Ex 10.3 Class 9 Maths Ganita Manjari Solutions | Sid Classes
Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.3 Advanced Weighted Averages and Mixtures. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this walkthrough covers profit percentages, daily distance averages, solution concentrations, weighted ratings, animal facility weight distributions, and Dead Sea salinity problems entirely clear.
🔑 Key Concepts & Formulas:
- Profit Percentage on Total Sales: \(\text{Profit } \% = \frac{\text{Total Profit Amount}}{\text{Total Sales Amount}} \times 100\).
- Weighted Average of Two Groups: If group 1 has \(x\) members with mean \(M_1\) and group 2 has \(y\) members with mean \(M_2\), the combined mean is \(\frac{x \cdot M_1 + y \cdot M_2}{x + y}\).
NCERT Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.3 Solutions
Question 1
A stationery shop owner made ₹8000 selling books, of which 30% is the profit amount, and ₹1000 selling book covers, of which 50% is the profit amount. What is the percentage of profit on the total sales?
Solution:
* Total sales \(= ₹8000 + ₹1000 = ₹9000\).
* Profit from books \(= 30\% \text{ of } ₹8000 = 0.30 \times 8000 = ₹2400\).
* Profit from book covers \(= 50\% \text{ of } ₹1000 = 0.50 \times 1000 = ₹500\).
* Total profit \(= ₹2400 + ₹500 = ₹2900\).
Percentage of profit on total sales \(= \frac{\text{Total Profit}}{\text{Total Sales}} \times 100 = \frac{2900}{9000} \times 100 = \frac{290}{9} \approx \mathbf{32.22\%}\).
Answer: The profit percentage on total sales is \(\mathbf{\frac{290}{9}\%}\) (or approximately \(\mathbf{32.22\%}\)).
Question 2
A white stork's migration is tracked. The average daily distance travelled, calculated over 20 days, is 44.5 km. On the 21st day, it flew 55 km. What is the average daily distance travelled over these 21 days? Make a guess before you calculate.
Solution:
* Total distance for first 20 days \(= 20 \times 44.5 = 890\) km.
* Distance on 21st day \(= 55\) km.
* Total distance for 21 days \(= 890 + 55 = 945\) km.
* New average daily distance \(= \frac{945}{21} = \mathbf{45}\) km.
Answer: The average daily distance over 21 days is exactly \(\mathbf{45}\) km. (Since the 21st day's distance of 55 km is greater than the previous average of 44.5 km, the overall average increases slightly).
Question 3
A 600 mL solution with 5% salt is mixed with a 300 mL solution with 8% sugar. What are the concentrations of salt and sugar in the mixture?
(i) Salt: 5%, Sugar: 8% (ii) Salt: 13%, Sugar: 3%
(iii) Salt: 6%, Sugar: 6% (iv) Salt: 5.55%, Sugar: 8.88%
(v) Salt: 3.33%, Sugar: 2.67% (vi) Salt: 4.1%, Sugar: 7.08%
Solution:
Total volume of the combined mixture \(= 600\text{ mL} + 300\text{ mL} = 900\text{ mL}\).
* Amount of salt \(= 5\% \text{ of } 600\text{ mL} = 0.05 \times 600 = 30\text{ mL}\).
* Concentration of salt in mixture \(= \frac{30}{900} \times 100 = \frac{100}{3}\% \approx \mathbf{3.33\%}\).
* Amount of sugar \(= 8\% \text{ of } 300\text{ mL} = 0.08 \times 300 = 24\text{ mL}\).
* Concentration of sugar in mixture \(= \frac{24}{900} \times 100 = \frac{8}{3}\% \approx \mathbf{2.67\%}\).
Answer: Option (v) Salt: 3.33%, Sugar: 2.67%.
Question 4
At a panipuri (golgappa) stall, the concentration of spice in the pani was 8%. What quantity of regular water should be mixed into the 10 litres of pani so that the spice level is reduced to \(\left(\frac{3}{4}\right)^{\text{th}}\) of the original concentration?
Solution:
* Original amount of spice \(= 8\% \text{ of } 10\text{ litres} = 0.08 \times 10 = 0.8\text{ litres}\).
* Target concentration \(= \frac{3}{4} \times 8\% = 6\% = 0.06\).
Let \(V\) be the total final volume after adding regular water.
$$\frac{0.8}{V} = 0.06 \implies V = \frac{0.8}{0.06} = \frac{80}{6} = \frac{40}{3}\text{ litres} = 13.33\text{ litres}$$
* Quantity of regular water to be added \(= \frac{40}{3} - 10 = \frac{10}{3}\text{ litres} = 3\frac{1}{3}\text{ litres}\) (or approx \(3.33\) litres / \(3333\) mL).
Answer: \(\mathbf{3\frac{1}{3}}\) litres (or \(\mathbf{3.33}\) litres) of regular water should be added.
Question 5
A physical fitness evaluation is undertaken. Final marks combine strength, flexibility, and agility in ratio \(4:5:6\). Rashi scored \(60, 65, 70\) and Keerthi scored \(55, 65, 75\) respectively.
(i) Find out whose total is more without calculating.
(ii) What are their final marks?
Solution:
Weights for strength, flexibility, and agility \(= 4, 5, 6\).
* Rashi's scores: \(60, 65, 70\)
* Keerthi's scores: \(55, 65, 75\)
(i) Without calculating exact totals:
Notice that both have identical middle scores (\(65\) weighted by \(5\)). Keerthi scores higher on agility (\(75\) vs \(70\)) which has the highest weight (\(6\)), whereas Rashi scores higher on strength (\(60\) vs \(55\)) which has the lowest weight (\(4\)). Therefore, Keerthi's total is more.
(ii) Calculate final marks:
* Rashi's final mark \(= \frac{60(4) + 65(5) + 70(6)}{4 + 5 + 6} = \frac{240 + 325 + 420}{15} = \frac{985}{15} = \mathbf{65.67}\).
* Keerthi's final mark \(= \frac{55(4) + 65(5) + 75(6)}{4 + 5 + 6} = \frac{220 + 325 + 450}{15} = \frac{995}{15} = \mathbf{66.33}\).
Answer: (i) Keerthi's total is higher; (ii) Rashi: \(\mathbf{65.67}\), Keerthi: \(\mathbf{66.33}\).
Question 6
A restaurant collected ratings from 10 customers on a scale of 1 to 5.
Food: \(5\star: 5, 4\star: 3, 3\star: 2, 2\star: 0, 1\star: 0\)
Ambience: \(5\star: 0, 4\star: 4, 3\star: 5, 2\star: 1, 1\star: 0\)
Service: \(5\star: 1, 4\star: 2, 3\star: 2, 2\star: 4, 1\star: 1\)
What is the average rating if weights are food : ambience : service = \(6 : 5 : 4\)?
Solution:
First, compute the mean score for each category out of 10 customers:
* Food mean \(= \frac{5(5) + 3(4) + 2(3) + 0(2) + 0(1)}{10} = \frac{25 + 12 + 6}{10} = \frac{43}{10} = 4.3\).
* Ambience mean \(= \frac{0(5) + 4(4) + 5(3) + 1(2) + 0(1)}{10} = \frac{0 + 16 + 15 + 2}{10} = \frac{33}{10} = 3.3\).
* Service mean \(= \frac{1(5) + 2(4) + 2(3) + 4(2) + 1(1)}{10} = \frac{5 + 8 + 6 + 8 + 1}{10} = \frac{28}{10} = 2.8\).
Now apply the weighted average with weights \(6 : 5 : 4\):
$$\text{Combined Average Rating} = \frac{4.3(6) + 3.3(5) + 2.8(4)}{6 + 5 + 4} = \frac{25.8 + 16.5 + 11.2}{15} = \frac{53.5}{15} \approx \mathbf{3.57}$$
Answer: The combined average rating is \(\mathbf{3.57}\).
Question 7
Table shows weight data of langurs: Male mean = 16.5 kg, Female mean = 13.8 kg, All langurs mean = 14.925 kg, Total langurs = 60.
(i) Which expression describes the scenario?
(ii) Find the number of male langurs.
(iii)-(v) Various updates to langur weights.
Solution:
(i) Correct Expression:
Let \(x\) be male langurs and \(y\) be female langurs. Since total langurs \(x + y = 60\), the weighted average formula gives expression (b) \(\frac{16.5x + 13.8y}{60} = 14.925\).
(ii) Find number of male langurs (\(x\)):
Given \(x + y = 60 \implies y = 60 - x\).
\(\frac{16.5x + 13.8(60 - x)}{60} = 14.925 \implies 16.5x + 828 - 13.8x = 895.5 \implies 2.7x = 67.5 \implies x = \mathbf{25}\) (and \(y = 35\)).
(iii) Female langur weighing 15.2 kg is admitted:
New female count \(= 36\). Previous total female weight \(= 35 \times 13.8 = 483\).
New total female weight \(= 483 + 15.2 = 498.2\). New mean \(= \frac{498.2}{36} \approx \mathbf{13.84}\) kg.
(iv) Two males weighing 16.9 kg and 16.1 kg are released:
Remaining males \(= 25 - 2 = 23\). Previous total male weight \(= 25 \times 16.5 = 412.5\).
Remaining weight \(= 412.5 - (16.9 + 16.1) = 412.5 - 33 = 379.5\). New mean \(= \frac{379.5}{23} = \mathbf{16.5}\) kg.
(v) One male lost 1 kg of weight:
Total male weight decreases by 1 kg from \(412.5\) to \(411.5\). Male count remains \(25\).
New average \(= \frac{411.5}{25} = \mathbf{16.46}\) kg.
Answer: (i) Option (b); (ii) 25 male langurs; (iii) 13.84 kg; (iv) 16.5 kg; (v) 16.46 kg.
Question 8
Dorjee collected 1 litre of water from the Dead Sea (salinity \(\approx 34\%\)). Other seas \(\approx 3.5\%\), Groundwater \(\approx 0.01\%\), Purified drinking water \(\approx 0.001\%\).
(i) Salinity of mixing 1 litre Dead Sea water with 2 litres purified drinking water?
(ii) Is it possible to mix Dead Sea water and purified water to get groundwater salinity? Quantity needed?
(iii) Is it possible to mix Dead Sea water and groundwater to get purified water salinity? Quantity needed?
Solution:
(i) Mixture of 1L Dead Sea (\(34\%\)) + 2L Purified water (\(0.001\%\)):
$$\text{Salinity} = \frac{1(34) + 2(0.001)}{1 + 2} = \frac{34.002}{3} \approx \mathbf{11.334\%}$$
(ii) Mix Dead Sea water and purified water to get groundwater salinity (\(0.01\%\)):
Yes, possible because groundwater salinity lies between Dead Sea (\(34\%\)) and purified water (\(0.001\%\)).
Let \(V\) be the volume of purified water needed for 1 litre of Dead Sea water:
$$\frac{1(34) + V(0.00001)}{1 + V} = 0.0001 \implies 34 \approx 0.0001V \implies V \approx \mathbf{340,000}\text{ litres}$$
(iii) Mix Dead Sea water and groundwater to get purified water salinity (\(0.001\%\)):
No, it is not possible. Both Dead Sea water (\(34\%\)) and groundwater (\(0.01\%\)) have higher salinity than purified water (\(0.001\%\)). Adding them together can never yield a salinity lower than the lowest component.
Frequently Asked Questions (FAQs)
Q1: How do you handle mixture problems with different concentrations?
A: Calculate the absolute amount of solute in each component, sum them up, and divide by the total combined volume or mass.
Q2: Can a mixture have a concentration lower than all of its individual ingredients?
A: No, the concentration of a mixture always lies between the lowest and highest individual component concentrations.
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