Tales by Dots and Lines

(3+7)÷2=5.

(8+9)÷2=8.5.

Values below 12 are 10, 10, 11: distances 2+2+1=5.

Value above 12 is 17: distance 5.

At 12 the distances on each side total 5.

Move the proposed centre above 12: distances on the lower side grow and those on the upper side shrink.

Move it below 12: the reverse happens.

A value above the mean pulls the balance point upward.

A value below the mean pulls it downward.

A value equal to the mean adds equal balance and leaves the average unchanged.

Removing an above-mean value removes more than the current share, pulling the average down.

Removing a below-mean value pulls the average up.

Removing a value equal to the mean leaves the average unchanged.

For two values, choose values whose average is m; for example m−2 and m+2.

For three values, choose deviations from m that sum to zero, such as m−2, m−1, m+3.

Original sum=79 and there are 11 values, so mean=79/11≈7.18.

Adding 10 to all 11 values adds 110 to the total.

New mean=(79+110)/11=189/11≈17.18.

For n values with mean a, the total is na.

After subtracting 2 from each, the total is na−2n.

New mean=(na−2n)/n=a−2.

If the original total is na, the new total is k(na).

Divide by the same n values: k(na)/n=ka.

The new value changes both the count and the middle position.

If it is above the old median, the new median can move upward; if below, it can move downward.

The actual result also depends on the other ordered values.

Total of 10 weights=39.2×10=392 kg.

Known total=349 kg.

Missing weight=392−349.

Recorded total=25.6×15=384.

Correct total=384−3=381.

Correct mean=381÷15.

Total families=3+11+9+7+3+1+1+1=36.

Weighted total=3×3+4×11+5×9+6×7+7×3+8+9+10=188.

Mean=188/36≈5.22.

For 36 data values, the median is the average of positions 18 and 19.

Cumulative frequency through size 4 is 14; through size 5 it is 23.

Both positions 18 and 19 therefore have value 5.

Columns B–G are Odia, Telugu, English, Mathematics, Social Science and Science.

Rows 2–23 list students in the order shown.

Ashwin’s row is 29, 31, 33, 34, 30, 28.

Compare each score with 30; 30 itself is not more than 30.

Science is column G, students occupy rows 2–23: use =AVERAGE(G2:G23).

Odia total is 687, so average=687/22≈31.23.

Telugu total is 739, so average=739/22≈33.59.

Use =AVERAGE(B2:B23) through =AVERAGE(G2:G23).

The column totals are 687, 739, 718, 751, 690 and 729, each over 22 students.

For an arithmetic sequence, mean=(first+last)/2.

Natural numbers 1–50: (1+50)/2=25.5.

Odd numbers 1–99: (1+99)/2=50.

Multiples 4–200: (4+200)/2=102.

There are 10 shown values. Their sum is 4+7+8+8+(5×9)+11=83.

With the missing dot there are 11 values, so the required total is 11×9=99.

Missing value=99−83=16.

Each measurement is exactly 1 cm too large.

Subtract 1 cm from the mean: 150.2−1.

Album A has seven song lengths: 5, 5, 5.25, 5.5, 5.75, 6 and 6.5 minutes.

Their total is 39 minutes. 39÷7=5.5714… minutes.

The other two plots do not have this mean.

There are 16 values; the middle positions are 8 and 9.

Both values are 41, so median=(41+41)/2=41.

Adding one value makes a 17-value set whose ninth value remains 41; adding 41 and 41 preserves the middle pair.

Removing any one value leaves the middle value 41 because the original two middle entries are both 41.

(i) Removing a low value cannot force the median downward; it may stay or rise.

(ii) A new value below mean m pulls the mean down: (S+x)/(n+1)<m when x<m.

(iii) The result depends on the added values.

(iv) Four low values move the middle position down into lower data; they cannot make the median rise.

Required total=8×10.375=83.

Known total=70.

y=83−70.

Sum=mean×number of values.

134×15=2,010.

Sort the ten fixed values: 8,8,12,18,25,29,35,39,47,73.

With 11 values, the median is the sixth value.

Test each option by inserting it into the ordered list.

Frequency table (rides: students): 0:3, 1:1, 2:4, 3:7, 4:7, 5:5, 6:4, 7:6, 8:3, 10:2.

There are 42 students and 193 rides altogether, so mean=193÷42≈4.60 rides.

The 21st and 22nd values are both 4, so median=4 rides.

(a) False: 3 students rode zero times. (b) True: 38 of 42 rode at least twice. (c) True: anyone riding more than 7 times in a 7-day week must ride more than once on some day.

(d) False: at least 5 students (the 3 who rode 8 times and 2 who rode 10) did so, and the plot cannot tell the exact number.

Adding 1 to every value raises both measures by 1: next mean≈5.60 and median=5.

There are 62 students.

Total trials=1×1+4×1+5×4+6×9+7×12+8×15+9×10+10×10=473.

Mean=473/62≈7.63.

The 31st and 32nd values both fall among the 8-trial group.

Put weekdays on the horizontal axis and customer counts on the vertical axis.

Plot each series with different markers; connect each series in day order.

Add a key and label both axes.

Weather stations record rainy days each month over several years.

Monthly averages are summed for an approximate annual average.

Totals: Mangaluru 112.1; Port Blair 125.8; Rameswaram 42.3 days.

Rameswaram averages 8.1 rainy days in October, 10.4 in November and 7.8 in December.

Values are much lower in most earlier months.

The New Delhi line rises during the south-west monsoon months.

Its rainy-day counts are highest around July and August, with the wet period extending across the summer monsoon.

Read the time axis to identify the covered months and years.

Read the plotted point for July 2017 and each January.

Estimate 2019 by combining or averaging the 12 monthly points if the question means the annual total/average.

Karnataka is in the southern region, where the map shows positive values; a reasonable estimate for the hidden value is about +80.

The strongest rice preferences are in Manipur, Nagaland, Mizoram, Tripura and Meghalaya.

The strongest wheat preferences are in Rajasthan, Haryana, Punjab, Madhya Pradesh and Delhi.

Values near zero indicate a closer balance; examples include Bihar, Maharashtra, Uttarakhand and Himachal Pradesh.

Use the activity key or repeated blocks to identify each colour.

Compare sleeping, school, travel and leisure blocks across Friday–Sunday.

Locate a long leisure block for the movie and the midday break in the school strip.

At age 6, the average is about 9.5 hours per day.

It falls through teenage years and is about 8 hours between ages 30 and 50.

After 50, it rises to about 8.5 hours.

Use the Lo Shu magic square whose rows, columns and diagonals each sum to 15.

Add 5 to each entry; each line then sums to 15+3×5=30, so its mean is 10.

(i) Choose totals 24, e.g. 7,8,9 and 6,8,10.

(ii) Middle pair must total 31, e.g. 10,15,16,20 and 1,15,16,100.

(iii) Total must be 68, e.g. 10,12,14,15,17 and 8,12,14,16,18.

(iv) 1,2,3,4,5,6 has mean=median=3.5.

(v) 1,1,1,1,1,25 has mean 5 and median 1.

One example is 5,21,14,12,12,21; sorted: 5,12,12,14,21,21, median=(12+14)/2=13.

There are infinitely many possibilities: values outside the two middle positions may grow without bound.

Total needed=6×6.5=39.

Known total=3+11+15+6=35, so blanks must total 4.

Positive counting-number pairs summing to 4 are (1,3), (2,2), (3,1).

(i) Two even numbers are 2a and 2b; their mean=a+b, not always even (2 and 4 average 3).

(ii) 5 and 10 average 7.5, not a multiple of 5.

(iii) 0,0,0,0,5 average 1, not a multiple of 5.

Original total=24×150.2=3,604.8 cm.

New total=3,604.8+149+152=3,905.8 cm for 26 students.

New mean=3,905.8÷26≈150.223 cm, slightly above 150.2.

The new median depends on the ordered heights near the centre.

Count dots at each value.

Compute the weighted total and divide by the total dot count.

Net change in total weight=−2+1+1=0, with the same number of people.

Therefore the mean is unchanged.

The median depends on which people changed and their positions in the ordered data.

Gujarat: ₹16.50 to ₹19.20, increase ₹2.70 (about 16.4%).

Uttar Pradesh: ₹16.15 to ₹24.81, increase ₹8.66 (about 53.6%).

West Bengal: ₹9.47 to ₹23.99, increase ₹14.52, the largest listed increase.

(i) In 1983, most rural households used kerosene (about 85%), while most urban households used electricity (about 64%); valid.

(ii) The kerosene line falls over time in both rural and urban graphs, so its use as the primary lighting source decreased; valid.

(iii) In 2000, about 90% of urban households used electricity, not 10%; invalid.

(iv) The graph records primary lighting sources, not interruptions in supply, so it cannot show that there were no power cuts; invalid.

Read the urban value at age 10 from its curve.

Find the rural point where the curve reaches 1.5 hours.

Compare plotted values at ages 10 and 15; distinguish averages from claims about every child.

Record each activity in 30-minute intervals over 24 hours (48 boxes).

For each activity, total the time across recorded days and divide by the number of days.

Compare patterns only after using the same activity categories.

Collect daily sleep or start/end/break times from each participant for the same period.

Calculate mean and median separately for children, adults and elderly participants, or compare school class time and breaks.

Show the data in labelled strips or a table and mention sample limitations.

For January, compare the sunrise times across locations; earlier clock time means earlier sunrise.

Day length=sunset time−sunrise time, accounting for times crossing noon.

Compare this difference across all months and locations.

New moon and full moon can be investigated by comparing moonrise and moonset timing patterns with lunar phases.

Read the date axis and identify the graph’s phase-related pattern, then verify against a calendar.