Class 9 Mathematics Chapter 10: Combining Quantities

Weight each average by its group size: $\dfrac{(165.5 \times 8) + (149.33 \times 3)}{11} = \dfrac{1772}{11} = 161.09$ cm. The value $157.42$ is what you get by averaging the two averages, which ignores that there are nearly three times as many seniors.

Each average carries the weight of the group behind it. Averaging the averages silently gives the 3 juniors the same say as the 8 seniors.

All three weeks have the same number of cycling days, so the sizes cancel and the plain average is correct: $\dfrac{12.8 + 15.8 + 18}{3} = \dfrac{46.6}{3} = 15.53$ minutes.

Putting $p = q = r$ into $\dfrac{ap + bq + cr}{p + q + r}$ gives $\dfrac{(a + b + c)p}{3p} = \dfrac{a + b + c}{3}$. The sizes cancel only when they are equal.

$ap$ is the total of the first collection and $bq$ the total of the second, so their sum over the total count $p + q$ is the combined average.

$\dfrac{(0.10 \times 500) + (0.20 \times 200)}{700} = \dfrac{90}{700} = 12.86%$. It sits nearer 10% because the bowl is the larger quantity. The value $15%$ would be right only for equal quantities.

Equal quantities give equal weights, so the result is the plain average of the two concentrations.

$\dfrac{(60 \times 3) + (64 \times 2) + (73 \times 5)}{3 + 2 + 5} = \dfrac{673}{10} = 67.3%$. Treating the three equally would give $65.67%$, which is lower because the best score carries the heaviest weight.

Rehmat has three scores but a divisor of $3 + 2 + 5 = 10$. Dividing by 3 would discard the weighting entirely.

$\dfrac{(5 \times 4) + (3 \times 3) + (4 \times 2)}{4 + 3 + 2} = \dfrac{37}{9} = 4.11$.

$\dfrac{(400 \times 6) + (550 \times 4)}{10} = \dfrac{4600}{10} = 460$. The plain average ₹475 ignores that more workers are on the lower wage.

$\dfrac{(48 \times 20) + (43 \times 30)}{50} = \dfrac{2250}{50} = 45$ kg. The result leans towards 43 because the girls are the larger group.

$\dfrac{(62 \times 30) + (70 \times 25) + (58 \times 45)}{100} = \dfrac{6220}{100} = 62.2$.

$\dfrac{(30 \times 2) + (50 \times 3)}{5} = \dfrac{210}{5} = 42%$. It lies nearer 50% because more of the stronger solution was used.

$\dfrac{(80 \times 1) + (70 \times 2) + (90 \times 3)}{6} = \dfrac{490}{6} = 81.67$.

It is always between the extremes, which makes it a quick check on your arithmetic. It certainly can equal one of the values, be a decimal, or fall below the plain average.

Weight measures how much say a value has. More weight, more pull towards it.

Doubling every weight doubles the numerator and the denominator alike, so the quotient is unchanged. Weights fix relative importance only.

$\dfrac{(12 \times 250) + (4 \times 750)}{1000} = \dfrac{6000}{1000} = 6%$. It is close to 4% because three-quarters of the mass is the weaker alloy.

$\dfrac{(68 \times 12) + (63 \times 18)}{30} = \dfrac{1950}{30} = 65$.

It must lie between the two, and where exactly depends on the sizes. It is 50 only when the groups are equal.

$673 \div 3 = 224.33$ against the correct $67.3$. Dividing by the number of values rather than the total weight is the commonest error in this topic.

Only equal sizes cancel in $\dfrac{ap + bq + cr}{p + q + r}$.

The moment the group sizes differ, each weekly average must be weighted by its own number of days.

Both are true and R is exactly why A happens — ignoring the weights is what makes the shortcut fail.

A is true and is a useful check. R is false: the divisor is the sum of the weights, not the number of values.

Both statements are true, but R only places the answer somewhere between 10% and 20%. It does not explain why this one lands exactly at the midpoint, which it does because the equal quantities give equal weights.

A is false — doubling every weight leaves the quotient unchanged. R is true and is exactly why A is false.

Both statements are true, but R says only that the answer is somewhere between the two group averages. What makes it nearer the SENIOR figure is that there are more seniors than juniors.

Each average is multiplied by the size of its OWN group. Swapping the weights, as the third option does, gives $153.56$ cm.

The midpoint is reached exactly when the weights are equal, which here means equal quantities.

A ratio such as $3 : 2 : 5$ states how much say each component has in the final figure.

6% of 100 g is 6 g. Percentage concentration is grams of substance per 100 g of mixture.

$\dfrac{(80 \times 10) + (50 \times 40)}{50} = \dfrac{2800}{50} = 56$, which sits near 50 because four-fifths of the values are in that group.

Equal weights reduce the weighted mean to the plain average, so no weighting is needed. The other three all have unequal weights.

$\dfrac{(60 \times 8) + (45 \times 12)}{20} = \dfrac{1020}{20} = 51$ per kg.

The larger group holds most of the values, so it dominates the total and pulls the result to itself.

A weight of 0 removes that value from both the numerator and the divisor, leaving an equally weighted mean of the remaining two.

When the averages are equal, the sizes make no difference at all — any weighted mean of 70 and 70 is 70.

Substituting $p = q$ gives $\dfrac{(a + b)p}{2p} = \dfrac{a + b}{2}$. Note that $a = b$ makes both sides equal $a$, but it does not make the two EXPRESSIONS identical in general.

$\dfrac{314.83}{11} = 28.6$ cm. Adding the two averages gives the sum of two numbers, not the sum of eleven heights; each average must first be multiplied by its own group size.

$\dfrac{(5 \times 100) + (4 \times 200) + (1 \times 700)}{1000} = \dfrac{2000}{1000} = 2%$. It sits near 1% because the third machine makes 70% of the items.

Each average is weighted by how many values stand behind it, which for a group of people is simply its size.

Family A's column reaches $1940 + 1700 + 1280 + 1200 + 1770 + 535 = 8425$.

Housing is 1940 for A, 1750 for B and 950 for C.

Healthcare is 1770 for A, 1210 for B and 1300 for C.

$1940 + 1700 + 1280 + 1200 + 1770 + 535 = 8425$.

Totals are 8425, 7286 and 7040, so Family C spends least overall.

Stacking buys the total and the part-to-whole view; it sells the common baseline.

Only the bottom segment starts at zero. Every other one begins where the one below ended.

$\dfrac{1940}{8425} = 0.230$, so 23.0%.

The category remains in the chart; its segment simply has no length.

Clustering by category puts the three families' housing bars side by side on a common base.

Education is 1280 for A, 1500 for B and 1540 for C.

Without a shared starting point the eye has nothing to judge the two lengths against.

$14234 - 7851 = 6383$.

$\dfrac{6383}{7851} = 0.813$, a rise of about 81%.

Heights on a stacked chart are cumulative; a single value is a difference of two of them.

Only the bottom segment can be read directly as a height.

Recreation is 535 for A, 0 for B and 150 for C.

The six categories can group inside each family, or the three families inside each category.

$\dfrac{1400}{7040} = 0.199$, so about 20%.

Losing the common baseline for every segment above the first is the real cost.

Both statements are true, but the shared order helps you find a category, not compare totals. What makes totals easy is that the full height of a column is the sum of its segments.

A is true. R is false, and is in fact the opposite of the truth: only the bottom segment begins at zero, which is precisely why it is the only one readable directly.

A is false — each chart makes different questions easy. R is true and is why cluster charts compare well.

Both true, and R is the reason: a zero value draws as a segment of no length.

A is true, since each segment sits on top of whatever came before it. R is false: the eye needs a common baseline, which is exactly why these two cannot be compared.

Family B's row is 1750, 1546, 1500, 1280, 1210, 0, so housing is highest.

$8425 + 7286 + 7040 = 22751$.

Totals are exactly what stacking shows at a glance.

$\dfrac{535}{8425} = 0.064$, so about 6%.

The sum is unchanged, but whichever category is placed at the bottom gains the zero baseline.

Transportation is 1200 for A, 1280 for B and 1400 for C.

Part against whole is read by eye; the exact value still needs the scale.

It is a middle segment, so its length must be found as a difference of two cumulative readings.

$\dfrac{1750}{7286} = 0.240$, about 24%.

The totals are A 8425, B 7286 and C 7040.

The height is the sum, so adding a component adds its value to the total.

The shared baseline is what makes side-by-side bars comparable.

1540 is greater than 950, so the education segment is longer wherever it is placed.

Grouping by category places those three bars side by side on a common base.

The height of a stacked column is the total of all its segments.

$14234 - 12630 = 1604$.

A middle segment lacks a common base across groups, which is exactly the comparison stacking makes hard.

1940 is the largest of the six.

The bars show proportions, not amounts, so each one is the whole of its own family.

Healthcare shares are 21.0%, 16.6% and 18.5%.

Families spending ₹8425 and ₹7040 draw identical bars; size is invisible.

$\dfrac{1940}{8425} = 0.230$.

Food shares are 20.2%, 21.2% and 24.1%.

Each slice sits in its own circle with no shared edge to measure against.

Every component is a fraction of the same whole, so the fractions sum to one.

C's 24.1% of ₹7040 is ₹1700, the same rupees as A's 20.2% of ₹8425.

Equal-length bars put every proportion on the same scale.

A zero share takes none of the bar; the bar itself still reaches 100%.

Housing shares are 23.0%, 24.0% and 13.5%.

The share of the whole is $\dfrac{c}{T}$, and the bar is $L$ long.

21.0% of 8425 is 1770; 18.5% of 7040 is 1300. The shares alone would not be enough without the totals.

A pie shows one family's shares well; the bar aligns the same category across all three.

$24.0 + 21.2 + 20.6 + 17.6 + 16.6 + 0 = 100.0$.

Every bar is the same length by construction, whatever the totals behind it.

Only the ordinary stack carries the totals as a readable height.

$0.219 \times 7040 = 1542$, which is the ₹1540 in the table, to rounding.

A common length is what makes the lengths of the bands comparable.

Totals are exactly what the chart discards.

Both true, and R is why: 100% of anything fills the same bar.

A is true, since the totals are not drawn on that chart at all. R is false - nothing on a 100% bar carries the totals, which is exactly why A holds.

Both statements are true, but R is about how the chart is drawn and says nothing about why C combines a large share with a small total.

A is false — comparing angles across separate circles is hard — while R is true in its own right.

Both true, and R is the reason: the absent category claims no part of the bar, so the remaining five - 24.0, 21.2, 20.6, 17.6 and 16.6 - still total exactly 100.

A share is a fraction OF a particular whole, so it must be multiplied by that whole.

Ignoring B's zero, the smallest non-zero share is C's recreation at 2.1%.

The same fraction of two different wholes gives two different amounts.

Both show shares; the bar additionally lines up the same category across groups.

$21.0 - 16.6 = 4.4$, and a difference between two percentages is measured in percentage points.

Doubling every component doubles the total too, so every share is unchanged.

Family B's shares are 24.0, 21.2, 20.6, 17.6, 16.6 and 0.

Each chart in this chapter trades one for the other; combining them is a design choice, not a chart type.

$0.199 \times 7040 = 1400$ and $0.176 \times 7286 = 1282$.

Within one bar the lengths are proportional to that family's own shares.

Ignoring recreation, C's five remaining shares sit between 13.5% and 24.1%, the narrowest band of the three.

A pie shows the whole of one family, so its slices total 100%.

$0.152 \times 8425 = 1281$, the ₹1280 in the table, to rounding.

Aligning the same band across equal-length bars is what the pies cannot do.

$24.0 + 21.2 + 20.6 + 17.6 + 16.6 = 100.0$. A zero component takes nothing from the whole.

$24.0 - 13.5 = 10.5$.

That assumption is exactly the information the chart has removed.