NCERT Solutions for Class 9 Maths Chapter 10 — Exercise Set 10.3#

$$\text{Profit on books} = 30\% \text{ of } ₹8000 = ₹2400$$

$$\text{Profit on covers} = 50\% \text{ of } ₹1000 = ₹500$$$$\text{Profit percentage} = \dfrac{2400 + 500}{8000 + 1000} \times 100$$

$$\text{Profit percentage} = \dfrac{2900}{9000} \times 100$$

$$\text{Profit percentage} \approx 32.22\%$$

∴ The profit on total sales is about 32.22% — much closer to 30% than to 50%, because most of the sales were books.

Guess: one day of 55 km among 21 days should raise the average only a little above 44.5 km — to about 45 km.

$$\text{Total in 20 days} = 44.5 \times 20 = 890 \text{ km}$$$$\text{Average over 21 days} = \dfrac{890 + 55}{21}$$

$$\text{Average over 21 days} = \dfrac{945}{21}$$

$$\text{Average over 21 days} = 45 \text{ km}$$

∴ The average is 45 km per day.

The mixture has $600 + 300 = 900$ mL. The salt comes only from the first solution and the sugar only from the second:

$$\text{Salt} = 5\% \text{ of } 600 = 30 \text{ mL}$$

$$\text{Sugar} = 8\% \text{ of } 300 = 24 \text{ mL}$$$$\text{Salt concentration} = \dfrac{30}{900} \times 100 \approx 3.33\%$$

$$\text{Sugar concentration} = \dfrac{24}{900} \times 100 \approx 2.67\%$$

∴ The answer is (v) Salt: 3.33%, Sugar: 2.67%. Both concentrations go down, because each substance is now spread through a larger volume.

The spice in the pani is:

$$8\% \text{ of } 10 \text{ L} = 0.8 \text{ L}$$

The new concentration should be:

$$\dfrac{3}{4} \times 8\% = 6\%$$

Let $x$ litres of water be added. The spice stays 0.8 L, so:

$$\dfrac{0.8}{10 + x} = \dfrac{6}{100}$$

$$\Rightarrow 0.06(10 + x) = 0.8$$

$$\Rightarrow 10 + x = \dfrac{40}{3}$$

$$\Rightarrow x = \dfrac{10}{3}$$

∴ About $3\dfrac{1}{3}$ litres (≈ 3.33 L) of water should be added.

Both have 65 in flexibility. Compared with Rashi, Keerthi has 5 fewer in strength (weight 4) and 5 more in agility (weight 6). Agility has the larger weight, so Keerthi gains more than she loses.

∴ Keerthi’s final marks are more.

$$\text{Rashi} = \dfrac{60 \times 4 + 65 \times 5 + 70 \times 6}{4 + 5 + 6} = \dfrac{985}{15} \approx 65.67$$

$$\text{Keerthi} = \dfrac{55 \times 4 + 65 \times 5 + 75 \times 6}{15} = \dfrac{995}{15} \approx 66.33$$

∴ Rashi scores about 65.67 and Keerthi about 66.33.

First, the average rating for each metric (each row is from 10 customers):

$$\text{Food} = \dfrac{5 \times 5 + 4 \times 3 + 3 \times 2}{10} = \dfrac{43}{10} = 4.3$$

$$\text{Ambience} = \dfrac{4 \times 4 + 3 \times 5 + 2 \times 1}{10} = \dfrac{33}{10} = 3.3$$

$$\text{Service} = \dfrac{5 \times 1 + 4 \times 2 + 3 \times 2 + 2 \times 4 + 1 \times 1}{10} = \dfrac{28}{10} = 2.8$$

Then the weighted mean with weights 6 : 5 : 4:

$$\text{Rating} = \dfrac{4.3 \times 6 + 3.3 \times 5 + 2.8 \times 4}{6 + 5 + 4}$$

$$\text{Rating} = \dfrac{25.8 + 16.5 + 11.2}{15}$$

$$\text{Rating} = \dfrac{53.5}{15}$$

$$\text{Rating} \approx 3.57$$

∴ The average rating is about 3.57 stars.

If the numbers were equal, the overall average would be exactly halfway, $\dfrac{16.5 + 13.8}{2} = 15.15$ kg. The actual overall average, 14.925 kg, is closer to the female average, so the females carry more weight in the mean.

∴ There are more female langurs.

Let there be $x$ males and $y$ females. The total weight divided by the number of langurs is the overall average, so:

$$\dfrac{16.5x + 13.8y}{x + y} = 14.925$$

∴ (a) describes the scenario. (Since $x + y = 60$, $\dfrac{16.5x + 13.8y}{60} = 14.925$ would also be correct — option (b) has a stray minus sign, which makes it wrong as printed. Options (c) and (d) divide by the wrong number.)

We have:

$$x + y = 60 \quad \cdots (1)$$

$$16.5x + 13.8y = 14.925 \times 60 = 895.5 \quad \cdots (2)$$

From equation (1), $y = 60 - x$. Substituting in equation (2):

$$16.5x + 13.8(60 - x) = 895.5$$

$$\Rightarrow 16.5x + 828 - 13.8x = 895.5$$

$$\Rightarrow 2.7x = 67.5$$

$$\Rightarrow x = 25$$

∴ There are 25 male langurs (and $60 - 25 = 35$ female langurs).

$$\text{Total female weight} = 35 \times 13.8 + 15.2 = 483 + 15.2 = 498.2 \text{ kg}$$$$\text{New average} = \dfrac{498.2}{36}$$

$$\text{New average} \approx 13.84 \text{ kg}$$

$$\text{Total male weight} = 25 \times 16.5 - (16.9 + 16.1) = 412.5 - 33 = 379.5 \text{ kg}$$$$\text{New average} = \dfrac{379.5}{23}$$

$$\text{New average} = 16.5 \text{ kg}$$

The average does not change, because the two langurs released also average $\dfrac{16.9 + 16.1}{2} = 16.5$ kg.

$$\text{New average} = \dfrac{379.5 - 1}{23}$$

$$\text{New average} = \dfrac{378.5}{23}$$

$$\text{New average} \approx 16.46 \text{ kg}$$

(The 1 kg loss is shared among 23 langurs, so the average falls by only $\dfrac{1}{23} \approx 0.04$ kg.)

$$\text{Salinity} = \dfrac{34 \times 1 + 0.001 \times 2}{1 + 2}$$

$$\text{Salinity} = \dfrac{34.002}{3}$$

$$\text{Salinity} \approx 11.33\%$$

Yes. The salinity of a mixture always lies between the salinities of its parts. Groundwater (0.01%) lies between purified water (0.001%) and Dead Sea water (34%), so it can be reached by adding enough purified water.

Let $x$ litres of purified water be mixed with 1 litre of Dead Sea water:

$$\dfrac{34 \times 1 + 0.001x}{1 + x} = 0.01$$

$$\Rightarrow 34 + 0.001x = 0.01 + 0.01x$$

$$\Rightarrow 33.99 = 0.009x$$

$$\Rightarrow x \approx 3776.7$$

∴ He would need about 3777 litres of purified water — the Dead Sea is extremely salty!

No. A mixture’s salinity always lies between the salinities of the two waters mixed, here between 0.01% and 34%. Purified water’s salinity, 0.001%, is lower than both, so no amount of groundwater can bring the mixture down to it.

(Trying the calculation shows the same: $\dfrac{34 + 0.01x}{1 + x} = 0.001$ gives $33.999 = -0.009x$, which needs a negative amount of water — impossible.)