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NCERT Solutions · Class 9 Mathematics How Quantities Combine: Understanding Data

33 questions · 33 still being checked

Exercise Set 10.3 1–8 (part 3 of 7)

  1. Exercise 1

    A stationery shop owner made ₹8000\displaystyle 8000 selling books, of which 30\displaystyle 30% is the profit amount, and ₹1000\displaystyle 1000 selling book covers, of which 50\displaystyle 50% is the profit amount. What is the percentage of profit on the total sales?

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    Profit percentage on the total sales is the weighted mean of $\displaystyle 30$% and $\displaystyle 50$%, with the sales as weights (not the plain $\displaystyle 40$%). Amounts are in ₹.\[\text{profit} = 0.30 \times 8000 + 0.50 \times 1000 \]\[= 2400 + 500 = 2900 \]\[\text{total sales} = 8000 + 1000 = 9000 \]\[\text{profit \%} = \frac{2900}{9000} \times 100 = \frac{290}{9} \approx 32.2\% \]Answer: about \(\displaystyle 32.2\%\)

  2. Exercise 2

    A white stork's migration is tracked. The average daily distance travelled, calculated over 20\displaystyle 20 days, is 44.5\displaystyle 44.5 km. On the 21st day, it flew 55\displaystyle 55 km. What is the average daily distance travelled over these 21\displaystyle 21 days? Make a guess before you calculate.

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    Guess: one day at $\displaystyle 55$ km among $\displaystyle 21$ days pulls the average up only a little, so slightly above $\displaystyle 44.5$, about $\displaystyle 45$ km.\[\text{distance in 20 days} = 20 \times 44.5 = 890 \text{ km} \]\[\text{distance in 21 days} = 890 + 55 = 945 \text{ km} \]\[\text{average} = \frac{945}{21} = 45 \text{ km} \]Answer: \(\displaystyle 45\) km per day, matching the guess

  3. Exercise 3

    A 600\displaystyle 600 mL solution with 5\displaystyle 5% salt is mixed with a 300\displaystyle 300 mL solution with 8\displaystyle 8% sugar. What are the concentrations of salt and sugar in the mixture?
    (i)
    Salt: 5\displaystyle 5%, Sugar: 8\displaystyle 8%
    (ii)
    Salt: 13\displaystyle 13%, Sugar: 3\displaystyle 3%
    (iii)
    Salt: 6\displaystyle 6%, Sugar: 6\displaystyle 6%
    (iv)
    Salt: 5.55\displaystyle 5.55%, Sugar: 8.88\displaystyle 8.88%
    (v)
    Salt: 3.33\displaystyle 3.33%, Sugar: 2.67\displaystyle 2.67%
    (vi)
    Salt: 4.1\displaystyle 4.1%, Sugar: 7.08\displaystyle 7.08%

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    Salt and sugar each sit in the whole mixture, and the volumes add.\[\text{salt} = 600 \times 0.05 = 30 \text{ mL} \]\[\text{sugar} = 300 \times 0.08 = 24 \text{ mL} \]\[\text{mixture} = 600 + 300 = 900 \text{ mL} \]\[\text{salt \%} = \frac{30}{900} \times 100 = 3.33\% \]\[\text{sugar \%} = \frac{24}{900} \times 100 = 2.67\% \]Answer: option (v), salt \(\displaystyle 3.33\%\) and sugar \(\displaystyle 2.67\%\)

  4. Exercise 4

    At a panipuri (golgappa) stall, the concentration of spice in the pani (spiced water) was 8\displaystyle 8%. Many customers complained that it was too spicy. What quantity of regular water should be mixed into the 10\displaystyle 10 litres of pani so that the spice level is reduced to (34)th \displaystyle \left(\frac{3}{4}\right)^{\text {th }} of the original concentration?

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    The quantity of spice stays the same; only the volume grows. Let \(\displaystyle x\) litres of water be added.\[\text{spice} = 0.08 \times 10 = 0.8 \text{ L} \]\[\text{new concentration} = \frac{3}{4} \times 8\% = 6\% \]\[\frac{0.8}{10 + x} = 0.06 \]\[10 + x = \frac{0.8}{0.06} = \frac{40}{3} \]\[x = \frac{40}{3} - 10 = \frac{10}{3} \]Answer: \(\displaystyle \frac{10}{3}\) L \(\displaystyle \approx 3.33\) L of water

  5. Exercise 5

    A physical fitness evaluation is being undertaken. The final marks are calculated by combining the marks for strength, flexibility, and agility in the ratio 4\displaystyle 4:5\displaystyle 5:6. Rashi has scored 60\displaystyle 60, 65\displaystyle 65, and 70. Keerthi has scored 55\displaystyle 55, 65\displaystyle 65, and 75\displaystyle 75 respectively.
    (i)
    Find out whose total is more without calculating.
    (ii)
    What are their final marks?

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    (i) Both have the same plain total, 195. Keerthi scores $\displaystyle 5$ less in strength (weight $\displaystyle 4$) but $\displaystyle 5$ more in agility (weight $\displaystyle 6$), and flexibility is equal. The gain sits on the larger weight, so Keerthi's total is more.(ii)\[\text{Rashi} = \frac{60 \times 4 + 65 \times 5 + 70 \times 6}{4 + 5 + 6} = \frac{985}{15} \approx 65.67 \]\[\text{Keerthi} = \frac{55 \times 4 + 65 \times 5 + 75 \times 6}{4 + 5 + 6} = \frac{995}{15} \approx 66.33 \]Answer: (i) Keerthi; (ii) Rashi \(\displaystyle \approx 65.67\), Keerthi \(\displaystyle \approx 66.33\)

  6. Exercise 6

    A restaurant collected ratings from 10\displaystyle 10 customers on a scale of 1\displaystyle 1 to 5. The resulting data is shown in the table below.
    5\displaystyle 5★4\displaystyle 4★3\displaystyle 3★2\displaystyle 2★1\displaystyle 1★
    Food5\displaystyle 53\displaystyle 32\displaystyle 20\displaystyle 00\displaystyle 0
    Ambience0\displaystyle 04\displaystyle 45\displaystyle 51\displaystyle 10\displaystyle 0
    Service1\displaystyle 12\displaystyle 22\displaystyle 24\displaystyle 41\displaystyle 1
    What is the average rating if the metrics are combined with the weights food : ambience : service = 6\displaystyle 6:5\displaystyle 5:4\displaystyle 4?

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    Each metric's average rating is a weighted mean of the stars, weighted by the number of customers.\[\text{Food} = \frac{5 \times 5 + 3 \times 4 + 2 \times 3}{10} = \frac{43}{10} = 4.3 \]\[\text{Ambience} = \frac{0 \times 5 + 4 \times 4 + 5 \times 3 + 1 \times 2 + 0 \times 1}{10} = \frac{33}{10} = 3.3 \]\[\text{Service} = \frac{1 \times 5 + 2 \times 4 + 2 \times 3 + 4 \times 2 + 1 \times 1}{10} = \frac{28}{10} = 2.8 \]Now combine these with weights \(\displaystyle 6 : 5 : 4\).\[\text{average} = \frac{6 \times 4.3 + 5 \times 3.3 + 4 \times 2.8}{6 + 5 + 4} = \frac{53.5}{15} \approx 3.57 \]Answer: about \(\displaystyle 3.57\) out of $\displaystyle 5$

  7. Exercise 7

    The following table shows the weight data of langurs in an animal facility. Without doing any computations, can you tell whether there are more male langurs or more female langurs? Or are they equal in number?
    Male langursFemale langursAll langurs
    Average weight16.5\displaystyle 16.5 kg13.8\displaystyle 13.8 kg14.925\displaystyle 14.925 kg
    Number of langurs60\displaystyle 60
    (i)
    Which of the following expression(s) describes the given scenario?
    (a)
    16.5x+13.8yx+y=14.925\displaystyle \frac{16.5 x+13.8 y}{x+y}=14.925
    (b)
    16.5x+13.8y60=−14.925\displaystyle \frac{16.5 x+13.8 y}{60}=-14.925
    (c)
    16.5x+13.8y2=−14.925\displaystyle \frac{16.5 x+13.8 y}{2}=-14.925
    (d)
    16.5x+13.8y16.5+13.8=−14.925\displaystyle \frac{16.5 x+13.8 y}{16.5+13.8}=-14.925
    (ii)
    Find out how many male langurs are present.
    (iii)
    A female langur weighing 15.2\displaystyle 15.2 kg is admitted to the facility. What is the average weight of the female langurs after this?
    (iv)
    Two male langurs weighing 16.9\displaystyle 16.9 kg and 16.1\displaystyle 16.1 kg are released from the facility. What is the average weight of the male langurs after this?
    (v)
    Now, suppose one of the male langurs lost 1\displaystyle 1 kg of weight. What is the average weight of all the male langurs after this?

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    The overall average \(\displaystyle 14.925\) is closer to \(\displaystyle 13.8\) (female) than to \(\displaystyle 16.5\) (male), so there are more female langurs. Part (ii) confirms it.(i) (a) is total weight over the number of langurs, \(\displaystyle x+y\). As \(\displaystyle x+y=60\), (b) is the same expression, so it is also right if its minus sign is a misprint. (c) divides by $\displaystyle 2$ and (d) by \(\displaystyle 16.5+13.8\), not by the number of langurs.(ii)\[x + y = 60 \]\[16.5x + 13.8y = 14.925 \times 60 = 895.5 \]\[16.5x + 13.8(60 - x) = 895.5 \]\[2.7x + 828 = 895.5 \]\[x = \frac{67.5}{2.7} = 25 \]So $\displaystyle 25$ males and $\displaystyle 35$ females.(iii) Females weigh \(\displaystyle 13.8 \times 35 = 483\) kg in all.\[\frac{483 + 15.2}{35 + 1} = \frac{498.2}{36} \approx 13.84 \text{ kg} \](iv) Males weigh \(\displaystyle 16.5 \times 25 = 412.5\) kg in all.\[\frac{412.5 - (16.9 + 16.1)}{25 - 2} = \frac{379.5}{23} = 16.5 \text{ kg} \]The two released average \(\displaystyle 16.5\) themselves, so the mean does not change.(v) Taking on from (iv), $\displaystyle 23$ males weigh \(\displaystyle 379.5\) kg.\[\frac{379.5 - 1}{23} = \frac{378.5}{23} \approx 16.46 \text{ kg} \]Starting from all $\displaystyle 25$ males also gives \(\displaystyle 16.46\) kg.Answer: more females; (i) (a), and (b) if its minus sign is a misprint; (ii) $\displaystyle 25$; (iii) \(\displaystyle \approx 13.84\) kg; (iv) \(\displaystyle 16.5\) kg; (v) \(\displaystyle \approx 16.46\) kg

  8. Exercise 8

    Dorjee has collected 1\displaystyle 1 litre of water from the Dead Sea! Using the information in the table, answer the following questions. A calculator can be used if necessary.
    Water SourceSalinity
    Dead Sea≈34%\displaystyle \approx 34 \%
    Other seas and oceans≈3.5%\displaystyle \approx 3.5 \%
    Ground water≈0.01%\displaystyle \approx 0.01 \%
    Purified drinking water≈0.001%\displaystyle \approx 0.001 \%
    (i)
    What is the salinity of the mixture if he mixes 1\displaystyle 1 litre of water from the Dead Sea with 2\displaystyle 2 litres of purified drinking water?
    (ii)
    Is it possible to mix water from the Dead Sea and purified drinking water to get a mixture with the salinity of groundwater? Why/Why not? What quantity of purified drinking water should Dorjee mix with 1\displaystyle 1 litre of water from the Dead Sea to get a mixture having the salinity of groundwater?
    (iii)
    Is it possible to mix water from the Dead Sea and groundwater to get a mixture with the salinity of purified drinking water? Why/Why not? What quantity of groundwater should he mix with 1\displaystyle 1 litre of water from the Dead Sea to get a mixture having the salinity of purified drinking water?

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    (i)\[\frac{34 \times 1 + 0.001 \times 2}{1 + 2} = \frac{34.002}{3} = 11.334\% \](ii) Yes. A mixture's salinity lies between those of its two parts, and \(\displaystyle 0.01\%\) lies between \(\displaystyle 0.001\%\) and \(\displaystyle 34\%\). Let \(\displaystyle q\) litres of purified water be added.\[\frac{34 \times 1 + 0.001q}{1 + q} = 0.01 \]\[34 + 0.001q = 0.01 + 0.01q \]\[33.99 = 0.009q \]\[q = \frac{33.99}{0.009} \approx 3776.67 \text{ L} \](iii) No. \(\displaystyle 0.001\%\) is below both \(\displaystyle 34\%\) and \(\displaystyle 0.01\%\), so no mixture of the two can reach it. Trying \(\displaystyle g\) litres of groundwater:\[\frac{34 \times 1 + 0.01g}{1 + g} = 0.001 \]\[34 + 0.01g = 0.001 + 0.001g \]\[33.999 = -0.009g \]\[g \approx -3777.67 < 0 \]A negative quantity is impossible.Answer: (i) \(\displaystyle 11.334\%\); (ii) possible, about \(\displaystyle 3776.67\) L; (iii) not possible