dot represents one observation, and its horizontal position gives the price.
Two dots above ₹42 therefore mean that ₹42 occurs twice.
The vertical stack counts occurrences; it does not add the two prices to make ₹84.

The dot plot shows values and how often they occur. It sorts them by size rather than keeping their original month-wise order.
Without month labels, I cannot tell which dot belongs to January, although the minimum, maximum and frequencies remain visible.

First, the heights in increasing order are 118, 165, 170, 173, 175.
There are five values, so the median is the third value.
Median = 170 cm. Two values lie before it and two after it in the ordered list.

With six values, there are two middle positions: the third and fourth.
Median = (164 + 165) ÷ 2
= 329 ÷ 2
= 164.5 cm. The median need not be one of the actual heights in the collection.

Four heights lie between 165 cm and 175 cm, a span of only 10 cm.
The value 118 cm is 47 cm below the next smallest height, 165 cm.
It is far from the rest of the data, so it is an outlier.

Sum of the remaining heights = 170 + 173 + 165 + 175 = 683 cm.
Mean = 683 ÷ 4 = 170.75 cm.
Removing the unusually low value raises the mean, which shows how strongly an outlier can affect it.

The sorted page counts are 10, 16, 16, 18, 20, 22, 26.
There are seven values, so the fourth value is the median.
Median = 18 pages. Sorting is necessary because the middle value in the original day-wise list is not necessarily the median.

There are six matches actually played, including the match with a score of zero.
A zero is a recorded score, whereas the dash represents no match played.
Therefore, mean score = total runs ÷ 6, with the zero included among those six observations.

B’s total = 0 + 8 + 6 + 4 = 18; mean = 18 ÷ 4 = 4.5.
C’s total = 8 + 11 + 13 = 32; mean = 32 ÷ 3 = 10⅔.
A zero counts as a score; a game not played does not.

Each of the nine readings is at most 20.5 L.
Their total is therefore at most 9 × 20.5 L.
Dividing by 9 gives a mean at most 20.5 L. Since 25 > 20.5, a mean of 25 L is impossible.

Both sections have 30 students, but their students’ heights may differ.
Mean height = total of the students’ heights ÷ 30.
The second section’s total height is unknown, so having the same number of students does not determine its mean.

A mean describes the whole group using one value. A larger mean does not compare every individual in one group with every individual in another.
Therefore, the supported conclusion concerns the class averages; the individual heights could still overlap.

A double column graph places two related bars in each monthly group.
Using the same price scale makes their heights directly comparable.
The month labels preserve the time order, so I can compare the towns within a month and follow changes across months.

Difference in price = ₹38 − ₹30
= ₹8 per kg.
On a double column graph with a common scale, Wahapur’s May bar would be higher than Yahapur’s by the height representing ₹8 per kg.

Each unit of bar length represents 20 launches.
Launches represented = 3 × 20
= 60
The scale converts the measured length into a data value. Starting from zero ensures that all three units belong to the bar’s value.

Mean daily daylight = total daylight hours ÷ number of days.
= 564 ÷ 30
= 18.8 hours per day.
The monthly total must be divided by the actual number of days to obtain a daily average.

Day 1 range = 34°C − 16°C = 18°C.
Day 2 range = 43°C − 30°C = 13°C.
Since 18 > 13, Day 1 has the greater range. The higher maximum on Day 2 does not by itself mean a greater range.

The count changed from 89,000 to 78,000.
Decrease = 89,000 − 78,000 = 11,000.
The comparison uses the two specified years. Although the 2024 count remains above the 2022 count, it is below the 2023 count.

From 2022 to 2023, the increase is 74,000 − 62,000 = 12,000.
From 2023 to 2024, it is 81,000 − 74,000 = 7,000.
Both changes are positive, so registrations increased each year, although the increases were unequal.

Ratio of the watching count to the participating count = 1240 ÷ 620
= 2.
Thus the watching count is twice the participating count. This comparison concerns the two survey counts and does not show that the groups contain entirely different people.

Sorted times: 220, 230, 240, 270, 280, 290, 320, 380.
There are eight values, so the middle pair is 270 and 280.
Median = (270 + 280) ÷ 2
= 275 seconds. This value splits the ordered positions into two equal halves.

The graph records how many buildings exceed 150 m in each city, not the height of each building.
A city with more such buildings need not have the tallest one.
Therefore, identifying the tallest individual building requires different data.